contestId int64 0 1.01k | index stringclasses 57
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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
260 | A | Adding Digits | PROGRAMMING | 1,400 | [
"implementation",
"math"
] | null | null | Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times. | The first line contains three integers: *a*,<=*b*,<=*n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=105). | In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them. | [
"5 4 5\n",
"12 11 1\n",
"260 150 10\n"
] | [
"524848\n",
"121\n",
"-1\n"
] | none | 500 | [
{
"input": "5 4 5",
"output": "524848"
},
{
"input": "12 11 1",
"output": "121"
},
{
"input": "260 150 10",
"output": "-1"
},
{
"input": "78843 5684 42717",
"output": "-1"
},
{
"input": "93248 91435 1133",
"output": "-1"
},
{
"input": "100000 10 64479"... | 1,611,649,769 | 2,147,483,647 | PyPy 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <iostream>
using namespace std;
int op(int a, int b){
for (int j=0;j<10;j++){
if ( (a*10+j) % b == 0 )
return a*10+j;
}
return -1;
}
int main(){
int a,b,n;
cin >> a >> b >> n;
for (int i=0;i<n;i++){
if (op(a,b) == -1){
cout << -1;
return 0;
}
else a = op(a,b);
}
cout << a;
return 0;
}
| Title: Adding Digits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has got two number: *a* and *b*. However, Vasya finds number *a* too short. So he decided to repeat the operation of lengthening number *a* *n* times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number *b*. If it is impossible to obtain the number which is divisible by *b*, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number *a* *n* times.
Input Specification:
The first line contains three integers: *a*,<=*b*,<=*n* (1<=≤<=*a*,<=*b*,<=*n*<=≤<=105).
Output Specification:
In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number *a* *n* times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them.
Demo Input:
['5 4 5\n', '12 11 1\n', '260 150 10\n']
Demo Output:
['524848\n', '121\n', '-1\n']
Note:
none | ```python
#include <iostream>
using namespace std;
int op(int a, int b){
for (int j=0;j<10;j++){
if ( (a*10+j) % b == 0 )
return a*10+j;
}
return -1;
}
int main(){
int a,b,n;
cin >> a >> b >> n;
for (int i=0;i<n;i++){
if (op(a,b) == -1){
cout << -1;
return 0;
}
else a = op(a,b);
}
cout << a;
return 0;
}
``` | -1 | |
459 | B | Pashmak and Flowers | PROGRAMMING | 1,300 | [
"combinatorics",
"implementation",
"sortings"
] | null | null | Pashmak decided to give Parmida a pair of flowers from the garden. There are *n* flowers in the garden and the *i*-th of them has a beauty number *b**i*. Parmida is a very strange girl so she doesn't want to have the two most beautiful flowers necessarily. She wants to have those pairs of flowers that their beauty difference is maximal possible!
Your task is to write a program which calculates two things:
1. The maximum beauty difference of flowers that Pashmak can give to Parmida. 1. The number of ways that Pashmak can pick the flowers. Two ways are considered different if and only if there is at least one flower that is chosen in the first way and not chosen in the second way. | The first line of the input contains *n* (2<=≤<=*n*<=≤<=2·105). In the next line there are *n* space-separated integers *b*1, *b*2, ..., *b**n* (1<=≤<=*b**i*<=≤<=109). | The only line of output should contain two integers. The maximum beauty difference and the number of ways this may happen, respectively. | [
"2\n1 2\n",
"3\n1 4 5\n",
"5\n3 1 2 3 1\n"
] | [
"1 1",
"4 1",
"2 4"
] | In the third sample the maximum beauty difference is 2 and there are 4 ways to do this:
1. choosing the first and the second flowers; 1. choosing the first and the fifth flowers; 1. choosing the fourth and the second flowers; 1. choosing the fourth and the fifth flowers. | 500 | [
{
"input": "2\n1 2",
"output": "1 1"
},
{
"input": "3\n1 4 5",
"output": "4 1"
},
{
"input": "5\n3 1 2 3 1",
"output": "2 4"
},
{
"input": "2\n1 1",
"output": "0 1"
},
{
"input": "3\n1 1 1",
"output": "0 3"
},
{
"input": "4\n1 1 1 1",
"output": "0 ... | 1,697,478,609 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | int(input())
nums = [int(i) for i in input().split()]
_max = max(nums) - min(nums)
count = 0
for i in nums:
for j in nums:
diff = abs(i - j)
if diff == _max:
count += 1
print(_max, count / 2) | Title: Pashmak and Flowers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pashmak decided to give Parmida a pair of flowers from the garden. There are *n* flowers in the garden and the *i*-th of them has a beauty number *b**i*. Parmida is a very strange girl so she doesn't want to have the two most beautiful flowers necessarily. She wants to have those pairs of flowers that their beauty difference is maximal possible!
Your task is to write a program which calculates two things:
1. The maximum beauty difference of flowers that Pashmak can give to Parmida. 1. The number of ways that Pashmak can pick the flowers. Two ways are considered different if and only if there is at least one flower that is chosen in the first way and not chosen in the second way.
Input Specification:
The first line of the input contains *n* (2<=≤<=*n*<=≤<=2·105). In the next line there are *n* space-separated integers *b*1, *b*2, ..., *b**n* (1<=≤<=*b**i*<=≤<=109).
Output Specification:
The only line of output should contain two integers. The maximum beauty difference and the number of ways this may happen, respectively.
Demo Input:
['2\n1 2\n', '3\n1 4 5\n', '5\n3 1 2 3 1\n']
Demo Output:
['1 1', '4 1', '2 4']
Note:
In the third sample the maximum beauty difference is 2 and there are 4 ways to do this:
1. choosing the first and the second flowers; 1. choosing the first and the fifth flowers; 1. choosing the fourth and the second flowers; 1. choosing the fourth and the fifth flowers. | ```python
int(input())
nums = [int(i) for i in input().split()]
_max = max(nums) - min(nums)
count = 0
for i in nums:
for j in nums:
diff = abs(i - j)
if diff == _max:
count += 1
print(_max, count / 2)
``` | 0 | |
334 | A | Candy Bags | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Gerald has *n* younger brothers and their number happens to be even. One day he bought *n*2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer *k* from 1 to *n*2 he has exactly one bag with *k* candies.
Help him give *n* bags of candies to each brother so that all brothers got the same number of candies. | The single line contains a single integer *n* (*n* is even, 2<=≤<=*n*<=≤<=100) — the number of Gerald's brothers. | Let's assume that Gerald indexes his brothers with numbers from 1 to *n*. You need to print *n* lines, on the *i*-th line print *n* integers — the numbers of candies in the bags for the *i*-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to *n*2. You can print the numbers in the lines in any order.
It is guaranteed that the solution exists at the given limits. | [
"2\n"
] | [
"1 4\n2 3\n"
] | The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother. | 500 | [
{
"input": "2",
"output": "1 4\n2 3"
},
{
"input": "4",
"output": "1 16 2 15\n3 14 4 13\n5 12 6 11\n7 10 8 9"
},
{
"input": "6",
"output": "1 36 2 35 3 34\n4 33 5 32 6 31\n7 30 8 29 9 28\n10 27 11 26 12 25\n13 24 14 23 15 22\n16 21 17 20 18 19"
},
{
"input": "8",
"output"... | 1,624,341,140 | 2,147,483,647 | PyPy 3 | OK | TESTS | 21 | 280 | 3,993,600 | n = int(input())
n_sqr = n**2
r = range(1, n**2+1)
for i in range(n_sqr//2):
print(r[i], r[n_sqr-1-i])
| Title: Candy Bags
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Gerald has *n* younger brothers and their number happens to be even. One day he bought *n*2 candy bags. One bag has one candy, one bag has two candies, one bag has three candies and so on. In fact, for each integer *k* from 1 to *n*2 he has exactly one bag with *k* candies.
Help him give *n* bags of candies to each brother so that all brothers got the same number of candies.
Input Specification:
The single line contains a single integer *n* (*n* is even, 2<=≤<=*n*<=≤<=100) — the number of Gerald's brothers.
Output Specification:
Let's assume that Gerald indexes his brothers with numbers from 1 to *n*. You need to print *n* lines, on the *i*-th line print *n* integers — the numbers of candies in the bags for the *i*-th brother. Naturally, all these numbers should be distinct and be within limits from 1 to *n*2. You can print the numbers in the lines in any order.
It is guaranteed that the solution exists at the given limits.
Demo Input:
['2\n']
Demo Output:
['1 4\n2 3\n']
Note:
The sample shows Gerald's actions if he has two brothers. In this case, his bags contain 1, 2, 3 and 4 candies. He can give the bags with 1 and 4 candies to one brother and the bags with 2 and 3 to the other brother. | ```python
n = int(input())
n_sqr = n**2
r = range(1, n**2+1)
for i in range(n_sqr//2):
print(r[i], r[n_sqr-1-i])
``` | 3 | |
58 | A | Chat room | PROGRAMMING | 1,000 | [
"greedy",
"strings"
] | A. Chat room | 1 | 256 | Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. | The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. | If Vasya managed to say hello, print "YES", otherwise print "NO". | [
"ahhellllloou\n",
"hlelo\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymeda... | 1,668,789,162 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 40 | 62 | 0 | s = input()
temp = 'hello'
ind = 0
for i in s:
if i == temp[ind]:
ind += 1
if ind == len(temp):
break
if ind == len(temp):
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
s = input()
temp = 'hello'
ind = 0
for i in s:
if i == temp[ind]:
ind += 1
if ind == len(temp):
break
if ind == len(temp):
print("YES")
else:
print("NO")
``` | 3.969 |
766 | B | Mahmoud and a Triangle | PROGRAMMING | 1,000 | [
"constructive algorithms",
"geometry",
"greedy",
"math",
"number theory",
"sortings"
] | null | null | Mahmoud has *n* line segments, the *i*-th of them has length *a**i*. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.
Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area. | The first line contains single integer *n* (3<=≤<=*n*<=≤<=105) — the number of line segments Mahmoud has.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the lengths of line segments Mahmoud has. | In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise. | [
"5\n1 5 3 2 4\n",
"3\n4 1 2\n"
] | [
"YES\n",
"NO\n"
] | For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle. | 1,000 | [
{
"input": "5\n1 5 3 2 4",
"output": "YES"
},
{
"input": "3\n4 1 2",
"output": "NO"
},
{
"input": "30\n197 75 517 39724 7906061 1153471 3 15166 168284 3019844 272293 316 16 24548 42 118 5792 5 9373 1866366 4886214 24 2206 712886 104005 1363 836 64273 440585 3576",
"output": "NO"
},... | 1,670,704,114 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 187 | 7,987,200 | import math
from collections import *
def solve():
n = int(input())
a = [int(i) for i in input().split()]
a.sort()
l = 0
for i in range(n-2):
d = (a[i]+a[i+1] + a[i+2]) / 2
if a[i]>d or a[i+1]>d or a[i+2]>d:
l = 1
print("NO")
break
if l == 0:
print("YES")
# t = int(input())
# for _ in range(t):
# solve()
solve() | Title: Mahmoud and a Triangle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mahmoud has *n* line segments, the *i*-th of them has length *a**i*. Ehab challenged him to use exactly 3 line segments to form a non-degenerate triangle. Mahmoud doesn't accept challenges unless he is sure he can win, so he asked you to tell him if he should accept the challenge. Given the lengths of the line segments, check if he can choose exactly 3 of them to form a non-degenerate triangle.
Mahmoud should use exactly 3 line segments, he can't concatenate two line segments or change any length. A non-degenerate triangle is a triangle with positive area.
Input Specification:
The first line contains single integer *n* (3<=≤<=*n*<=≤<=105) — the number of line segments Mahmoud has.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the lengths of line segments Mahmoud has.
Output Specification:
In the only line print "YES" if he can choose exactly three line segments and form a non-degenerate triangle with them, and "NO" otherwise.
Demo Input:
['5\n1 5 3 2 4\n', '3\n4 1 2\n']
Demo Output:
['YES\n', 'NO\n']
Note:
For the first example, he can use line segments with lengths 2, 4 and 5 to form a non-degenerate triangle. | ```python
import math
from collections import *
def solve():
n = int(input())
a = [int(i) for i in input().split()]
a.sort()
l = 0
for i in range(n-2):
d = (a[i]+a[i+1] + a[i+2]) / 2
if a[i]>d or a[i+1]>d or a[i+2]>d:
l = 1
print("NO")
break
if l == 0:
print("YES")
# t = int(input())
# for _ in range(t):
# solve()
solve()
``` | 0 | |
258 | A | Little Elephant and Bits | PROGRAMMING | 1,100 | [
"greedy",
"math"
] | null | null | The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper.
To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes).
The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation. | The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits. | In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem. | [
"101\n",
"110010\n"
] | [
"11\n",
"11010\n"
] | In the first sample the best strategy is to delete the second digit. That results in number 11<sub class="lower-index">2</sub> = 3<sub class="lower-index">10</sub>.
In the second sample the best strategy is to delete the third or fourth digits — that results in number 11010<sub class="lower-index">2</sub> = 26<sub class="lower-index">10</sub>. | 500 | [
{
"input": "101",
"output": "11"
},
{
"input": "110010",
"output": "11010"
},
{
"input": "10000",
"output": "1000"
},
{
"input": "1111111110",
"output": "111111111"
},
{
"input": "10100101011110101",
"output": "1100101011110101"
},
{
"input": "11101001... | 1,670,267,323 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 37 | 124 | 0 | s = input()
if '0' in s:
e = s.split('0',1)
if len(e) == 1:
print(e[0])
else:
print(e[0]+e[1])
else:
print(s[1:])
| Title: Little Elephant and Bits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper.
To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes).
The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation.
Input Specification:
The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits.
Output Specification:
In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem.
Demo Input:
['101\n', '110010\n']
Demo Output:
['11\n', '11010\n']
Note:
In the first sample the best strategy is to delete the second digit. That results in number 11<sub class="lower-index">2</sub> = 3<sub class="lower-index">10</sub>.
In the second sample the best strategy is to delete the third or fourth digits — that results in number 11010<sub class="lower-index">2</sub> = 26<sub class="lower-index">10</sub>. | ```python
s = input()
if '0' in s:
e = s.split('0',1)
if len(e) == 1:
print(e[0])
else:
print(e[0]+e[1])
else:
print(s[1:])
``` | 3 | |
427 | A | Police Recruits | PROGRAMMING | 800 | [
"implementation"
] | null | null | The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated. | The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time. | Print a single integer, the number of crimes which will go untreated. | [
"3\n-1 -1 1\n",
"8\n1 -1 1 -1 -1 1 1 1\n",
"11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n"
] | [
"2\n",
"1\n",
"8\n"
] | Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated. | 500 | [
{
"input": "3\n-1 -1 1",
"output": "2"
},
{
"input": "8\n1 -1 1 -1 -1 1 1 1",
"output": "1"
},
{
"input": "11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1",
"output": "8"
},
{
"input": "7\n-1 -1 1 1 -1 -1 1",
"output": "2"
},
{
"input": "21\n-1 -1 -1 -1 -1 3 2 -1 6 -1 -1 2 1 ... | 1,681,128,238 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 30 | 0 | z = input()
a = input()
b = a.split(" ")
police = 0
tally = 0
for ins in b:
ins = int(ins)
if ins < 0:
if police == 0:
tally += ins
continue
if police > 0:
police -= ins
continue
if ins > 0:
police += ins
if tally < 0:
print(tally*-1)
else:
print(tally) | Title: Police Recruits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.
Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.
If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.
Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.
Input Specification:
The first line of input will contain an integer *n* (1<=≤<=*n*<=≤<=105), the number of events. The next line will contain *n* space-separated integers.
If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.
Output Specification:
Print a single integer, the number of crimes which will go untreated.
Demo Input:
['3\n-1 -1 1\n', '8\n1 -1 1 -1 -1 1 1 1\n', '11\n-1 -1 2 -1 -1 -1 -1 -1 -1 -1 -1\n']
Demo Output:
['2\n', '1\n', '8\n']
Note:
Lets consider the second example:
1. Firstly one person is hired. 1. Then crime appears, the last hired person will investigate this crime. 1. One more person is hired. 1. One more crime appears, the last hired person will investigate this crime. 1. Crime appears. There is no free policeman at the time, so this crime will go untreated. 1. One more person is hired. 1. One more person is hired. 1. One more person is hired.
The answer is one, as one crime (on step 5) will go untreated. | ```python
z = input()
a = input()
b = a.split(" ")
police = 0
tally = 0
for ins in b:
ins = int(ins)
if ins < 0:
if police == 0:
tally += ins
continue
if police > 0:
police -= ins
continue
if ins > 0:
police += ins
if tally < 0:
print(tally*-1)
else:
print(tally)
``` | 0 | |
937 | A | Olympiad | PROGRAMMING | 800 | [
"implementation",
"sortings"
] | null | null | The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points.
As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria:
- At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded.
Determine the number of ways to choose a subset of participants that will receive the diplomas. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants.
The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores.
It's guaranteed that at least one participant has non-zero score. | Print a single integer — the desired number of ways. | [
"4\n1 3 3 2\n",
"3\n1 1 1\n",
"4\n42 0 0 42\n"
] | [
"3\n",
"1\n",
"1\n"
] | There are three ways to choose a subset in sample case one.
1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma!
The only option in sample case two is to award everyone.
Note that in sample case three participants with zero scores cannot get anything. | 500 | [
{
"input": "4\n1 3 3 2",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "1"
},
{
"input": "4\n42 0 0 42",
"output": "1"
},
{
"input": "10\n1 0 1 0 1 0 0 0 0 1",
"output": "1"
},
{
"input": "10\n572 471 540 163 50 30 561 510 43 200",
"output": "10"
},
{
... | 1,575,397,740 | 2,147,483,647 | Python 3 | OK | TESTS | 21 | 124 | 0 | n=int(input())
l=list(map(int,input().split()))
s=set(l)
if 0 in s:print(len(s)-1)
else:print(len(s)) | Title: Olympiad
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points.
As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria:
- At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded.
Determine the number of ways to choose a subset of participants that will receive the diplomas.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants.
The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores.
It's guaranteed that at least one participant has non-zero score.
Output Specification:
Print a single integer — the desired number of ways.
Demo Input:
['4\n1 3 3 2\n', '3\n1 1 1\n', '4\n42 0 0 42\n']
Demo Output:
['3\n', '1\n', '1\n']
Note:
There are three ways to choose a subset in sample case one.
1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma!
The only option in sample case two is to award everyone.
Note that in sample case three participants with zero scores cannot get anything. | ```python
n=int(input())
l=list(map(int,input().split()))
s=set(l)
if 0 in s:print(len(s)-1)
else:print(len(s))
``` | 3 | |
368 | A | Sereja and Coat Rack | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Sereja owns a restaurant for *n* people. The restaurant hall has a coat rack with *n* hooks. Each restaurant visitor can use a hook to hang his clothes on it. Using the *i*-th hook costs *a**i* rubles. Only one person can hang clothes on one hook.
Tonight Sereja expects *m* guests in the restaurant. Naturally, each guest wants to hang his clothes on an available hook with minimum price (if there are multiple such hooks, he chooses any of them). However if the moment a guest arrives the rack has no available hooks, Sereja must pay a *d* ruble fine to the guest.
Help Sereja find out the profit in rubles (possibly negative) that he will get tonight. You can assume that before the guests arrive, all hooks on the rack are available, all guests come at different time, nobody besides the *m* guests is visiting Sereja's restaurant tonight. | The first line contains two integers *n* and *d* (1<=≤<=*n*,<=*d*<=≤<=100). The next line contains integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100). The third line contains integer *m* (1<=≤<=*m*<=≤<=100). | In a single line print a single integer — the answer to the problem. | [
"2 1\n2 1\n2\n",
"2 1\n2 1\n10\n"
] | [
"3\n",
"-5\n"
] | In the first test both hooks will be used, so Sereja gets 1 + 2 = 3 rubles.
In the second test both hooks will be used but Sereja pays a fine 8 times, so the answer is 3 - 8 = - 5. | 500 | [
{
"input": "2 1\n2 1\n2",
"output": "3"
},
{
"input": "2 1\n2 1\n10",
"output": "-5"
},
{
"input": "1 1\n1\n2",
"output": "0"
},
{
"input": "3 96\n83 22 17\n19",
"output": "-1414"
},
{
"input": "8 4\n27 72 39 70 13 68 100 36\n95",
"output": "77"
},
{
"... | 1,684,753,752 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 35 | 62 | 0 | # https://codeforces.com/contest/368
import sys
input = lambda: sys.stdin.readline().rstrip() # faster!
n, d = map(int, input().split())
a = list(map(int, input().split()))
m = int(input())
if m >= n:
ans = sum(a) - (m - n) * d
else:
a.sort()
ans = sum(a[:m])
print(ans)
| Title: Sereja and Coat Rack
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja owns a restaurant for *n* people. The restaurant hall has a coat rack with *n* hooks. Each restaurant visitor can use a hook to hang his clothes on it. Using the *i*-th hook costs *a**i* rubles. Only one person can hang clothes on one hook.
Tonight Sereja expects *m* guests in the restaurant. Naturally, each guest wants to hang his clothes on an available hook with minimum price (if there are multiple such hooks, he chooses any of them). However if the moment a guest arrives the rack has no available hooks, Sereja must pay a *d* ruble fine to the guest.
Help Sereja find out the profit in rubles (possibly negative) that he will get tonight. You can assume that before the guests arrive, all hooks on the rack are available, all guests come at different time, nobody besides the *m* guests is visiting Sereja's restaurant tonight.
Input Specification:
The first line contains two integers *n* and *d* (1<=≤<=*n*,<=*d*<=≤<=100). The next line contains integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100). The third line contains integer *m* (1<=≤<=*m*<=≤<=100).
Output Specification:
In a single line print a single integer — the answer to the problem.
Demo Input:
['2 1\n2 1\n2\n', '2 1\n2 1\n10\n']
Demo Output:
['3\n', '-5\n']
Note:
In the first test both hooks will be used, so Sereja gets 1 + 2 = 3 rubles.
In the second test both hooks will be used but Sereja pays a fine 8 times, so the answer is 3 - 8 = - 5. | ```python
# https://codeforces.com/contest/368
import sys
input = lambda: sys.stdin.readline().rstrip() # faster!
n, d = map(int, input().split())
a = list(map(int, input().split()))
m = int(input())
if m >= n:
ans = sum(a) - (m - n) * d
else:
a.sort()
ans = sum(a[:m])
print(ans)
``` | 3 | |
844 | A | Diversity | PROGRAMMING | 1,000 | [
"greedy",
"implementation",
"strings"
] | null | null | Calculate the minimum number of characters you need to change in the string *s*, so that it contains at least *k* different letters, or print that it is impossible.
String *s* consists only of lowercase Latin letters, and it is allowed to change characters only to lowercase Latin letters too. | First line of input contains string *s*, consisting only of lowercase Latin letters (1<=≤<=|*s*|<=≤<=1000, |*s*| denotes the length of *s*).
Second line of input contains integer *k* (1<=≤<=*k*<=≤<=26). | Print single line with a minimum number of necessary changes, or the word «impossible» (without quotes) if it is impossible. | [
"yandex\n6\n",
"yahoo\n5\n",
"google\n7\n"
] | [
"0\n",
"1\n",
"impossible\n"
] | In the first test case string contains 6 different letters, so we don't need to change anything.
In the second test case string contains 4 different letters: {'*a*', '*h*', '*o*', '*y*'}. To get 5 different letters it is necessary to change one occurrence of '*o*' to some letter, which doesn't occur in the string, for example, {'*b*'}.
In the third test case, it is impossible to make 7 different letters because the length of the string is 6. | 500 | [
{
"input": "yandex\n6",
"output": "0"
},
{
"input": "yahoo\n5",
"output": "1"
},
{
"input": "google\n7",
"output": "impossible"
},
{
"input": "a\n1",
"output": "0"
},
{
"input": "z\n2",
"output": "impossible"
},
{
"input": "fwgfrwgkuwghfiruhewgirueguhe... | 1,508,265,190 | 2,147,483,647 | Python 3 | OK | TESTS | 41 | 62 | 5,529,600 | s = input()
k = int(input())
n = 0
a = list()
for i in s:
test = False
for j in a:
if i == j:
test = True
if not test:
a.append(i)
if len(s) < k:
print('impossible')
else:
print(max(k - len(a),0))
| Title: Diversity
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Calculate the minimum number of characters you need to change in the string *s*, so that it contains at least *k* different letters, or print that it is impossible.
String *s* consists only of lowercase Latin letters, and it is allowed to change characters only to lowercase Latin letters too.
Input Specification:
First line of input contains string *s*, consisting only of lowercase Latin letters (1<=≤<=|*s*|<=≤<=1000, |*s*| denotes the length of *s*).
Second line of input contains integer *k* (1<=≤<=*k*<=≤<=26).
Output Specification:
Print single line with a minimum number of necessary changes, or the word «impossible» (without quotes) if it is impossible.
