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values | name stringlengths 2 58 | type stringclasses 2
values | rating int64 0 3.5k | tags listlengths 0 11 | title stringclasses 522
values | time-limit stringclasses 8
values | memory-limit stringclasses 8
values | problem-description stringlengths 0 7.15k | input-specification stringlengths 0 2.05k | output-specification stringlengths 0 1.5k | demo-input listlengths 0 7 | demo-output listlengths 0 7 | note stringlengths 0 5.24k | points float64 0 425k | test_cases listlengths 0 402 | creationTimeSeconds int64 1.37B 1.7B | relativeTimeSeconds int64 8 2.15B | programmingLanguage stringclasses 3
values | verdict stringclasses 14
values | testset stringclasses 12
values | passedTestCount int64 0 1k | timeConsumedMillis int64 0 15k | memoryConsumedBytes int64 0 805M | code stringlengths 3 65.5k | prompt stringlengths 262 8.2k | response stringlengths 17 65.5k | score float64 -1 3.99 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
119 | A | Epic Game | PROGRAMMING | 800 | [
"implementation"
] | null | null | Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game. | The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile. | If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes). | [
"3 5 9\n",
"1 1 100\n"
] | [
"0",
"1"
] | The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that. | 500 | [
{
"input": "3 5 9",
"output": "0"
},
{
"input": "1 1 100",
"output": "1"
},
{
"input": "23 12 16",
"output": "1"
},
{
"input": "95 26 29",
"output": "1"
},
{
"input": "73 32 99",
"output": "1"
},
{
"input": "1 1 1",
"output": "0"
},
{
"inpu... | 1,599,897,806 | 2,147,483,647 | PyPy 3 | OK | TESTS | 45 | 280 | 0 | from math import gcd
j=0
a,b,c=map(int,input().split())
while c!=0:
c-=gcd(a,c)
j+=1
if c!=0:
c-=gcd(b,c)
j+=1
if j%2==0:
print(1)
else:
print(0) | Title: Epic Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Simon and Antisimon play a game. Initially each player receives one fixed positive integer that doesn't change throughout the game. Simon receives number *a* and Antisimon receives number *b*. They also have a heap of *n* stones. The players take turns to make a move and Simon starts. During a move a player should take from the heap the number of stones equal to the greatest common divisor of the fixed number he has received and the number of stones left in the heap. A player loses when he cannot take the required number of stones (i. e. the heap has strictly less stones left than one needs to take).
Your task is to determine by the given *a*, *b* and *n* who wins the game.
Input Specification:
The only string contains space-separated integers *a*, *b* and *n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=100) — the fixed numbers Simon and Antisimon have received correspondingly and the initial number of stones in the pile.
Output Specification:
If Simon wins, print "0" (without the quotes), otherwise print "1" (without the quotes).
Demo Input:
['3 5 9\n', '1 1 100\n']
Demo Output:
['0', '1']
Note:
The greatest common divisor of two non-negative integers *a* and *b* is such maximum positive integer *k*, that *a* is divisible by *k* without remainder and similarly, *b* is divisible by *k* without remainder. Let *gcd*(*a*, *b*) represent the operation of calculating the greatest common divisor of numbers *a* and *b*. Specifically, *gcd*(*x*, 0) = *gcd*(0, *x*) = *x*.
In the first sample the game will go like that:
- Simon should take *gcd*(3, 9) = 3 stones from the heap. After his move the heap has 6 stones left.- Antisimon should take *gcd*(5, 6) = 1 stone from the heap. After his move the heap has 5 stones left.- Simon should take *gcd*(3, 5) = 1 stone from the heap. After his move the heap has 4 stones left.- Antisimon should take *gcd*(5, 4) = 1 stone from the heap. After his move the heap has 3 stones left.- Simon should take *gcd*(3, 3) = 3 stones from the heap. After his move the heap has 0 stones left.- Antisimon should take *gcd*(5, 0) = 5 stones from the heap. As 0 < 5, it is impossible and Antisimon loses.
In the second sample each player during each move takes one stone from the heap. As *n* is even, Antisimon takes the last stone and Simon can't make a move after that. | ```python
from math import gcd
j=0
a,b,c=map(int,input().split())
while c!=0:
c-=gcd(a,c)
j+=1
if c!=0:
c-=gcd(b,c)
j+=1
if j%2==0:
print(1)
else:
print(0)
``` | 3 | |
265 | A | Colorful Stones (Simplified Edition) | PROGRAMMING | 800 | [
"implementation"
] | null | null | There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence. | The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence. | Print the final 1-based position of Liss in a single line. | [
"RGB\nRRR\n",
"RRRBGBRBBB\nBBBRR\n",
"BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n"
] | [
"2\n",
"3\n",
"15\n"
] | none | 500 | [
{
"input": "RGB\nRRR",
"output": "2"
},
{
"input": "RRRBGBRBBB\nBBBRR",
"output": "3"
},
{
"input": "BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB",
"output": "15"
},
{
"input": "G\nRRBBRBRRBR",
"output": "1"
},
... | 1,651,953,193 | 2,147,483,647 | Python 3 | OK | TESTS | 19 | 122 | 0 | s, t = input(), input()
moves = 0
for i in range(len(t)):
c = t[i]
if c == s[moves]:
moves = moves + 1
print(moves + 1) | Title: Colorful Stones (Simplified Edition)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a sequence of colorful stones. The color of each stone is one of red, green, or blue. You are given a string *s*. The *i*-th (1-based) character of *s* represents the color of the *i*-th stone. If the character is "R", "G", or "B", the color of the corresponding stone is red, green, or blue, respectively.
Initially Squirrel Liss is standing on the first stone. You perform instructions one or more times.
Each instruction is one of the three types: "RED", "GREEN", or "BLUE". After an instruction *c*, if Liss is standing on a stone whose colors is *c*, Liss will move one stone forward, else she will not move.
You are given a string *t*. The number of instructions is equal to the length of *t*, and the *i*-th character of *t* represents the *i*-th instruction.
Calculate the final position of Liss (the number of the stone she is going to stand on in the end) after performing all the instructions, and print its 1-based position. It is guaranteed that Liss don't move out of the sequence.
Input Specification:
The input contains two lines. The first line contains the string *s* (1<=≤<=|*s*|<=≤<=50). The second line contains the string *t* (1<=≤<=|*t*|<=≤<=50). The characters of each string will be one of "R", "G", or "B". It is guaranteed that Liss don't move out of the sequence.
Output Specification:
Print the final 1-based position of Liss in a single line.
Demo Input:
['RGB\nRRR\n', 'RRRBGBRBBB\nBBBRR\n', 'BRRBGBRGRBGRGRRGGBGBGBRGBRGRGGGRBRRRBRBBBGRRRGGBBB\nBBRBGGRGRGBBBRBGRBRBBBBRBRRRBGBBGBBRRBBGGRBRRBRGRB\n']
Demo Output:
['2\n', '3\n', '15\n']
Note:
none | ```python
s, t = input(), input()
moves = 0
for i in range(len(t)):
c = t[i]
if c == s[moves]:
moves = moves + 1
print(moves + 1)
``` | 3 | |
34 | B | Sale | PROGRAMMING | 900 | [
"greedy",
"sortings"
] | B. Sale | 2 | 256 | Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn. | The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets. | Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets. | [
"5 3\n-6 0 35 -2 4\n",
"4 2\n7 0 0 -7\n"
] | [
"8\n",
"7\n"
] | none | 1,000 | [
{
"input": "5 3\n-6 0 35 -2 4",
"output": "8"
},
{
"input": "4 2\n7 0 0 -7",
"output": "7"
},
{
"input": "6 6\n756 -611 251 -66 572 -818",
"output": "1495"
},
{
"input": "5 5\n976 437 937 788 518",
"output": "0"
},
{
"input": "5 3\n-2 -2 -2 -2 -2",
"output": "... | 1,611,304,855 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 92 | 0 | n,m=map(int,input().split());
a=list(map(lambda x: -1*int(x),input().split()));
a.sort();
sum=0;
for i in range(m):
if(a[i]>0):
break;
sum=sum+a[i];
print(-1*sum);
# 6 6
# 756 -611 251 -66 572 -818
| Title: Sale
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
Output Specification:
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
Demo Input:
['5 3\n-6 0 35 -2 4\n', '4 2\n7 0 0 -7\n']
Demo Output:
['8\n', '7\n']
Note:
none | ```python
n,m=map(int,input().split());
a=list(map(lambda x: -1*int(x),input().split()));
a.sort();
sum=0;
for i in range(m):
if(a[i]>0):
break;
sum=sum+a[i];
print(-1*sum);
# 6 6
# 756 -611 251 -66 572 -818
``` | 0 |
0 | none | none | none | 0 | [
"none"
] | null | null | You are given an array *a*1,<=*a*2,<=...,<=*a**n* consisting of *n* integers, and an integer *k*. You have to split the array into exactly *k* non-empty subsegments. You'll then compute the minimum integer on each subsegment, and take the maximum integer over the *k* obtained minimums. What is the maximum possible integer you can get?
Definitions of subsegment and array splitting are given in notes. | The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=<=105) — the size of the array *a* and the number of subsegments you have to split the array to.
The second line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (<=-<=109<=<=≤<=<=*a**i*<=≤<=<=109). | Print single integer — the maximum possible integer you can get if you split the array into *k* non-empty subsegments and take maximum of minimums on the subsegments. | [
"5 2\n1 2 3 4 5\n",
"5 1\n-4 -5 -3 -2 -1\n"
] | [
"5\n",
"-5\n"
] | A subsegment [*l*, *r*] (*l* ≤ *r*) of array *a* is the sequence *a*<sub class="lower-index">*l*</sub>, *a*<sub class="lower-index">*l* + 1</sub>, ..., *a*<sub class="lower-index">*r*</sub>.
Splitting of array *a* of *n* elements into *k* subsegments [*l*<sub class="lower-index">1</sub>, *r*<sub class="lower-index">1</sub>], [*l*<sub class="lower-index">2</sub>, *r*<sub class="lower-index">2</sub>], ..., [*l*<sub class="lower-index">*k*</sub>, *r*<sub class="lower-index">*k*</sub>] (*l*<sub class="lower-index">1</sub> = 1, *r*<sub class="lower-index">*k*</sub> = *n*, *l*<sub class="lower-index">*i*</sub> = *r*<sub class="lower-index">*i* - 1</sub> + 1 for all *i* > 1) is *k* sequences (*a*<sub class="lower-index">*l*<sub class="lower-index">1</sub></sub>, ..., *a*<sub class="lower-index">*r*<sub class="lower-index">1</sub></sub>), ..., (*a*<sub class="lower-index">*l*<sub class="lower-index">*k*</sub></sub>, ..., *a*<sub class="lower-index">*r*<sub class="lower-index">*k*</sub></sub>).
In the first example you should split the array into subsegments [1, 4] and [5, 5] that results in sequences (1, 2, 3, 4) and (5). The minimums are *min*(1, 2, 3, 4) = 1 and *min*(5) = 5. The resulting maximum is *max*(1, 5) = 5. It is obvious that you can't reach greater result.
In the second example the only option you have is to split the array into one subsegment [1, 5], that results in one sequence ( - 4, - 5, - 3, - 2, - 1). The only minimum is *min*( - 4, - 5, - 3, - 2, - 1) = - 5. The resulting maximum is - 5. | 0 | [
{
"input": "5 2\n1 2 3 4 5",
"output": "5"
},
{
"input": "5 1\n-4 -5 -3 -2 -1",
"output": "-5"
},
{
"input": "10 2\n10 9 1 -9 -7 -9 3 8 -10 5",
"output": "10"
},
{
"input": "10 4\n-8 -1 2 -3 9 -8 4 -3 5 9",
"output": "9"
},
{
"input": "1 1\n504262064",
"output... | 1,508,113,170 | 870 | Python 3 | WRONG_ANSWER | TESTS | 9 | 93 | 8,396,800 | read = lambda: tuple(map(int, input().split()))
n, k = read()
nums = read()
if k == 1:
print(min(nums))
else:
print(max(nums))
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array *a*1,<=*a*2,<=...,<=*a**n* consisting of *n* integers, and an integer *k*. You have to split the array into exactly *k* non-empty subsegments. You'll then compute the minimum integer on each subsegment, and take the maximum integer over the *k* obtained minimums. What is the maximum possible integer you can get?
Definitions of subsegment and array splitting are given in notes.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=<=105) — the size of the array *a* and the number of subsegments you have to split the array to.
The second line contains *n* integers *a*1,<=<=*a*2,<=<=...,<=<=*a**n* (<=-<=109<=<=≤<=<=*a**i*<=≤<=<=109).
Output Specification:
Print single integer — the maximum possible integer you can get if you split the array into *k* non-empty subsegments and take maximum of minimums on the subsegments.
Demo Input:
['5 2\n1 2 3 4 5\n', '5 1\n-4 -5 -3 -2 -1\n']
Demo Output:
['5\n', '-5\n']
Note:
A subsegment [*l*, *r*] (*l* ≤ *r*) of array *a* is the sequence *a*<sub class="lower-index">*l*</sub>, *a*<sub class="lower-index">*l* + 1</sub>, ..., *a*<sub class="lower-index">*r*</sub>.
Splitting of array *a* of *n* elements into *k* subsegments [*l*<sub class="lower-index">1</sub>, *r*<sub class="lower-index">1</sub>], [*l*<sub class="lower-index">2</sub>, *r*<sub class="lower-index">2</sub>], ..., [*l*<sub class="lower-index">*k*</sub>, *r*<sub class="lower-index">*k*</sub>] (*l*<sub class="lower-index">1</sub> = 1, *r*<sub class="lower-index">*k*</sub> = *n*, *l*<sub class="lower-index">*i*</sub> = *r*<sub class="lower-index">*i* - 1</sub> + 1 for all *i* > 1) is *k* sequences (*a*<sub class="lower-index">*l*<sub class="lower-index">1</sub></sub>, ..., *a*<sub class="lower-index">*r*<sub class="lower-index">1</sub></sub>), ..., (*a*<sub class="lower-index">*l*<sub class="lower-index">*k*</sub></sub>, ..., *a*<sub class="lower-index">*r*<sub class="lower-index">*k*</sub></sub>).
In the first example you should split the array into subsegments [1, 4] and [5, 5] that results in sequences (1, 2, 3, 4) and (5). The minimums are *min*(1, 2, 3, 4) = 1 and *min*(5) = 5. The resulting maximum is *max*(1, 5) = 5. It is obvious that you can't reach greater result.
In the second example the only option you have is to split the array into one subsegment [1, 5], that results in one sequence ( - 4, - 5, - 3, - 2, - 1). The only minimum is *min*( - 4, - 5, - 3, - 2, - 1) = - 5. The resulting maximum is - 5. | ```python
read = lambda: tuple(map(int, input().split()))
n, k = read()
nums = read()
if k == 1:
print(min(nums))
else:
print(max(nums))
``` | 0 | |
243 | E | Matrix | PROGRAMMING | 3,000 | [
"data structures"
] | null | null | Let's consider an *n*<=×<=*n* square matrix, consisting of digits one and zero.
We'll consider a matrix good, if it meets the following condition: in each row of the matrix all ones go in one group. That is, each row of the matrix looks like that 00...0011...1100...00 (or simply consists of zeroes if it has no ones).
You are given matrix *a* of size *n*<=×<=*n*, consisting of zeroes and ones. Your task is to determine whether you can get a good matrix *b* from it by rearranging the columns or not. | The first line contains integer *n* (1<=≤<=*n*<=≤<=500) — the size of matrix *a*.
Each of *n* following lines contains *n* characters "0" and "1" — matrix *a*. Note that the characters are written without separators. | Print "YES" in the first line, if you can rearrange the matrix columns so as to get a good matrix *b*. In the next *n* lines print the good matrix *b*. If there are multiple answers, you are allowed to print any of them.
If it is impossible to get a good matrix, print "NO". | [
"6\n100010\n110110\n011001\n010010\n000100\n011001\n",
"3\n110\n101\n011\n"
] | [
"YES\n011000\n111100\n000111\n001100\n100000\n000111\n",
"NO\n"
] | none | 2,500 | [] | 1,584,958,351 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 1 | 218 | 512,000 | from collections import deque
class IsNotC1P(Exception):
"""The given instance does not have the all consecutive ones property"""
pass
P_shape = 0
Q_shape = 1
L_shape = 2
EMPTY = 0
FULL = 1
PARTIAL = 2
# automaton is used for pattern recognition when reducing a Q node
# -1 represents the result of a forbidden transition
# E F P
automaton = [[1, 5, 4], # 0 initial state
[1, 2, 2], # 1
[3, 2, 3], # 2
[3, -1, -1], # 3
[6, 2, 3], # 4
[6, 5, 6], # 5
[6, -1, -1]] # 6
class PQ_node:
def __init__(self, shape, value=None):
self.shape = shape
self.sons = []
self.parent = None
self.value = value
self.full_leafs = 0
self.processed_sons = 0
self.mark = EMPTY
def add(self, node):
"""Add one node as descendant
"""
self.sons.append(node)
node.parent = self
def add_all(self, L):
for x in L:
self.add(x)
def add_group(self, L):
"""Add elements of L as descendants of the node.
If there are several elements in L, group them in a P-node first
"""
if len(L) == 1:
self.add(L[0])
elif len(L) >= 2:
x = PQ_node(P_shape)
x.add_all(L)
self.add(x)
# nothing to be done for an empty list L
def border(self, L):
"""Append to L the border of the subtree.
"""
if self.shape == L_shape:
L.append(self.value)
else:
for x in self.sons:
x.border(L)
def __str__(self):
if self.shape == L_shape:
return str(self.value)
for x in self.sons:
assert x.parent == self
if self.shape == P_shape:
return "(" + ",".join(map(str, self.sons)) + ")"
else:
return "[" + ",".join(map(str, self.sons)) + "]"
class PQ_tree:
def __init__(self, universe):
self.tree = PQ_node(P_shape)
self.leafs = []
for i in universe:
x = PQ_node(L_shape, value=i)
self.tree.add(x)
self.leafs.append(x)
def __str__(self):
"""returns a string representation,
() for P nodes and [] for Q nodes
"""
return str(self.tree)
def border(self):
"""returns the list of the leafs in order
"""
L = []
self.tree.border(L)
return L
def reduce(self, S):
queue = deque(self.leafs)
cleanup = [] # we don't need to cleanup leafs
is_key_node = False
while queue and not is_key_node: # while there are nodes to be processed
x = queue.popleft()
is_key_node = (x.full_leafs == len(S))
x.mark = PARTIAL # default mark
if x.shape == P_shape:
E = [] # group descendants according to marks
F = []
P = []
for y in x.sons:
if y.mark == EMPTY:
E.append(y)
elif y.mark == FULL:
F.append(y)
else:
P.append(y)
if len(P) == 0: # start long case analysis
if len(E) == 0:
x.mark = FULL
else:
if len(F) == 0:
# template P1
x.mark = EMPTY
else:
if is_key_node:
# template P2
x.sons = E
x.add_group(F)
else: # is not root
# template P3
x.shape = Q_shape
x.sons = []
x.add_group(E)
x.add_group(F)
elif len(P) == 1:
assert P[0].shape == Q_shape
if is_key_node:
# template P4
x.sons = E + [P[0]]
P[0].add_group(F)
else: # is not root
# template P5
x.shape = Q_shape
x.sons = []
x.add_group(E)
x.add_all(P[0].sons)
x.add_group(F)
elif len(P) == 2:
if is_key_node:
# template P6
x.sons = E
z = P[0]
z.add_group(F)
z.add_all(reversed(P[1].sons))
else:
raise IsNotC1P
else: # more than 2 partial descendants
raise IsNotC1P
elif x.shape == Q_shape:
state = 0
L = []
for y in x.sons:
previous = state
state = automaton[state][y.mark]
if state == -1:
raise IsNotC1P
elif (state == 3 or state == 6) and y.mark == PARTIAL:
assert y.shape == Q_shape
L += reversed(y.sons)
elif state == 6 and previous == 4:
L = L[::-1]
L.append(y)
elif y.mark == PARTIAL:
L += y.sons
else:
L.append(y)
if state == 3 and not is_key_node:
raise IsNotC1P
elif state == 5:
x.mark = FULL
x.sons = []
if state == 6:
x.add_all(reversed(L))
else:
x.add_all(L)
else: # x is a leaf
if x.value in S:
x.mark = FULL
x.full_leafs = 1
else:
x.mark = EMPTY
x.full_leafs = 0
if not is_key_node: # propagate node processing
z = x.parent
z.full_leafs += x.full_leafs # cumulate bottom up full leaf numbers
if z.processed_sons == 0:
cleanup.append(z) # first time considered
z.processed_sons += 1
if z.processed_sons == len(z.sons):
queue.append(z) # otherwise prune tree at z
for x in cleanup:
x.full_leafs = 0
x.processed_sons = 0
x.mark = EMPTY
def consecutive_ones_property(sets, universe=None):
""" Check the consecutive ones property.
:param list sets: is a list of subsets of the ground set.
:param groundset: is the set of all elements,
by default it is the union of the given sets
:returns: returns a list of the ordered ground set where
every given set is consecutive,
or None if there is no solution.
:complexity: O(len(groundset) * len(sets))
:disclaimer: an optimal implementation would have complexity
O(len(groundset) + len(sets) + sum(map(len,sets))),
and there are more recent easier algorithms for this problem.
"""
if universe is None:
universe = set()
for S in sets:
universe |= set(S)
tree = PQ_tree(universe)
try:
for S in sets:
tree.reduce(S)
return tree.border()
except IsNotC1P:
return None
import sys
n, *a = sys.stdin.read().split()
n = int(n)
s = []
u = []
x = 0
for st in a:
cur = []
u.append(x)
x += 1
for i in range(n):
if st[i] == '1':
cur.append(i)
s.append(cur)
r = consecutive_ones_property(s, u)
if r is None:
print("NO")
else:
print("YES")
for i in range(n):
for j in range(n):
print(a[i][r[j]], end='')
print()
| Title: Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's consider an *n*<=×<=*n* square matrix, consisting of digits one and zero.
We'll consider a matrix good, if it meets the following condition: in each row of the matrix all ones go in one group. That is, each row of the matrix looks like that 00...0011...1100...00 (or simply consists of zeroes if it has no ones).
You are given matrix *a* of size *n*<=×<=*n*, consisting of zeroes and ones. Your task is to determine whether you can get a good matrix *b* from it by rearranging the columns or not.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=500) — the size of matrix *a*.
Each of *n* following lines contains *n* characters "0" and "1" — matrix *a*. Note that the characters are written without separators.
Output Specification:
Print "YES" in the first line, if you can rearrange the matrix columns so as to get a good matrix *b*. In the next *n* lines print the good matrix *b*. If there are multiple answers, you are allowed to print any of them.
If it is impossible to get a good matrix, print "NO".
Demo Input:
['6\n100010\n110110\n011001\n010010\n000100\n011001\n', '3\n110\n101\n011\n']
Demo Output:
['YES\n011000\n111100\n000111\n001100\n100000\n000111\n', 'NO\n']
Note:
none | ```python
from collections import deque
class IsNotC1P(Exception):
"""The given instance does not have the all consecutive ones property"""
pass
P_shape = 0
Q_shape = 1
L_shape = 2
EMPTY = 0
FULL = 1
PARTIAL = 2
# automaton is used for pattern recognition when reducing a Q node
# -1 represents the result of a forbidden transition
# E F P
automaton = [[1, 5, 4], # 0 initial state
[1, 2, 2], # 1
[3, 2, 3], # 2
[3, -1, -1], # 3
[6, 2, 3], # 4
[6, 5, 6], # 5
[6, -1, -1]] # 6
class PQ_node:
def __init__(self, shape, value=None):
self.shape = shape
self.sons = []
self.parent = None
self.value = value
self.full_leafs = 0
self.processed_sons = 0
self.mark = EMPTY
def add(self, node):
"""Add one node as descendant
"""
self.sons.append(node)
node.parent = self
def add_all(self, L):
for x in L:
self.add(x)
def add_group(self, L):
"""Add elements of L as descendants of the node.
If there are several elements in L, group them in a P-node first
"""
if len(L) == 1:
self.add(L[0])
elif len(L) >= 2:
x = PQ_node(P_shape)
x.add_all(L)
self.add(x)
# nothing to be done for an empty list L
def border(self, L):
"""Append to L the border of the subtree.
"""
if self.shape == L_shape:
L.append(self.value)
else:
for x in self.sons:
x.border(L)
def __str__(self):
if self.shape == L_shape:
return str(self.value)
for x in self.sons:
assert x.parent == self
if self.shape == P_shape:
return "(" + ",".join(map(str, self.sons)) + ")"
else:
return "[" + ",".join(map(str, self.sons)) + "]"
class PQ_tree:
def __init__(self, universe):
self.tree = PQ_node(P_shape)
self.leafs = []
for i in universe:
x = PQ_node(L_shape, value=i)
self.tree.add(x)
self.leafs.append(x)
def __str__(self):
"""returns a string representation,
() for P nodes and [] for Q nodes
"""
return str(self.tree)
def border(self):
"""returns the list of the leafs in order
"""
L = []
self.tree.border(L)
return L
def reduce(self, S):
queue = deque(self.leafs)
cleanup = [] # we don't need to cleanup leafs
is_key_node = False
while queue and not is_key_node: # while there are nodes to be processed
x = queue.popleft()
is_key_node = (x.full_leafs == len(S))
x.mark = PARTIAL # default mark
if x.shape == P_shape:
E = [] # group descendants according to marks
F = []
P = []
for y in x.sons:
if y.mark == EMPTY:
E.append(y)
elif y.mark == FULL:
F.append(y)
else:
P.append(y)
if len(P) == 0: # start long case analysis
if len(E) == 0:
x.mark = FULL
else:
if len(F) == 0:
# template P1
x.mark = EMPTY
else:
if is_key_node:
# template P2
x.sons = E
x.add_group(F)
else: # is not root
# template P3
x.shape = Q_shape
x.sons = []
x.add_group(E)
x.add_group(F)
elif len(P) == 1:
assert P[0].shape == Q_shape
if is_key_node:
# template P4
x.sons = E + [P[0]]
P[0].add_group(F)
else: # is not root
# template P5
x.shape = Q_shape
x.sons = []
x.add_group(E)
x.add_all(P[0].sons)
x.add_group(F)
elif len(P) == 2:
if is_key_node:
# template P6
x.sons = E
z = P[0]
z.add_group(F)
z.add_all(reversed(P[1].sons))
else:
raise IsNotC1P
else: # more than 2 partial descendants
raise IsNotC1P
elif x.shape == Q_shape:
state = 0
L = []
for y in x.sons:
previous = state
state = automaton[state][y.mark]
if state == -1:
raise IsNotC1P
elif (state == 3 or state == 6) and y.mark == PARTIAL:
assert y.shape == Q_shape
L += reversed(y.sons)
elif state == 6 and previous == 4:
L = L[::-1]
L.append(y)
elif y.mark == PARTIAL:
L += y.sons
else:
L.append(y)
if state == 3 and not is_key_node:
raise IsNotC1P
elif state == 5:
x.mark = FULL
x.sons = []
if state == 6:
x.add_all(reversed(L))
else:
x.add_all(L)
else: # x is a leaf
if x.value in S:
x.mark = FULL
x.full_leafs = 1
else:
x.mark = EMPTY
x.full_leafs = 0
if not is_key_node: # propagate node processing
z = x.parent
z.full_leafs += x.full_leafs # cumulate bottom up full leaf numbers
if z.processed_sons == 0:
cleanup.append(z) # first time considered
z.processed_sons += 1
if z.processed_sons == len(z.sons):
queue.append(z) # otherwise prune tree at z
for x in cleanup:
x.full_leafs = 0
x.processed_sons = 0
x.mark = EMPTY
def consecutive_ones_property(sets, universe=None):
""" Check the consecutive ones property.
:param list sets: is a list of subsets of the ground set.
:param groundset: is the set of all elements,
by default it is the union of the given sets
:returns: returns a list of the ordered ground set where
every given set is consecutive,
or None if there is no solution.
:complexity: O(len(groundset) * len(sets))
:disclaimer: an optimal implementation would have complexity
O(len(groundset) + len(sets) + sum(map(len,sets))),
and there are more recent easier algorithms for this problem.
"""
if universe is None:
universe = set()
for S in sets:
universe |= set(S)
tree = PQ_tree(universe)
try:
for S in sets:
tree.reduce(S)
return tree.border()
except IsNotC1P:
return None
import sys
n, *a = sys.stdin.read().split()
n = int(n)
s = []
u = []
x = 0
for st in a:
cur = []
u.append(x)
x += 1
for i in range(n):
if st[i] == '1':
cur.append(i)
s.append(cur)
r = consecutive_ones_property(s, u)
if r is None:
print("NO")
else:
print("YES")
for i in range(n):
for j in range(n):
print(a[i][r[j]], end='')
print()
``` | -1 | |
205 | A | Little Elephant and Rozdil | PROGRAMMING | 900 | [
"brute force",
"implementation"
] | null | null | The Little Elephant loves Ukraine very much. Most of all he loves town Rozdol (ukr. "Rozdil").
However, Rozdil is dangerous to settle, so the Little Elephant wants to go to some other town. The Little Elephant doesn't like to spend much time on travelling, so for his journey he will choose a town that needs minimum time to travel to. If there are multiple such cities, then the Little Elephant won't go anywhere.
For each town except for Rozdil you know the time needed to travel to this town. Find the town the Little Elephant will go to or print "Still Rozdil", if he stays in Rozdil. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of cities. The next line contains *n* integers, separated by single spaces: the *i*-th integer represents the time needed to go from town Rozdil to the *i*-th town. The time values are positive integers, not exceeding 109.
You can consider the cities numbered from 1 to *n*, inclusive. Rozdil is not among the numbered cities. | Print the answer on a single line — the number of the town the Little Elephant will go to. If there are multiple cities with minimum travel time, print "Still Rozdil" (without the quotes). | [
"2\n7 4\n",
"7\n7 4 47 100 4 9 12\n"
] | [
"2\n",
"Still Rozdil\n"
] | In the first sample there are only two cities where the Little Elephant can go. The travel time for the first town equals 7, to the second one — 4. The town which is closest to Rodzil (the only one) is the second one, so the answer is 2.
In the second sample the closest cities are cities two and five, the travelling time to both of them equals 4, so the answer is "Still Rozdil". | 500 | [
{
"input": "2\n7 4",
"output": "2"
},
{
"input": "7\n7 4 47 100 4 9 12",
"output": "Still Rozdil"
},
{
"input": "1\n47",
"output": "1"
},
{
"input": "2\n1000000000 1000000000",
"output": "Still Rozdil"
},
{
"input": "7\n7 6 5 4 3 2 1",
"output": "7"
},
{
... | 1,561,724,273 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 2 | 216 | 0 | n=int(input())
A=[int(x) for x in input().split()]
B=[x for x in range(n)]
C=sorted(zip(A,B))
if C[0][0]==C[1][0]:
print("Still Rozdil")
else:
print(C[0][1]+1) | Title: Little Elephant and Rozdil
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant loves Ukraine very much. Most of all he loves town Rozdol (ukr. "Rozdil").
However, Rozdil is dangerous to settle, so the Little Elephant wants to go to some other town. The Little Elephant doesn't like to spend much time on travelling, so for his journey he will choose a town that needs minimum time to travel to. If there are multiple such cities, then the Little Elephant won't go anywhere.
For each town except for Rozdil you know the time needed to travel to this town. Find the town the Little Elephant will go to or print "Still Rozdil", if he stays in Rozdil.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of cities. The next line contains *n* integers, separated by single spaces: the *i*-th integer represents the time needed to go from town Rozdil to the *i*-th town. The time values are positive integers, not exceeding 109.
You can consider the cities numbered from 1 to *n*, inclusive. Rozdil is not among the numbered cities.
Output Specification:
Print the answer on a single line — the number of the town the Little Elephant will go to. If there are multiple cities with minimum travel time, print "Still Rozdil" (without the quotes).
Demo Input:
['2\n7 4\n', '7\n7 4 47 100 4 9 12\n']
Demo Output:
['2\n', 'Still Rozdil\n']
Note:
In the first sample there are only two cities where the Little Elephant can go. The travel time for the first town equals 7, to the second one — 4. The town which is closest to Rodzil (the only one) is the second one, so the answer is 2.
In the second sample the closest cities are cities two and five, the travelling time to both of them equals 4, so the answer is "Still Rozdil". | ```python
n=int(input())
A=[int(x) for x in input().split()]
B=[x for x in range(n)]
C=sorted(zip(A,B))
if C[0][0]==C[1][0]:
print("Still Rozdil")
else:
print(C[0][1]+1)
``` | -1 | |
873 | D | Merge Sort | PROGRAMMING | 1,800 | [
"constructive algorithms",
"divide and conquer"
] | null | null | Merge sort is a well-known sorting algorithm. The main function that sorts the elements of array *a* with indices from [*l*,<=*r*) can be implemented as follows:
1. If the segment [*l*,<=*r*) is already sorted in non-descending order (that is, for any *i* such that *l*<=≤<=*i*<=<<=*r*<=-<=1 *a*[*i*]<=≤<=*a*[*i*<=+<=1]), then end the function call; 1. Let ; 1. Call *mergesort*(*a*,<=*l*,<=*mid*); 1. Call *mergesort*(*a*,<=*mid*,<=*r*); 1. Merge segments [*l*,<=*mid*) and [*mid*,<=*r*), making the segment [*l*,<=*r*) sorted in non-descending order. The merge algorithm doesn't call any other functions.
The array in this problem is 0-indexed, so to sort the whole array, you need to call *mergesort*(*a*,<=0,<=*n*).
The number of calls of function *mergesort* is very important, so Ivan has decided to calculate it while sorting the array. For example, if *a*<==<={1,<=2,<=3,<=4}, then there will be 1 call of *mergesort* — *mergesort*(0,<=4), which will check that the array is sorted and then end. If *a*<==<={2,<=1,<=3}, then the number of calls is 3: first of all, you call *mergesort*(0,<=3), which then sets *mid*<==<=1 and calls *mergesort*(0,<=1) and *mergesort*(1,<=3), which do not perform any recursive calls because segments (0,<=1) and (1,<=3) are sorted.
Ivan has implemented the program that counts the number of *mergesort* calls, but now he needs to test it. To do this, he needs to find an array *a* such that *a* is a permutation of size *n* (that is, the number of elements in *a* is *n*, and every integer number from [1,<=*n*] can be found in this array), and the number of *mergesort* calls when sorting the array is exactly *k*.
Help Ivan to find an array he wants! | The first line contains two numbers *n* and *k* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*k*<=≤<=200000) — the size of a desired permutation and the number of *mergesort* calls required to sort it. | If a permutation of size *n* such that there will be exactly *k* calls of *mergesort* while sorting it doesn't exist, output <=-<=1. Otherwise output *n* integer numbers *a*[0],<=*a*[1],<=...,<=*a*[*n*<=-<=1] — the elements of a permutation that would meet the required conditions. If there are multiple answers, print any of them. | [
"3 3\n",
"4 1\n",
"5 6\n"
] | [
"2 1 3 ",
"1 2 3 4 ",
"-1\n"
] | none | 0 | [
{
"input": "3 3",
"output": "2 1 3 "
},
{
"input": "4 1",
"output": "1 2 3 4 "
},
{
"input": "5 6",
"output": "-1"
},
{
"input": "100 100",
"output": "-1"
},
{
"input": "10000 10001",
"output": "3 1 5 2 7 4 8 10 6 12 9 13 15 11 17 14 18 20 16 22 19 23 25 21 27... | 1,645,782,962 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 307,200 | n, k = [int(i) for i in input().split(' ')]
test = [None] * 100010;
def unsort(l, r):
global k;
if (k == 1 or l + 1 == r):
return;
k -= 2;
mid = int((l + r) >> 1);
test[mid], test[mid - 1] = test[mid - 1], test[mid];
unsort(l,mid);
unsort(mid,r);
if (k % 2 == 0):
print("-1");
k /= 2;
for i in range(0, n, 1):
test[i] = i + 1;
unsort(0, n);
if (k != 1):
saida = "";
for j in range(0, n, 1):
saida += str(test[j]) + " ";
print(saida)
else:
print("-1")
| Title: Merge Sort
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Merge sort is a well-known sorting algorithm. The main function that sorts the elements of array *a* with indices from [*l*,<=*r*) can be implemented as follows:
1. If the segment [*l*,<=*r*) is already sorted in non-descending order (that is, for any *i* such that *l*<=≤<=*i*<=<<=*r*<=-<=1 *a*[*i*]<=≤<=*a*[*i*<=+<=1]), then end the function call; 1. Let ; 1. Call *mergesort*(*a*,<=*l*,<=*mid*); 1. Call *mergesort*(*a*,<=*mid*,<=*r*); 1. Merge segments [*l*,<=*mid*) and [*mid*,<=*r*), making the segment [*l*,<=*r*) sorted in non-descending order. The merge algorithm doesn't call any other functions.
The array in this problem is 0-indexed, so to sort the whole array, you need to call *mergesort*(*a*,<=0,<=*n*).
The number of calls of function *mergesort* is very important, so Ivan has decided to calculate it while sorting the array. For example, if *a*<==<={1,<=2,<=3,<=4}, then there will be 1 call of *mergesort* — *mergesort*(0,<=4), which will check that the array is sorted and then end. If *a*<==<={2,<=1,<=3}, then the number of calls is 3: first of all, you call *mergesort*(0,<=3), which then sets *mid*<==<=1 and calls *mergesort*(0,<=1) and *mergesort*(1,<=3), which do not perform any recursive calls because segments (0,<=1) and (1,<=3) are sorted.
Ivan has implemented the program that counts the number of *mergesort* calls, but now he needs to test it. To do this, he needs to find an array *a* such that *a* is a permutation of size *n* (that is, the number of elements in *a* is *n*, and every integer number from [1,<=*n*] can be found in this array), and the number of *mergesort* calls when sorting the array is exactly *k*.
Help Ivan to find an array he wants!
Input Specification:
The first line contains two numbers *n* and *k* (1<=≤<=*n*<=≤<=100000, 1<=≤<=*k*<=≤<=200000) — the size of a desired permutation and the number of *mergesort* calls required to sort it.
Output Specification:
If a permutation of size *n* such that there will be exactly *k* calls of *mergesort* while sorting it doesn't exist, output <=-<=1. Otherwise output *n* integer numbers *a*[0],<=*a*[1],<=...,<=*a*[*n*<=-<=1] — the elements of a permutation that would meet the required conditions. If there are multiple answers, print any of them.
Demo Input:
['3 3\n', '4 1\n', '5 6\n']
Demo Output:
['2 1 3 ', '1 2 3 4 ', '-1\n']
Note:
none | ```python
n, k = [int(i) for i in input().split(' ')]
test = [None] * 100010;
def unsort(l, r):
global k;
if (k == 1 or l + 1 == r):
return;
k -= 2;
mid = int((l + r) >> 1);
test[mid], test[mid - 1] = test[mid - 1], test[mid];
unsort(l,mid);
unsort(mid,r);
if (k % 2 == 0):
print("-1");
k /= 2;
for i in range(0, n, 1):
test[i] = i + 1;
unsort(0, n);
if (k != 1):
saida = "";
for j in range(0, n, 1):
saida += str(test[j]) + " ";
print(saida)
else:
print("-1")
``` | 0 | |
698 | A | Vacations | PROGRAMMING | 1,400 | [
"dp"
] | null | null | Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out. | Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days. | [
"4\n1 3 2 0\n",
"7\n1 3 3 2 1 2 3\n",
"2\n2 2\n"
] | [
"2\n",
"0\n",
"1\n"
] | In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day. | 500 | [
{
"input": "4\n1 3 2 0",
"output": "2"
},
{
"input": "7\n1 3 3 2 1 2 3",
"output": "0"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "10\n0 0 1 1 0 0 0 0 1 0",
"output": "8"
},
{
"input": "100\n3 2 3 3 3 2 3 1 ... | 1,622,742,129 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 93 | 0 | def fun(ls):
prv=None
count3=0
rest=0
count0=0
for i,val in enumerate(ls):
if(val==3):
count3+=1
else:
if((prv==1 and val==2 ) or (prv==2 and val==1 )):
if(count3%2==0):
rest+=0
else:
rest+=1
else:
if((prv==1 and val==1 ) or (prv==2 and val==2 )):
# 1 1 and 2 2
if(count3%2==0):
rest+=1
else:
rest+=0
else:
if(val==0):
count0+=1
if((prv==0 and val!=0 ) or (prv==0 and val!=0 )):
# 0 1 and 0 2
rest+=0
else:
if((prv==0 and val==0 )):
rest+=0
prv=val
count3=0
print(rest+count0)
# T = int(input())
T=1
for _ in range(T):
# var=input()
# st=input()
# val=int(input())
# lt= list(map(int, input().split()))
# ls= list(map(int, input().split()))
# st=input()
val=int(input())
# ls.append(n)
# mx=max(mx,n)
ls= list(map(int, input().split()))
# st=list(input())
# lt= list(map(int, input().split()))
fun(ls)
| Title: Vacations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out.
Output Specification:
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days.
Demo Input:
['4\n1 3 2 0\n', '7\n1 3 3 2 1 2 3\n', '2\n2 2\n']
Demo Output:
['2\n', '0\n', '1\n']
Note:
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day. | ```python
def fun(ls):
prv=None
count3=0
rest=0
count0=0
for i,val in enumerate(ls):
if(val==3):
count3+=1
else:
if((prv==1 and val==2 ) or (prv==2 and val==1 )):
if(count3%2==0):
rest+=0
else:
rest+=1
else:
if((prv==1 and val==1 ) or (prv==2 and val==2 )):
# 1 1 and 2 2
if(count3%2==0):
rest+=1
else:
rest+=0
else:
if(val==0):
count0+=1
if((prv==0 and val!=0 ) or (prv==0 and val!=0 )):
# 0 1 and 0 2
rest+=0
else:
if((prv==0 and val==0 )):
rest+=0
prv=val
count3=0
print(rest+count0)
# T = int(input())
T=1
for _ in range(T):
# var=input()
# st=input()
# val=int(input())
# lt= list(map(int, input().split()))
# ls= list(map(int, input().split()))
# st=input()
val=int(input())
# ls.append(n)
# mx=max(mx,n)
ls= list(map(int, input().split()))
# st=list(input())
# lt= list(map(int, input().split()))
fun(ls)
``` | 0 | |
609 | A | USB Flash Drives | PROGRAMMING | 800 | [
"greedy",
"implementation",
"sortings"
] | null | null | Sean is trying to save a large file to a USB flash drive. He has *n* USB flash drives with capacities equal to *a*1,<=*a*2,<=...,<=*a**n* megabytes. The file size is equal to *m* megabytes.
Find the minimum number of USB flash drives needed to write Sean's file, if he can split the file between drives. | The first line contains positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of USB flash drives.
The second line contains positive integer *m* (1<=≤<=*m*<=≤<=105) — the size of Sean's file.
Each of the next *n* lines contains positive integer *a**i* (1<=≤<=*a**i*<=≤<=1000) — the sizes of USB flash drives in megabytes.
It is guaranteed that the answer exists, i. e. the sum of all *a**i* is not less than *m*. | Print the minimum number of USB flash drives to write Sean's file, if he can split the file between drives. | [
"3\n5\n2\n1\n3\n",
"3\n6\n2\n3\n2\n",
"2\n5\n5\n10\n"
] | [
"2\n",
"3\n",
"1\n"
] | In the first example Sean needs only two USB flash drives — the first and the third.
In the second example Sean needs all three USB flash drives.
In the third example Sean needs only one USB flash drive and he can use any available USB flash drive — the first or the second. | 0 | [
{
"input": "3\n5\n2\n1\n3",
"output": "2"
},
{
"input": "3\n6\n2\n3\n2",
"output": "3"
},
{
"input": "2\n5\n5\n10",
"output": "1"
},
{
"input": "5\n16\n8\n1\n3\n4\n9",
"output": "2"
},
{
"input": "10\n121\n10\n37\n74\n56\n42\n39\n6\n68\n8\n100",
"output": "2"
... | 1,677,322,590 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 46 | 0 | import bisect as bi
from sys import stdin
input = stdin.readline
def main():
n = int(input())
m = int(input())
l = []
for _ in range(n):
q = int(input())
l.append(q)
l.sort(reverse = True)
w = 0
for i in range(n):
m -= l[i]
w += 1
if m <= 0:
print(w)
break
main()
| Title: USB Flash Drives
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sean is trying to save a large file to a USB flash drive. He has *n* USB flash drives with capacities equal to *a*1,<=*a*2,<=...,<=*a**n* megabytes. The file size is equal to *m* megabytes.
