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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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
269 | B | Greenhouse Effect | PROGRAMMING | 1,700 | [
"dp"
] | null | null | Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated *n* plants in his greenhouse, of *m* different plant species numbered from 1 to *m*. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange *m*<=-<=1 borders that would divide the greenhouse into *m* sections numbered from 1 to *m* from left to right with each section housing a single species. He is free to place the borders, but in the end all of the *i*-th species plants must reside in *i*-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders. | The first line of input contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=5000, *n*<=≥<=*m*), the number of plants and the number of different species. Each of the following *n* lines contain two space-separated numbers: one integer number *s**i* (1<=≤<=*s**i*<=≤<=*m*), and one real number *x**i* (0<=≤<=*x**i*<=≤<=109), the species and position of the *i*-th plant. Each *x**i* will contain no more than 6 digits after the decimal point.
It is guaranteed that all *x**i* are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their *x**i* coordinates (*x**i*<=<<=*x**i*<=+<=1,<=1<=≤<=*i*<=<<=*n*). | Output a single integer — the minimum number of plants to be replanted. | [
"3 2\n2 1\n1 2.0\n1 3.100\n",
"3 3\n1 5.0\n2 5.5\n3 6.0\n",
"6 3\n1 14.284235\n2 17.921382\n1 20.328172\n3 20.842331\n1 25.790145\n1 27.204125\n"
] | [
"1\n",
"0\n",
"2\n"
] | In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed. | 1,000 | [
{
"input": "3 2\n2 1\n1 2.0\n1 3.100",
"output": "1"
},
{
"input": "3 3\n1 5.0\n2 5.5\n3 6.0",
"output": "0"
},
{
"input": "6 3\n1 14.284235\n2 17.921382\n1 20.328172\n3 20.842331\n1 25.790145\n1 27.204125",
"output": "2"
},
{
"input": "1 1\n1 0",
"output": "0"
},
{
... | 1,692,552,391 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 40 | 342 | 3,788,800 | import sys
input = lambda: sys.stdin.readline().rstrip()
from collections import deque,defaultdict,Counter
from itertools import permutations,combinations
from bisect import *
from heapq import *
from math import ceil,gcd,lcm,floor,comb
alph = 'abcdefghijklmnopqrstuvwxyz'
#pow(x,mod-2,mod)
N,M = map(int, input().split())
A = []
for _ in range(N):
s,x = input().split()
s = int(s)
x = float(x)
A.append((x,s))
A.sort()
A.append((A[-1][0]+1, M+1))
dp = [0]*(N+1)
for i in range(N+1):
x,s = A[i]
for j in range(i):
if A[j][1]<=s:
dp[i] = max(dp[i], dp[j]+1)
print(N-dp[-1])
| Title: Greenhouse Effect
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Emuskald is an avid horticulturist and owns the world's longest greenhouse — it is effectively infinite in length.
Over the years Emuskald has cultivated *n* plants in his greenhouse, of *m* different plant species numbered from 1 to *m*. His greenhouse is very narrow and can be viewed as an infinite line, with each plant occupying a single point on that line.
Emuskald has discovered that each species thrives at a different temperature, so he wants to arrange *m*<=-<=1 borders that would divide the greenhouse into *m* sections numbered from 1 to *m* from left to right with each section housing a single species. He is free to place the borders, but in the end all of the *i*-th species plants must reside in *i*-th section from the left.
Of course, it is not always possible to place the borders in such way, so Emuskald needs to replant some of his plants. He can remove each plant from its position and place it anywhere in the greenhouse (at any real coordinate) with no plant already in it. Since replanting is a lot of stress for the plants, help Emuskald find the minimum number of plants he has to replant to be able to place the borders.
Input Specification:
The first line of input contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=5000, *n*<=≥<=*m*), the number of plants and the number of different species. Each of the following *n* lines contain two space-separated numbers: one integer number *s**i* (1<=≤<=*s**i*<=≤<=*m*), and one real number *x**i* (0<=≤<=*x**i*<=≤<=109), the species and position of the *i*-th plant. Each *x**i* will contain no more than 6 digits after the decimal point.
It is guaranteed that all *x**i* are different; there is at least one plant of each species; the plants are given in order "from left to the right", that is in the ascending order of their *x**i* coordinates (*x**i*<=<<=*x**i*<=+<=1,<=1<=≤<=*i*<=<<=*n*).
Output Specification:
Output a single integer — the minimum number of plants to be replanted.
Demo Input:
['3 2\n2 1\n1 2.0\n1 3.100\n', '3 3\n1 5.0\n2 5.5\n3 6.0\n', '6 3\n1 14.284235\n2 17.921382\n1 20.328172\n3 20.842331\n1 25.790145\n1 27.204125\n']
Demo Output:
['1\n', '0\n', '2\n']
Note:
In the first test case, Emuskald can replant the first plant to the right of the last plant, so the answer is 1.
In the second test case, the species are already in the correct order, so no replanting is needed. | ```python
import sys
input = lambda: sys.stdin.readline().rstrip()
from collections import deque,defaultdict,Counter
from itertools import permutations,combinations
from bisect import *
from heapq import *
from math import ceil,gcd,lcm,floor,comb
alph = 'abcdefghijklmnopqrstuvwxyz'
#pow(x,mod-2,mod)
N,M = map(int, input().split())
A = []
for _ in range(N):
s,x = input().split()
s = int(s)
x = float(x)
A.append((x,s))
A.sort()
A.append((A[-1][0]+1, M+1))
dp = [0]*(N+1)
for i in range(N+1):
x,s = A[i]
for j in range(i):
if A[j][1]<=s:
dp[i] = max(dp[i], dp[j]+1)
print(N-dp[-1])
``` | 3 | |
361 | B | Levko and Permutation | PROGRAMMING | 1,200 | [
"constructive algorithms",
"math",
"number theory"
] | null | null | Levko loves permutations very much. A permutation of length *n* is a sequence of distinct positive integers, each is at most *n*.
Let’s assume that value *gcd*(*a*,<=*b*) shows the greatest common divisor of numbers *a* and *b*. Levko assumes that element *p**i* of permutation *p*1,<=*p*2,<=... ,<=*p**n* is good if *gcd*(*i*,<=*p**i*)<=><=1. Levko considers a permutation beautiful, if it has exactly *k* good elements. Unfortunately, he doesn’t know any beautiful permutation. Your task is to help him to find at least one of them. | The single line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 0<=≤<=*k*<=≤<=*n*). | In a single line print either any beautiful permutation or -1, if such permutation doesn’t exist.
If there are multiple suitable permutations, you are allowed to print any of them. | [
"4 2\n",
"1 1\n"
] | [
"2 4 3 1",
"-1\n"
] | In the first sample elements 4 and 3 are good because *gcd*(2, 4) = 2 > 1 and *gcd*(3, 3) = 3 > 1. Elements 2 and 1 are not good because *gcd*(1, 2) = 1 and *gcd*(4, 1) = 1. As there are exactly 2 good elements, the permutation is beautiful.
The second sample has no beautiful permutations. | 1,000 | [
{
"input": "4 2",
"output": "2 1 3 4 "
},
{
"input": "1 1",
"output": "-1"
},
{
"input": "7 4",
"output": "3 1 2 4 5 6 7 "
},
{
"input": "10 9",
"output": "1 2 3 4 5 6 7 8 9 10 "
},
{
"input": "10000 5000",
"output": "5000 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1... | 1,613,038,674 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | a, b = map(int, input().split())
if a == b:
print(-1)
else :
print(b + 1, end = ' ')
for i in range(2, b + 2) :
print(i, end=' ')
for i in range(b + 2, a + 1):
print(i, end=' ')
print(1, end=' ') | Title: Levko and Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Levko loves permutations very much. A permutation of length *n* is a sequence of distinct positive integers, each is at most *n*.
Let’s assume that value *gcd*(*a*,<=*b*) shows the greatest common divisor of numbers *a* and *b*. Levko assumes that element *p**i* of permutation *p*1,<=*p*2,<=... ,<=*p**n* is good if *gcd*(*i*,<=*p**i*)<=><=1. Levko considers a permutation beautiful, if it has exactly *k* good elements. Unfortunately, he doesn’t know any beautiful permutation. Your task is to help him to find at least one of them.
Input Specification:
The single line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105, 0<=≤<=*k*<=≤<=*n*).
Output Specification:
In a single line print either any beautiful permutation or -1, if such permutation doesn’t exist.
If there are multiple suitable permutations, you are allowed to print any of them.
Demo Input:
['4 2\n', '1 1\n']
Demo Output:
['2 4 3 1', '-1\n']
Note:
In the first sample elements 4 and 3 are good because *gcd*(2, 4) = 2 > 1 and *gcd*(3, 3) = 3 > 1. Elements 2 and 1 are not good because *gcd*(1, 2) = 1 and *gcd*(4, 1) = 1. As there are exactly 2 good elements, the permutation is beautiful.
The second sample has no beautiful permutations. | ```python
a, b = map(int, input().split())
if a == b:
print(-1)
else :
print(b + 1, end = ' ')
for i in range(2, b + 2) :
print(i, end=' ')
for i in range(b + 2, a + 1):
print(i, end=' ')
print(1, end=' ')
``` | 0 | |
208 | A | Dubstep | PROGRAMMING | 900 | [
"strings"
] | null | null | Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song. | The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word. | Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space. | [
"WUBWUBABCWUB\n",
"WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n"
] | [
"ABC ",
"WE ARE THE CHAMPIONS MY FRIEND "
] | In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB". | 500 | [
{
"input": "WUBWUBABCWUB",
"output": "ABC "
},
{
"input": "WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB",
"output": "WE ARE THE CHAMPIONS MY FRIEND "
},
{
"input": "WUBWUBWUBSR",
"output": "SR "
},
{
"input": "RWUBWUBWUBLWUB",
"output": "R L "
},
{
"input": "... | 1,687,680,552 | 2,147,483,647 | Python 3 | OK | TESTS | 71 | 92 | 0 | x = input().split("WUB")
z = ''
for i in x:
if i == '':
continue
z += (i + ' ')
print(z)
| Title: Dubstep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya works as a DJ in the best Berland nightclub, and he often uses dubstep music in his performance. Recently, he has decided to take a couple of old songs and make dubstep remixes from them.
Let's assume that a song consists of some number of words. To make the dubstep remix of this song, Vasya inserts a certain number of words "WUB" before the first word of the song (the number may be zero), after the last word (the number may be zero), and between words (at least one between any pair of neighbouring words), and then the boy glues together all the words, including "WUB", in one string and plays the song at the club.
For example, a song with words "I AM X" can transform into a dubstep remix as "WUBWUBIWUBAMWUBWUBX" and cannot transform into "WUBWUBIAMWUBX".
Recently, Petya has heard Vasya's new dubstep track, but since he isn't into modern music, he decided to find out what was the initial song that Vasya remixed. Help Petya restore the original song.
Input Specification:
The input consists of a single non-empty string, consisting only of uppercase English letters, the string's length doesn't exceed 200 characters. It is guaranteed that before Vasya remixed the song, no word contained substring "WUB" in it; Vasya didn't change the word order. It is also guaranteed that initially the song had at least one word.
Output Specification:
Print the words of the initial song that Vasya used to make a dubsteb remix. Separate the words with a space.
Demo Input:
['WUBWUBABCWUB\n', 'WUBWEWUBAREWUBWUBTHEWUBCHAMPIONSWUBMYWUBFRIENDWUB\n']
Demo Output:
['ABC ', 'WE ARE THE CHAMPIONS MY FRIEND ']
Note:
In the first sample: "WUBWUBABCWUB" = "WUB" + "WUB" + "ABC" + "WUB". That means that the song originally consisted of a single word "ABC", and all words "WUB" were added by Vasya.
In the second sample Vasya added a single word "WUB" between all neighbouring words, in the beginning and in the end, except for words "ARE" and "THE" — between them Vasya added two "WUB". | ```python
x = input().split("WUB")
z = ''
for i in x:
if i == '':
continue
z += (i + ' ')
print(z)
``` | 3 | |
486 | A | Calculating Function | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | For a positive integer *n* let's define a function *f*:
*f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n*
Your task is to calculate *f*(*n*) for a given integer *n*. | The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015). | Print *f*(*n*) in a single line. | [
"4\n",
"5\n"
] | [
"2\n",
"-3\n"
] | *f*(4) = - 1 + 2 - 3 + 4 = 2
*f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3 | 500 | [
{
"input": "4",
"output": "2"
},
{
"input": "5",
"output": "-3"
},
{
"input": "1000000000",
"output": "500000000"
},
{
"input": "1000000001",
"output": "-500000001"
},
{
"input": "1000000000000000",
"output": "500000000000000"
},
{
"input": "100",
... | 1,693,927,038 | 2,147,483,647 | Python 3 | OK | TESTS | 39 | 46 | 0 | n=int(input())
if(n%2==0):
p=(int)(n/2)
else:
p= (int)(-(n+1)/2)
print(p) | Title: Calculating Function
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For a positive integer *n* let's define a function *f*:
*f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n*
Your task is to calculate *f*(*n*) for a given integer *n*.
Input Specification:
The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015).
Output Specification:
Print *f*(*n*) in a single line.
Demo Input:
['4\n', '5\n']
Demo Output:
['2\n', '-3\n']
Note:
*f*(4) = - 1 + 2 - 3 + 4 = 2
*f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3 | ```python
n=int(input())
if(n%2==0):
p=(int)(n/2)
else:
p= (int)(-(n+1)/2)
print(p)
``` | 3 | |
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... | 1,692,213,444 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 12 | 3,000 | 9,011,200 | n = int(input())
dms = []
for i in range(n):
name = input()
if name in dms:
dms.pop(dms.index(name))
dms.append(name)
for i in range(len(dms) - 1, - 1, - 1):
print(dms[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())
dms = []
for i in range(n):
name = input()
if name in dms:
dms.pop(dms.index(name))
dms.append(name)
for i in range(len(dms) - 1, - 1, - 1):
print(dms[i])
``` | 0 | |
486 | A | Calculating Function | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | For a positive integer *n* let's define a function *f*:
*f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n*
Your task is to calculate *f*(*n*) for a given integer *n*. | The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015). | Print *f*(*n*) in a single line. | [
"4\n",
"5\n"
] | [
"2\n",
"-3\n"
] | *f*(4) = - 1 + 2 - 3 + 4 = 2
*f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3 | 500 | [
{
"input": "4",
"output": "2"
},
{
"input": "5",
"output": "-3"
},
{
"input": "1000000000",
"output": "500000000"
},
{
"input": "1000000001",
"output": "-500000001"
},
{
"input": "1000000000000000",
"output": "500000000000000"
},
{
"input": "100",
... | 1,697,565,214 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | n = int(input()) # Read the input integer n
# Initialize the result to 0
result = 0
# Loop through numbers from 1 to n
for i in range(1, n + 1):
if i % 2 == 0:
# If i is even, subtract it from the result
result -= i
else:
# If i is odd, add it to the result
result += i
print(result) # Print the final result
| Title: Calculating Function
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For a positive integer *n* let's define a function *f*:
*f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n*
Your task is to calculate *f*(*n*) for a given integer *n*.
Input Specification:
The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015).
Output Specification:
Print *f*(*n*) in a single line.
Demo Input:
['4\n', '5\n']
Demo Output:
['2\n', '-3\n']
Note:
*f*(4) = - 1 + 2 - 3 + 4 = 2
*f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3 | ```python
n = int(input()) # Read the input integer n
# Initialize the result to 0
result = 0
# Loop through numbers from 1 to n
for i in range(1, n + 1):
if i % 2 == 0:
# If i is even, subtract it from the result
result -= i
else:
# If i is odd, add it to the result
result += i
print(result) # Print the final result
``` | 0 | |
102 | B | Sum of Digits | PROGRAMMING | 1,000 | [
"implementation"
] | B. Sum of Digits | 2 | 265 | Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit? | The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes. | Print the number of times a number can be replaced by the sum of its digits until it only contains one digit. | [
"0\n",
"10\n",
"991\n"
] | [
"0\n",
"1\n",
"3\n"
] | In the first sample the number already is one-digit — Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit. | 1,000 | [
{
"input": "0",
"output": "0"
},
{
"input": "10",
"output": "1"
},
{
"input": "991",
"output": "3"
},
{
"input": "99",
"output": "2"
},
{
"input": "100",
"output": "1"
},
{
"input": "123456789",
"output": "2"
},
{
"input": "32",
"output... | 1,674,611,029 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 51 | 124 | 1,638,400 | def fun(s) :
if len(s)==1 :
return 0
sum=0
for me in s :
sum+=int(me)
return 1+fun(str(sum))
s=input()
print(fun(s)) | Title: Sum of Digits
Time Limit: 2 seconds
Memory Limit: 265 megabytes
Problem Description:
Having watched the last Harry Potter film, little Gerald also decided to practice magic. He found in his father's magical book a spell that turns any number in the sum of its digits. At the moment Gerald learned that, he came across a number *n*. How many times can Gerald put a spell on it until the number becomes one-digit?
Input Specification:
The first line contains the only integer *n* (0<=≤<=*n*<=≤<=10100000). It is guaranteed that *n* doesn't contain any leading zeroes.
Output Specification:
Print the number of times a number can be replaced by the sum of its digits until it only contains one digit.
Demo Input:
['0\n', '10\n', '991\n']
Demo Output:
['0\n', '1\n', '3\n']
Note:
In the first sample the number already is one-digit — Herald can't cast a spell.
The second test contains number 10. After one casting of a spell it becomes 1, and here the process is completed. Thus, Gerald can only cast the spell once.
The third test contains number 991. As one casts a spell the following transformations take place: 991 → 19 → 10 → 1. After three transformations the number becomes one-digit. | ```python
def fun(s) :
if len(s)==1 :
return 0
sum=0
for me in s :
sum+=int(me)
return 1+fun(str(sum))
s=input()
print(fun(s))
``` | 3.966052 |
268 | A | Games | PROGRAMMING | 800 | [
"brute force"
] | null | null | Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different.
There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number.
You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question. | The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively. | In a single line print the number of games where the host team is going to play in the guest uniform. | [
"3\n1 2\n2 4\n3 4\n",
"4\n100 42\n42 100\n5 42\n100 5\n",
"2\n1 2\n1 2\n"
] | [
"1\n",
"5\n",
"0\n"
] | In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2.
In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first). | 500 | [
{
"input": "3\n1 2\n2 4\n3 4",
"output": "1"
},
{
"input": "4\n100 42\n42 100\n5 42\n100 5",
"output": "5"
},
{
"input": "2\n1 2\n1 2",
"output": "0"
},
{
"input": "7\n4 7\n52 55\n16 4\n55 4\n20 99\n3 4\n7 52",
"output": "6"
},
{
"input": "10\n68 42\n1 35\n25 70\n... | 1,695,948,164 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 92 | 0 | num_of_teams= eval(input())
home_wear=list()
guest_wear=list()
for x in range (0,num_of_teams):
home,guest=input().split(' ')
home_wear.append(home)
guest_wear.append(guest)
counter=0
for y in home_wear:
for z in guest_wear :
if y==z:
counter+=1
print(counter)
| Title: Games
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Manao works on a sports TV. He's spent much time watching the football games of some country. After a while he began to notice different patterns. For example, each team has two sets of uniforms: home uniform and guest uniform. When a team plays a game at home, the players put on the home uniform. When a team plays as a guest on somebody else's stadium, the players put on the guest uniform. The only exception to that rule is: when the home uniform color of the host team matches the guests' uniform, the host team puts on its guest uniform as well. For each team the color of the home and guest uniform is different.
There are *n* teams taking part in the national championship. The championship consists of *n*·(*n*<=-<=1) games: each team invites each other team to its stadium. At this point Manao wondered: how many times during the championship is a host team going to put on the guest uniform? Note that the order of the games does not affect this number.
You know the colors of the home and guest uniform for each team. For simplicity, the colors are numbered by integers in such a way that no two distinct colors have the same number. Help Manao find the answer to his question.
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=30). Each of the following *n* lines contains a pair of distinct space-separated integers *h**i*, *a**i* (1<=≤<=*h**i*,<=*a**i*<=≤<=100) — the colors of the *i*-th team's home and guest uniforms, respectively.
Output Specification:
In a single line print the number of games where the host team is going to play in the guest uniform.
Demo Input:
['3\n1 2\n2 4\n3 4\n', '4\n100 42\n42 100\n5 42\n100 5\n', '2\n1 2\n1 2\n']
Demo Output:
['1\n', '5\n', '0\n']
Note:
In the first test case the championship consists of 6 games. The only game with the event in question is the game between teams 2 and 1 on the stadium of team 2.
In the second test sample the host team will have to wear guest uniform in the games between teams: 1 and 2, 2 and 1, 2 and 3, 3 and 4, 4 and 2 (the host team is written first). | ```python
num_of_teams= eval(input())
home_wear=list()
guest_wear=list()
for x in range (0,num_of_teams):
home,guest=input().split(' ')
home_wear.append(home)
guest_wear.append(guest)
counter=0
for y in home_wear:
for z in guest_wear :
if y==z:
counter+=1
print(counter)
``` | 3 | |
379 | A | New Year Candles | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number. | The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000). | Print a single integer — the number of hours Vasily can light up the room for. | [
"4 2\n",
"6 3\n"
] | [
"7\n",
"8\n"
] | Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours. | 500 | [
{
"input": "4 2",
"output": "7"
},
{
"input": "6 3",
"output": "8"
},
{
"input": "1000 1000",
"output": "1001"
},
{
"input": "123 5",
"output": "153"
},
{
"input": "1000 2",
"output": "1999"
},
{
"input": "1 2",
"output": "1"
},
{
"input": ... | 1,681,970,719 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 46 | 0 | a,b=map(int,input().split())
hour=0
while a>0:
a=a-1
hour=hour+1
if hour%b==0:
a=a+1
print(hour)
| Title: New Year Candles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily the Programmer loves romance, so this year he decided to illuminate his room with candles.
Vasily has *a* candles.When Vasily lights up a new candle, it first burns for an hour and then it goes out. Vasily is smart, so he can make *b* went out candles into a new candle. As a result, this new candle can be used like any other new candle.
Now Vasily wonders: for how many hours can his candles light up the room if he acts optimally well? Help him find this number.
Input Specification:
The single line contains two integers, *a* and *b* (1<=≤<=*a*<=≤<=1000; 2<=≤<=*b*<=≤<=1000).
Output Specification:
Print a single integer — the number of hours Vasily can light up the room for.
Demo Input:
['4 2\n', '6 3\n']
Demo Output:
['7\n', '8\n']
Note:
Consider the first sample. For the first four hours Vasily lights up new candles, then he uses four burned out candles to make two new ones and lights them up. When these candles go out (stop burning), Vasily can make another candle. Overall, Vasily can light up the room for 7 hours. | ```python
a,b=map(int,input().split())
hour=0
while a>0:
a=a-1
hour=hour+1
if hour%b==0:
a=a+1
print(hour)
``` | 3 | |
892 | B | Wrath | PROGRAMMING | 1,200 | [
"greedy",
"implementation",
"two pointers"
] | null | null | Hands that shed innocent blood!
There are *n* guilty people in a line, the *i*-th of them holds a claw with length *L**i*. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the *i*-th person kills the *j*-th person if and only if *j*<=<<=*i* and *j*<=≥<=*i*<=-<=*L**i*.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=106) — the number of guilty people.
Second line contains *n* space-separated integers *L*1,<=*L*2,<=...,<=*L**n* (0<=≤<=*L**i*<=≤<=109), where *L**i* is the length of the *i*-th person's claw. | Print one integer — the total number of alive people after the bell rings. | [
"4\n0 1 0 10\n",
"2\n0 0\n",
"10\n1 1 3 0 0 0 2 1 0 3\n"
] | [
"1\n",
"2\n",
"3\n"
] | In first sample the last person kills everyone in front of him. | 1,000 | [
{
"input": "4\n0 1 0 10",
"output": "1"
},
{
"input": "2\n0 0",
"output": "2"
},
{
"input": "10\n1 1 3 0 0 0 2 1 0 3",
"output": "3"
},
{
"input": "10\n0 0 2 0 0 3 3 2 2 0",
"output": "2"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "5\n0 0 0 1 0"... | 1,528,148,144 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 11 | 1,029 | 114,995,200 | def main():
n = int(input())
arr = list(map(int, input().split()))
curIdx = len(arr) - 2
cnt = 1
maxIdx = curIdx + 1
while curIdx >= 0:
if arr[maxIdx] < arr[curIdx]:
maxIdx = curIdx
if arr[maxIdx] == 0:
cnt += 1
curIdx -= 1
arr[maxIdx] = max(arr[maxIdx] - 1, 0)
print(cnt)
if __name__ == "__main__":
main()
| Title: Wrath
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Hands that shed innocent blood!
There are *n* guilty people in a line, the *i*-th of them holds a claw with length *L**i*. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the *i*-th person kills the *j*-th person if and only if *j*<=<<=*i* and *j*<=≥<=*i*<=-<=*L**i*.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=106) — the number of guilty people.
Second line contains *n* space-separated integers *L*1,<=*L*2,<=...,<=*L**n* (0<=≤<=*L**i*<=≤<=109), where *L**i* is the length of the *i*-th person's claw.
Output Specification:
Print one integer — the total number of alive people after the bell rings.
Demo Input:
['4\n0 1 0 10\n', '2\n0 0\n', '10\n1 1 3 0 0 0 2 1 0 3\n']
Demo Output:
['1\n', '2\n', '3\n']
Note:
In first sample the last person kills everyone in front of him. | ```python
def main():
n = int(input())
arr = list(map(int, input().split()))
curIdx = len(arr) - 2
cnt = 1
maxIdx = curIdx + 1
while curIdx >= 0:
if arr[maxIdx] < arr[curIdx]:
maxIdx = curIdx
if arr[maxIdx] == 0:
cnt += 1
curIdx -= 1
arr[maxIdx] = max(arr[maxIdx] - 1, 0)
print(cnt)
if __name__ == "__main__":
main()
``` | 0 | |
218 | B | Airport | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Lolek and Bolek are about to travel abroad by plane. The local airport has a special "Choose Your Plane" offer. The offer's conditions are as follows:
- it is up to a passenger to choose a plane to fly on; - if the chosen plane has *x* (*x*<=><=0) empty seats at the given moment, then the ticket for such a plane costs *x* zlotys (units of Polish currency).
The only ticket office of the airport already has a queue of *n* passengers in front of it. Lolek and Bolek have not stood in the queue yet, but they are already wondering what is the maximum and the minimum number of zlotys the airport administration can earn if all *n* passengers buy tickets according to the conditions of this offer?
The passengers buy tickets in turn, the first person in the queue goes first, then goes the second one, and so on up to *n*-th person. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of passengers in the queue and the number of planes in the airport, correspondingly. The next line contains *m* integers *a*1,<=*a*2,<=...,<=*a**m* (1<=≤<=*a**i*<=≤<=1000) — *a**i* stands for the number of empty seats in the *i*-th plane before the ticket office starts selling tickets.
The numbers in the lines are separated by a space. It is guaranteed that there are at least *n* empty seats in total. | Print two integers — the maximum and the minimum number of zlotys that the airport administration can earn, correspondingly. | [
"4 3\n2 1 1\n",
"4 3\n2 2 2\n"
] | [
"5 5\n",
"7 6\n"
] | In the first test sample the number of passengers is equal to the number of empty seats, so regardless of the way the planes are chosen, the administration will earn the same sum.
In the second sample the sum is maximized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 2-nd plane, the 3-rd person — to the 3-rd plane, the 4-th person — to the 1-st plane. The sum is minimized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 1-st plane, the 3-rd person — to the 2-nd plane, the 4-th person — to the 2-nd plane. | 500 | [
{
"input": "4 3\n2 1 1",
"output": "5 5"
},
{
"input": "4 3\n2 2 2",
"output": "7 6"
},
{
"input": "10 5\n10 3 3 1 2",
"output": "58 26"
},
{
"input": "10 1\n10",
"output": "55 55"
},
{
"input": "10 1\n100",
"output": "955 955"
},
{
"input": "10 2\n4 7... | 1,694,948,737 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 33 | 154 | 2,252,800 | n, m = map(int, input().split())
AMAX = list(map(int, input().split()))
AMIN = []
for i in range(len(AMAX)):
AMIN.append(AMAX[i])
MAX = 0
MIN = 0
# MAX
for i in range(n):
max_index = 0
amax = AMAX[0]
for j in range(m):
if AMAX[j] > amax:
max_index = j
amax = AMAX[j]
AMAX[max_index] -= 1
MAX += amax
# MIN
for i in range(n):
amin = 999999999999
min_index = -1
for j in range(m):
if AMIN[j] < amin and AMIN[j] != 0:
min_index = j
amin = AMIN[j]
AMIN[min_index] -= 1
MIN += amin
print(MAX, MIN) | Title: Airport
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Lolek and Bolek are about to travel abroad by plane. The local airport has a special "Choose Your Plane" offer. The offer's conditions are as follows:
- it is up to a passenger to choose a plane to fly on; - if the chosen plane has *x* (*x*<=><=0) empty seats at the given moment, then the ticket for such a plane costs *x* zlotys (units of Polish currency).
The only ticket office of the airport already has a queue of *n* passengers in front of it. Lolek and Bolek have not stood in the queue yet, but they are already wondering what is the maximum and the minimum number of zlotys the airport administration can earn if all *n* passengers buy tickets according to the conditions of this offer?
The passengers buy tickets in turn, the first person in the queue goes first, then goes the second one, and so on up to *n*-th person.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of passengers in the queue and the number of planes in the airport, correspondingly. The next line contains *m* integers *a*1,<=*a*2,<=...,<=*a**m* (1<=≤<=*a**i*<=≤<=1000) — *a**i* stands for the number of empty seats in the *i*-th plane before the ticket office starts selling tickets.
The numbers in the lines are separated by a space. It is guaranteed that there are at least *n* empty seats in total.
Output Specification:
Print two integers — the maximum and the minimum number of zlotys that the airport administration can earn, correspondingly.
Demo Input:
['4 3\n2 1 1\n', '4 3\n2 2 2\n']
Demo Output:
['5 5\n', '7 6\n']
Note:
In the first test sample the number of passengers is equal to the number of empty seats, so regardless of the way the planes are chosen, the administration will earn the same sum.
In the second sample the sum is maximized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 2-nd plane, the 3-rd person — to the 3-rd plane, the 4-th person — to the 1-st plane. The sum is minimized if the 1-st person in the queue buys a ticket to the 1-st plane, the 2-nd person — to the 1-st plane, the 3-rd person — to the 2-nd plane, the 4-th person — to the 2-nd plane. | ```python
n, m = map(int, input().split())
AMAX = list(map(int, input().split()))
AMIN = []
for i in range(len(AMAX)):
AMIN.append(AMAX[i])
MAX = 0
MIN = 0
# MAX
for i in range(n):
max_index = 0
amax = AMAX[0]
for j in range(m):
if AMAX[j] > amax:
max_index = j
amax = AMAX[j]
AMAX[max_index] -= 1
MAX += amax
# MIN
for i in range(n):
amin = 999999999999
min_index = -1
for j in range(m):
if AMIN[j] < amin and AMIN[j] != 0:
min_index = j
amin = AMIN[j]
AMIN[min_index] -= 1
MIN += amin
print(MAX, MIN)
``` | 3 | |
892 | B | Wrath | PROGRAMMING | 1,200 | [
"greedy",
"implementation",
"two pointers"
] | null | null | Hands that shed innocent blood!
There are *n* guilty people in a line, the *i*-th of them holds a claw with length *L**i*. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the *i*-th person kills the *j*-th person if and only if *j*<=<<=*i* and *j*<=≥<=*i*<=-<=*L**i*.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=106) — the number of guilty people.
Second line contains *n* space-separated integers *L*1,<=*L*2,<=...,<=*L**n* (0<=≤<=*L**i*<=≤<=109), where *L**i* is the length of the *i*-th person's claw. | Print one integer — the total number of alive people after the bell rings. | [
"4\n0 1 0 10\n",
"2\n0 0\n",
"10\n1 1 3 0 0 0 2 1 0 3\n"
] | [
"1\n",
"2\n",
"3\n"
] | In first sample the last person kills everyone in front of him. | 1,000 | [
{
"input": "4\n0 1 0 10",
"output": "1"
},
{
"input": "2\n0 0",
"output": "2"
},
{
"input": "10\n1 1 3 0 0 0 2 1 0 3",
"output": "3"
},
{
"input": "10\n0 0 2 0 0 3 3 2 2 0",
"output": "2"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "5\n0 0 0 1 0"... | 1,611,819,149 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 8 | 2,000 | 108,134,400 | n= int(input())
t = list(map(int,input().split()))
Li = [i-t[i] for i in range(n) ]
p=0
for k in range(n):
if k<n-1:
a=min(Li[k+1:])
if k>=a:
p+=1
print(n-p)
| Title: Wrath
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Hands that shed innocent blood!
There are *n* guilty people in a line, the *i*-th of them holds a claw with length *L**i*. The bell rings and every person kills some of people in front of him. All people kill others at the same time. Namely, the *i*-th person kills the *j*-th person if and only if *j*<=<<=*i* and *j*<=≥<=*i*<=-<=*L**i*.
You are given lengths of the claws. You need to find the total number of alive people after the bell rings.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=106) — the number of guilty people.
Second line contains *n* space-separated integers *L*1,<=*L*2,<=...,<=*L**n* (0<=≤<=*L**i*<=≤<=109), where *L**i* is the length of the *i*-th person's claw.
Output Specification:
Print one integer — the total number of alive people after the bell rings.
Demo Input:
['4\n0 1 0 10\n', '2\n0 0\n', '10\n1 1 3 0 0 0 2 1 0 3\n']
Demo Output:
['1\n', '2\n', '3\n']
Note:
In first sample the last person kills everyone in front of him. | ```python
n= int(input())
t = list(map(int,input().split()))
Li = [i-t[i] for i in range(n) ]
p=0
for k in range(n):
if k<n-1:
a=min(Li[k+1:])
if k>=a:
p+=1
print(n-p)
``` | 0 | |
120 | C | Winnie-the-Pooh and honey | PROGRAMMING | 1,100 | [
"implementation",
"math"
] | null | null | As we all know, Winnie-the-Pooh just adores honey. Ones he and the Piglet found out that the Rabbit has recently gotten hold of an impressive amount of this sweet and healthy snack. As you may guess, Winnie and the Piglet asked to come at the Rabbit's place. Thus, there are *n* jars of honey lined up in front of Winnie-the-Pooh, jar number *i* contains *a**i* kilos of honey. Winnie-the-Pooh eats the honey like that: each time he chooses a jar containing most honey. If the jar has less that *k* kilos of honey or if Winnie-the-Pooh has already eaten from it three times, he gives the jar to Piglet. Otherwise he eats exactly *k* kilos of honey from the jar and puts it back. Winnie does so until he gives all jars to the Piglet. Count how much honey Piglet will overall get after Winnie satisfies his hunger. | The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*k*<=≤<=100). The second line contains *n* integers *a*1, *a*2, ..., *a**n*, separated by spaces (1<=≤<=*a**i*<=≤<=100). | Print a single number — how many kilos of honey gets Piglet. | [
"3 3\n15 8 10\n"
] | [
"9\n"
] | none | 0 | [
{
"input": "3 3\n15 8 10",
"output": "9"
},
{
"input": "1 3\n3",
"output": "0"
},
{
"input": "3 4\n3 8 2",
"output": "5"
},
{
"input": "3 2\n95 25 49",
"output": "151"
},
{
"input": "3 1\n8 3 2",
"output": "5"
},
{
"input": "5 1\n4 7 9 5 7",
"outpu... | 1,660,523,005 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 20 | 124 | 0 | with open("input.txt", "r") as fr:
n, k = map(int, fr.readline().split())
a = list(map(int, fr.readline().split()))
for i in range(n):
for j in range(3):
if a[i] < k:
break
a[i] -= k
with open("output.txt", "w") as fw:
fw.write(str(sum(a)))
| Title: Winnie-the-Pooh and honey
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As we all know, Winnie-the-Pooh just adores honey. Ones he and the Piglet found out that the Rabbit has recently gotten hold of an impressive amount of this sweet and healthy snack. As you may guess, Winnie and the Piglet asked to come at the Rabbit's place. Thus, there are *n* jars of honey lined up in front of Winnie-the-Pooh, jar number *i* contains *a**i* kilos of honey. Winnie-the-Pooh eats the honey like that: each time he chooses a jar containing most honey. If the jar has less that *k* kilos of honey or if Winnie-the-Pooh has already eaten from it three times, he gives the jar to Piglet. Otherwise he eats exactly *k* kilos of honey from the jar and puts it back. Winnie does so until he gives all jars to the Piglet. Count how much honey Piglet will overall get after Winnie satisfies his hunger.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*k*<=≤<=100). The second line contains *n* integers *a*1, *a*2, ..., *a**n*, separated by spaces (1<=≤<=*a**i*<=≤<=100).
Output Specification:
Print a single number — how many kilos of honey gets Piglet.
Demo Input:
['3 3\n15 8 10\n']
Demo Output:
['9\n']
Note:
none | ```python
with open("input.txt", "r") as fr:
n, k = map(int, fr.readline().split())
a = list(map(int, fr.readline().split()))
for i in range(n):
for j in range(3):
if a[i] < k:
break
a[i] -= k
with open("output.txt", "w") as fw:
fw.write(str(sum(a)))
``` | 3 | |
588 | A | Duff and Meat | PROGRAMMING | 900 | [
"greedy"
] | null | null | Duff is addicted to meat! Malek wants to keep her happy for *n* days. In order to be happy in *i*-th day, she needs to eat exactly *a**i* kilograms of meat.
There is a big shop uptown and Malek wants to buy meat for her from there. In *i*-th day, they sell meat for *p**i* dollars per kilogram. Malek knows all numbers *a*1,<=...,<=*a**n* and *p*1,<=...,<=*p**n*. In each day, he can buy arbitrary amount of meat, also he can keep some meat he has for the future.
Malek is a little tired from cooking meat, so he asked for your help. Help him to minimize the total money he spends to keep Duff happy for *n* days. | The first line of input contains integer *n* (1<=≤<=*n*<=≤<=105), the number of days.