Demo Input:
['yandex\n6\n', 'yahoo\n5\n', 'google\n7\n']
Demo Output:
['0\n', '1\n', 'impossible\n']
Note:
In the first test case string contains 6 different letters, so we don't need to change anything.
In the second test case string contains 4 different letters: {'*a*', '*h*', '*o*', '*y*'}. To get 5 different letters it is necessary to change one occurrence of '*o*' to some letter, which doesn't occur in the string, for example, {'*b*'}.
In the third test case, it is impossible to make 7 different letters because the length of the string is 6. | ```python
s = input()
k = int(input())
n = 0
a = list()
for i in s:
test = False
for j in a:
if i == j:
test = True
if not test:
a.append(i)
if len(s) < k:
print('impossible')
else:
print(max(k - len(a),0))
``` | 3 | |
811 | A | Vladik and Courtesy | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | At regular competition Vladik and Valera won *a* and *b* candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn.
More formally, the guys take turns giving each other one candy more than they received in the previous turn.
This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy. | Single line of input data contains two space-separated integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) — number of Vladik and Valera candies respectively. | Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise. | [
"1 1\n",
"7 6\n"
] | [
"Valera\n",
"Vladik\n"
] | Illustration for first test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/ad9b7d0e481208de8e3a585aa1d96b9e1dda4fd7.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Illustration for second test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/9f4836d2ccdffaee5a63898e5d4e6caf2ed4678c.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 500 | [
{
"input": "1 1",
"output": "Valera"
},
{
"input": "7 6",
"output": "Vladik"
},
{
"input": "25 38",
"output": "Vladik"
},
{
"input": "8311 2468",
"output": "Valera"
},
{
"input": "250708 857756",
"output": "Vladik"
},
{
"input": "957985574 24997558",
... | 1,634,493,512 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 61 | 0 | a,b=map(int,input().split())
if a%2==1:
if a>b:
print("Vladik")
else:
print("Valera")
elif b%2==0:
if b>a:
print("Valera")
else:
print("Vladik") | Title: Vladik and Courtesy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
At regular competition Vladik and Valera won *a* and *b* candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn.
More formally, the guys take turns giving each other one candy more than they received in the previous turn.
This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy.
Input Specification:
Single line of input data contains two space-separated integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) — number of Vladik and Valera candies respectively.
Output Specification:
Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise.
Demo Input:
['1 1\n', '7 6\n']
Demo Output:
['Valera\n', 'Vladik\n']
Note:
Illustration for first test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/ad9b7d0e481208de8e3a585aa1d96b9e1dda4fd7.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Illustration for second test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/9f4836d2ccdffaee5a63898e5d4e6caf2ed4678c.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
a,b=map(int,input().split())
if a%2==1:
if a>b:
print("Vladik")
else:
print("Valera")
elif b%2==0:
if b>a:
print("Valera")
else:
print("Vladik")
``` | 0 | |
808 | D | Array Division | PROGRAMMING | 1,900 | [
"binary search",
"data structures",
"implementation"
] | null | null | Vasya has an array *a* consisting of positive integer numbers. Vasya wants to divide this array into two non-empty consecutive parts (the prefix and the suffix) so that the sum of all elements in the first part equals to the sum of elements in the second part. It is not always possible, so Vasya will move some element before dividing the array (Vasya will erase some element and insert it into an arbitrary position).
Inserting an element in the same position he was erased from is also considered moving.
Can Vasya divide the array after choosing the right element to move and its new position? | The first line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2... *a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array. | Print YES if Vasya can divide the array after moving one element. Otherwise print NO. | [
"3\n1 3 2\n",
"5\n1 2 3 4 5\n",
"5\n2 2 3 4 5\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | In the first example Vasya can move the second element to the end of the array.
In the second example no move can make the division possible.
In the third example Vasya can move the fourth element by one position to the left. | 0 | [
{
"input": "3\n1 3 2",
"output": "YES"
},
{
"input": "5\n1 2 3 4 5",
"output": "NO"
},
{
"input": "5\n2 2 3 4 5",
"output": "YES"
},
{
"input": "5\n72 32 17 46 82",
"output": "NO"
},
{
"input": "6\n26 10 70 11 69 57",
"output": "NO"
},
{
"input": "7\n4... | 1,565,187,757 | 2,147,483,647 | PyPy 3 | OK | TESTS | 115 | 327 | 15,462,400 | def possible(a):
if len(a)==1:
return False
if len(a)==2:
return a[0]==a[1]
total = sum(a)
vals = set([a[0]])
running = a[0]
for el in a[1:]:
running += el
vals.add(el)
if running == total-running:
return True
if (running - (total-running))%2==0 and (running - (total-running))//2 in vals:
return True
return False
n = int(input())
a = list(map(int, input().split()))
if possible(a) or possible(a[::-1]):
print("YES")
else:
print("NO") | Title: Array Division
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has an array *a* consisting of positive integer numbers. Vasya wants to divide this array into two non-empty consecutive parts (the prefix and the suffix) so that the sum of all elements in the first part equals to the sum of elements in the second part. It is not always possible, so Vasya will move some element before dividing the array (Vasya will erase some element and insert it into an arbitrary position).
Inserting an element in the same position he was erased from is also considered moving.
Can Vasya divide the array after choosing the right element to move and its new position?
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2... *a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array.
Output Specification:
Print YES if Vasya can divide the array after moving one element. Otherwise print NO.
Demo Input:
['3\n1 3 2\n', '5\n1 2 3 4 5\n', '5\n2 2 3 4 5\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first example Vasya can move the second element to the end of the array.
In the second example no move can make the division possible.
In the third example Vasya can move the fourth element by one position to the left. | ```python
def possible(a):
if len(a)==1:
return False
if len(a)==2:
return a[0]==a[1]
total = sum(a)
vals = set([a[0]])
running = a[0]
for el in a[1:]:
running += el
vals.add(el)
if running == total-running:
return True
if (running - (total-running))%2==0 and (running - (total-running))//2 in vals:
return True
return False
n = int(input())
a = list(map(int, input().split()))
if possible(a) or possible(a[::-1]):
print("YES")
else:
print("NO")
``` | 3 | |
104 | A | Blackjack | PROGRAMMING | 800 | [
"implementation"
] | A. Blackjack | 2 | 256 | One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one!
Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture.
In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses.
The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*. | The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points. | Print the numbers of ways to get the second card in the required way if the first card is the queen of spades. | [
"12\n",
"20\n",
"10\n"
] | [
"4",
"15",
"0"
] | In the first sample only four two's of different suits can earn the required sum of points.
In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use.
In the third sample there is no card, that would add a zero to the current ten points. | 500 | [
{
"input": "12",
"output": "4"
},
{
"input": "20",
"output": "15"
},
{
"input": "10",
"output": "0"
},
{
"input": "11",
"output": "4"
},
{
"input": "15",
"output": "4"
},
{
"input": "18",
"output": "4"
},
{
"input": "25",
"output": "0"
... | 1,626,073,614 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 154 | 6,758,400 | n = int(input())
reqNumber = n - 10
if reqNumber in [2,3,4,5,6,7,8,9] :
print(4)
elif reqNumber == 10:
print(15)
elif reqNumber == 11:
print(4)
elif reqNumber == 1 :
print(4)
else :
print(0) | Title: Blackjack
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one!
Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture.
In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses.
The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*.
Input Specification:
The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points.
Output Specification:
Print the numbers of ways to get the second card in the required way if the first card is the queen of spades.
Demo Input:
['12\n', '20\n', '10\n']
Demo Output:
['4', '15', '0']
Note:
In the first sample only four two's of different suits can earn the required sum of points.
In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use.
In the third sample there is no card, that would add a zero to the current ten points. | ```python
n = int(input())
reqNumber = n - 10
if reqNumber in [2,3,4,5,6,7,8,9] :
print(4)
elif reqNumber == 10:
print(15)
elif reqNumber == 11:
print(4)
elif reqNumber == 1 :
print(4)
else :
print(0)
``` | 3.948911 |
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"
... | 1,663,272,435 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | n=int(input())
s=0
for i in range(1,n-1):
if (n-i)%i==0:
s+=1
print(s) | 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
n=int(input())
s=0
for i in range(1,n-1):
if (n-i)%i==0:
s+=1
print(s)
``` | 0 | |
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"
},
... | 1,678,300,347 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 30 | 0 | print(4, n-4) | 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
print(4, n-4)
``` | -1 | |
17 | B | Hierarchy | PROGRAMMING | 1,500 | [
"dfs and similar",
"dsu",
"greedy",
"shortest paths"
] | B. Hierarchy | 2 | 64 | Nick's company employed *n* people. Now Nick needs to build a tree hierarchy of «supervisor-surbodinate» relations in the company (this is to say that each employee, except one, has exactly one supervisor). There are *m* applications written in the following form: «employee *a**i* is ready to become a supervisor of employee *b**i* at extra cost *c**i*». The qualification *q**j* of each employee is known, and for each application the following is true: *q**a**i*<=><=*q**b**i*.
Would you help Nick calculate the minimum cost of such a hierarchy, or find out that it is impossible to build it. | The first input line contains integer *n* (1<=≤<=*n*<=≤<=1000) — amount of employees in the company. The following line contains *n* space-separated numbers *q**j* (0<=≤<=*q**j*<=≤<=106)— the employees' qualifications. The following line contains number *m* (0<=≤<=*m*<=≤<=10000) — amount of received applications. The following *m* lines contain the applications themselves, each of them in the form of three space-separated numbers: *a**i*, *b**i* and *c**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, 0<=≤<=*c**i*<=≤<=106). Different applications can be similar, i.e. they can come from one and the same employee who offered to become a supervisor of the same person but at a different cost. For each application *q**a**i*<=><=*q**b**i*. | Output the only line — the minimum cost of building such a hierarchy, or -1 if it is impossible to build it. | [
"4\n7 2 3 1\n4\n1 2 5\n2 4 1\n3 4 1\n1 3 5\n",
"3\n1 2 3\n2\n3 1 2\n3 1 3\n"
] | [
"11\n",
"-1\n"
] | In the first sample one of the possible ways for building a hierarchy is to take applications with indexes 1, 2 and 4, which give 11 as the minimum total cost. In the second sample it is impossible to build the required hierarchy, so the answer is -1. | 0 | [
{
"input": "4\n7 2 3 1\n4\n1 2 5\n2 4 1\n3 4 1\n1 3 5",
"output": "11"
},
{
"input": "3\n1 2 3\n2\n3 1 2\n3 1 3",
"output": "-1"
},
{
"input": "1\n2\n0",
"output": "0"
},
{
"input": "2\n5 3\n4\n1 2 0\n1 2 5\n1 2 0\n1 2 7",
"output": "0"
},
{
"input": "3\n9 4 5\n5\... | 1,639,556,259 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 62 | 0 | import sys
def reverse_graph(graph: dict):
r = {}
for src in graph:
for (dst, c) in graph[src].items():
if dst in r:
r[dst][src] = c
else:
r[dst] = {src: c}
return r
def get_cycle(n: int, g: dict, visited: set = None, cycle: list = None):
if visited is None:
visited = set()
if cycle is None:
cycle = []
visited.add(n)
cycle += [n]
if n not in g:
return cycle
for e in g[n]:
if e not in visited:
cycle = get_cycle(e, g, visited, cycle)
return cycle
def merge_cycles(cycle, rev_graph, g, rg):
all_in_edges = []
min_internal = None
min_internal_weight = sys.maxsize
# find minimal internal edge weight
for n in cycle:
for e in rev_graph[n]:
if e in cycle:
if min_internal is None or rev_graph[n][e] < min_internal_weight:
min_internal = (n, e)
min_internal_weight = rev_graph[n][e]
continue
else:
all_in_edges.append((n, e))
# find the incoming edge with minimum modified cost
min_external = None
min_modified_weight = 0
for s, t in all_in_edges:
u, v = rg[s].popitem()
rg[s][u] = v
w = rev_graph[s][t] - (v - min_internal_weight)
if min_external is None or min_modified_weight > w:
min_external = (s, t)
min_modified_weight = w
u, w = rg[min_external[0]].popitem()
rem = (min_external[0], u)
rg[min_external[0]].clear()
if min_external[1] in rg:
rg[min_external[1]][min_external[0]] = w
else:
rg[min_external[1]] = {min_external[0]: w}
if rem[1] in g:
if rem[0] in g[rem[1]]:
del g[rem[1]][rem[0]]
if min_external[1] in g:
g[min_external[1]][min_external[0]] = w
else:
g[min_external[1]] = {min_external[0]: w}
def mst(root: int, graph: dict):
reversed_graph = reverse_graph(graph)
if root in reversed_graph:
reversed_graph[root] = {}
g = {}
for n in reversed_graph:
if len(reversed_graph[n]) == 0:
continue
minimum = sys.maxsize
s, d = None, None
for e in reversed_graph[n]:
if reversed_graph[n][e] < minimum:
minimum = reversed_graph[n][e]
s, d = n, e
if d in g:
g[d][s] = reversed_graph[s][d]
else:
g[d] = {s: reversed_graph[s][d]}
cycles = []
visited = set()
for n in g:
if n not in visited:
cycle = get_cycle(n, g, visited)
cycles.append(cycle)
rg = reverse_graph(g)
for cycle in cycles:
if root in cycle:
continue
merge_cycles(cycle, reversed_graph, g, rg)
return g
def read_graph(edges: list) -> dict:
graph = {}
for edge in edges:
if edge[0] - 1 not in graph.keys():
node_neighbors = {edge[1] - 1: edge[2]}
graph.update({edge[0] - 1: node_neighbors})
else:
graph[edge[0] - 1].update({edge[1] - 1: edge[2]})
return graph
def task4(n: int, qualifications: list, statements: list) -> int:
"""
Method for solving a startup problem
Parameters:
----------
n: int
Number of company employees
qualifications: list
Staff qualifications
statements: list
List of statements
Returns:
-------
Minimum cost of creating a hierarchy
"""
if len(qualifications) == 1:
return 0
root = qualifications.index(max(qualifications))
graph = read_graph(statements)
h = mst(root, graph)
visited_nodes = set(list(h.keys()) + [key for dct in h.values() for key in dct.keys()])
return sum(h[s][t] for s in h for t in h[s]) if n == len(visited_nodes) else -1
n = int(input())
q = list(map(int,input().split()))
m = int(input())
s = []
for i in range(0,m):
s.append(list(map(int,input().split())))
print(task4(n,q,s)) | Title: Hierarchy
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Nick's company employed *n* people. Now Nick needs to build a tree hierarchy of «supervisor-surbodinate» relations in the company (this is to say that each employee, except one, has exactly one supervisor). There are *m* applications written in the following form: «employee *a**i* is ready to become a supervisor of employee *b**i* at extra cost *c**i*». The qualification *q**j* of each employee is known, and for each application the following is true: *q**a**i*<=><=*q**b**i*.
Would you help Nick calculate the minimum cost of such a hierarchy, or find out that it is impossible to build it.
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=1000) — amount of employees in the company. The following line contains *n* space-separated numbers *q**j* (0<=≤<=*q**j*<=≤<=106)— the employees' qualifications. The following line contains number *m* (0<=≤<=*m*<=≤<=10000) — amount of received applications. The following *m* lines contain the applications themselves, each of them in the form of three space-separated numbers: *a**i*, *b**i* and *c**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, 0<=≤<=*c**i*<=≤<=106). Different applications can be similar, i.e. they can come from one and the same employee who offered to become a supervisor of the same person but at a different cost. For each application *q**a**i*<=><=*q**b**i*.
Output Specification:
Output the only line — the minimum cost of building such a hierarchy, or -1 if it is impossible to build it.
Demo Input:
['4\n7 2 3 1\n4\n1 2 5\n2 4 1\n3 4 1\n1 3 5\n', '3\n1 2 3\n2\n3 1 2\n3 1 3\n']
Demo Output:
['11\n', '-1\n']
Note:
In the first sample one of the possible ways for building a hierarchy is to take applications with indexes 1, 2 and 4, which give 11 as the minimum total cost. In the second sample it is impossible to build the required hierarchy, so the answer is -1. | ```python
import sys
def reverse_graph(graph: dict):
r = {}
for src in graph:
for (dst, c) in graph[src].items():
if dst in r:
r[dst][src] = c
else:
r[dst] = {src: c}
return r
def get_cycle(n: int, g: dict, visited: set = None, cycle: list = None):
if visited is None:
visited = set()
if cycle is None:
cycle = []
visited.add(n)
cycle += [n]
if n not in g:
return cycle
for e in g[n]:
if e not in visited:
cycle = get_cycle(e, g, visited, cycle)
return cycle
def merge_cycles(cycle, rev_graph, g, rg):
all_in_edges = []
min_internal = None
min_internal_weight = sys.maxsize
# find minimal internal edge weight
for n in cycle:
for e in rev_graph[n]:
if e in cycle:
if min_internal is None or rev_graph[n][e] < min_internal_weight:
min_internal = (n, e)
min_internal_weight = rev_graph[n][e]
continue
else:
all_in_edges.append((n, e))
# find the incoming edge with minimum modified cost
min_external = None
min_modified_weight = 0
for s, t in all_in_edges:
u, v = rg[s].popitem()
rg[s][u] = v
w = rev_graph[s][t] - (v - min_internal_weight)
if min_external is None or min_modified_weight > w:
min_external = (s, t)
min_modified_weight = w
u, w = rg[min_external[0]].popitem()
rem = (min_external[0], u)
rg[min_external[0]].clear()
if min_external[1] in rg:
rg[min_external[1]][min_external[0]] = w
else:
rg[min_external[1]] = {min_external[0]: w}
if rem[1] in g:
if rem[0] in g[rem[1]]:
del g[rem[1]][rem[0]]
if min_external[1] in g:
g[min_external[1]][min_external[0]] = w
else:
g[min_external[1]] = {min_external[0]: w}
def mst(root: int, graph: dict):
reversed_graph = reverse_graph(graph)
if root in reversed_graph:
reversed_graph[root] = {}
g = {}
for n in reversed_graph:
if len(reversed_graph[n]) == 0:
continue
minimum = sys.maxsize
s, d = None, None
for e in reversed_graph[n]:
if reversed_graph[n][e] < minimum:
minimum = reversed_graph[n][e]
s, d = n, e
if d in g:
g[d][s] = reversed_graph[s][d]
else:
g[d] = {s: reversed_graph[s][d]}
cycles = []
visited = set()
for n in g:
if n not in visited:
cycle = get_cycle(n, g, visited)
cycles.append(cycle)
rg = reverse_graph(g)
for cycle in cycles:
if root in cycle:
continue
merge_cycles(cycle, reversed_graph, g, rg)
return g
def read_graph(edges: list) -> dict:
graph = {}
for edge in edges:
if edge[0] - 1 not in graph.keys():
node_neighbors = {edge[1] - 1: edge[2]}
graph.update({edge[0] - 1: node_neighbors})
else:
graph[edge[0] - 1].update({edge[1] - 1: edge[2]})
return graph
def task4(n: int, qualifications: list, statements: list) -> int:
"""
Method for solving a startup problem
Parameters:
----------
n: int
Number of company employees
qualifications: list
Staff qualifications
statements: list
List of statements
Returns:
-------
Minimum cost of creating a hierarchy
"""
if len(qualifications) == 1:
return 0
root = qualifications.index(max(qualifications))
graph = read_graph(statements)
h = mst(root, graph)
visited_nodes = set(list(h.keys()) + [key for dct in h.values() for key in dct.keys()])
return sum(h[s][t] for s in h for t in h[s]) if n == len(visited_nodes) else -1
n = int(input())
q = list(map(int,input().split()))
m = int(input())
s = []
for i in range(0,m):
s.append(list(map(int,input().split())))
print(task4(n,q,s))
``` | 0 |
652 | C | Foe Pairs | PROGRAMMING | 1,800 | [
"combinatorics",
"sortings",
"two pointers"
] | null | null | You are given a permutation *p* of length *n*. Also you are given *m* foe pairs (*a**i*,<=*b**i*) (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*).
Your task is to count the number of different intervals (*x*,<=*y*) (1<=≤<=*x*<=≤<=*y*<=≤<=*n*) that do not contain any foe pairs. So you shouldn't count intervals (*x*,<=*y*) that contain at least one foe pair in it (the positions and order of the values from the foe pair are not important).
Consider some example: *p*<==<=[1,<=3,<=2,<=4] and foe pairs are {(3,<=2),<=(4,<=2)}. The interval (1,<=3) is incorrect because it contains a foe pair (3,<=2). The interval (1,<=4) is also incorrect because it contains two foe pairs (3,<=2) and (4,<=2). But the interval (1,<=2) is correct because it doesn't contain any foe pair. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=3·105) — the length of the permutation *p* and the number of foe pairs.
The second line contains *n* distinct integers *p**i* (1<=≤<=*p**i*<=≤<=*n*) — the elements of the permutation *p*.
Each of the next *m* lines contains two integers (*a**i*,<=*b**i*) (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*) — the *i*-th foe pair. Note a foe pair can appear multiple times in the given list. | Print the only integer *c* — the number of different intervals (*x*,<=*y*) that does not contain any foe pairs.
Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type. | [
"4 2\n1 3 2 4\n3 2\n2 4\n",
"9 5\n9 7 2 3 1 4 6 5 8\n1 6\n4 5\n2 7\n7 2\n2 7\n"
] | [
"5\n",
"20\n"
] | In the first example the intervals from the answer are (1, 1), (1, 2), (2, 2), (3, 3) and (4, 4). | 0 | [
{
"input": "4 2\n1 3 2 4\n3 2\n2 4",
"output": "5"
},
{
"input": "9 5\n9 7 2 3 1 4 6 5 8\n1 6\n4 5\n2 7\n7 2\n2 7",
"output": "20"
},
{
"input": "2 1\n1 2\n1 2",
"output": "2"
},
{
"input": "10 3\n4 10 5 1 6 8 9 2 3 7\n10 5\n2 10\n4 1",
"output": "39"
},
{
"input"... | 1,660,722,775 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 61 | 0 | import sys
# import random
# from collections import Counter, defaultdict, deque
# from functools import lru_cache, reduce
# from itertools import accumulate,product
# from heapq import nsmallest, nlargest, heapify, heappop, heappush
# input = sys.stdin.buffer.readline
input = sys.stdin.readline
def mp():return list(map(int,input().split()))
def it():return int(input())
import math
def solve():
n,m=mp()
nums=mp()
ans=0
record=[0]*(n+1)
edge=[0]*(n+1)
for i in range(n):
record[nums[i]]=i
for i in range(m):
x,y=mp()
x,y=record[x],record[y]
if x>y:x,y=y,x
edge[y]=max(edge[y],x)
ptr=0
for i in range(n):
ptr=max(ptr,edge[i])
ans+=i-ptr
print(ans)
return
if __name__ == '__main__':
# t=it()
# for _ in range(t):
# solve()
# n=it()
# n,m,k=mp()
solve()
| Title: Foe Pairs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a permutation *p* of length *n*. Also you are given *m* foe pairs (*a**i*,<=*b**i*) (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*).
Your task is to count the number of different intervals (*x*,<=*y*) (1<=≤<=*x*<=≤<=*y*<=≤<=*n*) that do not contain any foe pairs. So you shouldn't count intervals (*x*,<=*y*) that contain at least one foe pair in it (the positions and order of the values from the foe pair are not important).
Consider some example: *p*<==<=[1,<=3,<=2,<=4] and foe pairs are {(3,<=2),<=(4,<=2)}. The interval (1,<=3) is incorrect because it contains a foe pair (3,<=2). The interval (1,<=4) is also incorrect because it contains two foe pairs (3,<=2) and (4,<=2). But the interval (1,<=2) is correct because it doesn't contain any foe pair.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=3·105) — the length of the permutation *p* and the number of foe pairs.
The second line contains *n* distinct integers *p**i* (1<=≤<=*p**i*<=≤<=*n*) — the elements of the permutation *p*.
Each of the next *m* lines contains two integers (*a**i*,<=*b**i*) (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*) — the *i*-th foe pair. Note a foe pair can appear multiple times in the given list.
Output Specification:
Print the only integer *c* — the number of different intervals (*x*,<=*y*) that does not contain any foe pairs.
Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Demo Input:
['4 2\n1 3 2 4\n3 2\n2 4\n', '9 5\n9 7 2 3 1 4 6 5 8\n1 6\n4 5\n2 7\n7 2\n2 7\n']
Demo Output:
['5\n', '20\n']
Note:
In the first example the intervals from the answer are (1, 1), (1, 2), (2, 2), (3, 3) and (4, 4). | ```python
import sys
# import random
# from collections import Counter, defaultdict, deque
# from functools import lru_cache, reduce
# from itertools import accumulate,product
# from heapq import nsmallest, nlargest, heapify, heappop, heappush
# input = sys.stdin.buffer.readline
input = sys.stdin.readline
def mp():return list(map(int,input().split()))
def it():return int(input())
import math
def solve():
n,m=mp()
nums=mp()
ans=0
record=[0]*(n+1)
edge=[0]*(n+1)
for i in range(n):
record[nums[i]]=i
for i in range(m):
x,y=mp()
x,y=record[x],record[y]
if x>y:x,y=y,x
edge[y]=max(edge[y],x)
ptr=0
for i in range(n):
ptr=max(ptr,edge[i])
ans+=i-ptr
print(ans)
return
if __name__ == '__main__':
# t=it()
# for _ in range(t):
# solve()
# n=it()
# n,m,k=mp()
solve()
``` | 0 | |
18 | C | Stripe | PROGRAMMING | 1,200 | [
"data structures",
"implementation"
] | C. Stripe | 2 | 64 | Once Bob took a paper stripe of *n* squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into two pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem? | The first input line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of squares in the stripe. The second line contains *n* space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value. | Output the amount of ways to cut the stripe into two non-empty pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only. | [
"9\n1 5 -6 7 9 -16 0 -2 2\n",
"3\n1 1 1\n",
"2\n0 0\n"
] | [
"3\n",
"0\n",
"1\n"
] | none | 0 | [
{
"input": "9\n1 5 -6 7 9 -16 0 -2 2",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "0"
},
{
"input": "2\n0 0",
"output": "1"
},
{
"input": "4\n100 1 10 111",
"output": "1"
},
{
"input": "10\n0 4 -3 0 -2 2 -3 -3 2 5",
"output": "3"
},
{
"input": "... | 1,578,180,795 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 218 | 0 | n=int(input())
l=list(map(int,input().split()))
ls=[0]
x=0
for i in range(n):
x+=l[i]
ls.append(x)
count=0
for i in range(1,n):
if (ls[n]-ls[i])==0:
count+=1
print(count)
| Title: Stripe
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Once Bob took a paper stripe of *n* squares (the height of the stripe is 1 square). In each square he wrote an integer number, possibly negative. He became interested in how many ways exist to cut this stripe into two pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece, and each piece contains positive integer amount of squares. Would you help Bob solve this problem?
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=105) — amount of squares in the stripe. The second line contains *n* space-separated numbers — they are the numbers written in the squares of the stripe. These numbers are integer and do not exceed 10000 in absolute value.
Output Specification:
Output the amount of ways to cut the stripe into two non-empty pieces so that the sum of numbers from one piece is equal to the sum of numbers from the other piece. Don't forget that it's allowed to cut the stripe along the squares' borders only.
Demo Input:
['9\n1 5 -6 7 9 -16 0 -2 2\n', '3\n1 1 1\n', '2\n0 0\n']
Demo Output:
['3\n', '0\n', '1\n']
Note:
none | ```python
n=int(input())
l=list(map(int,input().split()))
ls=[0]
x=0
for i in range(n):
x+=l[i]
ls.append(x)
count=0
for i in range(1,n):
if (ls[n]-ls[i])==0:
count+=1
print(count)
``` | 0 |
588 | B | Duff in Love | PROGRAMMING | 1,300 | [
"math"
] | null | null | Duff is in love with lovely numbers! A positive integer *x* is called lovely if and only if there is no such positive integer *a*<=><=1 such that *a*2 is a divisor of *x*.
Malek has a number store! In his store, he has only divisors of positive integer *n* (and he has all of them). As a birthday present, Malek wants to give her a lovely number from his store. He wants this number to be as big as possible.
Malek always had issues in math, so he asked for your help. Please tell him what is the biggest lovely number in his store. | The first and only line of input contains one integer, *n* (1<=≤<=*n*<=≤<=1012). | Print the answer in one line. | [
"10\n",
"12\n"
] | [
"10\n",
"6\n"
] | In first sample case, there are numbers 1, 2, 5 and 10 in the shop. 10 isn't divisible by any perfect square, so 10 is lovely.
In second sample case, there are numbers 1, 2, 3, 4, 6 and 12 in the shop. 12 is divisible by 4 = 2<sup class="upper-index">2</sup>, so 12 is not lovely, while 6 is indeed lovely. | 1,000 | [
{
"input": "10",
"output": "10"
},
{
"input": "12",
"output": "6"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "2"
},
{
"input": "4",
"output": "2"
},
{
"input": "8",
"output": "2"
},
{
"input": "3",
"output": "3"
},
... | 1,575,478,207 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 9 | 2,000 | 0 | n=int(input())
for i in range(2,n+1):
while n%(i*i)==0:
n//=i
print(n)
| Title: Duff in Love
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Duff is in love with lovely numbers! A positive integer *x* is called lovely if and only if there is no such positive integer *a*<=><=1 such that *a*2 is a divisor of *x*.
Malek has a number store! In his store, he has only divisors of positive integer *n* (and he has all of them). As a birthday present, Malek wants to give her a lovely number from his store. He wants this number to be as big as possible.
Malek always had issues in math, so he asked for your help. Please tell him what is the biggest lovely number in his store.
Input Specification:
The first and only line of input contains one integer, *n* (1<=≤<=*n*<=≤<=1012).
Output Specification:
Print the answer in one line.
Demo Input:
['10\n', '12\n']
Demo Output:
['10\n', '6\n']
Note:
In first sample case, there are numbers 1, 2, 5 and 10 in the shop. 10 isn't divisible by any perfect square, so 10 is lovely.
In second sample case, there are numbers 1, 2, 3, 4, 6 and 12 in the shop. 12 is divisible by 4 = 2<sup class="upper-index">2</sup>, so 12 is not lovely, while 6 is indeed lovely. | ```python
n=int(input())
for i in range(2,n+1):
while n%(i*i)==0:
n//=i
print(n)
``` | 0 | |
985 | C | Liebig's Barrels | PROGRAMMING | 1,500 | [
"greedy"
] | null | null | You have *m*<==<=*n*·*k* wooden staves. The *i*-th stave has length *a**i*. You have to assemble *n* barrels consisting of *k* staves each, you can use any *k* staves to construct a barrel. Each stave must belong to exactly one barrel.
Let volume *v**j* of barrel *j* be equal to the length of the minimal stave in it.
You want to assemble exactly *n* barrels with the maximal total sum of volumes. But you have to make them equal enough, so a difference between volumes of any pair of the resulting barrels must not exceed *l*, i.e. |*v**x*<=-<=*v**y*|<=≤<=*l* for any 1<=≤<=*x*<=≤<=*n* and 1<=≤<=*y*<=≤<=*n*.
Print maximal total sum of volumes of equal enough barrels or 0 if it's impossible to satisfy the condition above. | The first line contains three space-separated integers *n*, *k* and *l* (1<=≤<=*n*,<=*k*<=≤<=105, 1<=≤<=*n*·*k*<=≤<=105, 0<=≤<=*l*<=≤<=109).
The second line contains *m*<==<=*n*·*k* space-separated integers *a*1,<=*a*2,<=...,<=*a**m* (1<=≤<=*a**i*<=≤<=109) — lengths of staves. | Print single integer — maximal total sum of the volumes of barrels or 0 if it's impossible to construct exactly *n* barrels satisfying the condition |*v**x*<=-<=*v**y*|<=≤<=*l* for any 1<=≤<=*x*<=≤<=*n* and 1<=≤<=*y*<=≤<=*n*. | [
"4 2 1\n2 2 1 2 3 2 2 3\n",
"2 1 0\n10 10\n",
"1 2 1\n5 2\n",
"3 2 1\n1 2 3 4 5 6\n"
] | [
"7\n",
"20\n",
"2\n",
"0\n"
] | In the first example you can form the following barrels: [1, 2], [2, 2], [2, 3], [2, 3].
In the second example you can form the following barrels: [10], [10].
In the third example you can form the following barrels: [2, 5].
In the fourth example difference between volumes of barrels in any partition is at least 2 so it is impossible to make barrels equal enough. | 0 | [
{
"input": "4 2 1\n2 2 1 2 3 2 2 3",
"output": "7"
},
{
"input": "2 1 0\n10 10",
"output": "20"
},
{
"input": "1 2 1\n5 2",
"output": "2"
},
{
"input": "3 2 1\n1 2 3 4 5 6",
"output": "0"
},
{
"input": "10 3 189\n267 697 667 4 52 128 85 616 142 344 413 660 962 194... | 1,527,154,002 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 171 | 7,884,800 | def binarySearchCount(arr,n,key):
left = 0
right = n
mid=0
while (left < right):
mid = left + (right-left)//2
if (arr[mid] == key):
while (mid+1<n and arr[mid+1] == key):
mid+=1
break
elif (arr[mid] > key):
right = mid
else:
left = mid + 1
while (arr[mid] > key):
mid-=1
return mid
n,k,l=map(int,input().split())
a=list(map(int,input().split()))
a=sorted(a)
s=binarySearchCount(a,n*k,a[0]+l)
v=0
temp=s//k
i=0
count=0
if s > k or n==1 or k==1:
while i<temp:
v+=a[i]
i+=k
count+=1
for i in range(n-count):
v+=a[s-i]
print(v) | Title: Liebig's Barrels
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have *m*<==<=*n*·*k* wooden staves. The *i*-th stave has length *a**i*. You have to assemble *n* barrels consisting of *k* staves each, you can use any *k* staves to construct a barrel. Each stave must belong to exactly one barrel.
Let volume *v**j* of barrel *j* be equal to the length of the minimal stave in it.
You want to assemble exactly *n* barrels with the maximal total sum of volumes. But you have to make them equal enough, so a difference between volumes of any pair of the resulting barrels must not exceed *l*, i.e. |*v**x*<=-<=*v**y*|<=≤<=*l* for any 1<=≤<=*x*<=≤<=*n* and 1<=≤<=*y*<=≤<=*n*.
Print maximal total sum of volumes of equal enough barrels or 0 if it's impossible to satisfy the condition above.
Input Specification:
The first line contains three space-separated integers *n*, *k* and *l* (1<=≤<=*n*,<=*k*<=≤<=105, 1<=≤<=*n*·*k*<=≤<=105, 0<=≤<=*l*<=≤<=109).
The second line contains *m*<==<=*n*·*k* space-separated integers *a*1,<=*a*2,<=...,<=*a**m* (1<=≤<=*a**i*<=≤<=109) — lengths of staves.
Output Specification:
Print single integer — maximal total sum of the volumes of barrels or 0 if it's impossible to construct exactly *n* barrels satisfying the condition |*v**x*<=-<=*v**y*|<=≤<=*l* for any 1<=≤<=*x*<=≤<=*n* and 1<=≤<=*y*<=≤<=*n*.
Demo Input:
['4 2 1\n2 2 1 2 3 2 2 3\n', '2 1 0\n10 10\n', '1 2 1\n5 2\n', '3 2 1\n1 2 3 4 5 6\n']
Demo Output:
['7\n', '20\n', '2\n', '0\n']
Note:
In the first example you can form the following barrels: [1, 2], [2, 2], [2, 3], [2, 3].
In the second example you can form the following barrels: [10], [10].
In the third example you can form the following barrels: [2, 5].
In the fourth example difference between volumes of barrels in any partition is at least 2 so it is impossible to make barrels equal enough. | ```python
def binarySearchCount(arr,n,key):
left = 0
right = n
mid=0
while (left < right):
mid = left + (right-left)//2
if (arr[mid] == key):
while (mid+1<n and arr[mid+1] == key):
mid+=1
break
elif (arr[mid] > key):
right = mid
else:
left = mid + 1
while (arr[mid] > key):
mid-=1
return mid
n,k,l=map(int,input().split())
a=list(map(int,input().split()))
a=sorted(a)
s=binarySearchCount(a,n*k,a[0]+l)
v=0
temp=s//k
i=0
count=0
if s > k or n==1 or k==1:
while i<temp:
v+=a[i]
i+=k
count+=1
for i in range(n-count):
v+=a[s-i]
print(v)
``` | 0 | |
327 | B | Hungry Sequence | PROGRAMMING | 1,200 | [
"math"
] | null | null | Iahub and Iahubina went to a date at a luxury restaurant. Everything went fine until paying for the food. Instead of money, the waiter wants Iahub to write a Hungry sequence consisting of *n* integers.
A sequence *a*1, *a*2, ..., *a**n*, consisting of *n* integers, is Hungry if and only if:
- Its elements are in increasing order. That is an inequality *a**i*<=<<=*a**j* holds for any two indices *i*,<=*j* (*i*<=<<=*j*). - For any two indices *i* and *j* (*i*<=<<=*j*), *a**j* must not be divisible by *a**i*.
Iahub is in trouble, so he asks you for help. Find a Hungry sequence with *n* elements. | The input contains a single integer: *n* (1<=≤<=*n*<=≤<=105). | Output a line that contains *n* space-separated integers *a*1 *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=107), representing a possible Hungry sequence. Note, that each *a**i* must not be greater than 10000000 (107) and less than 1.
If there are multiple solutions you can output any one. | [
"3\n",
"5\n"
] | [
"2 9 15\n",
"11 14 20 27 31\n"
] | none | 500 | [
{
"input": "3",
"output": "2 9 15"
},
{
"input": "5",
"output": "11 14 20 27 31"
},
{
"input": "1",
"output": "3"
},
{
"input": "1000",
"output": "3000 3001 3002 3003 3004 3005 3006 3007 3008 3009 3010 3011 3012 3013 3014 3015 3016 3017 3018 3019 3020 3021 3022 3023 3024 ... | 1,655,744,160 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 3 | 1,000 | 0 | l=[11,14,20,27,31]
n=int(input())
s=[]
if n<=5:
s=l[:n]
else:
s=l
c=0
d=n-5
e=31
while(c!=d):
e+=1
if e%11!=0 and e%14!=0 and e%20!=0 and e%27!=0 and e%31!=0:
f=0
for i in s:
if e%i!=0:
f+=1
if f==len(s):
s.append(e)
c+=1
for i in s:
print(i, end=" ") | Title: Hungry Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub and Iahubina went to a date at a luxury restaurant. Everything went fine until paying for the food. Instead of money, the waiter wants Iahub to write a Hungry sequence consisting of *n* integers.
A sequence *a*1, *a*2, ..., *a**n*, consisting of *n* integers, is Hungry if and only if:
- Its elements are in increasing order. That is an inequality *a**i*<=<<=*a**j* holds for any two indices *i*,<=*j* (*i*<=<<=*j*). - For any two indices *i* and *j* (*i*<=<<=*j*), *a**j* must not be divisible by *a**i*.
Iahub is in trouble, so he asks you for help. Find a Hungry sequence with *n* elements.
Input Specification:
The input contains a single integer: *n* (1<=≤<=*n*<=≤<=105).
Output Specification:
Output a line that contains *n* space-separated integers *a*1 *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=107), representing a possible Hungry sequence. Note, that each *a**i* must not be greater than 10000000 (107) and less than 1.
If there are multiple solutions you can output any one.
Demo Input:
['3\n', '5\n']
Demo Output:
['2 9 15\n', '11 14 20 27 31\n']
Note:
none | ```python
l=[11,14,20,27,31]
n=int(input())
s=[]
if n<=5:
s=l[:n]
else:
s=l
c=0
d=n-5
e=31
while(c!=d):
e+=1
if e%11!=0 and e%14!=0 and e%20!=0 and e%27!=0 and e%31!=0:
f=0
for i in s:
if e%i!=0:
f+=1
if f==len(s):
s.append(e)
c+=1
for i in s:
print(i, end=" ")
``` | 0 | |
651 | A | Joysticks | PROGRAMMING | 1,100 | [
"dp",
"greedy",
"implementation",
"math"
] | null | null | Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at *a*1 percent and second one is charged at *a*2 percent. You can connect charger to a joystick only at the beginning of each minute. In one minute joystick either discharges by 2 percent (if not connected to a charger) or charges by 1 percent (if connected to a charger).
Game continues while both joysticks have a positive charge. Hence, if at the beginning of minute some joystick is charged by 1 percent, it has to be connected to a charger, otherwise the game stops. If some joystick completely discharges (its charge turns to 0), the game also stops.
Determine the maximum number of minutes that game can last. It is prohibited to pause the game, i. e. at each moment both joysticks should be enabled. It is allowed for joystick to be charged by more than 100 percent. | The first line of the input contains two positive integers *a*1 and *a*2 (1<=≤<=*a*1,<=*a*2<=≤<=100), the initial charge level of first and second joystick respectively. | Output the only integer, the maximum number of minutes that the game can last. Game continues until some joystick is discharged. | [
"3 5\n",
"4 4\n"
] | [
"6\n",
"5\n"
] | In the first sample game lasts for 6 minute by using the following algorithm:
- at the beginning of the first minute connect first joystick to the charger, by the end of this minute first joystick is at 4%, second is at 3%; - continue the game without changing charger, by the end of the second minute the first joystick is at 5%, second is at 1%; - at the beginning of the third minute connect second joystick to the charger, after this minute the first joystick is at 3%, the second one is at 2%; - continue the game without changing charger, by the end of the fourth minute first joystick is at 1%, second one is at 3%; - at the beginning of the fifth minute connect first joystick to the charger, after this minute the first joystick is at 2%, the second one is at 1%; - at the beginning of the sixth minute connect second joystick to the charger, after this minute the first joystick is at 0%, the second one is at 2%.
After that the first joystick is completely discharged and the game is stopped. | 500 | [
{
"input": "3 5",
"output": "6"
},
{
"input": "4 4",
"output": "5"
},
{
"input": "100 100",
"output": "197"
},
{
"input": "1 100",
"output": "98"
},
{
"input": "100 1",
"output": "98"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "1 1",
... | 1,569,006,051 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 6 | 124 | 0 | def main():
a, b = [int(v) for v in input().split()]
c= 0
while a>0 and b>0:
c+=1
if a>b:
b+=1
a-=2
else:
b-=2
a+=1
print(c)
if __name__ == "__main__":
main()
| Title: Joysticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at *a*1 percent and second one is charged at *a*2 percent. You can connect charger to a joystick only at the beginning of each minute. In one minute joystick either discharges by 2 percent (if not connected to a charger) or charges by 1 percent (if connected to a charger).
Game continues while both joysticks have a positive charge. Hence, if at the beginning of minute some joystick is charged by 1 percent, it has to be connected to a charger, otherwise the game stops. If some joystick completely discharges (its charge turns to 0), the game also stops.
Determine the maximum number of minutes that game can last. It is prohibited to pause the game, i. e. at each moment both joysticks should be enabled. It is allowed for joystick to be charged by more than 100 percent.
Input Specification:
The first line of the input contains two positive integers *a*1 and *a*2 (1<=≤<=*a*1,<=*a*2<=≤<=100), the initial charge level of first and second joystick respectively.
Output Specification:
Output the only integer, the maximum number of minutes that the game can last. Game continues until some joystick is discharged.
Demo Input:
['3 5\n', '4 4\n']
Demo Output:
['6\n', '5\n']
Note:
In the first sample game lasts for 6 minute by using the following algorithm:
- at the beginning of the first minute connect first joystick to the charger, by the end of this minute first joystick is at 4%, second is at 3%; - continue the game without changing charger, by the end of the second minute the first joystick is at 5%, second is at 1%; - at the beginning of the third minute connect second joystick to the charger, after this minute the first joystick is at 3%, the second one is at 2%; - continue the game without changing charger, by the end of the fourth minute first joystick is at 1%, second one is at 3%; - at the beginning of the fifth minute connect first joystick to the charger, after this minute the first joystick is at 2%, the second one is at 1%; - at the beginning of the sixth minute connect second joystick to the charger, after this minute the first joystick is at 0%, the second one is at 2%.
After that the first joystick is completely discharged and the game is stopped. | ```python
def main():
a, b = [int(v) for v in input().split()]
c= 0
while a>0 and b>0:
c+=1
if a>b:
b+=1
a-=2
else:
b-=2
a+=1
print(c)
if __name__ == "__main__":
main()
``` | 0 | |
913 | B | Christmas Spruce | PROGRAMMING | 1,200 | [
"implementation",
"trees"
] | null | null | Consider a rooted tree. A rooted tree has one special vertex called the root. All edges are directed from the root. Vertex *u* is called a child of vertex *v* and vertex *v* is called a parent of vertex *u* if there exists a directed edge from *v* to *u*. A vertex is called a leaf if it doesn't have children and has a parent.
Let's call a rooted tree a spruce if its every non-leaf vertex has at least 3 leaf children. You are given a rooted tree, check whether it's a spruce.
The definition of a rooted tree can be found [here](https://goo.gl/1dqvzz). | The first line contains one integer *n* — the number of vertices in the tree (3<=≤<=*n*<=≤<=1<=000). Each of the next *n*<=-<=1 lines contains one integer *p**i* (1<=≤<=*i*<=≤<=*n*<=-<=1) — the index of the parent of the *i*<=+<=1-th vertex (1<=≤<=*p**i*<=≤<=*i*).
Vertex 1 is the root. It's guaranteed that the root has at least 2 children. | Print "Yes" if the tree is a spruce and "No" otherwise. | [
"4\n1\n1\n1\n",
"7\n1\n1\n1\n2\n2\n2\n",
"8\n1\n1\n1\n1\n3\n3\n3\n"
] | [
"Yes\n",
"No\n",
"Yes\n"
] | The first example:
<img class="tex-graphics" src="https://espresso.codeforces.com/8dd976913226df83d535dfa66193f5525f8471bc.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The second example:
<img class="tex-graphics" src="https://espresso.codeforces.com/44dad5804f5290a2e026c9c41a15151562df8682.png" style="max-width: 100.0%;max-height: 100.0%;"/>
It is not a spruce, because the non-leaf vertex 1 has only 2 leaf children.
The third example:
<img class="tex-graphics" src="https://espresso.codeforces.com/cf84a9e1585707f4ab06eff8eb1120a49b5e1ef7.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 750 | [
{
"input": "4\n1\n1\n1",
"output": "Yes"
},
{
"input": "7\n1\n1\n1\n2\n2\n2",
"output": "No"
},
{
"input": "8\n1\n1\n1\n1\n3\n3\n3",
"output": "Yes"
},
{
"input": "3\n1\n1",
"output": "No"
},
{
"input": "13\n1\n2\n2\n2\n1\n6\n6\n6\n1\n10\n10\n10",
"output": "N... | 1,690,438,982 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 46 | 0 | vert = int(input())
vocab = {}
list = []
list_t = []
for i in range(1,vert ):
list.append([i,int(input())])
user_tipiks = []
list_ind = []
list_used_ind = []
count_feed_vert = len(list)
tupik_vetv = 1
for i in range(0,len(list)):
ind = list[i][0]
if ind == 1:
continue
list_ind.append(ind)
item = list[i][1]
if item != 1 and item not in user_tipiks:
user_tipiks.append(item)
tupik_vetv += 1
for i in range(0,len(list)):
item = list[i][1]
if item in list_ind and item not in list_used_ind:
list_used_ind.append(item)
count_feed_vert-=1
tupik_vetv = tupik_vetv*3
if(count_feed_vert == tupik_vetv):
print("Yes")
else:
print("No")
| Title: Christmas Spruce
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Consider a rooted tree. A rooted tree has one special vertex called the root. All edges are directed from the root. Vertex *u* is called a child of vertex *v* and vertex *v* is called a parent of vertex *u* if there exists a directed edge from *v* to *u*. A vertex is called a leaf if it doesn't have children and has a parent.
Let's call a rooted tree a spruce if its every non-leaf vertex has at least 3 leaf children. You are given a rooted tree, check whether it's a spruce.
The definition of a rooted tree can be found [here](https://goo.gl/1dqvzz).
Input Specification:
The first line contains one integer *n* — the number of vertices in the tree (3<=≤<=*n*<=≤<=1<=000). Each of the next *n*<=-<=1 lines contains one integer *p**i* (1<=≤<=*i*<=≤<=*n*<=-<=1) — the index of the parent of the *i*<=+<=1-th vertex (1<=≤<=*p**i*<=≤<=*i*).
Vertex 1 is the root. It's guaranteed that the root has at least 2 children.
Output Specification:
Print "Yes" if the tree is a spruce and "No" otherwise.
Demo Input:
['4\n1\n1\n1\n', '7\n1\n1\n1\n2\n2\n2\n', '8\n1\n1\n1\n1\n3\n3\n3\n']
Demo Output:
['Yes\n', 'No\n', 'Yes\n']
Note:
The first example:
<img class="tex-graphics" src="https://espresso.codeforces.com/8dd976913226df83d535dfa66193f5525f8471bc.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The second example:
<img class="tex-graphics" src="https://espresso.codeforces.com/44dad5804f5290a2e026c9c41a15151562df8682.png" style="max-width: 100.0%;max-height: 100.0%;"/>
It is not a spruce, because the non-leaf vertex 1 has only 2 leaf children.