Find the minimum number of USB flash drives needed to write Sean's file, if he can split the file between drives.
Input Specification:
The first line contains positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of USB flash drives.
The second line contains positive integer *m* (1<=≤<=*m*<=≤<=105) — the size of Sean's file.
Each of the next *n* lines contains positive integer *a**i* (1<=≤<=*a**i*<=≤<=1000) — the sizes of USB flash drives in megabytes.
It is guaranteed that the answer exists, i. e. the sum of all *a**i* is not less than *m*.
Output Specification:
Print the minimum number of USB flash drives to write Sean's file, if he can split the file between drives.
Demo Input:
['3\n5\n2\n1\n3\n', '3\n6\n2\n3\n2\n', '2\n5\n5\n10\n']
Demo Output:
['2\n', '3\n', '1\n']
Note:
In the first example Sean needs only two USB flash drives — the first and the third.
In the second example Sean needs all three USB flash drives.
In the third example Sean needs only one USB flash drive and he can use any available USB flash drive — the first or the second. | ```python
import bisect as bi
from sys import stdin
input = stdin.readline
def main():
n = int(input())
m = int(input())
l = []
for _ in range(n):
q = int(input())
l.append(q)
l.sort(reverse = True)
w = 0
for i in range(n):
m -= l[i]
w += 1
if m <= 0:
print(w)
break
main()
``` | 3 | |
322 | B | Ciel and Flowers | PROGRAMMING | 1,600 | [
"combinatorics",
"math"
] | null | null | Fox Ciel has some flowers: *r* red flowers, *g* green flowers and *b* blue flowers. She wants to use these flowers to make several bouquets. There are 4 types of bouquets:
- To make a "red bouquet", it needs 3 red flowers. - To make a "green bouquet", it needs 3 green flowers. - To make a "blue bouquet", it needs 3 blue flowers. - To make a "mixing bouquet", it needs 1 red, 1 green and 1 blue flower.
Help Fox Ciel to find the maximal number of bouquets she can make. | The first line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=109) — the number of red, green and blue flowers. | Print the maximal number of bouquets Fox Ciel can make. | [
"3 6 9\n",
"4 4 4\n",
"0 0 0\n"
] | [
"6\n",
"4\n",
"0\n"
] | In test case 1, we can make 1 red bouquet, 2 green bouquets and 3 blue bouquets.
In test case 2, we can make 1 red, 1 green, 1 blue and 1 mixing bouquet. | 1,000 | [
{
"input": "3 6 9",
"output": "6"
},
{
"input": "4 4 4",
"output": "4"
},
{
"input": "0 0 0",
"output": "0"
},
{
"input": "0 3 6",
"output": "3"
},
{
"input": "7 8 9",
"output": "7"
},
{
"input": "8 8 9",
"output": "8"
},
{
"input": "15 3 9... | 1,618,426,645 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 33 | 154 | 0 | # def islucky(n):
# if(len(n)%2!=0):
# return False
# n4=0
# n7=0
# for c in n:
# if c=='4':
# n4+=1
# elif c=='7':
# n7+=1
# else:
# return False
# if n4==n7:
# return True
# else:
# return False
#
# n = input()
# t = int(n)
#
# i = t
# while True:
# if islucky(str(i)):
# print(i)
# break
# else:
# i+=1
#
#
r,b,g = [int(x) for x in input().split()]
s1 = r//3 + b//3 + g//3
s2 = 1 + (r-1)//3 + (b-1)//3 + (g-1)//3
s3 = 2 + (r-2)//3 + (b-2)//3 + (g-2)//3
# print(r,g,b)
print(max([s1,s2,s3]))
| Title: Ciel and Flowers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel has some flowers: *r* red flowers, *g* green flowers and *b* blue flowers. She wants to use these flowers to make several bouquets. There are 4 types of bouquets:
- To make a "red bouquet", it needs 3 red flowers. - To make a "green bouquet", it needs 3 green flowers. - To make a "blue bouquet", it needs 3 blue flowers. - To make a "mixing bouquet", it needs 1 red, 1 green and 1 blue flower.
Help Fox Ciel to find the maximal number of bouquets she can make.
Input Specification:
The first line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=109) — the number of red, green and blue flowers.
Output Specification:
Print the maximal number of bouquets Fox Ciel can make.
Demo Input:
['3 6 9\n', '4 4 4\n', '0 0 0\n']
Demo Output:
['6\n', '4\n', '0\n']
Note:
In test case 1, we can make 1 red bouquet, 2 green bouquets and 3 blue bouquets.
In test case 2, we can make 1 red, 1 green, 1 blue and 1 mixing bouquet. | ```python
# def islucky(n):
# if(len(n)%2!=0):
# return False
# n4=0
# n7=0
# for c in n:
# if c=='4':
# n4+=1
# elif c=='7':
# n7+=1
# else:
# return False
# if n4==n7:
# return True
# else:
# return False
#
# n = input()
# t = int(n)
#
# i = t
# while True:
# if islucky(str(i)):
# print(i)
# break
# else:
# i+=1
#
#
r,b,g = [int(x) for x in input().split()]
s1 = r//3 + b//3 + g//3
s2 = 1 + (r-1)//3 + (b-1)//3 + (g-1)//3
s3 = 2 + (r-2)//3 + (b-2)//3 + (g-2)//3
# print(r,g,b)
print(max([s1,s2,s3]))
``` | 0 | |
246 | A | Buggy Sorting | PROGRAMMING | 900 | [
"constructive algorithms",
"greedy",
"sortings"
] | null | null | Little boy Valera studies an algorithm of sorting an integer array. After studying the theory, he went on to the practical tasks. As a result, he wrote a program that sorts an array of *n* integers *a*1,<=*a*2,<=...,<=*a**n* in the non-decreasing order. The pseudocode of the program, written by Valera, is given below. The input of the program gets number *n* and array *a*.
But Valera could have made a mistake, because he hasn't yet fully learned the sorting algorithm. If Valera made a mistake in his program, you need to give a counter-example that makes his program work improperly (that is, the example that makes the program sort the array not in the non-decreasing order). If such example for the given value of *n* doesn't exist, print -1. | You've got a single integer *n* (1<=≤<=*n*<=≤<=50) — the size of the sorted array. | Print *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the counter-example, for which Valera's algorithm won't work correctly. If the counter-example that meets the described conditions is impossible to give, print -1.
If there are several counter-examples, consisting of *n* numbers, you are allowed to print any of them. | [
"1\n"
] | [
"-1\n"
] | none | 500 | [
{
"input": "1",
"output": "-1"
},
{
"input": "2",
"output": "-1"
},
{
"input": "3",
"output": "3 2 1 "
},
{
"input": "4",
"output": "4 3 2 1 "
},
{
"input": "5",
"output": "5 4 3 2 1 "
},
{
"input": "6",
"output": "6 5 4 3 2 1 "
},
{
"input... | 1,636,453,031 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 46 | 4,198,400 | n = int( input())
if n >= 3:
nums = [n]
for i in range (2, n):
nums.append(i)
nums.append(1)
print(nums)
else:
print(-1) | Title: Buggy Sorting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little boy Valera studies an algorithm of sorting an integer array. After studying the theory, he went on to the practical tasks. As a result, he wrote a program that sorts an array of *n* integers *a*1,<=*a*2,<=...,<=*a**n* in the non-decreasing order. The pseudocode of the program, written by Valera, is given below. The input of the program gets number *n* and array *a*.
But Valera could have made a mistake, because he hasn't yet fully learned the sorting algorithm. If Valera made a mistake in his program, you need to give a counter-example that makes his program work improperly (that is, the example that makes the program sort the array not in the non-decreasing order). If such example for the given value of *n* doesn't exist, print -1.
Input Specification:
You've got a single integer *n* (1<=≤<=*n*<=≤<=50) — the size of the sorted array.
Output Specification:
Print *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the counter-example, for which Valera's algorithm won't work correctly. If the counter-example that meets the described conditions is impossible to give, print -1.
If there are several counter-examples, consisting of *n* numbers, you are allowed to print any of them.
Demo Input:
['1\n']
Demo Output:
['-1\n']
Note:
none | ```python
n = int( input())
if n >= 3:
nums = [n]
for i in range (2, n):
nums.append(i)
nums.append(1)
print(nums)
else:
print(-1)
``` | 0 | |
789 | B | Masha and geometric depression | PROGRAMMING | 1,700 | [
"brute force",
"implementation",
"math"
] | null | null | Masha really loves algebra. On the last lesson, her strict teacher Dvastan gave she new exercise.
You are given geometric progression *b* defined by two integers *b*1 and *q*. Remind that a geometric progression is a sequence of integers *b*1,<=*b*2,<=*b*3,<=..., where for each *i*<=><=1 the respective term satisfies the condition *b**i*<==<=*b**i*<=-<=1·*q*, where *q* is called the common ratio of the progression. Progressions in Uzhlyandia are unusual: both *b*1 and *q* can equal 0. Also, Dvastan gave Masha *m* "bad" integers *a*1,<=*a*2,<=...,<=*a**m*, and an integer *l*.
Masha writes all progression terms one by one onto the board (including repetitive) while condition |*b**i*|<=≤<=*l* is satisfied (|*x*| means absolute value of *x*). There is an exception: if a term equals one of the "bad" integers, Masha skips it (doesn't write onto the board) and moves forward to the next term.
But the lesson is going to end soon, so Masha has to calculate how many integers will be written on the board. In order not to get into depression, Masha asked you for help: help her calculate how many numbers she will write, or print "inf" in case she needs to write infinitely many integers. | The first line of input contains four integers *b*1, *q*, *l*, *m* (-109<=≤<=*b*1,<=*q*<=≤<=109, 1<=≤<=*l*<=≤<=109, 1<=≤<=*m*<=≤<=105) — the initial term and the common ratio of progression, absolute value of maximal number that can be written on the board and the number of "bad" integers, respectively.
The second line contains *m* distinct integers *a*1,<=*a*2,<=...,<=*a**m* (-109<=≤<=*a**i*<=≤<=109) — numbers that will never be written on the board. | Print the only integer, meaning the number of progression terms that will be written on the board if it is finite, or "inf" (without quotes) otherwise. | [
"3 2 30 4\n6 14 25 48\n",
"123 1 2143435 4\n123 11 -5453 141245\n",
"123 1 2143435 4\n54343 -13 6 124\n"
] | [
"3",
"0",
"inf"
] | In the first sample case, Masha will write integers 3, 12, 24. Progression term 6 will be skipped because it is a "bad" integer. Terms bigger than 24 won't be written because they exceed *l* by absolute value.
In the second case, Masha won't write any number because all terms are equal 123 and this is a "bad" integer.
In the third case, Masha will write infinitely integers 123. | 1,000 | [
{
"input": "3 2 30 4\n6 14 25 48",
"output": "3"
},
{
"input": "123 1 2143435 4\n123 11 -5453 141245",
"output": "0"
},
{
"input": "123 1 2143435 4\n54343 -13 6 124",
"output": "inf"
},
{
"input": "3 2 25 2\n379195692 -69874783",
"output": "4"
},
{
"input": "3 2 3... | 1,490,818,285 | 2,147,483,647 | Python 3 | OK | TESTS | 116 | 93 | 12,902,400 | b, q, l, m = list(map(int, input().split()))
a = set(list(map(int, input().split())))
ans = 0
boo = False
i = 0
while (i < 34) and (abs(b) <= l):
if (b not in a):
ans += 1
if i > 31:
boo = True
b *= q
i += 1
if boo:
print('inf')
else:
print(ans) | Title: Masha and geometric depression
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Masha really loves algebra. On the last lesson, her strict teacher Dvastan gave she new exercise.
You are given geometric progression *b* defined by two integers *b*1 and *q*. Remind that a geometric progression is a sequence of integers *b*1,<=*b*2,<=*b*3,<=..., where for each *i*<=><=1 the respective term satisfies the condition *b**i*<==<=*b**i*<=-<=1·*q*, where *q* is called the common ratio of the progression. Progressions in Uzhlyandia are unusual: both *b*1 and *q* can equal 0. Also, Dvastan gave Masha *m* "bad" integers *a*1,<=*a*2,<=...,<=*a**m*, and an integer *l*.
Masha writes all progression terms one by one onto the board (including repetitive) while condition |*b**i*|<=≤<=*l* is satisfied (|*x*| means absolute value of *x*). There is an exception: if a term equals one of the "bad" integers, Masha skips it (doesn't write onto the board) and moves forward to the next term.
But the lesson is going to end soon, so Masha has to calculate how many integers will be written on the board. In order not to get into depression, Masha asked you for help: help her calculate how many numbers she will write, or print "inf" in case she needs to write infinitely many integers.
Input Specification:
The first line of input contains four integers *b*1, *q*, *l*, *m* (-109<=≤<=*b*1,<=*q*<=≤<=109, 1<=≤<=*l*<=≤<=109, 1<=≤<=*m*<=≤<=105) — the initial term and the common ratio of progression, absolute value of maximal number that can be written on the board and the number of "bad" integers, respectively.
The second line contains *m* distinct integers *a*1,<=*a*2,<=...,<=*a**m* (-109<=≤<=*a**i*<=≤<=109) — numbers that will never be written on the board.
Output Specification:
Print the only integer, meaning the number of progression terms that will be written on the board if it is finite, or "inf" (without quotes) otherwise.
Demo Input:
['3 2 30 4\n6 14 25 48\n', '123 1 2143435 4\n123 11 -5453 141245\n', '123 1 2143435 4\n54343 -13 6 124\n']
Demo Output:
['3', '0', 'inf']
Note:
In the first sample case, Masha will write integers 3, 12, 24. Progression term 6 will be skipped because it is a "bad" integer. Terms bigger than 24 won't be written because they exceed *l* by absolute value.
In the second case, Masha won't write any number because all terms are equal 123 and this is a "bad" integer.
In the third case, Masha will write infinitely integers 123. | ```python
b, q, l, m = list(map(int, input().split()))
a = set(list(map(int, input().split())))
ans = 0
boo = False
i = 0
while (i < 34) and (abs(b) <= l):
if (b not in a):
ans += 1
if i > 31:
boo = True
b *= q
i += 1
if boo:
print('inf')
else:
print(ans)
``` | 3 | |
1,000 | B | Light It Up | PROGRAMMING | 1,500 | [
"greedy"
] | null | null | Recently, you bought a brand new smart lamp with programming features. At first, you set up a schedule to the lamp. Every day it will turn power on at moment $0$ and turn power off at moment $M$. Moreover, the lamp allows you to set a program of switching its state (states are "lights on" and "lights off"). Unfortunately, some program is already installed into the lamp.
The lamp allows only good programs. Good program can be represented as a non-empty array $a$, where $0 < a_1 < a_2 < \dots < a_{|a|} < M$. All $a_i$ must be integers. Of course, preinstalled program is a good program.
The lamp follows program $a$ in next manner: at moment $0$ turns power and light on. Then at moment $a_i$ the lamp flips its state to opposite (if it was lit, it turns off, and vice versa). The state of the lamp flips instantly: for example, if you turn the light off at moment $1$ and then do nothing, the total time when the lamp is lit will be $1$. Finally, at moment $M$ the lamp is turning its power off regardless of its state.
Since you are not among those people who read instructions, and you don't understand the language it's written in, you realize (after some testing) the only possible way to alter the preinstalled program. You can insert at most one element into the program $a$, so it still should be a good program after alteration. Insertion can be done between any pair of consecutive elements of $a$, or even at the begining or at the end of $a$.
Find such a way to alter the program that the total time when the lamp is lit is maximum possible. Maybe you should leave program untouched. If the lamp is lit from $x$ till moment $y$, then its lit for $y - x$ units of time. Segments of time when the lamp is lit are summed up. | First line contains two space separated integers $n$ and $M$ ($1 \le n \le 10^5$, $2 \le M \le 10^9$) — the length of program $a$ and the moment when power turns off.
Second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($0 < a_1 < a_2 < \dots < a_n < M$) — initially installed program $a$. | Print the only integer — maximum possible total time when the lamp is lit. | [
"3 10\n4 6 7\n",
"2 12\n1 10\n",
"2 7\n3 4\n"
] | [
"8\n",
"9\n",
"6\n"
] | In the first example, one of possible optimal solutions is to insert value $x = 3$ before $a_1$, so program will be $[3, 4, 6, 7]$ and time of lamp being lit equals $(3 - 0) + (6 - 4) + (10 - 7) = 8$. Other possible solution is to insert $x = 5$ in appropriate place.
In the second example, there is only one optimal solution: to insert $x = 2$ between $a_1$ and $a_2$. Program will become $[1, 2, 10]$, and answer will be $(1 - 0) + (10 - 2) = 9$.
In the third example, optimal answer is to leave program untouched, so answer will be $(3 - 0) + (7 - 4) = 6$. | 0 | [
{
"input": "3 10\n4 6 7",
"output": "8"
},
{
"input": "2 12\n1 10",
"output": "9"
},
{
"input": "2 7\n3 4",
"output": "6"
},
{
"input": "1 2\n1",
"output": "1"
},
{
"input": "5 10\n1 3 5 6 8",
"output": "6"
},
{
"input": "7 1000000000\n1 10001 10011 20... | 1,635,300,150 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 1,000 | 4,505,600 | n,M=map(int,input().split())
a=[0]
adjust=[]
s=input().split()
for i in range(n):
a.append(int(s[i]))
a.append(int(M))
#print(a)
lit_time_initial=0
if len(a)%2!=0:
for i in range(0,len(a)-1,2):
lit_time_initial+=int(a[i+1])-int(a[i])
else:
for i in range(0,len(a),2):
lit_time_initial+=int(a[i+1])-int(a[i])
#print(lit_time_initial)
for i in range(M):
a_set=set(a)
a_set.add(i)
b=list(a_set)
adjust_time = 0
if len(b)%2!=0:
for i in range(0, len(b) - 1, 2):
adjust_time += int(b[i + 1]) - int(b[i])
else:
for i in range(0, len(b), 2):
adjust_time += int(b[i + 1]) - int(b[i])
if adjust_time>lit_time_initial:
adjust.append(adjust_time)
#print(adjust)
if len(adjust)==0:
print(lit_time_initial)
else:
print(max(adjust))
| Title: Light It Up
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, you bought a brand new smart lamp with programming features. At first, you set up a schedule to the lamp. Every day it will turn power on at moment $0$ and turn power off at moment $M$. Moreover, the lamp allows you to set a program of switching its state (states are "lights on" and "lights off"). Unfortunately, some program is already installed into the lamp.
The lamp allows only good programs. Good program can be represented as a non-empty array $a$, where $0 < a_1 < a_2 < \dots < a_{|a|} < M$. All $a_i$ must be integers. Of course, preinstalled program is a good program.
The lamp follows program $a$ in next manner: at moment $0$ turns power and light on. Then at moment $a_i$ the lamp flips its state to opposite (if it was lit, it turns off, and vice versa). The state of the lamp flips instantly: for example, if you turn the light off at moment $1$ and then do nothing, the total time when the lamp is lit will be $1$. Finally, at moment $M$ the lamp is turning its power off regardless of its state.
Since you are not among those people who read instructions, and you don't understand the language it's written in, you realize (after some testing) the only possible way to alter the preinstalled program. You can insert at most one element into the program $a$, so it still should be a good program after alteration. Insertion can be done between any pair of consecutive elements of $a$, or even at the begining or at the end of $a$.
Find such a way to alter the program that the total time when the lamp is lit is maximum possible. Maybe you should leave program untouched. If the lamp is lit from $x$ till moment $y$, then its lit for $y - x$ units of time. Segments of time when the lamp is lit are summed up.
Input Specification:
First line contains two space separated integers $n$ and $M$ ($1 \le n \le 10^5$, $2 \le M \le 10^9$) — the length of program $a$ and the moment when power turns off.
Second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($0 < a_1 < a_2 < \dots < a_n < M$) — initially installed program $a$.
Output Specification:
Print the only integer — maximum possible total time when the lamp is lit.
Demo Input:
['3 10\n4 6 7\n', '2 12\n1 10\n', '2 7\n3 4\n']
Demo Output:
['8\n', '9\n', '6\n']
Note:
In the first example, one of possible optimal solutions is to insert value $x = 3$ before $a_1$, so program will be $[3, 4, 6, 7]$ and time of lamp being lit equals $(3 - 0) + (6 - 4) + (10 - 7) = 8$. Other possible solution is to insert $x = 5$ in appropriate place.
In the second example, there is only one optimal solution: to insert $x = 2$ between $a_1$ and $a_2$. Program will become $[1, 2, 10]$, and answer will be $(1 - 0) + (10 - 2) = 9$.
In the third example, optimal answer is to leave program untouched, so answer will be $(3 - 0) + (7 - 4) = 6$. | ```python
n,M=map(int,input().split())
a=[0]
adjust=[]
s=input().split()
for i in range(n):
a.append(int(s[i]))
a.append(int(M))
#print(a)
lit_time_initial=0
if len(a)%2!=0:
for i in range(0,len(a)-1,2):
lit_time_initial+=int(a[i+1])-int(a[i])
else:
for i in range(0,len(a),2):
lit_time_initial+=int(a[i+1])-int(a[i])
#print(lit_time_initial)
for i in range(M):
a_set=set(a)
a_set.add(i)
b=list(a_set)
adjust_time = 0
if len(b)%2!=0:
for i in range(0, len(b) - 1, 2):
adjust_time += int(b[i + 1]) - int(b[i])
else:
for i in range(0, len(b), 2):
adjust_time += int(b[i + 1]) - int(b[i])
if adjust_time>lit_time_initial:
adjust.append(adjust_time)
#print(adjust)
if len(adjust)==0:
print(lit_time_initial)
else:
print(max(adjust))
``` | 0 | |
798 | A | Mike and palindrome | PROGRAMMING | 1,000 | [
"brute force",
"constructive algorithms",
"strings"
] | null | null | Mike has a string *s* consisting of only lowercase English letters. He wants to change exactly one character from the string so that the resulting one is a palindrome.
A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not. | The first and single line contains string *s* (1<=≤<=|*s*|<=≤<=15). | Print "YES" (without quotes) if Mike can change exactly one character so that the resulting string is palindrome or "NO" (without quotes) otherwise. | [
"abccaa\n",
"abbcca\n",
"abcda\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "abccaa",
"output": "YES"
},
{
"input": "abbcca",
"output": "NO"
},
{
"input": "abcda",
"output": "YES"
},
{
"input": "kyw",
"output": "YES"
},
{
"input": "fccf",
"output": "NO"
},
{
"input": "mnlm",
"output": "YES"
},
{
"input":... | 1,667,042,171 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 46 | 0 | a=input()
f=False
h=False
a1=a[:len(a)//2]
a2=a[len(a)//2:][::-1]
for i in range(len(a)//2):
if f==False and a1[i]!=a2[i]:
f=True
elif f==True and a1[i]!=a2[i]:
h=True
break
if h:
print('NO')
elif f and not h:
print('YES') | Title: Mike and palindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mike has a string *s* consisting of only lowercase English letters. He wants to change exactly one character from the string so that the resulting one is a palindrome.
A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not.
Input Specification:
The first and single line contains string *s* (1<=≤<=|*s*|<=≤<=15).
Output Specification:
Print "YES" (without quotes) if Mike can change exactly one character so that the resulting string is palindrome or "NO" (without quotes) otherwise.
Demo Input:
['abccaa\n', 'abbcca\n', 'abcda\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none | ```python
a=input()
f=False
h=False
a1=a[:len(a)//2]
a2=a[len(a)//2:][::-1]
for i in range(len(a)//2):
if f==False and a1[i]!=a2[i]:
f=True
elif f==True and a1[i]!=a2[i]:
h=True
break
if h:
print('NO')
elif f and not h:
print('YES')
``` | 0 | |
474 | A | Keyboard | PROGRAMMING | 900 | [
"implementation"
] | null | null | Our good friend Mole is trying to code a big message. He is typing on an unusual keyboard with characters arranged in following way:
Unfortunately Mole is blind, so sometimes it is problem for him to put his hands accurately. He accidentally moved both his hands with one position to the left or to the right. That means that now he presses not a button he wants, but one neighboring button (left or right, as specified in input).
We have a sequence of characters he has typed and we want to find the original message. | First line of the input contains one letter describing direction of shifting ('L' or 'R' respectively for left or right).
Second line contains a sequence of characters written by Mole. The size of this sequence will be no more than 100. Sequence contains only symbols that appear on Mole's keyboard. It doesn't contain spaces as there is no space on Mole's keyboard.
It is guaranteed that even though Mole hands are moved, he is still pressing buttons on keyboard and not hitting outside it. | Print a line that contains the original message. | [
"R\ns;;upimrrfod;pbr\n"
] | [
"allyouneedislove\n"
] | none | 500 | [
{
"input": "R\ns;;upimrrfod;pbr",
"output": "allyouneedislove"
},
{
"input": "R\nwertyuiop;lkjhgfdsxcvbnm,.",
"output": "qwertyuiolkjhgfdsazxcvbnm,"
},
{
"input": "L\nzxcvbnm,kjhgfdsaqwertyuio",
"output": "xcvbnm,.lkjhgfdswertyuiop"
},
{
"input": "R\nbubbuduppudup",
"outp... | 1,697,117,843 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 31 | 62 | 0 | string = "qwertyuiopasdfghjkl;zxcvbnm,./"
char = input()
inputstring = input()
answerstring = ""
addition = 1 if char == 'L' else -1
for x in inputstring :
num = string.index(x)+addition
answerstring += string[num]
print(answerstring) | Title: Keyboard
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Our good friend Mole is trying to code a big message. He is typing on an unusual keyboard with characters arranged in following way:
Unfortunately Mole is blind, so sometimes it is problem for him to put his hands accurately. He accidentally moved both his hands with one position to the left or to the right. That means that now he presses not a button he wants, but one neighboring button (left or right, as specified in input).
We have a sequence of characters he has typed and we want to find the original message.
Input Specification:
First line of the input contains one letter describing direction of shifting ('L' or 'R' respectively for left or right).
Second line contains a sequence of characters written by Mole. The size of this sequence will be no more than 100. Sequence contains only symbols that appear on Mole's keyboard. It doesn't contain spaces as there is no space on Mole's keyboard.
It is guaranteed that even though Mole hands are moved, he is still pressing buttons on keyboard and not hitting outside it.
Output Specification:
Print a line that contains the original message.
Demo Input:
['R\ns;;upimrrfod;pbr\n']
Demo Output:
['allyouneedislove\n']
Note:
none | ```python
string = "qwertyuiopasdfghjkl;zxcvbnm,./"
char = input()
inputstring = input()
answerstring = ""
addition = 1 if char == 'L' else -1
for x in inputstring :
num = string.index(x)+addition
answerstring += string[num]
print(answerstring)
``` | 3 | |
368 | B | Sereja and Suffixes | PROGRAMMING | 1,100 | [
"data structures",
"dp"
] | null | null | Sereja has an array *a*, consisting of *n* integers *a*1, *a*2, ..., *a**n*. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote out *m* integers *l*1,<=*l*2,<=...,<=*l**m* (1<=≤<=*l**i*<=≤<=*n*). For each number *l**i* he wants to know how many distinct numbers are staying on the positions *l**i*, *l**i*<=+<=1, ..., *n*. Formally, he want to find the number of distinct numbers among *a**l**i*,<=*a**l**i*<=+<=1,<=...,<=*a**n*.?
Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each *l**i*. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105) — the array elements.
Next *m* lines contain integers *l*1,<=*l*2,<=...,<=*l**m*. The *i*-th line contains integer *l**i* (1<=≤<=*l**i*<=≤<=*n*). | Print *m* lines — on the *i*-th line print the answer to the number *l**i*. | [
"10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n"
] | [
"6\n6\n6\n6\n6\n5\n4\n3\n2\n1\n"
] | none | 1,000 | [
{
"input": "10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10",
"output": "6\n6\n6\n6\n6\n5\n4\n3\n2\n1"
},
{
"input": "8 3\n8 6 4 3 4 2 4 8\n6\n4\n2",
"output": "3\n4\n5"
},
{
"input": "7 10\n1 3 8 6 2 2 7\n4\n2\n6\n3\n4\n4\n6\n2\n7\n4",
"output": "3\n5\n2\n4\n3\n3\... | 1,698,383,100 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 31 | 0 | n, m = map(int, input().split())
arr = list(map(int, input().split()))
# Step 1: Count unique
# Step 2: Create the dict
# Step 3: Loop through the array only ones
Unique = len(set(arr))
Data = {}
for i in set(arr):
Data[i] = arr.count(i)
Data[arr[0]] -= 1
if Data[arr[0]] == 0:
Unique -= 1
print(Unique)
for i in range(1, m):
print(Unique)
Data[arr[i]] -= 1
if Data[arr[i]] == 0:
Unique -= 1 | Title: Sereja and Suffixes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja has an array *a*, consisting of *n* integers *a*1, *a*2, ..., *a**n*. The boy cannot sit and do nothing, he decided to study an array. Sereja took a piece of paper and wrote out *m* integers *l*1,<=*l*2,<=...,<=*l**m* (1<=≤<=*l**i*<=≤<=*n*). For each number *l**i* he wants to know how many distinct numbers are staying on the positions *l**i*, *l**i*<=+<=1, ..., *n*. Formally, he want to find the number of distinct numbers among *a**l**i*,<=*a**l**i*<=+<=1,<=...,<=*a**n*.?
Sereja wrote out the necessary array elements but the array was so large and the boy was so pressed for time. Help him, find the answer for the described question for each *l**i*.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105) — the array elements.
Next *m* lines contain integers *l*1,<=*l*2,<=...,<=*l**m*. The *i*-th line contains integer *l**i* (1<=≤<=*l**i*<=≤<=*n*).
Output Specification:
Print *m* lines — on the *i*-th line print the answer to the number *l**i*.
Demo Input:
['10 10\n1 2 3 4 1 2 3 4 100000 99999\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n']
Demo Output:
['6\n6\n6\n6\n6\n5\n4\n3\n2\n1\n']
Note:
none | ```python
n, m = map(int, input().split())
arr = list(map(int, input().split()))
# Step 1: Count unique
# Step 2: Create the dict
# Step 3: Loop through the array only ones
Unique = len(set(arr))
Data = {}
for i in set(arr):
Data[i] = arr.count(i)
Data[arr[0]] -= 1
if Data[arr[0]] == 0:
Unique -= 1
print(Unique)
for i in range(1, m):
print(Unique)
Data[arr[i]] -= 1
if Data[arr[i]] == 0:
Unique -= 1
``` | 0 | |
80 | A | Panoramix's Prediction | PROGRAMMING | 800 | [
"brute force"
] | A. Panoramix's Prediction | 2 | 256 | A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix? | The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4. | Print YES, if *m* is the next prime number after *n*, or NO otherwise. | [
"3 5\n",
"7 11\n",
"7 9\n"
] | [
"YES",
"YES",
"NO"
] | none | 500 | [
{
"input": "3 5",
"output": "YES"
},
{
"input": "7 11",
"output": "YES"
},
{
"input": "7 9",
"output": "NO"
},
{
"input": "2 3",
"output": "YES"
},
{
"input": "2 4",
"output": "NO"
},
{
"input": "3 4",
"output": "NO"
},
{
"input": "3 5",
... | 1,526,322,736 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 156 | 7,065,600 | x=[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47]
a,b=map(int,input().split())
if b in x and x.index(b)-x.index(a)==1:
print('YES')
else:
print('NO') | Title: Panoramix's Prediction
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix?
Input Specification:
The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4.
Output Specification:
Print YES, if *m* is the next prime number after *n*, or NO otherwise.
Demo Input:
['3 5\n', '7 11\n', '7 9\n']
Demo Output:
['YES', 'YES', 'NO']
Note:
none | ```python
x=[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47]
a,b=map(int,input().split())
if b in x and x.index(b)-x.index(a)==1:
print('YES')
else:
print('NO')
``` | 3.947839 |
339 | A | Helpful Maths | PROGRAMMING | 800 | [
"greedy",
"implementation",
"sortings",
"strings"
] | null | null | Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum. | The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long. | Print the new sum that Xenia can count. | [
"3+2+1\n",
"1+1+3+1+3\n",
"2\n"
] | [
"1+2+3\n",
"1+1+1+3+3\n",
"2\n"
] | none | 500 | [
{
"input": "3+2+1",
"output": "1+2+3"
},
{
"input": "1+1+3+1+3",
"output": "1+1+1+3+3"
},
{
"input": "2",
"output": "2"
},
{
"input": "2+2+1+1+3",
"output": "1+1+2+2+3"
},
{
"input": "2+1+2+2+2+3+1+3+1+2",
"output": "1+1+1+2+2+2+2+2+3+3"
},
{
"input": ... | 1,697,132,204 | 2,147,483,647 | Python 3 | OK | TESTS | 21 | 62 | 0 | s=input()
l=[]
for i in s:
if i.isdigit():
l.append(i)
l.sort()
print('+'.join(l)) | Title: Helpful Maths
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Xenia the beginner mathematician is a third year student at elementary school. She is now learning the addition operation.
The teacher has written down the sum of multiple numbers. Pupils should calculate the sum. To make the calculation easier, the sum only contains numbers 1, 2 and 3. Still, that isn't enough for Xenia. She is only beginning to count, so she can calculate a sum only if the summands follow in non-decreasing order. For example, she can't calculate sum 1+3+2+1 but she can calculate sums 1+1+2 and 3+3.
You've got the sum that was written on the board. Rearrange the summans and print the sum in such a way that Xenia can calculate the sum.
Input Specification:
The first line contains a non-empty string *s* — the sum Xenia needs to count. String *s* contains no spaces. It only contains digits and characters "+". Besides, string *s* is a correct sum of numbers 1, 2 and 3. String *s* is at most 100 characters long.
Output Specification:
Print the new sum that Xenia can count.
Demo Input:
['3+2+1\n', '1+1+3+1+3\n', '2\n']
Demo Output:
['1+2+3\n', '1+1+1+3+3\n', '2\n']
Note:
none | ```python
s=input()
l=[]
for i in s:
if i.isdigit():
l.append(i)
l.sort()
print('+'.join(l))
``` | 3 | |
754 | A | Lesha and array splitting | PROGRAMMING | 1,200 | [
"constructive algorithms",
"greedy",
"implementation"
] | null | null | One spring day on his way to university Lesha found an array *A*. Lesha likes to split arrays into several parts. This time Lesha decided to split the array *A* into several, possibly one, new arrays so that the sum of elements in each of the new arrays is not zero. One more condition is that if we place the new arrays one after another they will form the old array *A*.
Lesha is tired now so he asked you to split the array. Help Lesha! | The first line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array *A*.
The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=103<=≤<=*a**i*<=≤<=103) — the elements of the array *A*. | If it is not possible to split the array *A* and satisfy all the constraints, print single line containing "NO" (without quotes).
Otherwise in the first line print "YES" (without quotes). In the next line print single integer *k* — the number of new arrays. In each of the next *k* lines print two integers *l**i* and *r**i* which denote the subarray *A*[*l**i*... *r**i*] of the initial array *A* being the *i*-th new array. Integers *l**i*, *r**i* should satisfy the following conditions:
- *l*1<==<=1 - *r**k*<==<=*n* - *r**i*<=+<=1<==<=*l**i*<=+<=1 for each 1<=≤<=*i*<=<<=*k*.
If there are multiple answers, print any of them. | [
"3\n1 2 -3\n",
"8\n9 -12 3 4 -4 -10 7 3\n",
"1\n0\n",
"4\n1 2 3 -5\n"
] | [
"YES\n2\n1 2\n3 3\n",
"YES\n2\n1 2\n3 8\n",
"NO\n",
"YES\n4\n1 1\n2 2\n3 3\n4 4\n"
] | none | 500 | [
{
"input": "3\n1 2 -3",
"output": "YES\n3\n1 1\n2 2\n3 3"
},
{
"input": "8\n9 -12 3 4 -4 -10 7 3",
"output": "YES\n8\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8"
},
{
"input": "1\n0",
"output": "NO"
},
{
"input": "4\n1 2 3 -5",
"output": "YES\n4\n1 1\n2 2\n3 3\n4 4"
},
{
... | 1,636,363,181 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 31 | 4,505,600 | n = int(input())
list1 = list(map(int,input().split()))
count = 0
count = 0
g = g.split(' ')
for i in range(len(g)-1):
if g[i]+g[i+1] !=0:
count+=1
if count == 0:
print('NO')
else:
print('YES')
print(count+1)
for i in range(1,count+2):
print(i,i) | Title: Lesha and array splitting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One spring day on his way to university Lesha found an array *A*. Lesha likes to split arrays into several parts. This time Lesha decided to split the array *A* into several, possibly one, new arrays so that the sum of elements in each of the new arrays is not zero. One more condition is that if we place the new arrays one after another they will form the old array *A*.
Lesha is tired now so he asked you to split the array. Help Lesha!
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array *A*.
The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=103<=≤<=*a**i*<=≤<=103) — the elements of the array *A*.
Output Specification:
If it is not possible to split the array *A* and satisfy all the constraints, print single line containing "NO" (without quotes).
Otherwise in the first line print "YES" (without quotes). In the next line print single integer *k* — the number of new arrays. In each of the next *k* lines print two integers *l**i* and *r**i* which denote the subarray *A*[*l**i*... *r**i*] of the initial array *A* being the *i*-th new array. Integers *l**i*, *r**i* should satisfy the following conditions:
- *l*1<==<=1 - *r**k*<==<=*n* - *r**i*<=+<=1<==<=*l**i*<=+<=1 for each 1<=≤<=*i*<=<<=*k*.
If there are multiple answers, print any of them.
Demo Input:
['3\n1 2 -3\n', '8\n9 -12 3 4 -4 -10 7 3\n', '1\n0\n', '4\n1 2 3 -5\n']
Demo Output:
['YES\n2\n1 2\n3 3\n', 'YES\n2\n1 2\n3 8\n', 'NO\n', 'YES\n4\n1 1\n2 2\n3 3\n4 4\n']
Note:
none | ```python
n = int(input())
list1 = list(map(int,input().split()))
count = 0
count = 0
g = g.split(' ')
for i in range(len(g)-1):
if g[i]+g[i+1] !=0:
count+=1
if count == 0:
print('NO')
else:
print('YES')
print(count+1)
for i in range(1,count+2):
print(i,i)
``` | -1 | |
812 | C | Sagheer and Nubian Market | PROGRAMMING | 1,500 | [
"binary search",
"sortings"
] | null | null | On his trip to Luxor and Aswan, Sagheer went to a Nubian market to buy some souvenirs for his friends and relatives. The market has some strange rules. It contains *n* different items numbered from 1 to *n*. The *i*-th item has base cost *a**i* Egyptian pounds. If Sagheer buys *k* items with indices *x*1,<=*x*2,<=...,<=*x**k*, then the cost of item *x**j* is *a**x**j*<=+<=*x**j*·*k* for 1<=≤<=*j*<=≤<=*k*. In other words, the cost of an item is equal to its base cost in addition to its index multiplied by the factor *k*.
Sagheer wants to buy as many souvenirs as possible without paying more than *S* Egyptian pounds. Note that he cannot buy a souvenir more than once. If there are many ways to maximize the number of souvenirs, he will choose the way that will minimize the total cost. Can you help him with this task? | The first line contains two integers *n* and *S* (1<=≤<=*n*<=≤<=105 and 1<=≤<=*S*<=≤<=109) — the number of souvenirs in the market and Sagheer's budget.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the base costs of the souvenirs. | On a single line, print two integers *k*, *T* — the maximum number of souvenirs Sagheer can buy and the minimum total cost to buy these *k* souvenirs. | [
"3 11\n2 3 5\n",
"4 100\n1 2 5 6\n",
"1 7\n7\n"
] | [
"2 11\n",
"4 54\n",
"0 0\n"
] | In the first example, he cannot take the three items because they will cost him [5, 9, 14] with total cost 28. If he decides to take only two items, then the costs will be [4, 7, 11]. So he can afford the first and second items.
In the second example, he can buy all items as they will cost him [5, 10, 17, 22].
In the third example, there is only one souvenir in the market which will cost him 8 pounds, so he cannot buy it. | 1,500 | [
{
"input": "3 11\n2 3 5",
"output": "2 11"
},
{
"input": "4 100\n1 2 5 6",
"output": "4 54"
},
{
"input": "1 7\n7",
"output": "0 0"
},
{
"input": "1 7\n5",
"output": "1 6"
},
{
"input": "1 1\n1",
"output": "0 0"
},
{
"input": "4 33\n4 3 2 1",
"outp... | 1,678,799,520 | 2,147,483,647 | PyPy 3 | OK | TESTS | 57 | 358 | 15,872,000 | n,s = map(int,input().split())
a = list(map(int,input().split()))
def find(k):
# ans = 0
b = [(i+1)*k+a[i] for i in range(n)]
b = sorted(b)
return sum(b[:k])
l,r = 0,n
ans = 0
while l<r:
md = (l+r+1)//2
ans = find(md)
# print(ans)
if ans<=s:
l = md
else:
r = md-1
print(l,find(l)) | Title: Sagheer and Nubian Market
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On his trip to Luxor and Aswan, Sagheer went to a Nubian market to buy some souvenirs for his friends and relatives. The market has some strange rules. It contains *n* different items numbered from 1 to *n*. The *i*-th item has base cost *a**i* Egyptian pounds. If Sagheer buys *k* items with indices *x*1,<=*x*2,<=...,<=*x**k*, then the cost of item *x**j* is *a**x**j*<=+<=*x**j*·*k* for 1<=≤<=*j*<=≤<=*k*. In other words, the cost of an item is equal to its base cost in addition to its index multiplied by the factor *k*.
Sagheer wants to buy as many souvenirs as possible without paying more than *S* Egyptian pounds. Note that he cannot buy a souvenir more than once. If there are many ways to maximize the number of souvenirs, he will choose the way that will minimize the total cost. Can you help him with this task?
Input Specification:
The first line contains two integers *n* and *S* (1<=≤<=*n*<=≤<=105 and 1<=≤<=*S*<=≤<=109) — the number of souvenirs in the market and Sagheer's budget.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the base costs of the souvenirs.
Output Specification:
On a single line, print two integers *k*, *T* — the maximum number of souvenirs Sagheer can buy and the minimum total cost to buy these *k* souvenirs.
Demo Input:
['3 11\n2 3 5\n', '4 100\n1 2 5 6\n', '1 7\n7\n']
Demo Output:
['2 11\n', '4 54\n', '0 0\n']
Note:
In the first example, he cannot take the three items because they will cost him [5, 9, 14] with total cost 28. If he decides to take only two items, then the costs will be [4, 7, 11]. So he can afford the first and second items.
In the second example, he can buy all items as they will cost him [5, 10, 17, 22].
In the third example, there is only one souvenir in the market which will cost him 8 pounds, so he cannot buy it. | ```python
n,s = map(int,input().split())
a = list(map(int,input().split()))
def find(k):
# ans = 0
b = [(i+1)*k+a[i] for i in range(n)]
b = sorted(b)
return sum(b[:k])
l,r = 0,n
ans = 0
while l<r:
md = (l+r+1)//2
ans = find(md)
# print(ans)
if ans<=s:
l = md
else:
r = md-1
print(l,find(l))
``` | 3 | |
432 | A | Choosing Teams | PROGRAMMING | 800 | [
"greedy",
"implementation",
"sortings"
] | null | null | The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times? | The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship. | Print a single number — the answer to the problem. | [
"5 2\n0 4 5 1 0\n",
"6 4\n0 1 2 3 4 5\n",
"6 5\n0 0 0 0 0 0\n"
] | [
"1\n",
"0\n",
"2\n"
] | In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits. | 500 | [
{
"input": "5 2\n0 4 5 1 0",
"output": "1"
},
{
"input": "6 4\n0 1 2 3 4 5",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0 0",
"output": "2"
},
{
"input": "3 4\n0 1 0",
"output": "1"
},
{
"input": "3 4\n0 2 0",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0... | 1,668,888,357 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 46 | 0 | n, m = [int(i) for i in input().split()]
k = [int(i) for i in input().split()]
a = []
for i in k:
if i +m <= 5:
a.append(i)
if len(a) < 3:
print(0)
else:
print(len(a)//3) | Title: Choosing Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times?
Input Specification:
The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship.
Output Specification:
Print a single number — the answer to the problem.