In the next *n* lines, *i*-th line contains two integers *a**i* and *p**i* (1<=≤<=*a**i*,<=*p**i*<=≤<=100), the amount of meat Duff needs and the cost of meat in that day. | Print the minimum money needed to keep Duff happy for *n* days, in one line. | [
"3\n1 3\n2 2\n3 1\n",
"3\n1 3\n2 1\n3 2\n"
] | [
"10\n",
"8\n"
] | In the first sample case: An optimal way would be to buy 1 kg on the first day, 2 kg on the second day and 3 kg on the third day.
In the second sample case: An optimal way would be to buy 1 kg on the first day and 5 kg (needed meat for the second and third day) on the second day. | 750 | [
{
"input": "3\n1 3\n2 2\n3 1",
"output": "10"
},
{
"input": "3\n1 3\n2 1\n3 2",
"output": "8"
},
{
"input": "1\n39 52",
"output": "2028"
},
{
"input": "2\n25 56\n94 17",
"output": "2998"
},
{
"input": "5\n39 21\n95 89\n73 90\n9 55\n85 32",
"output": "6321"
}... | 1,549,450,009 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 389 | 0 | N = int(input())
Lowest = 100000
SUM = 0
for i in range(N):
X = list(map(int, input().split()))
Lowest = min(Lowest, X[1])
SUM += X[0] * Lowest
print(SUM)
| Title: Duff and Meat
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Duff is addicted to meat! Malek wants to keep her happy for *n* days. In order to be happy in *i*-th day, she needs to eat exactly *a**i* kilograms of meat.
There is a big shop uptown and Malek wants to buy meat for her from there. In *i*-th day, they sell meat for *p**i* dollars per kilogram. Malek knows all numbers *a*1,<=...,<=*a**n* and *p*1,<=...,<=*p**n*. In each day, he can buy arbitrary amount of meat, also he can keep some meat he has for the future.
Malek is a little tired from cooking meat, so he asked for your help. Help him to minimize the total money he spends to keep Duff happy for *n* days.
Input Specification:
The first line of input contains integer *n* (1<=≤<=*n*<=≤<=105), the number of days.
In the next *n* lines, *i*-th line contains two integers *a**i* and *p**i* (1<=≤<=*a**i*,<=*p**i*<=≤<=100), the amount of meat Duff needs and the cost of meat in that day.
Output Specification:
Print the minimum money needed to keep Duff happy for *n* days, in one line.
Demo Input:
['3\n1 3\n2 2\n3 1\n', '3\n1 3\n2 1\n3 2\n']
Demo Output:
['10\n', '8\n']
Note:
In the first sample case: An optimal way would be to buy 1 kg on the first day, 2 kg on the second day and 3 kg on the third day.
In the second sample case: An optimal way would be to buy 1 kg on the first day and 5 kg (needed meat for the second and third day) on the second day. | ```python
N = int(input())
Lowest = 100000
SUM = 0
for i in range(N):
X = list(map(int, input().split()))
Lowest = min(Lowest, X[1])
SUM += X[0] * Lowest
print(SUM)
``` | 3 | |
82 | A | Double Cola | PROGRAMMING | 1,100 | [
"implementation",
"math"
] | A. Double Cola | 1 | 256 | Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon. | The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers. | Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially. | [
"1\n",
"6\n",
"1802\n"
] | [
"Sheldon\n",
"Sheldon\n",
"Penny\n"
] | none | 500 | [
{
"input": "1",
"output": "Sheldon"
},
{
"input": "6",
"output": "Sheldon"
},
{
"input": "1802",
"output": "Penny"
},
{
"input": "1",
"output": "Sheldon"
},
{
"input": "2",
"output": "Leonard"
},
{
"input": "3",
"output": "Penny"
},
{
"inpu... | 1,629,212,623 | 2,147,483,647 | Python 3 | OK | TESTS | 41 | 62 | 6,758,400 | n = int(input())
peeps = ["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
if n<6:
print(peeps[n-1])
else:
r = 1
while r*5 < n:
n -= r*5
r *= 2
print(peeps[(n-1)//r])
| Title: Double Cola
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input Specification:
The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output Specification:
Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Demo Input:
['1\n', '6\n', '1802\n']
Demo Output:
['Sheldon\n', 'Sheldon\n', 'Penny\n']
Note:
none | ```python
n = int(input())
peeps = ["Sheldon", "Leonard", "Penny", "Rajesh", "Howard"]
if n<6:
print(peeps[n-1])
else:
r = 1
while r*5 < n:
n -= r*5
r *= 2
print(peeps[(n-1)//r])
``` | 3.956411 |
928 | C | Dependency management | PROGRAMMING | 1,900 | [
"*special",
"graphs",
"implementation"
] | null | null | Polycarp is currently developing a project in Vaja language and using a popular dependency management system called Vamen. From Vamen's point of view both Vaja project and libraries are treated projects for simplicity.
A project in Vaja has its own uniqie non-empty name consisting of lowercase latin letters with length not exceeding 10 and version — positive integer from 1 to 106. Each project (keep in mind that it is determined by both its name and version) might depend on other projects. For sure, there are no cyclic dependencies.
You're given a list of project descriptions. The first of the given projects is the one being developed by Polycarp at this moment. Help Polycarp determine all projects that his project depends on (directly or via a certain chain).
It's possible that Polycarp's project depends on two different versions of some project. In this case collision resolving is applied, i.e. for each such project the system chooses the version that minimizes the distance from it to Polycarp's project. If there are several options, the newer (with the maximum version) is preferred. This version is considered actual; other versions and their dependencies are ignored.
More formal, choose such a set of projects of minimum possible size that the following conditions hold:
- Polycarp's project is chosen; - Polycarp's project depends (directly or indirectly) on all other projects in the set; - no two projects share the name; - for each project *x* that some other project in the set depends on we have either *x* or some *y* with other version and shorter chain to Polycarp's project chosen. In case of ties the newer one is chosen.
Output all Polycarp's project's dependencies (Polycarp's project itself should't be printed) in lexicographical order. | The first line contains an only integer *n* (1<=≤<=*n*<=≤<=1<=000) — the number of projects in Vaja.
The following lines contain the project descriptions. Each project is described by a line consisting of its name and version separated by space. The next line gives the number of direct dependencies (from 0 to *n*<=-<=1) and the dependencies themselves (one in a line) in arbitrary order. Each dependency is specified by its name and version. The projects are also given in arbitrary order, but the first of them is always Polycarp's. Project descriptions are separated by one empty line. Refer to samples for better understanding.
It's guaranteed that there are no cyclic dependencies. | Output all Polycarp's project's dependencies in lexicographical order. | [
"4\na 3\n2\nb 1\nc 1\n \nb 2\n0\n \nb 1\n1\nb 2\n \nc 1\n1\nb 2\n",
"9\ncodehorses 5\n3\nwebfrmk 6\nmashadb 1\nmashadb 2\n \ncommons 2\n0\n \nmashadb 3\n0\n \nwebfrmk 6\n2\nmashadb 3\ncommons 2\n \nextra 4\n1\nextra 3\n \nextra 3\n0\n \nextra 1\n0\n \nmashadb 1\n1\nextra 3\n \nmashadb 2\n1\nextra 1\n",
"3\nabc ... | [
"2\nb 1\nc 1\n",
"4\ncommons 2\nextra 1\nmashadb 2\nwebfrmk 6\n",
"1\ncba 2\n"
] | The first sample is given in the pic below. Arrow from *A* to *B* means that *B* directly depends on *A*. Projects that Polycarp's project «a» (version 3) depends on are painted black.
The second sample is again given in the pic below. Arrow from *A* to *B* means that *B* directly depends on *A*. Projects that Polycarp's project «codehorses» (version 5) depends on are paint it black. Note that «extra 1» is chosen instead of «extra 3» since «mashadb 1» and all of its dependencies are ignored due to «mashadb 2». | 2,000 | [
{
"input": "4\na 3\n2\nb 1\nc 1\n\nb 2\n0\n\nb 1\n1\nb 2\n\nc 1\n1\nb 2",
"output": "2\nb 1\nc 1"
},
{
"input": "9\ncodehorses 5\n3\nwebfrmk 6\nmashadb 1\nmashadb 2\n\ncommons 2\n0\n\nmashadb 3\n0\n\nwebfrmk 6\n2\nmashadb 3\ncommons 2\n\nextra 4\n1\nextra 3\n\nextra 3\n0\n\nextra 1\n0\n\nmashadb 1\n... | 1,519,551,517 | 65,017 | Python 3 | WRONG_ANSWER | PRETESTS | 8 | 62 | 5,632,000 | def get_project(s):
s = s.split(' ')
s[1] = int(s[1])
return tuple(s)
n = int(input())
projects = {}
main = get_project(input())
projects[main] = []
m = int(input())
for j in range(m):
projects[main].append(get_project(input()))
for i in range(1, n):
input()
project = get_project(input())
projects[project] = []
m = int(input())
for j in range(m):
projects[project].append(get_project(input()))
# print(projects)
def BFS(graph, start):
depends = []
INF = sum([len(graph[i]) for i in graph.keys()]) // 2 + 1
weight = dict()
weight[start] = -1
mark = {i: False for i in graph.keys()}
mark[start] = True
path = [(start, start)]
while path:
vertex, prev = path.pop(0)
weight[vertex] = weight[prev] + 1
exist = None
for i in depends:
if i[0] == vertex[0]:
exist = i
# print(vertex, exist)
if not exist:
depends.append(vertex)
elif weight[exist] == weight[vertex]:
if exist[1] < vertex[1]:
depends.remove(exist)
for i in projects[exist]:
path.remove((i, exist))
projects[exist] = []
depends.append(vertex)
else:
graph[vertex] = []
for v in graph[vertex]:
if not mark[v]:
mark[v] = True
path.append((v, vertex))
return depends
depends = BFS(projects, main)
depends.remove(main)
depends.sort(key = lambda item: item[0])
print(len(depends))
for i in depends:
print(i[0], str(i[1]))
| Title: Dependency management
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is currently developing a project in Vaja language and using a popular dependency management system called Vamen. From Vamen's point of view both Vaja project and libraries are treated projects for simplicity.
A project in Vaja has its own uniqie non-empty name consisting of lowercase latin letters with length not exceeding 10 and version — positive integer from 1 to 106. Each project (keep in mind that it is determined by both its name and version) might depend on other projects. For sure, there are no cyclic dependencies.
You're given a list of project descriptions. The first of the given projects is the one being developed by Polycarp at this moment. Help Polycarp determine all projects that his project depends on (directly or via a certain chain).
It's possible that Polycarp's project depends on two different versions of some project. In this case collision resolving is applied, i.e. for each such project the system chooses the version that minimizes the distance from it to Polycarp's project. If there are several options, the newer (with the maximum version) is preferred. This version is considered actual; other versions and their dependencies are ignored.
More formal, choose such a set of projects of minimum possible size that the following conditions hold:
- Polycarp's project is chosen; - Polycarp's project depends (directly or indirectly) on all other projects in the set; - no two projects share the name; - for each project *x* that some other project in the set depends on we have either *x* or some *y* with other version and shorter chain to Polycarp's project chosen. In case of ties the newer one is chosen.
Output all Polycarp's project's dependencies (Polycarp's project itself should't be printed) in lexicographical order.
Input Specification:
The first line contains an only integer *n* (1<=≤<=*n*<=≤<=1<=000) — the number of projects in Vaja.
The following lines contain the project descriptions. Each project is described by a line consisting of its name and version separated by space. The next line gives the number of direct dependencies (from 0 to *n*<=-<=1) and the dependencies themselves (one in a line) in arbitrary order. Each dependency is specified by its name and version. The projects are also given in arbitrary order, but the first of them is always Polycarp's. Project descriptions are separated by one empty line. Refer to samples for better understanding.
It's guaranteed that there are no cyclic dependencies.
Output Specification:
Output all Polycarp's project's dependencies in lexicographical order.
Demo Input:
['4\na 3\n2\nb 1\nc 1\n\xa0\nb 2\n0\n\xa0\nb 1\n1\nb 2\n\xa0\nc 1\n1\nb 2\n', '9\ncodehorses 5\n3\nwebfrmk 6\nmashadb 1\nmashadb 2\n\xa0\ncommons 2\n0\n\xa0\nmashadb 3\n0\n\xa0\nwebfrmk 6\n2\nmashadb 3\ncommons 2\n\xa0\nextra 4\n1\nextra 3\n\xa0\nextra 3\n0\n\xa0\nextra 1\n0\n\xa0\nmashadb 1\n1\nextra 3\n\xa0\nmashadb 2\n1\nextra 1\n', '3\nabc 1\n2\nabc 3\ncba 2\n\nabc 3\n0\n\ncba 2\n0\n']
Demo Output:
['2\nb 1\nc 1\n', '4\ncommons 2\nextra 1\nmashadb 2\nwebfrmk 6\n', '1\ncba 2\n']
Note:
The first sample is given in the pic below. Arrow from *A* to *B* means that *B* directly depends on *A*. Projects that Polycarp's project «a» (version 3) depends on are painted black.
The second sample is again given in the pic below. Arrow from *A* to *B* means that *B* directly depends on *A*. Projects that Polycarp's project «codehorses» (version 5) depends on are paint it black. Note that «extra 1» is chosen instead of «extra 3» since «mashadb 1» and all of its dependencies are ignored due to «mashadb 2». | ```python
def get_project(s):
s = s.split(' ')
s[1] = int(s[1])
return tuple(s)
n = int(input())
projects = {}
main = get_project(input())
projects[main] = []
m = int(input())
for j in range(m):
projects[main].append(get_project(input()))
for i in range(1, n):
input()
project = get_project(input())
projects[project] = []
m = int(input())
for j in range(m):
projects[project].append(get_project(input()))
# print(projects)
def BFS(graph, start):
depends = []
INF = sum([len(graph[i]) for i in graph.keys()]) // 2 + 1
weight = dict()
weight[start] = -1
mark = {i: False for i in graph.keys()}
mark[start] = True
path = [(start, start)]
while path:
vertex, prev = path.pop(0)
weight[vertex] = weight[prev] + 1
exist = None
for i in depends:
if i[0] == vertex[0]:
exist = i
# print(vertex, exist)
if not exist:
depends.append(vertex)
elif weight[exist] == weight[vertex]:
if exist[1] < vertex[1]:
depends.remove(exist)
for i in projects[exist]:
path.remove((i, exist))
projects[exist] = []
depends.append(vertex)
else:
graph[vertex] = []
for v in graph[vertex]:
if not mark[v]:
mark[v] = True
path.append((v, vertex))
return depends
depends = BFS(projects, main)
depends.remove(main)
depends.sort(key = lambda item: item[0])
print(len(depends))
for i in depends:
print(i[0], str(i[1]))
``` | 0 | |
316 | A1 | Special Task | PROGRAMMING | 1,100 | [
"greedy"
] | null | null | Special Agent Smart Beaver works in a secret research department of ABBYY. He's been working there for a long time and is satisfied with his job, as it allows him to eat out in the best restaurants and order the most expensive and exotic wood types there.
The content special agent has got an important task: to get the latest research by British scientists on the English Language. These developments are encoded and stored in a large safe. The Beaver's teeth are strong enough, so the authorities assured that upon arriving at the place the beaver won't have any problems with opening the safe.
And he finishes his aspen sprig and leaves for this important task. Of course, the Beaver arrived at the location without any problems, but alas. He can't open the safe with his strong and big teeth. At this point, the Smart Beaver get a call from the headquarters and learns that opening the safe with the teeth is not necessary, as a reliable source has sent the following information: the safe code consists of digits and has no leading zeroes. There also is a special hint, which can be used to open the safe. The hint is string *s* with the following structure:
- if *s**i* = "?", then the digit that goes *i*-th in the safe code can be anything (between 0 to 9, inclusively); - if *s**i* is a digit (between 0 to 9, inclusively), then it means that there is digit *s**i* on position *i* in code; - if the string contains letters from "A" to "J", then all positions with the same letters must contain the same digits and the positions with distinct letters must contain distinct digits. - The length of the safe code coincides with the length of the hint.
For example, hint "?JGJ9" has such matching safe code variants: "51919", "55959", "12329", "93539" and so on, and has wrong variants such as: "56669", "00111", "03539" and "13666".
After receiving such information, the authorities change the plan and ask the special agents to work quietly and gently and not to try to open the safe by mechanical means, and try to find the password using the given hint.
At a special agent school the Smart Beaver was the fastest in his platoon finding codes for such safes, but now he is not in that shape: the years take their toll ... Help him to determine the number of possible variants of the code to the safe, matching the given hint. After receiving this information, and knowing his own speed of entering codes, the Smart Beaver will be able to determine whether he will have time for tonight's show "Beavers are on the trail" on his favorite TV channel, or he should work for a sleepless night... | The first line contains string *s* — the hint to the safe code. String *s* consists of the following characters: ?, 0-9, A-J. It is guaranteed that the first character of string *s* doesn't equal to character 0.
The input limits for scoring 30 points are (subproblem A1):
- 1<=≤<=|*s*|<=≤<=5.
The input limits for scoring 100 points are (subproblems A1+A2):
- 1<=≤<=|*s*|<=≤<=105.
Here |*s*| means the length of string *s*. | Print the number of codes that match the given hint. | [
"AJ\n",
"1?AA\n"
] | [
"81\n",
"100\n"
] | none | 30 | [
{
"input": "AJ",
"output": "81"
},
{
"input": "1?AA",
"output": "100"
},
{
"input": "?",
"output": "9"
},
{
"input": "7",
"output": "1"
},
{
"input": "A",
"output": "9"
},
{
"input": "BBB?",
"output": "90"
},
{
"input": "BC??",
"output"... | 1,371,707,526 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS1 | 0 | 0 | 0 | def p(x,n):
ans=1
if n==0:
return ans
for i in range(n):
ans*=x-i
return ans;
lest={'A','B','C','D','E','F','G','H','I','J'}
str=raw_input("")
k=0;l=0;ans1=0;
if str[0] in lest:
is_it=True
for c in str:
for L in lest:
if L==c:
l+=1
lest.remove(L)
break
for c in str:
if c=='?':
k+=1
if is_it:
ans1=p(9,l-1)
ans=(10**k)*p(10,l)-ans1
if str[0]== "?":
ans=ans*0.9
print ans | Title: Special Task
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Special Agent Smart Beaver works in a secret research department of ABBYY. He's been working there for a long time and is satisfied with his job, as it allows him to eat out in the best restaurants and order the most expensive and exotic wood types there.
The content special agent has got an important task: to get the latest research by British scientists on the English Language. These developments are encoded and stored in a large safe. The Beaver's teeth are strong enough, so the authorities assured that upon arriving at the place the beaver won't have any problems with opening the safe.
And he finishes his aspen sprig and leaves for this important task. Of course, the Beaver arrived at the location without any problems, but alas. He can't open the safe with his strong and big teeth. At this point, the Smart Beaver get a call from the headquarters and learns that opening the safe with the teeth is not necessary, as a reliable source has sent the following information: the safe code consists of digits and has no leading zeroes. There also is a special hint, which can be used to open the safe. The hint is string *s* with the following structure:
- if *s**i* = "?", then the digit that goes *i*-th in the safe code can be anything (between 0 to 9, inclusively); - if *s**i* is a digit (between 0 to 9, inclusively), then it means that there is digit *s**i* on position *i* in code; - if the string contains letters from "A" to "J", then all positions with the same letters must contain the same digits and the positions with distinct letters must contain distinct digits. - The length of the safe code coincides with the length of the hint.
For example, hint "?JGJ9" has such matching safe code variants: "51919", "55959", "12329", "93539" and so on, and has wrong variants such as: "56669", "00111", "03539" and "13666".
After receiving such information, the authorities change the plan and ask the special agents to work quietly and gently and not to try to open the safe by mechanical means, and try to find the password using the given hint.
At a special agent school the Smart Beaver was the fastest in his platoon finding codes for such safes, but now he is not in that shape: the years take their toll ... Help him to determine the number of possible variants of the code to the safe, matching the given hint. After receiving this information, and knowing his own speed of entering codes, the Smart Beaver will be able to determine whether he will have time for tonight's show "Beavers are on the trail" on his favorite TV channel, or he should work for a sleepless night...
Input Specification:
The first line contains string *s* — the hint to the safe code. String *s* consists of the following characters: ?, 0-9, A-J. It is guaranteed that the first character of string *s* doesn't equal to character 0.
The input limits for scoring 30 points are (subproblem A1):
- 1<=≤<=|*s*|<=≤<=5.
The input limits for scoring 100 points are (subproblems A1+A2):
- 1<=≤<=|*s*|<=≤<=105.
Here |*s*| means the length of string *s*.
Output Specification:
Print the number of codes that match the given hint.
Demo Input:
['AJ\n', '1?AA\n']
Demo Output:
['81\n', '100\n']
Note:
none | ```python
def p(x,n):
ans=1
if n==0:
return ans
for i in range(n):
ans*=x-i
return ans;
lest={'A','B','C','D','E','F','G','H','I','J'}
str=raw_input("")
k=0;l=0;ans1=0;
if str[0] in lest:
is_it=True
for c in str:
for L in lest:
if L==c:
l+=1
lest.remove(L)
break
for c in str:
if c=='?':
k+=1
if is_it:
ans1=p(9,l-1)
ans=(10**k)*p(10,l)-ans1
if str[0]== "?":
ans=ans*0.9
print ans
``` | -1 | |
110 | A | Nearly Lucky Number | PROGRAMMING | 800 | [
"implementation"
] | A. Nearly Lucky Number | 2 | 256 | Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Unfortunately, not all numbers are lucky. Petya calls a number nearly lucky if the number of lucky digits in it is a lucky number. He wonders whether number *n* is a nearly lucky number. | The only line contains an integer *n* (1<=≤<=*n*<=≤<=1018).
Please do not use the %lld specificator to read or write 64-bit numbers in С++. It is preferred to use the cin, cout streams or the %I64d specificator. | Print on the single line "YES" if *n* is a nearly lucky number. Otherwise, print "NO" (without the quotes). | [
"40047\n",
"7747774\n",
"1000000000000000000\n"
] | [
"NO\n",
"YES\n",
"NO\n"
] | In the first sample there are 3 lucky digits (first one and last two), so the answer is "NO".
In the second sample there are 7 lucky digits, 7 is lucky number, so the answer is "YES".
In the third sample there are no lucky digits, so the answer is "NO". | 500 | [
{
"input": "40047",
"output": "NO"
},
{
"input": "7747774",
"output": "YES"
},
{
"input": "1000000000000000000",
"output": "NO"
},
{
"input": "7",
"output": "NO"
},
{
"input": "4",
"output": "NO"
},
{
"input": "474404774",
"output": "NO"
},
{
... | 1,695,317,659 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 34 | 154 | 0 | def n_lucky(n):
c = 0
for d in str(n):
if d == '4' or d == '7':
c += 1
return lucky(c)
def lucky(n):
lucky_d = ['4', '7']
for d in str(n):
if d not in lucky_d:
return "NO"
return "YES"
n = int(input())
result = n_lucky(n)
print(result) | Title: Nearly Lucky Number
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Unfortunately, not all numbers are lucky. Petya calls a number nearly lucky if the number of lucky digits in it is a lucky number. He wonders whether number *n* is a nearly lucky number.
Input Specification:
The only line contains an integer *n* (1<=≤<=*n*<=≤<=1018).
Please do not use the %lld specificator to read or write 64-bit numbers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output Specification:
Print on the single line "YES" if *n* is a nearly lucky number. Otherwise, print "NO" (without the quotes).
Demo Input:
['40047\n', '7747774\n', '1000000000000000000\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
In the first sample there are 3 lucky digits (first one and last two), so the answer is "NO".
In the second sample there are 7 lucky digits, 7 is lucky number, so the answer is "YES".
In the third sample there are no lucky digits, so the answer is "NO". | ```python
def n_lucky(n):
c = 0
for d in str(n):
if d == '4' or d == '7':
c += 1
return lucky(c)
def lucky(n):
lucky_d = ['4', '7']
for d in str(n):
if d not in lucky_d:
return "NO"
return "YES"
n = int(input())
result = n_lucky(n)
print(result)
``` | 3.9615 |
656 | G | You're a Professional | PROGRAMMING | 1,900 | [
"*special"
] | null | null | A simple recommendation system would recommend a user things liked by a certain number of their friends. In this problem you will implement part of such a system.
You are given user's friends' opinions about a list of items. You are also given a threshold *T* — the minimal number of "likes" necessary for an item to be recommended to the user.
Output the number of items in the list liked by at least *T* of user's friends. | The first line of the input will contain three space-separated integers: the number of friends *F* (1<=≤<=*F*<=≤<=10), the number of items *I* (1<=≤<=*I*<=≤<=10) and the threshold *T* (1<=≤<=*T*<=≤<=*F*).
The following *F* lines of input contain user's friends' opinions. *j*-th character of *i*-th line is 'Y' if *i*-th friend likes *j*-th item, and 'N' otherwise. | Output an integer — the number of items liked by at least *T* of user's friends. | [
"3 3 2\nYYY\nNNN\nYNY\n",
"4 4 1\nNNNY\nNNYN\nNYNN\nYNNN\n"
] | [
"2\n",
"4\n"
] | none | 0 | [
{
"input": "3 3 2\nYYY\nNNN\nYNY",
"output": "2"
},
{
"input": "4 4 1\nNNNY\nNNYN\nNYNN\nYNNN",
"output": "4"
},
{
"input": "3 5 2\nNYNNY\nYNNNN\nNNYYN",
"output": "0"
},
{
"input": "1 10 1\nYYYNYNNYNN",
"output": "5"
},
{
"input": "10 1 5\nY\nN\nN\nN\nY\nN\nN\nY\... | 1,459,533,476 | 7,076 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <iostream>
#include <cmath>
#include <cstdio>
#include <vector>
#include <map>
#include <string>
#include <cstring>
#include <algorithm>
#include <cstdlib>
using namespace std;
struct ty
{
int x, y;
};
long long i, j, n, m, t, ans, ct;
char r;
bool a[1002][1002];
int main() {
cin >> n >> m >> t;
for (i = 1; i <= n; ++i)
for (j = 1; j <= m; ++j) {
cin >> r;
if (r == 'Y') a[i][j] = true;
else a[i][j] = false;
}
ans = 0;
for (j = 1; j <= m; ++j) {
ct = 0;
for (i = 1; i <= n; ++i)
if (a[i][j]) ct++;
if (ct >= t) ans++;
}
cout << ans;
return 0;
}
| Title: You're a Professional
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A simple recommendation system would recommend a user things liked by a certain number of their friends. In this problem you will implement part of such a system.
You are given user's friends' opinions about a list of items. You are also given a threshold *T* — the minimal number of "likes" necessary for an item to be recommended to the user.
Output the number of items in the list liked by at least *T* of user's friends.
Input Specification:
The first line of the input will contain three space-separated integers: the number of friends *F* (1<=≤<=*F*<=≤<=10), the number of items *I* (1<=≤<=*I*<=≤<=10) and the threshold *T* (1<=≤<=*T*<=≤<=*F*).
The following *F* lines of input contain user's friends' opinions. *j*-th character of *i*-th line is 'Y' if *i*-th friend likes *j*-th item, and 'N' otherwise.
Output Specification:
Output an integer — the number of items liked by at least *T* of user's friends.
Demo Input:
['3 3 2\nYYY\nNNN\nYNY\n', '4 4 1\nNNNY\nNNYN\nNYNN\nYNNN\n']
Demo Output:
['2\n', '4\n']
Note:
none | ```python
#include <iostream>
#include <cmath>
#include <cstdio>
#include <vector>
#include <map>
#include <string>
#include <cstring>
#include <algorithm>
#include <cstdlib>
using namespace std;
struct ty
{
int x, y;
};
long long i, j, n, m, t, ans, ct;
char r;
bool a[1002][1002];
int main() {
cin >> n >> m >> t;
for (i = 1; i <= n; ++i)
for (j = 1; j <= m; ++j) {
cin >> r;
if (r == 'Y') a[i][j] = true;
else a[i][j] = false;
}
ans = 0;
for (j = 1; j <= m; ++j) {
ct = 0;
for (i = 1; i <= n; ++i)
if (a[i][j]) ct++;
if (ct >= t) ans++;
}
cout << ans;
return 0;
}
``` | -1 | |
327 | A | Flipping Game | PROGRAMMING | 1,200 | [
"brute force",
"dp",
"implementation"
] | null | null | Iahub got bored, so he invented a game to be played on paper.
He writes *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices *i* and *j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) and flips all values *a**k* for which their positions are in range [*i*,<=*j*] (that is *i*<=≤<=*k*<=≤<=*j*). Flip the value of *x* means to apply operation *x*<==<=1 - *x*.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub. | The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100). In the second line of the input there are *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. It is guaranteed that each of those *n* values is either 0 or 1. | Print an integer — the maximal number of 1s that can be obtained after exactly one move. | [
"5\n1 0 0 1 0\n",
"4\n1 0 0 1\n"
] | [
"4\n",
"4\n"
] | In the first case, flip the segment from 2 to 5 (*i* = 2, *j* = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (*i* = 2, *j* = 3) will turn all numbers into 1. | 500 | [
{
"input": "5\n1 0 0 1 0",
"output": "4"
},
{
"input": "4\n1 0 0 1",
"output": "4"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "8\n1 0 0 0 1 0 0 0",
"output": "7"
},
{
"input": "18\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 ... | 1,666,620,397 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 7 | 154 | 1,228,800 | n=int(input())
arr=[int(x) for x in input().split()]
prefix=[0]
for i in arr:
prefix.append(prefix[-1]+i)
prefix.append(prefix[-1])
maxi=0
for i in range(1,len(prefix)-1):
for j in range(i):
value=(i-j)-(prefix[i]-prefix[j])
maxi=max(value+prefix[j]+prefix[i+1]-prefix[i],maxi)
print(maxi)
| Title: Flipping Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub got bored, so he invented a game to be played on paper.
He writes *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices *i* and *j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) and flips all values *a**k* for which their positions are in range [*i*,<=*j*] (that is *i*<=≤<=*k*<=≤<=*j*). Flip the value of *x* means to apply operation *x*<==<=1 - *x*.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100). In the second line of the input there are *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. It is guaranteed that each of those *n* values is either 0 or 1.
Output Specification:
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Demo Input:
['5\n1 0 0 1 0\n', '4\n1 0 0 1\n']
Demo Output:
['4\n', '4\n']
Note:
In the first case, flip the segment from 2 to 5 (*i* = 2, *j* = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (*i* = 2, *j* = 3) will turn all numbers into 1. | ```python
n=int(input())
arr=[int(x) for x in input().split()]
prefix=[0]
for i in arr:
prefix.append(prefix[-1]+i)
prefix.append(prefix[-1])
maxi=0
for i in range(1,len(prefix)-1):
for j in range(i):
value=(i-j)-(prefix[i]-prefix[j])
maxi=max(value+prefix[j]+prefix[i+1]-prefix[i],maxi)
print(maxi)
``` | 0 | |
675 | C | Money Transfers | PROGRAMMING | 2,100 | [
"constructive algorithms",
"data structures",
"greedy",
"sortings"
] | null | null | There are *n* banks in the city where Vasya lives, they are located in a circle, such that any two banks are neighbouring if their indices differ by no more than 1. Also, bank 1 and bank *n* are neighbours if *n*<=><=1. No bank is a neighbour of itself.
Vasya has an account in each bank. Its balance may be negative, meaning Vasya owes some money to this bank.
There is only one type of operations available: transfer some amount of money from any bank to account in any neighbouring bank. There are no restrictions on the size of the sum being transferred or balance requirements to perform this operation.
Vasya doesn't like to deal with large numbers, so he asks you to determine the minimum number of operations required to change the balance of each bank account to zero. It's guaranteed, that this is possible to achieve, that is, the total balance of Vasya in all banks is equal to zero. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of banks.
The second line contains *n* integers *a**i* (<=-<=109<=≤<=*a**i*<=≤<=109), the *i*-th of them is equal to the initial balance of the account in the *i*-th bank. It's guaranteed that the sum of all *a**i* is equal to 0. | Print the minimum number of operations required to change balance in each bank to zero. | [
"3\n5 0 -5\n",
"4\n-1 0 1 0\n",
"4\n1 2 3 -6\n"
] | [
"1\n",
"2\n",
"3\n"
] | In the first sample, Vasya may transfer 5 from the first bank to the third.
In the second sample, Vasya may first transfer 1 from the third bank to the second, and then 1 from the second to the first.
In the third sample, the following sequence provides the optimal answer:
1. transfer 1 from the first bank to the second bank; 1. transfer 3 from the second bank to the third; 1. transfer 6 from the third bank to the fourth. | 1,500 | [
{
"input": "3\n5 0 -5",
"output": "1"
},
{
"input": "4\n-1 0 1 0",
"output": "2"
},
{
"input": "4\n1 2 3 -6",
"output": "3"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "50\n108431864 128274949 -554057370 -384620666 -202862975 -803855410 -482167063 -55139054 ... | 1,535,897,803 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 3 | 108 | 0 | n = int(input())
a = list(map(int, input().split()))
c = []
b = 0
for i in a:
if i == 0:
b += 1
else:
c.append(b)
b = 0
if b != 0:
if len(a) == 0 or a[0] != 0:
c.append(b)
else:
c[0] += b
print(n - 1 - sorted(c)[-1]) | Title: Money Transfers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* banks in the city where Vasya lives, they are located in a circle, such that any two banks are neighbouring if their indices differ by no more than 1. Also, bank 1 and bank *n* are neighbours if *n*<=><=1. No bank is a neighbour of itself.
Vasya has an account in each bank. Its balance may be negative, meaning Vasya owes some money to this bank.
There is only one type of operations available: transfer some amount of money from any bank to account in any neighbouring bank. There are no restrictions on the size of the sum being transferred or balance requirements to perform this operation.
Vasya doesn't like to deal with large numbers, so he asks you to determine the minimum number of operations required to change the balance of each bank account to zero. It's guaranteed, that this is possible to achieve, that is, the total balance of Vasya in all banks is equal to zero.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of banks.
The second line contains *n* integers *a**i* (<=-<=109<=≤<=*a**i*<=≤<=109), the *i*-th of them is equal to the initial balance of the account in the *i*-th bank. It's guaranteed that the sum of all *a**i* is equal to 0.
Output Specification:
Print the minimum number of operations required to change balance in each bank to zero.
Demo Input:
['3\n5 0 -5\n', '4\n-1 0 1 0\n', '4\n1 2 3 -6\n']
Demo Output:
['1\n', '2\n', '3\n']
Note:
In the first sample, Vasya may transfer 5 from the first bank to the third.
In the second sample, Vasya may first transfer 1 from the third bank to the second, and then 1 from the second to the first.
In the third sample, the following sequence provides the optimal answer:
1. transfer 1 from the first bank to the second bank; 1. transfer 3 from the second bank to the third; 1. transfer 6 from the third bank to the fourth. | ```python
n = int(input())
a = list(map(int, input().split()))
c = []
b = 0
for i in a:
if i == 0:
b += 1
else:
c.append(b)
b = 0
if b != 0:
if len(a) == 0 or a[0] != 0:
c.append(b)
else:
c[0] += b
print(n - 1 - sorted(c)[-1])
``` | -1 | |
0 | none | none | none | 0 | [
"none"
] | null | null | An atom of element X can exist in *n* distinct states with energies *E*1<=<<=*E*2<=<<=...<=<<=*E**n*. Arkady wants to build a laser on this element, using a three-level scheme. Here is a simplified description of the scheme.
Three distinct states *i*, *j* and *k* are selected, where *i*<=<<=*j*<=<<=*k*. After that the following process happens:
1. initially the atom is in the state *i*,1. we spend *E**k*<=-<=*E**i* energy to put the atom in the state *k*,1. the atom emits a photon with useful energy *E**k*<=-<=*E**j* and changes its state to the state *j*,1. the atom spontaneously changes its state to the state *i*, losing energy *E**j*<=-<=*E**i*,1. the process repeats from step 1.
Let's define the energy conversion efficiency as , i. e. the ration between the useful energy of the photon and spent energy.
Due to some limitations, Arkady can only choose such three states that *E**k*<=-<=*E**i*<=≤<=*U*.
Help Arkady to find such the maximum possible energy conversion efficiency within the above constraints. | The first line contains two integers *n* and *U* (3<=≤<=*n*<=≤<=105, 1<=≤<=*U*<=≤<=109) — the number of states and the maximum possible difference between *E**k* and *E**i*.
The second line contains a sequence of integers *E*1,<=*E*2,<=...,<=*E**n* (1<=≤<=*E*1<=<<=*E*2...<=<<=*E**n*<=≤<=109). It is guaranteed that all *E**i* are given in increasing order. | If it is not possible to choose three states that satisfy all constraints, print -1.
Otherwise, print one real number η — the maximum possible energy conversion efficiency. Your answer is considered correct its absolute or relative error does not exceed 10<=-<=9.
Formally, let your answer be *a*, and the jury's answer be *b*. Your answer is considered correct if . | [
"4 4\n1 3 5 7\n",
"10 8\n10 13 15 16 17 19 20 22 24 25\n",
"3 1\n2 5 10\n"
] | [
"0.5\n",
"0.875\n",
"-1\n"
] | In the first example choose states 1, 2 and 3, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/147ae7a830722917b0aa37d064df8eb74cfefb97.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second example choose states 4, 5 and 9, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f68f268de4eb2242167e6ec64e6b8aa60a5703ae.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 0 | [
{
"input": "4 4\n1 3 5 7",
"output": "0.5"
},
{
"input": "10 8\n10 13 15 16 17 19 20 22 24 25",
"output": "0.875"
},
{
"input": "3 1\n2 5 10",
"output": "-1"
},
{
"input": "5 3\n4 6 8 9 10",
"output": "0.5"
},
{
"input": "10 128\n110 121 140 158 174 188 251 271 27... | 1,521,912,599 | 6,899 | Python 3 | WRONG_ANSWER | PRETESTS | 8 | 78 | 7,065,600 | aa= (input()).split()
n = int(aa[0])
u = int(aa[1])
bb =input().split()
for i in range(0,n):
bb[i] = int(bb[i])
index = 0
temp = [0]*n
for i in range(2,n):
while(bb[i]-bb[index]>u):
index=index+1
if i-index<2:
temp[i]=-1
else:
temp[i] = index
min = -1
for i in range(2,n):
if temp[i]!=-1:
test = (bb[i]-bb[temp[i]+1])/(bb[i]-bb[temp[i]])
if test > min:
min = test
print(min)
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An atom of element X can exist in *n* distinct states with energies *E*1<=<<=*E*2<=<<=...<=<<=*E**n*. Arkady wants to build a laser on this element, using a three-level scheme. Here is a simplified description of the scheme.