The third example:
<img class="tex-graphics" src="https://espresso.codeforces.com/cf84a9e1585707f4ab06eff8eb1120a49b5e1ef7.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
vert = int(input())
vocab = {}
list = []
list_t = []
for i in range(1,vert ):
list.append([i,int(input())])
user_tipiks = []
list_ind = []
list_used_ind = []
count_feed_vert = len(list)
tupik_vetv = 1
for i in range(0,len(list)):
ind = list[i][0]
if ind == 1:
continue
list_ind.append(ind)
item = list[i][1]
if item != 1 and item not in user_tipiks:
user_tipiks.append(item)
tupik_vetv += 1
for i in range(0,len(list)):
item = list[i][1]
if item in list_ind and item not in list_used_ind:
list_used_ind.append(item)
count_feed_vert-=1
tupik_vetv = tupik_vetv*3
if(count_feed_vert == tupik_vetv):
print("Yes")
else:
print("No")
``` | 0 | |
143 | A | Help Vasilisa the Wise 2 | PROGRAMMING | 1,000 | [
"brute force",
"math"
] | null | null | Vasilisa the Wise from the Kingdom of Far Far Away got a magic box with a secret as a present from her friend Hellawisa the Wise from the Kingdom of A Little Closer. However, Vasilisa the Wise does not know what the box's secret is, since she cannot open it again. She hopes that you will help her one more time with that.
The box's lock looks as follows: it contains 4 identical deepenings for gems as a 2<=×<=2 square, and some integer numbers are written at the lock's edge near the deepenings. The example of a lock is given on the picture below.
The box is accompanied with 9 gems. Their shapes match the deepenings' shapes and each gem contains one number from 1 to 9 (each number is written on exactly one gem). The box will only open after it is decorated with gems correctly: that is, each deepening in the lock should be filled with exactly one gem. Also, the sums of numbers in the square's rows, columns and two diagonals of the square should match the numbers written at the lock's edge. For example, the above lock will open if we fill the deepenings with gems with numbers as is shown on the picture below.
Now Vasilisa the Wise wants to define, given the numbers on the box's lock, which gems she should put in the deepenings to open the box. Help Vasilisa to solve this challenging task. | The input contains numbers written on the edges of the lock of the box. The first line contains space-separated integers *r*1 and *r*2 that define the required sums of numbers in the rows of the square. The second line contains space-separated integers *c*1 and *c*2 that define the required sums of numbers in the columns of the square. The third line contains space-separated integers *d*1 and *d*2 that define the required sums of numbers on the main and on the side diagonals of the square (1<=≤<=*r*1,<=*r*2,<=*c*1,<=*c*2,<=*d*1,<=*d*2<=≤<=20). Correspondence between the above 6 variables and places where they are written is shown on the picture below. For more clarifications please look at the second sample test that demonstrates the example given in the problem statement. | Print the scheme of decorating the box with stones: two lines containing two space-separated integers from 1 to 9. The numbers should be pairwise different. If there is no solution for the given lock, then print the single number "-1" (without the quotes).
If there are several solutions, output any. | [
"3 7\n4 6\n5 5\n",
"11 10\n13 8\n5 16\n",
"1 2\n3 4\n5 6\n",
"10 10\n10 10\n10 10\n"
] | [
"1 2\n3 4\n",
"4 7\n9 1\n",
"-1\n",
"-1\n"
] | Pay attention to the last test from the statement: it is impossible to open the box because for that Vasilisa the Wise would need 4 identical gems containing number "5". However, Vasilisa only has one gem with each number from 1 to 9. | 500 | [
{
"input": "3 7\n4 6\n5 5",
"output": "1 2\n3 4"
},
{
"input": "11 10\n13 8\n5 16",
"output": "4 7\n9 1"
},
{
"input": "1 2\n3 4\n5 6",
"output": "-1"
},
{
"input": "10 10\n10 10\n10 10",
"output": "-1"
},
{
"input": "5 13\n8 10\n11 7",
"output": "3 2\n5 8"
... | 1,610,075,930 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 186 | 307,200 | r,r2 = [int(x) for x in input().split()]
c,c2 = [int(x) for x in input().split()]
d,d2 = [int(x) for x in input().split()]
res = []
for x in range(1,10):
for y in range(1,10):
for z in range(1,10):
for k in range(1,10):
if x!=y and x!=z and x!=k and y!=z and y!=k and x+y==r and z+k==r2 and x+z==c and y+k==c2:
res = [x,y,z,k]
break
print(res[0],res[1])
print(res[2],res[3])
| Title: Help Vasilisa the Wise 2
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasilisa the Wise from the Kingdom of Far Far Away got a magic box with a secret as a present from her friend Hellawisa the Wise from the Kingdom of A Little Closer. However, Vasilisa the Wise does not know what the box's secret is, since she cannot open it again. She hopes that you will help her one more time with that.
The box's lock looks as follows: it contains 4 identical deepenings for gems as a 2<=×<=2 square, and some integer numbers are written at the lock's edge near the deepenings. The example of a lock is given on the picture below.
The box is accompanied with 9 gems. Their shapes match the deepenings' shapes and each gem contains one number from 1 to 9 (each number is written on exactly one gem). The box will only open after it is decorated with gems correctly: that is, each deepening in the lock should be filled with exactly one gem. Also, the sums of numbers in the square's rows, columns and two diagonals of the square should match the numbers written at the lock's edge. For example, the above lock will open if we fill the deepenings with gems with numbers as is shown on the picture below.
Now Vasilisa the Wise wants to define, given the numbers on the box's lock, which gems she should put in the deepenings to open the box. Help Vasilisa to solve this challenging task.
Input Specification:
The input contains numbers written on the edges of the lock of the box. The first line contains space-separated integers *r*1 and *r*2 that define the required sums of numbers in the rows of the square. The second line contains space-separated integers *c*1 and *c*2 that define the required sums of numbers in the columns of the square. The third line contains space-separated integers *d*1 and *d*2 that define the required sums of numbers on the main and on the side diagonals of the square (1<=≤<=*r*1,<=*r*2,<=*c*1,<=*c*2,<=*d*1,<=*d*2<=≤<=20). Correspondence between the above 6 variables and places where they are written is shown on the picture below. For more clarifications please look at the second sample test that demonstrates the example given in the problem statement.
Output Specification:
Print the scheme of decorating the box with stones: two lines containing two space-separated integers from 1 to 9. The numbers should be pairwise different. If there is no solution for the given lock, then print the single number "-1" (without the quotes).
If there are several solutions, output any.
Demo Input:
['3 7\n4 6\n5 5\n', '11 10\n13 8\n5 16\n', '1 2\n3 4\n5 6\n', '10 10\n10 10\n10 10\n']
Demo Output:
['1 2\n3 4\n', '4 7\n9 1\n', '-1\n', '-1\n']
Note:
Pay attention to the last test from the statement: it is impossible to open the box because for that Vasilisa the Wise would need 4 identical gems containing number "5". However, Vasilisa only has one gem with each number from 1 to 9. | ```python
r,r2 = [int(x) for x in input().split()]
c,c2 = [int(x) for x in input().split()]
d,d2 = [int(x) for x in input().split()]
res = []
for x in range(1,10):
for y in range(1,10):
for z in range(1,10):
for k in range(1,10):
if x!=y and x!=z and x!=k and y!=z and y!=k and x+y==r and z+k==r2 and x+z==c and y+k==c2:
res = [x,y,z,k]
break
print(res[0],res[1])
print(res[2],res[3])
``` | 0 | |
214 | A | System of Equations | PROGRAMMING | 800 | [
"brute force"
] | null | null | Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
You should count, how many there are pairs of integers (*a*,<=*b*) (0<=≤<=*a*,<=*b*) which satisfy the system. | A single line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the parameters of the system. The numbers on the line are separated by a space. | On a single line print the answer to the problem. | [
"9 3\n",
"14 28\n",
"4 20\n"
] | [
"1\n",
"1\n",
"0\n"
] | In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair. | 500 | [
{
"input": "9 3",
"output": "1"
},
{
"input": "14 28",
"output": "1"
},
{
"input": "4 20",
"output": "0"
},
{
"input": "18 198",
"output": "1"
},
{
"input": "22 326",
"output": "1"
},
{
"input": "26 104",
"output": "1"
},
{
"input": "14 10"... | 1,667,490,038 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 54 | 124 | 0 | def solve(n, m):
valid_pairs_count = 0
valid_pairs = {}
a = 0
while (a ** 2) <= n:
b = n - (a ** 2)
if int(b) == b: # checks to see if b is an integer
valid_pairs[(a, b)] = True
a += 1
b = 0
while (b ** 2) <= m:
a = m - (b ** 2)
if int(a) == a:
if (a, b) in valid_pairs:
valid_pairs_count += 1
b += 1
return valid_pairs_count
if __name__ == "__main__":
n, m = [int(_) for _ in input().split(sep=" ")]
result = solve(n, m)
print(result)
| Title: System of Equations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Furik loves math lessons very much, so he doesn't attend them, unlike Rubik. But now Furik wants to get a good mark for math. For that Ms. Ivanova, his math teacher, gave him a new task. Furik solved the task immediately. Can you?
You are given a system of equations:
You should count, how many there are pairs of integers (*a*,<=*b*) (0<=≤<=*a*,<=*b*) which satisfy the system.
Input Specification:
A single line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the parameters of the system. The numbers on the line are separated by a space.
Output Specification:
On a single line print the answer to the problem.
Demo Input:
['9 3\n', '14 28\n', '4 20\n']
Demo Output:
['1\n', '1\n', '0\n']
Note:
In the first sample the suitable pair is integers (3, 0). In the second sample the suitable pair is integers (3, 5). In the third sample there is no suitable pair. | ```python
def solve(n, m):
valid_pairs_count = 0
valid_pairs = {}
a = 0
while (a ** 2) <= n:
b = n - (a ** 2)
if int(b) == b: # checks to see if b is an integer
valid_pairs[(a, b)] = True
a += 1
b = 0
while (b ** 2) <= m:
a = m - (b ** 2)
if int(a) == a:
if (a, b) in valid_pairs:
valid_pairs_count += 1
b += 1
return valid_pairs_count
if __name__ == "__main__":
n, m = [int(_) for _ in input().split(sep=" ")]
result = solve(n, m)
print(result)
``` | 3 | |
500 | A | New Year Transportation | PROGRAMMING | 1,000 | [
"dfs and similar",
"graphs",
"implementation"
] | null | null | New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system. | The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World. | If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO". | [
"8 4\n1 2 1 2 1 2 1\n",
"8 5\n1 2 1 2 1 1 1\n"
] | [
"YES\n",
"NO\n"
] | In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | 500 | [
{
"input": "8 4\n1 2 1 2 1 2 1",
"output": "YES"
},
{
"input": "8 5\n1 2 1 2 1 1 1",
"output": "NO"
},
{
"input": "20 19\n13 16 7 6 12 1 5 7 8 6 5 7 5 5 3 3 2 2 1",
"output": "YES"
},
{
"input": "50 49\n11 7 1 41 26 36 19 16 38 14 36 35 37 27 20 27 3 6 21 2 27 11 18 17 19 16 ... | 1,645,420,224 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 12 | 93 | 4,812,800 | n,k=map(int,input().split())
a=[int(x)for x in input().split()]
cur=1
for i in range(n-1):
if cur>=n-1:print('NO');break
cur += a[cur-1]
if cur==k:
print('YES');break | Title: New Year Transportation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system.
Input Specification:
The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output Specification:
If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO".
Demo Input:
['8 4\n1 2 1 2 1 2 1\n', '8 5\n1 2 1 2 1 1 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | ```python
n,k=map(int,input().split())
a=[int(x)for x in input().split()]
cur=1
for i in range(n-1):
if cur>=n-1:print('NO');break
cur += a[cur-1]
if cur==k:
print('YES');break
``` | 0 | |
999 | B | Reversing Encryption | PROGRAMMING | 900 | [
"implementation"
] | null | null | A string $s$ of length $n$ can be encrypted by the following algorithm:
- iterate over all divisors of $n$ in decreasing order (i.e. from $n$ to $1$), - for each divisor $d$, reverse the substring $s[1 \dots d]$ (i.e. the substring which starts at position $1$ and ends at position $d$).
For example, the above algorithm applied to the string $s$="codeforces" leads to the following changes: "codeforces" $\to$ "secrofedoc" $\to$ "orcesfedoc" $\to$ "rocesfedoc" $\to$ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because $d=1$).
You are given the encrypted string $t$. Your task is to decrypt this string, i.e., to find a string $s$ such that the above algorithm results in string $t$. It can be proven that this string $s$ always exists and is unique. | The first line of input consists of a single integer $n$ ($1 \le n \le 100$) — the length of the string $t$. The second line of input consists of the string $t$. The length of $t$ is $n$, and it consists only of lowercase Latin letters. | Print a string $s$ such that the above algorithm results in $t$. | [
"10\nrocesfedoc\n",
"16\nplmaetwoxesisiht\n",
"1\nz\n"
] | [
"codeforces\n",
"thisisexampletwo\n",
"z\n"
] | The first example is described in the problem statement. | 0 | [
{
"input": "10\nrocesfedoc",
"output": "codeforces"
},
{
"input": "16\nplmaetwoxesisiht",
"output": "thisisexampletwo"
},
{
"input": "1\nz",
"output": "z"
},
{
"input": "2\nir",
"output": "ri"
},
{
"input": "3\nilj",
"output": "jli"
},
{
"input": "4\nj... | 1,613,939,704 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 62 | 0 | n=int(input())
s=input()
for i in range(2, n+1):
if n%i==0:
z=str()
for j in range(i-1, -1, -1):
z=z+s[j]
for j in range(i, n):
z=z+s[j]
s=z
print(s) | Title: Reversing Encryption
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string $s$ of length $n$ can be encrypted by the following algorithm:
- iterate over all divisors of $n$ in decreasing order (i.e. from $n$ to $1$), - for each divisor $d$, reverse the substring $s[1 \dots d]$ (i.e. the substring which starts at position $1$ and ends at position $d$).
For example, the above algorithm applied to the string $s$="codeforces" leads to the following changes: "codeforces" $\to$ "secrofedoc" $\to$ "orcesfedoc" $\to$ "rocesfedoc" $\to$ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because $d=1$).
You are given the encrypted string $t$. Your task is to decrypt this string, i.e., to find a string $s$ such that the above algorithm results in string $t$. It can be proven that this string $s$ always exists and is unique.
Input Specification:
The first line of input consists of a single integer $n$ ($1 \le n \le 100$) — the length of the string $t$. The second line of input consists of the string $t$. The length of $t$ is $n$, and it consists only of lowercase Latin letters.
Output Specification:
Print a string $s$ such that the above algorithm results in $t$.
Demo Input:
['10\nrocesfedoc\n', '16\nplmaetwoxesisiht\n', '1\nz\n']
Demo Output:
['codeforces\n', 'thisisexampletwo\n', 'z\n']
Note:
The first example is described in the problem statement. | ```python
n=int(input())
s=input()
for i in range(2, n+1):
if n%i==0:
z=str()
for j in range(i-1, -1, -1):
z=z+s[j]
for j in range(i, n):
z=z+s[j]
s=z
print(s)
``` | 3 | |
515 | B | Drazil and His Happy Friends | PROGRAMMING | 1,300 | [
"brute force",
"dsu",
"meet-in-the-middle",
"number theory"
] | null | null | Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.
There are *n* boys and *m* girls among his friends. Let's number them from 0 to *n*<=-<=1 and 0 to *m*<=-<=1 separately. In *i*-th day, Drazil invites -th boy and -th girl to have dinner together (as Drazil is programmer, *i* starts from 0). If one of those two people is happy, the other one will also become happy. Otherwise, those two people remain in their states. Once a person becomes happy (or if he/she was happy originally), he stays happy forever.
Drazil wants to know whether he can use this plan to make all his friends become happy at some moment. | The first line contains two integer *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
The second line contains integer *b* (0<=≤<=*b*<=≤<=*n*), denoting the number of happy boys among friends of Drazil, and then follow *b* distinct integers *x*1,<=*x*2,<=...,<=*x**b* (0<=≤<=*x**i*<=<<=*n*), denoting the list of indices of happy boys.
The third line conatins integer *g* (0<=≤<=*g*<=≤<=*m*), denoting the number of happy girls among friends of Drazil, and then follow *g* distinct integers *y*1,<=*y*2,<=... ,<=*y**g* (0<=≤<=*y**j*<=<<=*m*), denoting the list of indices of happy girls.
It is guaranteed that there is at least one person that is unhappy among his friends. | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | [
"2 3\n0\n1 0\n",
"2 4\n1 0\n1 2\n",
"2 3\n1 0\n1 1\n"
] | [
"Yes\n",
"No\n",
"Yes\n"
] | By <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/169ade208e6eb4f9263c57aaff716529d59c3288.png" style="max-width: 100.0%;max-height: 100.0%;"/> we define the remainder of integer division of *i* by *k*.
In first sample case:
- On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. - On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes at this day. - On the 2-nd day, Drazil invites 0-th boy and 2-nd girl. Because 0-th boy is already happy he makes 2-nd girl become happy at this day. - On the 3-rd day, Drazil invites 1-st boy and 0-th girl. 0-th girl is happy, so she makes 1-st boy happy. - On the 4-th day, Drazil invites 0-th boy and 1-st girl. 0-th boy is happy, so he makes the 1-st girl happy. So, all friends become happy at this moment. | 1,000 | [
{
"input": "2 3\n0\n1 0",
"output": "Yes"
},
{
"input": "2 4\n1 0\n1 2",
"output": "No"
},
{
"input": "2 3\n1 0\n1 1",
"output": "Yes"
},
{
"input": "16 88\n6 5 14 2 0 12 7\n30 21 64 35 79 74 39 63 44 81 73 0 27 33 69 12 86 46 20 25 55 52 7 58 23 5 60 32 41 50 82",
"outpu... | 1,599,329,127 | 2,147,483,647 | PyPy 3 | OK | TESTS | 56 | 140 | 1,331,200 | n, m = [int(p) for p in input().split()]
arrb = [int(p) for p in input().split()]
arrg = [int(p) for p in input().split()]
arrb.pop(0)
arrg.pop(0)
cb = [False]*n
cg = [False]*m
for i in arrb:
cb[i] = True
for i in arrg:
cg[i] = True
i = 0
while i < 10**5:
b = cb[i%n]
g = cg[i%m]
if b and g:
i += 1
continue
elif b and not g:
cg[i%m] = True
elif not b and g:
cb[i%n] = True
else:
i += 1
continue
i += 1
if all(cb) and all(cg):
print('Yes')
else:
print('No') | Title: Drazil and His Happy Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.
There are *n* boys and *m* girls among his friends. Let's number them from 0 to *n*<=-<=1 and 0 to *m*<=-<=1 separately. In *i*-th day, Drazil invites -th boy and -th girl to have dinner together (as Drazil is programmer, *i* starts from 0). If one of those two people is happy, the other one will also become happy. Otherwise, those two people remain in their states. Once a person becomes happy (or if he/she was happy originally), he stays happy forever.
Drazil wants to know whether he can use this plan to make all his friends become happy at some moment.
Input Specification:
The first line contains two integer *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
The second line contains integer *b* (0<=≤<=*b*<=≤<=*n*), denoting the number of happy boys among friends of Drazil, and then follow *b* distinct integers *x*1,<=*x*2,<=...,<=*x**b* (0<=≤<=*x**i*<=<<=*n*), denoting the list of indices of happy boys.
The third line conatins integer *g* (0<=≤<=*g*<=≤<=*m*), denoting the number of happy girls among friends of Drazil, and then follow *g* distinct integers *y*1,<=*y*2,<=... ,<=*y**g* (0<=≤<=*y**j*<=<<=*m*), denoting the list of indices of happy girls.
It is guaranteed that there is at least one person that is unhappy among his friends.
Output Specification:
If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No".
Demo Input:
['2 3\n0\n1 0\n', '2 4\n1 0\n1 2\n', '2 3\n1 0\n1 1\n']
Demo Output:
['Yes\n', 'No\n', 'Yes\n']
Note:
By <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/169ade208e6eb4f9263c57aaff716529d59c3288.png" style="max-width: 100.0%;max-height: 100.0%;"/> we define the remainder of integer division of *i* by *k*.
In first sample case:
- On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day. - On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes at this day. - On the 2-nd day, Drazil invites 0-th boy and 2-nd girl. Because 0-th boy is already happy he makes 2-nd girl become happy at this day. - On the 3-rd day, Drazil invites 1-st boy and 0-th girl. 0-th girl is happy, so she makes 1-st boy happy. - On the 4-th day, Drazil invites 0-th boy and 1-st girl. 0-th boy is happy, so he makes the 1-st girl happy. So, all friends become happy at this moment. | ```python
n, m = [int(p) for p in input().split()]
arrb = [int(p) for p in input().split()]
arrg = [int(p) for p in input().split()]
arrb.pop(0)
arrg.pop(0)
cb = [False]*n
cg = [False]*m
for i in arrb:
cb[i] = True
for i in arrg:
cg[i] = True
i = 0
while i < 10**5:
b = cb[i%n]
g = cg[i%m]
if b and g:
i += 1
continue
elif b and not g:
cg[i%m] = True
elif not b and g:
cb[i%n] = True
else:
i += 1
continue
i += 1
if all(cb) and all(cg):
print('Yes')
else:
print('No')
``` | 3 | |
4 | C | Registration System | PROGRAMMING | 1,300 | [
"data structures",
"hashing",
"implementation"
] | C. Registration system | 5 | 64 | A new e-mail service "Berlandesk" is going to be opened in Berland in the near future. The site administration wants to launch their project as soon as possible, that's why they ask you to help. You're suggested to implement the prototype of site registration system. The system should work on the following principle.
Each time a new user wants to register, he sends to the system a request with his name. If such a name does not exist in the system database, it is inserted into the database, and the user gets the response OK, confirming the successful registration. If the name already exists in the system database, the system makes up a new user name, sends it to the user as a prompt and also inserts the prompt into the database. The new name is formed by the following rule. Numbers, starting with 1, are appended one after another to name (name1, name2, ...), among these numbers the least *i* is found so that name*i* does not yet exist in the database. | The first line contains number *n* (1<=≤<=*n*<=≤<=105). The following *n* lines contain the requests to the system. Each request is a non-empty line, and consists of not more than 32 characters, which are all lowercase Latin letters. | Print *n* lines, which are system responses to the requests: OK in case of successful registration, or a prompt with a new name, if the requested name is already taken. | [
"4\nabacaba\nacaba\nabacaba\nacab\n",
"6\nfirst\nfirst\nsecond\nsecond\nthird\nthird\n"
] | [
"OK\nOK\nabacaba1\nOK\n",
"OK\nfirst1\nOK\nsecond1\nOK\nthird1\n"
] | none | 0 | [
{
"input": "4\nabacaba\nacaba\nabacaba\nacab",
"output": "OK\nOK\nabacaba1\nOK"
},
{
"input": "6\nfirst\nfirst\nsecond\nsecond\nthird\nthird",
"output": "OK\nfirst1\nOK\nsecond1\nOK\nthird1"
},
{
"input": "1\nn",
"output": "OK"
},
{
"input": "2\nu\nu",
"output": "OK\nu1"
... | 1,687,379,603 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 6 | 5,000 | 1,126,400 | a = int(input())
x1 = input()
dataBase = ["OK"]
repeats = [x1]
for i in range(a-1):
x = input()
c=0
int(c)
for j in range(len(repeats)):
if x == repeats[j]:
c+=1
c = str(c)
repeats.append(x)
x+=c
if c == "0" :
dataBase.append("OK")
else:
dataBase.append(x)
for i in range(len(dataBase)):
print(dataBase[i])
| Title: Registration System
Time Limit: 5 seconds
Memory Limit: 64 megabytes
Problem Description:
A new e-mail service "Berlandesk" is going to be opened in Berland in the near future. The site administration wants to launch their project as soon as possible, that's why they ask you to help. You're suggested to implement the prototype of site registration system. The system should work on the following principle.
Each time a new user wants to register, he sends to the system a request with his name. If such a name does not exist in the system database, it is inserted into the database, and the user gets the response OK, confirming the successful registration. If the name already exists in the system database, the system makes up a new user name, sends it to the user as a prompt and also inserts the prompt into the database. The new name is formed by the following rule. Numbers, starting with 1, are appended one after another to name (name1, name2, ...), among these numbers the least *i* is found so that name*i* does not yet exist in the database.
Input Specification:
The first line contains number *n* (1<=≤<=*n*<=≤<=105). The following *n* lines contain the requests to the system. Each request is a non-empty line, and consists of not more than 32 characters, which are all lowercase Latin letters.
Output Specification:
Print *n* lines, which are system responses to the requests: OK in case of successful registration, or a prompt with a new name, if the requested name is already taken.
Demo Input:
['4\nabacaba\nacaba\nabacaba\nacab\n', '6\nfirst\nfirst\nsecond\nsecond\nthird\nthird\n']
Demo Output:
['OK\nOK\nabacaba1\nOK\n', 'OK\nfirst1\nOK\nsecond1\nOK\nthird1\n']
Note:
none | ```python
a = int(input())
x1 = input()
dataBase = ["OK"]
repeats = [x1]
for i in range(a-1):
x = input()
c=0
int(c)
for j in range(len(repeats)):
if x == repeats[j]:
c+=1
c = str(c)
repeats.append(x)
x+=c
if c == "0" :
dataBase.append("OK")
else:
dataBase.append(x)
for i in range(len(dataBase)):
print(dataBase[i])
``` | 0 |
817 | C | Really Big Numbers | PROGRAMMING | 1,600 | [
"binary search",
"brute force",
"dp",
"math"
] | null | null | Ivan likes to learn different things about numbers, but he is especially interested in really big numbers. Ivan thinks that a positive integer number *x* is really big if the difference between *x* and the sum of its digits (in decimal representation) is not less than *s*. To prove that these numbers may have different special properties, he wants to know how rare (or not rare) they are — in fact, he needs to calculate the quantity of really big numbers that are not greater than *n*.