Demo Input:
['5 2\n0 4 5 1 0\n', '6 4\n0 1 2 3 4 5\n', '6 5\n0 0 0 0 0 0\n']
Demo Output:
['1\n', '0\n', '2\n']
Note:
In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits. | ```python
n, m = [int(i) for i in input().split()]
k = [int(i) for i in input().split()]
a = []
for i in k:
if i +m <= 5:
a.append(i)
if len(a) < 3:
print(0)
else:
print(len(a)//3)
``` | 3 | |
239 | A | Two Bags of Potatoes | PROGRAMMING | 1,200 | [
"greedy",
"implementation",
"math"
] | null | null | Valera had two bags of potatoes, the first of these bags contains *x* (*x*<=≥<=1) potatoes, and the second — *y* (*y*<=≥<=1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains *x* potatoes) Valera lost. Valera remembers that the total amount of potatoes (*x*<=+<=*y*) in the two bags, firstly, was not gerater than *n*, and, secondly, was divisible by *k*.
Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order. | The first line of input contains three integers *y*, *k*, *n* (1<=≤<=*y*,<=*k*,<=*n*<=≤<=109; <=≤<=105). | Print the list of whitespace-separated integers — all possible values of *x* in ascending order. You should print each possible value of *x* exactly once.
If there are no such values of *x* print a single integer -1. | [
"10 1 10\n",
"10 6 40\n"
] | [
"-1\n",
"2 8 14 20 26 \n"
] | none | 500 | [
{
"input": "10 1 10",
"output": "-1"
},
{
"input": "10 6 40",
"output": "2 8 14 20 26 "
},
{
"input": "10 1 20",
"output": "1 2 3 4 5 6 7 8 9 10 "
},
{
"input": "1 10000 1000000000",
"output": "9999 19999 29999 39999 49999 59999 69999 79999 89999 99999 109999 119999 12999... | 1,461,759,462 | 2,147,483,647 | Python 3 | OK | TESTS | 48 | 592 | 8,499,200 | '''
Created on Apr 27, 2016
Gmail : [email protected]
@author: Md. Rezwanul Haque
'''
'''
Created on Apr 27, 2016
Gmail : [email protected]
@author: Md. Rezwanul Haque
'''
import sys
f = sys.stdin
# f = open("input.txt", "r")
y, k, n = map(int, f.readline().strip().split())
if y >= n:
first = -1
else:
t = k
while t <= y:
t += k
first = t-y
if first == -1:
print(-1)
else:
if first+y > n:
print(-1)
else:
res = []
for i in range(first, n+1-y, k):
res.append(i)
print(*res) | Title: Two Bags of Potatoes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera had two bags of potatoes, the first of these bags contains *x* (*x*<=≥<=1) potatoes, and the second — *y* (*y*<=≥<=1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains *x* potatoes) Valera lost. Valera remembers that the total amount of potatoes (*x*<=+<=*y*) in the two bags, firstly, was not gerater than *n*, and, secondly, was divisible by *k*.
Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order.
Input Specification:
The first line of input contains three integers *y*, *k*, *n* (1<=≤<=*y*,<=*k*,<=*n*<=≤<=109; <=≤<=105).
Output Specification:
Print the list of whitespace-separated integers — all possible values of *x* in ascending order. You should print each possible value of *x* exactly once.
If there are no such values of *x* print a single integer -1.
Demo Input:
['10 1 10\n', '10 6 40\n']
Demo Output:
['-1\n', '2 8 14 20 26 \n']
Note:
none | ```python
'''
Created on Apr 27, 2016
Gmail : [email protected]
@author: Md. Rezwanul Haque
'''
'''
Created on Apr 27, 2016
Gmail : [email protected]
@author: Md. Rezwanul Haque
'''
import sys
f = sys.stdin
# f = open("input.txt", "r")
y, k, n = map(int, f.readline().strip().split())
if y >= n:
first = -1
else:
t = k
while t <= y:
t += k
first = t-y
if first == -1:
print(-1)
else:
if first+y > n:
print(-1)
else:
res = []
for i in range(first, n+1-y, k):
res.append(i)
print(*res)
``` | 3 | |
749 | A | Bachgold Problem | PROGRAMMING | 800 | [
"greedy",
"implementation",
"math",
"number theory"
] | null | null | Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*. | The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000). | The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them. | [
"5\n",
"6\n"
] | [
"2\n2 3\n",
"3\n2 2 2\n"
] | none | 500 | [
{
"input": "5",
"output": "2\n2 3"
},
{
"input": "6",
"output": "3\n2 2 2"
},
{
"input": "2",
"output": "1\n2"
},
{
"input": "3",
"output": "1\n3"
},
{
"input": "99999",
"output": "49999\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ... | 1,627,756,606 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 46 | 6,963,200 | resheto = [0] * 999
resheto[0] = 1
resheto[1] = 1
for i in range(2, len(resheto)):
if resheto[i] == 0:
for j in range(i*2, len(resheto), i):
resheto[j] += 1
n = int(input())
prosto = []
for i in range(n):
if resheto[i] == 0:
prosto.append(i)
for i in prosto:
if n - i in prosto:
break
print(i, n - i) | Title: Bachgold Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*.
Input Specification:
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000).
Output Specification:
The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them.
Demo Input:
['5\n', '6\n']
Demo Output:
['2\n2 3\n', '3\n2 2 2\n']
Note:
none | ```python
resheto = [0] * 999
resheto[0] = 1
resheto[1] = 1
for i in range(2, len(resheto)):
if resheto[i] == 0:
for j in range(i*2, len(resheto), i):
resheto[j] += 1
n = int(input())
prosto = []
for i in range(n):
if resheto[i] == 0:
prosto.append(i)
for i in prosto:
if n - i in prosto:
break
print(i, n - i)
``` | 0 | |
965 | D | Single-use Stones | PROGRAMMING | 1,900 | [
"binary search",
"flows",
"greedy",
"two pointers"
] | null | null | A lot of frogs want to cross a river. A river is $w$ units width, but frogs can only jump $l$ units long, where $l < w$. Frogs can also jump on lengths shorter than $l$. but can't jump longer. Hopefully, there are some stones in the river to help them.
The stones are located at integer distances from the banks. There are $a_i$ stones at the distance of $i$ units from the bank the frogs are currently at. Each stone can only be used once by one frog, after that it drowns in the water.
What is the maximum number of frogs that can cross the river, given that then can only jump on the stones? | The first line contains two integers $w$ and $l$ ($1 \le l < w \le 10^5$) — the width of the river and the maximum length of a frog's jump.
The second line contains $w - 1$ integers $a_1, a_2, \ldots, a_{w-1}$ ($0 \le a_i \le 10^4$), where $a_i$ is the number of stones at the distance $i$ from the bank the frogs are currently at. | Print a single integer — the maximum number of frogs that can cross the river. | [
"10 5\n0 0 1 0 2 0 0 1 0\n",
"10 3\n1 1 1 1 2 1 1 1 1\n"
] | [
"3\n",
"3\n"
] | In the first sample two frogs can use the different stones at the distance $5$, and one frog can use the stones at the distances $3$ and then $8$.
In the second sample although there are two stones at the distance $5$, that does not help. The three paths are: $0 \to 3 \to 6 \to 9 \to 10$, $0 \to 2 \to 5 \to 8 \to 10$, $0 \to 1 \to 4 \to 7 \to 10$. | 2,000 | [
{
"input": "10 5\n0 0 1 0 2 0 0 1 0",
"output": "3"
},
{
"input": "10 3\n1 1 1 1 2 1 1 1 1",
"output": "3"
},
{
"input": "2 1\n0",
"output": "0"
},
{
"input": "2 1\n5",
"output": "5"
},
{
"input": "10 4\n0 0 6 2 7 1 6 4 0",
"output": "8"
},
{
"input": ... | 1,524,865,683 | 2,147,483,647 | Python 3 | OK | TESTS | 16 | 217 | 16,793,600 | input_string = input().split()
n = int(input_string[0])
l = int(input_string[1])
input_string = input().split()
v = []
for i in range(0, len(input_string)):
v.append(int(input_string[i]))
dp = [0]
for i in range(1, len(v) + 1):
dp.append(dp[i-1] + v[i-1])
ans = 1000000000000000000000000000
for i in range(l, len(dp)):
ans = min(ans, dp[i] - dp[i - l])
print(ans) | Title: Single-use Stones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A lot of frogs want to cross a river. A river is $w$ units width, but frogs can only jump $l$ units long, where $l < w$. Frogs can also jump on lengths shorter than $l$. but can't jump longer. Hopefully, there are some stones in the river to help them.
The stones are located at integer distances from the banks. There are $a_i$ stones at the distance of $i$ units from the bank the frogs are currently at. Each stone can only be used once by one frog, after that it drowns in the water.
What is the maximum number of frogs that can cross the river, given that then can only jump on the stones?
Input Specification:
The first line contains two integers $w$ and $l$ ($1 \le l < w \le 10^5$) — the width of the river and the maximum length of a frog's jump.
The second line contains $w - 1$ integers $a_1, a_2, \ldots, a_{w-1}$ ($0 \le a_i \le 10^4$), where $a_i$ is the number of stones at the distance $i$ from the bank the frogs are currently at.
Output Specification:
Print a single integer — the maximum number of frogs that can cross the river.
Demo Input:
['10 5\n0 0 1 0 2 0 0 1 0\n', '10 3\n1 1 1 1 2 1 1 1 1\n']
Demo Output:
['3\n', '3\n']
Note:
In the first sample two frogs can use the different stones at the distance $5$, and one frog can use the stones at the distances $3$ and then $8$.
In the second sample although there are two stones at the distance $5$, that does not help. The three paths are: $0 \to 3 \to 6 \to 9 \to 10$, $0 \to 2 \to 5 \to 8 \to 10$, $0 \to 1 \to 4 \to 7 \to 10$. | ```python
input_string = input().split()
n = int(input_string[0])
l = int(input_string[1])
input_string = input().split()
v = []
for i in range(0, len(input_string)):
v.append(int(input_string[i]))
dp = [0]
for i in range(1, len(v) + 1):
dp.append(dp[i-1] + v[i-1])
ans = 1000000000000000000000000000
for i in range(l, len(dp)):
ans = min(ans, dp[i] - dp[i - l])
print(ans)
``` | 3 | |
55 | A | Flea travel | PROGRAMMING | 1,200 | [
"implementation",
"math"
] | A. Flea travel | 1 | 256 | A flea is sitting at one of the *n* hassocks, arranged in a circle, at the moment. After minute number *k* the flea jumps through *k*<=-<=1 hassoсks (clockwise). For example, after the first minute the flea jumps to the neighboring hassock. You should answer: will the flea visit all the hassocks or not. We assume that flea has infinitely much time for this jumping. | The only line contains single integer: 1<=≤<=*n*<=≤<=1000 — number of hassocks. | Output "YES" if all the hassocks will be visited and "NO" otherwise. | [
"1\n",
"3\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "1",
"output": "YES"
},
{
"input": "3",
"output": "NO"
},
{
"input": "2",
"output": "YES"
},
{
"input": "4",
"output": "YES"
},
{
"input": "5",
"output": "NO"
},
{
"input": "6",
"output": "NO"
},
{
"input": "7",
"output": "NO... | 1,450,134,758 | 2,147,483,647 | Python 3 | OK | TESTS | 83 | 93 | 4,403,200 | n, c, k = int(input()), 0, 0
v = [[False] * n for i in range(n)]
while not v[c][k]:
v[c][k] = True
k = (k + 1) % n
c = (c + k) % n
print('YES' if all(any(x) for x in v) else 'NO') | Title: Flea travel
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
A flea is sitting at one of the *n* hassocks, arranged in a circle, at the moment. After minute number *k* the flea jumps through *k*<=-<=1 hassoсks (clockwise). For example, after the first minute the flea jumps to the neighboring hassock. You should answer: will the flea visit all the hassocks or not. We assume that flea has infinitely much time for this jumping.
Input Specification:
The only line contains single integer: 1<=≤<=*n*<=≤<=1000 — number of hassocks.
Output Specification:
Output "YES" if all the hassocks will be visited and "NO" otherwise.
Demo Input:
['1\n', '3\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
n, c, k = int(input()), 0, 0
v = [[False] * n for i in range(n)]
while not v[c][k]:
v[c][k] = True
k = (k + 1) % n
c = (c + k) % n
print('YES' if all(any(x) for x in v) else 'NO')
``` | 3.945298 |
903 | G | Yet Another Maxflow Problem | PROGRAMMING | 2,700 | [
"data structures",
"flows",
"graphs"
] | null | null | In this problem you will have to deal with a very special network.
The network consists of two parts: part *A* and part *B*. Each part consists of *n* vertices; *i*-th vertex of part *A* is denoted as *A**i*, and *i*-th vertex of part *B* is denoted as *B**i*.
For each index *i* (1<=≤<=*i*<=<<=*n*) there is a directed edge from vertex *A**i* to vertex *A**i*<=+<=1, and from *B**i* to *B**i*<=+<=1, respectively. Capacities of these edges are given in the input. Also there might be several directed edges going from part *A* to part *B* (but never from *B* to *A*).
You have to calculate the [maximum flow value](https://en.wikipedia.org/wiki/Maximum_flow_problem) from *A*1 to *B**n* in this network. Capacities of edges connecting *A**i* to *A**i*<=+<=1 might sometimes change, and you also have to maintain the maximum flow value after these changes. Apart from that, the network is fixed (there are no changes in part *B*, no changes of edges going from *A* to *B*, and no edge insertions or deletions).
Take a look at the example and the notes to understand the structure of the network better. | The first line contains three integer numbers *n*, *m* and *q* (2<=≤<=*n*,<=*m*<=≤<=2·105, 0<=≤<=*q*<=≤<=2·105) — the number of vertices in each part, the number of edges going from *A* to *B* and the number of changes, respectively.
Then *n*<=-<=1 lines follow, *i*-th line contains two integers *x**i* and *y**i* denoting that the edge from *A**i* to *A**i*<=+<=1 has capacity *x**i* and the edge from *B**i* to *B**i*<=+<=1 has capacity *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=109).
Then *m* lines follow, describing the edges from *A* to *B*. Each line contains three integers *x*, *y* and *z* denoting an edge from *A**x* to *B**y* with capacity *z* (1<=≤<=*x*,<=*y*<=≤<=*n*, 1<=≤<=*z*<=≤<=109). There might be multiple edges from *A**x* to *B**y*.
And then *q* lines follow, describing a sequence of changes to the network. *i*-th line contains two integers *v**i* and *w**i*, denoting that the capacity of the edge from *A**v**i* to *A**v**i*<=+<=1 is set to *w**i* (1<=≤<=*v**i*<=<<=*n*, 1<=≤<=*w**i*<=≤<=109). | Firstly, print the maximum flow value in the original network. Then print *q* integers, *i*-th of them must be equal to the maximum flow value after *i*-th change. | [
"4 3 2\n1 2\n3 4\n5 6\n2 2 7\n1 4 8\n4 3 9\n1 100\n2 100\n"
] | [
"9\n14\n14\n"
] | This is the original network in the example: | 0 | [
{
"input": "4 3 2\n1 2\n3 4\n5 6\n2 2 7\n1 4 8\n4 3 9\n1 100\n2 100",
"output": "9\n14\n14"
},
{
"input": "10 10 10\n291546518 199012865\n327731857 137263959\n145140225 631959974\n559674936 815057131\n677050070 949982094\n839693202 160045764\n967872826 489258292\n706535160 594950620\n230389718 27478... | 1,683,662,340 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 62 | 841 | 88,473,600 | import heapq
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def make_graph(n, m):
x, y, s = [0] * (2 * m), [0] * m, [0] * (n + 3)
for i in range(0, 2 * m, 2):
u, v, w = map(int, input().split())
s[u + 2] += 1
x[i], x[i + 1] = u, v
y[i >> 1] = w
for i in range(3, n + 3):
s[i] += s[i - 1]
G, W = [0] * m, [0] * m
for i in range(0, 2 * m, 2):
j = x[i] + 1
G[s[j]] = x[i ^ 1]
W[s[j]] = y[i >> 1]
s[j] += 1
return G, W, s
def f(i):
if not lazy[i]:
return
tree[i] += lazy[i]
if i < l1:
lazy[i << 1] += lazy[i]
lazy[i << 1 ^ 1] += lazy[i]
lazy[i] = 0
return
def update(l, r, s):
q, ll, rr, i = [1], [0], [l1 - 1], 0
while len(q) ^ i:
j = q[i]
l0, r0 = ll[i], rr[i]
if l <= l0 and r0 <= r:
if s:
lazy[j] += s
f(j)
i += 1
continue
f(j)
m0 = (l0 + r0) >> 1
if l <= m0 and l0 <= r:
q.append(j << 1)
ll.append(l0)
rr.append(m0)
if l <= r0 and m0 + 1 <= r:
q.append(j << 1 ^ 1)
ll.append(m0 + 1)
rr.append(r0)
i += 1
for i in q[::-1]:
if i < l1:
j, k = i << 1, i << 1 ^ 1
f(j)
f(k)
tree[i] = min(tree[j], tree[k])
return
def get_min(s, t):
update(s, t, 0)
s += l1
t += l1
ans = inf
while s <= t:
if s % 2:
ans = min(ans, tree[s])
s += 1
s >>= 1
if not t % 2:
ans = min(ans, tree[t])
t -= 1
t >>= 1
return ans
n, m, q = map(int, input().split())
a, b = [0] * (n + 1), [0] * n
for i in range(1, n):
x, y = map(int, input().split())
a[i], b[i] = x, y
G, W, s0 = make_graph(n, m)
l1 = pow(2, (n + 1).bit_length())
l2 = 2 * l1
inf = pow(10, 15) + 1
tree, lazy = [inf] * l2, [0] * l2
mi = inf
for i in range(n - 1, -1, -1):
mi = min(mi, b[i])
tree[i + l1] = mi
for i in range(l1 - 1, 0, -1):
tree[i] = min(tree[2 * i], tree[2 * i + 1])
c = [0] * (n + 1)
for i in range(1, n + 1):
for j in range(s0[i], s0[i + 1]):
update(0, G[j] - 1, W[j])
c[i] = tree[1]
h = []
for i in range(1, n + 1):
heapq.heappush(h, (a[i] + c[i], i))
ans = [h[0][0]]
for _ in range(q):
v, w = map(int, input().split())
heapq.heappush(h, (w + c[v], v))
a[v] = w
while (a[h[0][1]] + c[h[0][1]]) ^ h[0][0]:
heapq.heappop(h)
ans0 = h[0][0]
ans.append(ans0)
sys.stdout.write("\n".join(map(str, ans))) | Title: Yet Another Maxflow Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem you will have to deal with a very special network.
The network consists of two parts: part *A* and part *B*. Each part consists of *n* vertices; *i*-th vertex of part *A* is denoted as *A**i*, and *i*-th vertex of part *B* is denoted as *B**i*.
For each index *i* (1<=≤<=*i*<=<<=*n*) there is a directed edge from vertex *A**i* to vertex *A**i*<=+<=1, and from *B**i* to *B**i*<=+<=1, respectively. Capacities of these edges are given in the input. Also there might be several directed edges going from part *A* to part *B* (but never from *B* to *A*).
You have to calculate the [maximum flow value](https://en.wikipedia.org/wiki/Maximum_flow_problem) from *A*1 to *B**n* in this network. Capacities of edges connecting *A**i* to *A**i*<=+<=1 might sometimes change, and you also have to maintain the maximum flow value after these changes. Apart from that, the network is fixed (there are no changes in part *B*, no changes of edges going from *A* to *B*, and no edge insertions or deletions).
Take a look at the example and the notes to understand the structure of the network better.
Input Specification:
The first line contains three integer numbers *n*, *m* and *q* (2<=≤<=*n*,<=*m*<=≤<=2·105, 0<=≤<=*q*<=≤<=2·105) — the number of vertices in each part, the number of edges going from *A* to *B* and the number of changes, respectively.
Then *n*<=-<=1 lines follow, *i*-th line contains two integers *x**i* and *y**i* denoting that the edge from *A**i* to *A**i*<=+<=1 has capacity *x**i* and the edge from *B**i* to *B**i*<=+<=1 has capacity *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=109).
Then *m* lines follow, describing the edges from *A* to *B*. Each line contains three integers *x*, *y* and *z* denoting an edge from *A**x* to *B**y* with capacity *z* (1<=≤<=*x*,<=*y*<=≤<=*n*, 1<=≤<=*z*<=≤<=109). There might be multiple edges from *A**x* to *B**y*.
And then *q* lines follow, describing a sequence of changes to the network. *i*-th line contains two integers *v**i* and *w**i*, denoting that the capacity of the edge from *A**v**i* to *A**v**i*<=+<=1 is set to *w**i* (1<=≤<=*v**i*<=<<=*n*, 1<=≤<=*w**i*<=≤<=109).
Output Specification:
Firstly, print the maximum flow value in the original network. Then print *q* integers, *i*-th of them must be equal to the maximum flow value after *i*-th change.
Demo Input:
['4 3 2\n1 2\n3 4\n5 6\n2 2 7\n1 4 8\n4 3 9\n1 100\n2 100\n']
Demo Output:
['9\n14\n14\n']
Note:
This is the original network in the example: | ```python
import heapq
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def make_graph(n, m):
x, y, s = [0] * (2 * m), [0] * m, [0] * (n + 3)
for i in range(0, 2 * m, 2):
u, v, w = map(int, input().split())
s[u + 2] += 1
x[i], x[i + 1] = u, v
y[i >> 1] = w
for i in range(3, n + 3):
s[i] += s[i - 1]
G, W = [0] * m, [0] * m
for i in range(0, 2 * m, 2):
j = x[i] + 1
G[s[j]] = x[i ^ 1]
W[s[j]] = y[i >> 1]
s[j] += 1
return G, W, s
def f(i):
if not lazy[i]:
return
tree[i] += lazy[i]
if i < l1:
lazy[i << 1] += lazy[i]
lazy[i << 1 ^ 1] += lazy[i]
lazy[i] = 0
return
def update(l, r, s):
q, ll, rr, i = [1], [0], [l1 - 1], 0
while len(q) ^ i:
j = q[i]
l0, r0 = ll[i], rr[i]
if l <= l0 and r0 <= r:
if s:
lazy[j] += s
f(j)
i += 1
continue
f(j)
m0 = (l0 + r0) >> 1
if l <= m0 and l0 <= r:
q.append(j << 1)
ll.append(l0)
rr.append(m0)
if l <= r0 and m0 + 1 <= r:
q.append(j << 1 ^ 1)
ll.append(m0 + 1)
rr.append(r0)
i += 1
for i in q[::-1]:
if i < l1:
j, k = i << 1, i << 1 ^ 1
f(j)
f(k)
tree[i] = min(tree[j], tree[k])
return
def get_min(s, t):
update(s, t, 0)
s += l1
t += l1
ans = inf
while s <= t:
if s % 2:
ans = min(ans, tree[s])
s += 1
s >>= 1
if not t % 2:
ans = min(ans, tree[t])
t -= 1
t >>= 1
return ans
n, m, q = map(int, input().split())
a, b = [0] * (n + 1), [0] * n
for i in range(1, n):
x, y = map(int, input().split())
a[i], b[i] = x, y
G, W, s0 = make_graph(n, m)
l1 = pow(2, (n + 1).bit_length())
l2 = 2 * l1
inf = pow(10, 15) + 1
tree, lazy = [inf] * l2, [0] * l2
mi = inf
for i in range(n - 1, -1, -1):
mi = min(mi, b[i])
tree[i + l1] = mi
for i in range(l1 - 1, 0, -1):
tree[i] = min(tree[2 * i], tree[2 * i + 1])
c = [0] * (n + 1)
for i in range(1, n + 1):
for j in range(s0[i], s0[i + 1]):
update(0, G[j] - 1, W[j])
c[i] = tree[1]
h = []
for i in range(1, n + 1):
heapq.heappush(h, (a[i] + c[i], i))
ans = [h[0][0]]
for _ in range(q):
v, w = map(int, input().split())
heapq.heappush(h, (w + c[v], v))
a[v] = w
while (a[h[0][1]] + c[h[0][1]]) ^ h[0][0]:
heapq.heappop(h)
ans0 = h[0][0]
ans.append(ans0)
sys.stdout.write("\n".join(map(str, ans)))
``` | 3 | |
1,011 | A | Stages | PROGRAMMING | 900 | [
"greedy",
"implementation",
"sortings"
] | null | null | Natasha is going to fly to Mars. She needs to build a rocket, which consists of several stages in some order. Each of the stages is defined by a lowercase Latin letter. This way, the rocket can be described by the string — concatenation of letters, which correspond to the stages.
There are $n$ stages available. The rocket must contain exactly $k$ of them. Stages in the rocket should be ordered by their weight. So, after the stage with some letter can go only stage with a letter, which is at least two positions after in the alphabet (skipping one letter in between, or even more). For example, after letter 'c' can't go letters 'a', 'b', 'c' and 'd', but can go letters 'e', 'f', ..., 'z'.
For the rocket to fly as far as possible, its weight should be minimal. The weight of the rocket is equal to the sum of the weights of its stages. The weight of the stage is the number of its letter in the alphabet. For example, the stage 'a 'weighs one ton,' b 'weighs two tons, and' z' — $26$ tons.
Build the rocket with the minimal weight or determine, that it is impossible to build a rocket at all. Each stage can be used at most once. | The first line of input contains two integers — $n$ and $k$ ($1 \le k \le n \le 50$) – the number of available stages and the number of stages to use in the rocket.
The second line contains string $s$, which consists of exactly $n$ lowercase Latin letters. Each letter defines a new stage, which can be used to build the rocket. Each stage can be used at most once. | Print a single integer — the minimal total weight of the rocket or -1, if it is impossible to build the rocket at all. | [
"5 3\nxyabd\n",
"7 4\nproblem\n",
"2 2\nab\n",
"12 1\nabaabbaaabbb\n"
] | [
"29",
"34",
"-1",
"1"
] | In the first example, the following rockets satisfy the condition:
- "adx" (weight is $1+4+24=29$);- "ady" (weight is $1+4+25=30$);- "bdx" (weight is $2+4+24=30$);- "bdy" (weight is $2+4+25=31$).
Rocket "adx" has the minimal weight, so the answer is $29$.
In the second example, target rocket is "belo". Its weight is $2+5+12+15=34$.
In the third example, $n=k=2$, so the rocket must have both stages: 'a' and 'b'. This rocket doesn't satisfy the condition, because these letters are adjacent in the alphabet. Answer is -1. | 500 | [
{
"input": "5 3\nxyabd",
"output": "29"
},
{
"input": "7 4\nproblem",
"output": "34"
},
{
"input": "2 2\nab",
"output": "-1"
},
{
"input": "12 1\nabaabbaaabbb",
"output": "1"
},
{
"input": "50 13\nqwertyuiopasdfghjklzxcvbnmaaaaaaaaaaaaaaaaaaaaaaaa",
"output": ... | 1,533,707,179 | 2,147,483,647 | PyPy 3 | OK | TESTS | 29 | 140 | 0 | n,k = map(int,input().split())
s = input().rstrip()
l = list(s)
sum1 = 0
d = {}
for i in range(26):
d[chr(97 + i)] = i + 1
l.sort()
sum1 += d[l[0]]
count = 1
i,j = 1,0
while (i < len(l)) and (j < i) and count < k:
if d[l[i]] < d[l[j]] + 2:
i += 1
continue
else:
sum1 += d[l[i]]
count += 1
j = i
i += 1
if count < k:
print(-1)
else:
print(sum1) | Title: Stages
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Natasha is going to fly to Mars. She needs to build a rocket, which consists of several stages in some order. Each of the stages is defined by a lowercase Latin letter. This way, the rocket can be described by the string — concatenation of letters, which correspond to the stages.
There are $n$ stages available. The rocket must contain exactly $k$ of them. Stages in the rocket should be ordered by their weight. So, after the stage with some letter can go only stage with a letter, which is at least two positions after in the alphabet (skipping one letter in between, or even more). For example, after letter 'c' can't go letters 'a', 'b', 'c' and 'd', but can go letters 'e', 'f', ..., 'z'.
For the rocket to fly as far as possible, its weight should be minimal. The weight of the rocket is equal to the sum of the weights of its stages. The weight of the stage is the number of its letter in the alphabet. For example, the stage 'a 'weighs one ton,' b 'weighs two tons, and' z' — $26$ tons.
Build the rocket with the minimal weight or determine, that it is impossible to build a rocket at all. Each stage can be used at most once.
Input Specification:
The first line of input contains two integers — $n$ and $k$ ($1 \le k \le n \le 50$) – the number of available stages and the number of stages to use in the rocket.
The second line contains string $s$, which consists of exactly $n$ lowercase Latin letters. Each letter defines a new stage, which can be used to build the rocket. Each stage can be used at most once.
Output Specification:
Print a single integer — the minimal total weight of the rocket or -1, if it is impossible to build the rocket at all.
Demo Input:
['5 3\nxyabd\n', '7 4\nproblem\n', '2 2\nab\n', '12 1\nabaabbaaabbb\n']
Demo Output:
['29', '34', '-1', '1']
Note:
In the first example, the following rockets satisfy the condition:
- "adx" (weight is $1+4+24=29$);- "ady" (weight is $1+4+25=30$);- "bdx" (weight is $2+4+24=30$);- "bdy" (weight is $2+4+25=31$).
Rocket "adx" has the minimal weight, so the answer is $29$.
In the second example, target rocket is "belo". Its weight is $2+5+12+15=34$.
In the third example, $n=k=2$, so the rocket must have both stages: 'a' and 'b'. This rocket doesn't satisfy the condition, because these letters are adjacent in the alphabet. Answer is -1. | ```python
n,k = map(int,input().split())
s = input().rstrip()
l = list(s)
sum1 = 0
d = {}
for i in range(26):
d[chr(97 + i)] = i + 1
l.sort()
sum1 += d[l[0]]
count = 1
i,j = 1,0
while (i < len(l)) and (j < i) and count < k:
if d[l[i]] < d[l[j]] + 2:
i += 1
continue
else:
sum1 += d[l[i]]
count += 1
j = i
i += 1
if count < k:
print(-1)
else:
print(sum1)
``` | 3 | |
59 | A | Word | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Word | 2 | 256 | Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. | The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. | Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. | [
"HoUse\n",
"ViP\n",
"maTRIx\n"
] | [
"house\n",
"VIP\n",
"matrix\n"
] | none | 500 | [
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chn... | 1,563,478,354 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 248 | 0 | string = input()
up = 0
low = 0
i = 0
while(i<len(string)):
if ord(string[i])>=65 and ord(string[i])<=90:
up+=1
else:
low+=1
i+=1
if up>low:
ok = string.upper()
else:
ok = string.lower()
print(ok) | Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none | ```python
string = input()
up = 0
low = 0
i = 0
while(i<len(string)):
if ord(string[i])>=65 and ord(string[i])<=90:
up+=1
else:
low+=1
i+=1
if up>low:
ok = string.upper()
else:
ok = string.lower()
print(ok)
``` | 3.938 |
928 | B | Chat | PROGRAMMING | 1,400 | [
"*special",
"dp"
] | null | null | There are times you recall a good old friend and everything you've come through together. Luckily there are social networks — they store all your message history making it easy to know what you argued over 10 years ago.
More formal, your message history is a sequence of messages ordered by time sent numbered from 1 to *n* where *n* is the total number of messages in the chat.
Each message might contain a link to an earlier message which it is a reply to. When opening a message *x* or getting a link to it, the dialogue is shown in such a way that *k* previous messages, message *x* and *k* next messages are visible (with respect to message *x*). In case there are less than *k* messages somewhere, they are yet all shown.
Digging deep into your message history, you always read all visible messages and then go by the link in the current message *x* (if there is one) and continue reading in the same manner.
Determine the number of messages you'll read if your start from message number *t* for all *t* from 1 to *n*. Calculate these numbers independently. If you start with message *x*, the initial configuration is *x* itself, *k* previous and *k* next messages. Messages read multiple times are considered as one. | The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 0<=≤<=*k*<=≤<=*n*) — the total amount of messages and the number of previous and next messages visible.
The second line features a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=<<=*i*), where *a**i* denotes the *i*-th message link destination or zero, if there's no link from *i*. All messages are listed in chronological order. It's guaranteed that the link from message *x* goes to message with number strictly less than *x*. | Print *n* integers with *i*-th denoting the number of distinct messages you can read starting from message *i* and traversing the links while possible. | [
"6 0\n0 1 1 2 3 2\n",
"10 1\n0 1 0 3 4 5 2 3 7 0\n",
"2 2\n0 1\n"
] | [
"1 2 2 3 3 3 \n",
"2 3 3 4 5 6 6 6 8 2 \n",
"2 2 \n"
] | Consider *i* = 6 in sample case one. You will read message 6, then 2, then 1 and then there will be no link to go.
In the second sample case *i* = 6 gives you messages 5, 6, 7 since *k* = 1, then 4, 5, 6, then 2, 3, 4 and then the link sequence breaks. The number of distinct messages here is equal to 6. | 1,250 | [
{
"input": "6 0\n0 1 1 2 3 2",
"output": "1 2 2 3 3 3 "
},
{
"input": "10 1\n0 1 0 3 4 5 2 3 7 0",
"output": "2 3 3 4 5 6 6 6 8 2 "
},
{
"input": "2 2\n0 1",
"output": "2 2 "
},
{
"input": "1 1\n0",
"output": "1 "
},
{
"input": "5 2\n0 1 2 3 1",
"output": "3 4... | 1,519,502,903 | 16,403 | Python 3 | TIME_LIMIT_EXCEEDED | PRETESTS | 8 | 1,000 | 11,673,600 | a,b=map(int,input().split(' '))
A=list(map(int,input().split(' ')))
c=[]
for i in range(a):
s=1
B=[]
k=i
while A[k]!=0:
B.append(k + 1)
if k + 1 - b > 0:
for j in range(k, k - b, -1):
B.append(j)
else:
for j in range(k, 0, -1):
B.append(j)
if a - b < k + 1:
for j in range(k + 2, a + 1):
B.append(j)
else:
for j in range(k + 2, k + b + 2):
B.append(j)
k=A[k]-1
B.append(k + 1)
if k + 1 - b > 0:
for j in range(k, k - b, -1):
B.append(j)
else:
for j in range(k, 0, -1):
B.append(j)
if a - b < k + 1:
for j in range(k + 2, a + 1):
B.append(j)
else:
for j in range(k + 2, k + b + 2):
B.append(j)
c.append(len(set(B)))
print(*c)
| Title: Chat
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are times you recall a good old friend and everything you've come through together. Luckily there are social networks — they store all your message history making it easy to know what you argued over 10 years ago.
More formal, your message history is a sequence of messages ordered by time sent numbered from 1 to *n* where *n* is the total number of messages in the chat.
Each message might contain a link to an earlier message which it is a reply to. When opening a message *x* or getting a link to it, the dialogue is shown in such a way that *k* previous messages, message *x* and *k* next messages are visible (with respect to message *x*). In case there are less than *k* messages somewhere, they are yet all shown.
Digging deep into your message history, you always read all visible messages and then go by the link in the current message *x* (if there is one) and continue reading in the same manner.
Determine the number of messages you'll read if your start from message number *t* for all *t* from 1 to *n*. Calculate these numbers independently. If you start with message *x*, the initial configuration is *x* itself, *k* previous and *k* next messages. Messages read multiple times are considered as one.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 0<=≤<=*k*<=≤<=*n*) — the total amount of messages and the number of previous and next messages visible.
The second line features a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=<<=*i*), where *a**i* denotes the *i*-th message link destination or zero, if there's no link from *i*. All messages are listed in chronological order. It's guaranteed that the link from message *x* goes to message with number strictly less than *x*.
Output Specification:
Print *n* integers with *i*-th denoting the number of distinct messages you can read starting from message *i* and traversing the links while possible.
Demo Input:
['6 0\n0 1 1 2 3 2\n', '10 1\n0 1 0 3 4 5 2 3 7 0\n', '2 2\n0 1\n']
Demo Output:
['1 2 2 3 3 3 \n', '2 3 3 4 5 6 6 6 8 2 \n', '2 2 \n']
Note:
Consider *i* = 6 in sample case one. You will read message 6, then 2, then 1 and then there will be no link to go.
In the second sample case *i* = 6 gives you messages 5, 6, 7 since *k* = 1, then 4, 5, 6, then 2, 3, 4 and then the link sequence breaks. The number of distinct messages here is equal to 6. | ```python
a,b=map(int,input().split(' '))
A=list(map(int,input().split(' ')))
c=[]
for i in range(a):
s=1
B=[]
k=i
while A[k]!=0:
B.append(k + 1)
if k + 1 - b > 0:
for j in range(k, k - b, -1):
B.append(j)
else:
for j in range(k, 0, -1):
B.append(j)
if a - b < k + 1:
for j in range(k + 2, a + 1):
B.append(j)
else:
for j in range(k + 2, k + b + 2):
B.append(j)
k=A[k]-1
B.append(k + 1)
if k + 1 - b > 0:
for j in range(k, k - b, -1):
B.append(j)
else:
for j in range(k, 0, -1):
B.append(j)
if a - b < k + 1:
for j in range(k + 2, a + 1):
B.append(j)
else:
for j in range(k + 2, k + b + 2):
B.append(j)
c.append(len(set(B)))
print(*c)
``` | 0 | |
978 | A | Remove Duplicates | PROGRAMMING | 800 | [
"implementation"
] | null | null | Petya has an array $a$ consisting of $n$ integers. He wants to remove duplicate (equal) elements.
Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed. | The first line contains a single integer $n$ ($1 \le n \le 50$) — the number of elements in Petya's array.
The following line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1\,000$) — the Petya's array. | In the first line print integer $x$ — the number of elements which will be left in Petya's array after he removed the duplicates.
In the second line print $x$ integers separated with a space — Petya's array after he removed the duplicates. For each unique element only the rightmost entry should be left. | [
"6\n1 5 5 1 6 1\n",
"5\n2 4 2 4 4\n",
"5\n6 6 6 6 6\n"
] | [
"3\n5 6 1 \n",
"2\n2 4 \n",
"1\n6 \n"
] | In the first example you should remove two integers $1$, which are in the positions $1$ and $4$. Also you should remove the integer $5$, which is in the position $2$.
In the second example you should remove integer $2$, which is in the position $1$, and two integers $4$, which are in the positions $2$ and $4$.
In the third example you should remove four integers $6$, which are in the positions $1$, $2$, $3$ and $4$. | 0 | [
{
"input": "6\n1 5 5 1 6 1",
"output": "3\n5 6 1 "
},
{
"input": "5\n2 4 2 4 4",
"output": "2\n2 4 "
},
{
"input": "5\n6 6 6 6 6",
"output": "1\n6 "
},
{
"input": "7\n1 2 3 4 2 2 3",
"output": "4\n1 4 2 3 "
},
{
"input": "9\n100 100 100 99 99 99 100 100 100",
... | 1,697,153,764 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 30 | 0 | x = [1, 5, 5, 1, 6, 1]
# get the last insatnce of the number in the list
ans = []
for i in range(len(x) - 1, 0, -1):
if x[i] not in ans:
ans.insert(0, x[i])
else:
continue
print(len(ans))
print(*ans)
| Title: Remove Duplicates
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya has an array $a$ consisting of $n$ integers. He wants to remove duplicate (equal) elements.
Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 50$) — the number of elements in Petya's array.
The following line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1\,000$) — the Petya's array.
Output Specification:
In the first line print integer $x$ — the number of elements which will be left in Petya's array after he removed the duplicates.
In the second line print $x$ integers separated with a space — Petya's array after he removed the duplicates. For each unique element only the rightmost entry should be left.
Demo Input:
['6\n1 5 5 1 6 1\n', '5\n2 4 2 4 4\n', '5\n6 6 6 6 6\n']
Demo Output:
['3\n5 6 1 \n', '2\n2 4 \n', '1\n6 \n']
Note:
In the first example you should remove two integers $1$, which are in the positions $1$ and $4$. Also you should remove the integer $5$, which is in the position $2$.
In the second example you should remove integer $2$, which is in the position $1$, and two integers $4$, which are in the positions $2$ and $4$.
In the third example you should remove four integers $6$, which are in the positions $1$, $2$, $3$ and $4$. | ```python
x = [1, 5, 5, 1, 6, 1]
# get the last insatnce of the number in the list
ans = []
for i in range(len(x) - 1, 0, -1):
if x[i] not in ans:
ans.insert(0, x[i])
else:
continue
print(len(ans))
print(*ans)
``` | 0 | |
573 | B | Bear and Blocks | PROGRAMMING | 1,600 | [
"binary search",
"data structures",
"dp",
"math"
] | null | null | Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built *n* towers in a row. The *i*-th tower is made of *h**i* identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers. | The first line contains single integer *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=109) — sizes of towers. | Print the number of operations needed to destroy all towers. | [
"6\n2 1 4 6 2 2\n",
"7\n3 3 3 1 3 3 3\n"
] | [
"3\n",
"2\n"
] | The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. | 1,000 | [
{
"input": "6\n2 1 4 6 2 2",
"output": "3"
},
{
"input": "7\n3 3 3 1 3 3 3",
"output": "2"
},
{
"input": "7\n5128 5672 5805 5452 5882 5567 5032",
"output": "4"
},
{
"input": "10\n1 2 2 3 5 5 5 4 2 1",
"output": "5"
},
{
"input": "14\n20 20 20 20 20 20 3 20 20 20 2... | 1,463,984,004 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 6 | 1,000 | 13,824,000 | n = int(input())
heights = [0] + [int(h) for h in input().split(' ')] + [0]
tmp = [0 for i in range(len(heights))]
step = 0
while True:
done = True
for i in range(1, n):
tmp[i] = max(0, min(heights[i - 1], heights[i + 1], heights[i] - 1))
if tmp[i] > 0:
done = False
heights = tmp[:]
step += 1
if done:
break
print(step)
| Title: Bear and Blocks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is a little bear who loves to play. Today he is playing by destroying block towers. He built *n* towers in a row. The *i*-th tower is made of *h**i* identical blocks. For clarification see picture for the first sample.
Limak will repeat the following operation till everything is destroyed.
Block is called internal if it has all four neighbors, i.e. it has each side (top, left, down and right) adjacent to other block or to the floor. Otherwise, block is boundary. In one operation Limak destroys all boundary blocks. His paws are very fast and he destroys all those blocks at the same time.
Limak is ready to start. You task is to count how many operations will it take him to destroy all towers.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=109) — sizes of towers.
Output Specification:
Print the number of operations needed to destroy all towers.
Demo Input:
['6\n2 1 4 6 2 2\n', '7\n3 3 3 1 3 3 3\n']
Demo Output:
['3\n', '2\n']
Note:
The picture below shows all three operations for the first sample test. Each time boundary blocks are marked with red color. | ```python
n = int(input())
heights = [0] + [int(h) for h in input().split(' ')] + [0]
tmp = [0 for i in range(len(heights))]
step = 0
while True:
done = True
for i in range(1, n):
tmp[i] = max(0, min(heights[i - 1], heights[i + 1], heights[i] - 1))
if tmp[i] > 0:
done = False
heights = tmp[:]
step += 1
if done:
break
print(step)
``` | 0 | |
58 | A | Chat room | PROGRAMMING | 1,000 | [
"greedy",
"strings"
] | A. Chat room | 1 | 256 | Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. | The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. | If Vasya managed to say hello, print "YES", otherwise print "NO". | [
"ahhellllloou\n",
"hlelo\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymeda... | 1,571,157,928 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 108 | 0 | st = input()
strr = st[::-1]
if st.index('h')<st.index('e') and st.index('e')<st.index('l') and strr.index('l')<strr.index('e') and st.index('l')>st.index('o'):
print('YES')
else:
print('NO') | Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
st = input()
strr = st[::-1]
if st.index('h')<st.index('e') and st.index('e')<st.index('l') and strr.index('l')<strr.index('e') and st.index('l')>st.index('o'):
print('YES')
else:
print('NO')
``` | 0 |
248 | A | Cupboards | PROGRAMMING | 800 | [
"implementation"
] | null | null | One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*. | The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces. | In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs. | [
"5\n0 1\n1 0\n0 1\n1 1\n0 1\n"
] | [
"3\n"
] | none | 500 | [
{
"input": "5\n0 1\n1 0\n0 1\n1 1\n0 1",
"output": "3"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1\n1 1\n1 1",
"output": "1"
},
{
"input": "8\n0 1\n1 0\n0 1\n1 1\n0 1\n1 0\n0 1\n1 0",
"output": "7"
},
{
"input": "8\n1 0\n1 0\n1 0\n0 1\n0 1\n1 ... | 1,545,310,154 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 20 | 248 | 0 | ro = 0
lc = 0
lo = 0
rc = 0
r = []
l = []
n = int(input())
for i in range(0,n):
l = list(map(int,input().split()))
if l[0] == 1:
lo +=1
else:
lc+=1
if l[1] == 1:
ro+=1
else:
rc+=1
maxl = max(lo,lc)
maxr = max(ro,lc)
print(n-maxl+n-maxr)
| Title: Cupboards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*.