Three distinct states *i*, *j* and *k* are selected, where *i*<=<<=*j*<=<<=*k*. After that the following process happens:
1. initially the atom is in the state *i*,1. we spend *E**k*<=-<=*E**i* energy to put the atom in the state *k*,1. the atom emits a photon with useful energy *E**k*<=-<=*E**j* and changes its state to the state *j*,1. the atom spontaneously changes its state to the state *i*, losing energy *E**j*<=-<=*E**i*,1. the process repeats from step 1.
Let's define the energy conversion efficiency as , i. e. the ration between the useful energy of the photon and spent energy.
Due to some limitations, Arkady can only choose such three states that *E**k*<=-<=*E**i*<=≤<=*U*.
Help Arkady to find such the maximum possible energy conversion efficiency within the above constraints.
Input Specification:
The first line contains two integers *n* and *U* (3<=≤<=*n*<=≤<=105, 1<=≤<=*U*<=≤<=109) — the number of states and the maximum possible difference between *E**k* and *E**i*.
The second line contains a sequence of integers *E*1,<=*E*2,<=...,<=*E**n* (1<=≤<=*E*1<=<<=*E*2...<=<<=*E**n*<=≤<=109). It is guaranteed that all *E**i* are given in increasing order.
Output Specification:
If it is not possible to choose three states that satisfy all constraints, print -1.
Otherwise, print one real number η — the maximum possible energy conversion efficiency. Your answer is considered correct its absolute or relative error does not exceed 10<=-<=9.
Formally, let your answer be *a*, and the jury's answer be *b*. Your answer is considered correct if .
Demo Input:
['4 4\n1 3 5 7\n', '10 8\n10 13 15 16 17 19 20 22 24 25\n', '3 1\n2 5 10\n']
Demo Output:
['0.5\n', '0.875\n', '-1\n']
Note:
In the first example choose states 1, 2 and 3, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/147ae7a830722917b0aa37d064df8eb74cfefb97.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second example choose states 4, 5 and 9, so that the energy conversion efficiency becomes equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/f68f268de4eb2242167e6ec64e6b8aa60a5703ae.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
aa= (input()).split()
n = int(aa[0])
u = int(aa[1])
bb =input().split()
for i in range(0,n):
bb[i] = int(bb[i])
index = 0
temp = [0]*n
for i in range(2,n):
while(bb[i]-bb[index]>u):
index=index+1
if i-index<2:
temp[i]=-1
else:
temp[i] = index
min = -1
for i in range(2,n):
if temp[i]!=-1:
test = (bb[i]-bb[temp[i]+1])/(bb[i]-bb[temp[i]])
if test > min:
min = test
print(min)
``` | 0 | |
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... | 1,616,950,019 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 154 | 0 | y, k, n = map(int, input().split())
res = []
for num in range(0, n, k):
if num-y > 0:
res.append(num-y)
res1 = [str(bruh) for bruh in res]
if len(res) == 0:
print(-1)
else:
print(' '.join(res1))
| 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
y, k, n = map(int, input().split())
res = []
for num in range(0, n, k):
if num-y > 0:
res.append(num-y)
res1 = [str(bruh) for bruh in res]
if len(res) == 0:
print(-1)
else:
print(' '.join(res1))
``` | 0 | |
633 | B | A Trivial Problem | PROGRAMMING | 1,300 | [
"brute force",
"constructive algorithms",
"math",
"number theory"
] | null | null | Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer *m* and asks for the number of positive integers *n*, such that the factorial of *n* ends with exactly *m* zeroes. Are you among those great programmers who can solve this problem? | The only line of input contains an integer *m* (1<=≤<=*m*<=≤<=100<=000) — the required number of trailing zeroes in factorial. | First print *k* — the number of values of *n* such that the factorial of *n* ends with *m* zeroes. Then print these *k* integers in increasing order. | [
"1\n",
"5\n"
] | [
"5\n5 6 7 8 9 ",
"0"
] | The factorial of *n* is equal to the product of all integers from 1 to *n* inclusive, that is *n*! = 1·2·3·...·*n*.
In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880. | 500 | [
{
"input": "1",
"output": "5\n5 6 7 8 9 "
},
{
"input": "5",
"output": "0"
},
{
"input": "2",
"output": "5\n10 11 12 13 14 "
},
{
"input": "3",
"output": "5\n15 16 17 18 19 "
},
{
"input": "7",
"output": "5\n30 31 32 33 34 "
},
{
"input": "12",
"ou... | 1,456,508,621 | 1,721 | Python 3 | CHALLENGED | CHALLENGES | 8 | 217 | 11,366,400 | from bisect import bisect_left
from functools import lru_cache
@lru_cache(maxsize=None)
def trailingZeroes(n):
if n >= 5:
return trailingZeroes(n // 5) + n // 5
else:
return 0
z = [trailingZeroes(i) for i in range(0, 100000 + 1)]
n = int(input())
r = z.count(n)
print(r)
if r:
idx = bisect_left(z, n, 0, 100001)
nums = [x for x in range(idx, idx + r)]
for x in nums[:-1]:
print(x, end=' ')
print(nums[-1])
| Title: A Trivial Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mr. Santa asks all the great programmers of the world to solve a trivial problem. He gives them an integer *m* and asks for the number of positive integers *n*, such that the factorial of *n* ends with exactly *m* zeroes. Are you among those great programmers who can solve this problem?
Input Specification:
The only line of input contains an integer *m* (1<=≤<=*m*<=≤<=100<=000) — the required number of trailing zeroes in factorial.
Output Specification:
First print *k* — the number of values of *n* such that the factorial of *n* ends with *m* zeroes. Then print these *k* integers in increasing order.
Demo Input:
['1\n', '5\n']
Demo Output:
['5\n5 6 7 8 9 ', '0']
Note:
The factorial of *n* is equal to the product of all integers from 1 to *n* inclusive, that is *n*! = 1·2·3·...·*n*.
In the first sample, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 362880. | ```python
from bisect import bisect_left
from functools import lru_cache
@lru_cache(maxsize=None)
def trailingZeroes(n):
if n >= 5:
return trailingZeroes(n // 5) + n // 5
else:
return 0
z = [trailingZeroes(i) for i in range(0, 100000 + 1)]
n = int(input())
r = z.count(n)
print(r)
if r:
idx = bisect_left(z, n, 0, 100001)
nums = [x for x in range(idx, idx + r)]
for x in nums[:-1]:
print(x, end=' ')
print(nums[-1])
``` | -1 | |
1,003 | A | Polycarp's Pockets | PROGRAMMING | 800 | [
"implementation"
] | null | null | Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket.
For example, if Polycarp has got six coins represented as an array $a = [1, 2, 4, 3, 3, 2]$, he can distribute the coins into two pockets as follows: $[1, 2, 3], [2, 3, 4]$.
Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that. | The first line of the input contains one integer $n$ ($1 \le n \le 100$) — the number of coins.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$) — values of coins. | Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket. | [
"6\n1 2 4 3 3 2\n",
"1\n100\n"
] | [
"2\n",
"1\n"
] | none | 0 | [
{
"input": "6\n1 2 4 3 3 2",
"output": "2"
},
{
"input": "1\n100",
"output": "1"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100... | 1,679,841,667 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 93 | 2,150,400 | # LUOGU_RID: 106063571
n = int(input())
a = list(map(int, input().split()))
c = Counter(a)
print(max(c.values()))
| Title: Polycarp's Pockets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. Polycarp wants to distribute all the coins between his pockets, but he cannot put two coins with the same value into the same pocket.
For example, if Polycarp has got six coins represented as an array $a = [1, 2, 4, 3, 3, 2]$, he can distribute the coins into two pockets as follows: $[1, 2, 3], [2, 3, 4]$.
Polycarp wants to distribute all the coins with the minimum number of used pockets. Help him to do that.
Input Specification:
The first line of the input contains one integer $n$ ($1 \le n \le 100$) — the number of coins.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 100$) — values of coins.
Output Specification:
Print only one integer — the minimum number of pockets Polycarp needs to distribute all the coins so no two coins with the same value are put into the same pocket.
Demo Input:
['6\n1 2 4 3 3 2\n', '1\n100\n']
Demo Output:
['2\n', '1\n']
Note:
none | ```python
# LUOGU_RID: 106063571
n = int(input())
a = list(map(int, input().split()))
c = Counter(a)
print(max(c.values()))
``` | -1 | |
363 | B | Fence | PROGRAMMING | 1,100 | [
"brute force",
"dp"
] | null | null | There is a fence in front of Polycarpus's home. The fence consists of *n* planks of the same width which go one after another from left to right. The height of the *i*-th plank is *h**i* meters, distinct planks can have distinct heights.
Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly *k* consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such *k* consecutive planks that the sum of their heights is minimal possible.
Write the program that finds the indexes of *k* consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic). | The first line of the input contains integers *n* and *k* (1<=≤<=*n*<=≤<=1.5·105,<=1<=≤<=*k*<=≤<=*n*) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=100), where *h**i* is the height of the *i*-th plank of the fence. | Print such integer *j* that the sum of the heights of planks *j*, *j*<=+<=1, ..., *j*<=+<=*k*<=-<=1 is the minimum possible. If there are multiple such *j*'s, print any of them. | [
"7 3\n1 2 6 1 1 7 1\n"
] | [
"3\n"
] | In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8. | 1,000 | [
{
"input": "7 3\n1 2 6 1 1 7 1",
"output": "3"
},
{
"input": "1 1\n100",
"output": "1"
},
{
"input": "2 1\n10 20",
"output": "1"
},
{
"input": "10 5\n1 2 3 1 2 2 3 1 4 5",
"output": "1"
},
{
"input": "10 2\n3 1 4 1 4 6 2 1 4 6",
"output": "7"
},
{
"inp... | 1,676,522,875 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 30 | 0 | n,k = map(int,input().split())
list=[int(x) for x in input().split()]
total = sum(list[:k])
res=[]
res.append(tol)
for i in range(n-k):
total += (-list[i]+list[i+k])
res.append(tol)
print(res.index(min(res))+1) | Title: Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a fence in front of Polycarpus's home. The fence consists of *n* planks of the same width which go one after another from left to right. The height of the *i*-th plank is *h**i* meters, distinct planks can have distinct heights.
Polycarpus has bought a posh piano and is thinking about how to get it into the house. In order to carry out his plan, he needs to take exactly *k* consecutive planks from the fence. Higher planks are harder to tear off the fence, so Polycarpus wants to find such *k* consecutive planks that the sum of their heights is minimal possible.
Write the program that finds the indexes of *k* consecutive planks with minimal total height. Pay attention, the fence is not around Polycarpus's home, it is in front of home (in other words, the fence isn't cyclic).
Input Specification:
The first line of the input contains integers *n* and *k* (1<=≤<=*n*<=≤<=1.5·105,<=1<=≤<=*k*<=≤<=*n*) — the number of planks in the fence and the width of the hole for the piano. The second line contains the sequence of integers *h*1,<=*h*2,<=...,<=*h**n* (1<=≤<=*h**i*<=≤<=100), where *h**i* is the height of the *i*-th plank of the fence.
Output Specification:
Print such integer *j* that the sum of the heights of planks *j*, *j*<=+<=1, ..., *j*<=+<=*k*<=-<=1 is the minimum possible. If there are multiple such *j*'s, print any of them.
Demo Input:
['7 3\n1 2 6 1 1 7 1\n']
Demo Output:
['3\n']
Note:
In the sample, your task is to find three consecutive planks with the minimum sum of heights. In the given case three planks with indexes 3, 4 and 5 have the required attribute, their total height is 8. | ```python
n,k = map(int,input().split())
list=[int(x) for x in input().split()]
total = sum(list[:k])
res=[]
res.append(tol)
for i in range(n-k):
total += (-list[i]+list[i+k])
res.append(tol)
print(res.index(min(res))+1)
``` | -1 | |
797 | A | k-Factorization | PROGRAMMING | 1,100 | [
"implementation",
"math",
"number theory"
] | null | null | Given a positive integer *n*, find *k* integers (not necessary distinct) such that all these integers are strictly greater than 1, and their product is equal to *n*. | The first line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=100000, 1<=≤<=*k*<=≤<=20). | If it's impossible to find the representation of *n* as a product of *k* numbers, print -1.
Otherwise, print *k* integers in any order. Their product must be equal to *n*. If there are multiple answers, print any of them. | [
"100000 2\n",
"100000 20\n",
"1024 5\n"
] | [
"2 50000 \n",
"-1\n",
"2 64 2 2 2 \n"
] | none | 0 | [
{
"input": "100000 2",
"output": "2 50000 "
},
{
"input": "100000 20",
"output": "-1"
},
{
"input": "1024 5",
"output": "2 64 2 2 2 "
},
{
"input": "100000 10",
"output": "2 2 2 2 2 5 5 5 5 5 "
},
{
"input": "99999 3",
"output": "3 813 41 "
},
{
"input... | 1,604,942,659 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 6 | 156 | 1,331,200 | import sys
LI=lambda:list(map(int, sys.stdin.readline().split()))
MI=lambda:map(int, sys.stdin.readline().split())
SI=lambda:sys.stdin.readline().strip('\n')
II=lambda:int(sys.stdin.readline())
ok=[True]*1001
primes=[]
for i in range(2, 1001):
if ok[i]:
primes.append(i)
for j in range(i+i, 1001, i):
ok[j]=False
n, k=MI()
ans=[]
for p in primes:
while n%p==0 and k>1:
n//=p
ans.append(p)
k-=1
ans.append(n)
if k>1:
print(-1)
else:
print(*ans)
| Title: k-Factorization
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Given a positive integer *n*, find *k* integers (not necessary distinct) such that all these integers are strictly greater than 1, and their product is equal to *n*.
Input Specification:
The first line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=100000, 1<=≤<=*k*<=≤<=20).
Output Specification:
If it's impossible to find the representation of *n* as a product of *k* numbers, print -1.
Otherwise, print *k* integers in any order. Their product must be equal to *n*. If there are multiple answers, print any of them.
Demo Input:
['100000 2\n', '100000 20\n', '1024 5\n']
Demo Output:
['2 50000 \n', '-1\n', '2 64 2 2 2 \n']
Note:
none | ```python
import sys
LI=lambda:list(map(int, sys.stdin.readline().split()))
MI=lambda:map(int, sys.stdin.readline().split())
SI=lambda:sys.stdin.readline().strip('\n')
II=lambda:int(sys.stdin.readline())
ok=[True]*1001
primes=[]
for i in range(2, 1001):
if ok[i]:
primes.append(i)
for j in range(i+i, 1001, i):
ok[j]=False
n, k=MI()
ans=[]
for p in primes:
while n%p==0 and k>1:
n//=p
ans.append(p)
k-=1
ans.append(n)
if k>1:
print(-1)
else:
print(*ans)
``` | 0 | |
939 | A | Love Triangle | PROGRAMMING | 800 | [
"graphs"
] | null | null | As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are *n* planes on Earth, numbered from 1 to *n*, and the plane with number *i* likes the plane with number *f**i*, where 1<=≤<=*f**i*<=≤<=*n* and *f**i*<=≠<=*i*.
We call a love triangle a situation in which plane *A* likes plane *B*, plane *B* likes plane *C* and plane *C* likes plane *A*. Find out if there is any love triangle on Earth. | The first line contains a single integer *n* (2<=≤<=*n*<=≤<=5000) — the number of planes.
The second line contains *n* integers *f*1,<=*f*2,<=...,<=*f**n* (1<=≤<=*f**i*<=≤<=*n*, *f**i*<=≠<=*i*), meaning that the *i*-th plane likes the *f**i*-th. | Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case. | [
"5\n2 4 5 1 3\n",
"5\n5 5 5 5 1\n"
] | [
"YES\n",
"NO\n"
] | In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles. | 500 | [
{
"input": "5\n2 4 5 1 3",
"output": "YES"
},
{
"input": "5\n5 5 5 5 1",
"output": "NO"
},
{
"input": "3\n3 1 2",
"output": "YES"
},
{
"input": "10\n4 10 9 5 3 1 5 10 6 4",
"output": "NO"
},
{
"input": "10\n5 5 4 9 10 9 9 5 3 1",
"output": "YES"
},
{
"... | 1,649,156,116 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 16 | 61 | 2,560,000 |
a = int(input())
b = list(map(int,input().split()))
c = []
for i in b:
if b[b[b[i-1]-1]-1]==i:
c.append("YES")
c.append("NO")
print(c[0]) | Title: Love Triangle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are *n* planes on Earth, numbered from 1 to *n*, and the plane with number *i* likes the plane with number *f**i*, where 1<=≤<=*f**i*<=≤<=*n* and *f**i*<=≠<=*i*.
We call a love triangle a situation in which plane *A* likes plane *B*, plane *B* likes plane *C* and plane *C* likes plane *A*. Find out if there is any love triangle on Earth.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=5000) — the number of planes.
The second line contains *n* integers *f*1,<=*f*2,<=...,<=*f**n* (1<=≤<=*f**i*<=≤<=*n*, *f**i*<=≠<=*i*), meaning that the *i*-th plane likes the *f**i*-th.
Output Specification:
Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case.
Demo Input:
['5\n2 4 5 1 3\n', '5\n5 5 5 5 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles. | ```python
a = int(input())
b = list(map(int,input().split()))
c = []
for i in b:
if b[b[b[i-1]-1]-1]==i:
c.append("YES")
c.append("NO")
print(c[0])
``` | 3 | |
716 | A | Crazy Computer | PROGRAMMING | 800 | [
"implementation"
] | null | null | ZS the Coder is coding on a crazy computer. If you don't type in a word for a *c* consecutive seconds, everything you typed disappear!
More formally, if you typed a word at second *a* and then the next word at second *b*, then if *b*<=-<=*a*<=≤<=*c*, just the new word is appended to other words on the screen. If *b*<=-<=*a*<=><=*c*, then everything on the screen disappears and after that the word you have typed appears on the screen.
For example, if *c*<==<=5 and you typed words at seconds 1,<=3,<=8,<=14,<=19,<=20 then at the second 8 there will be 3 words on the screen. After that, everything disappears at the second 13 because nothing was typed. At the seconds 14 and 19 another two words are typed, and finally, at the second 20, one more word is typed, and a total of 3 words remain on the screen.
You're given the times when ZS the Coder typed the words. Determine how many words remain on the screen after he finished typing everything. | The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*c*<=≤<=109) — the number of words ZS the Coder typed and the crazy computer delay respectively.
The next line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=...<=<<=*t**n*<=≤<=109), where *t**i* denotes the second when ZS the Coder typed the *i*-th word. | Print a single positive integer, the number of words that remain on the screen after all *n* words was typed, in other words, at the second *t**n*. | [
"6 5\n1 3 8 14 19 20\n",
"6 1\n1 3 5 7 9 10\n"
] | [
"3",
"2"
] | The first sample is already explained in the problem statement.
For the second sample, after typing the first word at the second 1, it disappears because the next word is typed at the second 3 and 3 - 1 > 1. Similarly, only 1 word will remain at the second 9. Then, a word is typed at the second 10, so there will be two words on the screen, as the old word won't disappear because 10 - 9 ≤ 1. | 500 | [
{
"input": "6 5\n1 3 8 14 19 20",
"output": "3"
},
{
"input": "6 1\n1 3 5 7 9 10",
"output": "2"
},
{
"input": "1 1\n1000000000",
"output": "1"
},
{
"input": "5 5\n1 7 12 13 14",
"output": "4"
},
{
"input": "2 1000000000\n1 1000000000",
"output": "2"
},
{
... | 1,585,575,058 | 2,147,483,647 | PyPy 3 | OK | TESTS | 81 | 218 | 10,547,200 | c = int(input().split(' ')[1])
words = list(map(int, input().split(' ')))
l = 0; last = 0
for i in words:
if i - last > c:
l = 1
else:
l += 1
last = i
print(l) | Title: Crazy Computer
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
ZS the Coder is coding on a crazy computer. If you don't type in a word for a *c* consecutive seconds, everything you typed disappear!
More formally, if you typed a word at second *a* and then the next word at second *b*, then if *b*<=-<=*a*<=≤<=*c*, just the new word is appended to other words on the screen. If *b*<=-<=*a*<=><=*c*, then everything on the screen disappears and after that the word you have typed appears on the screen.
For example, if *c*<==<=5 and you typed words at seconds 1,<=3,<=8,<=14,<=19,<=20 then at the second 8 there will be 3 words on the screen. After that, everything disappears at the second 13 because nothing was typed. At the seconds 14 and 19 another two words are typed, and finally, at the second 20, one more word is typed, and a total of 3 words remain on the screen.
You're given the times when ZS the Coder typed the words. Determine how many words remain on the screen after he finished typing everything.
Input Specification:
The first line contains two integers *n* and *c* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*c*<=≤<=109) — the number of words ZS the Coder typed and the crazy computer delay respectively.
The next line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t*1<=<<=*t*2<=<<=...<=<<=*t**n*<=≤<=109), where *t**i* denotes the second when ZS the Coder typed the *i*-th word.
Output Specification:
Print a single positive integer, the number of words that remain on the screen after all *n* words was typed, in other words, at the second *t**n*.
Demo Input:
['6 5\n1 3 8 14 19 20\n', '6 1\n1 3 5 7 9 10\n']
Demo Output:
['3', '2']
Note:
The first sample is already explained in the problem statement.
For the second sample, after typing the first word at the second 1, it disappears because the next word is typed at the second 3 and 3 - 1 > 1. Similarly, only 1 word will remain at the second 9. Then, a word is typed at the second 10, so there will be two words on the screen, as the old word won't disappear because 10 - 9 ≤ 1. | ```python
c = int(input().split(' ')[1])
words = list(map(int, input().split(' ')))
l = 0; last = 0
for i in words:
if i - last > c:
l = 1
else:
l += 1
last = i
print(l)
``` | 3 | |
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... | 1,681,501,521 | 2,147,483,647 | Python 3 | OK | TESTS | 46 | 1,326 | 14,336,000 | import collections
n, m = list(map(int, input().split()))
nums = list(map(int, input().split()))
INF = 10 ** 9 + 13
arr = []
for index in range(n):
arr.append([])
for index in range(m):
v, u = list(map(int, input().split()))
v -= 1
u -= 1
arr[v].append(u)
arr[u].append(v)
used = [False] * n
res = INF
def dfs(v):
global arr, used, res
used[v] = True
res = min(res,nums[v])
for u in arr[v]:
if not used[u]:
dfs(u)
def bfs(v):
global arr, used, res
q = collections.deque([v])
used[v] = True
while q:
u = q.pop()
res = min(res, nums[u])
for w in arr[u]:
if not used[w]:
q.append(w)
used[w] = True
ans = 0
for index in range(n):
if not used[index]:
res = INF
bfs(index)
ans += res
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
import collections
n, m = list(map(int, input().split()))
nums = list(map(int, input().split()))
INF = 10 ** 9 + 13
arr = []
for index in range(n):
arr.append([])
for index in range(m):
v, u = list(map(int, input().split()))
v -= 1
u -= 1
arr[v].append(u)
arr[u].append(v)
used = [False] * n
res = INF
def dfs(v):
global arr, used, res
used[v] = True
res = min(res,nums[v])
for u in arr[v]:
if not used[u]:
dfs(u)
def bfs(v):
global arr, used, res
q = collections.deque([v])
used[v] = True
while q:
u = q.pop()
res = min(res, nums[u])
for w in arr[u]:
if not used[w]:
q.append(w)
used[w] = True
ans = 0
for index in range(n):
if not used[index]:
res = INF
bfs(index)
ans += res
print(ans)
``` | 3 | |
834 | A | The Useless Toy | PROGRAMMING | 900 | [
"implementation"
] | null | null | Walking through the streets of Marshmallow City, Slastyona have spotted some merchants selling a kind of useless toy which is very popular nowadays – caramel spinner! Wanting to join the craze, she has immediately bought the strange contraption.
Spinners in Sweetland have the form of V-shaped pieces of caramel. Each spinner can, well, spin around an invisible magic axis. At a specific point in time, a spinner can take 4 positions shown below (each one rotated 90 degrees relative to the previous, with the fourth one followed by the first one):
After the spinner was spun, it starts its rotation, which is described by a following algorithm: the spinner maintains its position for a second then majestically switches to the next position in clockwise or counter-clockwise order, depending on the direction the spinner was spun in.
Slastyona managed to have spinner rotating for exactly *n* seconds. Being fascinated by elegance of the process, she completely forgot the direction the spinner was spun in! Lucky for her, she managed to recall the starting position, and wants to deduct the direction given the information she knows. Help her do this. | There are two characters in the first string – the starting and the ending position of a spinner. The position is encoded with one of the following characters: v (ASCII code 118, lowercase v), < (ASCII code 60), ^ (ASCII code 94) or > (ASCII code 62) (see the picture above for reference). Characters are separated by a single space.
In the second strings, a single number *n* is given (0<=≤<=*n*<=≤<=109) – the duration of the rotation.
It is guaranteed that the ending position of a spinner is a result of a *n* second spin in any of the directions, assuming the given starting position. | Output cw, if the direction is clockwise, ccw – if counter-clockwise, and undefined otherwise. | [
"^ >\n1\n",
"< ^\n3\n",
"^ v\n6\n"
] | [
"cw\n",
"ccw\n",
"undefined\n"
] | none | 500 | [
{
"input": "^ >\n1",
"output": "cw"
},
{
"input": "< ^\n3",
"output": "ccw"
},
{
"input": "^ v\n6",
"output": "undefined"
},
{
"input": "^ >\n999999999",
"output": "ccw"
},
{
"input": "> v\n1",
"output": "cw"
},
{
"input": "v <\n1",
"output": "cw"
... | 1,502,384,288 | 2,147,483,647 | Python 3 | OK | TESTS | 47 | 62 | 4,608,000 | char1, char2 = input().split()
n = int(input())%4
cw = "v<^>v<^>"
ccw = "v>^<v>^<"
if n%2==1 and cw[cw.find(char1)+n]==char2:
print("cw")
elif n%2==1 and ccw[ccw.find(char1)+n]==char2:
print("ccw")
else:
print("undefined") | Title: The Useless Toy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Walking through the streets of Marshmallow City, Slastyona have spotted some merchants selling a kind of useless toy which is very popular nowadays – caramel spinner! Wanting to join the craze, she has immediately bought the strange contraption.
Spinners in Sweetland have the form of V-shaped pieces of caramel. Each spinner can, well, spin around an invisible magic axis. At a specific point in time, a spinner can take 4 positions shown below (each one rotated 90 degrees relative to the previous, with the fourth one followed by the first one):
After the spinner was spun, it starts its rotation, which is described by a following algorithm: the spinner maintains its position for a second then majestically switches to the next position in clockwise or counter-clockwise order, depending on the direction the spinner was spun in.
Slastyona managed to have spinner rotating for exactly *n* seconds. Being fascinated by elegance of the process, she completely forgot the direction the spinner was spun in! Lucky for her, she managed to recall the starting position, and wants to deduct the direction given the information she knows. Help her do this.
Input Specification:
There are two characters in the first string – the starting and the ending position of a spinner. The position is encoded with one of the following characters: v (ASCII code 118, lowercase v), < (ASCII code 60), ^ (ASCII code 94) or > (ASCII code 62) (see the picture above for reference). Characters are separated by a single space.
In the second strings, a single number *n* is given (0<=≤<=*n*<=≤<=109) – the duration of the rotation.
It is guaranteed that the ending position of a spinner is a result of a *n* second spin in any of the directions, assuming the given starting position.
Output Specification:
Output cw, if the direction is clockwise, ccw – if counter-clockwise, and undefined otherwise.
Demo Input:
['^ >\n1\n', '< ^\n3\n', '^ v\n6\n']
Demo Output:
['cw\n', 'ccw\n', 'undefined\n']
Note:
none | ```python
char1, char2 = input().split()
n = int(input())%4
cw = "v<^>v<^>"
ccw = "v>^<v>^<"
if n%2==1 and cw[cw.find(char1)+n]==char2:
print("cw")
elif n%2==1 and ccw[ccw.find(char1)+n]==char2:
print("ccw")
else:
print("undefined")
``` | 3 | |
841 | A | Generous Kefa | PROGRAMMING | 900 | [
"brute force",
"implementation"
] | null | null | One day Kefa found *n* baloons. For convenience, we denote color of *i*-th baloon as *s**i* — lowercase letter of the Latin alphabet. Also Kefa has *k* friends. Friend will be upset, If he get two baloons of the same color. Kefa want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print «YES», if he can, and «NO», otherwise. Note, that Kefa's friend will not upset, if he doesn't get baloons at all. | The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of baloons and friends.
Next line contains string *s* — colors of baloons. | Answer to the task — «YES» or «NO» in a single line.
You can choose the case (lower or upper) for each letter arbitrary. | [
"4 2\naabb\n",
"6 3\naacaab\n"
] | [
"YES\n",
"NO\n"
] | In the first sample Kefa can give 1-st and 3-rd baloon to the first friend, and 2-nd and 4-th to the second.
In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is «NO». | 500 | [
{
"input": "4 2\naabb",
"output": "YES"
},
{
"input": "6 3\naacaab",
"output": "NO"
},
{
"input": "2 2\nlu",
"output": "YES"
},
{
"input": "5 3\novvoo",
"output": "YES"
},
{
"input": "36 13\nbzbzcffczzcbcbzzfzbbfzfzzbfbbcbfccbf",
"output": "YES"
},
{
"... | 1,606,027,679 | 2,147,483,647 | Python 3 | OK | TESTS | 114 | 108 | 0 | n, k = map(int, input().split())
balloons_color = sorted(input())
dic = {}
for x in balloons_color:
if x in dic:
dic[x] += 1
else:
dic[x] = 1
if int(max(dic.values())) > k:
print("NO")
else:
print("YES")
| Title: Generous Kefa
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Kefa found *n* baloons. For convenience, we denote color of *i*-th baloon as *s**i* — lowercase letter of the Latin alphabet. Also Kefa has *k* friends. Friend will be upset, If he get two baloons of the same color. Kefa want to give out all baloons to his friends. Help Kefa to find out, can he give out all his baloons, such that no one of his friens will be upset — print «YES», if he can, and «NO», otherwise. Note, that Kefa's friend will not upset, if he doesn't get baloons at all.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of baloons and friends.
Next line contains string *s* — colors of baloons.
Output Specification:
Answer to the task — «YES» or «NO» in a single line.
You can choose the case (lower or upper) for each letter arbitrary.
Demo Input:
['4 2\naabb\n', '6 3\naacaab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample Kefa can give 1-st and 3-rd baloon to the first friend, and 2-nd and 4-th to the second.
In the second sample Kefa needs to give to all his friends baloons of color a, but one baloon will stay, thats why answer is «NO». | ```python
n, k = map(int, input().split())
balloons_color = sorted(input())
dic = {}
for x in balloons_color:
if x in dic:
dic[x] += 1
else:
dic[x] = 1
if int(max(dic.values())) > k:
print("NO")
else:
print("YES")
``` | 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",
"... | 1,459,587,433 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 92 | 5,017,600 | n,d = [int(i) for i in input().split()]
x = [int(i) for i in input().split()]
from collections import deque
D =deque()
total = 0
s = 0
for i in range(n-1):
while len(D)>0 and s+x[i+1]-x[i] > m:
s -= D.popleft()
if s + x[i+1] -x[i] <= m:
D.append(x[i+1]-x[i])
total += (len(D)*(len(D)-z1))//2
s += D[-1]
print(total) | 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
n,d = [int(i) for i in input().split()]
x = [int(i) for i in input().split()]
from collections import deque
D =deque()
total = 0
s = 0
for i in range(n-1):
while len(D)>0 and s+x[i+1]-x[i] > m:
s -= D.popleft()
if s + x[i+1] -x[i] <= m:
D.append(x[i+1]-x[i])
total += (len(D)*(len(D)-z1))//2
s += D[-1]
print(total)
``` | -1 | |
127 | A | Wasted Time | PROGRAMMING | 900 | [
"geometry"
] | null | null | Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline *A*1*A*2... *A**n*. Scrooge signs like that: first it places a pen at the point *A*1, then draws a segment from point *A*1 to point *A*2, then he draws a segment from point *A*2 to point *A*3 and so on to point *A**n*, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly *k* papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers. | The first line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=1000). Each of the following *n* lines contains the coordinates of the polyline's endpoints. The *i*-th one contains coordinates of the point *A**i* — integers *x**i* and *y**i*, separated by a space.
All points *A**i* are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters. | Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10<=-<=6. | [
"2 1\n0 0\n10 0\n",
"5 10\n3 1\n-5 6\n-2 -1\n3 2\n10 0\n",
"6 10\n5 0\n4 0\n6 0\n3 0\n7 0\n2 0\n"
] | [
"0.200000000",
"6.032163204",
"3.000000000"
] | none | 500 | [
{
"input": "2 1\n0 0\n10 0",
"output": "0.200000000"
},
{
"input": "5 10\n3 1\n-5 6\n-2 -1\n3 2\n10 0",
"output": "6.032163204"
},
{
"input": "6 10\n5 0\n4 0\n6 0\n3 0\n7 0\n2 0",
"output": "3.000000000"
},
{
"input": "10 95\n-20 -5\n2 -8\n14 13\n10 3\n17 11\n13 -12\n-6 11\n1... | 1,556,536,017 | 2,147,483,647 | Python 3 | OK | TESTS | 42 | 216 | 0 | from math import sqrt
X, Last, SUM = list(map(int, input().split())), list(map(int, input().split())), 0
for i in range(X[0] - 1):
Temp = list(map(int, input().split()))
SUM += sqrt((Last[0] - Temp[0]) ** 2 + (Last[1] - Temp[1]) ** 2)
Last = Temp
print((SUM * X[1]) / 50)
| Title: Wasted Time
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline *A*1*A*2... *A**n*. Scrooge signs like that: first it places a pen at the point *A*1, then draws a segment from point *A*1 to point *A*2, then he draws a segment from point *A*2 to point *A*3 and so on to point *A**n*, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly *k* papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input Specification:
The first line contains two integers *n* and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=1000). Each of the following *n* lines contains the coordinates of the polyline's endpoints. The *i*-th one contains coordinates of the point *A**i* — integers *x**i* and *y**i*, separated by a space.
All points *A**i* are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output Specification:
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10<=-<=6.
Demo Input:
['2 1\n0 0\n10 0\n', '5 10\n3 1\n-5 6\n-2 -1\n3 2\n10 0\n', '6 10\n5 0\n4 0\n6 0\n3 0\n7 0\n2 0\n']
Demo Output:
['0.200000000', '6.032163204', '3.000000000']
Note:
none | ```python
from math import sqrt
X, Last, SUM = list(map(int, input().split())), list(map(int, input().split())), 0
for i in range(X[0] - 1):
Temp = list(map(int, input().split()))
SUM += sqrt((Last[0] - Temp[0]) ** 2 + (Last[1] - Temp[1]) ** 2)
Last = Temp
print((SUM * X[1]) / 50)
``` | 3 | |
576 | A | Vasya and Petya's Game | PROGRAMMING | 1,500 | [
"math",
"number theory"
] | null | null | Vasya and Petya are playing a simple game. Vasya thought of number *x* between 1 and *n*, and Petya tries to guess the number.
Petya can ask questions like: "Is the unknown number divisible by number *y*?".
The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of.
Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers *y**i*, he should ask the questions about. | A single line contains number *n* (1<=≤<=*n*<=≤<=103). | Print the length of the sequence of questions *k* (0<=≤<=*k*<=≤<=*n*), followed by *k* numbers — the questions *y**i* (1<=≤<=*y**i*<=≤<=*n*).
If there are several correct sequences of questions of the minimum length, you are allowed to print any of them. | [
"4\n",
"6\n"
] | [
"3\n2 4 3 \n",
"4\n2 4 3 5 \n"
] | The sequence from the answer to the first sample test is actually correct.
If the unknown number is not divisible by one of the sequence numbers, it is equal to 1.
If the unknown number is divisible by 4, it is 4.
If the unknown number is divisible by 3, then the unknown number is 3.
Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter. | 500 | [
{
"input": "4",
"output": "3\n2 4 3 "
},
{
"input": "6",
"output": "4\n2 4 3 5 "
},
{
"input": "1",
"output": "0"
},
{
"input": "15",
"output": "9\n2 4 8 3 9 5 7 11 13 "
},
{
"input": "19",
"output": "12\n2 4 8 16 3 9 5 7 11 13 17 19 "
},
{
"input": "2... | 1,682,875,256 | 2,147,483,647 | PyPy 3-64 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define ff first
#define ss second
#define ll int64_t
#define ld long double
#define nl cout<<"\n"
#define i128 __int128_t
#define all(v) v.begin(),v.end()
#define mset(a,v) memset((a),(v),sizeof(a))
#define forn(i,a,b) for(int64_t i=int64_t(a);i<int64_t(b);++i)
#define forb(i,a,b) for(int64_t i=int64_t(a);i>=int64_t(b);--i)
#define fastio() ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define mod 1'000'000'007
#define mod2 998'244'353
#define inf 1'000'000'000'000'007
#define pi 3.14159265358979323846
template<class key,class cmp=std::less<key>>
using ordered_set=tree<key,null_type,cmp,rb_tree_tag,tree_order_statistics_node_update>;
template<class L,class R> ostream& operator<<(ostream& out,pair<L,R> &p) {return out<<"("<<p.ff<<", "<<p.ss<<")";}
template<class T> ostream& operator<<(ostream& out,vector<T> &v) {out<<"[";for(auto it=v.begin();it!=v.end();++it){if(it!=v.begin())out<<", ";out<<*it;}return out<<"]";}
template<class T> ostream& operator<<(ostream& out,deque<T> &v) {out<<"[";for(auto it=v.begin();it!=v.end();++it){if(it!=v.begin())out<<", ";out<<*it;}return out<<"]";}
template<class T> ostream& operator<<(ostream& out,set<T> &s) {out<<"{";for(auto it=s.begin();it!=s.end();++it){if(it!=s.begin())out<<", ";out<<*it;}return out<<"}";}
template<class T> ostream& operator<<(ostream& out,ordered_set<T> &s) {out<<"{";for(auto it=s.begin();it!=s.end();++it){if(it!=s.begin())out<<", ";out<<*it;}return out<<"}";}
template<class L,class R> ostream& operator<<(ostream& out,map<L,R> &m) {out<<"{";for(auto it=m.begin();it!=m.end();++it){if(it!=m.begin())out<<", ";out<<*it;}return out<<"}";}
void dbg_out() {cerr<<"]\n";}
template<typename Head,typename... Tail>
void dbg_out(Head H,Tail... T) {cerr<<H;if(sizeof...(Tail))cerr<<", ";dbg_out(T...);}
#ifdef LOCAL
#define dbg(...) cerr<<"["<<#__VA_ARGS__<<"] = [",dbg_out(__VA_ARGS__)
#else
#define dbg(...)