Ivan tried to do the calculations himself, but soon realized that it's too difficult for him. So he asked you to help him in calculations. | The first (and the only) line contains two integers *n* and *s* (1<=≤<=*n*,<=*s*<=≤<=1018). | Print one integer — the quantity of really big numbers that are not greater than *n*. | [
"12 1\n",
"25 20\n",
"10 9\n"
] | [
"3\n",
"0\n",
"1\n"
] | In the first example numbers 10, 11 and 12 are really big.
In the second example there are no really big numbers that are not greater than 25 (in fact, the first really big number is 30: 30 - 3 ≥ 20).
In the third example 10 is the only really big number (10 - 1 ≥ 9). | 0 | [
{
"input": "12 1",
"output": "3"
},
{
"input": "25 20",
"output": "0"
},
{
"input": "10 9",
"output": "1"
},
{
"input": "300 1000",
"output": "0"
},
{
"input": "500 1000",
"output": "0"
},
{
"input": "1000 2000",
"output": "0"
},
{
"input":... | 1,497,543,367 | 4,267 | Python 3 | OK | TESTS | 68 | 62 | 0 | def f(x):
ans = x
while x > 0:
ans -= x % 10
x //= 10
return ans
n, s = map(int, input().split())
if f(n) < s:
print(0)
exit(0)
l, r = 1, n
while l < r:
m = (l + r) // 2
if f(m) >= s:
r = (m // 10) * 10
else:
l = m + 1
print(n - l + 1)
| Title: Really Big Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ivan likes to learn different things about numbers, but he is especially interested in really big numbers. Ivan thinks that a positive integer number *x* is really big if the difference between *x* and the sum of its digits (in decimal representation) is not less than *s*. To prove that these numbers may have different special properties, he wants to know how rare (or not rare) they are — in fact, he needs to calculate the quantity of really big numbers that are not greater than *n*.
Ivan tried to do the calculations himself, but soon realized that it's too difficult for him. So he asked you to help him in calculations.
Input Specification:
The first (and the only) line contains two integers *n* and *s* (1<=≤<=*n*,<=*s*<=≤<=1018).
Output Specification:
Print one integer — the quantity of really big numbers that are not greater than *n*.
Demo Input:
['12 1\n', '25 20\n', '10 9\n']
Demo Output:
['3\n', '0\n', '1\n']
Note:
In the first example numbers 10, 11 and 12 are really big.
In the second example there are no really big numbers that are not greater than 25 (in fact, the first really big number is 30: 30 - 3 ≥ 20).
In the third example 10 is the only really big number (10 - 1 ≥ 9). | ```python
def f(x):
ans = x
while x > 0:
ans -= x % 10
x //= 10
return ans
n, s = map(int, input().split())
if f(n) < s:
print(0)
exit(0)
l, r = 1, n
while l < r:
m = (l + r) // 2
if f(m) >= s:
r = (m // 10) * 10
else:
l = m + 1
print(n - l + 1)
``` | 3 | |
807 | A | Is it rated? | PROGRAMMING | 900 | [
"implementation",
"sortings"
] | null | null | Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not. | The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants.
Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings. | If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe". | [
"6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n",
"4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n",
"5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n"
] | [
"rated\n",
"unrated\n",
"maybe\n"
] | In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not. | 500 | [
{
"input": "6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884",
"output": "rated"
},
{
"input": "4\n1500 1500\n1300 1300\n1200 1200\n1400 1400",
"output": "unrated"
},
{
"input": "5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699",
"output": "maybe"
},
{
... | 1,598,376,402 | 2,147,483,647 | Python 3 | OK | TESTS | 150 | 124 | 307,200 | n = int(input())
a=[[int(i) for i in input().split()] for i in range(n)]
ans = 'maybe'
for i in range(n):
if a[i][0] != a[i][1]:
ans = 'rated'
break
else:
for i in range(n-1):
if a[i][0] < a[i+1][0]:
ans = 'unrated'
break
print(ans)
| Title: Is it rated?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=1000) — the number of round participants.
Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=4126) — the rating of the *i*-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output Specification:
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Demo Input:
['6\n3060 3060\n2194 2194\n2876 2903\n2624 2624\n3007 2991\n2884 2884\n', '4\n1500 1500\n1300 1300\n1200 1200\n1400 1400\n', '5\n3123 3123\n2777 2777\n2246 2246\n2246 2246\n1699 1699\n']
Demo Output:
['rated\n', 'unrated\n', 'maybe\n']
Note:
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not. | ```python
n = int(input())
a=[[int(i) for i in input().split()] for i in range(n)]
ans = 'maybe'
for i in range(n):
if a[i][0] != a[i][1]:
ans = 'rated'
break
else:
for i in range(n-1):
if a[i][0] < a[i+1][0]:
ans = 'unrated'
break
print(ans)
``` | 3 | |
570 | A | Elections | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | The country of Byalechinsk is running elections involving *n* candidates. The country consists of *m* cities. We know how many people in each city voted for each candidate.
The electoral system in the country is pretty unusual. At the first stage of elections the votes are counted for each city: it is assumed that in each city won the candidate who got the highest number of votes in this city, and if several candidates got the maximum number of votes, then the winner is the one with a smaller index.
At the second stage of elections the winner is determined by the same principle over the cities: the winner of the elections is the candidate who won in the maximum number of cities, and among those who got the maximum number of cities the winner is the one with a smaller index.
Determine who will win the elections. | The first line of the input contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of candidates and of cities, respectively.
Each of the next *m* lines contains *n* non-negative integers, the *j*-th number in the *i*-th line *a**ij* (1<=≤<=*j*<=≤<=*n*, 1<=≤<=*i*<=≤<=*m*, 0<=≤<=*a**ij*<=≤<=109) denotes the number of votes for candidate *j* in city *i*.
It is guaranteed that the total number of people in all the cities does not exceed 109. | Print a single number — the index of the candidate who won the elections. The candidates are indexed starting from one. | [
"3 3\n1 2 3\n2 3 1\n1 2 1\n",
"3 4\n10 10 3\n5 1 6\n2 2 2\n1 5 7\n"
] | [
"2",
"1"
] | Note to the first sample test. At the first stage city 1 chosen candidate 3, city 2 chosen candidate 2, city 3 chosen candidate 2. The winner is candidate 2, he gained 2 votes.
Note to the second sample test. At the first stage in city 1 candidates 1 and 2 got the same maximum number of votes, but candidate 1 has a smaller index, so the city chose candidate 1. City 2 chosen candidate 3. City 3 chosen candidate 1, due to the fact that everyone has the same number of votes, and 1 has the smallest index. City 4 chosen the candidate 3. On the second stage the same number of cities chose candidates 1 and 3. The winner is candidate 1, the one with the smaller index. | 500 | [
{
"input": "3 3\n1 2 3\n2 3 1\n1 2 1",
"output": "2"
},
{
"input": "3 4\n10 10 3\n5 1 6\n2 2 2\n1 5 7",
"output": "1"
},
{
"input": "1 3\n5\n3\n2",
"output": "1"
},
{
"input": "3 1\n1 2 3",
"output": "3"
},
{
"input": "3 1\n100 100 100",
"output": "1"
},
{... | 1,472,980,234 | 2,147,483,647 | Python 3 | OK | TESTS | 62 | 62 | 102,400 | lineInput = list(map(int,str(input()).split()))
m = lineInput[0]
n = lineInput[1]
city = [0] * 110
for i in range(0,n):
cnt = list(map(int,str(input()).split()))
winner = 0
for j in range(0,m):
if(cnt[j] > cnt[winner]):
winner = j
city[winner] += 1
answer = int(0)
for i in range(0,m):
if(city[i] > city[answer]):
answer = i
answer += 1
print(answer) | Title: Elections
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The country of Byalechinsk is running elections involving *n* candidates. The country consists of *m* cities. We know how many people in each city voted for each candidate.
The electoral system in the country is pretty unusual. At the first stage of elections the votes are counted for each city: it is assumed that in each city won the candidate who got the highest number of votes in this city, and if several candidates got the maximum number of votes, then the winner is the one with a smaller index.
At the second stage of elections the winner is determined by the same principle over the cities: the winner of the elections is the candidate who won in the maximum number of cities, and among those who got the maximum number of cities the winner is the one with a smaller index.
Determine who will win the elections.
Input Specification:
The first line of the input contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of candidates and of cities, respectively.
Each of the next *m* lines contains *n* non-negative integers, the *j*-th number in the *i*-th line *a**ij* (1<=≤<=*j*<=≤<=*n*, 1<=≤<=*i*<=≤<=*m*, 0<=≤<=*a**ij*<=≤<=109) denotes the number of votes for candidate *j* in city *i*.
It is guaranteed that the total number of people in all the cities does not exceed 109.
Output Specification:
Print a single number — the index of the candidate who won the elections. The candidates are indexed starting from one.
Demo Input:
['3 3\n1 2 3\n2 3 1\n1 2 1\n', '3 4\n10 10 3\n5 1 6\n2 2 2\n1 5 7\n']
Demo Output:
['2', '1']
Note:
Note to the first sample test. At the first stage city 1 chosen candidate 3, city 2 chosen candidate 2, city 3 chosen candidate 2. The winner is candidate 2, he gained 2 votes.
Note to the second sample test. At the first stage in city 1 candidates 1 and 2 got the same maximum number of votes, but candidate 1 has a smaller index, so the city chose candidate 1. City 2 chosen candidate 3. City 3 chosen candidate 1, due to the fact that everyone has the same number of votes, and 1 has the smallest index. City 4 chosen the candidate 3. On the second stage the same number of cities chose candidates 1 and 3. The winner is candidate 1, the one with the smaller index. | ```python
lineInput = list(map(int,str(input()).split()))
m = lineInput[0]
n = lineInput[1]
city = [0] * 110
for i in range(0,n):
cnt = list(map(int,str(input()).split()))
winner = 0
for j in range(0,m):
if(cnt[j] > cnt[winner]):
winner = j
city[winner] += 1
answer = int(0)
for i in range(0,m):
if(city[i] > city[answer]):
answer = i
answer += 1
print(answer)
``` | 3 | |
3 | C | Tic-tac-toe | PROGRAMMING | 1,800 | [
"brute force",
"games",
"implementation"
] | C. Tic-tac-toe | 1 | 64 | Certainly, everyone is familiar with tic-tac-toe game. The rules are very simple indeed. Two players take turns marking the cells in a 3<=×<=3 grid (one player always draws crosses, the other — noughts). The player who succeeds first in placing three of his marks in a horizontal, vertical or diagonal line wins, and the game is finished. The player who draws crosses goes first. If the grid is filled, but neither Xs, nor 0s form the required line, a draw is announced.
You are given a 3<=×<=3 grid, each grid cell is empty, or occupied by a cross or a nought. You have to find the player (first or second), whose turn is next, or print one of the verdicts below:
- illegal — if the given board layout can't appear during a valid game; - the first player won — if in the given board layout the first player has just won; - the second player won — if in the given board layout the second player has just won; - draw — if the given board layout has just let to a draw. | The input consists of three lines, each of the lines contains characters ".", "X" or "0" (a period, a capital letter X, or a digit zero). | Print one of the six verdicts: first, second, illegal, the first player won, the second player won or draw. | [
"X0X\n.0.\n.X.\n"
] | [
"second\n"
] | none | 0 | [
{
"input": "X0X\n.0.\n.X.",
"output": "second"
},
{
"input": "0.X\nXX.\n000",
"output": "illegal"
},
{
"input": "XXX\n.0.\n000",
"output": "illegal"
},
{
"input": "XXX\n...\n000",
"output": "illegal"
},
{
"input": "X.X\nX..\n00.",
"output": "second"
},
{
... | 1,690,647,386 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 0 | 156 | 2,867,200 | cx = 0
c0 = 0
wx = 0
w0 = 0
b = [' '] * 9
def ch(c, i):
return (b[i] == c and b[i + 3] == c and b[i + 6] == c) or (b[i * 3 + 0] == c and b[i * 3 + 1] == c and b[i * 3 + 2] == c) or (b[0] == c and b[4] == c and b[8] == c) or (b[2] == c and b[4] == c and b[6] == c)
for i in range(9):
b[i] = input()
if b[i] == 'X':
cx += 1
elif b[i] == '0':
c0 += 1
for i in range(3):
wx += ch('X', i)
w0 += ch('0', i)
if (w0 > 0 and cx != c0) or (cx != c0 and cx != c0 + 1) or (wx > 0 and cx != c0 + 1) or (w0 > 0 and wx > 0):
print("illegal")
elif wx:
print("the first player won")
elif w0:
print("the second player won")
elif cx + c0 == 9:
print("draw")
elif cx == c0:
print("first")
else:
print("second")
| Title: Tic-tac-toe
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Certainly, everyone is familiar with tic-tac-toe game. The rules are very simple indeed. Two players take turns marking the cells in a 3<=×<=3 grid (one player always draws crosses, the other — noughts). The player who succeeds first in placing three of his marks in a horizontal, vertical or diagonal line wins, and the game is finished. The player who draws crosses goes first. If the grid is filled, but neither Xs, nor 0s form the required line, a draw is announced.
You are given a 3<=×<=3 grid, each grid cell is empty, or occupied by a cross or a nought. You have to find the player (first or second), whose turn is next, or print one of the verdicts below:
- illegal — if the given board layout can't appear during a valid game; - the first player won — if in the given board layout the first player has just won; - the second player won — if in the given board layout the second player has just won; - draw — if the given board layout has just let to a draw.
Input Specification:
The input consists of three lines, each of the lines contains characters ".", "X" or "0" (a period, a capital letter X, or a digit zero).
Output Specification:
Print one of the six verdicts: first, second, illegal, the first player won, the second player won or draw.
Demo Input:
['X0X\n.0.\n.X.\n']
Demo Output:
['second\n']
Note:
none | ```python
cx = 0
c0 = 0
wx = 0
w0 = 0
b = [' '] * 9
def ch(c, i):
return (b[i] == c and b[i + 3] == c and b[i + 6] == c) or (b[i * 3 + 0] == c and b[i * 3 + 1] == c and b[i * 3 + 2] == c) or (b[0] == c and b[4] == c and b[8] == c) or (b[2] == c and b[4] == c and b[6] == c)
for i in range(9):
b[i] = input()
if b[i] == 'X':
cx += 1
elif b[i] == '0':
c0 += 1
for i in range(3):
wx += ch('X', i)
w0 += ch('0', i)
if (w0 > 0 and cx != c0) or (cx != c0 and cx != c0 + 1) or (wx > 0 and cx != c0 + 1) or (w0 > 0 and wx > 0):
print("illegal")
elif wx:
print("the first player won")
elif w0:
print("the second player won")
elif cx + c0 == 9:
print("draw")
elif cx == c0:
print("first")
else:
print("second")
``` | -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": "... | 1,695,924,308 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | def notSameColor(s):
l = list(s)
cnt =0
for i in range(len(l)-1):
if l[i+1]==l[i]:
cnt +=1
return cnt
print(notSameColor(input())) | 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
def notSameColor(s):
l = list(s)
cnt =0
for i in range(len(l)-1):
if l[i+1]==l[i]:
cnt +=1
return cnt
print(notSameColor(input()))
``` | 0 | |
80 | A | Panoramix's Prediction | PROGRAMMING | 800 | [
"brute force"
] | A. Panoramix's Prediction | 2 | 256 | A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix? | The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4. | Print YES, if *m* is the next prime number after *n*, or NO otherwise. | [
"3 5\n",
"7 11\n",
"7 9\n"
] | [
"YES",
"YES",
"NO"
] | none | 500 | [
{
"input": "3 5",
"output": "YES"
},
{
"input": "7 11",
"output": "YES"
},
{
"input": "7 9",
"output": "NO"
},
{
"input": "2 3",
"output": "YES"
},
{
"input": "2 4",
"output": "NO"
},
{
"input": "3 4",
"output": "NO"
},
{
"input": "3 5",
... | 1,687,972,524 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 13 | 122 | 0 | a=input('')
n=int(a.split(' ')[0])
m=int(a.split(' ')[1])
c=n+1
for i in range(2,c):
if c%i!=0:
continue
else:
c+=1
if c==m:
print('YES')
else:
print('NO')
| Title: Panoramix's Prediction
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix?
Input Specification:
The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4.
Output Specification:
Print YES, if *m* is the next prime number after *n*, or NO otherwise.
Demo Input:
['3 5\n', '7 11\n', '7 9\n']
Demo Output:
['YES', 'YES', 'NO']
Note:
none | ```python
a=input('')
n=int(a.split(' ')[0])
m=int(a.split(' ')[1])
c=n+1
for i in range(2,c):
if c%i!=0:
continue
else:
c+=1
if c==m:
print('YES')
else:
print('NO')
``` | 0 |
735 | B | Urbanization | PROGRAMMING | 1,100 | [
"greedy",
"number theory",
"sortings"
] | null | null | Local authorities have heard a lot about combinatorial abilities of Ostap Bender so they decided to ask his help in the question of urbanization. There are *n* people who plan to move to the cities. The wealth of the *i* of them is equal to *a**i*. Authorities plan to build two cities, first for *n*1 people and second for *n*2 people. Of course, each of *n* candidates can settle in only one of the cities. Thus, first some subset of candidates of size *n*1 settle in the first city and then some subset of size *n*2 is chosen among the remaining candidates and the move to the second city. All other candidates receive an official refuse and go back home.
To make the statistic of local region look better in the eyes of their bosses, local authorities decided to pick subsets of candidates in such a way that the sum of arithmetic mean of wealth of people in each of the cities is as large as possible. Arithmetic mean of wealth in one city is the sum of wealth *a**i* among all its residents divided by the number of them (*n*1 or *n*2 depending on the city). The division should be done in real numbers without any rounding.
Please, help authorities find the optimal way to pick residents for two cities. | The first line of the input contains three integers *n*, *n*1 and *n*2 (1<=≤<=*n*,<=*n*1,<=*n*2<=≤<=100<=000, *n*1<=+<=*n*2<=≤<=*n*) — the number of candidates who want to move to the cities, the planned number of residents of the first city and the planned number of residents of the second city.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000), the *i*-th of them is equal to the wealth of the *i*-th candidate. | Print one real value — the maximum possible sum of arithmetic means of wealth of cities' residents. You answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if . | [
"2 1 1\n1 5\n",
"4 2 1\n1 4 2 3\n"
] | [
"6.00000000\n",
"6.50000000\n"
] | In the first sample, one of the optimal solutions is to move candidate 1 to the first city and candidate 2 to the second.
In the second sample, the optimal solution is to pick candidates 3 and 4 for the first city, and candidate 2 for the second one. Thus we obtain (*a*<sub class="lower-index">3</sub> + *a*<sub class="lower-index">4</sub>) / 2 + *a*<sub class="lower-index">2</sub> = (3 + 2) / 2 + 4 = 6.5 | 1,000 | [
{
"input": "2 1 1\n1 5",
"output": "6.00000000"
},
{
"input": "4 2 1\n1 4 2 3",
"output": "6.50000000"
},
{
"input": "3 1 2\n1 2 3",
"output": "4.50000000"
},
{
"input": "10 4 6\n3 5 7 9 12 25 67 69 83 96",
"output": "88.91666667"
},
{
"input": "19 7 12\n1 2 4 8 1... | 1,614,342,549 | 2,147,483,647 | PyPy 3 | OK | TESTS | 37 | 171 | 10,649,600 | n,n1,n2=map(int,input().split())
n1,n2=min(n1,n2),max(n1,n2)
l=sorted(list(map(int,input().split())))
l.reverse()
print(sum(l[:n1])/n1+sum(l[n1:n1+n2])/n2)
| Title: Urbanization
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Local authorities have heard a lot about combinatorial abilities of Ostap Bender so they decided to ask his help in the question of urbanization. There are *n* people who plan to move to the cities. The wealth of the *i* of them is equal to *a**i*. Authorities plan to build two cities, first for *n*1 people and second for *n*2 people. Of course, each of *n* candidates can settle in only one of the cities. Thus, first some subset of candidates of size *n*1 settle in the first city and then some subset of size *n*2 is chosen among the remaining candidates and the move to the second city. All other candidates receive an official refuse and go back home.
To make the statistic of local region look better in the eyes of their bosses, local authorities decided to pick subsets of candidates in such a way that the sum of arithmetic mean of wealth of people in each of the cities is as large as possible. Arithmetic mean of wealth in one city is the sum of wealth *a**i* among all its residents divided by the number of them (*n*1 or *n*2 depending on the city). The division should be done in real numbers without any rounding.
Please, help authorities find the optimal way to pick residents for two cities.
Input Specification:
The first line of the input contains three integers *n*, *n*1 and *n*2 (1<=≤<=*n*,<=*n*1,<=*n*2<=≤<=100<=000, *n*1<=+<=*n*2<=≤<=*n*) — the number of candidates who want to move to the cities, the planned number of residents of the first city and the planned number of residents of the second city.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000), the *i*-th of them is equal to the wealth of the *i*-th candidate.
Output Specification:
Print one real value — the maximum possible sum of arithmetic means of wealth of cities' residents. You answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
Demo Input:
['2 1 1\n1 5\n', '4 2 1\n1 4 2 3\n']
Demo Output:
['6.00000000\n', '6.50000000\n']
Note:
In the first sample, one of the optimal solutions is to move candidate 1 to the first city and candidate 2 to the second.
In the second sample, the optimal solution is to pick candidates 3 and 4 for the first city, and candidate 2 for the second one. Thus we obtain (*a*<sub class="lower-index">3</sub> + *a*<sub class="lower-index">4</sub>) / 2 + *a*<sub class="lower-index">2</sub> = (3 + 2) / 2 + 4 = 6.5 | ```python
n,n1,n2=map(int,input().split())
n1,n2=min(n1,n2),max(n1,n2)
l=sorted(list(map(int,input().split())))
l.reverse()
print(sum(l[:n1])/n1+sum(l[n1:n1+n2])/n2)
``` | 3 | |
779 | B | Weird Rounding | PROGRAMMING | 1,100 | [
"brute force",
"greedy"
] | null | null | Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10*k*.
In the given number of *n* Polycarp wants to remove the least number of digits to get a number that is divisible by 10*k*. For example, if *k*<==<=3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103<==<=1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number *n*, so that the result is divisible by 10*k*. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists. | The only line of the input contains two integer numbers *n* and *k* (0<=≤<=*n*<=≤<=2<=000<=000<=000, 1<=≤<=*k*<=≤<=9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros. | Print *w* — the required minimal number of digits to erase. After removing the appropriate *w* digits from the number *n*, the result should have a value that is divisible by 10*k*. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0). | [
"30020 3\n",
"100 9\n",
"10203049 2\n"
] | [
"1\n",
"2\n",
"3\n"
] | In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number. | 1,000 | [
{
"input": "30020 3",
"output": "1"
},
{
"input": "100 9",
"output": "2"
},
{
"input": "10203049 2",
"output": "3"
},
{
"input": "0 1",
"output": "0"
},
{
"input": "0 9",
"output": "0"
},
{
"input": "100 2",
"output": "0"
},
{
"input": "102... | 1,488,102,806 | 6,506 | Python 3 | WRONG_ANSWER | PRETESTS | 8 | 77 | 5,120,000 | import collections
s, k = input().split()
arr = list(map(int,list(s[1:])))
n = 0;
def cor(n,arr):
for i in range(0,n):
if arr[-(i+1)]!=0:
return False
return True
if len(s)<int(k):
print(len(s)-1)
else:
for i in range(len(arr)-1,-1,-1):
if arr[i]!=0:
arr.pop(i)
n+=1
if len(arr)>=int(k) and cor(int(k),arr):
break
if len(arr)<int(k):
n = len(s)-1
break
print(n)
| Title: Weird Rounding
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10*k*.
In the given number of *n* Polycarp wants to remove the least number of digits to get a number that is divisible by 10*k*. For example, if *k*<==<=3, in the number 30020 it is enough to delete a single digit (2). In this case, the result is 3000 that is divisible by 103<==<=1000.
Write a program that prints the minimum number of digits to be deleted from the given integer number *n*, so that the result is divisible by 10*k*. The result should not start with the unnecessary leading zero (i.e., zero can start only the number 0, which is required to be written as exactly one digit).
It is guaranteed that the answer exists.
Input Specification:
The only line of the input contains two integer numbers *n* and *k* (0<=≤<=*n*<=≤<=2<=000<=000<=000, 1<=≤<=*k*<=≤<=9).
It is guaranteed that the answer exists. All numbers in the input are written in traditional notation of integers, that is, without any extra leading zeros.