Input Specification:
The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces.
Output Specification:
In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs.
Demo Input:
['5\n0 1\n1 0\n0 1\n1 1\n0 1\n']
Demo Output:
['3\n']
Note:
none | ```python
ro = 0
lc = 0
lo = 0
rc = 0
r = []
l = []
n = int(input())
for i in range(0,n):
l = list(map(int,input().split()))
if l[0] == 1:
lo +=1
else:
lc+=1
if l[1] == 1:
ro+=1
else:
rc+=1
maxl = max(lo,lc)
maxr = max(ro,lc)
print(n-maxl+n-maxr)
``` | 0 | |
1,006 | E | Military Problem | PROGRAMMING | 1,600 | [
"dfs and similar",
"graphs",
"trees"
] | null | null | In this problem you will have to help Berland army with organizing their command delivery system.
There are $n$ officers in Berland army. The first officer is the commander of the army, and he does not have any superiors. Every other officer has exactly one direct superior. If officer $a$ is the direct superior of officer $b$, then we also can say that officer $b$ is a direct subordinate of officer $a$.
Officer $x$ is considered to be a subordinate (direct or indirect) of officer $y$ if one of the following conditions holds:
- officer $y$ is the direct superior of officer $x$; - the direct superior of officer $x$ is a subordinate of officer $y$.
For example, on the picture below the subordinates of the officer $3$ are: $5, 6, 7, 8, 9$.
The structure of Berland army is organized in such a way that every officer, except for the commander, is a subordinate of the commander of the army.
Formally, let's represent Berland army as a tree consisting of $n$ vertices, in which vertex $u$ corresponds to officer $u$. The parent of vertex $u$ corresponds to the direct superior of officer $u$. The root (which has index $1$) corresponds to the commander of the army.
Berland War Ministry has ordered you to give answers on $q$ queries, the $i$-th query is given as $(u_i, k_i)$, where $u_i$ is some officer, and $k_i$ is a positive integer.
To process the $i$-th query imagine how a command from $u_i$ spreads to the subordinates of $u_i$. Typical DFS (depth first search) algorithm is used here.
Suppose the current officer is $a$ and he spreads a command. Officer $a$ chooses $b$ — one of his direct subordinates (i.e. a child in the tree) who has not received this command yet. If there are many such direct subordinates, then $a$ chooses the one having minimal index. Officer $a$ gives a command to officer $b$. Afterwards, $b$ uses exactly the same algorithm to spread the command to its subtree. After $b$ finishes spreading the command, officer $a$ chooses the next direct subordinate again (using the same strategy). When officer $a$ cannot choose any direct subordinate who still hasn't received this command, officer $a$ finishes spreading the command.
Let's look at the following example:
If officer $1$ spreads a command, officers receive it in the following order: $[1, 2, 3, 5 ,6, 8, 7, 9, 4]$.
If officer $3$ spreads a command, officers receive it in the following order: $[3, 5, 6, 8, 7, 9]$.
If officer $7$ spreads a command, officers receive it in the following order: $[7, 9]$.
If officer $9$ spreads a command, officers receive it in the following order: $[9]$.
To answer the $i$-th query $(u_i, k_i)$, construct a sequence which describes the order in which officers will receive the command if the $u_i$-th officer spreads it. Return the $k_i$-th element of the constructed list or -1 if there are fewer than $k_i$ elements in it.
You should process queries independently. A query doesn't affect the following queries. | The first line of the input contains two integers $n$ and $q$ ($2 \le n \le 2 \cdot 10^5, 1 \le q \le 2 \cdot 10^5$) — the number of officers in Berland army and the number of queries.
The second line of the input contains $n - 1$ integers $p_2, p_3, \dots, p_n$ ($1 \le p_i < i$), where $p_i$ is the index of the direct superior of the officer having the index $i$. The commander has index $1$ and doesn't have any superiors.
The next $q$ lines describe the queries. The $i$-th query is given as a pair ($u_i, k_i$) ($1 \le u_i, k_i \le n$), where $u_i$ is the index of the officer which starts spreading a command, and $k_i$ is the index of the required officer in the command spreading sequence. | Print $q$ numbers, where the $i$-th number is the officer at the position $k_i$ in the list which describes the order in which officers will receive the command if it starts spreading from officer $u_i$. Print "-1" if the number of officers which receive the command is less than $k_i$.
You should process queries independently. They do not affect each other. | [
"9 6\n1 1 1 3 5 3 5 7\n3 1\n1 5\n3 4\n7 3\n1 8\n1 9\n"
] | [
"3\n6\n8\n-1\n9\n4\n"
] | none | 0 | [
{
"input": "9 6\n1 1 1 3 5 3 5 7\n3 1\n1 5\n3 4\n7 3\n1 8\n1 9",
"output": "3\n6\n8\n-1\n9\n4"
},
{
"input": "2 1\n1\n1 1",
"output": "1"
},
{
"input": "13 12\n1 1 1 1 1 1 1 1 1 1 1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "1\n1\n1\n1\n1\n1\n1\n1\n1\n... | 1,590,358,131 | 2,147,483,647 | PyPy 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <bits/stdc++.h>
using namespace std;
//simple for loop
#define f(n) for(int i=0; i<n; i++)
typedef long long int ll;
typedef vector<ll> vll;
ll inf=numeric_limits<ll>::max();
ll findNumofSUbords(ll v, vector<vll>& adjList, vll& numofSUbords)
{
if (adjList[v].empty())
{
numofSUbords[v]=0;
return(0);
}
ll subords=0;
for(auto neig: adjList[v])
{
subords+=findNumofSUbords(neig, adjList, numofSUbords)+1;
}
numofSUbords[v]=subords;
return(subords);
}
void dfs(ll v, vector<vll>& adjList, vll& orderOfVisit)
{
orderOfVisit.push_back(v);
for(auto neig: adjList[v])
{
dfs(neig, adjList, orderOfVisit);
}
}
int main()
{
ll n, q, a, b;
cin>>n>>q;
vector<vll> adjList(n+1);
vll numofSubords(n+1, 0), orderOfVisit;
f(n-1){ cin>>a; adjList[a].push_back(i+2);}
// for(int i=0; i<n+1; i++)
// {
// cout<<i<<": ";
// for(auto y: adjList[i]) cout<<y<<" ";
// cout<<"\n";
// }
dfs(1, adjList, orderOfVisit);
// for(auto x: orderOfVisit) cout<<x<<" ";
findNumofSUbords(1, adjList, numofSubords);
// for(auto x: numofSubords) cout<<x<<" ";
f(q)
{
cin>>a>>b;
if (numofSubords[a]+1<b)cout<<-1<<'\n';
else cout<<orderOfVisit[a-1+b-1]<<'\n';
}
return(0);
} | Title: Military Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem you will have to help Berland army with organizing their command delivery system.
There are $n$ officers in Berland army. The first officer is the commander of the army, and he does not have any superiors. Every other officer has exactly one direct superior. If officer $a$ is the direct superior of officer $b$, then we also can say that officer $b$ is a direct subordinate of officer $a$.
Officer $x$ is considered to be a subordinate (direct or indirect) of officer $y$ if one of the following conditions holds:
- officer $y$ is the direct superior of officer $x$; - the direct superior of officer $x$ is a subordinate of officer $y$.
For example, on the picture below the subordinates of the officer $3$ are: $5, 6, 7, 8, 9$.
The structure of Berland army is organized in such a way that every officer, except for the commander, is a subordinate of the commander of the army.
Formally, let's represent Berland army as a tree consisting of $n$ vertices, in which vertex $u$ corresponds to officer $u$. The parent of vertex $u$ corresponds to the direct superior of officer $u$. The root (which has index $1$) corresponds to the commander of the army.
Berland War Ministry has ordered you to give answers on $q$ queries, the $i$-th query is given as $(u_i, k_i)$, where $u_i$ is some officer, and $k_i$ is a positive integer.
To process the $i$-th query imagine how a command from $u_i$ spreads to the subordinates of $u_i$. Typical DFS (depth first search) algorithm is used here.
Suppose the current officer is $a$ and he spreads a command. Officer $a$ chooses $b$ — one of his direct subordinates (i.e. a child in the tree) who has not received this command yet. If there are many such direct subordinates, then $a$ chooses the one having minimal index. Officer $a$ gives a command to officer $b$. Afterwards, $b$ uses exactly the same algorithm to spread the command to its subtree. After $b$ finishes spreading the command, officer $a$ chooses the next direct subordinate again (using the same strategy). When officer $a$ cannot choose any direct subordinate who still hasn't received this command, officer $a$ finishes spreading the command.
Let's look at the following example:
If officer $1$ spreads a command, officers receive it in the following order: $[1, 2, 3, 5 ,6, 8, 7, 9, 4]$.
If officer $3$ spreads a command, officers receive it in the following order: $[3, 5, 6, 8, 7, 9]$.
If officer $7$ spreads a command, officers receive it in the following order: $[7, 9]$.
If officer $9$ spreads a command, officers receive it in the following order: $[9]$.
To answer the $i$-th query $(u_i, k_i)$, construct a sequence which describes the order in which officers will receive the command if the $u_i$-th officer spreads it. Return the $k_i$-th element of the constructed list or -1 if there are fewer than $k_i$ elements in it.
You should process queries independently. A query doesn't affect the following queries.
Input Specification:
The first line of the input contains two integers $n$ and $q$ ($2 \le n \le 2 \cdot 10^5, 1 \le q \le 2 \cdot 10^5$) — the number of officers in Berland army and the number of queries.
The second line of the input contains $n - 1$ integers $p_2, p_3, \dots, p_n$ ($1 \le p_i < i$), where $p_i$ is the index of the direct superior of the officer having the index $i$. The commander has index $1$ and doesn't have any superiors.
The next $q$ lines describe the queries. The $i$-th query is given as a pair ($u_i, k_i$) ($1 \le u_i, k_i \le n$), where $u_i$ is the index of the officer which starts spreading a command, and $k_i$ is the index of the required officer in the command spreading sequence.
Output Specification:
Print $q$ numbers, where the $i$-th number is the officer at the position $k_i$ in the list which describes the order in which officers will receive the command if it starts spreading from officer $u_i$. Print "-1" if the number of officers which receive the command is less than $k_i$.
You should process queries independently. They do not affect each other.
Demo Input:
['9 6\n1 1 1 3 5 3 5 7\n3 1\n1 5\n3 4\n7 3\n1 8\n1 9\n']
Demo Output:
['3\n6\n8\n-1\n9\n4\n']
Note:
none | ```python
#include <bits/stdc++.h>
using namespace std;
//simple for loop
#define f(n) for(int i=0; i<n; i++)
typedef long long int ll;
typedef vector<ll> vll;
ll inf=numeric_limits<ll>::max();
ll findNumofSUbords(ll v, vector<vll>& adjList, vll& numofSUbords)
{
if (adjList[v].empty())
{
numofSUbords[v]=0;
return(0);
}
ll subords=0;
for(auto neig: adjList[v])
{
subords+=findNumofSUbords(neig, adjList, numofSUbords)+1;
}
numofSUbords[v]=subords;
return(subords);
}
void dfs(ll v, vector<vll>& adjList, vll& orderOfVisit)
{
orderOfVisit.push_back(v);
for(auto neig: adjList[v])
{
dfs(neig, adjList, orderOfVisit);
}
}
int main()
{
ll n, q, a, b;
cin>>n>>q;
vector<vll> adjList(n+1);
vll numofSubords(n+1, 0), orderOfVisit;
f(n-1){ cin>>a; adjList[a].push_back(i+2);}
// for(int i=0; i<n+1; i++)
// {
// cout<<i<<": ";
// for(auto y: adjList[i]) cout<<y<<" ";
// cout<<"\n";
// }
dfs(1, adjList, orderOfVisit);
// for(auto x: orderOfVisit) cout<<x<<" ";
findNumofSUbords(1, adjList, numofSubords);
// for(auto x: numofSubords) cout<<x<<" ";
f(q)
{
cin>>a>>b;
if (numofSubords[a]+1<b)cout<<-1<<'\n';
else cout<<orderOfVisit[a-1+b-1]<<'\n';
}
return(0);
}
``` | -1 | |
732 | A | Buy a Shovel | PROGRAMMING | 800 | [
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null | Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel. | The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels. | Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change. | [
"117 3\n",
"237 7\n",
"15 2\n"
] | [
"9\n",
"1\n",
"2\n"
] | In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change. | 500 | [
{
"input": "117 3",
"output": "9"
},
{
"input": "237 7",
"output": "1"
},
{
"input": "15 2",
"output": "2"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1000 3",
"output": "1"
},
{
"input": "1000 1",
... | 1,688,905,992 | 2,147,483,647 | Python 3 | OK | TESTS | 54 | 46 | 0 | k, r = map(int, input().split())
shovels = 1
price = k
while price % 10 != 0 and price % 10 != r:
shovels += 1
price = k * shovels
print(shovels) | Title: Buy a Shovel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input Specification:
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output Specification:
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Demo Input:
['117 3\n', '237 7\n', '15 2\n']
Demo Output:
['9\n', '1\n', '2\n']
Note:
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change. | ```python
k, r = map(int, input().split())
shovels = 1
price = k
while price % 10 != 0 and price % 10 != r:
shovels += 1
price = k * shovels
print(shovels)
``` | 3 | |
199 | A | Hexadecimal's theorem | PROGRAMMING | 900 | [
"brute force",
"constructive algorithms",
"implementation",
"number theory"
] | null | null | Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers.
Let's remember how Fibonacci numbers can be calculated. *F*0<==<=0, *F*1<==<=1, and all the next numbers are *F**i*<==<=*F**i*<=-<=2<=+<=*F**i*<=-<=1.
So, Fibonacci numbers make a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, ...
If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number *n* by three not necessary different Fibonacci numbers or say that it is impossible. | The input contains of a single integer *n* (0<=≤<=*n*<=<<=109) — the number that should be represented by the rules described above. It is guaranteed that *n* is a Fibonacci number. | Output three required numbers: *a*, *b* and *c*. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes.
If there are multiple answers, print any of them. | [
"3\n",
"13\n"
] | [
"1 1 1\n",
"2 3 8\n"
] | none | 500 | [
{
"input": "3",
"output": "1 1 1"
},
{
"input": "13",
"output": "2 3 8"
},
{
"input": "0",
"output": "0 0 0"
},
{
"input": "1",
"output": "1 0 0"
},
{
"input": "2",
"output": "1 1 0"
},
{
"input": "1597",
"output": "233 377 987"
},
{
"input... | 1,637,301,308 | 2,147,483,647 | Python 3 | OK | TESTS | 54 | 92 | 0 | n = int(input())
if n == 0:
print("0 0 0")
elif n == 1:
print("1 0 0")
elif n == 2:
print("1 1 0")
elif n == 3:
print("1 1 1")
else:
s = [0, 1];c = 1;i = 2
while (c != n):
s.append(c)
c = s[i - 1] + s[i]
i += 1
print(s[i - 1], s[i - 3], s[i - 4]) | Title: Hexadecimal's theorem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, a chaotic virus Hexadecimal advanced a new theorem which will shake the Universe. She thinks that each Fibonacci number can be represented as sum of three not necessary different Fibonacci numbers.
Let's remember how Fibonacci numbers can be calculated. *F*0<==<=0, *F*1<==<=1, and all the next numbers are *F**i*<==<=*F**i*<=-<=2<=+<=*F**i*<=-<=1.
So, Fibonacci numbers make a sequence of numbers: 0, 1, 1, 2, 3, 5, 8, 13, ...
If you haven't run away from the PC in fear, you have to help the virus. Your task is to divide given Fibonacci number *n* by three not necessary different Fibonacci numbers or say that it is impossible.
Input Specification:
The input contains of a single integer *n* (0<=≤<=*n*<=<<=109) — the number that should be represented by the rules described above. It is guaranteed that *n* is a Fibonacci number.
Output Specification:
Output three required numbers: *a*, *b* and *c*. If there is no answer for the test you have to print "I'm too stupid to solve this problem" without the quotes.
If there are multiple answers, print any of them.
Demo Input:
['3\n', '13\n']
Demo Output:
['1 1 1\n', '2 3 8\n']
Note:
none | ```python
n = int(input())
if n == 0:
print("0 0 0")
elif n == 1:
print("1 0 0")
elif n == 2:
print("1 1 0")
elif n == 3:
print("1 1 1")
else:
s = [0, 1];c = 1;i = 2
while (c != n):
s.append(c)
c = s[i - 1] + s[i]
i += 1
print(s[i - 1], s[i - 3], s[i - 4])
``` | 3 | |
47 | B | Coins | PROGRAMMING | 1,200 | [
"implementation"
] | B. Coins | 2 | 256 | One day Vasya came across three Berland coins. They didn't have any numbers that's why Vasya didn't understand how their denominations differ. He supposed that if one coin is heavier than the other one, then it should be worth more. Vasya weighed all the three pairs of coins on pan balance scales and told you the results. Find out how the deminations of the coins differ or if Vasya has a mistake in the weighting results. No two coins are equal. | The input data contains the results of all the weighting, one result on each line. It is guaranteed that every coin pair was weighted exactly once. Vasya labelled the coins with letters «A», «B» and «C». Each result is a line that appears as (letter)(> or < sign)(letter). For example, if coin "A" proved lighter than coin "B", the result of the weighting is A<B. | It the results are contradictory, print Impossible. Otherwise, print without spaces the rearrangement of letters «A», «B» and «C» which represent the coins in the increasing order of their weights. | [
"A>B\nC<B\nA>C\n",
"A<B\nB>C\nC>A\n"
] | [
"CBA",
"ACB"
] | none | 1,000 | [
{
"input": "A>B\nC<B\nA>C",
"output": "CBA"
},
{
"input": "A<B\nB>C\nC>A",
"output": "ACB"
},
{
"input": "A<C\nB<A\nB>C",
"output": "Impossible"
},
{
"input": "A<B\nA<C\nB>C",
"output": "ACB"
},
{
"input": "B>A\nC<B\nC>A",
"output": "ACB"
},
{
"input":... | 1,586,485,532 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 218 | 307,200 | x=input()
y=input()
z=input()
if (x=="A>B" or x=="B<A"):
if (y=="B>C" or y=="C<B"):
if (z=="C<A" or z=="A>C"):
print("CBA")
else:
print("Impossible")
else:
if (z=="C<A" or z=="A>C"):
print("BCA")
else:
print("BAC")
else:
if (y=="C<B" or y=="B>C"):
if (z=="C<A" or z=="A>C"):
print("CAB")
else:
print("ACB")
else:
if (z=="C<A" or z=="A>C"):
print("Impossible")
else:
print("ABC")
| Title: Coins
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya came across three Berland coins. They didn't have any numbers that's why Vasya didn't understand how their denominations differ. He supposed that if one coin is heavier than the other one, then it should be worth more. Vasya weighed all the three pairs of coins on pan balance scales and told you the results. Find out how the deminations of the coins differ or if Vasya has a mistake in the weighting results. No two coins are equal.
Input Specification:
The input data contains the results of all the weighting, one result on each line. It is guaranteed that every coin pair was weighted exactly once. Vasya labelled the coins with letters «A», «B» and «C». Each result is a line that appears as (letter)(> or < sign)(letter). For example, if coin "A" proved lighter than coin "B", the result of the weighting is A<B.
Output Specification:
It the results are contradictory, print Impossible. Otherwise, print without spaces the rearrangement of letters «A», «B» and «C» which represent the coins in the increasing order of their weights.
Demo Input:
['A>B\nC<B\nA>C\n', 'A<B\nB>C\nC>A\n']
Demo Output:
['CBA', 'ACB']
Note:
none | ```python
x=input()
y=input()
z=input()
if (x=="A>B" or x=="B<A"):
if (y=="B>C" or y=="C<B"):
if (z=="C<A" or z=="A>C"):
print("CBA")
else:
print("Impossible")
else:
if (z=="C<A" or z=="A>C"):
print("BCA")
else:
print("BAC")
else:
if (y=="C<B" or y=="B>C"):
if (z=="C<A" or z=="A>C"):
print("CAB")
else:
print("ACB")
else:
if (z=="C<A" or z=="A>C"):
print("Impossible")
else:
print("ABC")
``` | 0 |
682 | A | Alyona and Numbers | PROGRAMMING | 1,100 | [
"constructive algorithms",
"math",
"number theory"
] | null | null | After finishing eating her bun, Alyona came up with two integers *n* and *m*. She decided to write down two columns of integers — the first column containing integers from 1 to *n* and the second containing integers from 1 to *m*. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and equals 0.
As usual, Alyona has some troubles and asks you to help. | The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1<=000<=000). | Print the only integer — the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and (*x*<=+<=*y*) is divisible by 5. | [
"6 12\n",
"11 14\n",
"1 5\n",
"3 8\n",
"5 7\n",
"21 21\n"
] | [
"14\n",
"31\n",
"1\n",
"5\n",
"7\n",
"88\n"
] | Following pairs are suitable in the first sample case:
- for *x* = 1 fits *y* equal to 4 or 9; - for *x* = 2 fits *y* equal to 3 or 8; - for *x* = 3 fits *y* equal to 2, 7 or 12; - for *x* = 4 fits *y* equal to 1, 6 or 11; - for *x* = 5 fits *y* equal to 5 or 10; - for *x* = 6 fits *y* equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case. | 500 | [
{
"input": "6 12",
"output": "14"
},
{
"input": "11 14",
"output": "31"
},
{
"input": "1 5",
"output": "1"
},
{
"input": "3 8",
"output": "5"
},
{
"input": "5 7",
"output": "7"
},
{
"input": "21 21",
"output": "88"
},
{
"input": "10 15",
... | 1,558,780,890 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 109 | 0 | n, m = map(int, input().split())
a = [0] * 5
b = [0] * 5
for i in range(5):
a[i] = (n-i)//5
b[i] = (m-i)//5
ans = 0
for i in range(5):
for j in range(5):
if i + j==5 or i + j == 0:
ans += a[i] * b[j]
print(ans) | Title: Alyona and Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After finishing eating her bun, Alyona came up with two integers *n* and *m*. She decided to write down two columns of integers — the first column containing integers from 1 to *n* and the second containing integers from 1 to *m*. Now the girl wants to count how many pairs of integers she can choose, one from the first column and the other from the second column, such that their sum is divisible by 5.
Formally, Alyona wants to count the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and equals 0.
As usual, Alyona has some troubles and asks you to help.
Input Specification:
The only line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1<=000<=000).
Output Specification:
Print the only integer — the number of pairs of integers (*x*,<=*y*) such that 1<=≤<=*x*<=≤<=*n*, 1<=≤<=*y*<=≤<=*m* and (*x*<=+<=*y*) is divisible by 5.
Demo Input:
['6 12\n', '11 14\n', '1 5\n', '3 8\n', '5 7\n', '21 21\n']
Demo Output:
['14\n', '31\n', '1\n', '5\n', '7\n', '88\n']
Note:
Following pairs are suitable in the first sample case:
- for *x* = 1 fits *y* equal to 4 or 9; - for *x* = 2 fits *y* equal to 3 or 8; - for *x* = 3 fits *y* equal to 2, 7 or 12; - for *x* = 4 fits *y* equal to 1, 6 or 11; - for *x* = 5 fits *y* equal to 5 or 10; - for *x* = 6 fits *y* equal to 4 or 9.
Only the pair (1, 4) is suitable in the third sample case. | ```python
n, m = map(int, input().split())
a = [0] * 5
b = [0] * 5
for i in range(5):
a[i] = (n-i)//5
b[i] = (m-i)//5
ans = 0
for i in range(5):
for j in range(5):
if i + j==5 or i + j == 0:
ans += a[i] * b[j]
print(ans)
``` | 0 | |
305 | B | Continued Fractions | PROGRAMMING | 1,700 | [
"brute force",
"implementation",
"math"
] | null | null | A continued fraction of height *n* is a fraction of form . You are given two rational numbers, one is represented as and the other one is represented as a finite fraction of height *n*. Check if they are equal. | The first line contains two space-separated integers *p*,<=*q* (1<=≤<=*q*<=≤<=*p*<=≤<=1018) — the numerator and the denominator of the first fraction.
The second line contains integer *n* (1<=≤<=*n*<=≤<=90) — the height of the second fraction. The third line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1018) — the continued fraction.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. | Print "YES" if these fractions are equal and "NO" otherwise. | [
"9 4\n2\n2 4\n",
"9 4\n3\n2 3 1\n",
"9 4\n3\n1 2 4\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | In the first sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5ff92f27aebea2560d99ad61202d20bab5ee5390.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/221368c79c05fc0ecad4e5f7a64f30b832fd99f5.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4fb4b411afc0fbad27a1c8fdd08ba88ec3830ef5.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 1,000 | [
{
"input": "9 4\n2\n2 4",
"output": "YES"
},
{
"input": "9 4\n3\n2 3 1",
"output": "YES"
},
{
"input": "9 4\n3\n1 2 4",
"output": "NO"
},
{
"input": "39088169 24157817\n36\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2",
"output": "YES"
},
{
... | 1,576,728,664 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 46 | 310 | 0 |
p,q=map(int,input().split())
n=int(input())
w=[int(x) for x in input().split()]
if p%q!=0:
r=[0]*n
i=0
while i+1<=n:
if q!=0:
r[i] = int(p / q)
p,q=q,p-int(p/q)*q
i+=1
else:
break
if r==w or (r[n-1]==0 and w[n-1]==1 and w[n-2]==r[n-2]-1):
print('YES')
else:
print('NO')
else:
if w==[int(p/q)] or w==[int(p/q)-1,1]:
print('YES')
else:
print('NO')
| Title: Continued Fractions
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A continued fraction of height *n* is a fraction of form . You are given two rational numbers, one is represented as and the other one is represented as a finite fraction of height *n*. Check if they are equal.
Input Specification:
The first line contains two space-separated integers *p*,<=*q* (1<=≤<=*q*<=≤<=*p*<=≤<=1018) — the numerator and the denominator of the first fraction.
The second line contains integer *n* (1<=≤<=*n*<=≤<=90) — the height of the second fraction. The third line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1018) — the continued fraction.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print "YES" if these fractions are equal and "NO" otherwise.
Demo Input:
['9 4\n2\n2 4\n', '9 4\n3\n2 3 1\n', '9 4\n3\n1 2 4\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/5ff92f27aebea2560d99ad61202d20bab5ee5390.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/221368c79c05fc0ecad4e5f7a64f30b832fd99f5.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/4fb4b411afc0fbad27a1c8fdd08ba88ec3830ef5.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
p,q=map(int,input().split())
n=int(input())
w=[int(x) for x in input().split()]
if p%q!=0:
r=[0]*n
i=0
while i+1<=n:
if q!=0:
r[i] = int(p / q)
p,q=q,p-int(p/q)*q
i+=1
else:
break
if r==w or (r[n-1]==0 and w[n-1]==1 and w[n-2]==r[n-2]-1):
print('YES')
else:
print('NO')
else:
if w==[int(p/q)] or w==[int(p/q)-1,1]:
print('YES')
else:
print('NO')
``` | 0 | |
312 | A | Whose sentence is it? | PROGRAMMING | 1,100 | [
"implementation",
"strings"
] | null | null | One day, liouzhou_101 got a chat record of Freda and Rainbow. Out of curiosity, he wanted to know which sentences were said by Freda, and which were said by Rainbow. According to his experience, he thought that Freda always said "lala." at the end of her sentences, while Rainbow always said "miao." at the beginning of his sentences. For each sentence in the chat record, help liouzhou_101 find whose sentence it is. | The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=10), number of sentences in the chat record. Each of the next *n* lines contains a sentence. A sentence is a string that contains only Latin letters (A-Z, a-z), underline (_), comma (,), point (.) and space ( ). Its length doesn’t exceed 100. | For each sentence, output "Freda's" if the sentence was said by Freda, "Rainbow's" if the sentence was said by Rainbow, or "OMG>.< I don't know!" if liouzhou_101 can’t recognize whose sentence it is. He can’t recognize a sentence if it begins with "miao." and ends with "lala.", or satisfies neither of the conditions. | [
"5\nI will go to play with you lala.\nwow, welcome.\nmiao.lala.\nmiao.\nmiao .\n"
] | [
"Freda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\n"
] | none | 500 | [
{
"input": "5\nI will go to play with you lala.\nwow, welcome.\nmiao.lala.\nmiao.\nmiao .",
"output": "Freda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!"
},
{
"input": "10\nLpAEKiHVJrzSZqBVSSyY\nYECGBlala.\nUZeGpeM.UCwiHmmA\nqt_,.b_.LSwJtJ.\nFAnXZtHlala.\nmiao.iape... | 1,562,097,280 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 9 | 93 | 0 | class CodeforcesTask312ASolution:
def __init__(self):
self.result = ''
self.mess_count = 0
self.messages = []
def read_input(self):
self.mess_count = int(input())
self.messages = [input() for x in range(self.mess_count)]
def process_task(self):
for message in self.messages:
if "miao." in message:
if "lala." in message:
self.result += "OMG>.< I don't know!\n"
else:
self.result += "Rainbow's\n"
else:
if "lala." in message:
self.result += "Freda's\n"
else:
self.result += "OMG>.< I don't know!\n"
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask312ASolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
| Title: Whose sentence is it?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day, liouzhou_101 got a chat record of Freda and Rainbow. Out of curiosity, he wanted to know which sentences were said by Freda, and which were said by Rainbow. According to his experience, he thought that Freda always said "lala." at the end of her sentences, while Rainbow always said "miao." at the beginning of his sentences. For each sentence in the chat record, help liouzhou_101 find whose sentence it is.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=10), number of sentences in the chat record. Each of the next *n* lines contains a sentence. A sentence is a string that contains only Latin letters (A-Z, a-z), underline (_), comma (,), point (.) and space ( ). Its length doesn’t exceed 100.
Output Specification:
For each sentence, output "Freda's" if the sentence was said by Freda, "Rainbow's" if the sentence was said by Rainbow, or "OMG>.< I don't know!" if liouzhou_101 can’t recognize whose sentence it is. He can’t recognize a sentence if it begins with "miao." and ends with "lala.", or satisfies neither of the conditions.
Demo Input:
['5\nI will go to play with you lala.\nwow, welcome.\nmiao.lala.\nmiao.\nmiao .\n']
Demo Output:
["Freda's\nOMG>.< I don't know!\nOMG>.< I don't know!\nRainbow's\nOMG>.< I don't know!\n"]
Note:
none | ```python
class CodeforcesTask312ASolution:
def __init__(self):
self.result = ''
self.mess_count = 0
self.messages = []
def read_input(self):
self.mess_count = int(input())
self.messages = [input() for x in range(self.mess_count)]
def process_task(self):
for message in self.messages:
if "miao." in message:
if "lala." in message:
self.result += "OMG>.< I don't know!\n"
else:
self.result += "Rainbow's\n"
else:
if "lala." in message:
self.result += "Freda's\n"
else:
self.result += "OMG>.< I don't know!\n"
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask312ASolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | 0 | |
432 | A | Choosing Teams | PROGRAMMING | 800 | [
"greedy",
"implementation",
"sortings"
] | null | null | The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times? | The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship. | Print a single number — the answer to the problem. | [
"5 2\n0 4 5 1 0\n",
"6 4\n0 1 2 3 4 5\n",
"6 5\n0 0 0 0 0 0\n"
] | [
"1\n",
"0\n",
"2\n"
] | In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits. | 500 | [
{
"input": "5 2\n0 4 5 1 0",
"output": "1"
},
{
"input": "6 4\n0 1 2 3 4 5",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0 0",
"output": "2"
},
{
"input": "3 4\n0 1 0",
"output": "1"
},
{
"input": "3 4\n0 2 0",
"output": "0"
},
{
"input": "6 5\n0 0 0 0 0... | 1,648,227,616 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 46 | 0 | from math import floor
n,k=[int(i1) for i1 in input().split()]
students=[int(i2) for i2 in input().split()]
eligible_students=0
max_rounds=5-k
for i3 in students:
if i3<=max_rounds:
eligible_students+=1
print(floor(eligible_students/3)) | Title: Choosing Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has *n* students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.
The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least *k* times?
Input Specification:
The first line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=2000; 1<=≤<=*k*<=≤<=5). The next line contains *n* integers: *y*1,<=*y*2,<=...,<=*y**n* (0<=≤<=*y**i*<=≤<=5), where *y**i* shows the number of times the *i*-th person participated in the ACM ICPC world championship.
Output Specification:
Print a single number — the answer to the problem.
Demo Input:
['5 2\n0 4 5 1 0\n', '6 4\n0 1 2 3 4 5\n', '6 5\n0 0 0 0 0 0\n']
Demo Output:
['1\n', '0\n', '2\n']
Note:
In the first sample only one team could be made: the first, the fourth and the fifth participants.
In the second sample no teams could be created.
In the third sample two teams could be created. Any partition into two teams fits. | ```python
from math import floor
n,k=[int(i1) for i1 in input().split()]
students=[int(i2) for i2 in input().split()]
eligible_students=0
max_rounds=5-k
for i3 in students:
if i3<=max_rounds:
eligible_students+=1
print(floor(eligible_students/3))
``` | 3 | |
27 | A | Next Test | PROGRAMMING | 1,200 | [
"implementation",
"sortings"
] | A. Next Test | 2 | 256 | «Polygon» is a system which allows to create programming tasks in a simple and professional way. When you add a test to the problem, the corresponding form asks you for the test index. As in most cases it is clear which index the next test will have, the system suggests the default value of the index. It is calculated as the smallest positive integer which is not used as an index for some previously added test.
You are to implement this feature. Create a program which determines the default index of the next test, given the indexes of the previously added tests. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=3000) — the amount of previously added tests. The second line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3000) — indexes of these tests. | Output the required default value for the next test index. | [
"3\n1 7 2\n"
] | [
"3\n"
] | none | 500 | [
{
"input": "1\n1",
"output": "2"
},
{
"input": "2\n2 1",
"output": "3"
},
{
"input": "3\n3 4 1",
"output": "2"
},
{
"input": "4\n6 4 3 5",
"output": "1"
},
{
"input": "5\n3 2 1 7 4",
"output": "5"
},
{
"input": "6\n4 1 2 5 3 7",
"output": "6"
},
... | 1,494,584,443 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 218 | 307,200 | '''input
3
1 7 2
'''
n = int(input())
a = sorted(map(int, input().split()))
x = 1
while x in a:
x += 1
print(x) | Title: Next Test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
«Polygon» is a system which allows to create programming tasks in a simple and professional way. When you add a test to the problem, the corresponding form asks you for the test index. As in most cases it is clear which index the next test will have, the system suggests the default value of the index. It is calculated as the smallest positive integer which is not used as an index for some previously added test.
You are to implement this feature. Create a program which determines the default index of the next test, given the indexes of the previously added tests.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=3000) — the amount of previously added tests. The second line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3000) — indexes of these tests.
Output Specification:
Output the required default value for the next test index.
Demo Input:
['3\n1 7 2\n']
Demo Output:
['3\n']
Note:
none | ```python
'''input
3
1 7 2
'''
n = int(input())
a = sorted(map(int, input().split()))
x = 1
while x in a:
x += 1
print(x)
``` | 3.944928 |
981 | B | Businessmen Problems | PROGRAMMING | 1,000 | [
"sortings"
] | null | null | Two famous competing companies ChemForces and TopChemist decided to show their sets of recently discovered chemical elements on an exhibition. However they know that no element should be present in the sets of both companies.
In order to avoid this representatives of both companies decided to make an agreement on the sets the companies should present. The sets should be chosen in the way that maximizes the total income of the companies.
All elements are enumerated with integers. The ChemForces company has discovered $n$ distinct chemical elements with indices $a_1, a_2, \ldots, a_n$, and will get an income of $x_i$ Berland rubles if the $i$-th element from this list is in the set of this company.
The TopChemist company discovered $m$ distinct chemical elements with indices $b_1, b_2, \ldots, b_m$, and it will get an income of $y_j$ Berland rubles for including the $j$-th element from this list to its set.
In other words, the first company can present any subset of elements from $\{a_1, a_2, \ldots, a_n\}$ (possibly empty subset), the second company can present any subset of elements from $\{b_1, b_2, \ldots, b_m\}$ (possibly empty subset). There shouldn't be equal elements in the subsets.
Help the representatives select the sets in such a way that no element is presented in both sets and the total income is the maximum possible. | The first line contains a single integer $n$ ($1 \leq n \leq 10^5$) — the number of elements discovered by ChemForces.
The $i$-th of the next $n$ lines contains two integers $a_i$ and $x_i$ ($1 \leq a_i \leq 10^9$, $1 \leq x_i \leq 10^9$) — the index of the $i$-th element and the income of its usage on the exhibition. It is guaranteed that all $a_i$ are distinct.
The next line contains a single integer $m$ ($1 \leq m \leq 10^5$) — the number of chemicals invented by TopChemist.
The $j$-th of the next $m$ lines contains two integers $b_j$ and $y_j$, ($1 \leq b_j \leq 10^9$, $1 \leq y_j \leq 10^9$) — the index of the $j$-th element and the income of its usage on the exhibition. It is guaranteed that all $b_j$ are distinct. | Print the maximum total income you can obtain by choosing the sets for both companies in such a way that no element is presented in both sets. | [
"3\n1 2\n7 2\n3 10\n4\n1 4\n2 4\n3 4\n4 4\n",
"1\n1000000000 239\n3\n14 15\n92 65\n35 89\n"
] | [
"24\n",
"408\n"
] | In the first example ChemForces can choose the set ($3, 7$), while TopChemist can choose ($1, 2, 4$). This way the total income is $(10 + 2) + (4 + 4 + 4) = 24$.
In the second example ChemForces can choose the only element $10^9$, while TopChemist can choose ($14, 92, 35$). This way the total income is $(239) + (15 + 65 + 89) = 408$. | 750 | [
{
"input": "3\n1 2\n7 2\n3 10\n4\n1 4\n2 4\n3 4\n4 4",
"output": "24"
},
{
"input": "1\n1000000000 239\n3\n14 15\n92 65\n35 89",
"output": "408"
},
{
"input": "10\n598654597 488228616\n544064902 21923894\n329635457 980089248\n988262691 654502493\n967529230 543358150\n835120075 128123793\... | 1,527,435,178 | 2,578 | Python 3 | OK | TESTS | 33 | 842 | 20,582,400 | n=int(input())
CF={}
for i in range(n):
x=list(map(int,input().split()))
CF[x[0]]=x[1]
m=int(input())
TC={}
for i in range(m):
x=list(map(int,input().split()))
TC[x[0]]=x[1]
keys_Tc=set(TC.keys())
keys_Cf=set(CF.keys())
common=keys_Cf&keys_Tc
s=0
for x in TC:
if x not in common:
s+=TC[x]
for x in CF:
if x not in common:
s+=CF[x]
for x in common:
s+=max(CF[x],TC[x])
print(s) | Title: Businessmen Problems
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two famous competing companies ChemForces and TopChemist decided to show their sets of recently discovered chemical elements on an exhibition. However they know that no element should be present in the sets of both companies.
In order to avoid this representatives of both companies decided to make an agreement on the sets the companies should present. The sets should be chosen in the way that maximizes the total income of the companies.
All elements are enumerated with integers. The ChemForces company has discovered $n$ distinct chemical elements with indices $a_1, a_2, \ldots, a_n$, and will get an income of $x_i$ Berland rubles if the $i$-th element from this list is in the set of this company.
The TopChemist company discovered $m$ distinct chemical elements with indices $b_1, b_2, \ldots, b_m$, and it will get an income of $y_j$ Berland rubles for including the $j$-th element from this list to its set.
In other words, the first company can present any subset of elements from $\{a_1, a_2, \ldots, a_n\}$ (possibly empty subset), the second company can present any subset of elements from $\{b_1, b_2, \ldots, b_m\}$ (possibly empty subset). There shouldn't be equal elements in the subsets.
Help the representatives select the sets in such a way that no element is presented in both sets and the total income is the maximum possible.
Input Specification:
The first line contains a single integer $n$ ($1 \leq n \leq 10^5$) — the number of elements discovered by ChemForces.
The $i$-th of the next $n$ lines contains two integers $a_i$ and $x_i$ ($1 \leq a_i \leq 10^9$, $1 \leq x_i \leq 10^9$) — the index of the $i$-th element and the income of its usage on the exhibition. It is guaranteed that all $a_i$ are distinct.
The next line contains a single integer $m$ ($1 \leq m \leq 10^5$) — the number of chemicals invented by TopChemist.
The $j$-th of the next $m$ lines contains two integers $b_j$ and $y_j$, ($1 \leq b_j \leq 10^9$, $1 \leq y_j \leq 10^9$) — the index of the $j$-th element and the income of its usage on the exhibition. It is guaranteed that all $b_j$ are distinct.
Output Specification:
Print the maximum total income you can obtain by choosing the sets for both companies in such a way that no element is presented in both sets.
Demo Input:
['3\n1 2\n7 2\n3 10\n4\n1 4\n2 4\n3 4\n4 4\n', '1\n1000000000 239\n3\n14 15\n92 65\n35 89\n']
Demo Output:
['24\n', '408\n']
Note:
In the first example ChemForces can choose the set ($3, 7$), while TopChemist can choose ($1, 2, 4$). This way the total income is $(10 + 2) + (4 + 4 + 4) = 24$.
In the second example ChemForces can choose the only element $10^9$, while TopChemist can choose ($14, 92, 35$). This way the total income is $(239) + (15 + 65 + 89) = 408$. | ```python
n=int(input())
CF={}
for i in range(n):
x=list(map(int,input().split()))
CF[x[0]]=x[1]
m=int(input())
TC={}
for i in range(m):
x=list(map(int,input().split()))
TC[x[0]]=x[1]
keys_Tc=set(TC.keys())
keys_Cf=set(CF.keys())
common=keys_Cf&keys_Tc
s=0
for x in TC:
if x not in common:
s+=TC[x]
for x in CF:
if x not in common:
s+=CF[x]
for x in common:
s+=max(CF[x],TC[x])
print(s)
``` | 3 | |
936 | B | Sleepy Game | PROGRAMMING | 2,100 | [
"dfs and similar",
"dp",
"games",
"graphs"
] | null | null | Petya and Vasya arranged a game. The game runs by the following rules. Players have a directed graph consisting of *n* vertices and *m* edges. One of the vertices contains a chip. Initially the chip is located at vertex *s*. Players take turns moving the chip along some edge of the graph. Petya goes first. Player who can't move the chip loses. If the game lasts for 106 turns the draw is announced.
Vasya was performing big laboratory work in "Spelling and parts of speech" at night before the game, so he fell asleep at the very beginning of the game. Petya decided to take the advantage of this situation and make both Petya's and Vasya's moves.
Your task is to help Petya find out if he can win the game or at least draw a tie. | The first line of input contain two integers *n* and *m* — the number of vertices and the number of edges in the graph (2<=≤<=*n*<=≤<=105, 0<=≤<=*m*<=≤<=2·105).
The next *n* lines contain the information about edges of the graph. *i*-th line (1<=≤<=*i*<=≤<=*n*) contains nonnegative integer *c**i* — number of vertices such that there is an edge from *i* to these vertices and *c**i* distinct integers *a**i*,<=*j* — indices of these vertices (1<=≤<=*a**i*,<=*j*<=≤<=*n*, *a**i*,<=*j*<=≠<=*i*).
It is guaranteed that the total sum of *c**i* equals to *m*.
The next line contains index of vertex *s* — the initial position of the chip (1<=≤<=*s*<=≤<=*n*). | If Petya can win print «Win» in the first line. In the next line print numbers *v*1,<=*v*2,<=...,<=*v**k* (1<=≤<=*k*<=≤<=106) — the sequence of vertices Petya should visit for the winning. Vertex *v*1 should coincide with *s*. For *i*<==<=1... *k*<=-<=1 there should be an edge from *v**i* to *v**i*<=+<=1 in the graph. There must be no possible move from vertex *v**k*. The sequence should be such that Petya wins the game.