#endif
//---------------------------------mars4---------------------------------
int main()
{
fastio();
ll z,n,m,t,k,i,j,l,d,h,r;
cin>>n;
vector<pair<ll,ll>> a(n);
for(auto &[l,r]:a)
{
cin>>l>>r;
}
vector<ll> ind(n);
iota(all(ind),(ll)0);
sort(all(ind),[&](ll i,ll j){
ll iB=a[i].ff/1000;
ll jB=a[j].ff/1000;
if(iB==jB)
{
if(iB&1)
{
return a[i].ss>a[j].ss;
}
else
{
return a[i].ss<a[j].ss;
}
}
return iB<jB;
});
for(ll i:ind)
{
cout<<i+1<<" ";
}
nl;
cerr<<"\nTime elapsed: "<<1000*clock()/CLOCKS_PER_SEC<<"ms\n";
return 0;
}
| Title: Vasya and Petya's Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya and Petya are playing a simple game. Vasya thought of number *x* between 1 and *n*, and Petya tries to guess the number.
Petya can ask questions like: "Is the unknown number divisible by number *y*?".
The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of.
Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers *y**i*, he should ask the questions about.
Input Specification:
A single line contains number *n* (1<=≤<=*n*<=≤<=103).
Output Specification:
Print the length of the sequence of questions *k* (0<=≤<=*k*<=≤<=*n*), followed by *k* numbers — the questions *y**i* (1<=≤<=*y**i*<=≤<=*n*).
If there are several correct sequences of questions of the minimum length, you are allowed to print any of them.
Demo Input:
['4\n', '6\n']
Demo Output:
['3\n2 4 3 \n', '4\n2 4 3 5 \n']
Note:
The sequence from the answer to the first sample test is actually correct.
If the unknown number is not divisible by one of the sequence numbers, it is equal to 1.
If the unknown number is divisible by 4, it is 4.
If the unknown number is divisible by 3, then the unknown number is 3.
Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter. | ```python
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define ff first
#define ss second
#define ll int64_t
#define ld long double
#define nl cout<<"\n"
#define i128 __int128_t
#define all(v) v.begin(),v.end()
#define mset(a,v) memset((a),(v),sizeof(a))
#define forn(i,a,b) for(int64_t i=int64_t(a);i<int64_t(b);++i)
#define forb(i,a,b) for(int64_t i=int64_t(a);i>=int64_t(b);--i)
#define fastio() ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define mod 1'000'000'007
#define mod2 998'244'353
#define inf 1'000'000'000'000'007
#define pi 3.14159265358979323846
template<class key,class cmp=std::less<key>>
using ordered_set=tree<key,null_type,cmp,rb_tree_tag,tree_order_statistics_node_update>;
template<class L,class R> ostream& operator<<(ostream& out,pair<L,R> &p) {return out<<"("<<p.ff<<", "<<p.ss<<")";}
template<class T> ostream& operator<<(ostream& out,vector<T> &v) {out<<"[";for(auto it=v.begin();it!=v.end();++it){if(it!=v.begin())out<<", ";out<<*it;}return out<<"]";}
template<class T> ostream& operator<<(ostream& out,deque<T> &v) {out<<"[";for(auto it=v.begin();it!=v.end();++it){if(it!=v.begin())out<<", ";out<<*it;}return out<<"]";}
template<class T> ostream& operator<<(ostream& out,set<T> &s) {out<<"{";for(auto it=s.begin();it!=s.end();++it){if(it!=s.begin())out<<", ";out<<*it;}return out<<"}";}
template<class T> ostream& operator<<(ostream& out,ordered_set<T> &s) {out<<"{";for(auto it=s.begin();it!=s.end();++it){if(it!=s.begin())out<<", ";out<<*it;}return out<<"}";}
template<class L,class R> ostream& operator<<(ostream& out,map<L,R> &m) {out<<"{";for(auto it=m.begin();it!=m.end();++it){if(it!=m.begin())out<<", ";out<<*it;}return out<<"}";}
void dbg_out() {cerr<<"]\n";}
template<typename Head,typename... Tail>
void dbg_out(Head H,Tail... T) {cerr<<H;if(sizeof...(Tail))cerr<<", ";dbg_out(T...);}
#ifdef LOCAL
#define dbg(...) cerr<<"["<<#__VA_ARGS__<<"] = [",dbg_out(__VA_ARGS__)
#else
#define dbg(...)
#endif
//---------------------------------mars4---------------------------------
int main()
{
fastio();
ll z,n,m,t,k,i,j,l,d,h,r;
cin>>n;
vector<pair<ll,ll>> a(n);
for(auto &[l,r]:a)
{
cin>>l>>r;
}
vector<ll> ind(n);
iota(all(ind),(ll)0);
sort(all(ind),[&](ll i,ll j){
ll iB=a[i].ff/1000;
ll jB=a[j].ff/1000;
if(iB==jB)
{
if(iB&1)
{
return a[i].ss>a[j].ss;
}
else
{
return a[i].ss<a[j].ss;
}
}
return iB<jB;
});
for(ll i:ind)
{
cout<<i+1<<" ";
}
nl;
cerr<<"\nTime elapsed: "<<1000*clock()/CLOCKS_PER_SEC<<"ms\n";
return 0;
}
``` | -1 | |
146 | A | Lucky Ticket | PROGRAMMING | 800 | [
"implementation"
] | null | null | Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya loves tickets very much. As we know, each ticket has a number that is a positive integer. Its length equals *n* (*n* is always even). Petya calls a ticket lucky if the ticket's number is a lucky number and the sum of digits in the first half (the sum of the first *n*<=/<=2 digits) equals the sum of digits in the second half (the sum of the last *n*<=/<=2 digits). Check if the given ticket is lucky. | The first line contains an even integer *n* (2<=≤<=*n*<=≤<=50) — the length of the ticket number that needs to be checked. The second line contains an integer whose length equals exactly *n* — the ticket number. The number may contain leading zeros. | On the first line print "YES" if the given ticket number is lucky. Otherwise, print "NO" (without the quotes). | [
"2\n47\n",
"4\n4738\n",
"4\n4774\n"
] | [
"NO\n",
"NO\n",
"YES\n"
] | In the first sample the sum of digits in the first half does not equal the sum of digits in the second half (4 ≠ 7).
In the second sample the ticket number is not the lucky number. | 500 | [
{
"input": "2\n47",
"output": "NO"
},
{
"input": "4\n4738",
"output": "NO"
},
{
"input": "4\n4774",
"output": "YES"
},
{
"input": "4\n4570",
"output": "NO"
},
{
"input": "6\n477477",
"output": "YES"
},
{
"input": "6\n777777",
"output": "YES"
},
... | 1,657,713,521 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 46 | 154 | 0 | n=int(input())
a=input()
k=0
for i in range(n):
if a[i]!='4' and a[i]!='7':
k+=1
if k!=0:
print('NO')
else:
k=int(n/2)
b=a[:k]
c=a[-k:]
s=0
s1=0
for i in range(k):
s+=int(b[i])
s1+=int(c[i])
if s==s1:
print('YES')
else:
print("NO") | Title: Lucky Ticket
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya loves tickets very much. As we know, each ticket has a number that is a positive integer. Its length equals *n* (*n* is always even). Petya calls a ticket lucky if the ticket's number is a lucky number and the sum of digits in the first half (the sum of the first *n*<=/<=2 digits) equals the sum of digits in the second half (the sum of the last *n*<=/<=2 digits). Check if the given ticket is lucky.
Input Specification:
The first line contains an even integer *n* (2<=≤<=*n*<=≤<=50) — the length of the ticket number that needs to be checked. The second line contains an integer whose length equals exactly *n* — the ticket number. The number may contain leading zeros.
Output Specification:
On the first line print "YES" if the given ticket number is lucky. Otherwise, print "NO" (without the quotes).
Demo Input:
['2\n47\n', '4\n4738\n', '4\n4774\n']
Demo Output:
['NO\n', 'NO\n', 'YES\n']
Note:
In the first sample the sum of digits in the first half does not equal the sum of digits in the second half (4 ≠ 7).
In the second sample the ticket number is not the lucky number. | ```python
n=int(input())
a=input()
k=0
for i in range(n):
if a[i]!='4' and a[i]!='7':
k+=1
if k!=0:
print('NO')
else:
k=int(n/2)
b=a[:k]
c=a[-k:]
s=0
s1=0
for i in range(k):
s+=int(b[i])
s1+=int(c[i])
if s==s1:
print('YES')
else:
print("NO")
``` | 3 | |
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"outpu... | 1,657,457,033 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 92 | 0 | a, b = input().split()
a, b = int(a), int(b)
print((a*b)//2) | Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
a, b = input().split()
a, b = int(a), int(b)
print((a*b)//2)
``` | 3.977 |
25 | A | IQ test | PROGRAMMING | 1,300 | [
"brute force"
] | A. IQ test | 2 | 256 | Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness. | The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness. | Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order. | [
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] | [
"3\n",
"2\n"
] | none | 0 | [
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output"... | 1,662,294,940 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 32 | 124 | 0 | x=input()
y=list(map(int,input().split()))
a=[]
b=[]
for i in y:
if i%2==0:
a.append(i)
else:
b.append(i)
if len(a)<len(b):
print(y.index(a[0])+1)
else:
print(y.index(b[0])+1) | Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none | ```python
x=input()
y=list(map(int,input().split()))
a=[]
b=[]
for i in y:
if i%2==0:
a.append(i)
else:
b.append(i)
if len(a)<len(b):
print(y.index(a[0])+1)
else:
print(y.index(b[0])+1)
``` | 3.969 |
186 | A | Comparing Strings | PROGRAMMING | 1,100 | [
"implementation",
"strings"
] | null | null | Some dwarves that are finishing the StUDY (State University for Dwarven Youngsters) Bachelor courses, have been told "no genome, no degree". That means that all dwarves should write a thesis on genome. Dwarven genome is far from simple. It is represented by a string that consists of lowercase Latin letters.
Dwarf Misha has already chosen the subject for his thesis: determining by two dwarven genomes, whether they belong to the same race. Two dwarves belong to the same race if we can swap two characters in the first dwarf's genome and get the second dwarf's genome as a result. Help Dwarf Misha and find out whether two gnomes belong to the same race or not. | The first line contains the first dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The second line contains the second dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The number of letters in each genome doesn't exceed 105. It is guaranteed that the strings that correspond to the genomes are different. The given genomes may have different length. | Print "YES", if the dwarves belong to the same race. Otherwise, print "NO". | [
"ab\nba\n",
"aa\nab\n"
] | [
"YES\n",
"NO\n"
] | - First example: you can simply swap two letters in string "ab". So we get "ba". - Second example: we can't change string "aa" into string "ab", because "aa" does not contain letter "b". | 500 | [
{
"input": "ab\nba",
"output": "YES"
},
{
"input": "aa\nab",
"output": "NO"
},
{
"input": "a\nza",
"output": "NO"
},
{
"input": "vvea\nvvae",
"output": "YES"
},
{
"input": "rtfabanpc\natfabrnpc",
"output": "YES"
},
{
"input": "mt\ntm",
"output": "Y... | 1,633,450,318 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 9 | 154 | 20,172,800 | s1=str(input())
s2=str(input())
l_1=[]
l_2=[]
l_1[:0]=s1
l_2[:0]=s2
p_1=sorted(l_1)
p_2=sorted(l_2)
if(len(l_1)!=len(l_2)):
print("NO")
else:
flag=False
for i in range (0,len(l_1)):
if(p_1[i]!=p_2[i]):
print("NO")
flag=True
break
if(flag==False):
for i in range (0,len(l_1)):
if(l_1[i]!=l_2[i]):
p=l_1[i]
q=l_2.index(p)
if l_1[q]==l_2[i]:
l_2[i]=p
l_2[q]=l_1[q]
break
for i in range (0,len(l_1)):
if(l_1[i]!=l_2[i]):
print("NO")
flag =True
break
if(flag==False):
print("YES")
| Title: Comparing Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Some dwarves that are finishing the StUDY (State University for Dwarven Youngsters) Bachelor courses, have been told "no genome, no degree". That means that all dwarves should write a thesis on genome. Dwarven genome is far from simple. It is represented by a string that consists of lowercase Latin letters.
Dwarf Misha has already chosen the subject for his thesis: determining by two dwarven genomes, whether they belong to the same race. Two dwarves belong to the same race if we can swap two characters in the first dwarf's genome and get the second dwarf's genome as a result. Help Dwarf Misha and find out whether two gnomes belong to the same race or not.
Input Specification:
The first line contains the first dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The second line contains the second dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The number of letters in each genome doesn't exceed 105. It is guaranteed that the strings that correspond to the genomes are different. The given genomes may have different length.
Output Specification:
Print "YES", if the dwarves belong to the same race. Otherwise, print "NO".
Demo Input:
['ab\nba\n', 'aa\nab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
- First example: you can simply swap two letters in string "ab". So we get "ba". - Second example: we can't change string "aa" into string "ab", because "aa" does not contain letter "b". | ```python
s1=str(input())
s2=str(input())
l_1=[]
l_2=[]
l_1[:0]=s1
l_2[:0]=s2
p_1=sorted(l_1)
p_2=sorted(l_2)
if(len(l_1)!=len(l_2)):
print("NO")
else:
flag=False
for i in range (0,len(l_1)):
if(p_1[i]!=p_2[i]):
print("NO")
flag=True
break
if(flag==False):
for i in range (0,len(l_1)):
if(l_1[i]!=l_2[i]):
p=l_1[i]
q=l_2.index(p)
if l_1[q]==l_2[i]:
l_2[i]=p
l_2[q]=l_1[q]
break
for i in range (0,len(l_1)):
if(l_1[i]!=l_2[i]):
print("NO")
flag =True
break
if(flag==False):
print("YES")
``` | 0 | |
3 | B | Lorry | PROGRAMMING | 1,900 | [
"greedy",
"sortings"
] | B. Lorry | 2 | 64 | A group of tourists is going to kayak and catamaran tour. A rented lorry has arrived to the boat depot to take kayaks and catamarans to the point of departure. It's known that all kayaks are of the same size (and each of them occupies the space of 1 cubic metre), and all catamarans are of the same size, but two times bigger than kayaks (and occupy the space of 2 cubic metres).
Each waterborne vehicle has a particular carrying capacity, and it should be noted that waterborne vehicles that look the same can have different carrying capacities. Knowing the truck body volume and the list of waterborne vehicles in the boat depot (for each one its type and carrying capacity are known), find out such set of vehicles that can be taken in the lorry, and that has the maximum total carrying capacity. The truck body volume of the lorry can be used effectively, that is to say you can always put into the lorry a waterborne vehicle that occupies the space not exceeding the free space left in the truck body. | The first line contains a pair of integer numbers *n* and *v* (1<=≤<=*n*<=≤<=105; 1<=≤<=*v*<=≤<=109), where *n* is the number of waterborne vehicles in the boat depot, and *v* is the truck body volume of the lorry in cubic metres. The following *n* lines contain the information about the waterborne vehicles, that is a pair of numbers *t**i*,<=*p**i* (1<=≤<=*t**i*<=≤<=2; 1<=≤<=*p**i*<=≤<=104), where *t**i* is the vehicle type (1 – a kayak, 2 – a catamaran), and *p**i* is its carrying capacity. The waterborne vehicles are enumerated in order of their appearance in the input file. | In the first line print the maximum possible carrying capacity of the set. In the second line print a string consisting of the numbers of the vehicles that make the optimal set. If the answer is not unique, print any of them. | [
"3 2\n1 2\n2 7\n1 3\n"
] | [
"7\n2\n"
] | none | 0 | [
{
"input": "3 2\n1 2\n2 7\n1 3",
"output": "7\n2"
},
{
"input": "5 3\n1 9\n2 9\n1 9\n2 10\n1 6",
"output": "24\n3 1 5"
},
{
"input": "10 10\n1 14\n2 15\n2 11\n2 12\n2 9\n1 14\n2 15\n1 9\n2 11\n2 6",
"output": "81\n6 1 7 2 4 9"
},
{
"input": "20 19\n2 47\n1 37\n1 48\n2 42\n2 4... | 1,635,669,251 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 62 | 4,505,600 | if __name__=='__main__':
a={}
b={}
dp={}
pre={}
n,v=map(int,input().split())
for i in range(1,n+1):
a[i],b[i]=map(int,input().split())
for i in range(1,n+1):
for j in range(v,a[i]-1,-1):
dp[j]=dp.get(j,0)
dp[j-a[i]]=dp.get(j-a[i],0)
if dp[j]<dp[j-a[i]]+b[i]:
dp[j]=dp[j-a[i]]+b[i]
pre[j]=i
print(dp[v])
ans=[]
while v!=0:
item=pre.get(v,0)
ans.append(item)
v=v-a[item]
ans.sort()
for i in ans:
print(i,end=' ')
| Title: Lorry
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
A group of tourists is going to kayak and catamaran tour. A rented lorry has arrived to the boat depot to take kayaks and catamarans to the point of departure. It's known that all kayaks are of the same size (and each of them occupies the space of 1 cubic metre), and all catamarans are of the same size, but two times bigger than kayaks (and occupy the space of 2 cubic metres).
Each waterborne vehicle has a particular carrying capacity, and it should be noted that waterborne vehicles that look the same can have different carrying capacities. Knowing the truck body volume and the list of waterborne vehicles in the boat depot (for each one its type and carrying capacity are known), find out such set of vehicles that can be taken in the lorry, and that has the maximum total carrying capacity. The truck body volume of the lorry can be used effectively, that is to say you can always put into the lorry a waterborne vehicle that occupies the space not exceeding the free space left in the truck body.
Input Specification:
The first line contains a pair of integer numbers *n* and *v* (1<=≤<=*n*<=≤<=105; 1<=≤<=*v*<=≤<=109), where *n* is the number of waterborne vehicles in the boat depot, and *v* is the truck body volume of the lorry in cubic metres. The following *n* lines contain the information about the waterborne vehicles, that is a pair of numbers *t**i*,<=*p**i* (1<=≤<=*t**i*<=≤<=2; 1<=≤<=*p**i*<=≤<=104), where *t**i* is the vehicle type (1 – a kayak, 2 – a catamaran), and *p**i* is its carrying capacity. The waterborne vehicles are enumerated in order of their appearance in the input file.
Output Specification:
In the first line print the maximum possible carrying capacity of the set. In the second line print a string consisting of the numbers of the vehicles that make the optimal set. If the answer is not unique, print any of them.
Demo Input:
['3 2\n1 2\n2 7\n1 3\n']
Demo Output:
['7\n2\n']
Note:
none | ```python
if __name__=='__main__':
a={}
b={}
dp={}
pre={}
n,v=map(int,input().split())
for i in range(1,n+1):
a[i],b[i]=map(int,input().split())
for i in range(1,n+1):
for j in range(v,a[i]-1,-1):
dp[j]=dp.get(j,0)
dp[j-a[i]]=dp.get(j-a[i],0)
if dp[j]<dp[j-a[i]]+b[i]:
dp[j]=dp[j-a[i]]+b[i]
pre[j]=i
print(dp[v])
ans=[]
while v!=0:
item=pre.get(v,0)
ans.append(item)
v=v-a[item]
ans.sort()
for i in ans:
print(i,end=' ')
``` | 0 |
33 | A | What is for dinner? | PROGRAMMING | 1,200 | [
"greedy",
"implementation"
] | A. What is for dinner? | 2 | 256 | In one little known, but very beautiful country called Waterland, lives a lovely shark Valerie. Like all the sharks, she has several rows of teeth, and feeds on crucians. One of Valerie's distinguishing features is that while eating one crucian she uses only one row of her teeth, the rest of the teeth are "relaxing".
For a long time our heroine had been searching the sea for crucians, but a great misfortune happened. Her teeth started to ache, and she had to see the local dentist, lobster Ashot. As a professional, Ashot quickly relieved Valerie from her toothache. Moreover, he managed to determine the cause of Valerie's developing caries (for what he was later nicknamed Cap).
It turned that Valerie eats too many crucians. To help Valerie avoid further reoccurrence of toothache, Ashot found for each Valerie's tooth its residual viability. Residual viability of a tooth is a value equal to the amount of crucians that Valerie can eat with this tooth. Every time Valerie eats a crucian, viability of all the teeth used for it will decrease by one. When the viability of at least one tooth becomes negative, the shark will have to see the dentist again.
Unhappy, Valerie came back home, where a portion of crucians was waiting for her. For sure, the shark couldn't say no to her favourite meal, but she had no desire to go back to the dentist. That's why she decided to eat the maximum amount of crucians from the portion but so that the viability of no tooth becomes negative.
As Valerie is not good at mathematics, she asked you to help her to find out the total amount of crucians that she can consume for dinner.
We should remind you that while eating one crucian Valerie uses exactly one row of teeth and the viability of each tooth from this row decreases by one. | The first line contains three integers *n*, *m*, *k* (1<=≤<=*m*<=≤<=*n*<=≤<=1000,<=0<=≤<=*k*<=≤<=106) — total amount of Valerie's teeth, amount of tooth rows and amount of crucians in Valerie's portion for dinner. Then follow *n* lines, each containing two integers: *r* (1<=≤<=*r*<=≤<=*m*) — index of the row, where belongs the corresponding tooth, and *c* (0<=≤<=*c*<=≤<=106) — its residual viability.
It's guaranteed that each tooth row has positive amount of teeth. | In the first line output the maximum amount of crucians that Valerie can consume for dinner. | [
"4 3 18\n2 3\n1 2\n3 6\n2 3\n",
"2 2 13\n1 13\n2 12\n"
] | [
"11\n",
"13\n"
] | none | 500 | [
{
"input": "4 3 18\n2 3\n1 2\n3 6\n2 3",
"output": "11"
},
{
"input": "2 2 13\n1 13\n2 12",
"output": "13"
},
{
"input": "5 4 8\n4 6\n4 5\n1 3\n2 0\n3 3",
"output": "8"
},
{
"input": "1 1 0\n1 3",
"output": "0"
},
{
"input": "7 1 30\n1 8\n1 15\n1 5\n1 17\n1 9\n1 1... | 1,652,198,339 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 60 | 0 | n,k=map(int,input().split())
l1=list(range(1,(n+1),2))
l2=list(range(2,(n+1),2))
l=l1+l2
print(l[k-1]) | Title: What is for dinner?
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
In one little known, but very beautiful country called Waterland, lives a lovely shark Valerie. Like all the sharks, she has several rows of teeth, and feeds on crucians. One of Valerie's distinguishing features is that while eating one crucian she uses only one row of her teeth, the rest of the teeth are "relaxing".
For a long time our heroine had been searching the sea for crucians, but a great misfortune happened. Her teeth started to ache, and she had to see the local dentist, lobster Ashot. As a professional, Ashot quickly relieved Valerie from her toothache. Moreover, he managed to determine the cause of Valerie's developing caries (for what he was later nicknamed Cap).
It turned that Valerie eats too many crucians. To help Valerie avoid further reoccurrence of toothache, Ashot found for each Valerie's tooth its residual viability. Residual viability of a tooth is a value equal to the amount of crucians that Valerie can eat with this tooth. Every time Valerie eats a crucian, viability of all the teeth used for it will decrease by one. When the viability of at least one tooth becomes negative, the shark will have to see the dentist again.
Unhappy, Valerie came back home, where a portion of crucians was waiting for her. For sure, the shark couldn't say no to her favourite meal, but she had no desire to go back to the dentist. That's why she decided to eat the maximum amount of crucians from the portion but so that the viability of no tooth becomes negative.
As Valerie is not good at mathematics, she asked you to help her to find out the total amount of crucians that she can consume for dinner.
We should remind you that while eating one crucian Valerie uses exactly one row of teeth and the viability of each tooth from this row decreases by one.
Input Specification:
The first line contains three integers *n*, *m*, *k* (1<=≤<=*m*<=≤<=*n*<=≤<=1000,<=0<=≤<=*k*<=≤<=106) — total amount of Valerie's teeth, amount of tooth rows and amount of crucians in Valerie's portion for dinner. Then follow *n* lines, each containing two integers: *r* (1<=≤<=*r*<=≤<=*m*) — index of the row, where belongs the corresponding tooth, and *c* (0<=≤<=*c*<=≤<=106) — its residual viability.
It's guaranteed that each tooth row has positive amount of teeth.
Output Specification:
In the first line output the maximum amount of crucians that Valerie can consume for dinner.
Demo Input:
['4 3 18\n2 3\n1 2\n3 6\n2 3\n', '2 2 13\n1 13\n2 12\n']
Demo Output:
['11\n', '13\n']
Note:
none | ```python
n,k=map(int,input().split())
l1=list(range(1,(n+1),2))
l2=list(range(2,(n+1),2))
l=l1+l2
print(l[k-1])
``` | -1 |
472 | A | Design Tutorial: Learn from Math | PROGRAMMING | 800 | [
"math",
"number theory"
] | null | null | One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.
For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem.
You are given an integer *n* no less than 12, express it as a sum of two composite numbers. | The only line contains an integer *n* (12<=≤<=*n*<=≤<=106). | Output two composite integers *x* and *y* (1<=<<=*x*,<=*y*<=<<=*n*) such that *x*<=+<=*y*<==<=*n*. If there are multiple solutions, you can output any of them. | [
"12\n",
"15\n",
"23\n",
"1000000\n"
] | [
"4 8\n",
"6 9\n",
"8 15\n",
"500000 500000\n"
] | In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well.
In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number. | 500 | [
{
"input": "12",
"output": "4 8"
},
{
"input": "15",
"output": "6 9"
},
{
"input": "23",
"output": "8 15"
},
{
"input": "1000000",
"output": "500000 500000"
},
{
"input": "63874",
"output": "4 63870"
},
{
"input": "14568",
"output": "4 14564"
},
... | 1,678,829,507 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 1 | 46 | 0 | x=int(input())
print(x//2,x//2+x%2) | Title: Design Tutorial: Learn from Math
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.
For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem.
You are given an integer *n* no less than 12, express it as a sum of two composite numbers.
Input Specification:
The only line contains an integer *n* (12<=≤<=*n*<=≤<=106).
Output Specification:
Output two composite integers *x* and *y* (1<=<<=*x*,<=*y*<=<<=*n*) such that *x*<=+<=*y*<==<=*n*. If there are multiple solutions, you can output any of them.
Demo Input:
['12\n', '15\n', '23\n', '1000000\n']
Demo Output:
['4 8\n', '6 9\n', '8 15\n', '500000 500000\n']
Note:
In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well.
In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number. | ```python
x=int(input())
print(x//2,x//2+x%2)
``` | 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... | 1,644,543,116 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 3 | 62 | 0 | n = int(input())
a = list(map(int, input().split()))
p = a.count(100)
s = a.count(200)
if (p-2*(s%2))%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()))
p = a.count(100)
s = a.count(200)
if (p-2*(s%2))%2 == 0:
print("YES")
else :
print("NO")
``` | 0 | |
870 | A | Search for Pretty Integers | PROGRAMMING | 900 | [
"brute force",
"implementation"
] | null | null | You are given two lists of non-zero digits.
Let's call an integer pretty if its (base 10) representation has at least one digit from the first list and at least one digit from the second list. What is the smallest positive pretty integer? | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=9) — the lengths of the first and the second lists, respectively.
The second line contains *n* distinct digits *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=9) — the elements of the first list.
The third line contains *m* distinct digits *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=9) — the elements of the second list. | Print the smallest pretty integer. | [
"2 3\n4 2\n5 7 6\n",
"8 8\n1 2 3 4 5 6 7 8\n8 7 6 5 4 3 2 1\n"
] | [
"25\n",
"1\n"
] | In the first example 25, 46, 24567 are pretty, as well as many other integers. The smallest among them is 25. 42 and 24 are not pretty because they don't have digits from the second list.
In the second example all integers that have at least one digit different from 9 are pretty. It's obvious that the smallest among them is 1, because it's the smallest positive integer. | 500 | [
{
"input": "2 3\n4 2\n5 7 6",
"output": "25"
},
{
"input": "8 8\n1 2 3 4 5 6 7 8\n8 7 6 5 4 3 2 1",
"output": "1"
},
{
"input": "1 1\n9\n1",
"output": "19"
},
{
"input": "9 1\n5 4 2 3 6 1 7 9 8\n9",
"output": "9"
},
{
"input": "5 3\n7 2 5 8 6\n3 1 9",
"output"... | 1,659,428,961 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 15 | 0 | l1 = []
l2 = []
n = int(input())
k = int(input())
for i in range(0, n):
c = int(input())
l1.append()
for i in range(0, k):
c = int(input())
l2.append()
x = min(l1)
y = min(l2)
if x == y:
print(x)
else:
print(str(x)+str(y)) | Title: Search for Pretty Integers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two lists of non-zero digits.
Let's call an integer pretty if its (base 10) representation has at least one digit from the first list and at least one digit from the second list. What is the smallest positive pretty integer?
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=9) — the lengths of the first and the second lists, respectively.
The second line contains *n* distinct digits *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=9) — the elements of the first list.
The third line contains *m* distinct digits *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=9) — the elements of the second list.
Output Specification:
Print the smallest pretty integer.
Demo Input:
['2 3\n4 2\n5 7 6\n', '8 8\n1 2 3 4 5 6 7 8\n8 7 6 5 4 3 2 1\n']
Demo Output:
['25\n', '1\n']
Note:
In the first example 25, 46, 24567 are pretty, as well as many other integers. The smallest among them is 25. 42 and 24 are not pretty because they don't have digits from the second list.
In the second example all integers that have at least one digit different from 9 are pretty. It's obvious that the smallest among them is 1, because it's the smallest positive integer. | ```python
l1 = []
l2 = []
n = int(input())
k = int(input())
for i in range(0, n):
c = int(input())
l1.append()
for i in range(0, k):
c = int(input())
l2.append()
x = min(l1)
y = min(l2)
if x == y:
print(x)
else:
print(str(x)+str(y))
``` | -1 | |
266 | A | Stones on the Table | PROGRAMMING | 800 | [
"implementation"
] | null | null | There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them. | The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stones on the table.
The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue. | Print a single integer — the answer to the problem. | [
"3\nRRG\n",
"5\nRRRRR\n",
"4\nBRBG\n"
] | [
"1\n",
"4\n",
"0\n"
] | none | 500 | [
{
"input": "3\nRRG",
"output": "1"
},
{
"input": "5\nRRRRR",
"output": "4"
},
{
"input": "4\nBRBG",
"output": "0"
},
{
"input": "1\nB",
"output": "0"
},
{
"input": "2\nBG",
"output": "0"
},
{
"input": "3\nBGB",
"output": "0"
},
{
"input": "... | 1,699,620,289 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 60 | 0 | n = int(input())
stones = input()
count=0
for i in range(n):
if (stones[i] == stones[i-1]):
count+=1
print(count) | Title: Stones on the Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* stones on the table in a row, each of them can be red, green or blue. Count the minimum number of stones to take from the table so that any two neighboring stones had different colors. Stones in a row are considered neighboring if there are no other stones between them.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=50) — the number of stones on the table.
The next line contains string *s*, which represents the colors of the stones. We'll consider the stones in the row numbered from 1 to *n* from left to right. Then the *i*-th character *s* equals "R", if the *i*-th stone is red, "G", if it's green and "B", if it's blue.
Output Specification:
Print a single integer — the answer to the problem.
Demo Input:
['3\nRRG\n', '5\nRRRRR\n', '4\nBRBG\n']
Demo Output:
['1\n', '4\n', '0\n']
Note:
none | ```python
n = int(input())
stones = input()
count=0
for i in range(n):
if (stones[i] == stones[i-1]):
count+=1
print(count)
``` | 0 | |
961 | B | Lecture Sleep | PROGRAMMING | 1,200 | [
"data structures",
"dp",
"implementation",
"two pointers"
] | null | null | Your friend Mishka and you attend a calculus lecture. Lecture lasts *n* minutes. Lecturer tells *a**i* theorems during the *i*-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array *t* of Mishka's behavior. If Mishka is asleep during the *i*-th minute of the lecture then *t**i* will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — *a**i* during the *i*-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for *k* minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and *n*<=-<=*k*<=+<=1. If you use it on some minute *i* then Mishka will be awake during minutes *j* such that and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up. | The first line of the input contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains *n* integer numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=104) — the number of theorems lecturer tells during the *i*-th minute.
The third line of the input contains *n* integer numbers *t*1,<=*t*2,<=... *t**n* (0<=≤<=*t**i*<=≤<=1) — type of Mishka's behavior at the *i*-th minute of the lecture. | Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up. | [
"6 3\n1 3 5 2 5 4\n1 1 0 1 0 0\n"
] | [
"16\n"
] | In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16. | 0 | [
{
"input": "6 3\n1 3 5 2 5 4\n1 1 0 1 0 0",
"output": "16"
},
{
"input": "5 3\n1 9999 10000 10000 10000\n0 0 0 0 0",
"output": "30000"
},
{
"input": "3 3\n10 10 10\n1 1 0",
"output": "30"
},
{
"input": "1 1\n423\n0",
"output": "423"
},
{
"input": "6 6\n1 3 5 2 5 4... | 1,662,396,574 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 218 | 7,372,800 | from collections import deque,Counter
from math import *
import sys
import random
from bisect import *
from functools import reduce
from sys import stdin
import copy
n,k = map(int,input().split())
arr = list(map(int,input().split()))
sleep = list(map(int,input().split()))
pre = [0 for i in range(n+1)]
one = 0
for i in range(n):
if sleep[i] == 1:
one+=arr[i]
arr[i] = 0
for i in range(0,n):
pre[i] = arr[i]+pre[i-1]
start = None
m = 0
for i in range(n):
if sleep[i] == 0:
x = n-1 if (i+k-1)>=n else i+k-1
val = (pre[x]-pre[i-1])
if val>m:
m = val
start = i
print(m+one)
| Title: Lecture Sleep
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend Mishka and you attend a calculus lecture. Lecture lasts *n* minutes. Lecturer tells *a**i* theorems during the *i*-th minute.
Mishka is really interested in calculus, though it is so hard to stay awake for all the time of lecture. You are given an array *t* of Mishka's behavior. If Mishka is asleep during the *i*-th minute of the lecture then *t**i* will be equal to 0, otherwise it will be equal to 1. When Mishka is awake he writes down all the theorems he is being told — *a**i* during the *i*-th minute. Otherwise he writes nothing.
You know some secret technique to keep Mishka awake for *k* minutes straight. However you can use it only once. You can start using it at the beginning of any minute between 1 and *n*<=-<=*k*<=+<=1. If you use it on some minute *i* then Mishka will be awake during minutes *j* such that and will write down all the theorems lecturer tells.
You task is to calculate the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Input Specification:
The first line of the input contains two integer numbers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105) — the duration of the lecture in minutes and the number of minutes you can keep Mishka awake.
The second line of the input contains *n* integer numbers *a*1,<=*a*2,<=... *a**n* (1<=≤<=*a**i*<=≤<=104) — the number of theorems lecturer tells during the *i*-th minute.
The third line of the input contains *n* integer numbers *t*1,<=*t*2,<=... *t**n* (0<=≤<=*t**i*<=≤<=1) — type of Mishka's behavior at the *i*-th minute of the lecture.
Output Specification:
Print only one integer — the maximum number of theorems Mishka will be able to write down if you use your technique only once to wake him up.
Demo Input:
['6 3\n1 3 5 2 5 4\n1 1 0 1 0 0\n']
Demo Output:
['16\n']
Note:
In the sample case the better way is to use the secret technique at the beginning of the third minute. Then the number of theorems Mishka will be able to write down will be equal to 16. | ```python
from collections import deque,Counter
from math import *
import sys
import random
from bisect import *
from functools import reduce
from sys import stdin
import copy
n,k = map(int,input().split())
arr = list(map(int,input().split()))
sleep = list(map(int,input().split()))
pre = [0 for i in range(n+1)]
one = 0
for i in range(n):
if sleep[i] == 1:
one+=arr[i]
arr[i] = 0
for i in range(0,n):
pre[i] = arr[i]+pre[i-1]
start = None
m = 0
for i in range(n):
if sleep[i] == 0:
x = n-1 if (i+k-1)>=n else i+k-1
val = (pre[x]-pre[i-1])
if val>m:
m = val
start = i
print(m+one)
``` | 3 | |
935 | A | Fafa and his Company | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly. | The input consists of a single line containing a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of employees in Fafa's company. | Print a single integer representing the answer to the problem. | [
"2\n",
"10\n"
] | [
"1\n",
"3\n"
] | In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them. | 500 | [
{
"input": "2",
"output": "1"
},
{
"input": "10",
"output": "3"
},
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "2"
},
{
"input": "6",
"output": "3"
},
{
"input": "13",
"output": "1"
},
{
"input": "100000",
"output": "35"
... | 1,661,408,625 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 62 | 0 | a=int(input())
ans=0
for i in range(1,a):
b=a-i
if b%i==0:
ans+=1
print(ans) | Title: Fafa and his Company
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fafa owns a company that works on huge projects. There are *n* employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best *l* employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees *n*, find in how many ways Fafa could choose the number of team leaders *l* in such a way that it is possible to divide employees between them evenly.