Output Specification:
Print *w* — the required minimal number of digits to erase. After removing the appropriate *w* digits from the number *n*, the result should have a value that is divisible by 10*k*. The result can start with digit 0 in the single case (the result is zero and written by exactly the only digit 0).
Demo Input:
['30020 3\n', '100 9\n', '10203049 2\n']
Demo Output:
['1\n', '2\n', '3\n']
Note:
In the example 2 you can remove two digits: 1 and any 0. The result is number 0 which is divisible by any number. | ```python
import collections
s, k = input().split()
arr = list(map(int,list(s[1:])))
n = 0;
def cor(n,arr):
for i in range(0,n):
if arr[-(i+1)]!=0:
return False
return True
if len(s)<int(k):
print(len(s)-1)
else:
for i in range(len(arr)-1,-1,-1):
if arr[i]!=0:
arr.pop(i)
n+=1
if len(arr)>=int(k) and cor(int(k),arr):
break
if len(arr)<int(k):
n = len(s)-1
break
print(n)
``` | 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"... | 1,694,605,737 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 31 | 0 | s=list(map(int, input().split(" ")))
n=s[0]
m=s[1]
a=s[2]
print((int(n/a)+1)*(int(m/a)+1)) | 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
s=list(map(int, input().split(" ")))
n=s[0]
m=s[1]
a=s[2]
print((int(n/a)+1)*(int(m/a)+1))
``` | 0 |
525 | B | Pasha and String | PROGRAMMING | 1,400 | [
"constructive algorithms",
"greedy",
"math",
"strings"
] | null | null | Pasha got a very beautiful string *s* for his birthday, the string consists of lowercase Latin letters. The letters in the string are numbered from 1 to |*s*| from left to right, where |*s*| is the length of the given string.
Pasha didn't like his present very much so he decided to change it. After his birthday Pasha spent *m* days performing the following transformations on his string — each day he chose integer *a**i* and reversed a piece of string (a segment) from position *a**i* to position |*s*|<=-<=*a**i*<=+<=1. It is guaranteed that 2·*a**i*<=≤<=|*s*|.
You face the following task: determine what Pasha's string will look like after *m* days. | The first line of the input contains Pasha's string *s* of length from 2 to 2·105 characters, consisting of lowercase Latin letters.
The second line contains a single integer *m* (1<=≤<=*m*<=≤<=105) — the number of days when Pasha changed his string.
The third line contains *m* space-separated elements *a**i* (1<=≤<=*a**i*; 2·*a**i*<=≤<=|*s*|) — the position from which Pasha started transforming the string on the *i*-th day. | In the first line of the output print what Pasha's string *s* will look like after *m* days. | [
"abcdef\n1\n2\n",
"vwxyz\n2\n2 2\n",
"abcdef\n3\n1 2 3\n"
] | [
"aedcbf\n",
"vwxyz\n",
"fbdcea\n"
] | none | 750 | [
{
"input": "abcdef\n1\n2",
"output": "aedcbf"
},
{
"input": "vwxyz\n2\n2 2",
"output": "vwxyz"
},
{
"input": "abcdef\n3\n1 2 3",
"output": "fbdcea"
},
{
"input": "jc\n5\n1 1 1 1 1",
"output": "cj"
},
{
"input": "wljqgdlxyc\n13\n3 4 3 3 5 4 4 2 4 4 5 3 3",
"out... | 1,622,030,531 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 26 | 2,000 | 15,052,800 | def reverse(lis, i, j):
n = (j - i + 1)
while(i <= j):
lis[i], lis[j] = lis[j], lis[i]
j -= 1
i += 1
return lis
s = input()
lis = list(s)
nn = len(s)
m = int(input())
arr = list(map(int, input().split()))
dic = {}
for i in arr:
if i in dic:
dic[i] += 1
else:
dic[i] = 1
for i in dic:
if dic[i] & 1 != 0:
#print(i - 1, nn - i)
reverse(lis, i - 1, nn - i)
for i in lis:
print(i, end = '')
| Title: Pasha and String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha got a very beautiful string *s* for his birthday, the string consists of lowercase Latin letters. The letters in the string are numbered from 1 to |*s*| from left to right, where |*s*| is the length of the given string.
Pasha didn't like his present very much so he decided to change it. After his birthday Pasha spent *m* days performing the following transformations on his string — each day he chose integer *a**i* and reversed a piece of string (a segment) from position *a**i* to position |*s*|<=-<=*a**i*<=+<=1. It is guaranteed that 2·*a**i*<=≤<=|*s*|.
You face the following task: determine what Pasha's string will look like after *m* days.
Input Specification:
The first line of the input contains Pasha's string *s* of length from 2 to 2·105 characters, consisting of lowercase Latin letters.
The second line contains a single integer *m* (1<=≤<=*m*<=≤<=105) — the number of days when Pasha changed his string.
The third line contains *m* space-separated elements *a**i* (1<=≤<=*a**i*; 2·*a**i*<=≤<=|*s*|) — the position from which Pasha started transforming the string on the *i*-th day.
Output Specification:
In the first line of the output print what Pasha's string *s* will look like after *m* days.
Demo Input:
['abcdef\n1\n2\n', 'vwxyz\n2\n2 2\n', 'abcdef\n3\n1 2 3\n']
Demo Output:
['aedcbf\n', 'vwxyz\n', 'fbdcea\n']
Note:
none | ```python
def reverse(lis, i, j):
n = (j - i + 1)
while(i <= j):
lis[i], lis[j] = lis[j], lis[i]
j -= 1
i += 1
return lis
s = input()
lis = list(s)
nn = len(s)
m = int(input())
arr = list(map(int, input().split()))
dic = {}
for i in arr:
if i in dic:
dic[i] += 1
else:
dic[i] = 1
for i in dic:
if dic[i] & 1 != 0:
#print(i - 1, nn - i)
reverse(lis, i - 1, nn - i)
for i in lis:
print(i, end = '')
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Kyoya Ootori has a bag with *n* colored balls that are colored with *k* different colors. The colors are labeled from 1 to *k*. Balls of the same color are indistinguishable. He draws balls from the bag one by one until the bag is empty. He noticed that he drew the last ball of color *i* before drawing the last ball of color *i*<=+<=1 for all *i* from 1 to *k*<=-<=1. Now he wonders how many different ways this can happen. | The first line of input will have one integer *k* (1<=≤<=*k*<=≤<=1000) the number of colors.
Then, *k* lines will follow. The *i*-th line will contain *c**i*, the number of balls of the *i*-th color (1<=≤<=*c**i*<=≤<=1000).
The total number of balls doesn't exceed 1000. | A single integer, the number of ways that Kyoya can draw the balls from the bag as described in the statement, modulo 1<=000<=000<=007. | [
"3\n2\n2\n1\n",
"4\n1\n2\n3\n4\n"
] | [
"3\n",
"1680\n"
] | In the first sample, we have 2 balls of color 1, 2 balls of color 2, and 1 ball of color 3. The three ways for Kyoya are: | 0 | [
{
"input": "3\n2\n2\n1",
"output": "3"
},
{
"input": "4\n1\n2\n3\n4",
"output": "1680"
},
{
"input": "10\n100\n100\n100\n100\n100\n100\n100\n100\n100\n100",
"output": "12520708"
},
{
"input": "5\n10\n10\n10\n10\n10",
"output": "425711769"
},
{
"input": "11\n291\n3... | 1,533,046,157 | 2,147,483,647 | PyPy 3 | OK | TESTS | 27 | 171 | 0 | import math
MOD = 1000000007
MAXN = 1005
def fac(n, k):
res = 1
for i in range(n + 1 - k, n + 1):
res *= i
return res
k = int(input())
sum = 0
ans = 1
for i in range(0, k):
num = int(input())
sum += num
ans = ans * fac(sum - 1, num - 1) // fac(num - 1, num - 1)
print (ans % MOD) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kyoya Ootori has a bag with *n* colored balls that are colored with *k* different colors. The colors are labeled from 1 to *k*. Balls of the same color are indistinguishable. He draws balls from the bag one by one until the bag is empty. He noticed that he drew the last ball of color *i* before drawing the last ball of color *i*<=+<=1 for all *i* from 1 to *k*<=-<=1. Now he wonders how many different ways this can happen.
Input Specification:
The first line of input will have one integer *k* (1<=≤<=*k*<=≤<=1000) the number of colors.
Then, *k* lines will follow. The *i*-th line will contain *c**i*, the number of balls of the *i*-th color (1<=≤<=*c**i*<=≤<=1000).
The total number of balls doesn't exceed 1000.
Output Specification:
A single integer, the number of ways that Kyoya can draw the balls from the bag as described in the statement, modulo 1<=000<=000<=007.
Demo Input:
['3\n2\n2\n1\n', '4\n1\n2\n3\n4\n']
Demo Output:
['3\n', '1680\n']
Note:
In the first sample, we have 2 balls of color 1, 2 balls of color 2, and 1 ball of color 3. The three ways for Kyoya are: | ```python
import math
MOD = 1000000007
MAXN = 1005
def fac(n, k):
res = 1
for i in range(n + 1 - k, n + 1):
res *= i
return res
k = int(input())
sum = 0
ans = 1
for i in range(0, k):
num = int(input())
sum += num
ans = ans * fac(sum - 1, num - 1) // fac(num - 1, num - 1)
print (ans % MOD)
``` | 3 | |
145 | A | Lucky Conversion | PROGRAMMING | 1,200 | [
"greedy",
"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 has two strings *a* and *b* of the same length *n*. The strings consist only of lucky digits. Petya can perform operations of two types:
- replace any one digit from string *a* by its opposite (i.e., replace 4 by 7 and 7 by 4); - swap any pair of digits in string *a*.
Petya is interested in the minimum number of operations that are needed to make string *a* equal to string *b*. Help him with the task. | The first and the second line contains strings *a* and *b*, correspondingly. Strings *a* and *b* have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105. | Print on the single line the single number — the minimum number of operations needed to convert string *a* into string *b*. | [
"47\n74\n",
"774\n744\n",
"777\n444\n"
] | [
"1\n",
"1\n",
"3\n"
] | In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites. | 500 | [
{
"input": "47\n74",
"output": "1"
},
{
"input": "774\n744",
"output": "1"
},
{
"input": "777\n444",
"output": "3"
},
{
"input": "74747474\n77777777",
"output": "4"
},
{
"input": "444444444444\n777777777777",
"output": "12"
},
{
"input": "4744744447774... | 1,617,813,459 | 2,147,483,647 | Python 3 | OK | TESTS | 51 | 248 | 1,945,600 | a = list(input())
b = list(input())
a_copy = a[:]
a_copy.sort()
b_copy = b[:]
b_copy.sort()
replace_counter = 0
swap_counter = 0
for i in range(len(a_copy)):
if(a_copy[i] != b_copy[i]):
replace_counter += 1
for i in range(len(a)):
if (a[i] != b[i]):
swap_counter += 1
swap_counter = (swap_counter - replace_counter) // 2
print(replace_counter + swap_counter) | Title: Lucky Conversion
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 has two strings *a* and *b* of the same length *n*. The strings consist only of lucky digits. Petya can perform operations of two types:
- replace any one digit from string *a* by its opposite (i.e., replace 4 by 7 and 7 by 4); - swap any pair of digits in string *a*.
Petya is interested in the minimum number of operations that are needed to make string *a* equal to string *b*. Help him with the task.
Input Specification:
The first and the second line contains strings *a* and *b*, correspondingly. Strings *a* and *b* have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 105.
Output Specification:
Print on the single line the single number — the minimum number of operations needed to convert string *a* into string *b*.
Demo Input:
['47\n74\n', '774\n744\n', '777\n444\n']
Demo Output:
['1\n', '1\n', '3\n']
Note:
In the first sample it is enough simply to swap the first and the second digit.
In the second sample we should replace the second digit with its opposite.
In the third number we should replace all three digits with their opposites. | ```python
a = list(input())
b = list(input())
a_copy = a[:]
a_copy.sort()
b_copy = b[:]
b_copy.sort()
replace_counter = 0
swap_counter = 0
for i in range(len(a_copy)):
if(a_copy[i] != b_copy[i]):
replace_counter += 1
for i in range(len(a)):
if (a[i] != b[i]):
swap_counter += 1
swap_counter = (swap_counter - replace_counter) // 2
print(replace_counter + swap_counter)
``` | 3 | |
855 | A | Tom Riddle's Diary | PROGRAMMING | 800 | [
"brute force",
"implementation",
"strings"
] | null | null | Harry Potter is on a mission to destroy You-Know-Who's Horcruxes. The first Horcrux that he encountered in the Chamber of Secrets is Tom Riddle's diary. The diary was with Ginny and it forced her to open the Chamber of Secrets. Harry wants to know the different people who had ever possessed the diary to make sure they are not under its influence.
He has names of *n* people who possessed the diary in order. You need to tell, for each person, if he/she possessed the diary at some point before or not.
Formally, for a name *s**i* in the *i*-th line, output "YES" (without quotes) if there exists an index *j* such that *s**i*<==<=*s**j* and *j*<=<<=*i*, otherwise, output "NO" (without quotes). | First line of input contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of names in the list.
Next *n* lines each contain a string *s**i*, consisting of lowercase English letters. The length of each string is between 1 and 100. | Output *n* lines each containing either "YES" or "NO" (without quotes), depending on whether this string was already present in the stream or not.
You can print each letter in any case (upper or lower). | [
"6\ntom\nlucius\nginny\nharry\nginny\nharry\n",
"3\na\na\na\n"
] | [
"NO\nNO\nNO\nNO\nYES\nYES\n",
"NO\nYES\nYES\n"
] | In test case 1, for *i* = 5 there exists *j* = 3 such that *s*<sub class="lower-index">*i*</sub> = *s*<sub class="lower-index">*j*</sub> and *j* < *i*, which means that answer for *i* = 5 is "YES". | 500 | [
{
"input": "6\ntom\nlucius\nginny\nharry\nginny\nharry",
"output": "NO\nNO\nNO\nNO\nYES\nYES"
},
{
"input": "3\na\na\na",
"output": "NO\nYES\nYES"
},
{
"input": "1\nzn",
"output": "NO"
},
{
"input": "9\nliyzmbjwnzryjokufuxcqtzwworjeoxkbaqrujrhdidqdvwdfzilwszgnzglnnbogaclckfnb... | 1,646,909,372 | 2,147,483,647 | Python 3 | OK | TESTS | 55 | 46 | 0 | memory=dict()
for _ in [0]*int(input()):
s=input()
try:memory[s]+=1;print('YES')
except:memory[s]=0;print('NO') | Title: Tom Riddle's Diary
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Harry Potter is on a mission to destroy You-Know-Who's Horcruxes. The first Horcrux that he encountered in the Chamber of Secrets is Tom Riddle's diary. The diary was with Ginny and it forced her to open the Chamber of Secrets. Harry wants to know the different people who had ever possessed the diary to make sure they are not under its influence.
He has names of *n* people who possessed the diary in order. You need to tell, for each person, if he/she possessed the diary at some point before or not.
Formally, for a name *s**i* in the *i*-th line, output "YES" (without quotes) if there exists an index *j* such that *s**i*<==<=*s**j* and *j*<=<<=*i*, otherwise, output "NO" (without quotes).
Input Specification:
First line of input contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of names in the list.
Next *n* lines each contain a string *s**i*, consisting of lowercase English letters. The length of each string is between 1 and 100.
Output Specification:
Output *n* lines each containing either "YES" or "NO" (without quotes), depending on whether this string was already present in the stream or not.
You can print each letter in any case (upper or lower).
Demo Input:
['6\ntom\nlucius\nginny\nharry\nginny\nharry\n', '3\na\na\na\n']
Demo Output:
['NO\nNO\nNO\nNO\nYES\nYES\n', 'NO\nYES\nYES\n']
Note:
In test case 1, for *i* = 5 there exists *j* = 3 such that *s*<sub class="lower-index">*i*</sub> = *s*<sub class="lower-index">*j*</sub> and *j* < *i*, which means that answer for *i* = 5 is "YES". | ```python
memory=dict()
for _ in [0]*int(input()):
s=input()
try:memory[s]+=1;print('YES')
except:memory[s]=0;print('NO')
``` | 3 | |
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... | 1,661,021,671 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 15 | 0 | n, valera = map(int, input().split())
amn = 0
nums = []
for j in range (n):
lots, *price = map(int, input().split())
for i in range (price):
if valera > i:
amn += 1
nums += j + 1
break
print (amn)
for y in nums:
print (y + ' ', end='') | Title: Valera and Antique Items
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera is a collector. Once he wanted to expand his collection with exactly one antique item.
Valera knows *n* sellers of antiques, the *i*-th of them auctioned *k**i* items. Currently the auction price of the *j*-th object of the *i*-th seller is *s**ij*. Valera gets on well with each of the *n* sellers. He is perfectly sure that if he outbids the current price of one of the items in the auction (in other words, offers the seller the money that is strictly greater than the current price of the item at the auction), the seller of the object will immediately sign a contract with him.
Unfortunately, Valera has only *v* units of money. Help him to determine which of the *n* sellers he can make a deal with.
Input Specification:
The first line contains two space-separated integers *n*,<=*v* (1<=≤<=*n*<=≤<=50; 104<=≤<=*v*<=≤<=106) — the number of sellers and the units of money the Valera has.
Then *n* lines follow. The *i*-th line first contains integer *k**i* (1<=≤<=*k**i*<=≤<=50) the number of items of the *i*-th seller. Then go *k**i* space-separated integers *s**i*1,<=*s**i*2,<=...,<=*s**ik**i* (104<=≤<=*s**ij*<=≤<=106) — the current prices of the items of the *i*-th seller.
Output Specification:
In the first line, print integer *p* — the number of sellers with who Valera can make a deal.
In the second line print *p* space-separated integers *q*1,<=*q*2,<=...,<=*q**p* (1<=≤<=*q**i*<=≤<=*n*) — the numbers of the sellers with who Valera can make a deal. Print the numbers of the sellers in the increasing order.
Demo Input:
['3 50000\n1 40000\n2 20000 60000\n3 10000 70000 190000\n', '3 50000\n1 50000\n3 100000 120000 110000\n3 120000 110000 120000\n']
Demo Output:
['3\n1 2 3\n', '0\n\n']
Note:
In the first sample Valera can bargain with each of the sellers. He can outbid the following items: a 40000 item from the first seller, a 20000 item from the second seller, and a 10000 item from the third seller.
In the second sample Valera can not make a deal with any of the sellers, as the prices of all items in the auction too big for him. | ```python
n, valera = map(int, input().split())
amn = 0
nums = []
for j in range (n):
lots, *price = map(int, input().split())
for i in range (price):
if valera > i:
amn += 1
nums += j + 1
break
print (amn)
for y in nums:
print (y + ' ', end='')
``` | -1 | |
104 | A | Blackjack | PROGRAMMING | 800 | [
"implementation"
] | A. Blackjack | 2 | 256 | One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one!
Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture.
In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses.
The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*. | The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points. | Print the numbers of ways to get the second card in the required way if the first card is the queen of spades. | [
"12\n",
"20\n",
"10\n"
] | [
"4",
"15",
"0"
] | In the first sample only four two's of different suits can earn the required sum of points.
In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use.
In the third sample there is no card, that would add a zero to the current ten points. | 500 | [
{
"input": "12",
"output": "4"
},
{
"input": "20",
"output": "15"
},
{
"input": "10",
"output": "0"
},
{
"input": "11",
"output": "4"
},
{
"input": "15",
"output": "4"
},
{
"input": "18",
"output": "4"
},
{
"input": "25",
"output": "0"
... | 1,545,760,108 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 218 | 0 | c = int(input())
c = c-10
if (c >= 1 and c < 10) or (c is 11):
print(4)
elif c is 10:
print(15)
else:
print(0)
| Title: Blackjack
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One rainy gloomy evening when all modules hid in the nearby cafes to drink hot energetic cocktails, the Hexadecimal virus decided to fly over the Mainframe to look for a Great Idea. And she has found one!
Why not make her own Codeforces, with blackjack and other really cool stuff? Many people will surely be willing to visit this splendid shrine of high culture.
In Mainframe a standard pack of 52 cards is used to play blackjack. The pack contains cards of 13 values: 2, 3, 4, 5, 6, 7, 8, 9, 10, jacks, queens, kings and aces. Each value also exists in one of four suits: hearts, diamonds, clubs and spades. Also, each card earns some value in points assigned to it: cards with value from two to ten earn from 2 to 10 points, correspondingly. An ace can either earn 1 or 11, whatever the player wishes. The picture cards (king, queen and jack) earn 10 points. The number of points a card earns does not depend on the suit. The rules of the game are very simple. The player gets two cards, if the sum of points of those cards equals *n*, then the player wins, otherwise the player loses.
The player has already got the first card, it's the queen of spades. To evaluate chances for victory, you should determine how many ways there are to get the second card so that the sum of points exactly equals *n*.
Input Specification:
The only line contains *n* (1<=≤<=*n*<=≤<=25) — the required sum of points.
Output Specification:
Print the numbers of ways to get the second card in the required way if the first card is the queen of spades.
Demo Input:
['12\n', '20\n', '10\n']
Demo Output:
['4', '15', '0']
Note:
In the first sample only four two's of different suits can earn the required sum of points.
In the second sample we can use all tens, jacks, queens and kings; overall it's 15 cards, as the queen of spades (as any other card) is only present once in the pack of cards and it's already in use.
In the third sample there is no card, that would add a zero to the current ten points. | ```python
c = int(input())
c = c-10
if (c >= 1 and c < 10) or (c is 11):
print(4)
elif c is 10:
print(15)
else:
print(0)
``` | 3.9455 |
999 | B | Reversing Encryption | PROGRAMMING | 900 | [
"implementation"
] | null | null | A string $s$ of length $n$ can be encrypted by the following algorithm:
- iterate over all divisors of $n$ in decreasing order (i.e. from $n$ to $1$), - for each divisor $d$, reverse the substring $s[1 \dots d]$ (i.e. the substring which starts at position $1$ and ends at position $d$).
For example, the above algorithm applied to the string $s$="codeforces" leads to the following changes: "codeforces" $\to$ "secrofedoc" $\to$ "orcesfedoc" $\to$ "rocesfedoc" $\to$ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because $d=1$).
You are given the encrypted string $t$. Your task is to decrypt this string, i.e., to find a string $s$ such that the above algorithm results in string $t$. It can be proven that this string $s$ always exists and is unique. | The first line of input consists of a single integer $n$ ($1 \le n \le 100$) — the length of the string $t$. The second line of input consists of the string $t$. The length of $t$ is $n$, and it consists only of lowercase Latin letters. | Print a string $s$ such that the above algorithm results in $t$. | [
"10\nrocesfedoc\n",
"16\nplmaetwoxesisiht\n",
"1\nz\n"
] | [
"codeforces\n",
"thisisexampletwo\n",
"z\n"
] | The first example is described in the problem statement. | 0 | [
{
"input": "10\nrocesfedoc",
"output": "codeforces"
},
{
"input": "16\nplmaetwoxesisiht",
"output": "thisisexampletwo"
},
{
"input": "1\nz",
"output": "z"
},
{
"input": "2\nir",
"output": "ri"
},
{
"input": "3\nilj",
"output": "jli"
},
{
"input": "4\nj... | 1,641,105,044 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 46 | 0 | length = int(input())
string = input()
def find_divisors(length):
divisors = sorted([i for i in range(1, (length // 2) + 2) if length % i == 0])
divisors.append(length)
return divisors[1:]
def decode(divisors, string):
string_list = [i for i in string]
for i in divisors:
string_list[:i] = string_list[:i][::-1]
return ''.join(string_list)
divisors = find_divisors(length)
decoded_string = decode(divisors, string)
print(decoded_string) | Title: Reversing Encryption
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string $s$ of length $n$ can be encrypted by the following algorithm:
- iterate over all divisors of $n$ in decreasing order (i.e. from $n$ to $1$), - for each divisor $d$, reverse the substring $s[1 \dots d]$ (i.e. the substring which starts at position $1$ and ends at position $d$).
For example, the above algorithm applied to the string $s$="codeforces" leads to the following changes: "codeforces" $\to$ "secrofedoc" $\to$ "orcesfedoc" $\to$ "rocesfedoc" $\to$ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because $d=1$).
You are given the encrypted string $t$. Your task is to decrypt this string, i.e., to find a string $s$ such that the above algorithm results in string $t$. It can be proven that this string $s$ always exists and is unique.
Input Specification:
The first line of input consists of a single integer $n$ ($1 \le n \le 100$) — the length of the string $t$. The second line of input consists of the string $t$. The length of $t$ is $n$, and it consists only of lowercase Latin letters.
Output Specification:
Print a string $s$ such that the above algorithm results in $t$.