If Petya can't win but can draw a tie, print «Draw» in the only line. Otherwise print «Lose». | [
"5 6\n2 2 3\n2 4 5\n1 4\n1 5\n0\n1\n",
"3 2\n1 3\n1 1\n0\n2\n",
"2 2\n1 2\n1 1\n1\n"
] | [
"Win\n1 2 4 5 \n",
"Lose\n",
"Draw\n"
] | In the first example the graph is the following:
Initially the chip is located at vertex 1. In the first move Petya moves the chip to vertex 2, after that he moves it to vertex 4 for Vasya. After that he moves to vertex 5. Now it is Vasya's turn and there is no possible move, so Petya wins.
In the second example the graph is the following:
Initially the chip is located at vertex 2. The only possible Petya's move is to go to vertex 1. After that he has to go to 3 for Vasya. Now it's Petya's turn but he has no possible move, so Petya loses.
In the third example the graph is the following:
Petya can't win, but he can move along the cycle, so the players will draw a tie. | 1,000 | [
{
"input": "5 6\n2 2 3\n2 4 5\n1 4\n1 5\n0\n1",
"output": "Win\n1 2 4 5 "
},
{
"input": "3 2\n1 3\n1 1\n0\n2",
"output": "Lose"
},
{
"input": "2 2\n1 2\n1 1\n1",
"output": "Draw"
},
{
"input": "92 69\n1 76\n1 14\n1 9\n0\n1 46\n1 80\n0\n0\n1 77\n0\n1 53\n1 81\n1 61\n1 40\n0\n1... | 1,520,567,156 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 46 | 5,632,000 | draw = False
nodes = []
result = []
init = 0
class node:
def __init__(self, nei, num):
self.pe = None
self.po = None
self.visit = False
self.rec = False
self.neigh = nei
self.number = num
def search(self):
self.visit = True
if self.neigh[0] == 0:
if self.po:
self.ret([], False)
return
for i in self.neigh[1:]:
if not nodes[i].rec:
global draw
draw = True
if self.pe:
nodes[i].po = self
if self.po:
nodes[i].pe = self
if not nodes[i].visit:
node.search(nodes[i])
self.rec = True
def ret(self, tab, even):
if self.number == init and not even:
global result
result = tab
elif even:
nodes[self.pe.number].ret(tab + [self.number], False)
else:
nodes[self.po.number].ret(tab + [self.number], True)
def main():
n, m = [int(x) for x in input().split()]
nodes.append(n)
for i in range(n):
nodes.append(node([int(x) for x in input().split()], i + 1))
global init
init = int(input())
nodes[init].pe = nodes[init]
node.search(nodes[init])
if len(result) > 0:
print('Win')
print(*result[::-1])
elif draw:
print('Draw')
else:
print('Lose')
if __name__ == "__main__":
main()
| Title: Sleepy Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya and Vasya arranged a game. The game runs by the following rules. Players have a directed graph consisting of *n* vertices and *m* edges. One of the vertices contains a chip. Initially the chip is located at vertex *s*. Players take turns moving the chip along some edge of the graph. Petya goes first. Player who can't move the chip loses. If the game lasts for 106 turns the draw is announced.
Vasya was performing big laboratory work in "Spelling and parts of speech" at night before the game, so he fell asleep at the very beginning of the game. Petya decided to take the advantage of this situation and make both Petya's and Vasya's moves.
Your task is to help Petya find out if he can win the game or at least draw a tie.
Input Specification:
The first line of input contain two integers *n* and *m* — the number of vertices and the number of edges in the graph (2<=≤<=*n*<=≤<=105, 0<=≤<=*m*<=≤<=2·105).
The next *n* lines contain the information about edges of the graph. *i*-th line (1<=≤<=*i*<=≤<=*n*) contains nonnegative integer *c**i* — number of vertices such that there is an edge from *i* to these vertices and *c**i* distinct integers *a**i*,<=*j* — indices of these vertices (1<=≤<=*a**i*,<=*j*<=≤<=*n*, *a**i*,<=*j*<=≠<=*i*).
It is guaranteed that the total sum of *c**i* equals to *m*.
The next line contains index of vertex *s* — the initial position of the chip (1<=≤<=*s*<=≤<=*n*).
Output Specification:
If Petya can win print «Win» in the first line. In the next line print numbers *v*1,<=*v*2,<=...,<=*v**k* (1<=≤<=*k*<=≤<=106) — the sequence of vertices Petya should visit for the winning. Vertex *v*1 should coincide with *s*. For *i*<==<=1... *k*<=-<=1 there should be an edge from *v**i* to *v**i*<=+<=1 in the graph. There must be no possible move from vertex *v**k*. The sequence should be such that Petya wins the game.
If Petya can't win but can draw a tie, print «Draw» in the only line. Otherwise print «Lose».
Demo Input:
['5 6\n2 2 3\n2 4 5\n1 4\n1 5\n0\n1\n', '3 2\n1 3\n1 1\n0\n2\n', '2 2\n1 2\n1 1\n1\n']
Demo Output:
['Win\n1 2 4 5 \n', 'Lose\n', 'Draw\n']
Note:
In the first example the graph is the following:
Initially the chip is located at vertex 1. In the first move Petya moves the chip to vertex 2, after that he moves it to vertex 4 for Vasya. After that he moves to vertex 5. Now it is Vasya's turn and there is no possible move, so Petya wins.
In the second example the graph is the following:
Initially the chip is located at vertex 2. The only possible Petya's move is to go to vertex 1. After that he has to go to 3 for Vasya. Now it's Petya's turn but he has no possible move, so Petya loses.
In the third example the graph is the following:
Petya can't win, but he can move along the cycle, so the players will draw a tie. | ```python
draw = False
nodes = []
result = []
init = 0
class node:
def __init__(self, nei, num):
self.pe = None
self.po = None
self.visit = False
self.rec = False
self.neigh = nei
self.number = num
def search(self):
self.visit = True
if self.neigh[0] == 0:
if self.po:
self.ret([], False)
return
for i in self.neigh[1:]:
if not nodes[i].rec:
global draw
draw = True
if self.pe:
nodes[i].po = self
if self.po:
nodes[i].pe = self
if not nodes[i].visit:
node.search(nodes[i])
self.rec = True
def ret(self, tab, even):
if self.number == init and not even:
global result
result = tab
elif even:
nodes[self.pe.number].ret(tab + [self.number], False)
else:
nodes[self.po.number].ret(tab + [self.number], True)
def main():
n, m = [int(x) for x in input().split()]
nodes.append(n)
for i in range(n):
nodes.append(node([int(x) for x in input().split()], i + 1))
global init
init = int(input())
nodes[init].pe = nodes[init]
node.search(nodes[init])
if len(result) > 0:
print('Win')
print(*result[::-1])
elif draw:
print('Draw')
else:
print('Lose')
if __name__ == "__main__":
main()
``` | 0 | |
69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10... | 1,563,173,241 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 80 | 218 | 0 | import sys
import math
s = int(input())
ans = [0,0,0]
for i in range(s):
a = [int(x) for x in input().split()]
ans[0] += a[0]
ans[1] += a[1]
ans[2] += a[2]
if sum(ans) == 0:
print("YES")
else:
print("NO")
| Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
import sys
import math
s = int(input())
ans = [0,0,0]
for i in range(s):
a = [int(x) for x in input().split()]
ans[0] += a[0]
ans[1] += a[1]
ans[2] += a[2]
if sum(ans) == 0:
print("YES")
else:
print("NO")
``` | 0 |
155 | A | I_love_\%username\% | PROGRAMMING | 800 | [
"brute force"
] | null | null | Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him. | The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000. | Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests. | [
"5\n100 50 200 150 200\n",
"10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n"
] | [
"2\n",
"4\n"
] | In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing. | 500 | [
{
"input": "5\n100 50 200 150 200",
"output": "2"
},
{
"input": "10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242",
"output": "4"
},
{
"input": "1\n6",
"output": "0"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "5\n100 36 53 7 81",
"output": "2"
... | 1,694,932,052 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 92 | 0 | num, *data = map(int, open(0).read().split())
subset = data[:1]
for element in data:
num -= min(subset) <= element <= max(subset)
subset += element,
print(num)
| Title: I_love_\%username\%
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him.
Input Specification:
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000.
Output Specification:
Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests.
Demo Input:
['5\n100 50 200 150 200\n', '10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n']
Demo Output:
['2\n', '4\n']
Note:
In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing. | ```python
num, *data = map(int, open(0).read().split())
subset = data[:1]
for element in data:
num -= min(subset) <= element <= max(subset)
subset += element,
print(num)
``` | 3 | |
519 | B | A and B and Compilation Errors | PROGRAMMING | 1,100 | [
"data structures",
"implementation",
"sortings"
] | null | null | A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected? | The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one. | Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively. | [
"5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n",
"6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n"
] | [
"8\n123\n",
"1\n3\n"
] | In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | 1,000 | [
{
"input": "5\n1 5 8 123 7\n123 7 5 1\n5 1 7",
"output": "8\n123"
},
{
"input": "6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5",
"output": "1\n3"
},
{
"input": "3\n1 2 3\n3 2\n2",
"output": "1\n3"
},
{
"input": "10\n460626451 802090732 277246428 661369649 388684428 784303821 376287098 6... | 1,659,005,772 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 33 | 156 | 25,292,800 | a = int(input())
b = list(map(int,input().split(" ")))
b.sort()
c = list(map(int,input().split(" ")))
c.sort()
d = list(map(int,input().split(" ")))
d.sort()
first,second=0,0
for i in range(len(c)):
if b[i] != c[i]:
first = b[i]
break
if first == 0:
first = b[-1]
print(first)
for i in range(len(d)):
if c[i] != d[i]:
second = c[i]
break
if second == 0:
second = c[-1]
print(second) | Title: A and B and Compilation Errors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output Specification:
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Demo Input:
['5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n', '6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n']
Demo Output:
['8\n123\n', '1\n3\n']
Note:
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | ```python
a = int(input())
b = list(map(int,input().split(" ")))
b.sort()
c = list(map(int,input().split(" ")))
c.sort()
d = list(map(int,input().split(" ")))
d.sort()
first,second=0,0
for i in range(len(c)):
if b[i] != c[i]:
first = b[i]
break
if first == 0:
first = b[-1]
print(first)
for i in range(len(d)):
if c[i] != d[i]:
second = c[i]
break
if second == 0:
second = c[-1]
print(second)
``` | 3 | |
58 | A | Chat room | PROGRAMMING | 1,000 | [
"greedy",
"strings"
] | A. Chat room | 1 | 256 | Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. | The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. | If Vasya managed to say hello, print "YES", otherwise print "NO". | [
"ahhellllloou\n",
"hlelo\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymeda... | 1,540,149,624 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 93 | 0 | x=input()
z=0
while (z<len(x) and x[n]!='h'):
z+=1
z+=1
while (z<len(x) and x[z]!='e'):
z+=1
z+=1
while (z<len(x) and x[z]!='l'):
z+=1
z+=1
while (z<len(x) and x[z]!='l'):
z+=1
z+=1
while (z<len(x) and x[z]!='o'):
z+=1
if z<len(x):
print("YES")
else :
print ("NO") | Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
x=input()
z=0
while (z<len(x) and x[n]!='h'):
z+=1
z+=1
while (z<len(x) and x[z]!='e'):
z+=1
z+=1
while (z<len(x) and x[z]!='l'):
z+=1
z+=1
while (z<len(x) and x[z]!='l'):
z+=1
z+=1
while (z<len(x) and x[z]!='o'):
z+=1
if z<len(x):
print("YES")
else :
print ("NO")
``` | -1 |
637 | B | Chat Order | PROGRAMMING | 1,200 | [
"*special",
"binary search",
"constructive algorithms",
"data structures",
"sortings"
] | null | null | Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a message is sent to some friend, his friend's chat rises to the very top of the page. The relative order of the other chats doesn't change. If there was no chat with this friend before, then a new chat is simply inserted to the top of the list.
Assuming that the chat list is initially empty, given the sequence of Polycaprus' messages make a list of chats after all of his messages are processed. Assume that no friend wrote any message to Polycarpus. | The first line contains integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of Polycarpus' messages. Next *n* lines enlist the message recipients in the order in which the messages were sent. The name of each participant is a non-empty sequence of lowercase English letters of length at most 10. | Print all the recipients to who Polycarp talked to in the order of chats with them, from top to bottom. | [
"4\nalex\nivan\nroman\nivan\n",
"8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina\n"
] | [
"ivan\nroman\nalex\n",
"alina\nmaria\nekaterina\ndarya\n"
] | In the first test case Polycarpus first writes to friend by name "alex", and the list looks as follows:
1. alex
Then Polycarpus writes to friend by name "ivan" and the list looks as follows:
1. ivan 1. alex
Polycarpus writes the third message to friend by name "roman" and the list looks as follows:
1. roman 1. ivan 1. alex
Polycarpus writes the fourth message to friend by name "ivan", to who he has already sent a message, so the list of chats changes as follows:
1. ivan 1. roman 1. alex | 1,000 | [
{
"input": "4\nalex\nivan\nroman\nivan",
"output": "ivan\nroman\nalex"
},
{
"input": "8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina",
"output": "alina\nmaria\nekaterina\ndarya"
},
{
"input": "1\nwdi",
"output": "wdi"
},
{
"input": "2\nypg\nypg",
"outpu... | 1,458,238,052 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 12 | 3,000 | 8,601,600 | n = int(input())
a = []
for i in range(n) :
a.append(input())
c = []
for i in range(len(a)-1,-1,-1) :
for q in range(len(c)) :
if c[q] == a[i] :
break
else :
c.append(a[i])
for i in range(len(c)):
print(c[i])
| Title: Chat Order
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is a big lover of killing time in social networks. A page with a chatlist in his favourite network is made so that when a message is sent to some friend, his friend's chat rises to the very top of the page. The relative order of the other chats doesn't change. If there was no chat with this friend before, then a new chat is simply inserted to the top of the list.
Assuming that the chat list is initially empty, given the sequence of Polycaprus' messages make a list of chats after all of his messages are processed. Assume that no friend wrote any message to Polycarpus.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of Polycarpus' messages. Next *n* lines enlist the message recipients in the order in which the messages were sent. The name of each participant is a non-empty sequence of lowercase English letters of length at most 10.
Output Specification:
Print all the recipients to who Polycarp talked to in the order of chats with them, from top to bottom.
Demo Input:
['4\nalex\nivan\nroman\nivan\n', '8\nalina\nmaria\nekaterina\ndarya\ndarya\nekaterina\nmaria\nalina\n']
Demo Output:
['ivan\nroman\nalex\n', 'alina\nmaria\nekaterina\ndarya\n']
Note:
In the first test case Polycarpus first writes to friend by name "alex", and the list looks as follows:
1. alex
Then Polycarpus writes to friend by name "ivan" and the list looks as follows:
1. ivan 1. alex
Polycarpus writes the third message to friend by name "roman" and the list looks as follows:
1. roman 1. ivan 1. alex
Polycarpus writes the fourth message to friend by name "ivan", to who he has already sent a message, so the list of chats changes as follows:
1. ivan 1. roman 1. alex | ```python
n = int(input())
a = []
for i in range(n) :
a.append(input())
c = []
for i in range(len(a)-1,-1,-1) :
for q in range(len(c)) :
if c[q] == a[i] :
break
else :
c.append(a[i])
for i in range(len(c)):
print(c[i])
``` | 0 | |
58 | A | Chat room | PROGRAMMING | 1,000 | [
"greedy",
"strings"
] | A. Chat room | 1 | 256 | Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. | The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. | If Vasya managed to say hello, print "YES", otherwise print "NO". | [
"ahhellllloou\n",
"hlelo\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymeda... | 1,548,784,151 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 108 | 0 | b=input()
a='hello'
k=0
for j in range(len(b)):
try:
if a[j] not in b:
print('NO')
k=1
break
c=b.find(a[j])
b=b[c+1:]
if a[j+1] in b:
continue
else:
print('NO')
k=1
break
except IndexError:
pass
if k==0 :
print('YES')
| Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
b=input()
a='hello'
k=0
for j in range(len(b)):
try:
if a[j] not in b:
print('NO')
k=1
break
c=b.find(a[j])
b=b[c+1:]
if a[j+1] in b:
continue
else:
print('NO')
k=1
break
except IndexError:
pass
if k==0 :
print('YES')
``` | 3.946 |
433 | B | Kuriyama Mirai's Stones | PROGRAMMING | 1,200 | [
"dp",
"implementation",
"sortings"
] | null | null | Kuriyama Mirai has killed many monsters and got many (namely *n*) stones. She numbers the stones from 1 to *n*. The cost of the *i*-th stone is *v**i*. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . 1. Let *u**i* be the cost of the *i*-th cheapest stone (the cost that will be on the *i*-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her .
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — costs of the stones.
The third line contains an integer *m* (1<=≤<=*m*<=≤<=105) — the number of Kuriyama Mirai's questions. Then follow *m* lines, each line contains three integers *type*, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*type*<=≤<=2), describing a question. If *type* equal to 1, then you should output the answer for the first question, else you should output the answer for the second one. | Print *m* lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input. | [
"6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6\n",
"4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2\n"
] | [
"24\n9\n28\n",
"10\n15\n5\n15\n5\n5\n2\n12\n3\n5\n"
] | Please note that the answers to the questions may overflow 32-bit integer type. | 1,500 | [
{
"input": "6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6",
"output": "24\n9\n28"
},
{
"input": "4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2",
"output": "10\n15\n5\n15\n5\n5\n2\n12\n3\n5"
},
{
"input": "4\n2 2 3 6\n9\n2 2 3\n1 1 3\n2 2 3\n2 2 3\n2 2 2\n1... | 1,645,941,522 | 2,147,483,647 | Python 3 | OK | TESTS | 46 | 1,044 | 9,216,000 | n = int(input())
alist = list(map(int, input().split()))
asort = sorted(alist)
for i in range(1, n):
alist[i] += alist[i - 1]
asort[i] += asort[i - 1]
m = int(input())
for _ in range(m):
t, l, r = map(int, input().split())
if t == 2:
print(asort[r-1] - asort[l - 2] if l - 1 != 0 else asort[r-1])
else:
print(alist[r-1] - alist[l - 2] if l - 1 != 0 else alist[r-1]) | Title: Kuriyama Mirai's Stones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kuriyama Mirai has killed many monsters and got many (namely *n*) stones. She numbers the stones from 1 to *n*. The cost of the *i*-th stone is *v**i*. Kuriyama Mirai wants to know something about these stones so she will ask you two kinds of questions:
1. She will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her . 1. Let *u**i* be the cost of the *i*-th cheapest stone (the cost that will be on the *i*-th place if we arrange all the stone costs in non-decreasing order). This time she will tell you two numbers, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*), and you should tell her .
For every question you should give the correct answer, or Kuriyama Mirai will say "fuyukai desu" and then become unhappy.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains *n* integers: *v*1,<=*v*2,<=...,<=*v**n* (1<=≤<=*v**i*<=≤<=109) — costs of the stones.
The third line contains an integer *m* (1<=≤<=*m*<=≤<=105) — the number of Kuriyama Mirai's questions. Then follow *m* lines, each line contains three integers *type*, *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=*n*; 1<=≤<=*type*<=≤<=2), describing a question. If *type* equal to 1, then you should output the answer for the first question, else you should output the answer for the second one.
Output Specification:
Print *m* lines. Each line must contain an integer — the answer to Kuriyama Mirai's question. Print the answers to the questions in the order of input.
Demo Input:
['6\n6 4 2 7 2 7\n3\n2 3 6\n1 3 4\n1 1 6\n', '4\n5 5 2 3\n10\n1 2 4\n2 1 4\n1 1 1\n2 1 4\n2 1 2\n1 1 1\n1 3 3\n1 1 3\n1 4 4\n1 2 2\n']
Demo Output:
['24\n9\n28\n', '10\n15\n5\n15\n5\n5\n2\n12\n3\n5\n']
Note:
Please note that the answers to the questions may overflow 32-bit integer type. | ```python
n = int(input())
alist = list(map(int, input().split()))
asort = sorted(alist)
for i in range(1, n):
alist[i] += alist[i - 1]
asort[i] += asort[i - 1]
m = int(input())
for _ in range(m):
t, l, r = map(int, input().split())
if t == 2:
print(asort[r-1] - asort[l - 2] if l - 1 != 0 else asort[r-1])
else:
print(alist[r-1] - alist[l - 2] if l - 1 != 0 else alist[r-1])
``` | 3 | |
155 | A | I_love_\%username\% | PROGRAMMING | 800 | [
"brute force"
] | null | null | Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him. | The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000. | Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests. | [
"5\n100 50 200 150 200\n",
"10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n"
] | [
"2\n",
"4\n"
] | In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing. | 500 | [
{
"input": "5\n100 50 200 150 200",
"output": "2"
},
{
"input": "10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242",
"output": "4"
},
{
"input": "1\n6",
"output": "0"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "5\n100 36 53 7 81",
"output": "2"
... | 1,677,237,718 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 37 | 122 | 0 | n = int(input())
scores = list(map(int, input().split()))
min_score = max_score = scores[0]
amazing_count = 0
for score in scores[1:]:
if score > max_score:
max_score = score
amazing_count += 1
elif score < min_score:
min_score = score
amazing_count += 1
print(amazing_count)
| Title: I_love_\%username\%
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him.
Input Specification:
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000.
Output Specification:
Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests.
Demo Input:
['5\n100 50 200 150 200\n', '10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n']
Demo Output:
['2\n', '4\n']
Note:
In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing. | ```python
n = int(input())
scores = list(map(int, input().split()))
min_score = max_score = scores[0]
amazing_count = 0
for score in scores[1:]:
if score > max_score:
max_score = score
amazing_count += 1
elif score < min_score:
min_score = score
amazing_count += 1
print(amazing_count)
``` | 3 | |
279 | B | Books | PROGRAMMING | 1,400 | [
"binary search",
"brute force",
"implementation",
"two pointers"
] | null | null | When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read. | The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book. | Print a single integer — the maximum number of books Valera can read. | [
"4 5\n3 1 2 1\n",
"3 3\n2 2 3\n"
] | [
"3\n",
"1\n"
] | none | 1,000 | [
{
"input": "4 5\n3 1 2 1",
"output": "3"
},
{
"input": "3 3\n2 2 3",
"output": "1"
},
{
"input": "1 3\n5",
"output": "0"
},
{
"input": "1 10\n4",
"output": "1"
},
{
"input": "2 10\n6 4",
"output": "2"
},
{
"input": "6 10\n2 3 4 2 1 1",
"output": "4... | 1,669,650,556 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 62 | 0 | n,t=map(int,input().split())
lis=list(map(int,input().split()))
lis.sort()
c=0
i=0
for i in range(len(lis)):
t=t-lis[i]
if(t>0):
c=c+1
print(c)
| Title: Books
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read.
Input Specification:
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book.
Output Specification:
Print a single integer — the maximum number of books Valera can read.
Demo Input:
['4 5\n3 1 2 1\n', '3 3\n2 2 3\n']
Demo Output:
['3\n', '1\n']
Note:
none | ```python
n,t=map(int,input().split())
lis=list(map(int,input().split()))
lis.sort()
c=0
i=0
for i in range(len(lis)):
t=t-lis[i]
if(t>0):
c=c+1
print(c)
``` | 0 | |
59 | A | Word | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Word | 2 | 256 | Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. | The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. | Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. | [
"HoUse\n",
"ViP\n",
"maTRIx\n"
] | [
"house\n",
"VIP\n",
"matrix\n"
] | none | 500 | [
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chn... | 1,578,856,006 | 2,147,483,647 | PyPy 3 | OK | TESTS | 30 | 280 | 0 | import sys
import io
import os
total = 0
failed = 0
def run(test,res):
x = io.StringIO()
with io.StringIO(test) as sys.stdin:
with x as sys.stdout:
work()
z = x.getvalue().strip()
sys.stdout = sys.__stdout__
print("Passed?", z == res)
print("Expected: ", res)
print("Actual : ", z)
global total, failed
total += 1
failed += 1 if (z != res) else 0
def work():
s = input()
c = 0
for e in s:
if e>='a' and e<='z':
c += 1
print(s.lower() if (c >= len(s)/2) else s.upper())
def test():
run("""HoUse""", """house""")
run("""ViP""", """VIP""")
run("""maTRIx""", """matrix""")
if('LOCALTEST' in os.environ):
test()
print("\n Result: %s (%d total, %d failed)" % ("FAILED" if (failed>0) else "PASSED", total, failed))
else:
work()
| Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none | ```python
import sys
import io
import os
total = 0
failed = 0
def run(test,res):
x = io.StringIO()
with io.StringIO(test) as sys.stdin:
with x as sys.stdout:
work()
z = x.getvalue().strip()
sys.stdout = sys.__stdout__
print("Passed?", z == res)
print("Expected: ", res)
print("Actual : ", z)
global total, failed
total += 1
failed += 1 if (z != res) else 0
def work():
s = input()
c = 0
for e in s:
if e>='a' and e<='z':
c += 1
print(s.lower() if (c >= len(s)/2) else s.upper())
def test():
run("""HoUse""", """house""")
run("""ViP""", """VIP""")
run("""maTRIx""", """matrix""")
if('LOCALTEST' in os.environ):
test()
print("\n Result: %s (%d total, %d failed)" % ("FAILED" if (failed>0) else "PASSED", total, failed))
else:
work()
``` | 3.93 |
26 | D | Tickets | PROGRAMMING | 2,400 | [
"combinatorics",
"math",
"probabilities"
] | D. Tickets | 2 | 256 | As a big fan of Formula One, Charlie is really happy with the fact that he has to organize ticket sells for the next Grand Prix race in his own city. Unfortunately, the finacial crisis is striking everywhere and all the banknotes left in his country are valued either 10 euros or 20 euros. The price of all tickets for the race is 10 euros, so whenever someone comes to the ticket store only with 20 euro banknote Charlie must have a 10 euro banknote to give them change. Charlie realize that with the huge deficit of banknotes this could be a problem. Charlie has some priceless information but couldn't make use of it, so he needs your help. Exactly *n*<=+<=*m* people will come to buy a ticket. *n* of them will have only a single 10 euro banknote, and *m* of them will have only a single 20 euro banknote. Currently Charlie has *k* 10 euro banknotes, which he can use for change if needed. All *n*<=+<=*m* people will come to the ticket store in random order, all orders are equiprobable. Return the probability that the ticket selling process will run smoothly, i.e. Charlie will have change for every person with 20 euro banknote. | The input consist of a single line with three space separated integers, *n*, *m* and *k* (0<=≤<=*n*,<=*m*<=≤<=105, 0<=≤<=*k*<=≤<=10). | Output on a single line the desired probability with at least 4 digits after the decimal point. | [
"5 3 1\n",
"0 5 5\n",
"0 1 0\n"
] | [
"0.857143\n",
"1\n",
"0\n"
] | none | 2,000 | [
{
"input": "5 3 1",
"output": "0.857143"
},
{
"input": "0 5 5",
"output": "1"
},
{
"input": "0 1 0",
"output": "0"
},
{
"input": "95105 76851 10",
"output": "0.904215"
},
{
"input": "60503 53620 1",
"output": "0.214637"
},
{
"input": "25902 30390 2",
... | 1,499,391,581 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 124 | 5,529,600 | def find(A):
n,m,k=A
if m<k:
return 1
temp1=1
temp2=1
for i in range(k+1):
temp1*=(m-i)
temp2*=(n+i+1)
return 1-temp1/temp2
print(find(list(map(int,input().strip().split(' ')))))
| Title: Tickets
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
As a big fan of Formula One, Charlie is really happy with the fact that he has to organize ticket sells for the next Grand Prix race in his own city. Unfortunately, the finacial crisis is striking everywhere and all the banknotes left in his country are valued either 10 euros or 20 euros. The price of all tickets for the race is 10 euros, so whenever someone comes to the ticket store only with 20 euro banknote Charlie must have a 10 euro banknote to give them change. Charlie realize that with the huge deficit of banknotes this could be a problem. Charlie has some priceless information but couldn't make use of it, so he needs your help. Exactly *n*<=+<=*m* people will come to buy a ticket. *n* of them will have only a single 10 euro banknote, and *m* of them will have only a single 20 euro banknote. Currently Charlie has *k* 10 euro banknotes, which he can use for change if needed. All *n*<=+<=*m* people will come to the ticket store in random order, all orders are equiprobable. Return the probability that the ticket selling process will run smoothly, i.e. Charlie will have change for every person with 20 euro banknote.
Input Specification:
The input consist of a single line with three space separated integers, *n*, *m* and *k* (0<=≤<=*n*,<=*m*<=≤<=105, 0<=≤<=*k*<=≤<=10).
Output Specification:
Output on a single line the desired probability with at least 4 digits after the decimal point.
Demo Input:
['5 3 1\n', '0 5 5\n', '0 1 0\n']
Demo Output:
['0.857143\n', '1\n', '0\n']
Note:
none | ```python
def find(A):
n,m,k=A
if m<k:
return 1
temp1=1
temp2=1
for i in range(k+1):
temp1*=(m-i)
temp2*=(n+i+1)
return 1-temp1/temp2
print(find(list(map(int,input().strip().split(' ')))))
``` | 0 |
111 | A | Petya and Inequiations | PROGRAMMING | 1,400 | [
"greedy"
] | A. Petya and Inequiations | 2 | 256 | Little Petya loves inequations. Help him find *n* positive integers *a*1,<=*a*2,<=...,<=*a**n*, such that the following two conditions are satisfied:
- *a*12<=+<=*a*22<=+<=...<=+<=*a**n*2<=≥<=*x*- *a*1<=+<=*a*2<=+<=...<=+<=*a**n*<=≤<=*y* | The first line contains three space-separated integers *n*, *x* and *y* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*x*<=≤<=1012,<=1<=≤<=*y*<=≤<=106).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is recommended to use cin, cout streams or the %I64d specificator. | Print *n* positive integers that satisfy the conditions, one integer per line. If such numbers do not exist, print a single number "-1". If there are several solutions, print any of them. | [
"5 15 15\n",
"2 3 2\n",
"1 99 11\n"
] | [
"4\n4\n1\n1\n2\n",
"-1\n",
"11\n"
] | none | 500 | [
{
"input": "5 15 15",
"output": "11\n1\n1\n1\n1"
},
{
"input": "2 3 2",
"output": "-1"
},
{
"input": "1 99 11",
"output": "11"
},
{
"input": "100000 810000099998 1000000",
"output": "900001\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n... | 1,591,529,265 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 48 | 436 | 7,884,800 | n, x, y = map(int, input().split())
a = y - n + 1
if a * a + n - 1 < x: exit(print(-1))
print(a)
for _ in range(n - 1):
print(1)
| Title: Petya and Inequiations
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Petya loves inequations. Help him find *n* positive integers *a*1,<=*a*2,<=...,<=*a**n*, such that the following two conditions are satisfied:
- *a*12<=+<=*a*22<=+<=...<=+<=*a**n*2<=≥<=*x*- *a*1<=+<=*a*2<=+<=...<=+<=*a**n*<=≤<=*y*
Input Specification:
The first line contains three space-separated integers *n*, *x* and *y* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*x*<=≤<=1012,<=1<=≤<=*y*<=≤<=106).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is recommended to use cin, cout streams or the %I64d specificator.
Output Specification:
Print *n* positive integers that satisfy the conditions, one integer per line. If such numbers do not exist, print a single number "-1". If there are several solutions, print any of them.
Demo Input:
['5 15 15\n', '2 3 2\n', '1 99 11\n']
Demo Output:
['4\n4\n1\n1\n2\n', '-1\n', '11\n']
Note:
none | ```python
n, x, y = map(int, input().split())
a = y - n + 1
if a * a + n - 1 < x: exit(print(-1))
print(a)
for _ in range(n - 1):
print(1)
``` | 0 |
960 | B | Minimize the error | PROGRAMMING | 1,500 | [
"data structures",
"greedy",
"sortings"
] | null | null | You are given two arrays *A* and *B*, each of size *n*. The error, *E*, between these two arrays is defined . You have to perform exactly *k*1 operations on array *A* and exactly *k*2 operations on array *B*. In one operation, you have to choose one element of the array and increase or decrease it by 1.
Output the minimum possible value of error after *k*1 operations on array *A* and *k*2 operations on array *B* have been performed. | The first line contains three space-separated integers *n* (1<=≤<=*n*<=≤<=103), *k*1 and *k*2 (0<=≤<=*k*1<=+<=*k*2<=≤<=103, *k*1 and *k*2 are non-negative) — size of arrays and number of operations to perform on *A* and *B* respectively.
Second line contains *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=106<=≤<=*a**i*<=≤<=106) — array *A*.
Third line contains *n* space separated integers *b*1,<=*b*2,<=...,<=*b**n* (<=-<=106<=≤<=*b**i*<=≤<=106)— array *B*. | Output a single integer — the minimum possible value of after doing exactly *k*1 operations on array *A* and exactly *k*2 operations on array *B*. | [
"2 0 0\n1 2\n2 3\n",
"2 1 0\n1 2\n2 2\n",
"2 5 7\n3 4\n14 4\n"
] | [
"2",
"0",
"1"
] | In the first sample case, we cannot perform any operations on *A* or *B*. Therefore the minimum possible error *E* = (1 - 2)<sup class="upper-index">2</sup> + (2 - 3)<sup class="upper-index">2</sup> = 2.
In the second sample case, we are required to perform exactly one operation on *A*. In order to minimize error, we increment the first element of *A* by 1. Now, *A* = [2, 2]. The error is now *E* = (2 - 2)<sup class="upper-index">2</sup> + (2 - 2)<sup class="upper-index">2</sup> = 0. This is the minimum possible error obtainable.
In the third sample case, we can increase the first element of *A* to 8, using the all of the 5 moves available to us. Also, the first element of *B* can be reduced to 8 using the 6 of the 7 available moves. Now *A* = [8, 4] and *B* = [8, 4]. The error is now *E* = (8 - 8)<sup class="upper-index">2</sup> + (4 - 4)<sup class="upper-index">2</sup> = 0, but we are still left with 1 move for array *B*. Increasing the second element of *B* to 5 using the left move, we get *B* = [8, 5] and *E* = (8 - 8)<sup class="upper-index">2</sup> + (4 - 5)<sup class="upper-index">2</sup> = 1. | 1,000 | [
{
"input": "2 0 0\n1 2\n2 3",
"output": "2"
},
{
"input": "2 1 0\n1 2\n2 2",
"output": "0"
},
{
"input": "2 5 7\n3 4\n14 4",
"output": "1"
},
{
"input": "2 0 1\n1 2\n2 2",
"output": "0"
},
{
"input": "2 1 1\n0 0\n1 1",
"output": "0"
},
{
"input": "5 5 ... | 1,523,211,399 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 77 | 7,065,600 |
lenght, a, b = map(int, input().split())
masa = list(map(int, input().split()))
masb = list(map(int, input().split()))
masc = list(abs(masa[i] - masb[i]) for i in range(lenght) if abs(masa[i] - masb[i]) != 0)
moves = a + b
if masc:
while moves > 0:
masc.sort()
masc[~0] -= 1
moves -= 1
print(sum(i ** 2 for i in masc))
else:
print(moves % 2)
| Title: Minimize the error
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two arrays *A* and *B*, each of size *n*. The error, *E*, between these two arrays is defined . You have to perform exactly *k*1 operations on array *A* and exactly *k*2 operations on array *B*. In one operation, you have to choose one element of the array and increase or decrease it by 1.
Output the minimum possible value of error after *k*1 operations on array *A* and *k*2 operations on array *B* have been performed.
Input Specification:
The first line contains three space-separated integers *n* (1<=≤<=*n*<=≤<=103), *k*1 and *k*2 (0<=≤<=*k*1<=+<=*k*2<=≤<=103, *k*1 and *k*2 are non-negative) — size of arrays and number of operations to perform on *A* and *B* respectively.
Second line contains *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=106<=≤<=*a**i*<=≤<=106) — array *A*.
Third line contains *n* space separated integers *b*1,<=*b*2,<=...,<=*b**n* (<=-<=106<=≤<=*b**i*<=≤<=106)— array *B*.
Output Specification:
Output a single integer — the minimum possible value of after doing exactly *k*1 operations on array *A* and exactly *k*2 operations on array *B*.
Demo Input:
['2 0 0\n1 2\n2 3\n', '2 1 0\n1 2\n2 2\n', '2 5 7\n3 4\n14 4\n']
Demo Output:
['2', '0', '1']
Note:
In the first sample case, we cannot perform any operations on *A* or *B*. Therefore the minimum possible error *E* = (1 - 2)<sup class="upper-index">2</sup> + (2 - 3)<sup class="upper-index">2</sup> = 2.
In the second sample case, we are required to perform exactly one operation on *A*. In order to minimize error, we increment the first element of *A* by 1. Now, *A* = [2, 2]. The error is now *E* = (2 - 2)<sup class="upper-index">2</sup> + (2 - 2)<sup class="upper-index">2</sup> = 0. This is the minimum possible error obtainable.
In the third sample case, we can increase the first element of *A* to 8, using the all of the 5 moves available to us. Also, the first element of *B* can be reduced to 8 using the 6 of the 7 available moves. Now *A* = [8, 4] and *B* = [8, 4]. The error is now *E* = (8 - 8)<sup class="upper-index">2</sup> + (4 - 4)<sup class="upper-index">2</sup> = 0, but we are still left with 1 move for array *B*. Increasing the second element of *B* to 5 using the left move, we get *B* = [8, 5] and *E* = (8 - 8)<sup class="upper-index">2</sup> + (4 - 5)<sup class="upper-index">2</sup> = 1. | ```python
lenght, a, b = map(int, input().split())
masa = list(map(int, input().split()))
masb = list(map(int, input().split()))
masc = list(abs(masa[i] - masb[i]) for i in range(lenght) if abs(masa[i] - masb[i]) != 0)
moves = a + b
if masc:
while moves > 0:
masc.sort()
masc[~0] -= 1
moves -= 1
print(sum(i ** 2 for i in masc))
else:
print(moves % 2)
``` | 0 | |
276 | D | Little Girl and Maximum XOR | PROGRAMMING | 1,700 | [
"bitmasks",
"dp",
"greedy",
"implementation",
"math"
] | null | null | A little girl loves problems on bitwise operations very much. Here's one of them.
You are given two integers *l* and *r*. Let's consider the values of for all pairs of integers *a* and *b* (*l*<=≤<=*a*<=≤<=*b*<=≤<=*r*). Your task is to find the maximum value among all considered ones.
Expression means applying bitwise excluding or operation to integers *x* and *y*. The given operation exists in all modern programming languages, for example, in languages *C*++ and *Java* it is represented as "^", in *Pascal* — as "xor". | The single line contains space-separated integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier. | In a single line print a single integer — the maximum value of for all pairs of integers *a*, *b* (*l*<=≤<=*a*<=≤<=*b*<=≤<=*r*). | [
"1 2\n",
"8 16\n",
"1 1\n"
] | [
"3\n",
"31\n",
"0\n"
] | none | 2,000 | [
{
"input": "1 2",
"output": "3"
},
{
"input": "8 16",
"output": "31"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "506 677",
"output": "1023"
},
{
"input": "33 910",
"output": "1023"
},
{
"input": "36 94",
"output": "127"
},
{
"input": ... | 1,680,549,416 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 8 | 46 | 0 | l, r = map(int, input().split())
if l == r:
print(0)
exit()
num2 = str(bin(r))[2:]
num1 = str(bin(l))[2:]
temp = ([0] * (len(num2) - len(num1)))
temp.extend(num1)
# print(temp)
# print(num2)
total = 0
counter = 0
diff = 0
for i in range(len(num2)-1, -1, -1):
if temp[i] != num2[i]:
diff = i
# print(diff)
for i in range(len(num2)-1, -1, -1):
if diff <= i:
total += pow(2, counter)
elif num2[i] == "1":
total += pow(2, counter)
counter += 1
print(total) | Title: Little Girl and Maximum XOR
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A little girl loves problems on bitwise operations very much. Here's one of them.
You are given two integers *l* and *r*. Let's consider the values of for all pairs of integers *a* and *b* (*l*<=≤<=*a*<=≤<=*b*<=≤<=*r*). Your task is to find the maximum value among all considered ones.
Expression means applying bitwise excluding or operation to integers *x* and *y*. The given operation exists in all modern programming languages, for example, in languages *C*++ and *Java* it is represented as "^", in *Pascal* — as "xor".
Input Specification:
The single line contains space-separated integers *l* and *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output Specification:
In a single line print a single integer — the maximum value of for all pairs of integers *a*, *b* (*l*<=≤<=*a*<=≤<=*b*<=≤<=*r*).
Demo Input:
['1 2\n', '8 16\n', '1 1\n']
Demo Output:
['3\n', '31\n', '0\n']
Note:
none | ```python
l, r = map(int, input().split())
if l == r:
print(0)
exit()
num2 = str(bin(r))[2:]
num1 = str(bin(l))[2:]
temp = ([0] * (len(num2) - len(num1)))
temp.extend(num1)
# print(temp)
# print(num2)
total = 0
counter = 0
diff = 0
for i in range(len(num2)-1, -1, -1):
if temp[i] != num2[i]:
diff = i
# print(diff)
for i in range(len(num2)-1, -1, -1):
if diff <= i:
total += pow(2, counter)
elif num2[i] == "1":
total += pow(2, counter)
counter += 1
print(total)
``` | 0 | |
298 | B | Sail | PROGRAMMING | 1,200 | [
"brute force",
"greedy",
"implementation"
] | null | null | The polar bears are going fishing. They plan to sail from (*s**x*,<=*s**y*) to (*e**x*,<=*e**y*). However, the boat can only sail by wind. At each second, the wind blows in one of these directions: east, south, west or north. Assume the boat is currently at (*x*,<=*y*).
- If the wind blows to the east, the boat will move to (*x*<=+<=1,<=*y*). - If the wind blows to the south, the boat will move to (*x*,<=*y*<=-<=1). - If the wind blows to the west, the boat will move to (*x*<=-<=1,<=*y*). - If the wind blows to the north, the boat will move to (*x*,<=*y*<=+<=1).
Alternatively, they can hold the boat by the anchor. In this case, the boat stays at (*x*,<=*y*). Given the wind direction for *t* seconds, what is the earliest time they sail to (*e**x*,<=*e**y*)? | The first line contains five integers *t*,<=*s**x*,<=*s**y*,<=*e**x*,<=*e**y* (1<=≤<=*t*<=≤<=105,<=<=-<=109<=≤<=*s**x*,<=*s**y*,<=*e**x*,<=*e**y*<=≤<=109). The starting location and the ending location will be different.
The second line contains *t* characters, the *i*-th character is the wind blowing direction at the *i*-th second. It will be one of the four possibilities: "E" (east), "S" (south), "W" (west) and "N" (north). | If they can reach (*e**x*,<=*e**y*) within *t* seconds, print the earliest time they can achieve it. Otherwise, print "-1" (without quotes). | [
"5 0 0 1 1\nSESNW\n",
"10 5 3 3 6\nNENSWESNEE\n"
] | [
"4\n",
"-1\n"
] | In the first sample, they can stay at seconds 1, 3, and move at seconds 2, 4.
In the second sample, they cannot sail to the destination. | 500 | [
{
"input": "5 0 0 1 1\nSESNW",
"output": "4"
},
{
"input": "10 5 3 3 6\nNENSWESNEE",
"output": "-1"
},
{
"input": "19 -172106364 -468680119 -172106365 -468680119\nSSEEESSSESESWSEESSS",
"output": "13"
},
{
"input": "39 -1000000000 -1000000000 -999999997 -1000000000\nENEENWSWSS... | 1,571,518,503 | 2,147,483,647 | Python 3 | OK | TESTS | 43 | 372 | 1,331,200 | p=input()
t,sx,sy,ex,ey=map(int,p.split())
q=input()
q=q.replace('',' ')
q=q.split()
i=0
while (sx!=ex or sy!=ey) and i<t:
if q[i]=='E':
if sx<ex:
sx+=1
elif q[i]=='W':
if sx>ex:
sx-=1
elif q[i]=='N':
if sy<ey:
sy+=1
else:
if sy>ey:
sy-=1
i+=1
if sx==ex and sy==ey:
print(i)
else:
print(-1)
| Title: Sail
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The polar bears are going fishing. They plan to sail from (*s**x*,<=*s**y*) to (*e**x*,<=*e**y*). However, the boat can only sail by wind. At each second, the wind blows in one of these directions: east, south, west or north. Assume the boat is currently at (*x*,<=*y*).