Input Specification:
The input consists of a single line containing a positive integer *n* (2<=≤<=*n*<=≤<=105) — the number of employees in Fafa's company.
Output Specification:
Print a single integer representing the answer to the problem.
Demo Input:
['2\n', '10\n']
Demo Output:
['1\n', '3\n']
Note:
In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility. - choose 2 employees as team leaders with 4 employees under the responsibility of each of them. - choose 5 employees as team leaders with 1 employee under the responsibility of each of them. | ```python
a=int(input())
ans=0
for i in range(1,a):
b=a-i
if b%i==0:
ans+=1
print(ans)
``` | 3 | |
785 | A | Anton and Polyhedrons | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number! | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron. | Output one number — the total number of faces in all the polyhedrons in Anton's collection. | [
"4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n",
"3\nDodecahedron\nOctahedron\nOctahedron\n"
] | [
"42\n",
"28\n"
] | In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces. | 500 | [
{
"input": "4\nIcosahedron\nCube\nTetrahedron\nDodecahedron",
"output": "42"
},
{
"input": "3\nDodecahedron\nOctahedron\nOctahedron",
"output": "28"
},
{
"input": "25\nIcosahedron\nOctahedron\nTetrahedron\nDodecahedron\nCube\nIcosahedron\nOctahedron\nCube\nTetrahedron\nIcosahedron\nIcosa... | 1,699,524,634 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 218 | 0 | Tetrahedron=4
Cube=6
Octahedron=8
Dodecahedron=12
Icosahedron=20
n=int(input())
r=0
for i in range(n):
s=input()
if s=="Tetrahedron":
r+=Tetrahedron
elif s=="Cube":
r+=Cube
elif s=="Octahedron":
r+=Octahedron
elif s=="Dodecahedron":
r+=Dodecahedron
elif s=="Icosahedron":
r+=Icosahedron
else:
print("Try again.....")
print(r) | Title: Anton and Polyhedrons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.
Output Specification:
Output one number — the total number of faces in all the polyhedrons in Anton's collection.
Demo Input:
['4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n', '3\nDodecahedron\nOctahedron\nOctahedron\n']
Demo Output:
['42\n', '28\n']
Note:
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces. | ```python
Tetrahedron=4
Cube=6
Octahedron=8
Dodecahedron=12
Icosahedron=20
n=int(input())
r=0
for i in range(n):
s=input()
if s=="Tetrahedron":
r+=Tetrahedron
elif s=="Cube":
r+=Cube
elif s=="Octahedron":
r+=Octahedron
elif s=="Dodecahedron":
r+=Dodecahedron
elif s=="Icosahedron":
r+=Icosahedron
else:
print("Try again.....")
print(r)
``` | 3 | |
867 | A | Between the Offices | PROGRAMMING | 800 | [
"implementation"
] | null | null | As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last *n* days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last *n* days, or not. | The first line of input contains single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days.
The second line contains a string of length *n* consisting of only capital 'S' and 'F' letters. If the *i*-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence. | Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower). | [
"4\nFSSF\n",
"2\nSF\n",
"10\nFFFFFFFFFF\n",
"10\nSSFFSFFSFF\n"
] | [
"NO\n",
"YES\n",
"NO\n",
"YES\n"
] | In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though. | 500 | [
{
"input": "4\nFSSF",
"output": "NO"
},
{
"input": "2\nSF",
"output": "YES"
},
{
"input": "10\nFFFFFFFFFF",
"output": "NO"
},
{
"input": "10\nSSFFSFFSFF",
"output": "YES"
},
{
"input": "20\nSFSFFFFSSFFFFSSSSFSS",
"output": "NO"
},
{
"input": "20\nSSFFF... | 1,586,533,704 | 2,147,483,647 | PyPy 3 | OK | TESTS | 34 | 140 | 0 | n = int(input())
s = str(input())
f=0
sf=0
for i in range(len(s)-1):
if s[i] == 'S' and s[i+1] == 'F':
f=f+1
elif s[i] == 'F' and s[i+1] == 'S':
sf=sf+1
if f>sf:
print("YES")
else:
print("NO") | Title: Between the Offices
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you may know, MemSQL has American offices in both San Francisco and Seattle. Being a manager in the company, you travel a lot between the two cities, always by plane.
You prefer flying from Seattle to San Francisco than in the other direction, because it's warmer in San Francisco. You are so busy that you don't remember the number of flights you have made in either direction. However, for each of the last *n* days you know whether you were in San Francisco office or in Seattle office. You always fly at nights, so you never were at both offices on the same day. Given this information, determine if you flew more times from Seattle to San Francisco during the last *n* days, or not.
Input Specification:
The first line of input contains single integer *n* (2<=≤<=*n*<=≤<=100) — the number of days.
The second line contains a string of length *n* consisting of only capital 'S' and 'F' letters. If the *i*-th letter is 'S', then you were in Seattle office on that day. Otherwise you were in San Francisco. The days are given in chronological order, i.e. today is the last day in this sequence.
Output Specification:
Print "YES" if you flew more times from Seattle to San Francisco, and "NO" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['4\nFSSF\n', '2\nSF\n', '10\nFFFFFFFFFF\n', '10\nSSFFSFFSFF\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n', 'YES\n']
Note:
In the first example you were initially at San Francisco, then flew to Seattle, were there for two days and returned to San Francisco. You made one flight in each direction, so the answer is "NO".
In the second example you just flew from Seattle to San Francisco, so the answer is "YES".
In the third example you stayed the whole period in San Francisco, so the answer is "NO".
In the fourth example if you replace 'S' with ones, and 'F' with zeros, you'll get the first few digits of π in binary representation. Not very useful information though. | ```python
n = int(input())
s = str(input())
f=0
sf=0
for i in range(len(s)-1):
if s[i] == 'S' and s[i+1] == 'F':
f=f+1
elif s[i] == 'F' and s[i+1] == 'S':
sf=sf+1
if f>sf:
print("YES")
else:
print("NO")
``` | 3 | |
996 | A | Hit the Lottery | PROGRAMMING | 800 | [
"dp",
"greedy"
] | null | null | Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance? | The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$). | Output the minimum number of bills that Allen could receive. | [
"125\n",
"43\n",
"1000000000\n"
] | [
"3\n",
"5\n",
"10000000\n"
] | In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills.
In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills.
In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills. | 500 | [
{
"input": "125",
"output": "3"
},
{
"input": "43",
"output": "5"
},
{
"input": "1000000000",
"output": "10000000"
},
{
"input": "4",
"output": "4"
},
{
"input": "5",
"output": "1"
},
{
"input": "1",
"output": "1"
},
{
"input": "74",
"o... | 1,682,917,278 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 62 | 3,891,200 | n=int(input())
denom=[100,20,10,5,1]
def dp(n,idx=0):
global denom
if idx>len(denom)-1 or n<0:
return 0
return min(1+dp(n-denom[idx],idx),dp(n,idx+1))
print(dp(n)) | Title: Hit the Lottery
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance?
Input Specification:
The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$).
Output Specification:
Output the minimum number of bills that Allen could receive.
Demo Input:
['125\n', '43\n', '1000000000\n']
Demo Output:
['3\n', '5\n', '10000000\n']
Note:
In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills.
In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills.
In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills. | ```python
n=int(input())
denom=[100,20,10,5,1]
def dp(n,idx=0):
global denom
if idx>len(denom)-1 or n<0:
return 0
return min(1+dp(n-denom[idx],idx),dp(n,idx+1))
print(dp(n))
``` | 0 | |
509 | A | Maximum in Table | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | An *n*<=×<=*n* table *a* is defined as follows:
- The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1.
These conditions define all the values in the table.
You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above. | The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table. | Print a single line containing a positive integer *m* — the maximum value in the table. | [
"1\n",
"5\n"
] | [
"1",
"70"
] | In the second test the rows of the table look as follows: | 0 | [
{
"input": "1",
"output": "1"
},
{
"input": "5",
"output": "70"
},
{
"input": "2",
"output": "2"
},
{
"input": "3",
"output": "6"
},
{
"input": "4",
"output": "20"
},
{
"input": "6",
"output": "252"
},
{
"input": "7",
"output": "924"
... | 1,609,072,815 | 2,147,483,647 | PyPy 3 | OK | TESTS | 10 | 155 | 0 | n = int(input())
if n == 1:
print(n)
else:
l = [[1] * n] * n
for i in range(1, n):
for j in range(1, n):
l[i][j] = l[i][j - 1] + l[i - 1][j]
print(l[n-1][n-1])
| Title: Maximum in Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An *n*<=×<=*n* table *a* is defined as follows:
- The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1.
These conditions define all the values in the table.
You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above.
Input Specification:
The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table.
Output Specification:
Print a single line containing a positive integer *m* — the maximum value in the table.
Demo Input:
['1\n', '5\n']
Demo Output:
['1', '70']
Note:
In the second test the rows of the table look as follows: | ```python
n = int(input())
if n == 1:
print(n)
else:
l = [[1] * n] * n
for i in range(1, n):
for j in range(1, n):
l[i][j] = l[i][j - 1] + l[i - 1][j]
print(l[n-1][n-1])
``` | 3 | |
230 | B | T-primes | PROGRAMMING | 1,300 | [
"binary search",
"implementation",
"math",
"number theory"
] | null | null | We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors.
You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not. | The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier. | Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't. | [
"3\n4 5 6\n"
] | [
"YES\nNO\nNO\n"
] | The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO". | 500 | [
{
"input": "3\n4 5 6",
"output": "YES\nNO\nNO"
},
{
"input": "2\n48 49",
"output": "NO\nYES"
},
{
"input": "10\n10 9 8 7 6 5 4 3 2 1",
"output": "NO\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO"
},
{
"input": "1\n36",
"output": "NO"
},
{
"input": "1\n999966000289",
"... | 1,696,926,600 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | import math
n = int(input())
a = [int(i) for i in input().split()]
a1 = [i for i in a if int(math.sqrt(i))**2 == int(i)]
p1=[0]*n
m = max(a1)
p = [0]*(m+1)
for i in range(2,m+1):
if p[i] == 0:
j=2
while i*j <m:
p[i*j] = 1
j+=1
for i in range(n):
if a[i] in a1:
if p[int(math.sqrt(a[]))] == 0:
print('YES')
else:
print('NO')
else:
print('NO') | Title: T-primes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors.
You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not.
Input Specification:
The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't.
Demo Input:
['3\n4 5 6\n']
Demo Output:
['YES\nNO\nNO\n']
Note:
The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO". | ```python
import math
n = int(input())
a = [int(i) for i in input().split()]
a1 = [i for i in a if int(math.sqrt(i))**2 == int(i)]
p1=[0]*n
m = max(a1)
p = [0]*(m+1)
for i in range(2,m+1):
if p[i] == 0:
j=2
while i*j <m:
p[i*j] = 1
j+=1
for i in range(n):
if a[i] in a1:
if p[int(math.sqrt(a[]))] == 0:
print('YES')
else:
print('NO')
else:
print('NO')
``` | -1 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Some time ago Leonid have known about idempotent functions. Idempotent function defined on a set {1,<=2,<=...,<=*n*} is such function , that for any the formula *g*(*g*(*x*))<==<=*g*(*x*) holds.
Let's denote as *f*(*k*)(*x*) the function *f* applied *k* times to the value *x*. More formally, *f*(1)(*x*)<==<=*f*(*x*), *f*(*k*)(*x*)<==<=*f*(*f*(*k*<=-<=1)(*x*)) for each *k*<=><=1.
You are given some function . Your task is to find minimum positive integer *k* such that function *f*(*k*)(*x*) is idempotent. | In the first line of the input there is a single integer *n* (1<=≤<=*n*<=≤<=200) — the size of function *f* domain.
In the second line follow *f*(1),<=*f*(2),<=...,<=*f*(*n*) (1<=≤<=*f*(*i*)<=≤<=*n* for each 1<=≤<=*i*<=≤<=*n*), the values of a function. | Output minimum *k* such that function *f*(*k*)(*x*) is idempotent. | [
"4\n1 2 2 4\n",
"3\n2 3 3\n",
"3\n2 3 1\n"
] | [
"1\n",
"2\n",
"3\n"
] | In the first sample test function *f*(*x*) = *f*<sup class="upper-index">(1)</sup>(*x*) is already idempotent since *f*(*f*(1)) = *f*(1) = 1, *f*(*f*(2)) = *f*(2) = 2, *f*(*f*(3)) = *f*(3) = 2, *f*(*f*(4)) = *f*(4) = 4.
In the second sample test:
- function *f*(*x*) = *f*<sup class="upper-index">(1)</sup>(*x*) isn't idempotent because *f*(*f*(1)) = 3 but *f*(1) = 2; - function *f*(*x*) = *f*<sup class="upper-index">(2)</sup>(*x*) is idempotent since for any *x* it is true that *f*<sup class="upper-index">(2)</sup>(*x*) = 3, so it is also true that *f*<sup class="upper-index">(2)</sup>(*f*<sup class="upper-index">(2)</sup>(*x*)) = 3.
In the third sample test:
- function *f*(*x*) = *f*<sup class="upper-index">(1)</sup>(*x*) isn't idempotent because *f*(*f*(1)) = 3 but *f*(1) = 2; - function *f*(*f*(*x*)) = *f*<sup class="upper-index">(2)</sup>(*x*) isn't idempotent because *f*<sup class="upper-index">(2)</sup>(*f*<sup class="upper-index">(2)</sup>(1)) = 2 but *f*<sup class="upper-index">(2)</sup>(1) = 3; - function *f*(*f*(*f*(*x*))) = *f*<sup class="upper-index">(3)</sup>(*x*) is idempotent since it is identity function: *f*<sup class="upper-index">(3)</sup>(*x*) = *x* for any <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/46a8c73444c646004dfde04451775e7af924d108.png" style="max-width: 100.0%;max-height: 100.0%;"/> meaning that the formula *f*<sup class="upper-index">(3)</sup>(*f*<sup class="upper-index">(3)</sup>(*x*)) = *f*<sup class="upper-index">(3)</sup>(*x*) also holds. | 0 | [
{
"input": "4\n1 2 2 4",
"output": "1"
},
{
"input": "3\n2 3 3",
"output": "2"
},
{
"input": "3\n2 3 1",
"output": "3"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "16\n1 4 13 9 11 16 14 6 5 12 7 8 15 2 3 10",
"output": "105"
},
{
"input": "20\n1 ... | 1,430,670,463 | 1,663 | Python 3 | OK | TESTS | 65 | 62 | 0 | n = int(input())
f = list(map(int, input().split()))
for i in range(n):
f[i] -= 1
def gcd(a, b):
while (a != 0 and b != 0):
a, b = b, a % b
return a + b
def lcm(a, b):
return a * b // gcd(a, b)
ans = 1
minn = 0
for i in range(n):
vis = [False] * n
cur = i
st, pr = 0, 0
while not vis[cur]:
vis[cur] = True
cur = f[cur]
st += 1
fs = cur
cur = i
while cur != fs:
pr += 1
cur = f[cur]
minn = max(minn, pr)
ans = lcm(ans, st - pr)
print(((max(0, minn - 1))// ans + 1) * ans)
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Some time ago Leonid have known about idempotent functions. Idempotent function defined on a set {1,<=2,<=...,<=*n*} is such function , that for any the formula *g*(*g*(*x*))<==<=*g*(*x*) holds.
Let's denote as *f*(*k*)(*x*) the function *f* applied *k* times to the value *x*. More formally, *f*(1)(*x*)<==<=*f*(*x*), *f*(*k*)(*x*)<==<=*f*(*f*(*k*<=-<=1)(*x*)) for each *k*<=><=1.
You are given some function . Your task is to find minimum positive integer *k* such that function *f*(*k*)(*x*) is idempotent.
Input Specification:
In the first line of the input there is a single integer *n* (1<=≤<=*n*<=≤<=200) — the size of function *f* domain.
In the second line follow *f*(1),<=*f*(2),<=...,<=*f*(*n*) (1<=≤<=*f*(*i*)<=≤<=*n* for each 1<=≤<=*i*<=≤<=*n*), the values of a function.
Output Specification:
Output minimum *k* such that function *f*(*k*)(*x*) is idempotent.
Demo Input:
['4\n1 2 2 4\n', '3\n2 3 3\n', '3\n2 3 1\n']
Demo Output:
['1\n', '2\n', '3\n']
Note:
In the first sample test function *f*(*x*) = *f*<sup class="upper-index">(1)</sup>(*x*) is already idempotent since *f*(*f*(1)) = *f*(1) = 1, *f*(*f*(2)) = *f*(2) = 2, *f*(*f*(3)) = *f*(3) = 2, *f*(*f*(4)) = *f*(4) = 4.
In the second sample test:
- function *f*(*x*) = *f*<sup class="upper-index">(1)</sup>(*x*) isn't idempotent because *f*(*f*(1)) = 3 but *f*(1) = 2; - function *f*(*x*) = *f*<sup class="upper-index">(2)</sup>(*x*) is idempotent since for any *x* it is true that *f*<sup class="upper-index">(2)</sup>(*x*) = 3, so it is also true that *f*<sup class="upper-index">(2)</sup>(*f*<sup class="upper-index">(2)</sup>(*x*)) = 3.
In the third sample test:
- function *f*(*x*) = *f*<sup class="upper-index">(1)</sup>(*x*) isn't idempotent because *f*(*f*(1)) = 3 but *f*(1) = 2; - function *f*(*f*(*x*)) = *f*<sup class="upper-index">(2)</sup>(*x*) isn't idempotent because *f*<sup class="upper-index">(2)</sup>(*f*<sup class="upper-index">(2)</sup>(1)) = 2 but *f*<sup class="upper-index">(2)</sup>(1) = 3; - function *f*(*f*(*f*(*x*))) = *f*<sup class="upper-index">(3)</sup>(*x*) is idempotent since it is identity function: *f*<sup class="upper-index">(3)</sup>(*x*) = *x* for any <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/46a8c73444c646004dfde04451775e7af924d108.png" style="max-width: 100.0%;max-height: 100.0%;"/> meaning that the formula *f*<sup class="upper-index">(3)</sup>(*f*<sup class="upper-index">(3)</sup>(*x*)) = *f*<sup class="upper-index">(3)</sup>(*x*) also holds. | ```python
n = int(input())
f = list(map(int, input().split()))
for i in range(n):
f[i] -= 1
def gcd(a, b):
while (a != 0 and b != 0):
a, b = b, a % b
return a + b
def lcm(a, b):
return a * b // gcd(a, b)
ans = 1
minn = 0
for i in range(n):
vis = [False] * n
cur = i
st, pr = 0, 0
while not vis[cur]:
vis[cur] = True
cur = f[cur]
st += 1
fs = cur
cur = i
while cur != fs:
pr += 1
cur = f[cur]
minn = max(minn, pr)
ans = lcm(ans, st - pr)
print(((max(0, minn - 1))// ans + 1) * ans)
``` | 3 | |
535 | B | Tavas and SaDDas | PROGRAMMING | 1,100 | [
"bitmasks",
"brute force",
"combinatorics",
"implementation"
] | null | null | Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you."
The problem is:
You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
If we sort all lucky numbers in increasing order, what's the 1-based index of *n*?
Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back. | The first and only line of input contains a lucky number *n* (1<=≤<=*n*<=≤<=109). | Print the index of *n* among all lucky numbers. | [
"4\n",
"7\n",
"77\n"
] | [
"1\n",
"2\n",
"6\n"
] | none | 1,000 | [
{
"input": "4",
"output": "1"
},
{
"input": "7",
"output": "2"
},
{
"input": "77",
"output": "6"
},
{
"input": "4",
"output": "1"
},
{
"input": "474744",
"output": "83"
},
{
"input": "777774",
"output": "125"
},
{
"input": "447",
"outpu... | 1,692,316,580 | 2,147,483,647 | PyPy 3 | OK | TESTS | 33 | 93 | 0 | n = input().replace("4" , "0").replace("7" , "1")
length = len(n)
num = 2 ** (length) - 2
for index , i in enumerate(n):
if int(i):
num += 2 ** (length - index - 1)
print(num + 1)
| Title: Tavas and SaDDas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you."
The problem is:
You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
If we sort all lucky numbers in increasing order, what's the 1-based index of *n*?
Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back.
Input Specification:
The first and only line of input contains a lucky number *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
Print the index of *n* among all lucky numbers.
Demo Input:
['4\n', '7\n', '77\n']
Demo Output:
['1\n', '2\n', '6\n']
Note:
none | ```python
n = input().replace("4" , "0").replace("7" , "1")
length = len(n)
num = 2 ** (length) - 2
for index , i in enumerate(n):
if int(i):
num += 2 ** (length - index - 1)
print(num + 1)
``` | 3 | |
476 | A | Dreamoon and Stairs | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Dreamoon wants to climb up a stair of *n* steps. He can climb 1 or 2 steps at each move. Dreamoon wants the number of moves to be a multiple of an integer *m*.
What is the minimal number of moves making him climb to the top of the stairs that satisfies his condition? | The single line contains two space separated integers *n*, *m* (0<=<<=*n*<=≤<=10000,<=1<=<<=*m*<=≤<=10). | Print a single integer — the minimal number of moves being a multiple of *m*. If there is no way he can climb satisfying condition print <=-<=1 instead. | [
"10 2\n",
"3 5\n"
] | [
"6\n",
"-1\n"
] | For the first sample, Dreamoon could climb in 6 moves with following sequence of steps: {2, 2, 2, 2, 1, 1}.
For the second sample, there are only three valid sequence of steps {2, 1}, {1, 2}, {1, 1, 1} with 2, 2, and 3 steps respectively. All these numbers are not multiples of 5. | 500 | [
{
"input": "10 2",
"output": "6"
},
{
"input": "3 5",
"output": "-1"
},
{
"input": "29 7",
"output": "21"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "1 2",
"output": "-1"
},
{
"input": "10000 2",
"output": "5000"
},
{
"input": "10000 ... | 1,673,623,159 | 2,147,483,647 | PyPy 3 | OK | TESTS | 35 | 77 | 0 | import math
def main():
n,m=map(int,input().split())
if n<m:
print(-1)
return
temp=n/2
if temp%m==0:
print(int(temp))
else:
temp/=m
temp=math.ceil(temp)
temp*=m
print(temp)
main()
| Title: Dreamoon and Stairs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dreamoon wants to climb up a stair of *n* steps. He can climb 1 or 2 steps at each move. Dreamoon wants the number of moves to be a multiple of an integer *m*.
What is the minimal number of moves making him climb to the top of the stairs that satisfies his condition?
Input Specification:
The single line contains two space separated integers *n*, *m* (0<=<<=*n*<=≤<=10000,<=1<=<<=*m*<=≤<=10).
Output Specification:
Print a single integer — the minimal number of moves being a multiple of *m*. If there is no way he can climb satisfying condition print <=-<=1 instead.
Demo Input:
['10 2\n', '3 5\n']
Demo Output:
['6\n', '-1\n']
Note:
For the first sample, Dreamoon could climb in 6 moves with following sequence of steps: {2, 2, 2, 2, 1, 1}.
For the second sample, there are only three valid sequence of steps {2, 1}, {1, 2}, {1, 1, 1} with 2, 2, and 3 steps respectively. All these numbers are not multiples of 5. | ```python
import math
def main():
n,m=map(int,input().split())
if n<m:
print(-1)
return
temp=n/2
if temp%m==0:
print(int(temp))
else:
temp/=m
temp=math.ceil(temp)
temp*=m
print(temp)
main()
``` | 3 | |
439 | A | Devu, the Singer and Churu, the Joker | PROGRAMMING | 900 | [
"greedy",
"implementation"
] | null | null | Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited.
Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly.
The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly.
People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest.
You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions:
- The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible.
If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event. | The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100). | If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event. | [
"3 30\n2 2 1\n",
"3 20\n2 1 1\n"
] | [
"5\n",
"-1\n"
] | Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way:
- First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes.
Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes.
Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1. | 500 | [
{
"input": "3 30\n2 2 1",
"output": "5"
},
{
"input": "3 20\n2 1 1",
"output": "-1"
},
{
"input": "50 10000\n5 4 10 9 9 6 7 7 7 3 3 7 7 4 7 4 10 10 1 7 10 3 1 4 5 7 2 10 10 10 2 3 4 7 6 1 8 4 7 3 8 8 4 10 1 1 9 2 6 1",
"output": "1943"
},
{
"input": "50 10000\n4 7 15 9 11 12 ... | 1,640,778,164 | 2,147,483,647 | Python 3 | OK | TESTS | 26 | 62 | 0 | n,d=map(int,input().split())
num=list(map(int,input().split()))
sum=0
for i in num:
sum+=i
if(sum+(n-1)*10>d):
print(-1)
else:
print(int((d-sum)/5))
| Title: Devu, the Singer and Churu, the Joker
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Devu is a renowned classical singer. He is invited to many big functions/festivals. Recently he was invited to "All World Classical Singing Festival". Other than Devu, comedian Churu was also invited.
Devu has provided organizers a list of the songs and required time for singing them. He will sing *n* songs, *i**th* song will take *t**i* minutes exactly.
The Comedian, Churu will crack jokes. All his jokes are of 5 minutes exactly.
People have mainly come to listen Devu. But you know that he needs rest of 10 minutes after each song. On the other hand, Churu being a very active person, doesn't need any rest.
You as one of the organizers should make an optimal sсhedule for the event. For some reasons you must follow the conditions:
- The duration of the event must be no more than *d* minutes; - Devu must complete all his songs; - With satisfying the two previous conditions the number of jokes cracked by Churu should be as many as possible.
If it is not possible to find a way to conduct all the songs of the Devu, output -1. Otherwise find out maximum number of jokes that Churu can crack in the grand event.
Input Specification:
The first line contains two space separated integers *n*, *d* (1<=≤<=*n*<=≤<=100; 1<=≤<=*d*<=≤<=10000). The second line contains *n* space-separated integers: *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=100).
Output Specification:
If there is no way to conduct all the songs of Devu, output -1. Otherwise output the maximum number of jokes that Churu can crack in the grand event.
Demo Input:
['3 30\n2 2 1\n', '3 20\n2 1 1\n']
Demo Output:
['5\n', '-1\n']
Note:
Consider the first example. The duration of the event is 30 minutes. There could be maximum 5 jokes in the following way:
- First Churu cracks a joke in 5 minutes. - Then Devu performs the first song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now Devu performs second song for 2 minutes. - Then Churu cracks 2 jokes in 10 minutes. - Now finally Devu will perform his last song in 1 minutes.
Total time spent is 5 + 2 + 10 + 2 + 10 + 1 = 30 minutes.
Consider the second example. There is no way of organizing Devu's all songs. Hence the answer is -1. | ```python
n,d=map(int,input().split())
num=list(map(int,input().split()))
sum=0
for i in num:
sum+=i
if(sum+(n-1)*10>d):
print(-1)
else:
print(int((d-sum)/5))
``` | 3 | |
46 | B | T-shirts from Sponsor | PROGRAMMING | 1,100 | [
"implementation"
] | B. T-shirts from Sponsor | 2 | 256 | One day a well-known sponsor of a well-known contest decided to give every participant of the contest a T-shirt as a present. A natural problem occurred: on the one hand, it is not clear how many T-shirts of what sizes should be ordered, and on the other hand, one doesn't want to order too many T-shirts (and we do not exactly paper the walls with the oversupply). After considerable brain racking and some pre-estimating, the sponsor representatives ordered a certain number of T-shirts of sizes S, M, L, XL and XXL. The T-shirts turned out to bring good luck, that's why on the contest day there built up a line of *K* participants willing to get one. Every contestant is characterized by his/her desired T-shirt size (so it happens that for all the participants it is also one of the sizes S, M, L, XL and XXL). The participants come up to get a T-shirt one by one and try to choose the most suitable one, choosing it like this. If there is still a T-shirt of the optimal size left, that he/she takes it without further ado. Otherwise the contestant would prefer to choose a T-shirt with the size as close to the optimal one as possible (the distance between neighboring sizes is considered equal to one). If the variant of choice is not unique, the contestant will take a T-shirt of a bigger size (in case he/she grows more). For example, for a person whose optimal size is L the preference list looks like this: L, XL, M, XXL, S. Using the data on how many T-shirts of every size had been ordered by the organizers, on the size of contestants in the line determine who got a T-shirt of what size. | The first line contains five non-negative integers *N**S*,<=*N**M*,<=*N**L*,<=*N**XL*,<=*N**XXL* not exceeding 1000 which represent the number of T-shirts of the corresponding sizes. The second line contains an integer *K* (1<=≤<=*K*<=≤<=1000) which represents the number of participants. The next *K* lines contain the optimal T-shirt sizes for the contestants. The sizes are given in the order in which the participants stand in the line. It is guaranteed that *N**S*<=+<=*N**M*<=+<=*N**L*<=+<=*N**XL*<=+<=*N**XXL*<=≥<=*K*. | For each contestant, print a line containing the size of the T-shirt he/she got. | [
"1 0 2 0 1\n3\nXL\nXXL\nM\n"
] | [
"XXL\nL\nL\n"
] | none | 0 | [
{
"input": "1 0 2 0 1\n3\nXL\nXXL\nM",
"output": "XXL\nL\nL"
},
{
"input": "0 0 0 0 1\n1\nS",
"output": "XXL"
},
{
"input": "1 0 1 0 1\n1\nS",
"output": "S"
},
{
"input": "1 0 0 0 1\n2\nS\nL",
"output": "S\nXXL"
},
{
"input": "1 1 1 1 1\n2\nXL\nM",
"output": "... | 1,627,264,083 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 1 | 122 | 6,963,200 | sizes = {"S":0,"M":1,"L":2,"XL":3,"XXL":4}
invsizes = {0:"S",1:"M",2:"L",3:"XL",4:"XXL"}
'''
def getKey(value):
keys = list(sizes.keys())
values = list(sizes.values())
return keys[values.index(value)]
'''
n = list(map(int,input().split()))
k = int(input())
for person in range(k):
preference = sizes[input()]
larger , smaller = preference+1,preference
currentchoice = smaller
for choice in range(5):
if n[currentchoice]!=0:
print(invsizes[currentchoice])
#print(n)
n[currentchoice]-=1
break
if larger==5:
currentchoice=smaller
smaller -=1
continue
if smaller ==5:
currentchoice = larger
larger+=1
continue
if choice%2==0:
currentchoice=larger
smaller-=1
if choice%2==1:
currentchoice=smaller
larger+=1
| Title: T-shirts from Sponsor
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day a well-known sponsor of a well-known contest decided to give every participant of the contest a T-shirt as a present. A natural problem occurred: on the one hand, it is not clear how many T-shirts of what sizes should be ordered, and on the other hand, one doesn't want to order too many T-shirts (and we do not exactly paper the walls with the oversupply). After considerable brain racking and some pre-estimating, the sponsor representatives ordered a certain number of T-shirts of sizes S, M, L, XL and XXL. The T-shirts turned out to bring good luck, that's why on the contest day there built up a line of *K* participants willing to get one. Every contestant is characterized by his/her desired T-shirt size (so it happens that for all the participants it is also one of the sizes S, M, L, XL and XXL). The participants come up to get a T-shirt one by one and try to choose the most suitable one, choosing it like this. If there is still a T-shirt of the optimal size left, that he/she takes it without further ado. Otherwise the contestant would prefer to choose a T-shirt with the size as close to the optimal one as possible (the distance between neighboring sizes is considered equal to one). If the variant of choice is not unique, the contestant will take a T-shirt of a bigger size (in case he/she grows more). For example, for a person whose optimal size is L the preference list looks like this: L, XL, M, XXL, S. Using the data on how many T-shirts of every size had been ordered by the organizers, on the size of contestants in the line determine who got a T-shirt of what size.
Input Specification:
The first line contains five non-negative integers *N**S*,<=*N**M*,<=*N**L*,<=*N**XL*,<=*N**XXL* not exceeding 1000 which represent the number of T-shirts of the corresponding sizes. The second line contains an integer *K* (1<=≤<=*K*<=≤<=1000) which represents the number of participants. The next *K* lines contain the optimal T-shirt sizes for the contestants. The sizes are given in the order in which the participants stand in the line. It is guaranteed that *N**S*<=+<=*N**M*<=+<=*N**L*<=+<=*N**XL*<=+<=*N**XXL*<=≥<=*K*.
Output Specification:
For each contestant, print a line containing the size of the T-shirt he/she got.
Demo Input:
['1 0 2 0 1\n3\nXL\nXXL\nM\n']
Demo Output:
['XXL\nL\nL\n']
Note:
none | ```python
sizes = {"S":0,"M":1,"L":2,"XL":3,"XXL":4}
invsizes = {0:"S",1:"M",2:"L",3:"XL",4:"XXL"}
'''
def getKey(value):
keys = list(sizes.keys())
values = list(sizes.values())
return keys[values.index(value)]
'''
n = list(map(int,input().split()))
k = int(input())
for person in range(k):
preference = sizes[input()]
larger , smaller = preference+1,preference
currentchoice = smaller
for choice in range(5):
if n[currentchoice]!=0:
print(invsizes[currentchoice])
#print(n)
n[currentchoice]-=1
break
if larger==5:
currentchoice=smaller
smaller -=1
continue
if smaller ==5:
currentchoice = larger
larger+=1
continue
if choice%2==0:
currentchoice=larger
smaller-=1
if choice%2==1:
currentchoice=smaller
larger+=1
``` | -1 |
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":... | 1,612,703,157 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 124 | 307,200 | result = ""
lines = []
for x in range(0, 3):
lines.append(input())
# code
obj = {
"A": 0,
"B": 0,
"C": 0,
}
result=""
for line in lines:
letter1 = line[0]
letter2 = line[2]
symbol = line[1]
if symbol == ">":
obj[letter1] +=1
else:
obj[letter2]+=1
# for letter in obj:
print(obj)
for k, v in sorted(obj.items(), key=lambda item: item[1]):
result+=k
print(result)
| 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
result = ""
lines = []
for x in range(0, 3):
lines.append(input())
# code
obj = {
"A": 0,
"B": 0,
"C": 0,
}
result=""
for line in lines:
letter1 = line[0]
letter2 = line[2]
symbol = line[1]
if symbol == ">":
obj[letter1] +=1
else:
obj[letter2]+=1
# for letter in obj:
print(obj)
for k, v in sorted(obj.items(), key=lambda item: item[1]):
result+=k
print(result)
``` | 0 |
1,009 | C | Annoying Present | PROGRAMMING | 1,700 | [
"greedy",
"math"
] | null | null | Alice got an array of length $n$ as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen $m$ changes of the following form. For some integer numbers $x$ and $d$, he chooses an arbitrary position $i$ ($1 \le i \le n$) and for every $j \in [1, n]$ adds $x + d \cdot dist(i, j)$ to the value of the $j$-th cell. $dist(i, j)$ is the distance between positions $i$ and $j$ (i.e. $dist(i, j) = |i - j|$, where $|x|$ is an absolute value of $x$).
For example, if Alice currently has an array $[2, 1, 2, 2]$ and Bob chooses position $3$ for $x = -1$ and $d = 2$ then the array will become $[2 - 1 + 2 \cdot 2,~1 - 1 + 2 \cdot 1,~2 - 1 + 2 \cdot 0,~2 - 1 + 2 \cdot 1]$ = $[5, 2, 1, 3]$. Note that Bob can't choose position $i$ outside of the array (that is, smaller than $1$ or greater than $n$).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve? | The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^5$) — the number of elements of the array and the number of changes.
Each of the next $m$ lines contains two integers $x_i$ and $d_i$ ($-10^3 \le x_i, d_i \le 10^3$) — the parameters for the $i$-th change. | Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed $10^{-6}$. | [
"2 3\n-1 3\n0 0\n-1 -4\n",
"3 2\n0 2\n5 0\n"
] | [
"-2.500000000000000\n",
"7.000000000000000\n"
] | none | 0 | [
{
"input": "2 3\n-1 3\n0 0\n-1 -4",
"output": "-2.500000000000000"
},
{
"input": "3 2\n0 2\n5 0",
"output": "7.000000000000000"
},
{
"input": "8 8\n-21 -60\n-96 -10\n-4 -19\n-27 -4\n57 -15\n-95 62\n-42 1\n-17 64",
"output": "-16.500000000000000"
},
{
"input": "1 1\n0 0",
... | 1,659,008,673 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 8 | 873 | 7,680,000 |
from decimal import Decimal
n,m=map(int,input().split())#以空格间隔
ave1=Decimal(0.0)
ave2=[]
ave3=Decimal(0.0)
for i in range(m):
x,y=map(int,input().split())#以空格间隔
x=Decimal(x)
y=Decimal(y)
ave1+=x
if(y<=0):
if(n%2==0):
ave2.append(y*Decimal(Decimal(n/2+1)*Decimal(n/2)/Decimal(2)/Decimal(n)+Decimal(n/2)*Decimal(n/2-1)/Decimal(2)/Decimal(n)))
else:
ave2.append(Decimal(2)*y*Decimal(Decimal(n/2+1)/Decimal(n)*Decimal(n/2)/Decimal(2)))
else:
ave2.append(y*Decimal(n-1)/Decimal(2))
for i in range(m):
ave3+=ave2[i]
print(round(ave1+ave3,15)) | Title: Annoying Present
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Alice got an array of length $n$ as a birthday present once again! This is the third year in a row!
And what is more disappointing, it is overwhelmengly boring, filled entirely with zeros. Bob decided to apply some changes to the array to cheer up Alice.
Bob has chosen $m$ changes of the following form. For some integer numbers $x$ and $d$, he chooses an arbitrary position $i$ ($1 \le i \le n$) and for every $j \in [1, n]$ adds $x + d \cdot dist(i, j)$ to the value of the $j$-th cell. $dist(i, j)$ is the distance between positions $i$ and $j$ (i.e. $dist(i, j) = |i - j|$, where $|x|$ is an absolute value of $x$).
For example, if Alice currently has an array $[2, 1, 2, 2]$ and Bob chooses position $3$ for $x = -1$ and $d = 2$ then the array will become $[2 - 1 + 2 \cdot 2,~1 - 1 + 2 \cdot 1,~2 - 1 + 2 \cdot 0,~2 - 1 + 2 \cdot 1]$ = $[5, 2, 1, 3]$. Note that Bob can't choose position $i$ outside of the array (that is, smaller than $1$ or greater than $n$).