Demo Input:
['10\nrocesfedoc\n', '16\nplmaetwoxesisiht\n', '1\nz\n']
Demo Output:
['codeforces\n', 'thisisexampletwo\n', 'z\n']
Note:
The first example is described in the problem statement. | ```python
length = int(input())
string = input()
def find_divisors(length):
divisors = sorted([i for i in range(1, (length // 2) + 2) if length % i == 0])
divisors.append(length)
return divisors[1:]
def decode(divisors, string):
string_list = [i for i in string]
for i in divisors:
string_list[:i] = string_list[:i][::-1]
return ''.join(string_list)
divisors = find_divisors(length)
decoded_string = decode(divisors, string)
print(decoded_string)
``` | 0 | |
962 | D | Merge Equals | PROGRAMMING | 1,600 | [
"data structures",
"implementation"
] | null | null | You are given an array of positive integers. While there are at least two equal elements, we will perform the following operation. We choose the smallest value $x$ that occurs in the array $2$ or more times. Take the first two occurrences of $x$ in this array (the two leftmost occurrences). Remove the left of these two occurrences, and the right one is replaced by the sum of this two values (that is, $2 \cdot x$).
Determine how the array will look after described operations are performed.
For example, consider the given array looks like $[3, 4, 1, 2, 2, 1, 1]$. It will be changed in the following way: $[3, 4, 1, 2, 2, 1, 1]~\rightarrow~[3, 4, 2, 2, 2, 1]~\rightarrow~[3, 4, 4, 2, 1]~\rightarrow~[3, 8, 2, 1]$.
If the given array is look like $[1, 1, 3, 1, 1]$ it will be changed in the following way: $[1, 1, 3, 1, 1]~\rightarrow~[2, 3, 1, 1]~\rightarrow~[2, 3, 2]~\rightarrow~[3, 4]$. | The first line contains a single integer $n$ ($2 \le n \le 150\,000$) — the number of elements in the array.
The second line contains a sequence from $n$ elements $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^{9}$) — the elements of the array. | In the first line print an integer $k$ — the number of elements in the array after all the performed operations. In the second line print $k$ integers — the elements of the array after all the performed operations. | [
"7\n3 4 1 2 2 1 1\n",
"5\n1 1 3 1 1\n",
"5\n10 40 20 50 30\n"
] | [
"4\n3 8 2 1 \n",
"2\n3 4 \n",
"5\n10 40 20 50 30 \n"
] | The first two examples were considered in the statement.
In the third example all integers in the given array are distinct, so it will not change. | 0 | [
{
"input": "7\n3 4 1 2 2 1 1",
"output": "4\n3 8 2 1 "
},
{
"input": "5\n1 1 3 1 1",
"output": "2\n3 4 "
},
{
"input": "5\n10 40 20 50 30",
"output": "5\n10 40 20 50 30 "
},
{
"input": "100\n10 10 15 12 15 13 15 12 10 10 15 11 13 14 13 14 10 13 12 10 14 12 13 11 14 15 12 11 1... | 1,553,605,087 | 2,147,483,647 | PyPy 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <bits/stdc++.h>
using namespace std;
#define ll long long
#define push push_back
#define pop pop_back
// #define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define itr iterator
#define f first
#define s second
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
ll ans,i,j,ind,n;
ll a[150001];
map <ll,ll> mp;
int main(){
fast
cin>>n;
for (i=0;i<n;i++){
cin>>a[i];
while(1){
// cout<<a[i]<<" "<<mp[a[i]]<<endl;
if (mp[a[i]]==0){mp[a[i]]=i+1;break;}
ind=mp[a[i]]-1;
mp[a[i]]=0;
a[i]=2*a[i];
a[ind]=-1;
++ans;
// cout<<a[i]<<"ET"<<endl;
}
}
cout<<n-ans<<endl;
for (j=0;j<n;j++){
if (a[j]>0)
cout<<a[j]<<" ";
}
cout<<"\n";
return 0;
}
| Title: Merge Equals
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array of positive integers. While there are at least two equal elements, we will perform the following operation. We choose the smallest value $x$ that occurs in the array $2$ or more times. Take the first two occurrences of $x$ in this array (the two leftmost occurrences). Remove the left of these two occurrences, and the right one is replaced by the sum of this two values (that is, $2 \cdot x$).
Determine how the array will look after described operations are performed.
For example, consider the given array looks like $[3, 4, 1, 2, 2, 1, 1]$. It will be changed in the following way: $[3, 4, 1, 2, 2, 1, 1]~\rightarrow~[3, 4, 2, 2, 2, 1]~\rightarrow~[3, 4, 4, 2, 1]~\rightarrow~[3, 8, 2, 1]$.
If the given array is look like $[1, 1, 3, 1, 1]$ it will be changed in the following way: $[1, 1, 3, 1, 1]~\rightarrow~[2, 3, 1, 1]~\rightarrow~[2, 3, 2]~\rightarrow~[3, 4]$.
Input Specification:
The first line contains a single integer $n$ ($2 \le n \le 150\,000$) — the number of elements in the array.
The second line contains a sequence from $n$ elements $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^{9}$) — the elements of the array.
Output Specification:
In the first line print an integer $k$ — the number of elements in the array after all the performed operations. In the second line print $k$ integers — the elements of the array after all the performed operations.
Demo Input:
['7\n3 4 1 2 2 1 1\n', '5\n1 1 3 1 1\n', '5\n10 40 20 50 30\n']
Demo Output:
['4\n3 8 2 1 \n', '2\n3 4 \n', '5\n10 40 20 50 30 \n']
Note:
The first two examples were considered in the statement.
In the third example all integers in the given array are distinct, so it will not change. | ```python
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define push push_back
#define pop pop_back
// #define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define itr iterator
#define f first
#define s second
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
ll ans,i,j,ind,n;
ll a[150001];
map <ll,ll> mp;
int main(){
fast
cin>>n;
for (i=0;i<n;i++){
cin>>a[i];
while(1){
// cout<<a[i]<<" "<<mp[a[i]]<<endl;
if (mp[a[i]]==0){mp[a[i]]=i+1;break;}
ind=mp[a[i]]-1;
mp[a[i]]=0;
a[i]=2*a[i];
a[ind]=-1;
++ans;
// cout<<a[i]<<"ET"<<endl;
}
}
cout<<n-ans<<endl;
for (j=0;j<n;j++){
if (a[j]>0)
cout<<a[j]<<" ";
}
cout<<"\n";
return 0;
}
``` | -1 | |
305 | E | Playing with String | PROGRAMMING | 2,300 | [
"games"
] | null | null | Two people play the following string game. Initially the players have got some string *s*. The players move in turns, the player who cannot make a move loses.
Before the game began, the string is written on a piece of paper, one letter per cell.
A player's move is the sequence of actions:
1. The player chooses one of the available pieces of paper with some string written on it. Let's denote it is *t*. Note that initially, only one piece of paper is available. 1. The player chooses in the string *t*<==<=*t*1*t*2... *t*|*t*| character in position *i* (1<=≤<=*i*<=≤<=|*t*|) such that for some positive integer *l* (0<=<<=*i*<=-<=*l*; *i*<=+<=*l*<=≤<=|*t*|) the following equations hold: *t**i*<=-<=1<==<=*t**i*<=+<=1, *t**i*<=-<=2<==<=*t**i*<=+<=2, ..., *t**i*<=-<=*l*<==<=*t**i*<=+<=*l*. 1. Player cuts the cell with the chosen character. As a result of the operation, he gets three new pieces of paper, the first one will contain string *t*1*t*2... *t**i*<=-<=1, the second one will contain a string consisting of a single character *t**i*, the third one contains string *t**i*<=+<=1*t**i*<=+<=2... *t*|*t*|.
Your task is to determine the winner provided that both players play optimally well. If the first player wins, find the position of character that is optimal to cut in his first move. If there are multiple positions, print the minimal possible one. | The first line contains string *s* (1<=≤<=|*s*|<=≤<=5000). It is guaranteed that string *s* only contains lowercase English letters. | If the second player wins, print in the single line "Second" (without the quotes). Otherwise, print in the first line "First" (without the quotes), and in the second line print the minimal possible winning move — integer *i* (1<=≤<=*i*<=≤<=|*s*|). | [
"abacaba\n",
"abcde\n"
] | [
"First\n2\n",
"Second\n"
] | In the first sample the first player has multiple winning moves. But the minimum one is to cut the character in position 2.
In the second sample the first player has no available moves. | 2,500 | [
{
"input": "abacaba",
"output": "First\n2"
},
{
"input": "abcde",
"output": "Second"
},
{
"input": "aaaaa",
"output": "First\n3"
},
{
"input": "aaabbbbbbbbabaaabbaabbbbabbabaabaabbbaabbbbbbabbbabaabaaabaaaabbaaabbbbaabbbaaabababbbbabbabbabaaaaabababbbaabbbaabababaaabababbaaaa... | 1,371,123,563 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 78 | 0 | s = input()
d = [0 for i in s]
n = 0
m = 0
for i in range(1, len(s) - 1):
if s[i - 1] == s[i + 1]:
d[i] = d[i - 1] + 1
d[i - 1] = 0
for i in d:
if 1 <= i % 5 <= 2:
n ^= 1
elif i % 5 == 3:
m ^= 1
if n == m == 0:
print('Second')
else:
print('First')
for i in range(1, len(s) - 1):
if m == 0 and d[i] <= 2 or m == n and d[i] >= 3:
print(i - d[i] + 2)
break
if n == 0 and d[i] >= 3:
print(i - d[i] + 3)
break | Title: Playing with String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two people play the following string game. Initially the players have got some string *s*. The players move in turns, the player who cannot make a move loses.
Before the game began, the string is written on a piece of paper, one letter per cell.
A player's move is the sequence of actions:
1. The player chooses one of the available pieces of paper with some string written on it. Let's denote it is *t*. Note that initially, only one piece of paper is available. 1. The player chooses in the string *t*<==<=*t*1*t*2... *t*|*t*| character in position *i* (1<=≤<=*i*<=≤<=|*t*|) such that for some positive integer *l* (0<=<<=*i*<=-<=*l*; *i*<=+<=*l*<=≤<=|*t*|) the following equations hold: *t**i*<=-<=1<==<=*t**i*<=+<=1, *t**i*<=-<=2<==<=*t**i*<=+<=2, ..., *t**i*<=-<=*l*<==<=*t**i*<=+<=*l*. 1. Player cuts the cell with the chosen character. As a result of the operation, he gets three new pieces of paper, the first one will contain string *t*1*t*2... *t**i*<=-<=1, the second one will contain a string consisting of a single character *t**i*, the third one contains string *t**i*<=+<=1*t**i*<=+<=2... *t*|*t*|.
Your task is to determine the winner provided that both players play optimally well. If the first player wins, find the position of character that is optimal to cut in his first move. If there are multiple positions, print the minimal possible one.
Input Specification:
The first line contains string *s* (1<=≤<=|*s*|<=≤<=5000). It is guaranteed that string *s* only contains lowercase English letters.
Output Specification:
If the second player wins, print in the single line "Second" (without the quotes). Otherwise, print in the first line "First" (without the quotes), and in the second line print the minimal possible winning move — integer *i* (1<=≤<=*i*<=≤<=|*s*|).
Demo Input:
['abacaba\n', 'abcde\n']
Demo Output:
['First\n2\n', 'Second\n']
Note:
In the first sample the first player has multiple winning moves. But the minimum one is to cut the character in position 2.
In the second sample the first player has no available moves. | ```python
s = input()
d = [0 for i in s]
n = 0
m = 0
for i in range(1, len(s) - 1):
if s[i - 1] == s[i + 1]:
d[i] = d[i - 1] + 1
d[i - 1] = 0
for i in d:
if 1 <= i % 5 <= 2:
n ^= 1
elif i % 5 == 3:
m ^= 1
if n == m == 0:
print('Second')
else:
print('First')
for i in range(1, len(s) - 1):
if m == 0 and d[i] <= 2 or m == n and d[i] >= 3:
print(i - d[i] + 2)
break
if n == 0 and d[i] >= 3:
print(i - d[i] + 3)
break
``` | 0 | |
75 | C | Modified GCD | PROGRAMMING | 1,600 | [
"binary search",
"number theory"
] | C. Modified GCD | 2 | 256 | Well, here is another math class task. In mathematics, GCD is the greatest common divisor, and it's an easy task to calculate the GCD between two positive integers.
A common divisor for two positive numbers is a number which both numbers are divisible by.
But your teacher wants to give you a harder task, in this task you have to find the greatest common divisor *d* between two integers *a* and *b* that is in a given range from *low* to *high* (inclusive), i.e. *low*<=≤<=*d*<=≤<=*high*. It is possible that there is no common divisor in the given range.
You will be given the two integers *a* and *b*, then *n* queries. Each query is a range from *low* to *high* and you have to answer each query. | The first line contains two integers *a* and *b*, the two integers as described above (1<=≤<=*a*,<=*b*<=≤<=109). The second line contains one integer *n*, the number of queries (1<=≤<=*n*<=≤<=104). Then *n* lines follow, each line contains one query consisting of two integers, *low* and *high* (1<=≤<=*low*<=≤<=*high*<=≤<=109). | Print *n* lines. The *i*-th of them should contain the result of the *i*-th query in the input. If there is no common divisor in the given range for any query, you should print -1 as a result for this query. | [
"9 27\n3\n1 5\n10 11\n9 11\n"
] | [
"3\n-1\n9\n"
] | none | 1,500 | [
{
"input": "9 27\n3\n1 5\n10 11\n9 11",
"output": "3\n-1\n9"
},
{
"input": "48 72\n2\n8 29\n29 37",
"output": "24\n-1"
},
{
"input": "90 100\n10\n51 61\n6 72\n1 84\n33 63\n37 69\n18 21\n9 54\n49 90\n14 87\n37 90",
"output": "-1\n10\n10\n-1\n-1\n-1\n10\n-1\n-1\n-1"
},
{
"input... | 1,679,571,378 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 4 | 2,000 | 0 | def gcd(a, b):
if b == 0:
return a
return gcd(b, a % b)
a, b = map(int, input().split())
n = int(input())
for _ in range(n):
low, high = map(int, input().split())
if (gcd(a, b) < low):
print('-1')
else:
g = 1
for i in range(2, a + 1):
if a % i == 0 and b % i == 0:
g = i
break
d = gcd(a, b)
while d > high:
d = gcd(a // g, b)
print(d) | Title: Modified GCD
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Well, here is another math class task. In mathematics, GCD is the greatest common divisor, and it's an easy task to calculate the GCD between two positive integers.
A common divisor for two positive numbers is a number which both numbers are divisible by.
But your teacher wants to give you a harder task, in this task you have to find the greatest common divisor *d* between two integers *a* and *b* that is in a given range from *low* to *high* (inclusive), i.e. *low*<=≤<=*d*<=≤<=*high*. It is possible that there is no common divisor in the given range.
You will be given the two integers *a* and *b*, then *n* queries. Each query is a range from *low* to *high* and you have to answer each query.
Input Specification:
The first line contains two integers *a* and *b*, the two integers as described above (1<=≤<=*a*,<=*b*<=≤<=109). The second line contains one integer *n*, the number of queries (1<=≤<=*n*<=≤<=104). Then *n* lines follow, each line contains one query consisting of two integers, *low* and *high* (1<=≤<=*low*<=≤<=*high*<=≤<=109).
Output Specification:
Print *n* lines. The *i*-th of them should contain the result of the *i*-th query in the input. If there is no common divisor in the given range for any query, you should print -1 as a result for this query.
Demo Input:
['9 27\n3\n1 5\n10 11\n9 11\n']
Demo Output:
['3\n-1\n9\n']
Note:
none | ```python
def gcd(a, b):
if b == 0:
return a
return gcd(b, a % b)
a, b = map(int, input().split())
n = int(input())
for _ in range(n):
low, high = map(int, input().split())
if (gcd(a, b) < low):
print('-1')
else:
g = 1
for i in range(2, a + 1):
if a % i == 0 and b % i == 0:
g = i
break
d = gcd(a, b)
while d > high:
d = gcd(a // g, b)
print(d)
``` | 0 |
930 | D | Game with Tokens | PROGRAMMING | 2,500 | [
"data structures",
"games",
"implementation"
] | null | null | Consider the following game for two players. There is one white token and some number of black tokens. Each token is placed on a plane in a point with integer coordinates *x* and *y*.
The players take turn making moves, white starts. On each turn, a player moves all tokens of their color by 1 to up, down, left or right. Black player can choose directions for each token independently.
After a turn of the white player the white token can not be in a point where a black token is located. There are no other constraints on locations of the tokens: positions of black tokens can coincide, after a turn of the black player and initially the white token can be in the same point with some black point. If at some moment the white player can't make a move, he loses. If the white player makes 10100500 moves, he wins.
You are to solve the following problem. You are given initial positions of all black tokens. It is guaranteed that initially all these positions are distinct. In how many places can the white token be located initially so that if both players play optimally, the black player wins? | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of black points.
The (*i*<=+<=1)-th line contains two integers *x**i*, *y**i* (<=-<=105<=≤<=*x**i*,<=*y**i*,<=<=≤<=105) — the coordinates of the point where the *i*-th black token is initially located.
It is guaranteed that initial positions of black tokens are distinct. | Print the number of points where the white token can be located initially, such that if both players play optimally, the black player wins. | [
"4\n-2 -1\n0 1\n0 -3\n2 -1\n",
"4\n-2 0\n-1 1\n0 -2\n1 -1\n",
"16\n2 1\n1 2\n-1 1\n0 1\n0 0\n1 1\n2 -1\n2 0\n1 0\n-1 -1\n1 -1\n2 2\n0 -1\n-1 0\n0 2\n-1 2\n"
] | [
"4\n",
"2\n",
"4\n"
] | In the first and second examples initial positions of black tokens are shown with black points, possible positions of the white token (such that the black player wins) are shown with white points.
The first example: <img class="tex-graphics" src="https://espresso.codeforces.com/5054b8d2df2fac92c92f96fae82d21c365d12983.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The second example: <img class="tex-graphics" src="https://espresso.codeforces.com/eb795dd6abb95cfafb1d1cb7d8c8798825dcc180.png" style="max-width: 100.0%;max-height: 100.0%;"/>
In the third example the white tokens should be located in the inner square 2 × 2, to make the black player win. <img class="tex-graphics" src="https://espresso.codeforces.com/6dfd863f649b92860dfd6b446ea004abc01b71a6.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 2,000 | [
{
"input": "4\n-2 -1\n0 1\n0 -3\n2 -1",
"output": "4"
},
{
"input": "4\n-2 0\n-1 1\n0 -2\n1 -1",
"output": "2"
},
{
"input": "16\n2 1\n1 2\n-1 1\n0 1\n0 0\n1 1\n2 -1\n2 0\n1 0\n-1 -1\n1 -1\n2 2\n0 -1\n-1 0\n0 2\n-1 2",
"output": "4"
},
{
"input": "1\n1 2",
"output": "0"
... | 1,632,748,661 | 2,147,483,647 | PyPy 3 | OK | TESTS | 62 | 795 | 121,548,800 | #FB HACKERCUP
import os,sys,math
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from _builtin_ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
from bisect import bisect_left as lower_bound, bisect_right as upper_bound
def so(): return int(input())
def st(): return input()
def mj(): return map(int,input().strip().split(" "))
def msj(): return map(str,input().strip().split(" "))
def le(): return list(map(int,input().split()))
def lebe():return list(map(int, input()))
# def dmain():
# sys.setrecursionlimit(1000000)
# threading.stack_size(1024000)
# thread = threading.Thread(target=main)
# thread.start()
def joro(L):
return(''.join(map(str, L)))
def cheems(c,d):
a=1
while(d!=0):
if(d%2==1):
a*=c
c=c*c
d=d//2
return a
def decimalToBinary(n): return bin(n).replace("0b","")
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
# def npr(n, r):
# return factorial(n) // factorial(n - r) if n >= r else 0
# def ncr(n, r):
# return factorial(n) // (factorial(r) * factorial(n - r)) if n >= r else 0
def lower_bound(li, num):
answer = -1
start = 0
end = len(li) - 1
while (start <= end):
middle = (end + start) // 2
if li[middle] >= num:
answer = middle
end = middle - 1
else:
start = middle + 1
return answer # min index where x is not less than num
def upper_bound(li, num):
answer = -1
start = 0
end = len(li) - 1
while (start <= end):
middle = (end + start) // 2
if li[middle] <= num:
answer = middle
start = middle + 1
else:
end = middle - 1
return answer # max index where x is not greater than num
def tir(a,b,c):
if(0==c):
return 1
if(len(a)<=b):
return 0
if(c!=-1):
return (tir(a,1+b,c+a[b]) or tir(a,b+1,c-a[b]) or tir(a,1+b,c))
else:
return (tir(a,1+b,a[b]) or tir(a,b+1,-a[b]) or tir(a,1+b,-1))
def abs(x):
return x if x >= 0 else -x
doi=int(1e9+7)
boi=int(1e9-7)
koi=int(5+2e6)
moi=int(4e6+5+100)
def binary_search(li, val, lb, ub):
# print(lb, ub, li)
ans = -1
while (lb <= ub):
mid = (lb + ub) // 2
# print('mid is',mid, li[mid])
if li[mid] > val:
ub = mid - 1
elif val > li[mid]:
lb = mid + 1
else:
ans = mid # return index
break
return ans
def kadane(x): # maximum sum contiguous subarray
sum_so_far = 0
current_sum = 0
for i in x:
current_sum += i
if current_sum < 0:
current_sum = 0
else:
sum_so_far = max(sum_so_far, current_sum)
return sum_so_far
def wubu(m):
import math as my
d=0
while(not m%2):
m=m//2
d=1+d
for i in range(3,int(my.sqrt(m))+1,2):
while(not m%i):
m=m//i
d=1+d
return int(m>1)+d
def pref(li):
pref_sum = [0]
for i in li:
pref_sum.append(pref_sum[-1] + i)
return pref_sum
def SieveOfEratosthenes(n):
prime = [True for i in range(n + 1)]
p = 2
li = []
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n + 1, p):
prime[i] = False
p += 1
for p in range(2, len(prime)):
if prime[p]:
li.append(p)
return li
def primefactors(n):
import math as my
factors = []
while (n % 2 == 0):
factors.append(2)
n //= 2
for i in range(3, int(my.sqrt(n)) + 1, 2): # only odd factors left
while n % i == 0:
factors.append(i)
n //= i
if n > 2: # incase of prime
factors.append(n)
return factors
def read():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
def tr(n):
return n*(n+1)//2
def bro(q,r,c):
bec,pot,h=0,0,0
ds=[]
bs=[]
es=[]
for i in range(c):
while(h<len(r) and r[h]<=q[i][0]):
pot=0
ds.clear()
bs.clear()
h=1+h
for i in range(c):
ds.append(q[i][1])
es.append(q[i][1])
ds=list(set(ds))
for i in ds:
bs.append(es.count(i))
print(bs)
for i in range(c):
if(q[i][0]>r[h-1]):
bec=bec+pot-bs[i]
pot=1+pot
return bec
def iu():
import sys
input = sys.stdin.readline
import math as my
import bisect as by
m=so()
bec=0
A=[0]*moi
B=[0]*moi
C=[0]*moi
D=[0]*moi
E=[0]*moi
F=[0]*moi
for i in range(koi*2+1):
A[i]=sys.maxsize
B[i]=-sys.maxsize
for i in range(1,m+1):
p,q=mj()
r=p+q+koi
s=p-q
A[r]=min(A[r],s)
B[r]=max(B[r],s)
for i in range(2):
C[i]=A[i]
D[i]=B[i]
for i in range(2,1+2*koi):
C[i]=min(C[i-2],A[i])
D[i]=max(B[i],D[i-2])
for i in range(2*koi-2,2*koi+1):
E[i]=A[i]
F[i]=B[i]
for i in range(2*koi-2,-1,-1):
E[i]=min(E[2+i],A[i])
F[i]=max(B[i],F[2+i])
for i in range(1,2*koi+1,1):
se=min(D[i-1],F[1+i])-max(C[i-1],E[1+i])
bec=max(0,se//2)+bec
print(bec)
def main():
for i in range(1):
iu()
# region fastio
# template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
#read()
main()
#dmain()
# Comment Read() | Title: Game with Tokens
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Consider the following game for two players. There is one white token and some number of black tokens. Each token is placed on a plane in a point with integer coordinates *x* and *y*.
The players take turn making moves, white starts. On each turn, a player moves all tokens of their color by 1 to up, down, left or right. Black player can choose directions for each token independently.
After a turn of the white player the white token can not be in a point where a black token is located. There are no other constraints on locations of the tokens: positions of black tokens can coincide, after a turn of the black player and initially the white token can be in the same point with some black point. If at some moment the white player can't make a move, he loses. If the white player makes 10100500 moves, he wins.
You are to solve the following problem. You are given initial positions of all black tokens. It is guaranteed that initially all these positions are distinct. In how many places can the white token be located initially so that if both players play optimally, the black player wins?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of black points.
The (*i*<=+<=1)-th line contains two integers *x**i*, *y**i* (<=-<=105<=≤<=*x**i*,<=*y**i*,<=<=≤<=105) — the coordinates of the point where the *i*-th black token is initially located.
It is guaranteed that initial positions of black tokens are distinct.
Output Specification:
Print the number of points where the white token can be located initially, such that if both players play optimally, the black player wins.