- If the wind blows to the east, the boat will move to (*x*<=+<=1,<=*y*). - If the wind blows to the south, the boat will move to (*x*,<=*y*<=-<=1). - If the wind blows to the west, the boat will move to (*x*<=-<=1,<=*y*). - If the wind blows to the north, the boat will move to (*x*,<=*y*<=+<=1).
Alternatively, they can hold the boat by the anchor. In this case, the boat stays at (*x*,<=*y*). Given the wind direction for *t* seconds, what is the earliest time they sail to (*e**x*,<=*e**y*)?
Input Specification:
The first line contains five integers *t*,<=*s**x*,<=*s**y*,<=*e**x*,<=*e**y* (1<=≤<=*t*<=≤<=105,<=<=-<=109<=≤<=*s**x*,<=*s**y*,<=*e**x*,<=*e**y*<=≤<=109). The starting location and the ending location will be different.
The second line contains *t* characters, the *i*-th character is the wind blowing direction at the *i*-th second. It will be one of the four possibilities: "E" (east), "S" (south), "W" (west) and "N" (north).
Output Specification:
If they can reach (*e**x*,<=*e**y*) within *t* seconds, print the earliest time they can achieve it. Otherwise, print "-1" (without quotes).
Demo Input:
['5 0 0 1 1\nSESNW\n', '10 5 3 3 6\nNENSWESNEE\n']
Demo Output:
['4\n', '-1\n']
Note:
In the first sample, they can stay at seconds 1, 3, and move at seconds 2, 4.
In the second sample, they cannot sail to the destination. | ```python
p=input()
t,sx,sy,ex,ey=map(int,p.split())
q=input()
q=q.replace('',' ')
q=q.split()
i=0
while (sx!=ex or sy!=ey) and i<t:
if q[i]=='E':
if sx<ex:
sx+=1
elif q[i]=='W':
if sx>ex:
sx-=1
elif q[i]=='N':
if sy<ey:
sy+=1
else:
if sy>ey:
sy-=1
i+=1
if sx==ex and sy==ey:
print(i)
else:
print(-1)
``` | 3 | |
630 | G | Challenge Pennants | PROGRAMMING | 1,600 | [
"combinatorics",
"math"
] | null | null | Because of budget cuts one IT company established new non-financial reward system instead of bonuses.
Two kinds of actions are rewarded: fixing critical bugs and suggesting new interesting features. A man who fixed a critical bug gets "I fixed a critical bug" pennant on his table. A man who suggested a new interesting feature gets "I suggested a new feature" pennant on his table.
Because of the limited budget of the new reward system only 5 "I fixed a critical bug" pennants and 3 "I suggested a new feature" pennants were bought.
In order to use these pennants for a long time they were made challenge ones. When a man fixes a new critical bug one of the earlier awarded "I fixed a critical bug" pennants is passed on to his table. When a man suggests a new interesting feature one of the earlier awarded "I suggested a new feature" pennants is passed on to his table.
One man can have several pennants of one type and of course he can have pennants of both types on his table. There are *n* tables in the IT company. Find the number of ways to place the pennants on these tables given that each pennant is situated on one of the tables and each table is big enough to contain any number of pennants. | The only line of the input contains one integer *n* (1<=≤<=*n*<=≤<=500) — the number of tables in the IT company. | Output one integer — the amount of ways to place the pennants on *n* tables. | [
"2\n"
] | [
"24"
] | none | 0 | [
{
"input": "2",
"output": "24"
},
{
"input": "1",
"output": "1"
},
{
"input": "3",
"output": "210"
},
{
"input": "4",
"output": "1120"
},
{
"input": "5",
"output": "4410"
},
{
"input": "6",
"output": "14112"
},
{
"input": "7",
"output":... | 1,687,723,013 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 15 | 62 | 0 | import sys
input = lambda: sys.stdin.readline().rstrip()
import math
N = int(input())
a = math.comb(N+5-1,5)
b = math.comb(N+3-1,3)
print(a*b) | Title: Challenge Pennants
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Because of budget cuts one IT company established new non-financial reward system instead of bonuses.
Two kinds of actions are rewarded: fixing critical bugs and suggesting new interesting features. A man who fixed a critical bug gets "I fixed a critical bug" pennant on his table. A man who suggested a new interesting feature gets "I suggested a new feature" pennant on his table.
Because of the limited budget of the new reward system only 5 "I fixed a critical bug" pennants and 3 "I suggested a new feature" pennants were bought.
In order to use these pennants for a long time they were made challenge ones. When a man fixes a new critical bug one of the earlier awarded "I fixed a critical bug" pennants is passed on to his table. When a man suggests a new interesting feature one of the earlier awarded "I suggested a new feature" pennants is passed on to his table.
One man can have several pennants of one type and of course he can have pennants of both types on his table. There are *n* tables in the IT company. Find the number of ways to place the pennants on these tables given that each pennant is situated on one of the tables and each table is big enough to contain any number of pennants.
Input Specification:
The only line of the input contains one integer *n* (1<=≤<=*n*<=≤<=500) — the number of tables in the IT company.
Output Specification:
Output one integer — the amount of ways to place the pennants on *n* tables.
Demo Input:
['2\n']
Demo Output:
['24']
Note:
none | ```python
import sys
input = lambda: sys.stdin.readline().rstrip()
import math
N = int(input())
a = math.comb(N+5-1,5)
b = math.comb(N+3-1,3)
print(a*b)
``` | 3 | |
877 | A | Alex and broken contest | PROGRAMMING | 1,100 | [
"implementation",
"strings"
] | null | null | One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive. | The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 — the name of the problem. | Print "YES", if problem is from this contest, and "NO" otherwise. | [
"Alex_and_broken_contest\n",
"NikitaAndString\n",
"Danil_and_Olya\n"
] | [
"NO",
"YES",
"NO"
] | none | 500 | [
{
"input": "Alex_and_broken_contest",
"output": "NO"
},
{
"input": "NikitaAndString",
"output": "YES"
},
{
"input": "Danil_and_Olya",
"output": "NO"
},
{
"input": "Slava____and_the_game",
"output": "YES"
},
{
"input": "Olya_and_energy_drinks",
"output": "YES"
... | 1,521,712,898 | 2,147,483,647 | Python 3 | OK | TESTS | 55 | 93 | 7,065,600 | s=input()
D="Danil"
O="Olya"
S="Slava"
A="Ann"
N="Nikita"
if s.count(D)==1 and s.count(O)==0 and s.count(S)==0 and s.count(A)==0 and s.count(N)==0:
print("YES")
elif s.count(D)==0 and s.count(O)==1 and s.count(S)==0 and s.count(A)==0 and s.count(N)==0:
print("YES")
elif s.count(D)==0 and s.count(O)==0 and s.count(S)==1 and s.count(A)==0 and s.count(N)==0:
print("YES")
elif s.count(D)==0 and s.count(O)==0 and s.count(S)==0 and s.count(A)==1 and s.count(N)==0:
print("YES")
elif s.count(D)==0 and s.count(O)==0 and s.count(S)==0 and s.count(A)==0 and s.count(N)==1:
print("YES")
else:
print("NO")
| Title: Alex and broken contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.
But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.
It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".
Names are case sensitive.
Input Specification:
The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 — the name of the problem.
Output Specification:
Print "YES", if problem is from this contest, and "NO" otherwise.
Demo Input:
['Alex_and_broken_contest\n', 'NikitaAndString\n', 'Danil_and_Olya\n']
Demo Output:
['NO', 'YES', 'NO']
Note:
none | ```python
s=input()
D="Danil"
O="Olya"
S="Slava"
A="Ann"
N="Nikita"
if s.count(D)==1 and s.count(O)==0 and s.count(S)==0 and s.count(A)==0 and s.count(N)==0:
print("YES")
elif s.count(D)==0 and s.count(O)==1 and s.count(S)==0 and s.count(A)==0 and s.count(N)==0:
print("YES")
elif s.count(D)==0 and s.count(O)==0 and s.count(S)==1 and s.count(A)==0 and s.count(N)==0:
print("YES")
elif s.count(D)==0 and s.count(O)==0 and s.count(S)==0 and s.count(A)==1 and s.count(N)==0:
print("YES")
elif s.count(D)==0 and s.count(O)==0 and s.count(S)==0 and s.count(A)==0 and s.count(N)==1:
print("YES")
else:
print("NO")
``` | 3 | |
27 | E | Number With The Given Amount Of Divisors | PROGRAMMING | 2,000 | [
"brute force",
"dp",
"number theory"
] | E. Number With The Given Amount Of Divisors | 2 | 256 | Given the number *n*, find the smallest positive integer which has exactly *n* divisors. It is guaranteed that for the given *n* the answer will not exceed 1018. | The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=1000). | Output the smallest positive integer with exactly *n* divisors. | [
"4\n",
"6\n"
] | [
"6\n",
"12\n"
] | none | 2,500 | [
{
"input": "1",
"output": "1"
},
{
"input": "7",
"output": "64"
},
{
"input": "8",
"output": "24"
},
{
"input": "9",
"output": "36"
},
{
"input": "10",
"output": "48"
},
{
"input": "15",
"output": "144"
},
{
"input": "20",
"output": "24... | 1,617,574,239 | 2,147,483,647 | PyPy 3 | OK | TESTS | 22 | 248 | 3,891,200 | INF = 20**18
primes = []
a = [1]*100
def crivo():
for i in range(2, 100):
if a[i]:
primes.append(i)
for j in range(i+i, 100, i):
a[j] = 0
def dp(div, idx):
div = int(div)
idx = int(idx)
if div == 1:
return 1
resp = INF
for i in range(1, div):
if div%(i+1) == 0:
resp = min(resp, primes[idx]**i * dp(div/(i+1), idx+1))
return resp
crivo()
n = int(input())
print(dp(n, 0))
| Title: Number With The Given Amount Of Divisors
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Given the number *n*, find the smallest positive integer which has exactly *n* divisors. It is guaranteed that for the given *n* the answer will not exceed 1018.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=1000).
Output Specification:
Output the smallest positive integer with exactly *n* divisors.
Demo Input:
['4\n', '6\n']
Demo Output:
['6\n', '12\n']
Note:
none | ```python
INF = 20**18
primes = []
a = [1]*100
def crivo():
for i in range(2, 100):
if a[i]:
primes.append(i)
for j in range(i+i, 100, i):
a[j] = 0
def dp(div, idx):
div = int(div)
idx = int(idx)
if div == 1:
return 1
resp = INF
for i in range(1, div):
if div%(i+1) == 0:
resp = min(resp, primes[idx]**i * dp(div/(i+1), idx+1))
return resp
crivo()
n = int(input())
print(dp(n, 0))
``` | 3.930752 |
69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10... | 1,691,409,517 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 0 | 124 | 2,764,800 | n=int(input())
x,y,z=map(int,input().split())
static=False
for v in zip(x,y,z):
if sum(v) == 0:
print("YES")
static=True
else:
print("No") | Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
n=int(input())
x,y,z=map(int,input().split())
static=False
for v in zip(x,y,z):
if sum(v) == 0:
print("YES")
static=True
else:
print("No")
``` | -1 |
16 | B | Burglar and Matches | PROGRAMMING | 900 | [
"greedy",
"implementation",
"sortings"
] | B. Burglar and Matches | 0 | 64 | A burglar got into a matches warehouse and wants to steal as many matches as possible. In the warehouse there are *m* containers, in the *i*-th container there are *a**i* matchboxes, and each matchbox contains *b**i* matches. All the matchboxes are of the same size. The burglar's rucksack can hold *n* matchboxes exactly. Your task is to find out the maximum amount of matches that a burglar can carry away. He has no time to rearrange matches in the matchboxes, that's why he just chooses not more than *n* matchboxes so that the total amount of matches in them is maximal. | The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=2·108) and integer *m* (1<=≤<=*m*<=≤<=20). The *i*<=+<=1-th line contains a pair of numbers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=108,<=1<=≤<=*b**i*<=≤<=10). All the input numbers are integer. | Output the only number — answer to the problem. | [
"7 3\n5 10\n2 5\n3 6\n",
"3 3\n1 3\n2 2\n3 1\n"
] | [
"62\n",
"7\n"
] | none | 0 | [
{
"input": "7 3\n5 10\n2 5\n3 6",
"output": "62"
},
{
"input": "3 3\n1 3\n2 2\n3 1",
"output": "7"
},
{
"input": "1 1\n1 2",
"output": "2"
},
{
"input": "1 2\n1 9\n1 6",
"output": "9"
},
{
"input": "1 10\n1 1\n1 9\n1 3\n1 9\n1 7\n1 10\n1 4\n1 7\n1 3\n1 1",
"ou... | 1,641,817,941 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 92 | 0 | n,m = map(int,input().split())
matchboxes = {}
for _ in range(m):
a,b = map(int,input().split()) #a refers to number of matchboxes holding b matches
matchboxes[b] = a if b not in matchboxes else matchboxes[b] + a
highest = 0
for key,value in sorted(matchboxes.items(),key = lambda x: x[0],reverse = True):
if value <= n:
highest += key*value
n -= value
else:
highest += key*n
n = 0
if n == 0:
break
print(highest)
| Title: Burglar and Matches
Time Limit: 0 seconds
Memory Limit: 64 megabytes
Problem Description:
A burglar got into a matches warehouse and wants to steal as many matches as possible. In the warehouse there are *m* containers, in the *i*-th container there are *a**i* matchboxes, and each matchbox contains *b**i* matches. All the matchboxes are of the same size. The burglar's rucksack can hold *n* matchboxes exactly. Your task is to find out the maximum amount of matches that a burglar can carry away. He has no time to rearrange matches in the matchboxes, that's why he just chooses not more than *n* matchboxes so that the total amount of matches in them is maximal.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=2·108) and integer *m* (1<=≤<=*m*<=≤<=20). The *i*<=+<=1-th line contains a pair of numbers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=108,<=1<=≤<=*b**i*<=≤<=10). All the input numbers are integer.
Output Specification:
Output the only number — answer to the problem.
Demo Input:
['7 3\n5 10\n2 5\n3 6\n', '3 3\n1 3\n2 2\n3 1\n']
Demo Output:
['62\n', '7\n']
Note:
none | ```python
n,m = map(int,input().split())
matchboxes = {}
for _ in range(m):
a,b = map(int,input().split()) #a refers to number of matchboxes holding b matches
matchboxes[b] = a if b not in matchboxes else matchboxes[b] + a
highest = 0
for key,value in sorted(matchboxes.items(),key = lambda x: x[0],reverse = True):
if value <= n:
highest += key*value
n -= value
else:
highest += key*n
n = 0
if n == 0:
break
print(highest)
``` | 3 |
231 | A | Team | PROGRAMMING | 800 | [
"brute force",
"greedy"
] | null | null | One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution. | The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces. | Print a single integer — the number of problems the friends will implement on the contest. | [
"3\n1 1 0\n1 1 1\n1 0 0\n",
"2\n1 0 0\n0 1 1\n"
] | [
"2\n",
"1\n"
] | In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. | 500 | [
{
"input": "3\n1 1 0\n1 1 1\n1 0 0",
"output": "2"
},
{
"input": "2\n1 0 0\n0 1 1",
"output": "1"
},
{
"input": "1\n1 0 0",
"output": "0"
},
{
"input": "2\n1 0 0\n1 1 1",
"output": "1"
},
{
"input": "5\n1 0 0\n0 1 0\n1 1 1\n0 0 1\n0 0 0",
"output": "1"
},
... | 1,697,835,375 | 2,147,483,647 | Python 3 | OK | TESTS | 21 | 92 | 0 | n = int(input())
counter = 0
for i in range(0, n):
k = input()
if k.count("1") >= 2:
counter += 1
print(counter)
| Title: Team
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day three best friends Petya, Vasya and Tonya decided to form a team and take part in programming contests. Participants are usually offered several problems during programming contests. Long before the start the friends decided that they will implement a problem if at least two of them are sure about the solution. Otherwise, the friends won't write the problem's solution.
This contest offers *n* problems to the participants. For each problem we know, which friend is sure about the solution. Help the friends find the number of problems for which they will write a solution.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of problems in the contest. Then *n* lines contain three integers each, each integer is either 0 or 1. If the first number in the line equals 1, then Petya is sure about the problem's solution, otherwise he isn't sure. The second number shows Vasya's view on the solution, the third number shows Tonya's view. The numbers on the lines are separated by spaces.
Output Specification:
Print a single integer — the number of problems the friends will implement on the contest.
Demo Input:
['3\n1 1 0\n1 1 1\n1 0 0\n', '2\n1 0 0\n0 1 1\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample Petya and Vasya are sure that they know how to solve the first problem and all three of them know how to solve the second problem. That means that they will write solutions for these problems. Only Petya is sure about the solution for the third problem, but that isn't enough, so the friends won't take it.
In the second sample the friends will only implement the second problem, as Vasya and Tonya are sure about the solution. | ```python
n = int(input())
counter = 0
for i in range(0, n):
k = input()
if k.count("1") >= 2:
counter += 1
print(counter)
``` | 3 | |
393 | A | Nineteen | PROGRAMMING | 0 | [] | null | null | Alice likes word "nineteen" very much. She has a string *s* and wants the string to contain as many such words as possible. For that reason she can rearrange the letters of the string.
For example, if she has string "xiineteenppnnnewtnee", she can get string "xnineteenppnineteenw", containing (the occurrences marked) two such words. More formally, word "nineteen" occurs in the string the number of times you can read it starting from some letter of the string. Of course, you shouldn't skip letters.
Help her to find the maximum number of "nineteen"s that she can get in her string. | The first line contains a non-empty string *s*, consisting only of lowercase English letters. The length of string *s* doesn't exceed 100. | Print a single integer — the maximum number of "nineteen"s that she can get in her string. | [
"nniinneetteeeenn\n",
"nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii\n",
"nineteenineteen\n"
] | [
"2",
"2",
"2"
] | none | 500 | [
{
"input": "nniinneetteeeenn",
"output": "2"
},
{
"input": "nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii",
"output": "2"
},
{
"input": "nineteenineteen",
"output": "2"
},
{
"input": "nssemsnnsitjtihtthij",
"output": "0"
},
{
"input": "eehihnttehtherjsihihn... | 1,560,212,128 | 2,147,483,647 | Python 3 | OK | TESTS | 58 | 108 | 0 | n = input()
vet = [0,0,0,0];
for i in range(0,len(n)):
if(n[i] == 'n'):
vet[0]+=1
elif(n[i] == 'i'):
vet[1]+=1
elif(n[i] == 'e'):
vet[2]+=1
elif(n[i] == 't'):
vet[3]+=1
vet[0] = (vet[0] - 1) / 2
vet[2] /= 3
vet.sort()
print(int(vet[0])) | Title: Nineteen
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Alice likes word "nineteen" very much. She has a string *s* and wants the string to contain as many such words as possible. For that reason she can rearrange the letters of the string.
For example, if she has string "xiineteenppnnnewtnee", she can get string "xnineteenppnineteenw", containing (the occurrences marked) two such words. More formally, word "nineteen" occurs in the string the number of times you can read it starting from some letter of the string. Of course, you shouldn't skip letters.
Help her to find the maximum number of "nineteen"s that she can get in her string.
Input Specification:
The first line contains a non-empty string *s*, consisting only of lowercase English letters. The length of string *s* doesn't exceed 100.
Output Specification:
Print a single integer — the maximum number of "nineteen"s that she can get in her string.
Demo Input:
['nniinneetteeeenn\n', 'nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii\n', 'nineteenineteen\n']
Demo Output:
['2', '2', '2']
Note:
none | ```python
n = input()
vet = [0,0,0,0];
for i in range(0,len(n)):
if(n[i] == 'n'):
vet[0]+=1
elif(n[i] == 'i'):
vet[1]+=1
elif(n[i] == 'e'):
vet[2]+=1
elif(n[i] == 't'):
vet[3]+=1
vet[0] = (vet[0] - 1) / 2
vet[2] /= 3
vet.sort()
print(int(vet[0]))
``` | 3 | |
251 | A | Points on Line | PROGRAMMING | 1,300 | [
"binary search",
"combinatorics",
"two pointers"
] | null | null | Little Petya likes points a lot. Recently his mom has presented him *n* points lying on the line *OX*. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed *d*.
Note that the order of the points inside the group of three chosen points doesn't matter. | The first line contains two integers: *n* and *d* (1<=≤<=*n*<=≤<=105; 1<=≤<=*d*<=≤<=109). The next line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n*, their absolute value doesn't exceed 109 — the *x*-coordinates of the points that Petya has got.
It is guaranteed that the coordinates of the points in the input strictly increase. | Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed *d*.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. | [
"4 3\n1 2 3 4\n",
"4 2\n-3 -2 -1 0\n",
"5 19\n1 10 20 30 50\n"
] | [
"4\n",
"2\n",
"1\n"
] | In the first sample any group of three points meets our conditions.
In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}.
In the third sample only one group does: {1, 10, 20}. | 500 | [
{
"input": "4 3\n1 2 3 4",
"output": "4"
},
{
"input": "4 2\n-3 -2 -1 0",
"output": "2"
},
{
"input": "5 19\n1 10 20 30 50",
"output": "1"
},
{
"input": "10 5\n31 36 43 47 48 50 56 69 71 86",
"output": "2"
},
{
"input": "10 50\n1 4 20 27 65 79 82 83 99 100",
"... | 1,675,540,751 | 2,147,483,647 | PyPy 3 | OK | TESTS | 39 | 404 | 28,467,200 | num, choose = map(int, input().split())
lista = list(map(int, input().split()))
ans = 0
j = 0
for i in range(0,num):
while(lista[i] - lista[j] > choose): j+=1
ans += ((i - j) * (i - j - 1))/2
print(int(ans))
| Title: Points on Line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Petya likes points a lot. Recently his mom has presented him *n* points lying on the line *OX*. Now Petya is wondering in how many ways he can choose three distinct points so that the distance between the two farthest of them doesn't exceed *d*.
Note that the order of the points inside the group of three chosen points doesn't matter.
Input Specification:
The first line contains two integers: *n* and *d* (1<=≤<=*n*<=≤<=105; 1<=≤<=*d*<=≤<=109). The next line contains *n* integers *x*1,<=*x*2,<=...,<=*x**n*, their absolute value doesn't exceed 109 — the *x*-coordinates of the points that Petya has got.
It is guaranteed that the coordinates of the points in the input strictly increase.
Output Specification:
Print a single integer — the number of groups of three points, where the distance between two farthest points doesn't exceed *d*.
Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['4 3\n1 2 3 4\n', '4 2\n-3 -2 -1 0\n', '5 19\n1 10 20 30 50\n']
Demo Output:
['4\n', '2\n', '1\n']
Note:
In the first sample any group of three points meets our conditions.
In the seconds sample only 2 groups of three points meet our conditions: {-3, -2, -1} and {-2, -1, 0}.
In the third sample only one group does: {1, 10, 20}. | ```python
num, choose = map(int, input().split())
lista = list(map(int, input().split()))
ans = 0
j = 0
for i in range(0,num):
while(lista[i] - lista[j] > choose): j+=1
ans += ((i - j) * (i - j - 1))/2
print(int(ans))
``` | 3 | |
609 | A | USB Flash Drives | PROGRAMMING | 800 | [
"greedy",
"implementation",
"sortings"
] | null | null | Sean is trying to save a large file to a USB flash drive. He has *n* USB flash drives with capacities equal to *a*1,<=*a*2,<=...,<=*a**n* megabytes. The file size is equal to *m* megabytes.
Find the minimum number of USB flash drives needed to write Sean's file, if he can split the file between drives. | The first line contains positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of USB flash drives.
The second line contains positive integer *m* (1<=≤<=*m*<=≤<=105) — the size of Sean's file.
Each of the next *n* lines contains positive integer *a**i* (1<=≤<=*a**i*<=≤<=1000) — the sizes of USB flash drives in megabytes.
It is guaranteed that the answer exists, i. e. the sum of all *a**i* is not less than *m*. | Print the minimum number of USB flash drives to write Sean's file, if he can split the file between drives. | [
"3\n5\n2\n1\n3\n",
"3\n6\n2\n3\n2\n",
"2\n5\n5\n10\n"
] | [
"2\n",
"3\n",
"1\n"
] | In the first example Sean needs only two USB flash drives — the first and the third.
In the second example Sean needs all three USB flash drives.
In the third example Sean needs only one USB flash drive and he can use any available USB flash drive — the first or the second. | 0 | [
{
"input": "3\n5\n2\n1\n3",
"output": "2"
},
{
"input": "3\n6\n2\n3\n2",
"output": "3"
},
{
"input": "2\n5\n5\n10",
"output": "1"
},
{
"input": "5\n16\n8\n1\n3\n4\n9",
"output": "2"
},
{
"input": "10\n121\n10\n37\n74\n56\n42\n39\n6\n68\n8\n100",
"output": "2"
... | 1,562,078,765 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 108 | 0 | n = int(input())
m = int(input())
a = []
for i in range(n):
x = int(input())
a.append(x)
a.sort()
a = a[::-1]
count = 0
i = 0
res = 0
while i < n and res <= m:
if res + a[i] <= m:
res = res + a[i]
i += 1
if res == m:
print(i - 1)
else:
print(i)
| Title: USB Flash Drives
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sean is trying to save a large file to a USB flash drive. He has *n* USB flash drives with capacities equal to *a*1,<=*a*2,<=...,<=*a**n* megabytes. The file size is equal to *m* megabytes.
Find the minimum number of USB flash drives needed to write Sean's file, if he can split the file between drives.
Input Specification:
The first line contains positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of USB flash drives.
The second line contains positive integer *m* (1<=≤<=*m*<=≤<=105) — the size of Sean's file.
Each of the next *n* lines contains positive integer *a**i* (1<=≤<=*a**i*<=≤<=1000) — the sizes of USB flash drives in megabytes.
It is guaranteed that the answer exists, i. e. the sum of all *a**i* is not less than *m*.
Output Specification:
Print the minimum number of USB flash drives to write Sean's file, if he can split the file between drives.
Demo Input:
['3\n5\n2\n1\n3\n', '3\n6\n2\n3\n2\n', '2\n5\n5\n10\n']
Demo Output:
['2\n', '3\n', '1\n']
Note:
In the first example Sean needs only two USB flash drives — the first and the third.
In the second example Sean needs all three USB flash drives.
In the third example Sean needs only one USB flash drive and he can use any available USB flash drive — the first or the second. | ```python
n = int(input())
m = int(input())
a = []
for i in range(n):
x = int(input())
a.append(x)
a.sort()
a = a[::-1]
count = 0
i = 0
res = 0
while i < n and res <= m:
if res + a[i] <= m:
res = res + a[i]
i += 1
if res == m:
print(i - 1)
else:
print(i)
``` | 0 | |
598 | B | Queries on a String | PROGRAMMING | 1,300 | [
"implementation",
"strings"
] | null | null | You are given a string *s* and should process *m* queries. Each query is described by two 1-based indices *l**i*, *r**i* and integer *k**i*. It means that you should cyclically shift the substring *s*[*l**i*... *r**i*] *k**i* times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string *s* is abacaba and the query is *l*1<==<=3,<=*r*1<==<=6,<=*k*1<==<=1 then the answer is abbacaa. If after that we would process the query *l*2<==<=1,<=*r*2<==<=4,<=*k*2<==<=2 then we would get the string baabcaa. | The first line of the input contains the string *s* (1<=≤<=|*s*|<=≤<=10<=000) in its initial state, where |*s*| stands for the length of *s*. It contains only lowercase English letters.
Second line contains a single integer *m* (1<=≤<=*m*<=≤<=300) — the number of queries.
The *i*-th of the next *m* lines contains three integers *l**i*, *r**i* and *k**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=|*s*|,<=1<=≤<=*k**i*<=≤<=1<=000<=000) — the description of the *i*-th query. | Print the resulting string *s* after processing all *m* queries. | [
"abacaba\n2\n3 6 1\n1 4 2\n"
] | [
"baabcaa\n"
] | The sample is described in problem statement. | 0 | [
{
"input": "abacaba\n2\n3 6 1\n1 4 2",
"output": "baabcaa"
},
{
"input": "u\n1\n1 1 1",
"output": "u"
},
{
"input": "p\n5\n1 1 5\n1 1 9\n1 1 10\n1 1 10\n1 1 4",
"output": "p"
},
{
"input": "ssssssssss\n5\n5 7 9\n3 9 3\n2 7 1\n7 7 10\n1 9 6",
"output": "ssssssssss"
},
... | 1,636,131,187 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 46 | 28,467,200 | s = input()
q = int(input())
for i in range(q):
l, r, k = map(int, input().split())
if l==r:
continue
k%=(r-l)
b = s[l-1:r]
s = s[:l-1] + b[len(b)-k:] + b[:len(b)-k] + s[r:]
print(s) | Title: Queries on a String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string *s* and should process *m* queries. Each query is described by two 1-based indices *l**i*, *r**i* and integer *k**i*. It means that you should cyclically shift the substring *s*[*l**i*... *r**i*] *k**i* times. The queries should be processed one after another in the order they are given.
One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.
For example, if the string *s* is abacaba and the query is *l*1<==<=3,<=*r*1<==<=6,<=*k*1<==<=1 then the answer is abbacaa. If after that we would process the query *l*2<==<=1,<=*r*2<==<=4,<=*k*2<==<=2 then we would get the string baabcaa.
Input Specification:
The first line of the input contains the string *s* (1<=≤<=|*s*|<=≤<=10<=000) in its initial state, where |*s*| stands for the length of *s*. It contains only lowercase English letters.
Second line contains a single integer *m* (1<=≤<=*m*<=≤<=300) — the number of queries.
The *i*-th of the next *m* lines contains three integers *l**i*, *r**i* and *k**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=|*s*|,<=1<=≤<=*k**i*<=≤<=1<=000<=000) — the description of the *i*-th query.
Output Specification:
Print the resulting string *s* after processing all *m* queries.
Demo Input:
['abacaba\n2\n3 6 1\n1 4 2\n']
Demo Output:
['baabcaa\n']
Note:
The sample is described in problem statement. | ```python
s = input()
q = int(input())
for i in range(q):
l, r, k = map(int, input().split())
if l==r:
continue
k%=(r-l)
b = s[l-1:r]
s = s[:l-1] + b[len(b)-k:] + b[:len(b)-k] + s[r:]
print(s)
``` | 0 | |
227 | B | Effective Approach | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat. | Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier. | [
"2\n1 2\n1\n1\n",
"2\n2 1\n1\n1\n",
"3\n3 1 2\n3\n1 2 3\n"
] | [
"1 2\n",
"2 1\n",
"6 6\n"
] | In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | 1,000 | [
{
"input": "2\n1 2\n1\n1",
"output": "1 2"
},
{
"input": "2\n2 1\n1\n1",
"output": "2 1"
},
{
"input": "3\n3 1 2\n3\n1 2 3",
"output": "6 6"
},
{
"input": "9\n2 9 3 1 6 4 7 8 5\n9\n5 1 5 2 8 4 4 4 5",
"output": "58 32"
},
{
"input": "10\n3 10 9 2 7 6 5 8 4 1\n1\n4... | 1,633,399,584 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 2,000 | 7,372,800 | n = int(input())
x = list(map(int,input().split()))
m = int(input())
y = list(map(int,input().split()))
f = 0
s = 0
for i in range(m):
f += (x.index(y[i])+1)
x = x[::-1]
for i in range(m):
s+= (x.index(y[i])+1)
print(f,s)
| Title: Effective Approach
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
Output Specification:
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Demo Input:
['2\n1 2\n1\n1\n', '2\n2 1\n1\n1\n', '3\n3 1 2\n3\n1 2 3\n']
Demo Output:
['1 2\n', '2 1\n', '6 6\n']
Note:
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | ```python
n = int(input())
x = list(map(int,input().split()))
m = int(input())
y = list(map(int,input().split()))
f = 0
s = 0
for i in range(m):
f += (x.index(y[i])+1)
x = x[::-1]
for i in range(m):
s+= (x.index(y[i])+1)
print(f,s)
``` | 0 | |
931 | A | Friends Meeting | PROGRAMMING | 800 | [
"brute force",
"greedy",
"implementation",
"math"
] | null | null | Two friends are on the coordinate axis *Ox* in points with integer coordinates. One of them is in the point *x*1<==<=*a*, another one is in the point *x*2<==<=*b*.
Each of the friends can move by one along the line in any direction unlimited number of times. When a friend moves, the tiredness of a friend changes according to the following rules: the first move increases the tiredness by 1, the second move increases the tiredness by 2, the third — by 3 and so on. For example, if a friend moves first to the left, then to the right (returning to the same point), and then again to the left his tiredness becomes equal to 1<=+<=2<=+<=3<==<=6.
The friends want to meet in a integer point. Determine the minimum total tiredness they should gain, if they meet in the same point. | The first line contains a single integer *a* (1<=≤<=*a*<=≤<=1000) — the initial position of the first friend.
The second line contains a single integer *b* (1<=≤<=*b*<=≤<=1000) — the initial position of the second friend.
It is guaranteed that *a*<=≠<=*b*. | Print the minimum possible total tiredness if the friends meet in the same point. | [
"3\n4\n",
"101\n99\n",
"5\n10\n"
] | [
"1\n",
"2\n",
"9\n"
] | In the first example the first friend should move by one to the right (then the meeting happens at point 4), or the second friend should move by one to the left (then the meeting happens at point 3). In both cases, the total tiredness becomes 1.
In the second example the first friend should move by one to the left, and the second friend should move by one to the right. Then they meet in the point 100, and the total tiredness becomes 1 + 1 = 2.
In the third example one of the optimal ways is the following. The first friend should move three times to the right, and the second friend — two times to the left. Thus the friends meet in the point 8, and the total tiredness becomes 1 + 2 + 3 + 1 + 2 = 9. | 500 | [
{
"input": "3\n4",
"output": "1"
},
{
"input": "101\n99",
"output": "2"
},
{
"input": "5\n10",
"output": "9"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "1\n1000",
"output": "250000"
},
{
"input": "999\n1000",
"output": "1"
},
{
"inpu... | 1,570,327,331 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 124 | 0 | a=int(input())
b=int(input())
if abs(a-b)==1:
print("1")
else:
n=abs(a-b)
h=n//2
s=(h*(h+1))//2
if n%2==0:
print(2*s)
else:
r=(2*s)+(h+1)
print(r) | Title: Friends Meeting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two friends are on the coordinate axis *Ox* in points with integer coordinates. One of them is in the point *x*1<==<=*a*, another one is in the point *x*2<==<=*b*.
Each of the friends can move by one along the line in any direction unlimited number of times. When a friend moves, the tiredness of a friend changes according to the following rules: the first move increases the tiredness by 1, the second move increases the tiredness by 2, the third — by 3 and so on. For example, if a friend moves first to the left, then to the right (returning to the same point), and then again to the left his tiredness becomes equal to 1<=+<=2<=+<=3<==<=6.
The friends want to meet in a integer point. Determine the minimum total tiredness they should gain, if they meet in the same point.
Input Specification:
The first line contains a single integer *a* (1<=≤<=*a*<=≤<=1000) — the initial position of the first friend.
The second line contains a single integer *b* (1<=≤<=*b*<=≤<=1000) — the initial position of the second friend.
It is guaranteed that *a*<=≠<=*b*.
Output Specification:
Print the minimum possible total tiredness if the friends meet in the same point.
Demo Input:
['3\n4\n', '101\n99\n', '5\n10\n']
Demo Output:
['1\n', '2\n', '9\n']
Note:
In the first example the first friend should move by one to the right (then the meeting happens at point 4), or the second friend should move by one to the left (then the meeting happens at point 3). In both cases, the total tiredness becomes 1.
In the second example the first friend should move by one to the left, and the second friend should move by one to the right. Then they meet in the point 100, and the total tiredness becomes 1 + 1 = 2.
In the third example one of the optimal ways is the following. The first friend should move three times to the right, and the second friend — two times to the left. Thus the friends meet in the point 8, and the total tiredness becomes 1 + 2 + 3 + 1 + 2 = 9. | ```python
a=int(input())
b=int(input())
if abs(a-b)==1:
print("1")
else:
n=abs(a-b)
h=n//2
s=(h*(h+1))//2
if n%2==0:
print(2*s)
else:
r=(2*s)+(h+1)
print(r)
``` | 3 | |
699 | A | Launch of Collider | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | There will be a launch of a new, powerful and unusual collider very soon, which located along a straight line. *n* particles will be launched inside it. All of them are located in a straight line and there can not be two or more particles located in the same point. The coordinates of the particles coincide with the distance in meters from the center of the collider, *x**i* is the coordinate of the *i*-th particle and its position in the collider at the same time. All coordinates of particle positions are even integers.
You know the direction of each particle movement — it will move to the right or to the left after the collider's launch start. All particles begin to move simultaneously at the time of the collider's launch start. Each particle will move straight to the left or straight to the right with the constant speed of 1 meter per microsecond. The collider is big enough so particles can not leave it in the foreseeable time.
Write the program which finds the moment of the first collision of any two particles of the collider. In other words, find the number of microseconds before the first moment when any two particles are at the same point. | The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of particles.
The second line contains *n* symbols "L" and "R". If the *i*-th symbol equals "L", then the *i*-th particle will move to the left, otherwise the *i*-th symbol equals "R" and the *i*-th particle will move to the right.
The third line contains the sequence of pairwise distinct even integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=109) — the coordinates of particles in the order from the left to the right. It is guaranteed that the coordinates of particles are given in the increasing order. | In the first line print the only integer — the first moment (in microseconds) when two particles are at the same point and there will be an explosion.
Print the only integer -1, if the collision of particles doesn't happen. | [
"4\nRLRL\n2 4 6 10\n",
"3\nLLR\n40 50 60\n"
] | [
"1\n",
"-1\n"
] | In the first sample case the first explosion will happen in 1 microsecond because the particles number 1 and 2 will simultaneously be at the same point with the coordinate 3.
In the second sample case there will be no explosion because there are no particles which will simultaneously be at the same point. | 500 | [
{
"input": "4\nRLRL\n2 4 6 10",
"output": "1"
},
{
"input": "3\nLLR\n40 50 60",
"output": "-1"
},
{
"input": "4\nRLLR\n46 230 264 470",
"output": "92"
},
{
"input": "6\nLLRLLL\n446 492 650 844 930 970",
"output": "97"
},
{
"input": "8\nRRLLLLLL\n338 478 512 574 59... | 1,505,710,610 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 46 | 4,608,000 | n = int(input())
a = input()
arr = list(map(int,(input().split())))
arrL=[];arrR = []
for i in range(0,n):
if a[i]=='L':arrL.append(arr[i])
elif a[i]=='R':arrR.append(arr[i])
if arrL==[] or arrR ==[] :
print(-1);quit()
mL = max(arrL)
mR = min(arrR)
if mL > mR :
arrL =sorted(arrL)
arrR=sorted(arrR)
print((arrL[0]-arrR[len(arrR)-1])//2)
else : print(-1) | Title: Launch of Collider
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There will be a launch of a new, powerful and unusual collider very soon, which located along a straight line. *n* particles will be launched inside it. All of them are located in a straight line and there can not be two or more particles located in the same point. The coordinates of the particles coincide with the distance in meters from the center of the collider, *x**i* is the coordinate of the *i*-th particle and its position in the collider at the same time. All coordinates of particle positions are even integers.
You know the direction of each particle movement — it will move to the right or to the left after the collider's launch start. All particles begin to move simultaneously at the time of the collider's launch start. Each particle will move straight to the left or straight to the right with the constant speed of 1 meter per microsecond. The collider is big enough so particles can not leave it in the foreseeable time.
Write the program which finds the moment of the first collision of any two particles of the collider. In other words, find the number of microseconds before the first moment when any two particles are at the same point.
Input Specification:
The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of particles.
The second line contains *n* symbols "L" and "R". If the *i*-th symbol equals "L", then the *i*-th particle will move to the left, otherwise the *i*-th symbol equals "R" and the *i*-th particle will move to the right.
The third line contains the sequence of pairwise distinct even integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=109) — the coordinates of particles in the order from the left to the right. It is guaranteed that the coordinates of particles are given in the increasing order.
Output Specification:
In the first line print the only integer — the first moment (in microseconds) when two particles are at the same point and there will be an explosion.
Print the only integer -1, if the collision of particles doesn't happen.
Demo Input:
['4\nRLRL\n2 4 6 10\n', '3\nLLR\n40 50 60\n']
Demo Output:
['1\n', '-1\n']
Note:
In the first sample case the first explosion will happen in 1 microsecond because the particles number 1 and 2 will simultaneously be at the same point with the coordinate 3.
In the second sample case there will be no explosion because there are no particles which will simultaneously be at the same point. | ```python
n = int(input())
a = input()
arr = list(map(int,(input().split())))
arrL=[];arrR = []
for i in range(0,n):
if a[i]=='L':arrL.append(arr[i])
elif a[i]=='R':arrR.append(arr[i])
if arrL==[] or arrR ==[] :
print(-1);quit()
mL = max(arrL)
mR = min(arrR)
if mL > mR :
arrL =sorted(arrL)
arrR=sorted(arrR)
print((arrL[0]-arrR[len(arrR)-1])//2)
else : print(-1)
``` | 0 | |
893 | C | Rumor | PROGRAMMING | 1,300 | [
"dfs and similar",
"graphs",
"greedy"
] | null | null | Vova promised himself that he would never play computer games... But recently Firestorm — a well-known game developing company — published their newest game, World of Farcraft, and it became really popular. Of course, Vova started playing it.
Now he tries to solve a quest. The task is to come to a settlement named Overcity and spread a rumor in it.
Vova knows that there are *n* characters in Overcity. Some characters are friends to each other, and they share information they got. Also Vova knows that he can bribe each character so he or she starts spreading the rumor; *i*-th character wants *c**i* gold in exchange for spreading the rumor. When a character hears the rumor, he tells it to all his friends, and they start spreading the rumor to their friends (for free), and so on.
The quest is finished when all *n* characters know the rumor. What is the minimum amount of gold Vova needs to spend in order to finish the quest?
Take a look at the notes if you think you haven't understood the problem completely. | The first line contains two integer numbers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105) — the number of characters in Overcity and the number of pairs of friends.
The second line contains *n* integer numbers *c**i* (0<=≤<=*c**i*<=≤<=109) — the amount of gold *i*-th character asks to start spreading the rumor.
Then *m* lines follow, each containing a pair of numbers (*x**i*,<=*y**i*) which represent that characters *x**i* and *y**i* are friends (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*). It is guaranteed that each pair is listed at most once. | Print one number — the minimum amount of gold Vova has to spend in order to finish the quest. | [
"5 2\n2 5 3 4 8\n1 4\n4 5\n",
"10 0\n1 2 3 4 5 6 7 8 9 10\n",
"10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10\n"
] | [
"10\n",
"55\n",
"15\n"
] | In the first example the best decision is to bribe the first character (he will spread the rumor to fourth character, and the fourth one will spread it to fifth). Also Vova has to bribe the second and the third characters, so they know the rumor.
In the second example Vova has to bribe everyone.
In the third example the optimal decision is to bribe the first, the third, the fifth, the seventh and the ninth characters. | 0 | [
{
"input": "5 2\n2 5 3 4 8\n1 4\n4 5",
"output": "10"
},
{
"input": "10 0\n1 2 3 4 5 6 7 8 9 10",
"output": "55"
},
{
"input": "10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10",
"output": "15"
},
{
"input": "1 0\n0",
"output": "0"
},
{
"input": "1 0\n10000000... | 1,679,779,817 | 2,147,483,647 | PyPy 3 | OK | TESTS | 46 | 1,107 | 17,408,000 | from collections import deque
n, m=map(int, input().split())
gold=[0]+list(map(int, input().split()))
adjacency_list={i: [] for i in range(1, n+1)}
for i in range(m):
l, r=map(int, input().split())
adjacency_list[l].append(r)
adjacency_list[r].append(l)
marks=[0 for i in range(n+1)]
connected_components={}
def bfs(curr, marker):
marks[curr]=marker
connected_components[marker]=min(connected_components[marker], gold[curr])
q=deque([curr])
while len(q)!=0:
front=q.pop()
for i in adjacency_list[front]:
if marks[i]==0:
marks[i]=marker
connected_components[marker]=min(connected_components[marker], gold[i])
q.append(i)
marker=1
for i in range(1, n+1):
if marks[i]==0:
connected_components[marker]=float('inf')
bfs(i, marker)
marker+=1
ans=0
for i in connected_components:
ans+=connected_components[i]
print(ans) | Title: Rumor
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vova promised himself that he would never play computer games... But recently Firestorm — a well-known game developing company — published their newest game, World of Farcraft, and it became really popular. Of course, Vova started playing it.