Alice will be the happiest when the elements of the array are as big as possible. Bob claimed that the arithmetic mean value of the elements will work fine as a metric.
What is the maximum arithmetic mean value Bob can achieve?
Input Specification:
The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10^5$) — the number of elements of the array and the number of changes.
Each of the next $m$ lines contains two integers $x_i$ and $d_i$ ($-10^3 \le x_i, d_i \le 10^3$) — the parameters for the $i$-th change.
Output Specification:
Print the maximal average arithmetic mean of the elements Bob can achieve.
Your answer is considered correct if its absolute or relative error doesn't exceed $10^{-6}$.
Demo Input:
['2 3\n-1 3\n0 0\n-1 -4\n', '3 2\n0 2\n5 0\n']
Demo Output:
['-2.500000000000000\n', '7.000000000000000\n']
Note:
none | ```python
from decimal import Decimal
n,m=map(int,input().split())#以空格间隔
ave1=Decimal(0.0)
ave2=[]
ave3=Decimal(0.0)
for i in range(m):
x,y=map(int,input().split())#以空格间隔
x=Decimal(x)
y=Decimal(y)
ave1+=x
if(y<=0):
if(n%2==0):
ave2.append(y*Decimal(Decimal(n/2+1)*Decimal(n/2)/Decimal(2)/Decimal(n)+Decimal(n/2)*Decimal(n/2-1)/Decimal(2)/Decimal(n)))
else:
ave2.append(Decimal(2)*y*Decimal(Decimal(n/2+1)/Decimal(n)*Decimal(n/2)/Decimal(2)))
else:
ave2.append(y*Decimal(n-1)/Decimal(2))
for i in range(m):
ave3+=ave2[i]
print(round(ave1+ave3,15))
``` | 0 | |
617 | A | Elephant | PROGRAMMING | 800 | [
"math"
] | null | null | An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=><=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house. | The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house. | Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*. | [
"5\n",
"12\n"
] | [
"1\n",
"3\n"
] | In the first sample the elephant needs to make one step of length 5 to reach the point *x*.
In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves. | 500 | [
{
"input": "5",
"output": "1"
},
{
"input": "12",
"output": "3"
},
{
"input": "999999",
"output": "200000"
},
{
"input": "41",
"output": "9"
},
{
"input": "1000000",
"output": "200000"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
... | 1,698,158,299 | 2,147,483,647 | Python 3 | OK | TESTS | 34 | 46 | 0 | x = int(input())
steps = 0
for i in range(5, 0, -1):
single = x // i
x -= single * i
steps += single
if x == 0:
break
print(steps) | Title: Elephant
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An elephant decided to visit his friend. It turned out that the elephant's house is located at point 0 and his friend's house is located at point *x*(*x*<=><=0) of the coordinate line. In one step the elephant can move 1, 2, 3, 4 or 5 positions forward. Determine, what is the minimum number of steps he need to make in order to get to his friend's house.
Input Specification:
The first line of the input contains an integer *x* (1<=≤<=*x*<=≤<=1<=000<=000) — The coordinate of the friend's house.
Output Specification:
Print the minimum number of steps that elephant needs to make to get from point 0 to point *x*.
Demo Input:
['5\n', '12\n']
Demo Output:
['1\n', '3\n']
Note:
In the first sample the elephant needs to make one step of length 5 to reach the point *x*.
In the second sample the elephant can get to point *x* if he moves by 3, 5 and 4. There are other ways to get the optimal answer but the elephant cannot reach *x* in less than three moves. | ```python
x = int(input())
steps = 0
for i in range(5, 0, -1):
single = x // i
x -= single * i
steps += single
if x == 0:
break
print(steps)
``` | 3 | |
510 | A | Fox And Snake | PROGRAMMING | 800 | [
"implementation"
] | null | null | Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern. | The only line contains two integers: *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=50).
*n* is an odd number. | Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces. | [
"3 3\n",
"3 4\n",
"5 3\n",
"9 9\n"
] | [
"###\n..#\n###\n",
"####\n...#\n####\n",
"###\n..#\n###\n#..\n###\n",
"#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n"
] | none | 500 | [
{
"input": "3 3",
"output": "###\n..#\n###"
},
{
"input": "3 4",
"output": "####\n...#\n####"
},
{
"input": "5 3",
"output": "###\n..#\n###\n#..\n###"
},
{
"input": "9 9",
"output": "#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#... | 1,687,843,990 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 28 | 62 | 0 | n,m=map(int,input().split())
snake=[]
for _ in range(1,n+1):
if _%2!=0:
snake.append('#'*m)
elif _%4==2:
snake.append('.'*(m-1)+'#')
elif _%4==0:
snake.append('#'+'.'*(m-1))
for row in snake:
print(row) | Title: Fox And Snake
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern.
Input Specification:
The only line contains two integers: *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=50).
*n* is an odd number.
Output Specification:
Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces.
Demo Input:
['3 3\n', '3 4\n', '5 3\n', '9 9\n']
Demo Output:
['###\n..#\n###\n', '####\n...#\n####\n', '###\n..#\n###\n#..\n###\n', '#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n']
Note:
none | ```python
n,m=map(int,input().split())
snake=[]
for _ in range(1,n+1):
if _%2!=0:
snake.append('#'*m)
elif _%4==2:
snake.append('.'*(m-1)+'#')
elif _%4==0:
snake.append('#'+'.'*(m-1))
for row in snake:
print(row)
``` | 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"
},
... | 1,676,471,362 | 2,147,483,647 | Python 3 | OK | TESTS | 19 | 92 | 0 | stones1 = list(map(str,input()))
stones2 = list(map(str,input()))
i = 0
j = 0
while j < len(stones2):
if stones1[i] == stones2[j]:
i+=1
j += 1
print (i+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
stones1 = list(map(str,input()))
stones2 = list(map(str,input()))
i = 0
j = 0
while j < len(stones2):
if stones1[i] == stones2[j]:
i+=1
j += 1
print (i+1)
``` | 3 | |
291 | A | Spyke Talks | PROGRAMMING | 800 | [
"*special",
"implementation",
"sortings"
] | null | null | Polycarpus is the director of a large corporation. There are *n* secretaries working for the corporation, each of them corresponds via the famous Spyke VoIP system during the day. We know that when two people call each other via Spyke, the Spyke network assigns a unique ID to this call, a positive integer session number.
One day Polycarpus wondered which secretaries are talking via the Spyke and which are not. For each secretary, he wrote out either the session number of his call or a 0 if this secretary wasn't talking via Spyke at that moment.
Help Polycarpus analyze these data and find out the number of pairs of secretaries that are talking. If Polycarpus has made a mistake in the data and the described situation could not have taken place, say so.
Note that the secretaries can correspond via Spyke not only with each other, but also with the people from other places. Also, Spyke conferences aren't permitted — that is, one call connects exactly two people. | The first line contains integer *n* (1<=≤<=*n*<=≤<=103) — the number of secretaries in Polycarpus's corporation. The next line contains *n* space-separated integers: *id*1,<=*id*2,<=...,<=*id**n* (0<=≤<=*id**i*<=≤<=109). Number *id**i* equals the number of the call session of the *i*-th secretary, if the secretary is talking via Spyke, or zero otherwise.
Consider the secretaries indexed from 1 to *n* in some way. | Print a single integer — the number of pairs of chatting secretaries, or -1 if Polycarpus's got a mistake in his records and the described situation could not have taken place. | [
"6\n0 1 7 1 7 10\n",
"3\n1 1 1\n",
"1\n0\n"
] | [
"2\n",
"-1\n",
"0\n"
] | In the first test sample there are two Spyke calls between secretaries: secretary 2 and secretary 4, secretary 3 and secretary 5.
In the second test sample the described situation is impossible as conferences aren't allowed. | 500 | [
{
"input": "6\n0 1 7 1 7 10",
"output": "2"
},
{
"input": "3\n1 1 1",
"output": "-1"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "5\n2 2 1 1 3",
"output": "2"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "10\n4 21 3 21 21 1 1 2 2 3",
"ou... | 1,467,734,515 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 7 | 124 | 0 | n = int(input())
l = list(map(int,input().split()))
l.sort()
r = 0
bad = False
for i in range(n-1):
if l[i] == l[i+1]:
r += 1
if i < n-2 and l[i+2] == l[i]:
print(-1)
bad = True
break
if not bad :
print(r) | Title: Spyke Talks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus is the director of a large corporation. There are *n* secretaries working for the corporation, each of them corresponds via the famous Spyke VoIP system during the day. We know that when two people call each other via Spyke, the Spyke network assigns a unique ID to this call, a positive integer session number.
One day Polycarpus wondered which secretaries are talking via the Spyke and which are not. For each secretary, he wrote out either the session number of his call or a 0 if this secretary wasn't talking via Spyke at that moment.
Help Polycarpus analyze these data and find out the number of pairs of secretaries that are talking. If Polycarpus has made a mistake in the data and the described situation could not have taken place, say so.
Note that the secretaries can correspond via Spyke not only with each other, but also with the people from other places. Also, Spyke conferences aren't permitted — that is, one call connects exactly two people.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=103) — the number of secretaries in Polycarpus's corporation. The next line contains *n* space-separated integers: *id*1,<=*id*2,<=...,<=*id**n* (0<=≤<=*id**i*<=≤<=109). Number *id**i* equals the number of the call session of the *i*-th secretary, if the secretary is talking via Spyke, or zero otherwise.
Consider the secretaries indexed from 1 to *n* in some way.
Output Specification:
Print a single integer — the number of pairs of chatting secretaries, or -1 if Polycarpus's got a mistake in his records and the described situation could not have taken place.
Demo Input:
['6\n0 1 7 1 7 10\n', '3\n1 1 1\n', '1\n0\n']
Demo Output:
['2\n', '-1\n', '0\n']
Note:
In the first test sample there are two Spyke calls between secretaries: secretary 2 and secretary 4, secretary 3 and secretary 5.
In the second test sample the described situation is impossible as conferences aren't allowed. | ```python
n = int(input())
l = list(map(int,input().split()))
l.sort()
r = 0
bad = False
for i in range(n-1):
if l[i] == l[i+1]:
r += 1
if i < n-2 and l[i+2] == l[i]:
print(-1)
bad = True
break
if not bad :
print(r)
``` | 0 | |
706 | B | Interesting drink | PROGRAMMING | 1,100 | [
"binary search",
"dp",
"implementation"
] | null | null | Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in *n* different shops in the city. It's known that the price of one bottle in the shop *i* is equal to *x**i* coins.
Vasiliy plans to buy his favorite drink for *q* consecutive days. He knows, that on the *i*-th day he will be able to spent *m**i* coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola". | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains *n* integers *x**i* (1<=≤<=*x**i*<=≤<=100<=000) — prices of the bottles of the drink in the *i*-th shop.
The third line contains a single integer *q* (1<=≤<=*q*<=≤<=100<=000) — the number of days Vasiliy plans to buy the drink.
Then follow *q* lines each containing one integer *m**i* (1<=≤<=*m**i*<=≤<=109) — the number of coins Vasiliy can spent on the *i*-th day. | Print *q* integers. The *i*-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the *i*-th day. | [
"5\n3 10 8 6 11\n4\n1\n10\n3\n11\n"
] | [
"0\n4\n1\n5\n"
] | On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop. | 1,000 | [
{
"input": "5\n3 10 8 6 11\n4\n1\n10\n3\n11",
"output": "0\n4\n1\n5"
},
{
"input": "5\n868 987 714 168 123\n10\n424\n192\n795\n873\n117\n914\n735\n158\n631\n471",
"output": "2\n2\n3\n4\n0\n4\n3\n1\n2\n2"
},
{
"input": "3\n435 482 309\n7\n245\n241\n909\n745\n980\n29\n521",
"output": "... | 1,697,895,392 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 10 | 2,000 | 10,752,000 | n=int(input())
x=sorted(list(map(int,input().split())))
q=int(input())
ans=0
for i in range(q):
m=int(input())
for j in range(n):
if m>=x[j]:
ans+=1
else:
break
print(ans)
ans=0 | Title: Interesting drink
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in *n* different shops in the city. It's known that the price of one bottle in the shop *i* is equal to *x**i* coins.
Vasiliy plans to buy his favorite drink for *q* consecutive days. He knows, that on the *i*-th day he will be able to spent *m**i* coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains *n* integers *x**i* (1<=≤<=*x**i*<=≤<=100<=000) — prices of the bottles of the drink in the *i*-th shop.
The third line contains a single integer *q* (1<=≤<=*q*<=≤<=100<=000) — the number of days Vasiliy plans to buy the drink.
Then follow *q* lines each containing one integer *m**i* (1<=≤<=*m**i*<=≤<=109) — the number of coins Vasiliy can spent on the *i*-th day.
Output Specification:
Print *q* integers. The *i*-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the *i*-th day.
Demo Input:
['5\n3 10 8 6 11\n4\n1\n10\n3\n11\n']
Demo Output:
['0\n4\n1\n5\n']
Note:
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop. | ```python
n=int(input())
x=sorted(list(map(int,input().split())))
q=int(input())
ans=0
for i in range(q):
m=int(input())
for j in range(n):
if m>=x[j]:
ans+=1
else:
break
print(ans)
ans=0
``` | 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 ... | 1,383,640,325 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 92 | 204,800 | n = int(input())
a = b = 0
for i in range(n):
x, y = map(int, input().split())
a += x;
b += y;
print(min(n-a+b, n+a-b))
| 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
n = int(input())
a = b = 0
for i in range(n):
x, y = map(int, input().split())
a += x;
b += y;
print(min(n-a+b, n+a-b))
``` | 0 | |
540 | A | Combination Lock | PROGRAMMING | 800 | [
"implementation"
] | null | null | Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock.
The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that? | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of disks on the combination lock.
The second line contains a string of *n* digits — the original state of the disks.
The third line contains a string of *n* digits — Scrooge McDuck's combination that opens the lock. | Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock. | [
"5\n82195\n64723\n"
] | [
"13\n"
] | In the sample he needs 13 moves:
- 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 500 | [
{
"input": "5\n82195\n64723",
"output": "13"
},
{
"input": "12\n102021090898\n010212908089",
"output": "16"
},
{
"input": "1\n8\n1",
"output": "3"
},
{
"input": "2\n83\n57",
"output": "7"
},
{
"input": "10\n0728592530\n1362615763",
"output": "27"
},
{
... | 1,647,174,664 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 8 | 46 | 0 | def main(ar1, ar2):
res = 0
for i in range(len(ar1)):
a = ar1[i]
b = ar2[i]
if a > b:
k1 = a-b
k2 = 10 + b - a
elif b > a:
k1 = b-a
k2 = 10 + a - b
if k1 <= k2:
res += k1
else:
res += k2
return res
n = input()
arr1 = [int(i) for i in list(input())]
arr2 = [int(i) for i in list(input())]
print(main(arr1, arr2)) | Title: Combination Lock
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Scrooge McDuck keeps his most treasured savings in a home safe with a combination lock. Each time he wants to put there the treasures that he's earned fair and square, he has to open the lock.
The combination lock is represented by *n* rotating disks with digits from 0 to 9 written on them. Scrooge McDuck has to turn some disks so that the combination of digits on the disks forms a secret combination. In one move, he can rotate one disk one digit forwards or backwards. In particular, in one move he can go from digit 0 to digit 9 and vice versa. What minimum number of actions does he need for that?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of disks on the combination lock.
The second line contains a string of *n* digits — the original state of the disks.
The third line contains a string of *n* digits — Scrooge McDuck's combination that opens the lock.
Output Specification:
Print a single integer — the minimum number of moves Scrooge McDuck needs to open the lock.
Demo Input:
['5\n82195\n64723\n']
Demo Output:
['13\n']
Note:
In the sample he needs 13 moves:
- 1 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b8967f65a723782358b93eff9ce69f336817cf70.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 2 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/07fa58573ece0d32c4d555e498d2b24d2f70f36a.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 3 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/cc2275d9252aae35a6867c6a5b4ba7596e9a7626.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 4 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/b100aea470fcaaab4e9529b234ba0d7875943c10.png" style="max-width: 100.0%;max-height: 100.0%;"/> - 5 disk: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/eb2cbe4324cebca65b85816262a85e473cd65967.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
def main(ar1, ar2):
res = 0
for i in range(len(ar1)):
a = ar1[i]
b = ar2[i]
if a > b:
k1 = a-b
k2 = 10 + b - a
elif b > a:
k1 = b-a
k2 = 10 + a - b
if k1 <= k2:
res += k1
else:
res += k2
return res
n = input()
arr1 = [int(i) for i in list(input())]
arr2 = [int(i) for i in list(input())]
print(main(arr1, arr2))
``` | 0 | |
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"outpu... | 1,627,668,466 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 154 | 6,758,400 | l = list(map(int, input().split()))
Final=0
Final =(l[0]*l[1])//2
print(Final)
| Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
l = list(map(int, input().split()))
Final=0
Final =(l[0]*l[1])//2
print(Final)
``` | 3.948911 |
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... | 1,608,728,480 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #nineteen:n = 3,i = 1,e = 3,t = 1
s = input() #输入字符串
n = 0 #变量初始化
i = 0
e = 0
t = 0
for u in range(len(s)): #循环判断n,i,e,t的个数
if u == 'n' :
n += 1
elif u == 'i' :
i += 1
elif u == 'e' :
e += 1
elif u == 't' :
t += 1
else :
continue
if n < 3 or i < 1 or e < 3 or t < 1 : #特判
print(0)
else :
n1 = n // 3
i1 = i
e1 = e // 3
t1 = t
print(min(n1,i1,e1,t1)) | 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
#nineteen:n = 3,i = 1,e = 3,t = 1
s = input() #输入字符串
n = 0 #变量初始化
i = 0
e = 0
t = 0
for u in range(len(s)): #循环判断n,i,e,t的个数
if u == 'n' :
n += 1
elif u == 'i' :
i += 1
elif u == 'e' :
e += 1
elif u == 't' :
t += 1
else :
continue
if n < 3 or i < 1 or e < 3 or t < 1 : #特判
print(0)
else :
n1 = n // 3
i1 = i
e1 = e // 3
t1 = t
print(min(n1,i1,e1,t1))
``` | -1 | |
318 | A | Even Odds | PROGRAMMING | 900 | [
"math"
] | null | null | Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first *n*. He writes down the following sequence of numbers: firstly all odd integers from 1 to *n* (in ascending order), then all even integers from 1 to *n* (also in ascending order). Help our hero to find out which number will stand at the position number *k*. | The only line of input contains integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1012).
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. | Print the number that will stand at the position number *k* after Volodya's manipulations. | [
"10 3\n",
"7 7\n"
] | [
"5",
"6"
] | In the first sample Volodya's sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5. | 500 | [
{
"input": "10 3",
"output": "5"
},
{
"input": "7 7",
"output": "6"
},
{
"input": "7 1",
"output": "1"
},
{
"input": "7 2",
"output": "3"
},
{
"input": "8 5",
"output": "2"
},
{
"input": "8 3",
"output": "5"
},
{
"input": "8 4",
"output... | 1,693,225,605 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 62 | 0 | n , k = map(int , input().split())
if k<=(n+1)//2:
x = 2*k-1
else:
x = 2*(k-(n+1))//2
print(x) | Title: Even Odds
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first *n*. He writes down the following sequence of numbers: firstly all odd integers from 1 to *n* (in ascending order), then all even integers from 1 to *n* (also in ascending order). Help our hero to find out which number will stand at the position number *k*.
Input Specification:
The only line of input contains integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1012).
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:
Print the number that will stand at the position number *k* after Volodya's manipulations.
Demo Input:
['10 3\n', '7 7\n']
Demo Output:
['5', '6']
Note:
In the first sample Volodya's sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5. | ```python
n , k = map(int , input().split())
if k<=(n+1)//2:
x = 2*k-1
else:
x = 2*(k-(n+1))//2
print(x)
``` | 0 | |
404 | A | Valera and X | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Valera is a little boy. Yesterday he got a huge Math hometask at school, so Valera didn't have enough time to properly learn the English alphabet for his English lesson. Unfortunately, the English teacher decided to have a test on alphabet today. At the test Valera got a square piece of squared paper. The length of the side equals *n* squares (*n* is an odd number) and each unit square contains some small letter of the English alphabet.
Valera needs to know if the letters written on the square piece of paper form letter "X". Valera's teacher thinks that the letters on the piece of paper form an "X", if:
- on both diagonals of the square paper all letters are the same; - all other squares of the paper (they are not on the diagonals) contain the same letter that is different from the letters on the diagonals.
Help Valera, write the program that completes the described task for him. | The first line contains integer *n* (3<=≤<=*n*<=<<=300; *n* is odd). Each of the next *n* lines contains *n* small English letters — the description of Valera's paper. | Print string "YES", if the letters on the paper form letter "X". Otherwise, print string "NO". Print the strings without quotes. | [
"5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox\n",
"3\nwsw\nsws\nwsw\n",
"3\nxpx\npxp\nxpe\n"
] | [
"NO\n",
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox",
"output": "NO"
},
{
"input": "3\nwsw\nsws\nwsw",
"output": "YES"
},
{
"input": "3\nxpx\npxp\nxpe",
"output": "NO"
},
{
"input": "5\nliiil\nilili\niilii\nilili\nliiil",
"output": "YES"
},
{
"input": "7\nbwccccb\nck... | 1,664,355,470 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 46 | 0 | def solve(l):
c = l[0][0]
coord = [(0,0)]
for i in range(1,n):
coord.append((i,i))
if l[i][i] != c:
return False
for i in range(0, n):
coord.append( (n-1-i, i))
if l[n-1-i][i] != c:
return False
co = set(coord)
q = l[0][1]
for i in range(n):
for j in range(n):
if (i,j) not in co and l[i][j] != q:
return False
return True
n=int(input())
l = []
for i in range(n):
l.append(input())
if solve(l):
print ("YES")
else:
print ("NO")
| Title: Valera and X
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera is a little boy. Yesterday he got a huge Math hometask at school, so Valera didn't have enough time to properly learn the English alphabet for his English lesson. Unfortunately, the English teacher decided to have a test on alphabet today. At the test Valera got a square piece of squared paper. The length of the side equals *n* squares (*n* is an odd number) and each unit square contains some small letter of the English alphabet.
Valera needs to know if the letters written on the square piece of paper form letter "X". Valera's teacher thinks that the letters on the piece of paper form an "X", if:
- on both diagonals of the square paper all letters are the same; - all other squares of the paper (they are not on the diagonals) contain the same letter that is different from the letters on the diagonals.
Help Valera, write the program that completes the described task for him.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=<<=300; *n* is odd). Each of the next *n* lines contains *n* small English letters — the description of Valera's paper.
Output Specification:
Print string "YES", if the letters on the paper form letter "X". Otherwise, print string "NO". Print the strings without quotes.
Demo Input:
['5\nxooox\noxoxo\nsoxoo\noxoxo\nxooox\n', '3\nwsw\nsws\nwsw\n', '3\nxpx\npxp\nxpe\n']
Demo Output:
['NO\n', 'YES\n', 'NO\n']
Note:
none | ```python
def solve(l):
c = l[0][0]
coord = [(0,0)]
for i in range(1,n):
coord.append((i,i))
if l[i][i] != c:
return False
for i in range(0, n):
coord.append( (n-1-i, i))
if l[n-1-i][i] != c:
return False
co = set(coord)
q = l[0][1]
for i in range(n):
for j in range(n):
if (i,j) not in co and l[i][j] != q:
return False
return True
n=int(input())
l = []
for i in range(n):
l.append(input())
if solve(l):
print ("YES")
else:
print ("NO")
``` | 0 | |
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"
... | 1,687,369,329 | 2,147,483,647 | Python 3 | OK | TESTS | 55 | 46 | 0 | s = input()
c=0
fren = ["Danil", "Olya", "Slava", "Ann", "Nikita"]
for name in fren:
if name in s:
c += s.count(name)
if (c == 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()
c=0
fren = ["Danil", "Olya", "Slava", "Ann", "Nikita"]
for name in fren:
if name in s:
c += s.count(name)
if (c == 1):
print("Yes")
else:
print("No")
``` | 3 | |
5 | E | Bindian Signalizing | PROGRAMMING | 2,400 | [
"data structures"
] | E. Bindian Signalizing | 4 | 256 | Everyone knows that long ago on the territory of present-day Berland there lived Bindian tribes. Their capital was surrounded by *n* hills, forming a circle. On each hill there was a watchman, who watched the neighbourhood day and night.
In case of any danger the watchman could make a fire on the hill. One watchman could see the signal of another watchman, if on the circle arc connecting the two hills there was no hill higher than any of the two. As for any two hills there are two different circle arcs connecting them, the signal was seen if the above mentioned condition was satisfied on at least one of the arcs. For example, for any two neighbouring watchmen it is true that the signal of one will be seen by the other.
An important characteristics of this watch system was the amount of pairs of watchmen able to see each other's signals. You are to find this amount by the given heights of the hills. | The first line of the input data contains an integer number *n* (3<=≤<=*n*<=≤<=106), *n* — the amount of hills around the capital. The second line contains *n* numbers — heights of the hills in clockwise order. All height numbers are integer and lie between 1 and 109. | Print the required amount of pairs. | [
"5\n1 2 4 5 3\n"
] | [
"7\n"
] | none | 0 | [
{
"input": "5\n1 2 4 5 3",
"output": "7"
},
{
"input": "3\n2118 2118 2118",
"output": "3"
},
{
"input": "3\n2221 1976 2221",
"output": "3"
},
{
"input": "3\n140 989 2895",
"output": "3"
},
{
"input": "4\n2440 2440 2440 2440",
"output": "6"
},
{
"input"... | 1,691,562,422 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | def can_see_signal(heights, hill1, hill2):
if hill1 == hill2:
return True
min_height = min(heights[hill1], heights[hill2])
max_height = max(heights[hill1], heights[hill2])
for i in range(len(heights)):
if i != hill1 and i != hill2:
if heights[i] >= max_height:
return False
return True
def count_visible_signals(heights):
n = len(heights)
count = 0
for hill1 in range(n):
for hill2 in range(hill1 + 1, n):
if can_see_signal(heights, hill1, hill2):
count += 1
return count
| Title: Bindian Signalizing
Time Limit: 4 seconds
Memory Limit: 256 megabytes
Problem Description:
Everyone knows that long ago on the territory of present-day Berland there lived Bindian tribes. Their capital was surrounded by *n* hills, forming a circle. On each hill there was a watchman, who watched the neighbourhood day and night.
In case of any danger the watchman could make a fire on the hill. One watchman could see the signal of another watchman, if on the circle arc connecting the two hills there was no hill higher than any of the two. As for any two hills there are two different circle arcs connecting them, the signal was seen if the above mentioned condition was satisfied on at least one of the arcs. For example, for any two neighbouring watchmen it is true that the signal of one will be seen by the other.
An important characteristics of this watch system was the amount of pairs of watchmen able to see each other's signals. You are to find this amount by the given heights of the hills.
Input Specification:
The first line of the input data contains an integer number *n* (3<=≤<=*n*<=≤<=106), *n* — the amount of hills around the capital. The second line contains *n* numbers — heights of the hills in clockwise order. All height numbers are integer and lie between 1 and 109.
Output Specification:
Print the required amount of pairs.
Demo Input:
['5\n1 2 4 5 3\n']
Demo Output:
['7\n']
Note:
none | ```python
def can_see_signal(heights, hill1, hill2):
if hill1 == hill2:
return True
min_height = min(heights[hill1], heights[hill2])
max_height = max(heights[hill1], heights[hill2])
for i in range(len(heights)):
if i != hill1 and i != hill2:
if heights[i] >= max_height:
return False
return True
def count_visible_signals(heights):
n = len(heights)
count = 0
for hill1 in range(n):
for hill2 in range(hill1 + 1, n):
if can_see_signal(heights, hill1, hill2):
count += 1
return count
``` | 0 |
868 | A | Bark to Unlock | PROGRAMMING | 900 | [
"brute force",
"implementation",
"strings"
] | null | null | As technologies develop, manufacturers are making the process of unlocking a phone as user-friendly as possible. To unlock its new phone, Arkady's pet dog Mu-mu has to bark the password once. The phone represents a password as a string of two lowercase English letters.
Mu-mu's enemy Kashtanka wants to unlock Mu-mu's phone to steal some sensible information, but it can only bark *n* distinct words, each of which can be represented as a string of two lowercase English letters. Kashtanka wants to bark several words (not necessarily distinct) one after another to pronounce a string containing the password as a substring. Tell if it's possible to unlock the phone in this way, or not. | The first line contains two lowercase English letters — the password on the phone.
The second line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of words Kashtanka knows.
The next *n* lines contain two lowercase English letters each, representing the words Kashtanka knows. The words are guaranteed to be distinct. | Print "YES" if Kashtanka can bark several words in a line forming a string containing the password, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower). | [
"ya\n4\nah\noy\nto\nha\n",
"hp\n2\nht\ntp\n",
"ah\n1\nha\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | In the first example the password is "ya", and Kashtanka can bark "oy" and then "ah", and then "ha" to form the string "oyahha" which contains the password. So, the answer is "YES".
In the second example Kashtanka can't produce a string containing password as a substring. Note that it can bark "ht" and then "tp" producing "http", but it doesn't contain the password "hp" as a substring.
In the third example the string "hahahaha" contains "ah" as a substring. | 250 | [
{
"input": "ya\n4\nah\noy\nto\nha",
"output": "YES"
},
{
"input": "hp\n2\nht\ntp",
"output": "NO"
},
{
"input": "ah\n1\nha",
"output": "YES"
},
{
"input": "bb\n4\nba\nab\naa\nbb",
"output": "YES"
},
{
"input": "bc\n4\nca\nba\nbb\ncc",
"output": "YES"
},
{
... | 1,509,408,786 | 2,147,483,647 | Python 3 | OK | TESTS | 90 | 62 | 0 | #!/usr/bin/env
first, last = input()
num_words = int(input())
ending_first = False
starting_last = False
found = False
for _ in range(num_words):
wfirst, wlast = input()
if wfirst == first and wlast == last:
print('YES')
found = True
break
if wlast == first:
ending_first = True
if wfirst == last:
starting_last = True
if ending_first and starting_last:
print('YES')
found = True
break
if not found: print('NO')
| Title: Bark to Unlock
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As technologies develop, manufacturers are making the process of unlocking a phone as user-friendly as possible. To unlock its new phone, Arkady's pet dog Mu-mu has to bark the password once. The phone represents a password as a string of two lowercase English letters.
Mu-mu's enemy Kashtanka wants to unlock Mu-mu's phone to steal some sensible information, but it can only bark *n* distinct words, each of which can be represented as a string of two lowercase English letters. Kashtanka wants to bark several words (not necessarily distinct) one after another to pronounce a string containing the password as a substring. Tell if it's possible to unlock the phone in this way, or not.
Input Specification:
The first line contains two lowercase English letters — the password on the phone.
The second line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of words Kashtanka knows.
The next *n* lines contain two lowercase English letters each, representing the words Kashtanka knows. The words are guaranteed to be distinct.
Output Specification:
Print "YES" if Kashtanka can bark several words in a line forming a string containing the password, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower).
Demo Input:
['ya\n4\nah\noy\nto\nha\n', 'hp\n2\nht\ntp\n', 'ah\n1\nha\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first example the password is "ya", and Kashtanka can bark "oy" and then "ah", and then "ha" to form the string "oyahha" which contains the password. So, the answer is "YES".
In the second example Kashtanka can't produce a string containing password as a substring. Note that it can bark "ht" and then "tp" producing "http", but it doesn't contain the password "hp" as a substring.
In the third example the string "hahahaha" contains "ah" as a substring. | ```python
#!/usr/bin/env
first, last = input()
num_words = int(input())
ending_first = False
starting_last = False
found = False
for _ in range(num_words):
wfirst, wlast = input()
if wfirst == first and wlast == last:
print('YES')
found = True
break
if wlast == first:
ending_first = True
if wfirst == last:
starting_last = True
if ending_first and starting_last:
print('YES')
found = True
break
if not found: print('NO')
``` | 3 | |
991 | A | If at first you don't succeed... | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Each student eagerly awaits the day he would pass the exams successfully. Thus, Vasya was ready to celebrate, but, alas, he didn't pass it. However, many of Vasya's fellow students from the same group were more successful and celebrated after the exam.
Some of them celebrated in the BugDonalds restaurant, some of them — in the BeaverKing restaurant, the most successful ones were fast enough to celebrate in both of restaurants. Students which didn't pass the exam didn't celebrate in any of those restaurants and elected to stay home to prepare for their reexamination. However, this quickly bored Vasya and he started checking celebration photos on the Kilogramm. He found out that, in total, BugDonalds was visited by $A$ students, BeaverKing — by $B$ students and $C$ students visited both restaurants. Vasya also knows that there are $N$ students in his group.
Based on this info, Vasya wants to determine either if his data contradicts itself or, if it doesn't, how many students in his group didn't pass the exam. Can you help him so he won't waste his valuable preparation time? | The first line contains four integers — $A$, $B$, $C$ and $N$ ($0 \leq A, B, C, N \leq 100$). | If a distribution of $N$ students exists in which $A$ students visited BugDonalds, $B$ — BeaverKing, $C$ — both of the restaurants and at least one student is left home (it is known that Vasya didn't pass the exam and stayed at home), output one integer — amount of students (including Vasya) who did not pass the exam.
If such a distribution does not exist and Vasya made a mistake while determining the numbers $A$, $B$, $C$ or $N$ (as in samples 2 and 3), output $-1$. | [
"10 10 5 20\n",
"2 2 0 4\n",
"2 2 2 1\n"
] | [
"5",
"-1",
"-1"
] | The first sample describes following situation: $5$ only visited BugDonalds, $5$ students only visited BeaverKing, $5$ visited both of them and $5$ students (including Vasya) didn't pass the exam.
In the second sample $2$ students only visited BugDonalds and $2$ only visited BeaverKing, but that means all $4$ students in group passed the exam which contradicts the fact that Vasya didn't pass meaning that this situation is impossible.
The third sample describes a situation where $2$ students visited BugDonalds but the group has only $1$ which makes it clearly impossible. | 500 | [
{
"input": "10 10 5 20",
"output": "5"
},
{
"input": "2 2 0 4",
"output": "-1"
},
{
"input": "2 2 2 1",
"output": "-1"
},
{
"input": "98 98 97 100",
"output": "1"
},
{
"input": "1 5 2 10",
"output": "-1"
},
{
"input": "5 1 2 10",
"output": "-1"
}... | 1,601,637,640 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 109 | 0 | a,b,c,n=map(int,input().split())
print([-1,n+c-a-b][n-(a+b-c)>0]) | Title: If at first you don't succeed...
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Each student eagerly awaits the day he would pass the exams successfully. Thus, Vasya was ready to celebrate, but, alas, he didn't pass it. However, many of Vasya's fellow students from the same group were more successful and celebrated after the exam.
Some of them celebrated in the BugDonalds restaurant, some of them — in the BeaverKing restaurant, the most successful ones were fast enough to celebrate in both of restaurants. Students which didn't pass the exam didn't celebrate in any of those restaurants and elected to stay home to prepare for their reexamination. However, this quickly bored Vasya and he started checking celebration photos on the Kilogramm. He found out that, in total, BugDonalds was visited by $A$ students, BeaverKing — by $B$ students and $C$ students visited both restaurants. Vasya also knows that there are $N$ students in his group.
Based on this info, Vasya wants to determine either if his data contradicts itself or, if it doesn't, how many students in his group didn't pass the exam. Can you help him so he won't waste his valuable preparation time?
Input Specification:
The first line contains four integers — $A$, $B$, $C$ and $N$ ($0 \leq A, B, C, N \leq 100$).
Output Specification:
If a distribution of $N$ students exists in which $A$ students visited BugDonalds, $B$ — BeaverKing, $C$ — both of the restaurants and at least one student is left home (it is known that Vasya didn't pass the exam and stayed at home), output one integer — amount of students (including Vasya) who did not pass the exam.
If such a distribution does not exist and Vasya made a mistake while determining the numbers $A$, $B$, $C$ or $N$ (as in samples 2 and 3), output $-1$.
Demo Input:
['10 10 5 20\n', '2 2 0 4\n', '2 2 2 1\n']
Demo Output:
['5', '-1', '-1']
Note:
The first sample describes following situation: $5$ only visited BugDonalds, $5$ students only visited BeaverKing, $5$ visited both of them and $5$ students (including Vasya) didn't pass the exam.
In the second sample $2$ students only visited BugDonalds and $2$ only visited BeaverKing, but that means all $4$ students in group passed the exam which contradicts the fact that Vasya didn't pass meaning that this situation is impossible.
The third sample describes a situation where $2$ students visited BugDonalds but the group has only $1$ which makes it clearly impossible. | ```python
a,b,c,n=map(int,input().split())
print([-1,n+c-a-b][n-(a+b-c)>0])
``` | 0 | |
581 | A | Vasya the Hipster | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him? | The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got. | Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day. | [
"3 1\n",
"2 3\n",
"7 3\n"
] | [
"1 1\n",
"2 0\n",
"3 2\n"
] | In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day. | 500 | [
{
"input": "3 1",
"output": "1 1"
},
{
"input": "2 3",
"output": "2 0"
},
{
"input": "7 3",
"output": "3 2"
},
{
"input": "100 100",
"output": "100 0"
},
{
"input": "4 10",
"output": "4 3"
},
{
"input": "6 10",
"output": "6 2"
},
{
"input":... | 1,700,131,325 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 46 | 0 | a,b = map(int,input().split())
c = min(a,b)
d = max(a,b)
print(c,(d-c) // 2) | Title: Vasya the Hipster
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
Input Specification:
The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got.
Output Specification:
Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
Demo Input:
['3 1\n', '2 3\n', '7 3\n']
Demo Output:
['1 1\n', '2 0\n', '3 2\n']
Note:
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day. | ```python
a,b = map(int,input().split())
c = min(a,b)
d = max(a,b)
print(c,(d-c) // 2)
``` | 3 | |
591 | A | Wizards' Duel | PROGRAMMING | 900 | [
"implementation",
"math"
] | null | null | Harry Potter and He-Who-Must-Not-Be-Named engaged in a fight to the death once again. This time they are located at opposite ends of the corridor of length *l*. Two opponents simultaneously charge a deadly spell in the enemy. We know that the impulse of Harry's magic spell flies at a speed of *p* meters per second, and the impulse of You-Know-Who's magic spell flies at a speed of *q* meters per second.