Now he tries to solve a quest. The task is to come to a settlement named Overcity and spread a rumor in it.
Vova knows that there are *n* characters in Overcity. Some characters are friends to each other, and they share information they got. Also Vova knows that he can bribe each character so he or she starts spreading the rumor; *i*-th character wants *c**i* gold in exchange for spreading the rumor. When a character hears the rumor, he tells it to all his friends, and they start spreading the rumor to their friends (for free), and so on.
The quest is finished when all *n* characters know the rumor. What is the minimum amount of gold Vova needs to spend in order to finish the quest?
Take a look at the notes if you think you haven't understood the problem completely.
Input Specification:
The first line contains two integer numbers *n* and *m* (1<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105) — the number of characters in Overcity and the number of pairs of friends.
The second line contains *n* integer numbers *c**i* (0<=≤<=*c**i*<=≤<=109) — the amount of gold *i*-th character asks to start spreading the rumor.
Then *m* lines follow, each containing a pair of numbers (*x**i*,<=*y**i*) which represent that characters *x**i* and *y**i* are friends (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*). It is guaranteed that each pair is listed at most once.
Output Specification:
Print one number — the minimum amount of gold Vova has to spend in order to finish the quest.
Demo Input:
['5 2\n2 5 3 4 8\n1 4\n4 5\n', '10 0\n1 2 3 4 5 6 7 8 9 10\n', '10 5\n1 6 2 7 3 8 4 9 5 10\n1 2\n3 4\n5 6\n7 8\n9 10\n']
Demo Output:
['10\n', '55\n', '15\n']
Note:
In the first example the best decision is to bribe the first character (he will spread the rumor to fourth character, and the fourth one will spread it to fifth). Also Vova has to bribe the second and the third characters, so they know the rumor.
In the second example Vova has to bribe everyone.
In the third example the optimal decision is to bribe the first, the third, the fifth, the seventh and the ninth characters. | ```python
from collections import deque
n, m=map(int, input().split())
gold=[0]+list(map(int, input().split()))
adjacency_list={i: [] for i in range(1, n+1)}
for i in range(m):
l, r=map(int, input().split())
adjacency_list[l].append(r)
adjacency_list[r].append(l)
marks=[0 for i in range(n+1)]
connected_components={}
def bfs(curr, marker):
marks[curr]=marker
connected_components[marker]=min(connected_components[marker], gold[curr])
q=deque([curr])
while len(q)!=0:
front=q.pop()
for i in adjacency_list[front]:
if marks[i]==0:
marks[i]=marker
connected_components[marker]=min(connected_components[marker], gold[i])
q.append(i)
marker=1
for i in range(1, n+1):
if marks[i]==0:
connected_components[marker]=float('inf')
bfs(i, marker)
marker+=1
ans=0
for i in connected_components:
ans+=connected_components[i]
print(ans)
``` | 3 | |
158 | A | Next Round | PROGRAMMING | 800 | [
"*special",
"implementation"
] | null | null | "Contestant who earns a score equal to or greater than the *k*-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules.
A total of *n* participants took part in the contest (*n*<=≥<=*k*), and you already know their scores. Calculate how many participants will advance to the next round. | The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50) separated by a single space.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the score earned by the participant who got the *i*-th place. The given sequence is non-increasing (that is, for all *i* from 1 to *n*<=-<=1 the following condition is fulfilled: *a**i*<=≥<=*a**i*<=+<=1). | Output the number of participants who advance to the next round. | [
"8 5\n10 9 8 7 7 7 5 5\n",
"4 2\n0 0 0 0\n"
] | [
"6\n",
"0\n"
] | In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers.
In the second example nobody got a positive score. | 500 | [
{
"input": "8 5\n10 9 8 7 7 7 5 5",
"output": "6"
},
{
"input": "4 2\n0 0 0 0",
"output": "0"
},
{
"input": "5 1\n1 1 1 1 1",
"output": "5"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "5"
},
{
"input": "1 1\n10",
"output": "1"
},
{
"input": "17 14\n16 15... | 1,696,682,327 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 62 | 0 | n,k=map(int,input().split())
a=list(map(int,input().split()))
if a[k+1]>0:
print(k+1)
else:
print(0) | Title: Next Round
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Contestant who earns a score equal to or greater than the *k*-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules.
A total of *n* participants took part in the contest (*n*<=≥<=*k*), and you already know their scores. Calculate how many participants will advance to the next round.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50) separated by a single space.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the score earned by the participant who got the *i*-th place. The given sequence is non-increasing (that is, for all *i* from 1 to *n*<=-<=1 the following condition is fulfilled: *a**i*<=≥<=*a**i*<=+<=1).
Output Specification:
Output the number of participants who advance to the next round.
Demo Input:
['8 5\n10 9 8 7 7 7 5 5\n', '4 2\n0 0 0 0\n']
Demo Output:
['6\n', '0\n']
Note:
In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers.
In the second example nobody got a positive score. | ```python
n,k=map(int,input().split())
a=list(map(int,input().split()))
if a[k+1]>0:
print(k+1)
else:
print(0)
``` | 0 | |
1,008 | A | Romaji | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.
In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.
Help Vitya find out if a word $s$ is Berlanese. | The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters. | Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".
You can print each letter in any case (upper or lower). | [
"sumimasen\n",
"ninja\n",
"codeforces\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.
In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese. | 500 | [
{
"input": "sumimasen",
"output": "YES"
},
{
"input": "ninja",
"output": "YES"
},
{
"input": "codeforces",
"output": "NO"
},
{
"input": "auuaoonntanonnuewannnnpuuinniwoonennyolonnnvienonpoujinndinunnenannmuveoiuuhikucuziuhunnnmunzancenen",
"output": "YES"
},
{
"in... | 1,637,147,268 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 46 | 0 | a=input()
b="aeioun"
c="bcdfghjklmpqrstvwxyz"
d=len(a)
f=0
for i in range(d):
if i == d-1:
break
else:
if a[i] in c:
if a[i+1] in b:
f="YES"
pass
else:
f="NO"
break
if f=="YES":
print(f)
else:
print("NO")
| Title: Romaji
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.
In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.
Help Vitya find out if a word $s$ is Berlanese.
Input Specification:
The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters.
Output Specification:
Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".
You can print each letter in any case (upper or lower).
Demo Input:
['sumimasen\n', 'ninja\n', 'codeforces\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.
In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese. | ```python
a=input()
b="aeioun"
c="bcdfghjklmpqrstvwxyz"
d=len(a)
f=0
for i in range(d):
if i == d-1:
break
else:
if a[i] in c:
if a[i+1] in b:
f="YES"
pass
else:
f="NO"
break
if f=="YES":
print(f)
else:
print("NO")
``` | 0 | |
844 | B | Rectangles | PROGRAMMING | 1,300 | [
"combinatorics",
"math"
] | null | null | You are given *n*<=×<=*m* table. Each cell of the table is colored white or black. Find the number of non-empty sets of cells such that:
1. All cells in a set have the same color. 1. Every two cells in a set share row or column. | The first line of input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) — the number of rows and the number of columns correspondingly.
The next *n* lines of input contain descriptions of rows. There are *m* integers, separated by spaces, in each line. The number equals 0 if the corresponding cell is colored white and equals 1 if the corresponding cell is colored black. | Output single integer — the number of non-empty sets from the problem description. | [
"1 1\n0\n",
"2 3\n1 0 1\n0 1 0\n"
] | [
"1\n",
"8\n"
] | In the second example, there are six one-element sets. Additionally, there are two two-element sets, the first one consists of the first and the third cells of the first row, the second one consists of the first and the third cells of the second row. To sum up, there are 8 sets. | 1,000 | [
{
"input": "1 1\n0",
"output": "1"
},
{
"input": "2 3\n1 0 1\n0 1 0",
"output": "8"
},
{
"input": "2 2\n1 1\n1 1",
"output": "8"
},
{
"input": "1 10\n0 0 0 0 0 0 0 0 0 0",
"output": "1023"
},
{
"input": "11 1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1",
"output": "2047"... | 1,503,594,925 | 2,425 | Python 3 | WRONG_ANSWER | PRETESTS | 3 | 62 | 0 | def a_sum(n):
return (1+n)*n//2
n, m = map(int, input().split())
board = []
for _ in range(n):
board.append(''.join(input().split()))
res = []
for i in range(n):
a, b = board[i].count('0'), board[i].count('1')
if a > 1:
res.append(a)
if b > 1:
res.append(b)
board = list(zip(*board))
for i in range(n):
a, b = board[i].count('0'), board[i].count('1')
if a > 1:
res.append(a)
if b > 1:
res.append(b)
print(sum([a_sum(x-1) for x in res]) + n*m)
| Title: Rectangles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given *n*<=×<=*m* table. Each cell of the table is colored white or black. Find the number of non-empty sets of cells such that:
1. All cells in a set have the same color. 1. Every two cells in a set share row or column.
Input Specification:
The first line of input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) — the number of rows and the number of columns correspondingly.
The next *n* lines of input contain descriptions of rows. There are *m* integers, separated by spaces, in each line. The number equals 0 if the corresponding cell is colored white and equals 1 if the corresponding cell is colored black.
Output Specification:
Output single integer — the number of non-empty sets from the problem description.
Demo Input:
['1 1\n0\n', '2 3\n1 0 1\n0 1 0\n']
Demo Output:
['1\n', '8\n']
Note:
In the second example, there are six one-element sets. Additionally, there are two two-element sets, the first one consists of the first and the third cells of the first row, the second one consists of the first and the third cells of the second row. To sum up, there are 8 sets. | ```python
def a_sum(n):
return (1+n)*n//2
n, m = map(int, input().split())
board = []
for _ in range(n):
board.append(''.join(input().split()))
res = []
for i in range(n):
a, b = board[i].count('0'), board[i].count('1')
if a > 1:
res.append(a)
if b > 1:
res.append(b)
board = list(zip(*board))
for i in range(n):
a, b = board[i].count('0'), board[i].count('1')
if a > 1:
res.append(a)
if b > 1:
res.append(b)
print(sum([a_sum(x-1) for x in res]) + n*m)
``` | 0 | |
441 | A | Valera and Antique Items | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Valera is a collector. Once he wanted to expand his collection with exactly one antique item.
Valera knows *n* sellers of antiques, the *i*-th of them auctioned *k**i* items. Currently the auction price of the *j*-th object of the *i*-th seller is *s**ij*. Valera gets on well with each of the *n* sellers. He is perfectly sure that if he outbids the current price of one of the items in the auction (in other words, offers the seller the money that is strictly greater than the current price of the item at the auction), the seller of the object will immediately sign a contract with him.
Unfortunately, Valera has only *v* units of money. Help him to determine which of the *n* sellers he can make a deal with. | The first line contains two space-separated integers *n*,<=*v* (1<=≤<=*n*<=≤<=50; 104<=≤<=*v*<=≤<=106) — the number of sellers and the units of money the Valera has.
Then *n* lines follow. The *i*-th line first contains integer *k**i* (1<=≤<=*k**i*<=≤<=50) the number of items of the *i*-th seller. Then go *k**i* space-separated integers *s**i*1,<=*s**i*2,<=...,<=*s**ik**i* (104<=≤<=*s**ij*<=≤<=106) — the current prices of the items of the *i*-th seller. | In the first line, print integer *p* — the number of sellers with who Valera can make a deal.
In the second line print *p* space-separated integers *q*1,<=*q*2,<=...,<=*q**p* (1<=≤<=*q**i*<=≤<=*n*) — the numbers of the sellers with who Valera can make a deal. Print the numbers of the sellers in the increasing order. | [
"3 50000\n1 40000\n2 20000 60000\n3 10000 70000 190000\n",
"3 50000\n1 50000\n3 100000 120000 110000\n3 120000 110000 120000\n"
] | [
"3\n1 2 3\n",
"0\n\n"
] | In the first sample Valera can bargain with each of the sellers. He can outbid the following items: a 40000 item from the first seller, a 20000 item from the second seller, and a 10000 item from the third seller.
In the second sample Valera can not make a deal with any of the sellers, as the prices of all items in the auction too big for him. | 500 | [
{
"input": "3 50000\n1 40000\n2 20000 60000\n3 10000 70000 190000",
"output": "3\n1 2 3"
},
{
"input": "3 50000\n1 50000\n3 100000 120000 110000\n3 120000 110000 120000",
"output": "0"
},
{
"input": "2 100001\n1 895737\n1 541571",
"output": "0"
},
{
"input": "1 1000000\n1 100... | 1,677,771,506 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 61 | 1,433,600 | n , v = list(map(int , input().split()))
a = []
for el in range(n):
l = list(map(int , input().split()))
if(l[1] < v):
a.append(el+1)
print(len(a))
print(*a) | Title: Valera and Antique Items
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera is a collector. Once he wanted to expand his collection with exactly one antique item.
Valera knows *n* sellers of antiques, the *i*-th of them auctioned *k**i* items. Currently the auction price of the *j*-th object of the *i*-th seller is *s**ij*. Valera gets on well with each of the *n* sellers. He is perfectly sure that if he outbids the current price of one of the items in the auction (in other words, offers the seller the money that is strictly greater than the current price of the item at the auction), the seller of the object will immediately sign a contract with him.
Unfortunately, Valera has only *v* units of money. Help him to determine which of the *n* sellers he can make a deal with.
Input Specification:
The first line contains two space-separated integers *n*,<=*v* (1<=≤<=*n*<=≤<=50; 104<=≤<=*v*<=≤<=106) — the number of sellers and the units of money the Valera has.
Then *n* lines follow. The *i*-th line first contains integer *k**i* (1<=≤<=*k**i*<=≤<=50) the number of items of the *i*-th seller. Then go *k**i* space-separated integers *s**i*1,<=*s**i*2,<=...,<=*s**ik**i* (104<=≤<=*s**ij*<=≤<=106) — the current prices of the items of the *i*-th seller.
Output Specification:
In the first line, print integer *p* — the number of sellers with who Valera can make a deal.
In the second line print *p* space-separated integers *q*1,<=*q*2,<=...,<=*q**p* (1<=≤<=*q**i*<=≤<=*n*) — the numbers of the sellers with who Valera can make a deal. Print the numbers of the sellers in the increasing order.
Demo Input:
['3 50000\n1 40000\n2 20000 60000\n3 10000 70000 190000\n', '3 50000\n1 50000\n3 100000 120000 110000\n3 120000 110000 120000\n']
Demo Output:
['3\n1 2 3\n', '0\n\n']
Note:
In the first sample Valera can bargain with each of the sellers. He can outbid the following items: a 40000 item from the first seller, a 20000 item from the second seller, and a 10000 item from the third seller.
In the second sample Valera can not make a deal with any of the sellers, as the prices of all items in the auction too big for him. | ```python
n , v = list(map(int , input().split()))
a = []
for el in range(n):
l = list(map(int , input().split()))
if(l[1] < v):
a.append(el+1)
print(len(a))
print(*a)
``` | 0 | |
41 | A | Translation | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Translation | 2 | 256 | The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. | The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. | If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. | [
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | none | 500 | [
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk... | 1,591,177,603 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 218 | 0 | a = input()
b = input()
s1 = a[:len(a)//2]
s2 = a[len(a)//2:]
if(s2[::-1]+s1[::-1] == b):
print("YES")
else:
print("NO") | Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none | ```python
a = input()
b = input()
s1 = a[:len(a)//2]
s2 = a[len(a)//2:]
if(s2[::-1]+s1[::-1] == b):
print("YES")
else:
print("NO")
``` | 3.9455 |
0 | none | none | none | 0 | [
"none"
] | null | null | Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are *n* players (including Limak himself) and right now all of them have bids on the table. *i*-th of them has bid with size *a**i* dollars.
Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot? | First line of input contains an integer *n* (2<=≤<=*n*<=≤<=105), the number of players.
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the bids of players. | Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise. | [
"4\n75 150 75 50\n",
"3\n100 150 250\n"
] | [
"Yes\n",
"No\n"
] | In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid.
It can be shown that in the second sample test there is no way to make all bids equal. | 0 | [
{
"input": "4\n75 150 75 50",
"output": "Yes"
},
{
"input": "3\n100 150 250",
"output": "No"
},
{
"input": "7\n34 34 68 34 34 68 34",
"output": "Yes"
},
{
"input": "10\n72 96 12 18 81 20 6 2 54 1",
"output": "No"
},
{
"input": "20\n958692492 954966768 77387000 724... | 1,440,879,578 | 2,258 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 15 | 2,000 | 11,980,800 | def p(x):
res = []
i = 2
while x > 1:
while x % i == 0:
x //= i
if i != 2 and i != 3:
res.append(i)
i += 1
return res
read = lambda: map(int, input().split())
n = int(input())
a = [p(int(i)) for i in input().split()]
k = a[0]
f = sum(k == a[i] for i in range(1, n)) == n - 1
print('Yes' if f else 'No') | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is an old brown bear. He often plays poker with his friends. Today they went to a casino. There are *n* players (including Limak himself) and right now all of them have bids on the table. *i*-th of them has bid with size *a**i* dollars.
Each player can double his bid any number of times and triple his bid any number of times. The casino has a great jackpot for making all bids equal. Is it possible that Limak and his friends will win a jackpot?
Input Specification:
First line of input contains an integer *n* (2<=≤<=*n*<=≤<=105), the number of players.
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the bids of players.
Output Specification:
Print "Yes" (without the quotes) if players can make their bids become equal, or "No" otherwise.
Demo Input:
['4\n75 150 75 50\n', '3\n100 150 250\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first sample test first and third players should double their bids twice, second player should double his bid once and fourth player should both double and triple his bid.
It can be shown that in the second sample test there is no way to make all bids equal. | ```python
def p(x):
res = []
i = 2
while x > 1:
while x % i == 0:
x //= i
if i != 2 and i != 3:
res.append(i)
i += 1
return res
read = lambda: map(int, input().split())
n = int(input())
a = [p(int(i)) for i in input().split()]
k = a[0]
f = sum(k == a[i] for i in range(1, n)) == n - 1
print('Yes' if f else 'No')
``` | 0 | |
910 | A | The Way to Home | PROGRAMMING | 800 | [
"dfs and similar",
"dp",
"greedy",
"implementation"
] | null | null | A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*.
For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*.
Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1. | The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump.
The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one. | If the frog can not reach the home, print -1.
In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1. | [
"8 4\n10010101\n",
"4 2\n1001\n",
"8 4\n11100101\n",
"12 3\n101111100101\n"
] | [
"2\n",
"-1\n",
"3\n",
"4\n"
] | In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four).
In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two. | 500 | [
{
"input": "8 4\n10010101",
"output": "2"
},
{
"input": "4 2\n1001",
"output": "-1"
},
{
"input": "8 4\n11100101",
"output": "3"
},
{
"input": "12 3\n101111100101",
"output": "4"
},
{
"input": "5 4\n11011",
"output": "1"
},
{
"input": "5 4\n10001",
... | 1,614,950,671 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 307,200 |
def find_path(n, d, map_):
cur_i = 0
jump_count = 0
end_flag = False
while not end_flag:
max_jump = -1
for i in range(d, 0, -1):
if (cur_i + i) < n and fmap[cur_i + i] == '1':
max_jump = i
break
print(cur_i, max_jump)
if max_jump == -1:
end_flag = True
jump_count = -1
# print(-1)
elif cur_i + max_jump == (n - 1):
end_flag = True
cur_i += max_jump
jump_count += 1
else:
cur_i += max_jump
jump_count += 1
return jump_count if jump_count == -1 else jump_count + 1
if __name__ == '__main__':
n, d = [int(i) for i in input().split(' ')]
fmap = input()
# n = 8
# d = 4
# fmap = '10010101'
jumps = find_path(n, d, fmap)
print(jumps)
| Title: The Way to Home
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A frog lives on the axis *Ox* and needs to reach home which is in the point *n*. She starts from the point 1. The frog can jump to the right at a distance not more than *d*. So, after she jumped from the point *x* she can reach the point *x*<=+<=*a*, where *a* is an integer from 1 to *d*.
For each point from 1 to *n* is known if there is a lily flower in it. The frog can jump only in points with a lilies. Guaranteed that there are lilies in the points 1 and *n*.
Determine the minimal number of jumps that the frog needs to reach home which is in the point *n* from the point 1. Consider that initially the frog is in the point 1. If the frog can not reach home, print -1.
Input Specification:
The first line contains two integers *n* and *d* (2<=≤<=*n*<=≤<=100, 1<=≤<=*d*<=≤<=*n*<=-<=1) — the point, which the frog wants to reach, and the maximal length of the frog jump.
The second line contains a string *s* of length *n*, consisting of zeros and ones. If a character of the string *s* equals to zero, then in the corresponding point there is no lily flower. In the other case, in the corresponding point there is a lily flower. Guaranteed that the first and the last characters of the string *s* equal to one.
Output Specification:
If the frog can not reach the home, print -1.
In the other case, print the minimal number of jumps that the frog needs to reach the home which is in the point *n* from the point 1.
Demo Input:
['8 4\n10010101\n', '4 2\n1001\n', '8 4\n11100101\n', '12 3\n101111100101\n']
Demo Output:
['2\n', '-1\n', '3\n', '4\n']
Note:
In the first example the from can reach home in two jumps: the first jump from the point 1 to the point 4 (the length of the jump is three), and the second jump from the point 4 to the point 8 (the length of the jump is four).
In the second example the frog can not reach home, because to make it she need to jump on a distance three, but the maximum length of her jump equals to two. | ```python
def find_path(n, d, map_):
cur_i = 0
jump_count = 0
end_flag = False
while not end_flag:
max_jump = -1
for i in range(d, 0, -1):
if (cur_i + i) < n and fmap[cur_i + i] == '1':
max_jump = i
break
print(cur_i, max_jump)
if max_jump == -1:
end_flag = True
jump_count = -1
# print(-1)
elif cur_i + max_jump == (n - 1):
end_flag = True
cur_i += max_jump
jump_count += 1
else:
cur_i += max_jump
jump_count += 1
return jump_count if jump_count == -1 else jump_count + 1
if __name__ == '__main__':
n, d = [int(i) for i in input().split(' ')]
fmap = input()
# n = 8
# d = 4
# fmap = '10010101'
jumps = find_path(n, d, fmap)
print(jumps)
``` | 0 | |
18 | C | Stripe | PROGRAMMING | 1,200 | [
"data structures",
"implementation"
] | C. Stripe | 2 | 64 | Once Bob took a paper stripe of *n* squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into two pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem? | The first input line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of squares in the stripe. The second line contains *n* space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value. | Output the amount of ways to cut the stripe into two non-empty pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only. | [
"9\n1 5 -6 7 9 -16 0 -2 2\n",
"3\n1 1 1\n",
"2\n0 0\n"
] | [
"3\n",
"0\n",
"1\n"
] | none | 0 | [
{
"input": "9\n1 5 -6 7 9 -16 0 -2 2",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "0"
},
{
"input": "2\n0 0",
"output": "1"
},
{
"input": "4\n100 1 10 111",
"output": "1"
},
{
"input": "10\n0 4 -3 0 -2 2 -3 -3 2 5",
"output": "3"
},
{
"input": "... | 1,485,789,948 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 218 | 11,980,800 | # 18C
# θ(n) time
# θ(n) space
__author__ = 'artyom'
# SOLUTION
def main():
n = read()
a = read(3)
s = 0
t = sum(a)
count = 0
for i in range(n - 1):
s += a[i]
t -= a[i]
if s == t:
count += 1
return count
# HELPERS
def read(mode=1, size=None):
# 0: String
# 1: Integer
# 2: List of strings
# 3: List of integers
# 4: Matrix of integers
if mode == 0: return input().strip()
if mode == 1: return int(input().strip())
if mode == 2: return input().strip().split()
if mode == 3: return list(map(int, input().strip().split()))
a = []
for _ in range(size):
a.append(read(3))
return a
def write(s="\n"):
if s is None: s = ''
if isinstance(s, tuple) or isinstance(s, list): s = ' '.join(map(str, s))
s = str(s)
print(s, end="\n")
write(main()) | Title: Stripe
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Once Bob took a paper stripe of *n* squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into two pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem?
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of squares in the stripe. The second line contains *n* space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value.
Output Specification:
Output the amount of ways to cut the stripe into two non-empty pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only.
Demo Input:
['9\n1 5 -6 7 9 -16 0 -2 2\n', '3\n1 1 1\n', '2\n0 0\n']
Demo Output:
['3\n', '0\n', '1\n']
Note:
none | ```python
# 18C
# θ(n) time
# θ(n) space
__author__ = 'artyom'
# SOLUTION
def main():
n = read()
a = read(3)
s = 0
t = sum(a)
count = 0
for i in range(n - 1):
s += a[i]
t -= a[i]
if s == t:
count += 1
return count
# HELPERS
def read(mode=1, size=None):
# 0: String
# 1: Integer
# 2: List of strings
# 3: List of integers
# 4: Matrix of integers
if mode == 0: return input().strip()
if mode == 1: return int(input().strip())
if mode == 2: return input().strip().split()
if mode == 3: return list(map(int, input().strip().split()))
a = []
for _ in range(size):
a.append(read(3))
return a
def write(s="\n"):
if s is None: s = ''
if isinstance(s, tuple) or isinstance(s, list): s = ' '.join(map(str, s))
s = str(s)
print(s, end="\n")
write(main())
``` | 3.856236 |
495 | B | Modular Equations | PROGRAMMING | 1,600 | [
"math",
"number theory"
] | null | null | Last week, Hamed learned about a new type of equations in his math class called Modular Equations. Lets define *i* modulo *j* as the remainder of division of *i* by *j* and denote it by . A Modular Equation, as Hamed's teacher described, is an equation of the form in which *a* and *b* are two non-negative integers and *x* is a variable. We call a positive integer *x* for which a solution of our equation.
Hamed didn't pay much attention to the class since he was watching a movie. He only managed to understand the definitions of these equations.
Now he wants to write his math exercises but since he has no idea how to do that, he asked you for help. He has told you all he knows about Modular Equations and asked you to write a program which given two numbers *a* and *b* determines how many answers the Modular Equation has. | In the only line of the input two space-separated integers *a* and *b* (0<=≤<=*a*,<=*b*<=≤<=109) are given. | If there is an infinite number of answers to our equation, print "infinity" (without the quotes). Otherwise print the number of solutions of the Modular Equation . | [
"21 5\n",
"9435152 272\n",
"10 10\n"
] | [
"2\n",
"282\n",
"infinity\n"
] | In the first sample the answers of the Modular Equation are 8 and 16 since <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/6f5ff39ebd209bf990adaf91f4b82f9687097224.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 1,000 | [
{
"input": "21 5",
"output": "2"
},
{
"input": "9435152 272",
"output": "282"
},
{
"input": "10 10",
"output": "infinity"
},
{
"input": "0 1000000000",
"output": "0"
},
{
"input": "11 2",
"output": "2"
},
{
"input": "1 0",
"output": "1"
},
{
... | 1,585,016,684 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 1 | 1,000 | 0 | #ax =- b(mod n)
#a % x = b
def factorNumber(x, b):
factors = []
for i in range(1, x+1):
if x % i == 0 and i >= b: factors.append(i)
return factors
(a,b) = list(map(int,input().split()))
answers = 0
if a == b:
print("infinity")
elif a < b:
print(0)
else:
factors = factorNumber(a-b,b)
for i in factors:
if a % i == b:
answers += 1
print(answers)
| Title: Modular Equations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Last week, Hamed learned about a new type of equations in his math class called Modular Equations. Lets define *i* modulo *j* as the remainder of division of *i* by *j* and denote it by . A Modular Equation, as Hamed's teacher described, is an equation of the form in which *a* and *b* are two non-negative integers and *x* is a variable. We call a positive integer *x* for which a solution of our equation.
Hamed didn't pay much attention to the class since he was watching a movie. He only managed to understand the definitions of these equations.
Now he wants to write his math exercises but since he has no idea how to do that, he asked you for help. He has told you all he knows about Modular Equations and asked you to write a program which given two numbers *a* and *b* determines how many answers the Modular Equation has.
Input Specification:
In the only line of the input two space-separated integers *a* and *b* (0<=≤<=*a*,<=*b*<=≤<=109) are given.
Output Specification:
If there is an infinite number of answers to our equation, print "infinity" (without the quotes). Otherwise print the number of solutions of the Modular Equation .
Demo Input:
['21 5\n', '9435152 272\n', '10 10\n']
Demo Output:
['2\n', '282\n', 'infinity\n']
Note:
In the first sample the answers of the Modular Equation are 8 and 16 since <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/6f5ff39ebd209bf990adaf91f4b82f9687097224.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
#ax =- b(mod n)
#a % x = b
def factorNumber(x, b):
factors = []
for i in range(1, x+1):
if x % i == 0 and i >= b: factors.append(i)
return factors
(a,b) = list(map(int,input().split()))
answers = 0
if a == b:
print("infinity")
elif a < b:
print(0)
else:
factors = factorNumber(a-b,b)
for i in factors:
if a % i == b:
answers += 1
print(answers)
``` | 0 | |
604 | B | More Cowbell | PROGRAMMING | 1,400 | [
"binary search",
"greedy"
] | null | null | Kevin Sun wants to move his precious collection of *n* cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into *k* boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his *i*-th (1<=≤<=*i*<=≤<=*n*) cowbell is an integer *s**i*. In fact, he keeps his cowbells sorted by size, so *s**i*<=-<=1<=≤<=*s**i* for any *i*<=><=1. Also an expert packer, Kevin can fit one or two cowbells into a box of size *s* if and only if the sum of their sizes does not exceed *s*. Given this information, help Kevin determine the smallest *s* for which it is possible to put all of his cowbells into *k* boxes of size *s*. | The first line of the input contains two space-separated integers *n* and *k* (1<=≤<=*n*<=≤<=2·*k*<=≤<=100<=000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains *n* space-separated integers *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=*s*1<=≤<=*s*2<=≤<=...<=≤<=*s**n*<=≤<=1<=000<=000), the sizes of Kevin's cowbells. It is guaranteed that the sizes *s**i* are given in non-decreasing order. | Print a single integer, the smallest *s* for which it is possible for Kevin to put all of his cowbells into *k* boxes of size *s*. | [
"2 1\n2 5\n",
"4 3\n2 3 5 9\n",
"3 2\n3 5 7\n"
] | [
"7\n",
"9\n",
"8\n"
] | In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}. | 1,000 | [
{
"input": "2 1\n2 5",
"output": "7"
},
{
"input": "4 3\n2 3 5 9",
"output": "9"
},
{
"input": "3 2\n3 5 7",
"output": "8"
},
{
"input": "20 11\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "10 10\n3 15 31 61 63 63 68 94 98 100",
"outp... | 1,448,993,242 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | import math
def can_pack(m, b):
global n, k
cnt = 0
if b[-1] > m:
return False
p1 = 0
p2 = 0
while p1 <= p2:
if b[p2] + b[p1] <= m:
p1 += 1
p2 -= 1
else:
p2 -= 1
cnt += 1
return cnt <= k
n, k = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]
l = 1
r = 10 ** 9
while l != r - 1:
m = (l + r) // 2
if not can_pack(m, b):
l = m
else:
r = m
print(r) | Title: More Cowbell
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kevin Sun wants to move his precious collection of *n* cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into *k* boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection.
Kevin is a meticulous cowbell collector and knows that the size of his *i*-th (1<=≤<=*i*<=≤<=*n*) cowbell is an integer *s**i*. In fact, he keeps his cowbells sorted by size, so *s**i*<=-<=1<=≤<=*s**i* for any *i*<=><=1. Also an expert packer, Kevin can fit one or two cowbells into a box of size *s* if and only if the sum of their sizes does not exceed *s*. Given this information, help Kevin determine the smallest *s* for which it is possible to put all of his cowbells into *k* boxes of size *s*.
Input Specification:
The first line of the input contains two space-separated integers *n* and *k* (1<=≤<=*n*<=≤<=2·*k*<=≤<=100<=000), denoting the number of cowbells and the number of boxes, respectively.
The next line contains *n* space-separated integers *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=*s*1<=≤<=*s*2<=≤<=...<=≤<=*s**n*<=≤<=1<=000<=000), the sizes of Kevin's cowbells. It is guaranteed that the sizes *s**i* are given in non-decreasing order.
Output Specification:
Print a single integer, the smallest *s* for which it is possible for Kevin to put all of his cowbells into *k* boxes of size *s*.
Demo Input:
['2 1\n2 5\n', '4 3\n2 3 5 9\n', '3 2\n3 5 7\n']
Demo Output:
['7\n', '9\n', '8\n']
Note:
In the first sample, Kevin must pack his two cowbells into the same box.
In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.
In the third sample, the optimal solution is {3, 5} and {7}. | ```python
import math
def can_pack(m, b):
global n, k
cnt = 0
if b[-1] > m:
return False
p1 = 0
p2 = 0
while p1 <= p2:
if b[p2] + b[p1] <= m:
p1 += 1
p2 -= 1
else:
p2 -= 1
cnt += 1
return cnt <= k
n, k = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]
l = 1
r = 10 ** 9
while l != r - 1:
m = (l + r) // 2
if not can_pack(m, b):
l = m
else:
r = m
print(r)
``` | 0 | |
499 | A | Watching a movie | PROGRAMMING | 1,000 | [
"greedy",
"implementation"
] | null | null | You have decided to watch the best moments of some movie. There are two buttons on your player:
1. Watch the current minute of the movie. By pressing this button, you watch the current minute of the movie and the player automatically proceeds to the next minute of the movie. 1. Skip exactly *x* minutes of the movie (*x* is some fixed positive integer). If the player is now at the *t*-th minute of the movie, then as a result of pressing this button, it proceeds to the minute (*t*<=+<=*x*).
Initially the movie is turned on in the player on the first minute, and you want to watch exactly *n* best moments of the movie, the *i*-th best moment starts at the *l**i*-th minute and ends at the *r**i*-th minute (more formally, the *i*-th best moment consists of minutes: *l**i*,<=*l**i*<=+<=1,<=...,<=*r**i*).
Determine, what is the minimum number of minutes of the movie you have to watch if you want to watch all the best moments? | The first line contains two space-separated integers *n*, *x* (1<=≤<=*n*<=≤<=50, 1<=≤<=*x*<=≤<=105) — the number of the best moments of the movie and the value of *x* for the second button.
The following *n* lines contain the descriptions of the best moments of the movie, the *i*-th line of the description contains two integers separated by a space *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=105).
It is guaranteed that for all integers *i* from 2 to *n* the following condition holds: *r**i*<=-<=1<=<<=*l**i*. | Output a single number — the answer to the problem. | [
"2 3\n5 6\n10 12\n",
"1 1\n1 100000\n"
] | [
"6\n",
"100000\n"
] | In the first sample, the player was initially standing on the first minute. As the minutes from the 1-st to the 4-th one don't contain interesting moments, we press the second button. Now we can not press the second button and skip 3 more minutes, because some of them contain interesting moments. Therefore, we watch the movie from the 4-th to the 6-th minute, after that the current time is 7. Similarly, we again skip 3 minutes and then watch from the 10-th to the 12-th minute of the movie. In total, we watch 6 minutes of the movie.
In the second sample, the movie is very interesting, so you'll have to watch all 100000 minutes of the movie. | 500 | [
{
"input": "2 3\n5 6\n10 12",
"output": "6"
},
{
"input": "1 1\n1 100000",
"output": "100000"
},
{
"input": "10 1\n2156 3497\n4784 7775\n14575 31932\n33447 35902\n36426 47202\n48772 60522\n63982 68417\n78537 79445\n90081 90629\n94325 95728",
"output": "53974"
},
{
"input": "1... | 1,620,373,764 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 28 | 62 | 6,963,200 | def compute_curr(curr,x, l,r):
if x == l:
curr = l
elif x < l:
curr += ((l-curr)//x)*x
return curr
def solve(arr,n,x):
curr = 1
count = 0
for i,v in enumerate(arr):
curr = compute_curr(curr,x,*v)
count += v[1]-curr+1
curr = v[1]+1
return count
def main():
n,x = list(map(int, list(input().split(" "))))
arr = []
for _ in range(n):
arr.append(list(map(int, list(input().split(" ")))))
print(solve(arr,n,x))
main() | Title: Watching a movie
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have decided to watch the best moments of some movie. There are two buttons on your player:
1. Watch the current minute of the movie. By pressing this button, you watch the current minute of the movie and the player automatically proceeds to the next minute of the movie. 1. Skip exactly *x* minutes of the movie (*x* is some fixed positive integer). If the player is now at the *t*-th minute of the movie, then as a result of pressing this button, it proceeds to the minute (*t*<=+<=*x*).
Initially the movie is turned on in the player on the first minute, and you want to watch exactly *n* best moments of the movie, the *i*-th best moment starts at the *l**i*-th minute and ends at the *r**i*-th minute (more formally, the *i*-th best moment consists of minutes: *l**i*,<=*l**i*<=+<=1,<=...,<=*r**i*).
Determine, what is the minimum number of minutes of the movie you have to watch if you want to watch all the best moments?
Input Specification:
The first line contains two space-separated integers *n*, *x* (1<=≤<=*n*<=≤<=50, 1<=≤<=*x*<=≤<=105) — the number of the best moments of the movie and the value of *x* for the second button.
The following *n* lines contain the descriptions of the best moments of the movie, the *i*-th line of the description contains two integers separated by a space *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=105).
It is guaranteed that for all integers *i* from 2 to *n* the following condition holds: *r**i*<=-<=1<=<<=*l**i*.
Output Specification:
Output a single number — the answer to the problem.
Demo Input:
['2 3\n5 6\n10 12\n', '1 1\n1 100000\n']
Demo Output:
['6\n', '100000\n']
Note:
In the first sample, the player was initially standing on the first minute. As the minutes from the 1-st to the 4-th one don't contain interesting moments, we press the second button. Now we can not press the second button and skip 3 more minutes, because some of them contain interesting moments. Therefore, we watch the movie from the 4-th to the 6-th minute, after that the current time is 7. Similarly, we again skip 3 minutes and then watch from the 10-th to the 12-th minute of the movie. In total, we watch 6 minutes of the movie.
In the second sample, the movie is very interesting, so you'll have to watch all 100000 minutes of the movie. | ```python
def compute_curr(curr,x, l,r):
if x == l:
curr = l
elif x < l:
curr += ((l-curr)//x)*x
return curr
def solve(arr,n,x):
curr = 1
count = 0
for i,v in enumerate(arr):
curr = compute_curr(curr,x,*v)
count += v[1]-curr+1
curr = v[1]+1
return count
def main():
n,x = list(map(int, list(input().split(" "))))
arr = []
for _ in range(n):
arr.append(list(map(int, list(input().split(" ")))))
print(solve(arr,n,x))
main()
``` | 0 | |
58 | B | Coins | PROGRAMMING | 1,300 | [
"greedy"
] | B. Coins | 2 | 256 | In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly *n* Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations. | The first and only line contains an integer *n* (1<=≤<=*n*<=≤<=106) which represents the denomination of the most expensive coin. | Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination *n* of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them. | [
"10\n",
"4\n",
"3\n"
] | [
"10 5 1\n",
"4 2 1\n",
"3 1\n"
] | none | 1,000 | [
{
"input": "10",
"output": "10 5 1"
},
{
"input": "4",
"output": "4 2 1"
},
{
"input": "3",
"output": "3 1"
},
{
"input": "2",
"output": "2 1"
},
{
"input": "5",
"output": "5 1"
},
{
"input": "6",
"output": "6 3 1"
},
{
"input": "7",
"o... | 1,610,359,795 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 9 | 154 | 0 | def factors(n):
c=2
if n<=1000:
c=1
i=1
ans=[n]
while i**c<=n:
if n%i==0 and i!=n:
ans.append(i)
i+=1
return ans
def coins(n):
if n==1:
return 1
a=factors(n)
a.remove(n)
maxi=max(a,key=lambda s:len(factors(s)))
print(*([n]+sorted(factors(maxi),reverse=True)))
return ""
print(coins(int(input()))) | Title: Coins
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly *n* Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.
Input Specification:
The first and only line contains an integer *n* (1<=≤<=*n*<=≤<=106) which represents the denomination of the most expensive coin.
Output Specification:
Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination *n* of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.
Demo Input:
['10\n', '4\n', '3\n']
Demo Output:
['10 5 1\n', '4 2 1\n', '3 1\n']
Note:
none | ```python
def factors(n):
c=2
if n<=1000:
c=1
i=1
ans=[n]
while i**c<=n:
if n%i==0 and i!=n:
ans.append(i)
i+=1
return ans
def coins(n):
if n==1:
return 1
a=factors(n)
a.remove(n)
maxi=max(a,key=lambda s:len(factors(s)))
print(*([n]+sorted(factors(maxi),reverse=True)))
return ""
print(coins(int(input())))
``` | 0 |
433 | A | Kitahara Haruki's Gift | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | Kitahara Haruki has bought *n* apples for Touma Kazusa and Ogiso Setsuna. Now he wants to divide all the apples between the friends.
Each apple weights 100 grams or 200 grams. Of course Kitahara Haruki doesn't want to offend any of his friend. Therefore the total weight of the apples given to Touma Kazusa must be equal to the total weight of the apples given to Ogiso Setsuna.
But unfortunately Kitahara Haruki doesn't have a knife right now, so he cannot split any apple into some parts. Please, tell him: is it possible to divide all the apples in a fair way between his friends? | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of apples. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (*w**i*<==<=100 or *w**i*<==<=200), where *w**i* is the weight of the *i*-th apple. | In a single line print "YES" (without the quotes) if it is possible to divide all the apples between his friends. Otherwise print "NO" (without the quotes). | [
"3\n100 200 100\n",
"4\n100 100 100 200\n"
] | [
"YES\n",
"NO\n"
] | In the first test sample Kitahara Haruki can give the first and the last apple to Ogiso Setsuna and the middle apple to Touma Kazusa. | 500 | [
{
"input": "3\n100 200 100",
"output": "YES"
},
{
"input": "4\n100 100 100 200",
"output": "NO"
},
{
"input": "1\n100",
"output": "NO"
},
{
"input": "1\n200",
"output": "NO"
},
{
"input": "2\n100 100",
"output": "YES"
},
{
"input": "2\n200 200",
"o... | 1,680,019,533 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 3 | 46 | 0 | n=int(input())
a=list(map(int,input().split()))
if (sum(a)//100)%2==0:
print("YES")
else:
print("NO") | Title: Kitahara Haruki's Gift
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kitahara Haruki has bought *n* apples for Touma Kazusa and Ogiso Setsuna. Now he wants to divide all the apples between the friends.
Each apple weights 100 grams or 200 grams. Of course Kitahara Haruki doesn't want to offend any of his friend. Therefore the total weight of the apples given to Touma Kazusa must be equal to the total weight of the apples given to Ogiso Setsuna.
But unfortunately Kitahara Haruki doesn't have a knife right now, so he cannot split any apple into some parts. Please, tell him: is it possible to divide all the apples in a fair way between his friends?
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of apples. The second line contains *n* integers *w*1,<=*w*2,<=...,<=*w**n* (*w**i*<==<=100 or *w**i*<==<=200), where *w**i* is the weight of the *i*-th apple.
Output Specification:
In a single line print "YES" (without the quotes) if it is possible to divide all the apples between his friends. Otherwise print "NO" (without the quotes).
Demo Input:
['3\n100 200 100\n', '4\n100 100 100 200\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first test sample Kitahara Haruki can give the first and the last apple to Ogiso Setsuna and the middle apple to Touma Kazusa. | ```python
n=int(input())
a=list(map(int,input().split()))
if (sum(a)//100)%2==0:
print("YES")
else:
print("NO")
``` | 0 | |
545 | C | Woodcutters | PROGRAMMING | 1,500 | [
"dp",
"greedy"
] | null | null | Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are *n* trees located along the road at points with coordinates *x*1,<=*x*2,<=...,<=*x**n*. Each tree has its height *h**i*. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [*x**i*<=-<=*h**i*,<=*x**i*] or [*x**i*;*x**i*<=+<=*h**i*]. The tree that is not cut down occupies a single point with coordinate *x**i*. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of trees.
Next *n* lines contain pairs of integers *x**i*,<=*h**i* (1<=≤<=*x**i*,<=*h**i*<=≤<=109) — the coordinate and the height of the *і*-th tree.
The pairs are given in the order of ascending *x**i*. No two trees are located at the point with the same coordinate. | Print a single number — the maximum number of trees that you can cut down by the given rules. | [
"5\n1 2\n2 1\n5 10\n10 9\n19 1\n",
"5\n1 2\n2 1\n5 10\n10 9\n20 1\n"
] | [
"3\n",
"4\n"
] | In the first sample you can fell the trees like that:
- fell the 1-st tree to the left — now it occupies segment [ - 1;1] - fell the 2-nd tree to the right — now it occupies segment [2;3] - leave the 3-rd tree — it occupies point 5 - leave the 4-th tree — it occupies point 10 - fell the 5-th tree to the right — now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19]. | 1,750 | [
{
"input": "5\n1 2\n2 1\n5 10\n10 9\n19 1",
"output": "3"
},
{
"input": "5\n1 2\n2 1\n5 10\n10 9\n20 1",
"output": "4"
},
{
"input": "4\n10 4\n15 1\n19 3\n20 1",
"output": "4"
},
{
"input": "35\n1 7\n3 11\n6 12\n7 6\n8 5\n9 11\n15 3\n16 10\n22 2\n23 3\n25 7\n27 3\n34 5\n35 10... | 1,666,771,479 | 2,147,483,647 | Python 3 | OK | TESTS | 67 | 343 | 10,035,200 | n = int(input())
trees = []
for i in range(n):
trees.append(list(map(int,input().split())))
# trees[i][0]: position
# trees[i][1]: height
r_edge = -trees[0][1]
max_num = 1
for i in range(n-1):
# try to cut i-th tree
# if can be cut leftward, then rightward, then cannot be cut
if trees[i][0]-trees[i][1] > r_edge:
max_num += 1
r_edge = trees[i][0]
elif trees[i][0]+trees[i][1] < trees[i+1][0]:
max_num += 1
r_edge = trees[i][0]+trees[i][1]
else:
r_edge = trees[i][0]
print(max_num) | Title: Woodcutters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are *n* trees located along the road at points with coordinates *x*1,<=*x*2,<=...,<=*x**n*. Each tree has its height *h**i*. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [*x**i*<=-<=*h**i*,<=*x**i*] or [*x**i*;*x**i*<=+<=*h**i*]. The tree that is not cut down occupies a single point with coordinate *x**i*. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of trees.
Next *n* lines contain pairs of integers *x**i*,<=*h**i* (1<=≤<=*x**i*,<=*h**i*<=≤<=109) — the coordinate and the height of the *і*-th tree.
The pairs are given in the order of ascending *x**i*. No two trees are located at the point with the same coordinate.
Output Specification:
Print a single number — the maximum number of trees that you can cut down by the given rules.
Demo Input:
['5\n1 2\n2 1\n5 10\n10 9\n19 1\n', '5\n1 2\n2 1\n5 10\n10 9\n20 1\n']
Demo Output:
['3\n', '4\n']
Note:
In the first sample you can fell the trees like that:
- fell the 1-st tree to the left — now it occupies segment [ - 1;1] - fell the 2-nd tree to the right — now it occupies segment [2;3] - leave the 3-rd tree — it occupies point 5 - leave the 4-th tree — it occupies point 10 - fell the 5-th tree to the right — now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19]. | ```python
n = int(input())
trees = []
for i in range(n):
trees.append(list(map(int,input().split())))
# trees[i][0]: position
# trees[i][1]: height
r_edge = -trees[0][1]
max_num = 1
for i in range(n-1):
# try to cut i-th tree
# if can be cut leftward, then rightward, then cannot be cut
if trees[i][0]-trees[i][1] > r_edge:
max_num += 1
r_edge = trees[i][0]
elif trees[i][0]+trees[i][1] < trees[i+1][0]:
max_num += 1
r_edge = trees[i][0]+trees[i][1]
else:
r_edge = trees[i][0]
print(max_num)
``` | 3 | |
989 | A | A Blend of Springtime | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | "What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order. | The first and only line of input contains a non-empty string $s$ consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only ($\lvert s \rvert \leq 100$) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively. | Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower). | [
".BAC.\n",
"AA..CB\n"
] | [
"Yes\n",
"No\n"
] | In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell. | 500 | [
{
"input": ".BAC.",
"output": "Yes"
},
{
"input": "AA..CB",
"output": "No"
},
{
"input": ".",
"output": "No"
},
{
"input": "ACB.AAAAAA",
"output": "Yes"
},
{
"input": "B.BC.BBBCA",
"output": "Yes"
},
{
"input": "BA..CAB..B",
"output": "Yes"
},
... | 1,528,738,547 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 92 | 0 | l = input()
if "ABC" in l or "ACB" in l or "BCA" in l or "BAC" in l:
print("Yes")
elif "CAB" in l or "CBA" in l:
print("Yes")
else:
print("No")
| Title: A Blend of Springtime
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
Input Specification:
The first and only line of input contains a non-empty string $s$ consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only ($\lvert s \rvert \leq 100$) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output Specification:
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['.BAC.\n', 'AA..CB\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell. | ```python
l = input()
if "ABC" in l or "ACB" in l or "BCA" in l or "BAC" in l:
print("Yes")
elif "CAB" in l or "CBA" in l:
print("Yes")
else:
print("No")
``` | 3 | |
579 | A | Raising Bacteria | PROGRAMMING | 1,000 | [
"bitmasks"
] | null | null | You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days? | The only line containing one integer *x* (1<=≤<=*x*<=≤<=109). | The only line containing one integer: the answer. | [
"5\n",
"8\n"
] | [
"2\n",
"1\n"
] | For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1. | 250 | [
{
"input": "5",
"output": "2"
},
{
"input": "8",
"output": "1"
},
{
"input": "536870911",
"output": "29"
},
{
"input": "1",
"output": "1"
},
{
"input": "343000816",
"output": "14"
},
{
"input": "559980448",
"output": "12"
},
{
"input": "697... | 1,692,380,314 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 36 | 61 | 0 | def main():
x = int(input())
restos = 0
dividendo = x
while (dividendo != 1):
restos += dividendo % 2
dividendo = dividendo // 2
bacterias = restos + 1 # bacteria inicial
print(bacterias)
main()
| Title: Raising Bacteria
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are a lover of bacteria. You want to raise some bacteria in a box.
Initially, the box is empty. Each morning, you can put any number of bacteria into the box. And each night, every bacterium in the box will split into two bacteria. You hope to see exactly *x* bacteria in the box at some moment.
What is the minimum number of bacteria you need to put into the box across those days?
Input Specification:
The only line containing one integer *x* (1<=≤<=*x*<=≤<=109).
Output Specification:
The only line containing one integer: the answer.
Demo Input:
['5\n', '8\n']
Demo Output:
['2\n', '1\n']
Note:
For the first sample, we can add one bacterium in the box in the first day morning and at the third morning there will be 4 bacteria in the box. Now we put one more resulting 5 in the box. We added 2 bacteria in the process so the answer is 2.
For the second sample, we can put one in the first morning and in the 4-th morning there will be 8 in the box. So the answer is 1. | ```python
def main():
x = int(input())
restos = 0
dividendo = x
while (dividendo != 1):
restos += dividendo % 2
dividendo = dividendo // 2
bacterias = restos + 1 # bacteria inicial
print(bacterias)
main()
``` | 3 | |
791 | A | Bear and Big Brother | PROGRAMMING | 800 | [
"implementation"
] | null | null | Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob.
Right now, Limak and Bob weigh *a* and *b* respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight.
Limak eats a lot and his weight is tripled after every year, while Bob's weight is doubled after every year.
After how many full years will Limak become strictly larger (strictly heavier) than Bob? | The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10) — the weight of Limak and the weight of Bob respectively. | Print one integer, denoting the integer number of years after which Limak will become strictly larger than Bob. | [
"4 7\n",
"4 9\n",
"1 1\n"
] | [
"2\n",
"3\n",
"1\n"
] | In the first sample, Limak weighs 4 and Bob weighs 7 initially. After one year their weights are 4·3 = 12 and 7·2 = 14 respectively (one weight is tripled while the other one is doubled). Limak isn't larger than Bob yet. After the second year weights are 36 and 28, so the first weight is greater than the second one. Limak became larger than Bob after two years so you should print 2.
In the second sample, Limak's and Bob's weights in next years are: 12 and 18, then 36 and 36, and finally 108 and 72 (after three years). The answer is 3. Remember that Limak wants to be larger than Bob and he won't be satisfied with equal weights.
In the third sample, Limak becomes larger than Bob after the first year. Their weights will be 3 and 2 then. | 500 | [
{
"input": "4 7",
"output": "2"
},
{
"input": "4 9",
"output": "3"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "4 6",
"output": "2"
},
{
"input": "1 10",
"output": "6"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 2",
"output... | 1,693,669,884 | 2,147,483,647 | Python 3 | OK | TESTS | 62 | 46 | 0 | import math
m=input()
g=m.split()
alist=[]
alist.append((int(g[0]), int(g[1])))
count=0
for c in alist:
a,b=c
if a==b:
print(1)
else:
x = min(a, b)
y = max(a, b)
while x < y+0.0001:
x=x*3
y=y*2
count += 1
print(count)
| Title: Bear and Big Brother
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob.
Right now, Limak and Bob weigh *a* and *b* respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight.
Limak eats a lot and his weight is tripled after every year, while Bob's weight is doubled after every year.
After how many full years will Limak become strictly larger (strictly heavier) than Bob?
Input Specification:
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10) — the weight of Limak and the weight of Bob respectively.
Output Specification:
Print one integer, denoting the integer number of years after which Limak will become strictly larger than Bob.
Demo Input:
['4 7\n', '4 9\n', '1 1\n']
Demo Output:
['2\n', '3\n', '1\n']
Note:
In the first sample, Limak weighs 4 and Bob weighs 7 initially. After one year their weights are 4·3 = 12 and 7·2 = 14 respectively (one weight is tripled while the other one is doubled). Limak isn't larger than Bob yet. After the second year weights are 36 and 28, so the first weight is greater than the second one. Limak became larger than Bob after two years so you should print 2.
In the second sample, Limak's and Bob's weights in next years are: 12 and 18, then 36 and 36, and finally 108 and 72 (after three years). The answer is 3. Remember that Limak wants to be larger than Bob and he won't be satisfied with equal weights.
In the third sample, Limak becomes larger than Bob after the first year. Their weights will be 3 and 2 then. | ```python
import math
m=input()
g=m.split()
alist=[]
alist.append((int(g[0]), int(g[1])))
count=0
for c in alist:
a,b=c
if a==b:
print(1)
else:
x = min(a, b)
y = max(a, b)
while x < y+0.0001:
x=x*3
y=y*2
count += 1
print(count)
``` | 3 | |
699 | A | Launch of Collider | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | There will be a launch of a new, powerful and unusual collider very soon, which located along a straight line. *n* particles will be launched inside it. All of them are located in a straight line and there can not be two or more particles located in the same point. The coordinates of the particles coincide with the distance in meters from the center of the collider, *x**i* is the coordinate of the *i*-th particle and its position in the collider at the same time. All coordinates of particle positions are even integers.
You know the direction of each particle movement — it will move to the right or to the left after the collider's launch start. All particles begin to move simultaneously at the time of the collider's launch start. Each particle will move straight to the left or straight to the right with the constant speed of 1 meter per microsecond. The collider is big enough so particles can not leave it in the foreseeable time.
Write the program which finds the moment of the first collision of any two particles of the collider. In other words, find the number of microseconds before the first moment when any two particles are at the same point. | The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of particles.
The second line contains *n* symbols "L" and "R". If the *i*-th symbol equals "L", then the *i*-th particle will move to the left, otherwise the *i*-th symbol equals "R" and the *i*-th particle will move to the right.
The third line contains the sequence of pairwise distinct even integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=109) — the coordinates of particles in the order from the left to the right. It is guaranteed that the coordinates of particles are given in the increasing order. | In the first line print the only integer — the first moment (in microseconds) when two particles are at the same point and there will be an explosion.
Print the only integer -1, if the collision of particles doesn't happen. | [
"4\nRLRL\n2 4 6 10\n",
"3\nLLR\n40 50 60\n"
] | [
"1\n",
"-1\n"
] | In the first sample case the first explosion will happen in 1 microsecond because the particles number 1 and 2 will simultaneously be at the same point with the coordinate 3.
In the second sample case there will be no explosion because there are no particles which will simultaneously be at the same point. | 500 | [
{
"input": "4\nRLRL\n2 4 6 10",
"output": "1"
},
{
"input": "3\nLLR\n40 50 60",
"output": "-1"
},
{
"input": "4\nRLLR\n46 230 264 470",
"output": "92"
},
{
"input": "6\nLLRLLL\n446 492 650 844 930 970",
"output": "97"
},
{
"input": "8\nRRLLLLLL\n338 478 512 574 59... | 1,656,362,900 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 46 | 0 | n = int(input())
dirs = input()
p = [int(i) for i in input().split()]
r = -1
l = -1
for i in range(n - 1):
if dirs[i] + dirs[i + 1] == "RL":
r = i
l = i + 1
break
if r == -1 or l == -1:
print(-1)
else:
print((p[l] - p[r]) // 2) | Title: Launch of Collider
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There will be a launch of a new, powerful and unusual collider very soon, which located along a straight line. *n* particles will be launched inside it. All of them are located in a straight line and there can not be two or more particles located in the same point. The coordinates of the particles coincide with the distance in meters from the center of the collider, *x**i* is the coordinate of the *i*-th particle and its position in the collider at the same time. All coordinates of particle positions are even integers.
You know the direction of each particle movement — it will move to the right or to the left after the collider's launch start. All particles begin to move simultaneously at the time of the collider's launch start. Each particle will move straight to the left or straight to the right with the constant speed of 1 meter per microsecond. The collider is big enough so particles can not leave it in the foreseeable time.
Write the program which finds the moment of the first collision of any two particles of the collider. In other words, find the number of microseconds before the first moment when any two particles are at the same point.
Input Specification:
The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of particles.
The second line contains *n* symbols "L" and "R". If the *i*-th symbol equals "L", then the *i*-th particle will move to the left, otherwise the *i*-th symbol equals "R" and the *i*-th particle will move to the right.
The third line contains the sequence of pairwise distinct even integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=109) — the coordinates of particles in the order from the left to the right. It is guaranteed that the coordinates of particles are given in the increasing order.
Output Specification:
In the first line print the only integer — the first moment (in microseconds) when two particles are at the same point and there will be an explosion.
Print the only integer -1, if the collision of particles doesn't happen.
Demo Input:
['4\nRLRL\n2 4 6 10\n', '3\nLLR\n40 50 60\n']
Demo Output:
['1\n', '-1\n']
Note:
In the first sample case the first explosion will happen in 1 microsecond because the particles number 1 and 2 will simultaneously be at the same point with the coordinate 3.
In the second sample case there will be no explosion because there are no particles which will simultaneously be at the same point. | ```python
n = int(input())
dirs = input()
p = [int(i) for i in input().split()]
r = -1
l = -1
for i in range(n - 1):
if dirs[i] + dirs[i + 1] == "RL":
r = i
l = i + 1
break
if r == -1 or l == -1:
print(-1)
else:
print((p[l] - p[r]) // 2)
``` | 0 | |
962 | A | Equator | PROGRAMMING | 1,300 | [
"implementation"
] | null | null | Polycarp has created his own training plan to prepare for the programming contests. He will train for $n$ days, all days are numbered from $1$ to $n$, beginning from the first.
On the $i$-th day Polycarp will necessarily solve $a_i$ problems. One evening Polycarp plans to celebrate the equator. He will celebrate it on the first evening of such a day that from the beginning of the training and to this day inclusive he will solve half or more of all the problems.
Determine the index of day when Polycarp will celebrate the equator. | The first line contains a single integer $n$ ($1 \le n \le 200\,000$) — the number of days to prepare for the programming contests.
The second line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10\,000$), where $a_i$ equals to the number of problems, which Polycarp will solve on the $i$-th day. | Print the index of the day when Polycarp will celebrate the equator. | [
"4\n1 3 2 1\n",
"6\n2 2 2 2 2 2\n"
] | [
"2\n",
"3\n"
] | In the first example Polycarp will celebrate the equator on the evening of the second day, because up to this day (inclusive) he will solve $4$ out of $7$ scheduled problems on four days of the training.
In the second example Polycarp will celebrate the equator on the evening of the third day, because up to this day (inclusive) he will solve $6$ out of $12$ scheduled problems on six days of the training. | 0 | [
{
"input": "4\n1 3 2 1",
"output": "2"
},
{
"input": "6\n2 2 2 2 2 2",
"output": "3"
},
{
"input": "1\n10000",
"output": "1"
},
{
"input": "3\n2 1 1",
"output": "1"
},
{
"input": "2\n1 3",
"output": "2"
},
{
"input": "4\n2 1 1 3",
"output": "3"
}... | 1,524,313,310 | 530 | Python 3 | OK | TESTS | 106 | 187 | 20,787,200 | n = int(input())
a = list(map(int, input().split()))
sum1 = sum(a)
sum2 = 0
i = 0
while sum2 < (sum1 + 1) // 2:
sum2 += a[i]
i += 1
print(i) | Title: Equator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has created his own training plan to prepare for the programming contests. He will train for $n$ days, all days are numbered from $1$ to $n$, beginning from the first.
On the $i$-th day Polycarp will necessarily solve $a_i$ problems. One evening Polycarp plans to celebrate the equator. He will celebrate it on the first evening of such a day that from the beginning of the training and to this day inclusive he will solve half or more of all the problems.
Determine the index of day when Polycarp will celebrate the equator.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 200\,000$) — the number of days to prepare for the programming contests.
The second line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10\,000$), where $a_i$ equals to the number of problems, which Polycarp will solve on the $i$-th day.
Output Specification:
Print the index of the day when Polycarp will celebrate the equator.
Demo Input:
['4\n1 3 2 1\n', '6\n2 2 2 2 2 2\n']
Demo Output:
['2\n', '3\n']
Note:
In the first example Polycarp will celebrate the equator on the evening of the second day, because up to this day (inclusive) he will solve $4$ out of $7$ scheduled problems on four days of the training.
In the second example Polycarp will celebrate the equator on the evening of the third day, because up to this day (inclusive) he will solve $6$ out of $12$ scheduled problems on six days of the training. | ```python
n = int(input())
a = list(map(int, input().split()))
sum1 = sum(a)
sum2 = 0
i = 0
while sum2 < (sum1 + 1) // 2:
sum2 += a[i]
i += 1
print(i)
``` | 3 | |
820 | A | Mister B and Book Reading | PROGRAMMING | 900 | [
"implementation"
] | null | null | Mister B once received a gift: it was a book about aliens, which he started read immediately. This book had *c* pages.
At first day Mister B read *v*0 pages, but after that he started to speed up. Every day, starting from the second, he read *a* pages more than on the previous day (at first day he read *v*0 pages, at second — *v*0<=+<=*a* pages, at third — *v*0<=+<=2*a* pages, and so on). But Mister B is just a human, so he physically wasn't able to read more than *v*1 pages per day.
Also, to refresh his memory, every day, starting from the second, Mister B had to reread last *l* pages he read on the previous day. Mister B finished the book when he read the last page for the first time.
Help Mister B to calculate how many days he needed to finish the book. | First and only line contains five space-separated integers: *c*, *v*0, *v*1, *a* and *l* (1<=≤<=*c*<=≤<=1000, 0<=≤<=*l*<=<<=*v*0<=≤<=*v*1<=≤<=1000, 0<=≤<=*a*<=≤<=1000) — the length of the book in pages, the initial reading speed, the maximum reading speed, the acceleration in reading speed and the number of pages for rereading. | Print one integer — the number of days Mister B needed to finish the book. | [
"5 5 10 5 4\n",
"12 4 12 4 1\n",
"15 1 100 0 0\n"
] | [
"1\n",
"3\n",
"15\n"
] | In the first sample test the book contains 5 pages, so Mister B read it right at the first day.
In the second sample test at first day Mister B read pages number 1 - 4, at second day — 4 - 11, at third day — 11 - 12 and finished the book.
In third sample test every day Mister B read 1 page of the book, so he finished in 15 days. | 500 | [
{
"input": "5 5 10 5 4",
"output": "1"
},
{
"input": "12 4 12 4 1",
"output": "3"
},
{
"input": "15 1 100 0 0",
"output": "15"
},
{
"input": "1 1 1 0 0",
"output": "1"
},
{
"input": "1000 999 1000 1000 998",
"output": "2"
},
{
"input": "1000 2 2 5 1",
... | 1,498,647,317 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | l=raw_input()
arr=l.split()
for i in range(0,5):
arr[i]=int(arr[i])
count=1
i=1
left_pages=arr[0]
if arr[0]<arr[1]:
print count
else:
left_pages=left_pages-arr[1]
while left_pages>0:
count+=1
effective_pages_read_per_day = arr[1] + i * arr[3] - arr[4]
checker = arr[1] + i * arr[3] + arr[4]
if checker<=arr[2]:
left_pages=left_pages-effective_pages_read_per_day
else:
left_pages=left_pages-(arr[2]-arr[4])
i+=1
print count
| Title: Mister B and Book Reading
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mister B once received a gift: it was a book about aliens, which he started read immediately. This book had *c* pages.
At first day Mister B read *v*0 pages, but after that he started to speed up. Every day, starting from the second, he read *a* pages more than on the previous day (at first day he read *v*0 pages, at second — *v*0<=+<=*a* pages, at third — *v*0<=+<=2*a* pages, and so on). But Mister B is just a human, so he physically wasn't able to read more than *v*1 pages per day.
Also, to refresh his memory, every day, starting from the second, Mister B had to reread last *l* pages he read on the previous day. Mister B finished the book when he read the last page for the first time.
Help Mister B to calculate how many days he needed to finish the book.
Input Specification:
First and only line contains five space-separated integers: *c*, *v*0, *v*1, *a* and *l* (1<=≤<=*c*<=≤<=1000, 0<=≤<=*l*<=<<=*v*0<=≤<=*v*1<=≤<=1000, 0<=≤<=*a*<=≤<=1000) — the length of the book in pages, the initial reading speed, the maximum reading speed, the acceleration in reading speed and the number of pages for rereading.
Output Specification:
Print one integer — the number of days Mister B needed to finish the book.
Demo Input:
['5 5 10 5 4\n', '12 4 12 4 1\n', '15 1 100 0 0\n']
Demo Output:
['1\n', '3\n', '15\n']
Note:
In the first sample test the book contains 5 pages, so Mister B read it right at the first day.
In the second sample test at first day Mister B read pages number 1 - 4, at second day — 4 - 11, at third day — 11 - 12 and finished the book.
In third sample test every day Mister B read 1 page of the book, so he finished in 15 days. | ```python
l=raw_input()
arr=l.split()
for i in range(0,5):
arr[i]=int(arr[i])
count=1
i=1
left_pages=arr[0]
if arr[0]<arr[1]:
print count
else:
left_pages=left_pages-arr[1]
while left_pages>0:
count+=1
effective_pages_read_per_day = arr[1] + i * arr[3] - arr[4]
checker = arr[1] + i * arr[3] + arr[4]
if checker<=arr[2]:
left_pages=left_pages-effective_pages_read_per_day
else:
left_pages=left_pages-(arr[2]-arr[4])
i+=1
print count
``` | -1 | |
743 | B | Chloe and the sequence | PROGRAMMING | 1,200 | [
"binary search",
"bitmasks",
"constructive algorithms",
"implementation"
] | null | null | Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad.
Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (*n*<=-<=1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1,<=2,<=1] after the first step, the sequence [1,<=2,<=1,<=3,<=1,<=2,<=1] after the second step.
The task is to find the value of the element with index *k* (the elements are numbered from 1) in the obtained sequence, i. e. after (*n*<=-<=1) steps.
Please help Chloe to solve the problem! | The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=2*n*<=-<=1). | Print single integer — the integer at the *k*-th position in the obtained sequence. | [
"3 2\n",
"4 8\n"
] | [
"2",
"4"
] | In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2.
In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4. | 1,000 | [
{
"input": "3 2",
"output": "2"
},
{
"input": "4 8",
"output": "4"
},
{
"input": "5 27",
"output": "1"
},
{
"input": "7 44",
"output": "3"
},
{
"input": "15 18432",
"output": "12"
},
{
"input": "20 259676",
"output": "3"
},
{
"input": "30 6... | 1,559,724,082 | 2,147,483,647 | Python 3 | MEMORY_LIMIT_EXCEEDED | TESTS | 6 | 296 | 268,390,400 | n,k=map(int,input().split())
a=[1]
for i in range(2,n+1):
a.extend(a)
a.insert(len(a)//2,i)
print(a[k-1])
| Title: Chloe and the sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad.
Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (*n*<=-<=1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1,<=2,<=1] after the first step, the sequence [1,<=2,<=1,<=3,<=1,<=2,<=1] after the second step.
The task is to find the value of the element with index *k* (the elements are numbered from 1) in the obtained sequence, i. e. after (*n*<=-<=1) steps.
Please help Chloe to solve the problem!
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=2*n*<=-<=1).
Output Specification:
Print single integer — the integer at the *k*-th position in the obtained sequence.
Demo Input:
['3 2\n', '4 8\n']
Demo Output:
['2', '4']
Note:
In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2.
In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4. | ```python
n,k=map(int,input().split())
a=[1]
for i in range(2,n+1):
a.extend(a)
a.insert(len(a)//2,i)
print(a[k-1])
``` | 0 | |
381 | A | Sereja and Dima | PROGRAMMING | 800 | [
"greedy",
"implementation",
"two pointers"
] | null | null | Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her. | The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000. | On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game. | [
"4\n4 1 2 10\n",
"7\n1 2 3 4 5 6 7\n"
] | [
"12 5\n",
"16 12\n"
] | In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5. | 500 | [
{
"input": "4\n4 1 2 10",
"output": "12 5"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "16 12"
},
{
"input": "42\n15 29 37 22 16 5 26 31 6 32 19 3 45 36 33 14 25 20 48 7 42 11 24 28 9 18 8 21 47 17 38 40 44 4 35 1 43 39 41 27 12 13",
"output": "613 418"
},
{
"input": "43\n32 ... | 1,680,099,223 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 15 | 0 | cont=0
condi=True
n= int(input())
sere=0
dima=0
arr=[]
for i in range(n):
arr.append(int(input()))
for i in range(len(arr)):
if arr[0+cont]>arr[-1]:
if condi==True:
sere= sere+arr[0+cont]
condi=False
else:
dima= dima+arr[0+cont]
condi=True
cont+=1
else:
if condi==True:
sere= sere+arr[-1]
condi=False
else:
dima= dima+arr[-1]
condi=True
arr.pop()
print(sere, " ", dima) | Title: Sereja and Dima
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output Specification:
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
Demo Input:
['4\n4 1 2 10\n', '7\n1 2 3 4 5 6 7\n']
Demo Output:
['12 5\n', '16 12\n']
Note:
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5. | ```python
cont=0
condi=True
n= int(input())
sere=0
dima=0
arr=[]
for i in range(n):
arr.append(int(input()))
for i in range(len(arr)):
if arr[0+cont]>arr[-1]:
if condi==True:
sere= sere+arr[0+cont]
condi=False
else:
dima= dima+arr[0+cont]
condi=True
cont+=1
else:
if condi==True:
sere= sere+arr[-1]
condi=False
else:
dima= dima+arr[-1]
condi=True
arr.pop()
print(sere, " ", dima)
``` | -1 | |
691 | A | Fashion in Berland | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | According to rules of the Berland fashion, a jacket should be fastened by all the buttons except only one, but not necessarily it should be the last one. Also if the jacket has only one button, it should be fastened, so the jacket will not swinging open.
You are given a jacket with *n* buttons. Determine if it is fastened in a right way. | The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of buttons on the jacket.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1). The number *a**i*<==<=0 if the *i*-th button is not fastened. Otherwise *a**i*<==<=1. | In the only line print the word "YES" if the jacket is fastened in a right way. Otherwise print the word "NO". | [
"3\n1 0 1\n",
"3\n1 0 0\n"
] | [
"YES\n",
"NO\n"
] | none | 0 | [
{
"input": "3\n1 0 1",
"output": "YES"
},
{
"input": "3\n1 0 0",
"output": "NO"
},
{
"input": "3\n1 1 0",
"output": "YES"
},
{
"input": "3\n1 1 1",
"output": "NO"
},
{
"input": "3\n0 0 1",
"output": "NO"
},
{
"input": "3\n0 0 0",
"output": "NO"
}... | 1,597,781,880 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 13 | 124 | 0 | import sys
n = int(input())
a = input().split()
count = 0
num = "".join(a)
y = num.count("0")
if n == 1:
print("YES")
sys.exit()
if y == 1 and n > 1:
print("YES")
else:
print("NO") | Title: Fashion in Berland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
According to rules of the Berland fashion, a jacket should be fastened by all the buttons except only one, but not necessarily it should be the last one. Also if the jacket has only one button, it should be fastened, so the jacket will not swinging open.
You are given a jacket with *n* buttons. Determine if it is fastened in a right way.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of buttons on the jacket.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1). The number *a**i*<==<=0 if the *i*-th button is not fastened. Otherwise *a**i*<==<=1.
Output Specification:
In the only line print the word "YES" if the jacket is fastened in a right way. Otherwise print the word "NO".
Demo Input:
['3\n1 0 1\n', '3\n1 0 0\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
import sys
n = int(input())
a = input().split()
count = 0
num = "".join(a)
y = num.count("0")
if n == 1:
print("YES")
sys.exit()
if y == 1 and n > 1:
print("YES")
else:
print("NO")
``` | 0 | |
794 | A | Bank Robbery | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | A robber has attempted to rob a bank but failed to complete his task. However, he had managed to open all the safes.
Oleg the bank client loves money (who doesn't), and decides to take advantage of this failed robbery and steal some money from the safes. There are many safes arranged in a line, where the *i*-th safe from the left is called safe *i*. There are *n* banknotes left in all the safes in total. The *i*-th banknote is in safe *x**i*. Oleg is now at safe *a*. There are two security guards, one of which guards the safe *b* such that *b*<=<<=*a*, i.e. the first guard is to the left of Oleg. The other guard guards the safe *c* so that *c*<=><=*a*, i.e. he is to the right of Oleg.
The two guards are very lazy, so they do not move. In every second, Oleg can either take all the banknotes from the current safe or move to any of the neighboring safes. However, he cannot visit any safe that is guarded by security guards at any time, becaues he might be charged for stealing. Determine the maximum amount of banknotes Oleg can gather. | The first line of input contains three space-separated integers, *a*, *b* and *c* (1<=≤<=*b*<=<<=*a*<=<<=*c*<=≤<=109), denoting the positions of Oleg, the first security guard and the second security guard, respectively.
The next line of input contains a single integer *n* (1<=≤<=*n*<=≤<=105), denoting the number of banknotes.
The next line of input contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109), denoting that the *i*-th banknote is located in the *x**i*-th safe. Note that *x**i* are not guaranteed to be distinct. | Output a single integer: the maximum number of banknotes Oleg can take. | [
"5 3 7\n8\n4 7 5 5 3 6 2 8\n",
"6 5 7\n5\n1 5 7 92 3\n"
] | [
"4\n",
"0\n"
] | In the first example Oleg can take the banknotes in positions 4, 5, 6 (note that there are 2 banknotes at position 5). Oleg can't take the banknotes in safes 7 and 8 because he can't run into the second security guard. Similarly, Oleg cannot take the banknotes at positions 3 and 2 because he can't run into the first security guard. Thus, he can take a maximum of 4 banknotes.
For the second sample, Oleg can't take any banknotes without bumping into any of the security guards. | 500 | [
{
"input": "5 3 7\n8\n4 7 5 5 3 6 2 8",
"output": "4"
},
{
"input": "6 5 7\n5\n1 5 7 92 3",
"output": "0"
},
{
"input": "3 2 4\n1\n3",
"output": "1"
},
{
"input": "5 3 8\n12\n8 3 4 5 7 6 8 3 5 4 7 6",
"output": "8"
},
{
"input": "7 3 10\n5\n3 3 3 3 3",
"output... | 1,495,063,526 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 4 | 2,000 | 7,270,400 | line1 = input()
a,b,c = line1.split(' ')
a,b,c = int(a),int(b),int(c)
input()
safes = range(b+1,c)
line3 = input().split(' ')
c = 0
for sf in safes:
c += line3.count(str(sf))
print(c) | Title: Bank Robbery
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A robber has attempted to rob a bank but failed to complete his task. However, he had managed to open all the safes.
Oleg the bank client loves money (who doesn't), and decides to take advantage of this failed robbery and steal some money from the safes. There are many safes arranged in a line, where the *i*-th safe from the left is called safe *i*. There are *n* banknotes left in all the safes in total. The *i*-th banknote is in safe *x**i*. Oleg is now at safe *a*. There are two security guards, one of which guards the safe *b* such that *b*<=<<=*a*, i.e. the first guard is to the left of Oleg. The other guard guards the safe *c* so that *c*<=><=*a*, i.e. he is to the right of Oleg.
The two guards are very lazy, so they do not move. In every second, Oleg can either take all the banknotes from the current safe or move to any of the neighboring safes. However, he cannot visit any safe that is guarded by security guards at any time, becaues he might be charged for stealing. Determine the maximum amount of banknotes Oleg can gather.
Input Specification:
The first line of input contains three space-separated integers, *a*, *b* and *c* (1<=≤<=*b*<=<<=*a*<=<<=*c*<=≤<=109), denoting the positions of Oleg, the first security guard and the second security guard, respectively.
The next line of input contains a single integer *n* (1<=≤<=*n*<=≤<=105), denoting the number of banknotes.
The next line of input contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (1<=≤<=*x**i*<=≤<=109), denoting that the *i*-th banknote is located in the *x**i*-th safe. Note that *x**i* are not guaranteed to be distinct.
Output Specification:
Output a single integer: the maximum number of banknotes Oleg can take.
Demo Input:
['5 3 7\n8\n4 7 5 5 3 6 2 8\n', '6 5 7\n5\n1 5 7 92 3\n']
Demo Output:
['4\n', '0\n']
Note:
In the first example Oleg can take the banknotes in positions 4, 5, 6 (note that there are 2 banknotes at position 5). Oleg can't take the banknotes in safes 7 and 8 because he can't run into the second security guard. Similarly, Oleg cannot take the banknotes at positions 3 and 2 because he can't run into the first security guard. Thus, he can take a maximum of 4 banknotes.
For the second sample, Oleg can't take any banknotes without bumping into any of the security guards. | ```python
line1 = input()
a,b,c = line1.split(' ')
a,b,c = int(a),int(b),int(c)
input()
safes = range(b+1,c)
line3 = input().split(' ')
c = 0
for sf in safes:
c += line3.count(str(sf))
print(c)
``` | 0 | |
600 | B | Queries about less or equal elements | PROGRAMMING | 1,300 | [
"binary search",
"data structures",
"sortings",
"two pointers"
] | null | null | You are given two arrays of integers *a* and *b*. For each element of the second array *b**j* you should find the number of elements in array *a* that are less than or equal to the value *b**j*. | The first line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=2·105) — the sizes of arrays *a* and *b*.
The second line contains *n* integers — the elements of array *a* (<=-<=109<=≤<=*a**i*<=≤<=109).
The third line contains *m* integers — the elements of array *b* (<=-<=109<=≤<=*b**j*<=≤<=109). | Print *m* integers, separated by spaces: the *j*-th of which is equal to the number of such elements in array *a* that are less than or equal to the value *b**j*. | [
"5 4\n1 3 5 7 9\n6 4 2 8\n",
"5 5\n1 2 1 2 5\n3 1 4 1 5\n"
] | [
"3 2 1 4\n",
"4 2 4 2 5\n"
] | none | 0 | [
{
"input": "5 4\n1 3 5 7 9\n6 4 2 8",
"output": "3 2 1 4"
},
{
"input": "5 5\n1 2 1 2 5\n3 1 4 1 5",
"output": "4 2 4 2 5"
},
{
"input": "1 1\n-1\n-2",
"output": "0"
},
{
"input": "1 1\n-80890826\n686519510",
"output": "1"
},
{
"input": "11 11\n237468511 -77918754... | 1,655,034,456 | 2,147,483,647 | Python 3 | OK | TESTS | 60 | 1,076 | 27,545,600 | if __name__ == '__main__':
n, m = list(map(int, input().strip().split()))
a = list(map(int, input().strip().split()))
b = list(map(int, input().strip().split()))
c = []
for i in range(m):
c.append([b[i], i])
a.sort()
c = sorted(c, key=lambda k: k[0])
left = 0
res = [0] * m
for right in range(m):
while left < n and a[left] <= c[right][0]:
left += 1
res[c[right][1]] = left
for i in range(m):
print(res[i], end=" ") | Title: Queries about less or equal elements
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two arrays of integers *a* and *b*. For each element of the second array *b**j* you should find the number of elements in array *a* that are less than or equal to the value *b**j*.
Input Specification:
The first line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=2·105) — the sizes of arrays *a* and *b*.
The second line contains *n* integers — the elements of array *a* (<=-<=109<=≤<=*a**i*<=≤<=109).
The third line contains *m* integers — the elements of array *b* (<=-<=109<=≤<=*b**j*<=≤<=109).
Output Specification:
Print *m* integers, separated by spaces: the *j*-th of which is equal to the number of such elements in array *a* that are less than or equal to the value *b**j*.
Demo Input:
['5 4\n1 3 5 7 9\n6 4 2 8\n', '5 5\n1 2 1 2 5\n3 1 4 1 5\n']
Demo Output:
['3 2 1 4\n', '4 2 4 2 5\n']
Note:
none | ```python
if __name__ == '__main__':
n, m = list(map(int, input().strip().split()))
a = list(map(int, input().strip().split()))
b = list(map(int, input().strip().split()))
c = []
for i in range(m):
c.append([b[i], i])
a.sort()
c = sorted(c, key=lambda k: k[0])
left = 0
res = [0] * m
for right in range(m):
while left < n and a[left] <= c[right][0]:
left += 1
res[c[right][1]] = left
for i in range(m):
print(res[i], end=" ")
``` | 3 | |
779 | C | Dishonest Sellers | PROGRAMMING | 1,200 | [
"constructive algorithms",
"greedy",
"sortings"
] | null | null | Igor found out discounts in a shop and decided to buy *n* items. Discounts at the store will last for a week and Igor knows about each item that its price now is *a**i*, and after a week of discounts its price will be *b**i*.
Not all of sellers are honest, so now some products could be more expensive than after a week of discounts.
Igor decided that buy at least *k* of items now, but wait with the rest of the week in order to save money as much as possible. Your task is to determine the minimum money that Igor can spend to buy all *n* items. | In the first line there are two positive integer numbers *n* and *k* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*k*<=≤<=*n*) — total number of items to buy and minimal number of items Igor wants to by right now.
The second line contains sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) — prices of items during discounts (i.e. right now).
The third line contains sequence of integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=104) — prices of items after discounts (i.e. after a week). | Print the minimal amount of money Igor will spend to buy all *n* items. Remember, he should buy at least *k* items right now. | [
"3 1\n5 4 6\n3 1 5\n",
"5 3\n3 4 7 10 3\n4 5 5 12 5\n"
] | [
"10\n",
"25\n"
] | In the first example Igor should buy item 3 paying 6. But items 1 and 2 he should buy after a week. He will pay 3 and 1 for them. So in total he will pay 6 + 3 + 1 = 10.
In the second example Igor should buy right now items 1, 2, 4 and 5, paying for them 3, 4, 10 and 3, respectively. Item 3 he should buy after a week of discounts, he will pay 5 for it. In total he will spend 3 + 4 + 10 + 3 + 5 = 25. | 1,000 | [
{
"input": "3 1\n5 4 6\n3 1 5",
"output": "10"
},
{
"input": "5 3\n3 4 7 10 3\n4 5 5 12 5",
"output": "25"
},
{
"input": "1 0\n9\n8",
"output": "8"
},
{
"input": "2 0\n4 10\n1 2",
"output": "3"
},
{
"input": "4 2\n19 5 17 13\n3 18 8 10",
"output": "29"
},
... | 1,536,151,973 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 45 | 421 | 17,920,000 | a = [int(k) for k in input().split()]
b = [int(k) for k in input().split()]
c = [int(k) for k in input().split()]
d = [b[k] - c[k] for k in range(a[0])]
d.sort()
e = 0
while d[e] < 0 and e < a[0]:
e += 1
if e > a[1]:
print(sum(d[0:e])+sum(c))
else:
print(sum(d[0:a[1]])+sum(c)) | Title: Dishonest Sellers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Igor found out discounts in a shop and decided to buy *n* items. Discounts at the store will last for a week and Igor knows about each item that its price now is *a**i*, and after a week of discounts its price will be *b**i*.
Not all of sellers are honest, so now some products could be more expensive than after a week of discounts.
Igor decided that buy at least *k* of items now, but wait with the rest of the week in order to save money as much as possible. Your task is to determine the minimum money that Igor can spend to buy all *n* items.
Input Specification:
In the first line there are two positive integer numbers *n* and *k* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*k*<=≤<=*n*) — total number of items to buy and minimal number of items Igor wants to by right now.
The second line contains sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104) — prices of items during discounts (i.e. right now).
The third line contains sequence of integers *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=104) — prices of items after discounts (i.e. after a week).
Output Specification:
Print the minimal amount of money Igor will spend to buy all *n* items. Remember, he should buy at least *k* items right now.
Demo Input:
['3 1\n5 4 6\n3 1 5\n', '5 3\n3 4 7 10 3\n4 5 5 12 5\n']
Demo Output:
['10\n', '25\n']
Note:
In the first example Igor should buy item 3 paying 6. But items 1 and 2 he should buy after a week. He will pay 3 and 1 for them. So in total he will pay 6 + 3 + 1 = 10.
In the second example Igor should buy right now items 1, 2, 4 and 5, paying for them 3, 4, 10 and 3, respectively. Item 3 he should buy after a week of discounts, he will pay 5 for it. In total he will spend 3 + 4 + 10 + 3 + 5 = 25. | ```python
a = [int(k) for k in input().split()]
b = [int(k) for k in input().split()]
c = [int(k) for k in input().split()]
d = [b[k] - c[k] for k in range(a[0])]
d.sort()
e = 0
while d[e] < 0 and e < a[0]:
e += 1
if e > a[1]:
print(sum(d[0:e])+sum(c))
else:
print(sum(d[0:a[1]])+sum(c))
``` | -1 | |
556 | A | Case of the Zeros and Ones | PROGRAMMING | 900 | [
"greedy"
] | null | null | Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result.
Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number. | First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has.
The second line contains the string of length *n* consisting only from zeros and ones. | Output the minimum length of the string that may remain after applying the described operations several times. | [
"4\n1100\n",
"5\n01010\n",
"8\n11101111\n"
] | [
"0\n",
"1\n",
"6\n"
] | In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 250 | [
{
"input": "4\n1100",
"output": "0"
},
{
"input": "5\n01010",
"output": "1"
},
{
"input": "8\n11101111",
"output": "6"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "2\n00",
"output": "2"
},
{
"input"... | 1,640,629,192 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 61 | 204,800 | n=int(input())
s=input()
c0=s.count("0")
c1=s.count("1")
print(c0+c1-2*min(c0,c1))
| Title: Case of the Zeros and Ones
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length *n* consisting of zeroes and ones. Consider the following operation: we choose any two adjacent positions in the string, and if one them contains 0, and the other contains 1, then we are allowed to remove these two digits from the string, obtaining a string of length *n*<=-<=2 as a result.
Now Andreid thinks about what is the minimum length of the string that can remain after applying the described operation several times (possibly, zero)? Help him to calculate this number.
Input Specification:
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=2·105), the length of the string that Andreid has.
The second line contains the string of length *n* consisting only from zeros and ones.
Output Specification:
Output the minimum length of the string that may remain after applying the described operations several times.
Demo Input:
['4\n1100\n', '5\n01010\n', '8\n11101111\n']
Demo Output:
['0\n', '1\n', '6\n']
Note:
In the first sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/10df55364c21c6e8d5da31b6ab6f6294c4fc26b3.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/19ec5dcd85f0b5cf757aa076ace72df39634de2d.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the third sample test it is possible to change the string like the following: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/dc34a159e4230375fa325555527ebc748811f188.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
n=int(input())
s=input()
c0=s.count("0")
c1=s.count("1")
print(c0+c1-2*min(c0,c1))
``` | 3 |
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