The impulses are moving through the corridor toward each other, and at the time of the collision they turn round and fly back to those who cast them without changing their original speeds. Then, as soon as the impulse gets back to it's caster, the wizard reflects it and sends again towards the enemy, without changing the original speed of the impulse.
Since Harry has perfectly mastered the basics of magic, he knows that after the second collision both impulses will disappear, and a powerful explosion will occur exactly in the place of their collision. However, the young wizard isn't good at math, so he asks you to calculate the distance from his position to the place of the second meeting of the spell impulses, provided that the opponents do not change positions during the whole fight. | The first line of the input contains a single integer *l* (1<=≤<=*l*<=≤<=1<=000) — the length of the corridor where the fight takes place.
The second line contains integer *p*, the third line contains integer *q* (1<=≤<=*p*,<=*q*<=≤<=500) — the speeds of magical impulses for Harry Potter and He-Who-Must-Not-Be-Named, respectively. | Print a single real number — the distance from the end of the corridor, where Harry is located, to the place of the second meeting of the spell impulses. Your answer will be considered correct if its absolute or relative error will not exceed 10<=-<=4.
Namely: let's assume that your answer equals *a*, and the answer of the jury is *b*. The checker program will consider your answer correct if . | [
"100\n50\n50\n",
"199\n60\n40\n"
] | [
"50\n",
"119.4\n"
] | In the first sample the speeds of the impulses are equal, so both of their meetings occur exactly in the middle of the corridor. | 500 | [
{
"input": "100\n50\n50",
"output": "50"
},
{
"input": "199\n60\n40",
"output": "119.4"
},
{
"input": "1\n1\n1",
"output": "0.5"
},
{
"input": "1\n1\n500",
"output": "0.001996007984"
},
{
"input": "1\n500\n1",
"output": "0.998003992"
},
{
"input": "1\n... | 1,686,315,843 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 45 | 62 | 0 | import sys
input = sys.stdin.readline
n = int(input())
p = int(input())
q = int(input())
print(p / (p + q) * n) | Title: Wizards' Duel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Harry Potter and He-Who-Must-Not-Be-Named engaged in a fight to the death once again. This time they are located at opposite ends of the corridor of length *l*. Two opponents simultaneously charge a deadly spell in the enemy. We know that the impulse of Harry's magic spell flies at a speed of *p* meters per second, and the impulse of You-Know-Who's magic spell flies at a speed of *q* meters per second.
The impulses are moving through the corridor toward each other, and at the time of the collision they turn round and fly back to those who cast them without changing their original speeds. Then, as soon as the impulse gets back to it's caster, the wizard reflects it and sends again towards the enemy, without changing the original speed of the impulse.
Since Harry has perfectly mastered the basics of magic, he knows that after the second collision both impulses will disappear, and a powerful explosion will occur exactly in the place of their collision. However, the young wizard isn't good at math, so he asks you to calculate the distance from his position to the place of the second meeting of the spell impulses, provided that the opponents do not change positions during the whole fight.
Input Specification:
The first line of the input contains a single integer *l* (1<=≤<=*l*<=≤<=1<=000) — the length of the corridor where the fight takes place.
The second line contains integer *p*, the third line contains integer *q* (1<=≤<=*p*,<=*q*<=≤<=500) — the speeds of magical impulses for Harry Potter and He-Who-Must-Not-Be-Named, respectively.
Output Specification:
Print a single real number — the distance from the end of the corridor, where Harry is located, to the place of the second meeting of the spell impulses. Your answer will be considered correct if its absolute or relative error will not exceed 10<=-<=4.
Namely: let's assume that your answer equals *a*, and the answer of the jury is *b*. The checker program will consider your answer correct if .
Demo Input:
['100\n50\n50\n', '199\n60\n40\n']
Demo Output:
['50\n', '119.4\n']
Note:
In the first sample the speeds of the impulses are equal, so both of their meetings occur exactly in the middle of the corridor. | ```python
import sys
input = sys.stdin.readline
n = int(input())
p = int(input())
q = int(input())
print(p / (p + q) * n)
``` | 3 | |
493 | D | Vasya and Chess | PROGRAMMING | 1,700 | [
"constructive algorithms",
"games",
"math"
] | null | null | Vasya decided to learn to play chess. Classic chess doesn't seem interesting to him, so he plays his own sort of chess.
The queen is the piece that captures all squares on its vertical, horizontal and diagonal lines. If the cell is located on the same vertical, horizontal or diagonal line with queen, and the cell contains a piece of the enemy color, the queen is able to move to this square. After that the enemy's piece is removed from the board. The queen cannot move to a cell containing an enemy piece if there is some other piece between it and the queen.
There is an *n*<=×<=*n* chessboard. We'll denote a cell on the intersection of the *r*-th row and *c*-th column as (*r*,<=*c*). The square (1,<=1) contains the white queen and the square (1,<=*n*) contains the black queen. All other squares contain green pawns that don't belong to anyone.
The players move in turns. The player that moves first plays for the white queen, his opponent plays for the black queen.
On each move the player has to capture some piece with his queen (that is, move to a square that contains either a green pawn or the enemy queen). The player loses if either he cannot capture any piece during his move or the opponent took his queen during the previous move.
Help Vasya determine who wins if both players play with an optimal strategy on the board *n*<=×<=*n*. | The input contains a single number *n* (2<=≤<=*n*<=≤<=109) — the size of the board. | On the first line print the answer to problem — string "white" or string "black", depending on who wins if the both players play optimally.
If the answer is "white", then you should also print two integers *r* and *c* representing the cell (*r*,<=*c*), where the first player should make his first move to win. If there are multiple such cells, print the one with the minimum *r*. If there are still multiple squares, print the one with the minimum *c*. | [
"2\n",
"3\n"
] | [
"white\n1 2\n",
"black\n"
] | In the first sample test the white queen can capture the black queen at the first move, so the white player wins.
In the second test from the statement if the white queen captures the green pawn located on the central vertical line, then it will be captured by the black queen during the next move. So the only move for the white player is to capture the green pawn located at (2, 1).
Similarly, the black queen doesn't have any other options but to capture the green pawn located at (2, 3), otherwise if it goes to the middle vertical line, it will be captured by the white queen.
During the next move the same thing happens — neither the white, nor the black queen has other options rather than to capture green pawns situated above them. Thus, the white queen ends up on square (3, 1), and the black queen ends up on square (3, 3).
In this situation the white queen has to capture any of the green pawns located on the middle vertical line, after that it will be captured by the black queen. Thus, the player who plays for the black queen wins. | 1,500 | [
{
"input": "2",
"output": "white\n1 2"
},
{
"input": "3",
"output": "black"
},
{
"input": "4",
"output": "white\n1 2"
},
{
"input": "6",
"output": "white\n1 2"
},
{
"input": "10",
"output": "white\n1 2"
},
{
"input": "16",
"output": "white\n1 2"
... | 1,597,889,658 | 2,147,483,647 | Python 3 | OK | TESTS | 26 | 109 | 0 | space_numbers = int(input())
if (space_numbers % 2 != 0):
print("black")
else:
print("white")
print("{} {}".format(1, 2))
| Title: Vasya and Chess
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya decided to learn to play chess. Classic chess doesn't seem interesting to him, so he plays his own sort of chess.
The queen is the piece that captures all squares on its vertical, horizontal and diagonal lines. If the cell is located on the same vertical, horizontal or diagonal line with queen, and the cell contains a piece of the enemy color, the queen is able to move to this square. After that the enemy's piece is removed from the board. The queen cannot move to a cell containing an enemy piece if there is some other piece between it and the queen.
There is an *n*<=×<=*n* chessboard. We'll denote a cell on the intersection of the *r*-th row and *c*-th column as (*r*,<=*c*). The square (1,<=1) contains the white queen and the square (1,<=*n*) contains the black queen. All other squares contain green pawns that don't belong to anyone.
The players move in turns. The player that moves first plays for the white queen, his opponent plays for the black queen.
On each move the player has to capture some piece with his queen (that is, move to a square that contains either a green pawn or the enemy queen). The player loses if either he cannot capture any piece during his move or the opponent took his queen during the previous move.
Help Vasya determine who wins if both players play with an optimal strategy on the board *n*<=×<=*n*.
Input Specification:
The input contains a single number *n* (2<=≤<=*n*<=≤<=109) — the size of the board.
Output Specification:
On the first line print the answer to problem — string "white" or string "black", depending on who wins if the both players play optimally.
If the answer is "white", then you should also print two integers *r* and *c* representing the cell (*r*,<=*c*), where the first player should make his first move to win. If there are multiple such cells, print the one with the minimum *r*. If there are still multiple squares, print the one with the minimum *c*.
Demo Input:
['2\n', '3\n']
Demo Output:
['white\n1 2\n', 'black\n']
Note:
In the first sample test the white queen can capture the black queen at the first move, so the white player wins.
In the second test from the statement if the white queen captures the green pawn located on the central vertical line, then it will be captured by the black queen during the next move. So the only move for the white player is to capture the green pawn located at (2, 1).
Similarly, the black queen doesn't have any other options but to capture the green pawn located at (2, 3), otherwise if it goes to the middle vertical line, it will be captured by the white queen.
During the next move the same thing happens — neither the white, nor the black queen has other options rather than to capture green pawns situated above them. Thus, the white queen ends up on square (3, 1), and the black queen ends up on square (3, 3).
In this situation the white queen has to capture any of the green pawns located on the middle vertical line, after that it will be captured by the black queen. Thus, the player who plays for the black queen wins. | ```python
space_numbers = int(input())
if (space_numbers % 2 != 0):
print("black")
else:
print("white")
print("{} {}".format(1, 2))
``` | 3 | |
349 | A | Cinema Line | PROGRAMMING | 1,100 | [
"greedy",
"implementation"
] | null | null | The new "Die Hard" movie has just been released! There are *n* people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 ruble bill. A "Die Hard" ticket costs 25 rubles. Can the booking clerk sell a ticket to each person and give the change if he initially has no money and sells the tickets strictly in the order people follow in the line? | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of people in the line. The next line contains *n* integers, each of them equals 25, 50 or 100 — the values of the bills the people have. The numbers are given in the order from the beginning of the line (at the box office) to the end of the line. | Print "YES" (without the quotes) if the booking clerk can sell a ticket to each person and give the change. Otherwise print "NO". | [
"4\n25 25 50 50\n",
"2\n25 100\n",
"4\n50 50 25 25\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | none | 500 | [
{
"input": "4\n25 25 50 50",
"output": "YES"
},
{
"input": "2\n25 100",
"output": "NO"
},
{
"input": "4\n50 50 25 25",
"output": "NO"
},
{
"input": "3\n25 50 100",
"output": "NO"
},
{
"input": "10\n25 25 25 25 25 25 25 25 25 25",
"output": "YES"
},
{
"... | 1,621,695,725 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 280 | 5,324,800 | n = int(input())
l = [int(i) for i in input() .split()]
s1 = 0
s2 = 0
for i in range(n):
k = l[i] - 25
if(k == 75 and s1 >= 1 and s2 >= 1):
s1 -=1
s2-=1
elif(k == 75 and s1 >= 3):
s1 -=3
elif(k == 25 and s1 >= 1):
s1 -=1
elif(k == 0):
s1 = s1
s2 = s2
else:
print("NO")
exit()
if(l[i] == 50):
s2+=1
elif(l[i] == 25):
s1+=1
print("YES")
| Title: Cinema Line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The new "Die Hard" movie has just been released! There are *n* people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 ruble bill. A "Die Hard" ticket costs 25 rubles. Can the booking clerk sell a ticket to each person and give the change if he initially has no money and sells the tickets strictly in the order people follow in the line?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of people in the line. The next line contains *n* integers, each of them equals 25, 50 or 100 — the values of the bills the people have. The numbers are given in the order from the beginning of the line (at the box office) to the end of the line.
Output Specification:
Print "YES" (without the quotes) if the booking clerk can sell a ticket to each person and give the change. Otherwise print "NO".
Demo Input:
['4\n25 25 50 50\n', '2\n25 100\n', '4\n50 50 25 25\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none | ```python
n = int(input())
l = [int(i) for i in input() .split()]
s1 = 0
s2 = 0
for i in range(n):
k = l[i] - 25
if(k == 75 and s1 >= 1 and s2 >= 1):
s1 -=1
s2-=1
elif(k == 75 and s1 >= 3):
s1 -=3
elif(k == 25 and s1 >= 1):
s1 -=1
elif(k == 0):
s1 = s1
s2 = s2
else:
print("NO")
exit()
if(l[i] == 50):
s2+=1
elif(l[i] == 25):
s1+=1
print("YES")
``` | 3 | |
482 | A | Diverse Permutation | PROGRAMMING | 1,200 | [
"constructive algorithms",
"greedy"
] | null | null | Permutation *p* is an ordered set of integers *p*1,<=<=<=*p*2,<=<=<=...,<=<=<=*p**n*, consisting of *n* distinct positive integers not larger than *n*. We'll denote as *n* the length of permutation *p*1,<=<=<=*p*2,<=<=<=...,<=<=<=*p**n*.
Your task is to find such permutation *p* of length *n*, that the group of numbers |*p*1<=-<=*p*2|,<=|*p*2<=-<=*p*3|,<=...,<=|*p**n*<=-<=1<=-<=*p**n*| has exactly *k* distinct elements. | The single line of the input contains two space-separated positive integers *n*, *k* (1<=≤<=*k*<=<<=*n*<=≤<=105). | Print *n* integers forming the permutation. If there are multiple answers, print any of them. | [
"3 2\n",
"3 1\n",
"5 2\n"
] | [
"1 3 2\n",
"1 2 3\n",
"1 3 2 4 5\n"
] | By |*x*| we denote the absolute value of number *x*. | 500 | [
{
"input": "3 2",
"output": "1 3 2"
},
{
"input": "3 1",
"output": "1 2 3"
},
{
"input": "5 2",
"output": "1 3 2 4 5"
},
{
"input": "5 4",
"output": "1 5 2 4 3"
},
{
"input": "10 4",
"output": "1 10 2 9 8 7 6 5 4 3"
},
{
"input": "10 3",
"output": ... | 1,649,676,883 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 3 | 61 | 0 | n,k = map(int,input().split())
k = n-k
for i in range(k+1,0,-1):
print(i,end=' ')
lst=[ ]
for i in range(k+2,n+1):
lst.append(i)
while (len(lst)>0):
print(lst[-1],end=' ')
lst.pop(-1)
if (len(lst)<1):
break
print(lst[0],end=' ')
#lst.pop(-1)
lst.pop(0)
| Title: Diverse Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Permutation *p* is an ordered set of integers *p*1,<=<=<=*p*2,<=<=<=...,<=<=<=*p**n*, consisting of *n* distinct positive integers not larger than *n*. We'll denote as *n* the length of permutation *p*1,<=<=<=*p*2,<=<=<=...,<=<=<=*p**n*.
Your task is to find such permutation *p* of length *n*, that the group of numbers |*p*1<=-<=*p*2|,<=|*p*2<=-<=*p*3|,<=...,<=|*p**n*<=-<=1<=-<=*p**n*| has exactly *k* distinct elements.
Input Specification:
The single line of the input contains two space-separated positive integers *n*, *k* (1<=≤<=*k*<=<<=*n*<=≤<=105).
Output Specification:
Print *n* integers forming the permutation. If there are multiple answers, print any of them.
Demo Input:
['3 2\n', '3 1\n', '5 2\n']
Demo Output:
['1 3 2\n', '1 2 3\n', '1 3 2 4 5\n']
Note:
By |*x*| we denote the absolute value of number *x*. | ```python
n,k = map(int,input().split())
k = n-k
for i in range(k+1,0,-1):
print(i,end=' ')
lst=[ ]
for i in range(k+2,n+1):
lst.append(i)
while (len(lst)>0):
print(lst[-1],end=' ')
lst.pop(-1)
if (len(lst)<1):
break
print(lst[0],end=' ')
#lst.pop(-1)
lst.pop(0)
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question. | The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants | Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES"). | [
"1 3 2 1 2 1\n",
"1 1 1 1 1 99\n"
] | [
"YES\n",
"NO\n"
] | In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater. | 0 | [
{
"input": "1 3 2 1 2 1",
"output": "YES"
},
{
"input": "1 1 1 1 1 99",
"output": "NO"
},
{
"input": "1000 1000 1000 1000 1000 1000",
"output": "YES"
},
{
"input": "0 0 0 0 0 0",
"output": "YES"
},
{
"input": "633 609 369 704 573 416",
"output": "NO"
},
{
... | 1,519,754,089 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 4 | 77 | 6,041,600 | import sys
import copy
nums = [int(i) for i in input().split()]
# 6 choose 3
for i in range(4):
for j in range(i,5):
for k in range(j,6):
n = copy.deepcopy(nums)
n.remove(nums[i])
n.remove(nums[j])
n.remove(nums[k])
if nums[i] + nums[j] + nums[k] == sum(n):
print("YES")
sys.exit()
print("NO")
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a small but very proud high school it was decided to win ACM ICPC. This goal requires to compose as many teams of three as possible, but since there were only 6 students who wished to participate, the decision was to build exactly two teams.
After practice competition, participant number *i* got a score of *a**i*. Team score is defined as sum of scores of its participants. High school management is interested if it's possible to build two teams with equal scores. Your task is to answer that question.
Input Specification:
The single line contains six integers *a*1,<=...,<=*a*6 (0<=≤<=*a**i*<=≤<=1000) — scores of the participants
Output Specification:
Print "YES" (quotes for clarity), if it is possible to build teams with equal score, and "NO" otherwise.
You can print each character either upper- or lowercase ("YeS" and "yes" are valid when the answer is "YES").
Demo Input:
['1 3 2 1 2 1\n', '1 1 1 1 1 99\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, first team can be composed of 1st, 2nd and 6th participant, second — of 3rd, 4th and 5th: team scores are 1 + 3 + 1 = 2 + 1 + 2 = 5.
In the second sample, score of participant number 6 is too high: his team score will be definitely greater. | ```python
import sys
import copy
nums = [int(i) for i in input().split()]
# 6 choose 3
for i in range(4):
for j in range(i,5):
for k in range(j,6):
n = copy.deepcopy(nums)
n.remove(nums[i])
n.remove(nums[j])
n.remove(nums[k])
if nums[i] + nums[j] + nums[k] == sum(n):
print("YES")
sys.exit()
print("NO")
``` | -1 | |
365 | A | Good Number | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*). | The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109). | Print a single integer — the number of *k*-good numbers in *a*. | [
"10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n",
"2 1\n1\n10\n"
] | [
"10\n",
"1\n"
] | none | 500 | [
{
"input": "10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560",
"output": "10"
},
{
"input": "2 1\n1\n10",
"output": "1"
},
{
"input": "1 0\n1000000000",
"output": "1"
},
{
"input": "1 1\n1000000000",
"output": "1"
},
{
... | 1,687,538,663 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 30 | 0 | lengh,number=map(int,input().split())
c=0
for i in range(lengh):
s=(list(input()))
s.sort()
if len((s))>number:
if int(s[-1])==number:
c+=1
print(c) | Title: Good Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*).
Input Specification:
The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print a single integer — the number of *k*-good numbers in *a*.
Demo Input:
['10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n', '2 1\n1\n10\n']
Demo Output:
['10\n', '1\n']
Note:
none | ```python
lengh,number=map(int,input().split())
c=0
for i in range(lengh):
s=(list(input()))
s.sort()
if len((s))>number:
if int(s[-1])==number:
c+=1
print(c)
``` | 0 | |
1 | A | Theatre Square | PROGRAMMING | 1,000 | [
"math"
] | A. Theatre Square | 1 | 256 | Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square. | The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109). | Write the needed number of flagstones. | [
"6 6 4\n"
] | [
"4\n"
] | none | 0 | [
{
"input": "6 6 4",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 1 1",
"output": "2"
},
{
"input": "1 2 1",
"output": "2"
},
{
"input": "2 2 1",
"output": "4"
},
{
"input": "2 1 2",
"output": "1"
},
{
"input": "1 1 3"... | 1,664,822,021 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 46 | 0 | n,m,a = input().split()
n,m,a = [int(n), int(m),int(a)]
if n*m == 1 :
print("1")
else:
c = (m//a)+1
d = (n//a)+1
e = c*d
print(e) | Title: Theatre Square
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Theatre Square in the capital city of Berland has a rectangular shape with the size *n*<=×<=*m* meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size *a*<=×<=*a*.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input Specification:
The input contains three positive integer numbers in the first line: *n*,<=<=*m* and *a* (1<=≤<=<=*n*,<=*m*,<=*a*<=≤<=109).
Output Specification:
Write the needed number of flagstones.
Demo Input:
['6 6 4\n']
Demo Output:
['4\n']
Note:
none | ```python
n,m,a = input().split()
n,m,a = [int(n), int(m),int(a)]
if n*m == 1 :
print("1")
else:
c = (m//a)+1
d = (n//a)+1
e = c*d
print(e)
``` | 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... | 1,606,313,430 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 34 | 140 | 0 | string = input()
chars = [
'h',
'e',
'l',
'l',
'o'
]
currIndex = 0
for i in string:
if currIndex == 4:
break
if i == chars[currIndex]:
currIndex += 1
if currIndex == 4:
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
string = input()
chars = [
'h',
'e',
'l',
'l',
'o'
]
currIndex = 0
for i in string:
if currIndex == 4:
break
if i == chars[currIndex]:
currIndex += 1
if currIndex == 4:
print("YES")
else:
print("NO")
``` | 0 |
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":... | 1,600,051,669 | 2,147,483,647 | Python 3 | OK | TESTS | 50 | 218 | 307,200 | order = []
for i in range(3):
x = input()
if(x[1] == ">"):
order.append([x[2], x[0]])
else:
order.append([x[0], x[2]])
ls = []
for p, row in enumerate(order):
for q, col in enumerate(row):
ls.append([col, q])
d = {"A": 0, "B": 0, "C": 0}
for i in range(len(ls)):
d[ls[i][0]] += ls[i][1]
d = sorted(d.items(), key = lambda k: k[1])
if(d[2][1] == 2):
print(d[0][0]+ d[1][0]+ d[2][0])
else:
print("Impossible")
| 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
order = []
for i in range(3):
x = input()
if(x[1] == ">"):
order.append([x[2], x[0]])
else:
order.append([x[0], x[2]])
ls = []
for p, row in enumerate(order):
for q, col in enumerate(row):
ls.append([col, q])
d = {"A": 0, "B": 0, "C": 0}
for i in range(len(ls)):
d[ls[i][0]] += ls[i][1]
d = sorted(d.items(), key = lambda k: k[1])
if(d[2][1] == 2):
print(d[0][0]+ d[1][0]+ d[2][0])
else:
print("Impossible")
``` | 3.944928 |
38 | A | Army | PROGRAMMING | 800 | [
"implementation"
] | A. Army | 2 | 256 | The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank.
One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible.
Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream. | The first input line contains an integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=≤<=*d**i*<=≤<=100). The third input line contains two integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=*n*). The numbers on the lines are space-separated. | Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*. | [
"3\n5 6\n1 2\n",
"3\n5 6\n1 3\n"
] | [
"5\n",
"11\n"
] | none | 0 | [
{
"input": "3\n5 6\n1 2",
"output": "5"
},
{
"input": "3\n5 6\n1 3",
"output": "11"
},
{
"input": "2\n55\n1 2",
"output": "55"
},
{
"input": "3\n85 78\n1 3",
"output": "163"
},
{
"input": "4\n63 4 49\n2 3",
"output": "4"
},
{
"input": "5\n93 83 42 56\n... | 1,605,283,789 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 5 | 278 | 0 | n = int(input())
l = list(map(int,input().split()))
a,b = map(int,input().split())
y = b-a
sum1 = 0
if(a==1):
for i in range(y):
sum1 = sum1 + l[i]
print(sum1)
else:
print(l[a-1])
| Title: Army
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank.
One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible.
Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream.
Input Specification:
The first input line contains an integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=≤<=*d**i*<=≤<=100). The third input line contains two integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=*n*). The numbers on the lines are space-separated.
Output Specification:
Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*.
Demo Input:
['3\n5 6\n1 2\n', '3\n5 6\n1 3\n']
Demo Output:
['5\n', '11\n']
Note:
none | ```python
n = int(input())
l = list(map(int,input().split()))
a,b = map(int,input().split())
y = b-a
sum1 = 0
if(a==1):
for i in range(y):
sum1 = sum1 + l[i]
print(sum1)
else:
print(l[a-1])
``` | 0 |
644 | B | Processing Queries | PROGRAMMING | 1,700 | [
"*special",
"constructive algorithms",
"data structures",
"two pointers"
] | null | null | In this problem you have to simulate the workflow of one-thread server. There are *n* queries to process, the *i*-th will be received at moment *t**i* and needs to be processed for *d**i* units of time. All *t**i* are guaranteed to be distinct.
When a query appears server may react in three possible ways:
1. If server is free and query queue is empty, then server immediately starts to process this query. 1. If server is busy and there are less than *b* queries in the queue, then new query is added to the end of the queue. 1. If server is busy and there are already *b* queries pending in the queue, then new query is just rejected and will never be processed.
As soon as server finished to process some query, it picks new one from the queue (if it's not empty, of course). If a new query comes at some moment *x*, and the server finishes to process another query at exactly the same moment, we consider that first query is picked from the queue and only then new query appears.
For each query find the moment when the server will finish to process it or print -1 if this query will be rejected. | The first line of the input contains two integers *n* and *b* (1<=≤<=*n*,<=*b*<=≤<=200<=000) — the number of queries and the maximum possible size of the query queue.
Then follow *n* lines with queries descriptions (in chronological order). Each description consists of two integers *t**i* and *d**i* (1<=≤<=*t**i*,<=*d**i*<=≤<=109), where *t**i* is the moment of time when the *i*-th query appears and *d**i* is the time server needs to process it. It is guaranteed that *t**i*<=-<=1<=<<=*t**i* for all *i*<=><=1. | Print the sequence of *n* integers *e*1,<=*e*2,<=...,<=*e**n*, where *e**i* is the moment the server will finish to process the *i*-th query (queries are numbered in the order they appear in the input) or <=-<=1 if the corresponding query will be rejected. | [
"5 1\n2 9\n4 8\n10 9\n15 2\n19 1\n",
"4 1\n2 8\n4 8\n10 9\n15 2\n"
] | [
"11 19 -1 21 22 \n",
"10 18 27 -1 \n"
] | Consider the first sample.
1. The server will start to process first query at the moment 2 and will finish to process it at the moment 11. 1. At the moment 4 second query appears and proceeds to the queue. 1. At the moment 10 third query appears. However, the server is still busy with query 1, *b* = 1 and there is already query 2 pending in the queue, so third query is just rejected. 1. At the moment 11 server will finish to process first query and will take the second query from the queue. 1. At the moment 15 fourth query appears. As the server is currently busy it proceeds to the queue. 1. At the moment 19 two events occur simultaneously: server finishes to proceed the second query and the fifth query appears. As was said in the statement above, first server will finish to process the second query, then it will pick the fourth query from the queue and only then will the fifth query appear. As the queue is empty fifth query is proceed there. 1. Server finishes to process query number 4 at the moment 21. Query number 5 is picked from the queue. 1. Server finishes to process query number 5 at the moment 22. | 1,000 | [
{
"input": "5 1\n2 9\n4 8\n10 9\n15 2\n19 1",
"output": "11 19 -1 21 22 "
},
{
"input": "4 1\n2 8\n4 8\n10 9\n15 2",
"output": "10 18 27 -1 "
},
{
"input": "1 1\n1000000000 1000000000",
"output": "2000000000 "
},
{
"input": "4 3\n999999996 1000000000\n999999997 1000000000\n99... | 1,541,355,974 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 171 | 819,200 | '''
@Author: Pham T. Nhan
@Date: 4/11/2018
@Name: Processing Queries
@Link: http://codeforces.com/problemset/problem/644/B
'''
from queue import Queue
def main():
n, max_size = map(int, input().split())
queue = Queue(maxsize=max_size)
time_finish = 0
time_result = [0 for _ in range(n)]
place = 0
# Processing
# Read input until start time >= time_finish. Meanwhile, put to queue or -1
start = time_finish - 1
while start < time_finish and place < n:
start, process = map(int, input().split())
if queue.full() and start < time_finish:
time_result[place] = -1
place += 1
elif start < time_finish:
queue.put((start, process, place))
place += 1
# Server is free
elif start > time_finish:
while not queue.empty():
query = queue.get()
time_result[query[2]] = time_finish + query[1]
time_finish += query[1]
time_result[place] = max(time_finish, start) + process
place += 1
time_finish = max(time_finish, start) + process
elif start == time_finish:
while not queue.empty():
query = queue.get()
time_result[query[2]] = time_finish + query[1]
time_finish += query[1]
queue.put((start, process, place))
place += 1
# Process left after in queue
while not queue.empty():
query = queue.get()
time_result[query[2]] = time_finish + query[1]
time_finish += query[1]
for time in time_result:
print(time, end=' ')
if __name__ == '__main__':
main()
| Title: Processing Queries
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem you have to simulate the workflow of one-thread server. There are *n* queries to process, the *i*-th will be received at moment *t**i* and needs to be processed for *d**i* units of time. All *t**i* are guaranteed to be distinct.
When a query appears server may react in three possible ways:
1. If server is free and query queue is empty, then server immediately starts to process this query. 1. If server is busy and there are less than *b* queries in the queue, then new query is added to the end of the queue. 1. If server is busy and there are already *b* queries pending in the queue, then new query is just rejected and will never be processed.
As soon as server finished to process some query, it picks new one from the queue (if it's not empty, of course). If a new query comes at some moment *x*, and the server finishes to process another query at exactly the same moment, we consider that first query is picked from the queue and only then new query appears.
For each query find the moment when the server will finish to process it or print -1 if this query will be rejected.
Input Specification:
The first line of the input contains two integers *n* and *b* (1<=≤<=*n*,<=*b*<=≤<=200<=000) — the number of queries and the maximum possible size of the query queue.
Then follow *n* lines with queries descriptions (in chronological order). Each description consists of two integers *t**i* and *d**i* (1<=≤<=*t**i*,<=*d**i*<=≤<=109), where *t**i* is the moment of time when the *i*-th query appears and *d**i* is the time server needs to process it. It is guaranteed that *t**i*<=-<=1<=<<=*t**i* for all *i*<=><=1.
Output Specification:
Print the sequence of *n* integers *e*1,<=*e*2,<=...,<=*e**n*, where *e**i* is the moment the server will finish to process the *i*-th query (queries are numbered in the order they appear in the input) or <=-<=1 if the corresponding query will be rejected.
Demo Input:
['5 1\n2 9\n4 8\n10 9\n15 2\n19 1\n', '4 1\n2 8\n4 8\n10 9\n15 2\n']
Demo Output:
['11 19 -1 21 22 \n', '10 18 27 -1 \n']
Note:
Consider the first sample.
1. The server will start to process first query at the moment 2 and will finish to process it at the moment 11. 1. At the moment 4 second query appears and proceeds to the queue. 1. At the moment 10 third query appears. However, the server is still busy with query 1, *b* = 1 and there is already query 2 pending in the queue, so third query is just rejected. 1. At the moment 11 server will finish to process first query and will take the second query from the queue. 1. At the moment 15 fourth query appears. As the server is currently busy it proceeds to the queue. 1. At the moment 19 two events occur simultaneously: server finishes to proceed the second query and the fifth query appears. As was said in the statement above, first server will finish to process the second query, then it will pick the fourth query from the queue and only then will the fifth query appear. As the queue is empty fifth query is proceed there. 1. Server finishes to process query number 4 at the moment 21. Query number 5 is picked from the queue. 1. Server finishes to process query number 5 at the moment 22. | ```python
'''
@Author: Pham T. Nhan
@Date: 4/11/2018
@Name: Processing Queries
@Link: http://codeforces.com/problemset/problem/644/B
'''
from queue import Queue
def main():
n, max_size = map(int, input().split())
queue = Queue(maxsize=max_size)
time_finish = 0
time_result = [0 for _ in range(n)]
place = 0
# Processing
# Read input until start time >= time_finish. Meanwhile, put to queue or -1
start = time_finish - 1
while start < time_finish and place < n:
start, process = map(int, input().split())
if queue.full() and start < time_finish:
time_result[place] = -1
place += 1
elif start < time_finish:
queue.put((start, process, place))
place += 1
# Server is free
elif start > time_finish:
while not queue.empty():
query = queue.get()
time_result[query[2]] = time_finish + query[1]
time_finish += query[1]
time_result[place] = max(time_finish, start) + process
place += 1
time_finish = max(time_finish, start) + process
elif start == time_finish:
while not queue.empty():
query = queue.get()
time_result[query[2]] = time_finish + query[1]
time_finish += query[1]
queue.put((start, process, place))
place += 1
# Process left after in queue
while not queue.empty():
query = queue.get()
time_result[query[2]] = time_finish + query[1]
time_finish += query[1]
for time in time_result:
print(time, end=' ')
if __name__ == '__main__':
main()
``` | 0 | |
41 | B | Martian Dollar | PROGRAMMING | 1,400 | [
"brute force"
] | B. Martian Dollar | 2 | 256 | One day Vasya got hold of information on the Martian dollar course in bourles for the next *n* days. The buying prices and the selling prices for one dollar on day *i* are the same and are equal to *a**i*. Vasya has *b* bourles. He can buy a certain number of dollars and then sell it no more than once in *n* days. According to Martian laws, one can buy only an integer number of dollars. Which maximal sum of money in bourles can Vasya get by the end of day *n*? | The first line contains two integers *n* and *b* (1<=≤<=*n*,<=*b*<=≤<=2000) — the number of days and the initial number of money in bourles. The next line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2000) — the prices of Martian dollars. | Print the single number — which maximal sum of money in bourles can Vasya get by the end of day *n*. | [
"2 4\n3 7\n",
"4 10\n4 3 2 1\n",
"4 10\n4 2 3 1\n"
] | [
"8\n",
"10\n",
"15\n"
] | none | 1,000 | [
{
"input": "2 4\n3 7",
"output": "8"
},
{
"input": "4 10\n4 3 2 1",
"output": "10"
},
{
"input": "4 10\n4 2 3 1",
"output": "15"
},
{
"input": "2 755\n51 160",
"output": "2281"
},
{
"input": "3 385\n978 1604 1888",
"output": "385"
},
{
"input": "4 1663... | 1,420,656,079 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 12 | 716 | 0 | import sys
n, b = [int(x) for x in (sys.stdin.readline()).split()]
an = [int(x) for x in (sys.stdin.readline()).split()]
vmin = 2001
vmn = 0
vmx = 0
for i in range(n - 1):
for j in range(i + 1, n):
if(vmin > an[i] - an[j]):
vmin = an[i] - an[j]
vmn = an[i]
vmx = an[j]
elif(vmin == an[i] - an[j]):
if(an[i] < vmn):
vmn = an[i]
vmx = an[j]
if(vmin < 0):
t = int(b / vmn)
b = b % vmn
b += t * vmx
print(b)
else:
print(b)
| Title: Martian Dollar
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya got hold of information on the Martian dollar course in bourles for the next *n* days. The buying prices and the selling prices for one dollar on day *i* are the same and are equal to *a**i*. Vasya has *b* bourles. He can buy a certain number of dollars and then sell it no more than once in *n* days. According to Martian laws, one can buy only an integer number of dollars. Which maximal sum of money in bourles can Vasya get by the end of day *n*?
Input Specification:
The first line contains two integers *n* and *b* (1<=≤<=*n*,<=*b*<=≤<=2000) — the number of days and the initial number of money in bourles. The next line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2000) — the prices of Martian dollars.
Output Specification:
Print the single number — which maximal sum of money in bourles can Vasya get by the end of day *n*.
Demo Input:
['2 4\n3 7\n', '4 10\n4 3 2 1\n', '4 10\n4 2 3 1\n']
Demo Output:
['8\n', '10\n', '15\n']
Note:
none | ```python
import sys
n, b = [int(x) for x in (sys.stdin.readline()).split()]
an = [int(x) for x in (sys.stdin.readline()).split()]
vmin = 2001
vmn = 0
vmx = 0
for i in range(n - 1):
for j in range(i + 1, n):
if(vmin > an[i] - an[j]):
vmin = an[i] - an[j]
vmn = an[i]
vmx = an[j]
elif(vmin == an[i] - an[j]):
if(an[i] < vmn):
vmn = an[i]
vmx = an[j]
if(vmin < 0):
t = int(b / vmn)
b = b % vmn
b += t * vmx
print(b)
else:
print(b)
``` | 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... | 1,511,526,774 | 2,147,483,647 | Python 3 | OK | TESTS | 26 | 92 | 6,144,000 | # -*- coding: utf-8 -*-
import math
import collections
import bisect
import heapq
import time
import random
import itertools
import sys
"""
created by shhuan at 2017/11/24 20:29
"""
N, K = map(int, input().split())
ans = []
for i in range(N):
a = [int(x) for x in input().split()]
if any(K > v for v in a[1:]):
ans.append(i+1)
print(len(ans))
if ans:
print(" ".join(map(str, ans))) | 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
# -*- coding: utf-8 -*-
import math
import collections
import bisect
import heapq
import time
import random
import itertools
import sys
"""
created by shhuan at 2017/11/24 20:29
"""
N, K = map(int, input().split())
ans = []
for i in range(N):
a = [int(x) for x in input().split()]
if any(K > v for v in a[1:]):
ans.append(i+1)
print(len(ans))
if ans:
print(" ".join(map(str, ans)))
``` | 3 | |
233 | A | Perfect Permutation | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | A permutation is a sequence of integers *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. Let's denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size of permutation *p*1,<=*p*2,<=...,<=*p**n*.
Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation *p* that for any *i* (1<=≤<=*i*<=≤<=*n*) (*n* is the permutation size) the following equations hold *p**p**i*<==<=*i* and *p**i*<=≠<=*i*. Nickolas asks you to print any perfect permutation of size *n* for the given *n*. | A single line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the permutation size. | If a perfect permutation of size *n* doesn't exist, print a single integer -1. Otherwise print *n* distinct integers from 1 to *n*, *p*1,<=*p*2,<=...,<=*p**n* — permutation *p*, that is perfect. Separate printed numbers by whitespaces. | [
"1\n",
"2\n",
"4\n"
] | [
"-1\n",
"2 1 \n",
"2 1 4 3 \n"
] | none | 500 | [
{
"input": "1",
"output": "-1"
},
{
"input": "2",
"output": "2 1 "
},
{
"input": "4",
"output": "2 1 4 3 "
},
{
"input": "3",
"output": "-1"
},
{
"input": "5",
"output": "-1"
},
{
"input": "6",
"output": "2 1 4 3 6 5 "
},
{
"input": "7",
... | 1,647,205,461 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 60 | 0 | n = int(input())
l = [i for i in range(n,0,-1)]
print(*l) | Title: Perfect Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A permutation is a sequence of integers *p*1,<=*p*2,<=...,<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. Let's denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size of permutation *p*1,<=*p*2,<=...,<=*p**n*.
Nickolas adores permutations. He likes some permutations more than the others. He calls such permutations perfect. A perfect permutation is such permutation *p* that for any *i* (1<=≤<=*i*<=≤<=*n*) (*n* is the permutation size) the following equations hold *p**p**i*<==<=*i* and *p**i*<=≠<=*i*. Nickolas asks you to print any perfect permutation of size *n* for the given *n*.
Input Specification:
A single line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the permutation size.
Output Specification:
If a perfect permutation of size *n* doesn't exist, print a single integer -1. Otherwise print *n* distinct integers from 1 to *n*, *p*1,<=*p*2,<=...,<=*p**n* — permutation *p*, that is perfect. Separate printed numbers by whitespaces.
Demo Input:
['1\n', '2\n', '4\n']
Demo Output:
['-1\n', '2 1 \n', '2 1 4 3 \n']
Note:
none | ```python
n = int(input())
l = [i for i in range(n,0,-1)]
print(*l)
``` | 0 | |
750 | A | New Year and Hurry | PROGRAMMING | 800 | [
"binary search",
"brute force",
"implementation",
"math"
] | null | null | Limak is going to participate in a contest on the last day of the 2016. The contest will start at 20:00 and will last four hours, exactly until midnight. There will be *n* problems, sorted by difficulty, i.e. problem 1 is the easiest and problem *n* is the hardest. Limak knows it will take him 5·*i* minutes to solve the *i*-th problem.
Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier. He needs *k* minutes to get there from his house, where he will participate in the contest first.
How many problems can Limak solve if he wants to make it to the party? | The only line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=240) — the number of the problems in the contest and the number of minutes Limak needs to get to the party from his house. | Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier. | [
"3 222\n",
"4 190\n",
"7 1\n"
] | [
"2\n",
"4\n",
"7\n"
] | In the first sample, there are 3 problems and Limak needs 222 minutes to get to the party. The three problems require 5, 10 and 15 minutes respectively. Limak can spend 5 + 10 = 15 minutes to solve first two problems. Then, at 20:15 he can leave his house to get to the party at 23:57 (after 222 minutes). In this scenario Limak would solve 2 problems. He doesn't have enough time to solve 3 problems so the answer is 2.
In the second sample, Limak can solve all 4 problems in 5 + 10 + 15 + 20 = 50 minutes. At 20:50 he will leave the house and go to the party. He will get there exactly at midnight.
In the third sample, Limak needs only 1 minute to get to the party. He has enough time to solve all 7 problems. | 500 | [
{
"input": "3 222",
"output": "2"
},
{
"input": "4 190",
"output": "4"
},
{
"input": "7 1",
"output": "7"
},
{
"input": "10 135",
"output": "6"
},
{
"input": "10 136",
"output": "5"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 240",
... | 1,679,519,771 | 2,147,483,647 | Python 3 | OK | TESTS | 52 | 46 | 0 | n,k=input().split()
n=int(n);k=int(k)
contest_period= 4*60
minutes_used=0
problems=0
sm=0
for i in range(1,n+1):
minutes_used+=(i*5)
sm=k + minutes_used
if(sm <= contest_period):
problems+=1
else:
break
print(problems)
| Title: New Year and Hurry
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is going to participate in a contest on the last day of the 2016. The contest will start at 20:00 and will last four hours, exactly until midnight. There will be *n* problems, sorted by difficulty, i.e. problem 1 is the easiest and problem *n* is the hardest. Limak knows it will take him 5·*i* minutes to solve the *i*-th problem.
Limak's friends organize a New Year's Eve party and Limak wants to be there at midnight or earlier. He needs *k* minutes to get there from his house, where he will participate in the contest first.
How many problems can Limak solve if he wants to make it to the party?
Input Specification:
The only line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=10, 1<=≤<=*k*<=≤<=240) — the number of the problems in the contest and the number of minutes Limak needs to get to the party from his house.
Output Specification:
Print one integer, denoting the maximum possible number of problems Limak can solve so that he could get to the party at midnight or earlier.
Demo Input:
['3 222\n', '4 190\n', '7 1\n']
Demo Output:
['2\n', '4\n', '7\n']
Note:
In the first sample, there are 3 problems and Limak needs 222 minutes to get to the party. The three problems require 5, 10 and 15 minutes respectively. Limak can spend 5 + 10 = 15 minutes to solve first two problems. Then, at 20:15 he can leave his house to get to the party at 23:57 (after 222 minutes). In this scenario Limak would solve 2 problems. He doesn't have enough time to solve 3 problems so the answer is 2.
In the second sample, Limak can solve all 4 problems in 5 + 10 + 15 + 20 = 50 minutes. At 20:50 he will leave the house and go to the party. He will get there exactly at midnight.
In the third sample, Limak needs only 1 minute to get to the party. He has enough time to solve all 7 problems. | ```python
n,k=input().split()
n=int(n);k=int(k)
contest_period= 4*60
minutes_used=0
problems=0
sm=0
for i in range(1,n+1):
minutes_used+=(i*5)
sm=k + minutes_used
if(sm <= contest_period):
problems+=1
else:
break
print(problems)
``` | 3 | |
908 | A | New Year and Counting Cards | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | Your friend has *n* cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true. | The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit. | Print a single integer, the minimum number of cards you must turn over to verify your claim. | [
"ee\n",
"z\n",
"0ay1\n"
] | [
"2\n",
"0\n",
"2\n"
] | In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards. | 500 | [
{
"input": "ee",
"output": "2"
},
{
"input": "z",
"output": "0"
},
{
"input": "0ay1",
"output": "2"
},
{
"input": "0abcdefghijklmnopqrstuvwxyz1234567896",
"output": "10"
},
{
"input": "0a0a9e9e2i2i9o9o6u6u9z9z4x4x9b9b",
"output": "18"
},
{
"input": "01... | 1,583,077,662 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 93 | 0 | s=input()
c=0
for i in range(len(s)):
if s[i]=="a" or s[i]=="e" or s[i]=="i" or s[i]=="o" or s[i]=="u":
c=c+1
elif s[i]=="0" or s[i]=="2" or s[i]=="4" or s[i]=="6" or s[i]=="8":
c=c+1
else:
c=c
print(c) | Title: New Year and Counting Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend has *n* cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
Input Specification:
The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit.
Output Specification:
Print a single integer, the minimum number of cards you must turn over to verify your claim.
Demo Input:
['ee\n', 'z\n', '0ay1\n']
Demo Output:
['2\n', '0\n', '2\n']
Note:
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards. | ```python
s=input()
c=0
for i in range(len(s)):
if s[i]=="a" or s[i]=="e" or s[i]=="i" or s[i]=="o" or s[i]=="u":
c=c+1
elif s[i]=="0" or s[i]=="2" or s[i]=="4" or s[i]=="6" or s[i]=="8":
c=c+1
else:
c=c
print(c)
``` | 0 | |
985 | B | Switches and Lamps | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | You are given *n* switches and *m* lamps. The *i*-th switch turns on some subset of the lamps. This information is given as the matrix *a* consisting of *n* rows and *m* columns where *a**i*,<=*j*<==<=1 if the *i*-th switch turns on the *j*-th lamp and *a**i*,<=*j*<==<=0 if the *i*-th switch is not connected to the *j*-th lamp.
Initially all *m* lamps are turned off.
Switches change state only from "off" to "on". It means that if you press two or more switches connected to the same lamp then the lamp will be turned on after any of this switches is pressed and will remain its state even if any switch connected to this lamp is pressed afterwards.
It is guaranteed that if you push all *n* switches then all *m* lamps will be turned on.
Your think that you have too many switches and you would like to ignore one of them.
Your task is to say if there exists such a switch that if you will ignore (not use) it but press all the other *n*<=-<=1 switches then all the *m* lamps will be turned on. | The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=2000) — the number of the switches and the number of the lamps.
The following *n* lines contain *m* characters each. The character *a**i*,<=*j* is equal to '1' if the *i*-th switch turns on the *j*-th lamp and '0' otherwise.
It is guaranteed that if you press all *n* switches all *m* lamps will be turned on. | Print "YES" if there is a switch that if you will ignore it and press all the other *n*<=-<=1 switches then all *m* lamps will be turned on. Print "NO" if there is no such switch. | [
"4 5\n10101\n01000\n00111\n10000\n",
"4 5\n10100\n01000\n00110\n00101\n"
] | [
"YES\n",
"NO\n"
] | none | 0 | [
{
"input": "4 5\n10101\n01000\n00111\n10000",
"output": "YES"
},
{
"input": "4 5\n10100\n01000\n00110\n00101",
"output": "NO"
},
{
"input": "1 5\n11111",
"output": "NO"
},
{
"input": "10 1\n1\n0\n0\n0\n0\n0\n0\n0\n0\n1",
"output": "YES"
},
{
"input": "1 1\n1",
... | 1,527,050,462 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 2 | 3,000 | 4,608,000 | # -*- coding: utf-8 -*-
"""
Created on Wed May 23 06:30:24 2018
@author: Arsanuos
"""
def main():
n, m = [int(t) for t in input().split()]
arr = []
for i in range(n):
arr.append(input())
neg = 0
while neg != n - 1:
tmp = set()
for index in range(len(arr)):
if index == neg:
continue
item = arr[index]
for i in range(m):
if item[i] == '1':
tmp.add(i)
if len(tmp) == m:
print("YES")
return
neg +=1
print("NO")
if __name__ == "__main__":
main() | Title: Switches and Lamps
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given *n* switches and *m* lamps. The *i*-th switch turns on some subset of the lamps. This information is given as the matrix *a* consisting of *n* rows and *m* columns where *a**i*,<=*j*<==<=1 if the *i*-th switch turns on the *j*-th lamp and *a**i*,<=*j*<==<=0 if the *i*-th switch is not connected to the *j*-th lamp.
Initially all *m* lamps are turned off.
Switches change state only from "off" to "on". It means that if you press two or more switches connected to the same lamp then the lamp will be turned on after any of this switches is pressed and will remain its state even if any switch connected to this lamp is pressed afterwards.
It is guaranteed that if you push all *n* switches then all *m* lamps will be turned on.
Your think that you have too many switches and you would like to ignore one of them.
Your task is to say if there exists such a switch that if you will ignore (not use) it but press all the other *n*<=-<=1 switches then all the *m* lamps will be turned on.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=2000) — the number of the switches and the number of the lamps.
The following *n* lines contain *m* characters each. The character *a**i*,<=*j* is equal to '1' if the *i*-th switch turns on the *j*-th lamp and '0' otherwise.
It is guaranteed that if you press all *n* switches all *m* lamps will be turned on.
Output Specification:
Print "YES" if there is a switch that if you will ignore it and press all the other *n*<=-<=1 switches then all *m* lamps will be turned on. Print "NO" if there is no such switch.
Demo Input:
['4 5\n10101\n01000\n00111\n10000\n', '4 5\n10100\n01000\n00110\n00101\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
# -*- coding: utf-8 -*-
"""
Created on Wed May 23 06:30:24 2018
@author: Arsanuos
"""
def main():
n, m = [int(t) for t in input().split()]
arr = []
for i in range(n):
arr.append(input())
neg = 0
while neg != n - 1:
tmp = set()
for index in range(len(arr)):
if index == neg:
continue
item = arr[index]
for i in range(m):
if item[i] == '1':
tmp.add(i)
if len(tmp) == m:
print("YES")
return
neg +=1
print("NO")
if __name__ == "__main__":
main()
``` | 0 | |
677 | A | Vanya and Fence | PROGRAMMING | 800 | [
"implementation"
] | null | null | Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*.
Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard? | The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person. | Print a single integer — the minimum possible valid width of the road. | [
"3 7\n4 5 14\n",
"6 1\n1 1 1 1 1 1\n",
"6 5\n7 6 8 9 10 5\n"
] | [
"4\n",
"6\n",
"11\n"
] | In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4.
In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough.
In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11. | 500 | [
{
"input": "3 7\n4 5 14",
"output": "4"
},
{
"input": "6 1\n1 1 1 1 1 1",
"output": "6"
},
{
"input": "6 5\n7 6 8 9 10 5",
"output": "11"
},
{
"input": "10 420\n214 614 297 675 82 740 174 23 255 15",
"output": "13"
},
{
"input": "10 561\n657 23 1096 487 785 66 481... | 1,692,743,235 | 2,147,483,647 | Python 3 | OK | TESTS | 29 | 46 | 0 | # https://codeforces.com/problemset/problem/677/A
_, h_fence = map(int, input().split(" "))
heights = list(map(int, input().split(" ")))
width = 0
for h in heights:
if h <= h_fence:
width += 1
else:
width += 2
print(width)
| Title: Vanya and Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*.
Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard?
Input Specification:
The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person.
Output Specification:
Print a single integer — the minimum possible valid width of the road.
Demo Input:
['3 7\n4 5 14\n', '6 1\n1 1 1 1 1 1\n', '6 5\n7 6 8 9 10 5\n']
Demo Output:
['4\n', '6\n', '11\n']
Note:
In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4.
In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough.
In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11. | ```python
# https://codeforces.com/problemset/problem/677/A
_, h_fence = map(int, input().split(" "))
heights = list(map(int, input().split(" ")))
width = 0
for h in heights:
if h <= h_fence:
width += 1
else:
width += 2
print(width)
``` | 3 | |
731 | A | Night at the Museum | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he received embosser and was to take stock of the whole exposition.
Embosser is a special devise that allows to "print" the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it's allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter 'a'. Other letters are located as shown on the picture:
After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It's not required to return the wheel to its initial position with pointer on the letter 'a'.
Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it. | The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It's guaranteed that the string consists of only lowercase English letters. | Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input. | [
"zeus\n",
"map\n",
"ares\n"
] | [
"18\n",
"35\n",
"34\n"
] | To print the string from the first sample it would be optimal to perform the following sequence of rotations:
1. from 'a' to 'z' (1 rotation counterclockwise), 1. from 'z' to 'e' (5 clockwise rotations), 1. from 'e' to 'u' (10 rotations counterclockwise), 1. from 'u' to 's' (2 counterclockwise rotations). | 500 | [
{
"input": "zeus",
"output": "18"
},
{
"input": "map",
"output": "35"
},
{
"input": "ares",
"output": "34"
},
{
"input": "l",
"output": "11"
},
{
"input": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuv",
"... | 1,666,291,604 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | s=input()
c=0
for i in range(len(s)):
a=ord(s[i])-97
b=26-a
if a<b:
c+=a
else:
c+=b
print(c) | Title: Night at the Museum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Grigoriy, like the hero of one famous comedy film, found a job as a night security guard at the museum. At first night he received embosser and was to take stock of the whole exposition.
Embosser is a special devise that allows to "print" the text of a plastic tape. Text is printed sequentially, character by character. The device consists of a wheel with a lowercase English letters written in a circle, static pointer to the current letter and a button that print the chosen letter. At one move it's allowed to rotate the alphabetic wheel one step clockwise or counterclockwise. Initially, static pointer points to letter 'a'. Other letters are located as shown on the picture:
After Grigoriy add new item to the base he has to print its name on the plastic tape and attach it to the corresponding exhibit. It's not required to return the wheel to its initial position with pointer on the letter 'a'.
Our hero is afraid that some exhibits may become alive and start to attack him, so he wants to print the names as fast as possible. Help him, for the given string find the minimum number of rotations of the wheel required to print it.
Input Specification:
The only line of input contains the name of some exhibit — the non-empty string consisting of no more than 100 characters. It's guaranteed that the string consists of only lowercase English letters.
Output Specification:
Print one integer — the minimum number of rotations of the wheel, required to print the name given in the input.
Demo Input:
['zeus\n', 'map\n', 'ares\n']
Demo Output:
['18\n', '35\n', '34\n']
Note:
To print the string from the first sample it would be optimal to perform the following sequence of rotations:
1. from 'a' to 'z' (1 rotation counterclockwise), 1. from 'z' to 'e' (5 clockwise rotations), 1. from 'e' to 'u' (10 rotations counterclockwise), 1. from 'u' to 's' (2 counterclockwise rotations). | ```python
s=input()
c=0
for i in range(len(s)):
a=ord(s[i])-97
b=26-a
if a<b:
c+=a
else:
c+=b
print(c)
``` | 0 | |
785 | A | Anton and Polyhedrons | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number! | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron. | Output one number — the total number of faces in all the polyhedrons in Anton's collection. | [
"4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n",
"3\nDodecahedron\nOctahedron\nOctahedron\n"
] | [
"42\n",
"28\n"
] | In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces. | 500 | [
{
"input": "4\nIcosahedron\nCube\nTetrahedron\nDodecahedron",
"output": "42"
},
{
"input": "3\nDodecahedron\nOctahedron\nOctahedron",
"output": "28"
},
{
"input": "25\nIcosahedron\nOctahedron\nTetrahedron\nDodecahedron\nCube\nIcosahedron\nOctahedron\nCube\nTetrahedron\nIcosahedron\nIcosa... | 1,679,399,191 | 2,147,483,647 | PyPy 3 | OK | TESTS | 30 | 1,513 | 14,028,800 | a = int(input())
cnt = 0
l = []
for i in range(a):
b = input()
l.append(b)
p = l.count("Tetrahedron")
q = l.count("Cube")
r = l.count("Octahedron")
s = l.count("Dodecahedron")
t = l.count("Icosahedron")
cnt = p*4 + q*6 + r*8 + s*12 + t*20
print(cnt)
# "Tetrahedron" (without quotes), if the i-th polyhedron in Anton's collection is a tetrahedron.
# "Cube" (without quotes), if the i-th polyhedron in Anton's collection is a cube.
# "Octahedron" (without quotes), if the i-th polyhedron in Anton's collection is an octahedron.
# "Dodecahedron" (without quotes), if the i-th polyhedron in Anton's collection is a dodecahedron.
# "Icosahedron" (without quotes), if the i-th polyhedron in Anton's collection is an icosahedron.
| Title: Anton and Polyhedrons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:
- Tetrahedron. Tetrahedron has 4 triangular faces. - Cube. Cube has 6 square faces. - Octahedron. Octahedron has 8 triangular faces. - Dodecahedron. Dodecahedron has 12 pentagonal faces. - Icosahedron. Icosahedron has 20 triangular faces.
All five kinds of polyhedrons are shown on the picture below:
Anton has a collection of *n* polyhedrons. One day he decided to know, how many faces his polyhedrons have in total. Help Anton and find this number!
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of polyhedrons in Anton's collection.
Each of the following *n* lines of the input contains a string *s**i* — the name of the *i*-th polyhedron in Anton's collection. The string can look like this:
- "Tetrahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a tetrahedron. - "Cube" (without quotes), if the *i*-th polyhedron in Anton's collection is a cube. - "Octahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an octahedron. - "Dodecahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is a dodecahedron. - "Icosahedron" (without quotes), if the *i*-th polyhedron in Anton's collection is an icosahedron.
Output Specification:
Output one number — the total number of faces in all the polyhedrons in Anton's collection.
Demo Input:
['4\nIcosahedron\nCube\nTetrahedron\nDodecahedron\n', '3\nDodecahedron\nOctahedron\nOctahedron\n']
Demo Output:
['42\n', '28\n']
Note:
In the first sample Anton has one icosahedron, one cube, one tetrahedron and one dodecahedron. Icosahedron has 20 faces, cube has 6 faces, tetrahedron has 4 faces and dodecahedron has 12 faces. In total, they have 20 + 6 + 4 + 12 = 42 faces. | ```python
a = int(input())
cnt = 0
l = []
for i in range(a):
b = input()
l.append(b)
p = l.count("Tetrahedron")
q = l.count("Cube")
r = l.count("Octahedron")
s = l.count("Dodecahedron")
t = l.count("Icosahedron")
cnt = p*4 + q*6 + r*8 + s*12 + t*20
print(cnt)
# "Tetrahedron" (without quotes), if the i-th polyhedron in Anton's collection is a tetrahedron.
# "Cube" (without quotes), if the i-th polyhedron in Anton's collection is a cube.
# "Octahedron" (without quotes), if the i-th polyhedron in Anton's collection is an octahedron.
# "Dodecahedron" (without quotes), if the i-th polyhedron in Anton's collection is a dodecahedron.
# "Icosahedron" (without quotes), if the i-th polyhedron in Anton's collection is an icosahedron.
``` | 3 | |
115 | C | Plumber | PROGRAMMING | 2,200 | [
"math"
] | null | null | Little John aspires to become a plumber! Today he has drawn a grid consisting of *n* rows and *m* columns, consisting of *n*<=×<=*m* square cells.
In each cell he will draw a pipe segment. He can only draw four types of segments numbered from 1 to 4, illustrated as follows:
Each pipe segment has two ends, illustrated by the arrows in the picture above. For example, segment 1 has ends at top and left side of it.
Little John considers the piping system to be leaking if there is at least one pipe segment inside the grid whose end is not connected to another pipe's end or to the border of the grid. The image below shows an example of leaking and non-leaking systems of size 1<=×<=2.
Now, you will be given the grid that has been partially filled by Little John. Each cell will either contain one of the four segments above, or be empty. Find the number of possible different non-leaking final systems after Little John finishes filling all of the empty cells with pipe segments. Print this number modulo 1000003 (106<=+<=3).
Note that rotations or flipping of the grid are not allowed and so two configurations that are identical only when one of them has been rotated or flipped either horizontally or vertically are considered two different configurations. | The first line will contain two single-space separated integers *n* and *m* (1<=≤<=*n*,<=*m*,<=*n*·*m*<=≤<=5·105) — the number of rows and columns respectively. Then *n* lines follow, each contains exactly *m* characters — the description of the grid. Each character describes a cell and is either one of these:
- "1" - "4" — a pipe segment of one of four types as described above - "." — an empty cell | Print a single integer denoting the number of possible final non-leaking pipe systems modulo 1000003 (106<=+<=3). If there are no such configurations, print 0. | [
"2 2\n13\n..\n",
"3 1\n1\n4\n.\n",
"2 2\n3.\n.1\n"
] | [
"2\n",
"0\n",
"1\n"
] | For the first example, the initial configuration of the grid is as follows.
The only two possible final non-leaking pipe configurations are as follows:
For the second example, the initial grid is already leaking, so there will be no final grid that is non-leaking.
For the final example, there's only one possible non-leaking final grid as follows. | 1,500 | [
{
"input": "2 2\n13\n..",
"output": "2"
},
{
"input": "3 1\n1\n4\n.",
"output": "0"
},
{
"input": "2 2\n3.\n.1",
"output": "1"
},
{
"input": "2 2\n..\n..",
"output": "16"
},
{
"input": "2 2\n32\n41",
"output": "1"
},
{
"input": "4 4\n....\n.3.3\n....\n... | 1,681,924,550 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 31 | 3,000 | 35,840,000 | n, m = map(int, input().split())
s = ['']*n
h, v = [[0]*2 for _ in range(m)], [[0]*2 for _ in range(n)]
for i in range(n):
s[i] = input().strip()
for j in range(m):
if s[i][j] != '.':
x = int(s[i][j]) - 1
v[i][(x<2)^(j&1)] = 1
h[j][(x>0 and x<3)^(i&1)] = 1
w = 1
for i in range(n):
w = (w*(2-v[i][0]-v[i][1]))%1000003
for i in range(m):
w = (w*(2-h[i][0]-h[i][1]))%1000003
print(w)
| Title: Plumber
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little John aspires to become a plumber! Today he has drawn a grid consisting of *n* rows and *m* columns, consisting of *n*<=×<=*m* square cells.
In each cell he will draw a pipe segment. He can only draw four types of segments numbered from 1 to 4, illustrated as follows:
Each pipe segment has two ends, illustrated by the arrows in the picture above. For example, segment 1 has ends at top and left side of it.
Little John considers the piping system to be leaking if there is at least one pipe segment inside the grid whose end is not connected to another pipe's end or to the border of the grid. The image below shows an example of leaking and non-leaking systems of size 1<=×<=2.
Now, you will be given the grid that has been partially filled by Little John. Each cell will either contain one of the four segments above, or be empty. Find the number of possible different non-leaking final systems after Little John finishes filling all of the empty cells with pipe segments. Print this number modulo 1000003 (106<=+<=3).
Note that rotations or flipping of the grid are not allowed and so two configurations that are identical only when one of them has been rotated or flipped either horizontally or vertically are considered two different configurations.
Input Specification:
The first line will contain two single-space separated integers *n* and *m* (1<=≤<=*n*,<=*m*,<=*n*·*m*<=≤<=5·105) — the number of rows and columns respectively. Then *n* lines follow, each contains exactly *m* characters — the description of the grid. Each character describes a cell and is either one of these:
- "1" - "4" — a pipe segment of one of four types as described above - "." — an empty cell
Output Specification:
Print a single integer denoting the number of possible final non-leaking pipe systems modulo 1000003 (106<=+<=3). If there are no such configurations, print 0.
Demo Input:
['2 2\n13\n..\n', '3 1\n1\n4\n.\n', '2 2\n3.\n.1\n']
Demo Output:
['2\n', '0\n', '1\n']
Note:
For the first example, the initial configuration of the grid is as follows.
The only two possible final non-leaking pipe configurations are as follows:
For the second example, the initial grid is already leaking, so there will be no final grid that is non-leaking.
For the final example, there's only one possible non-leaking final grid as follows. | ```python
n, m = map(int, input().split())
s = ['']*n
h, v = [[0]*2 for _ in range(m)], [[0]*2 for _ in range(n)]
for i in range(n):
s[i] = input().strip()
for j in range(m):
if s[i][j] != '.':
x = int(s[i][j]) - 1
v[i][(x<2)^(j&1)] = 1
h[j][(x>0 and x<3)^(i&1)] = 1
w = 1
for i in range(n):
w = (w*(2-v[i][0]-v[i][1]))%1000003
for i in range(m):
w = (w*(2-h[i][0]-h[i][1]))%1000003
print(w)
``` | 0 | |
409 | H | A + B Strikes Back | PROGRAMMING | 1,500 | [
"*special",
"brute force",
"constructive algorithms",
"dsu",
"implementation"
] | null | null | A + B is often used as an example of the easiest problem possible to show some contest platform. However, some scientists have observed that sometimes this problem is not so easy to get accepted. Want to try? | The input contains two integers *a* and *b* (0<=≤<=*a*,<=*b*<=≤<=103), separated by a single space. | Output the sum of the given integers. | [
"5 14\n",
"381 492\n"
] | [
"19\n",
"873\n"
] | none | 0 | [
{
"input": "5 14",
"output": "19"
},
{
"input": "381 492",
"output": "873"
},
{
"input": "536 298",
"output": "834"
},
{
"input": "143 522",
"output": "665"
},
{
"input": "433 126",
"output": "559"
},
{
"input": "723 350",
"output": "1073"
},
{... | 1,616,266,542 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 0 | 0 | A, B = map(int, input().split())
print(A + B) | Title: A + B Strikes Back
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A + B is often used as an example of the easiest problem possible to show some contest platform. However, some scientists have observed that sometimes this problem is not so easy to get accepted. Want to try?
Input Specification:
The input contains two integers *a* and *b* (0<=≤<=*a*,<=*b*<=≤<=103), separated by a single space.
Output Specification:
Output the sum of the given integers.
Demo Input:
['5 14\n', '381 492\n']
Demo Output:
['19\n', '873\n']
Note:
none | ```python
A, B = map(int, input().split())
print(A + B)
``` | 0 | |
259 | A | Little Elephant and Chess | PROGRAMMING | 1,000 | [
"brute force",
"strings"
] | null | null | The Little Elephant loves chess very much.
One day the Little Elephant and his friend decided to play chess. They've got the chess pieces but the board is a problem. They've got an 8<=×<=8 checkered board, each square is painted either black or white. The Little Elephant and his friend know that a proper chessboard doesn't have any side-adjacent cells with the same color and the upper left cell is white. To play chess, they want to make the board they have a proper chessboard. For that the friends can choose any row of the board and cyclically shift the cells of the chosen row, that is, put the last (rightmost) square on the first place in the row and shift the others one position to the right. You can run the described operation multiple times (or not run it at all).
For example, if the first line of the board looks like that "BBBBBBWW" (the white cells of the line are marked with character "W", the black cells are marked with character "B"), then after one cyclic shift it will look like that "WBBBBBBW".
Help the Little Elephant and his friend to find out whether they can use any number of the described operations to turn the board they have into a proper chessboard. | The input consists of exactly eight lines. Each line contains exactly eight characters "W" or "B" without any spaces: the *j*-th character in the *i*-th line stands for the color of the *j*-th cell of the *i*-th row of the elephants' board. Character "W" stands for the white color, character "B" stands for the black color.
Consider the rows of the board numbered from 1 to 8 from top to bottom, and the columns — from 1 to 8 from left to right. The given board can initially be a proper chessboard. | In a single line print "YES" (without the quotes), if we can make the board a proper chessboard and "NO" (without the quotes) otherwise. | [
"WBWBWBWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\n",
"WBWBWBWB\nWBWBWBWB\nBBWBWWWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWWW\nBWBWBWBW\nBWBWBWBW\n"
] | [
"YES\n",
"NO\n"
] | In the first sample you should shift the following lines one position to the right: the 3-rd, the 6-th, the 7-th and the 8-th.
In the second sample there is no way you can achieve the goal. | 500 | [
{
"input": "WBWBWBWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB",
"output": "YES"
},
{
"input": "WBWBWBWB\nWBWBWBWB\nBBWBWWWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWWW\nBWBWBWBW\nBWBWBWBW",
"output": "NO"
},
{
"input": "BWBWBWBW\nWBWBWBWB\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBW... | 1,500,577,267 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 122 | 4,608,000 | def proper(line):
current = line[0]
for j in range(1, len(line)):
if line[j] == current:
return False
else:
current = line[j]
return True
answer = "YES"
for i in range(8):
if not proper(input()):
answer = "NO"
print(answer)
| Title: Little Elephant and Chess
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant loves chess very much.
One day the Little Elephant and his friend decided to play chess. They've got the chess pieces but the board is a problem. They've got an 8<=×<=8 checkered board, each square is painted either black or white. The Little Elephant and his friend know that a proper chessboard doesn't have any side-adjacent cells with the same color and the upper left cell is white. To play chess, they want to make the board they have a proper chessboard. For that the friends can choose any row of the board and cyclically shift the cells of the chosen row, that is, put the last (rightmost) square on the first place in the row and shift the others one position to the right. You can run the described operation multiple times (or not run it at all).
For example, if the first line of the board looks like that "BBBBBBWW" (the white cells of the line are marked with character "W", the black cells are marked with character "B"), then after one cyclic shift it will look like that "WBBBBBBW".
Help the Little Elephant and his friend to find out whether they can use any number of the described operations to turn the board they have into a proper chessboard.
Input Specification:
The input consists of exactly eight lines. Each line contains exactly eight characters "W" or "B" without any spaces: the *j*-th character in the *i*-th line stands for the color of the *j*-th cell of the *i*-th row of the elephants' board. Character "W" stands for the white color, character "B" stands for the black color.
Consider the rows of the board numbered from 1 to 8 from top to bottom, and the columns — from 1 to 8 from left to right. The given board can initially be a proper chessboard.
Output Specification:
In a single line print "YES" (without the quotes), if we can make the board a proper chessboard and "NO" (without the quotes) otherwise.
Demo Input:
['WBWBWBWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWBW\nWBWBWBWB\nWBWBWBWB\nBWBWBWBW\nWBWBWBWB\n', 'WBWBWBWB\nWBWBWBWB\nBBWBWWWB\nBWBWBWBW\nBWBWBWBW\nBWBWBWWW\nBWBWBWBW\nBWBWBWBW\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you should shift the following lines one position to the right: the 3-rd, the 6-th, the 7-th and the 8-th.
In the second sample there is no way you can achieve the goal. | ```python
def proper(line):
current = line[0]
for j in range(1, len(line)):
if line[j] == current:
return False
else:
current = line[j]
return True
answer = "YES"
for i in range(8):
if not proper(input()):
answer = "NO"
print(answer)
``` | 3 | |
6 | A | Triangle | PROGRAMMING | 900 | [
"brute force",
"geometry"
] | A. Triangle | 2 | 64 | Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him. | The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks. | Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length. | [
"4 2 1 3\n",
"7 2 2 4\n",
"3 5 9 1\n"
] | [
"TRIANGLE\n",
"SEGMENT\n",
"IMPOSSIBLE\n"
] | none | 0 | [
{
"input": "4 2 1 3",
"output": "TRIANGLE"
},
{
"input": "7 2 2 4",
"output": "SEGMENT"
},
{
"input": "3 5 9 1",
"output": "IMPOSSIBLE"
},
{
"input": "3 1 5 1",
"output": "IMPOSSIBLE"
},
{
"input": "10 10 10 10",
"output": "TRIANGLE"
},
{
"input": "11 ... | 1,622,023,762 | 2,147,483,647 | Python 3 | OK | TESTS | 83 | 154 | 0 | a,b,c,d=map(int,input().split())
def T(p,q,r):
if p+q>r and q+r>p and p+r>q :
return True
else:
return False
def S(p,q,r):
if p+q==r or q+r==p or p+r==q :
return True
else:
return False
if T(a,b,c) or T(b,c,d) or T(c,d,a) or T(a,b,d) :
print("TRIANGLE")
elif S(a,b,c) or S(b,c,d) or S(c,d,a) or S(a,b,d) :
print("SEGMENT")
else:
print("IMPOSSIBLE") | Title: Triangle
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input Specification:
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output Specification:
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Demo Input:
['4 2 1 3\n', '7 2 2 4\n', '3 5 9 1\n']
Demo Output:
['TRIANGLE\n', 'SEGMENT\n', 'IMPOSSIBLE\n']
Note:
none | ```python
a,b,c,d=map(int,input().split())
def T(p,q,r):
if p+q>r and q+r>p and p+r>q :
return True
else:
return False
def S(p,q,r):
if p+q==r or q+r==p or p+r==q :
return True
else:
return False
if T(a,b,c) or T(b,c,d) or T(c,d,a) or T(a,b,d) :
print("TRIANGLE")
elif S(a,b,c) or S(b,c,d) or S(c,d,a) or S(a,b,d) :
print("SEGMENT")
else:
print("IMPOSSIBLE")
``` | 3.9615 |
981 | A | Antipalindrome | PROGRAMMING | 900 | [
"brute force",
"implementation",
"strings"
] | null | null | A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not.
A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$.
Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all.
Some time ago Ann read the word $s$. What is the word she changed it into? | The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only. | If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$.
Note that there can be multiple longest substrings that are not palindromes, but their length is unique. | [
"mew\n",
"wuffuw\n",
"qqqqqqqq\n"
] | [
"3\n",
"5\n",
"0\n"
] | "mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$.
The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$.
All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$. | 500 | [
{
"input": "mew",
"output": "3"
},
{
"input": "wuffuw",
"output": "5"
},
{
"input": "qqqqqqqq",
"output": "0"
},
{
"input": "ijvji",
"output": "4"
},
{
"input": "iiiiiii",
"output": "0"
},
{
"input": "wobervhvvkihcuyjtmqhaaigvvgiaahqmtjyuchikvvhvrebow"... | 1,571,653,774 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 26 | 124 | 0 | a=input()
b=len(a)
x=b//2
c=0
for i in range(1,x+1):
if a[i-1]!=a[b-i]:
print(b)
break
else:
c=b-1
if a[0]==a[1]:
print(0)
elif c==b-1:
print(c)
| Title: Antipalindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, the strings "kek", "abacaba", "r" and "papicipap" are palindromes, while the strings "abb" and "iq" are not.
A substring $s[l \ldots r]$ ($1<=\leq<=l<=\leq<=r<=\leq<=|s|$) of a string $s<==<=s_{1}s_{2} \ldots s_{|s|}$ is the string $s_{l}s_{l<=+<=1} \ldots s_{r}$.
Anna does not like palindromes, so she makes her friends call her Ann. She also changes all the words she reads in a similar way. Namely, each word $s$ is changed into its longest substring that is not a palindrome. If all the substrings of $s$ are palindromes, she skips the word at all.
Some time ago Ann read the word $s$. What is the word she changed it into?
Input Specification:
The first line contains a non-empty string $s$ with length at most $50$ characters, containing lowercase English letters only.
Output Specification:
If there is such a substring in $s$ that is not a palindrome, print the maximum length of such a substring. Otherwise print $0$.
Note that there can be multiple longest substrings that are not palindromes, but their length is unique.
Demo Input:
['mew\n', 'wuffuw\n', 'qqqqqqqq\n']
Demo Output:
['3\n', '5\n', '0\n']
Note:
"mew" is not a palindrome, so the longest substring of it that is not a palindrome, is the string "mew" itself. Thus, the answer for the first example is $3$.
The string "uffuw" is one of the longest non-palindrome substrings (of length $5$) of the string "wuffuw", so the answer for the second example is $5$.
All substrings of the string "qqqqqqqq" consist of equal characters so they are palindromes. This way, there are no non-palindrome substrings. Thus, the answer for the third example is $0$. | ```python
a=input()
b=len(a)
x=b//2
c=0
for i in range(1,x+1):
if a[i-1]!=a[b-i]:
print(b)
break
else:
c=b-1
if a[0]==a[1]:
print(0)
elif c==b-1:
print(c)
``` | -1 |
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