Demo Input:
['4\n-2 -1\n0 1\n0 -3\n2 -1\n', '4\n-2 0\n-1 1\n0 -2\n1 -1\n', '16\n2 1\n1 2\n-1 1\n0 1\n0 0\n1 1\n2 -1\n2 0\n1 0\n-1 -1\n1 -1\n2 2\n0 -1\n-1 0\n0 2\n-1 2\n']
Demo Output:
['4\n', '2\n', '4\n']
Note:
In the first and second examples initial positions of black tokens are shown with black points, possible positions of the white token (such that the black player wins) are shown with white points.
The first example: <img class="tex-graphics" src="https://espresso.codeforces.com/5054b8d2df2fac92c92f96fae82d21c365d12983.png" style="max-width: 100.0%;max-height: 100.0%;"/>
The second example: <img class="tex-graphics" src="https://espresso.codeforces.com/eb795dd6abb95cfafb1d1cb7d8c8798825dcc180.png" style="max-width: 100.0%;max-height: 100.0%;"/>
In the third example the white tokens should be located in the inner square 2 × 2, to make the black player win. <img class="tex-graphics" src="https://espresso.codeforces.com/6dfd863f649b92860dfd6b446ea004abc01b71a6.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
#FB HACKERCUP
import os,sys,math
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from _builtin_ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
from bisect import bisect_left as lower_bound, bisect_right as upper_bound
def so(): return int(input())
def st(): return input()
def mj(): return map(int,input().strip().split(" "))
def msj(): return map(str,input().strip().split(" "))
def le(): return list(map(int,input().split()))
def lebe():return list(map(int, input()))
# def dmain():
# sys.setrecursionlimit(1000000)
# threading.stack_size(1024000)
# thread = threading.Thread(target=main)
# thread.start()
def joro(L):
return(''.join(map(str, L)))
def cheems(c,d):
a=1
while(d!=0):
if(d%2==1):
a*=c
c=c*c
d=d//2
return a
def decimalToBinary(n): return bin(n).replace("0b","")
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
# def npr(n, r):
# return factorial(n) // factorial(n - r) if n >= r else 0
# def ncr(n, r):
# return factorial(n) // (factorial(r) * factorial(n - r)) if n >= r else 0
def lower_bound(li, num):
answer = -1
start = 0
end = len(li) - 1
while (start <= end):
middle = (end + start) // 2
if li[middle] >= num:
answer = middle
end = middle - 1
else:
start = middle + 1
return answer # min index where x is not less than num
def upper_bound(li, num):
answer = -1
start = 0
end = len(li) - 1
while (start <= end):
middle = (end + start) // 2
if li[middle] <= num:
answer = middle
start = middle + 1
else:
end = middle - 1
return answer # max index where x is not greater than num
def tir(a,b,c):
if(0==c):
return 1
if(len(a)<=b):
return 0
if(c!=-1):
return (tir(a,1+b,c+a[b]) or tir(a,b+1,c-a[b]) or tir(a,1+b,c))
else:
return (tir(a,1+b,a[b]) or tir(a,b+1,-a[b]) or tir(a,1+b,-1))
def abs(x):
return x if x >= 0 else -x
doi=int(1e9+7)
boi=int(1e9-7)
koi=int(5+2e6)
moi=int(4e6+5+100)
def binary_search(li, val, lb, ub):
# print(lb, ub, li)
ans = -1
while (lb <= ub):
mid = (lb + ub) // 2
# print('mid is',mid, li[mid])
if li[mid] > val:
ub = mid - 1
elif val > li[mid]:
lb = mid + 1
else:
ans = mid # return index
break
return ans
def kadane(x): # maximum sum contiguous subarray
sum_so_far = 0
current_sum = 0
for i in x:
current_sum += i
if current_sum < 0:
current_sum = 0
else:
sum_so_far = max(sum_so_far, current_sum)
return sum_so_far
def wubu(m):
import math as my
d=0
while(not m%2):
m=m//2
d=1+d
for i in range(3,int(my.sqrt(m))+1,2):
while(not m%i):
m=m//i
d=1+d
return int(m>1)+d
def pref(li):
pref_sum = [0]
for i in li:
pref_sum.append(pref_sum[-1] + i)
return pref_sum
def SieveOfEratosthenes(n):
prime = [True for i in range(n + 1)]
p = 2
li = []
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n + 1, p):
prime[i] = False
p += 1
for p in range(2, len(prime)):
if prime[p]:
li.append(p)
return li
def primefactors(n):
import math as my
factors = []
while (n % 2 == 0):
factors.append(2)
n //= 2
for i in range(3, int(my.sqrt(n)) + 1, 2): # only odd factors left
while n % i == 0:
factors.append(i)
n //= i
if n > 2: # incase of prime
factors.append(n)
return factors
def read():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
def tr(n):
return n*(n+1)//2
def bro(q,r,c):
bec,pot,h=0,0,0
ds=[]
bs=[]
es=[]
for i in range(c):
while(h<len(r) and r[h]<=q[i][0]):
pot=0
ds.clear()
bs.clear()
h=1+h
for i in range(c):
ds.append(q[i][1])
es.append(q[i][1])
ds=list(set(ds))
for i in ds:
bs.append(es.count(i))
print(bs)
for i in range(c):
if(q[i][0]>r[h-1]):
bec=bec+pot-bs[i]
pot=1+pot
return bec
def iu():
import sys
input = sys.stdin.readline
import math as my
import bisect as by
m=so()
bec=0
A=[0]*moi
B=[0]*moi
C=[0]*moi
D=[0]*moi
E=[0]*moi
F=[0]*moi
for i in range(koi*2+1):
A[i]=sys.maxsize
B[i]=-sys.maxsize
for i in range(1,m+1):
p,q=mj()
r=p+q+koi
s=p-q
A[r]=min(A[r],s)
B[r]=max(B[r],s)
for i in range(2):
C[i]=A[i]
D[i]=B[i]
for i in range(2,1+2*koi):
C[i]=min(C[i-2],A[i])
D[i]=max(B[i],D[i-2])
for i in range(2*koi-2,2*koi+1):
E[i]=A[i]
F[i]=B[i]
for i in range(2*koi-2,-1,-1):
E[i]=min(E[2+i],A[i])
F[i]=max(B[i],F[2+i])
for i in range(1,2*koi+1,1):
se=min(D[i-1],F[1+i])-max(C[i-1],E[1+i])
bec=max(0,se//2)+bec
print(bec)
def main():
for i in range(1):
iu()
# region fastio
# template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
#read()
main()
#dmain()
# Comment Read()
``` | 3 | |
976 | A | Minimum Binary Number | PROGRAMMING | 800 | [
"implementation"
] | null | null | String can be called correct if it consists of characters "0" and "1" and there are no redundant leading zeroes. Here are some examples: "0", "10", "1001".
You are given a correct string *s*.
You can perform two different operations on this string:
1. swap any pair of adjacent characters (for example, "101" "110"); 1. replace "11" with "1" (for example, "110" "10").
Let *val*(*s*) be such a number that *s* is its binary representation.
Correct string *a* is less than some other correct string *b* iff *val*(*a*)<=<<=*val*(*b*).
Your task is to find the minimum correct string that you can obtain from the given one using the operations described above. You can use these operations any number of times in any order (or even use no operations at all). | The first line contains integer number *n* (1<=≤<=*n*<=≤<=100) — the length of string *s*.
The second line contains the string *s* consisting of characters "0" and "1". It is guaranteed that the string *s* is correct. | Print one string — the minimum correct string that you can obtain from the given one. | [
"4\n1001\n",
"1\n1\n"
] | [
"100\n",
"1\n"
] | In the first example you can obtain the answer by the following sequence of operations: "1001" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1010" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1100" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "100".
In the second example you can't obtain smaller answer no matter what operations you use. | 0 | [
{
"input": "4\n1001",
"output": "100"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "100\n1110111100001111011111111010110011111111011110000111101101011100110110001011000000101010110101011100",
"output": "1000000000000000000000000000000000000000"
},
{
"input": "100\n100000... | 1,525,437,646 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 62 | 7,065,600 | a = input()
if (a == '0'):
print(a)
else:
print('1' + a.count('0') * '0') | Title: Minimum Binary Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
String can be called correct if it consists of characters "0" and "1" and there are no redundant leading zeroes. Here are some examples: "0", "10", "1001".
You are given a correct string *s*.
You can perform two different operations on this string:
1. swap any pair of adjacent characters (for example, "101" "110"); 1. replace "11" with "1" (for example, "110" "10").
Let *val*(*s*) be such a number that *s* is its binary representation.
Correct string *a* is less than some other correct string *b* iff *val*(*a*)<=<<=*val*(*b*).
Your task is to find the minimum correct string that you can obtain from the given one using the operations described above. You can use these operations any number of times in any order (or even use no operations at all).
Input Specification:
The first line contains integer number *n* (1<=≤<=*n*<=≤<=100) — the length of string *s*.
The second line contains the string *s* consisting of characters "0" and "1". It is guaranteed that the string *s* is correct.
Output Specification:
Print one string — the minimum correct string that you can obtain from the given one.
Demo Input:
['4\n1001\n', '1\n1\n']
Demo Output:
['100\n', '1\n']
Note:
In the first example you can obtain the answer by the following sequence of operations: "1001" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1010" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1100" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "100".
In the second example you can't obtain smaller answer no matter what operations you use. | ```python
a = input()
if (a == '0'):
print(a)
else:
print('1' + a.count('0') * '0')
``` | 0 | |
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"
}... | 1,606,661,164 | 2,147,483,647 | Python 3 | OK | TESTS | 76 | 124 | 0 | a,b,c,d=list(map(int,input().split()))
print(-1 if d-a-b+c<=0 or a<c or b<c else d-a-b+c) | 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,d=list(map(int,input().split()))
print(-1 if d-a-b+c<=0 or a<c or b<c else d-a-b+c)
``` | 3 | |
343 | B | Alternating Current | PROGRAMMING | 1,600 | [
"data structures",
"greedy",
"implementation"
] | null | null | Mad scientist Mike has just finished constructing a new device to search for extraterrestrial intelligence! He was in such a hurry to launch it for the first time that he plugged in the power wires without giving it a proper glance and started experimenting right away. After a while Mike observed that the wires ended up entangled and now have to be untangled again.
The device is powered by two wires "plus" and "minus". The wires run along the floor from the wall (on the left) to the device (on the right). Both the wall and the device have two contacts in them on the same level, into which the wires are plugged in some order. The wires are considered entangled if there are one or more places where one wire runs above the other one. For example, the picture below has four such places (top view):
Mike knows the sequence in which the wires run above each other. Mike also noticed that on the left side, the "plus" wire is always plugged into the top contact (as seen on the picture). He would like to untangle the wires without unplugging them and without moving the device. Determine if it is possible to do that. A wire can be freely moved and stretched on the floor, but cannot be cut.
To understand the problem better please read the notes to the test samples. | The single line of the input contains a sequence of characters "+" and "-" of length *n* (1<=≤<=*n*<=≤<=100000). The *i*-th (1<=≤<=*i*<=≤<=*n*) position of the sequence contains the character "+", if on the *i*-th step from the wall the "plus" wire runs above the "minus" wire, and the character "-" otherwise. | Print either "Yes" (without the quotes) if the wires can be untangled or "No" (without the quotes) if the wires cannot be untangled. | [
"-++-\n",
"+-\n",
"++\n",
"-\n"
] | [
"Yes\n",
"No\n",
"Yes\n",
"No\n"
] | The first testcase corresponds to the picture in the statement. To untangle the wires, one can first move the "plus" wire lower, thus eliminating the two crosses in the middle, and then draw it under the "minus" wire, eliminating also the remaining two crosses.
In the second testcase the "plus" wire makes one full revolution around the "minus" wire. Thus the wires cannot be untangled:
In the third testcase the "plus" wire simply runs above the "minus" wire twice in sequence. The wires can be untangled by lifting "plus" and moving it higher:
In the fourth testcase the "minus" wire runs above the "plus" wire once. The wires cannot be untangled without moving the device itself: | 1,000 | [
{
"input": "-++-",
"output": "Yes"
},
{
"input": "+-",
"output": "No"
},
{
"input": "++",
"output": "Yes"
},
{
"input": "-",
"output": "No"
},
{
"input": "+-+-",
"output": "No"
},
{
"input": "-+-",
"output": "No"
},
{
"input": "-++-+--+",
... | 1,602,432,838 | 2,147,483,647 | PyPy 3 | OK | TESTS | 62 | 280 | 5,017,600 |
s = input()
ss = []
for i in s:
if len(ss) > 0 and ss[-1] == i:
ss.pop()
else:
ss.append(i)
if len(ss) == 0:
print("Yes")
else:
print("No")
| Title: Alternating Current
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mad scientist Mike has just finished constructing a new device to search for extraterrestrial intelligence! He was in such a hurry to launch it for the first time that he plugged in the power wires without giving it a proper glance and started experimenting right away. After a while Mike observed that the wires ended up entangled and now have to be untangled again.
The device is powered by two wires "plus" and "minus". The wires run along the floor from the wall (on the left) to the device (on the right). Both the wall and the device have two contacts in them on the same level, into which the wires are plugged in some order. The wires are considered entangled if there are one or more places where one wire runs above the other one. For example, the picture below has four such places (top view):
Mike knows the sequence in which the wires run above each other. Mike also noticed that on the left side, the "plus" wire is always plugged into the top contact (as seen on the picture). He would like to untangle the wires without unplugging them and without moving the device. Determine if it is possible to do that. A wire can be freely moved and stretched on the floor, but cannot be cut.
To understand the problem better please read the notes to the test samples.
Input Specification:
The single line of the input contains a sequence of characters "+" and "-" of length *n* (1<=≤<=*n*<=≤<=100000). The *i*-th (1<=≤<=*i*<=≤<=*n*) position of the sequence contains the character "+", if on the *i*-th step from the wall the "plus" wire runs above the "minus" wire, and the character "-" otherwise.
Output Specification:
Print either "Yes" (without the quotes) if the wires can be untangled or "No" (without the quotes) if the wires cannot be untangled.
Demo Input:
['-++-\n', '+-\n', '++\n', '-\n']
Demo Output:
['Yes\n', 'No\n', 'Yes\n', 'No\n']
Note:
The first testcase corresponds to the picture in the statement. To untangle the wires, one can first move the "plus" wire lower, thus eliminating the two crosses in the middle, and then draw it under the "minus" wire, eliminating also the remaining two crosses.
In the second testcase the "plus" wire makes one full revolution around the "minus" wire. Thus the wires cannot be untangled:
In the third testcase the "plus" wire simply runs above the "minus" wire twice in sequence. The wires can be untangled by lifting "plus" and moving it higher:
In the fourth testcase the "minus" wire runs above the "plus" wire once. The wires cannot be untangled without moving the device itself: | ```python
s = input()
ss = []
for i in s:
if len(ss) > 0 and ss[-1] == i:
ss.pop()
else:
ss.append(i)
if len(ss) == 0:
print("Yes")
else:
print("No")
``` | 3 | |
253 | B | Physics Practical | PROGRAMMING | 1,400 | [
"binary search",
"dp",
"sortings",
"two pointers"
] | null | null | One day Vasya was on a physics practical, performing the task on measuring the capacitance. He followed the teacher's advice and did as much as *n* measurements, and recorded the results in the notebook. After that he was about to show the results to the teacher, but he remembered that at the last lesson, the teacher had made his friend Petya redo the experiment because the largest and the smallest results differed by more than two times. Vasya is lazy, and he does not want to redo the experiment. He wants to do the task and go home play computer games. So he decided to cheat: before Vasya shows the measurements to the teacher, he will erase some of them, so as to make the largest and the smallest results of the remaining measurements differ in no more than two times. In other words, if the remaining measurements have the smallest result *x*, and the largest result *y*, then the inequality *y*<=≤<=2·*x* must fulfill. Of course, to avoid the teacher's suspicion, Vasya wants to remove as few measurement results as possible from his notes.
Help Vasya, find what minimum number of measurement results he will have to erase from his notes so that the largest and the smallest of the remaining results of the measurements differed in no more than two times. | The first line contains integer *n* (2<=≤<=*n*<=≤<=105) — the number of measurements Vasya made. The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=5000) — the results of the measurements. The numbers on the second line are separated by single spaces. | Print a single integer — the minimum number of results Vasya will have to remove. | [
"6\n4 5 3 8 3 7\n",
"4\n4 3 2 4\n"
] | [
"2\n",
"0\n"
] | In the first sample you can remove the fourth and the sixth measurement results (values 8 and 7). Then the maximum of the remaining values will be 5, and the minimum one will be 3. Or else, you can remove the third and fifth results (both equal 3). After that the largest remaining result will be 8, and the smallest one will be 4. | 1,000 | [
{
"input": "6\n4 5 3 8 3 7",
"output": "2"
},
{
"input": "4\n4 3 2 4",
"output": "0"
},
{
"input": "6\n5 6 4 9 4 8",
"output": "1"
},
{
"input": "4\n5 4 1 5",
"output": "1"
},
{
"input": "2\n3 2",
"output": "0"
},
{
"input": "10\n39 9 18 13 6 16 47 15 ... | 1,614,400,454 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 216 | 2,150,400 | from collections import Counter as cc
inp = open("input.txt","r+")
f = inp.readlines()
for i in range(len(f)):
f[i].replace("\n","")
n = int(f[0])
l = list(map(int, f[1].split()))
d = dict(cc(l))
g = sorted(list(d.items()), key = lambda x:x[0])
# print(g)
start = 0
end = len(g)-1
ans=0
while start<end:
if g[start][0]*2>=g[end][0]:
break
else:
if g[start][1]<g[end][1]:
start+=1
ans+=g[start][1]
else:
end-=1
ans+=g[end][1]
outp = open("output.txt","w+")
outp.write(str(ans))
| Title: Physics Practical
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya was on a physics practical, performing the task on measuring the capacitance. He followed the teacher's advice and did as much as *n* measurements, and recorded the results in the notebook. After that he was about to show the results to the teacher, but he remembered that at the last lesson, the teacher had made his friend Petya redo the experiment because the largest and the smallest results differed by more than two times. Vasya is lazy, and he does not want to redo the experiment. He wants to do the task and go home play computer games. So he decided to cheat: before Vasya shows the measurements to the teacher, he will erase some of them, so as to make the largest and the smallest results of the remaining measurements differ in no more than two times. In other words, if the remaining measurements have the smallest result *x*, and the largest result *y*, then the inequality *y*<=≤<=2·*x* must fulfill. Of course, to avoid the teacher's suspicion, Vasya wants to remove as few measurement results as possible from his notes.
Help Vasya, find what minimum number of measurement results he will have to erase from his notes so that the largest and the smallest of the remaining results of the measurements differed in no more than two times.
Input Specification:
The first line contains integer *n* (2<=≤<=*n*<=≤<=105) — the number of measurements Vasya made. The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=5000) — the results of the measurements. The numbers on the second line are separated by single spaces.
Output Specification:
Print a single integer — the minimum number of results Vasya will have to remove.
Demo Input:
['6\n4 5 3 8 3 7\n', '4\n4 3 2 4\n']
Demo Output:
['2\n', '0\n']
Note:
In the first sample you can remove the fourth and the sixth measurement results (values 8 and 7). Then the maximum of the remaining values will be 5, and the minimum one will be 3. Or else, you can remove the third and fifth results (both equal 3). After that the largest remaining result will be 8, and the smallest one will be 4. | ```python
from collections import Counter as cc
inp = open("input.txt","r+")
f = inp.readlines()
for i in range(len(f)):
f[i].replace("\n","")
n = int(f[0])
l = list(map(int, f[1].split()))
d = dict(cc(l))
g = sorted(list(d.items()), key = lambda x:x[0])
# print(g)
start = 0
end = len(g)-1
ans=0
while start<end:
if g[start][0]*2>=g[end][0]:
break
else:
if g[start][1]<g[end][1]:
start+=1
ans+=g[start][1]
else:
end-=1
ans+=g[end][1]
outp = open("output.txt","w+")
outp.write(str(ans))
``` | -1 | |
126 | B | Password | PROGRAMMING | 1,700 | [
"binary search",
"dp",
"hashing",
"string suffix structures",
"strings"
] | null | null | Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them.
A little later they found a string *s*, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring *t* of the string *s*.
Prefix supposed that the substring *t* is the beginning of the string *s*; Suffix supposed that the substring *t* should be the end of the string *s*; and Obelix supposed that *t* should be located somewhere inside the string *s*, that is, *t* is neither its beginning, nor its end.
Asterix chose the substring *t* so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring *t* aloud, the temple doors opened.
You know the string *s*. Find the substring *t* or determine that such substring does not exist and all that's been written above is just a nice legend. | You are given the string *s* whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters. | Print the string *t*. If a suitable *t* string does not exist, then print "Just a legend" without the quotes. | [
"fixprefixsuffix\n",
"abcdabc\n"
] | [
"fix",
"Just a legend"
] | none | 1,000 | [
{
"input": "fixprefixsuffix",
"output": "fix"
},
{
"input": "abcdabc",
"output": "Just a legend"
},
{
"input": "qwertyqwertyqwerty",
"output": "qwerty"
},
{
"input": "papapapap",
"output": "papap"
},
{
"input": "aaaaaaaaaa",
"output": "aaaaaaaa"
},
{
"... | 1,651,942,179 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 1 | 124 | 3,276,800 | from string import ascii_lowercase as lc
s=input()
n=len(s)
dt1=dict()
dt2=dict()
dt3=dict()
for i in range(26):
dt1[lc[i]]=i+1
dt2[lc[i]]=(i+1)*2
dt3[lc[i]]=(i+1)*3
l1,l2,l3=[],[],[]
cl1,cl2,cl3=[dt1[s[0]]**2+0],[dt2[s[0]]**3+0],[dt3[s[0]]+0]
for i in range(n):
l1.append(dt1[s[i]]**2)
l2.append(dt2[s[i]]**3)
l3.append(dt3[s[i]])
if i>0:
cl1.append(l1[-1]+cl1[-1])
cl2.append(l2[-1]+cl2[-1])
cl3.append(l3[-1]+cl3[-1])
s1=''
for i in range(n):
s1+=s[-1-i]
rl1,rl2,rl3=[],[],[]
rcl1,rcl2,rcl3=[dt1[s1[0]]**2+0],[dt2[s1[0]]**3+0],[dt3[s1[0]]+0]
for i in range(n):
rl1.append(dt1[s1[i]]**2)
rl2.append(dt2[s1[i]]**3)
rl3.append(dt3[s1[i]])
if i>0:
rcl1.append(rl1[-1]+rcl1[-1])
rcl2.append(rl2[-1]+rcl2[-1])
rcl3.append(rl3[-1]+rcl3[-1])
s1=''
j=0
for i in range(n):
if (cl1[i]==rcl1[i] and cl2[i]==rcl2[i]
and cl3[i]==rcl3[i]):
s1=cl1[i]
s2,s3=cl2[i],cl3[i]
j=i+1
break
ans=''
for i in range(1,n-j-1):
if (cl1[i+j]-cl1[i]==s1 and cl2[i+j]-cl2[i]==s2
and cl3[i+j]-cl3[i]==s3):
ans=s[:j]
break
if ans=='':
ans='just a legend'
print(ans)
| Title: Password
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Asterix, Obelix and their temporary buddies Suffix and Prefix has finally found the Harmony temple. However, its doors were firmly locked and even Obelix had no luck opening them.
A little later they found a string *s*, carved on a rock below the temple's gates. Asterix supposed that that's the password that opens the temple and read the string aloud. However, nothing happened. Then Asterix supposed that a password is some substring *t* of the string *s*.
Prefix supposed that the substring *t* is the beginning of the string *s*; Suffix supposed that the substring *t* should be the end of the string *s*; and Obelix supposed that *t* should be located somewhere inside the string *s*, that is, *t* is neither its beginning, nor its end.
Asterix chose the substring *t* so as to please all his companions. Besides, from all acceptable variants Asterix chose the longest one (as Asterix loves long strings). When Asterix read the substring *t* aloud, the temple doors opened.
You know the string *s*. Find the substring *t* or determine that such substring does not exist and all that's been written above is just a nice legend.
Input Specification:
You are given the string *s* whose length can vary from 1 to 106 (inclusive), consisting of small Latin letters.
Output Specification:
Print the string *t*. If a suitable *t* string does not exist, then print "Just a legend" without the quotes.
Demo Input:
['fixprefixsuffix\n', 'abcdabc\n']
Demo Output:
['fix', 'Just a legend']
Note:
none | ```python
from string import ascii_lowercase as lc
s=input()
n=len(s)
dt1=dict()
dt2=dict()
dt3=dict()
for i in range(26):
dt1[lc[i]]=i+1
dt2[lc[i]]=(i+1)*2
dt3[lc[i]]=(i+1)*3
l1,l2,l3=[],[],[]
cl1,cl2,cl3=[dt1[s[0]]**2+0],[dt2[s[0]]**3+0],[dt3[s[0]]+0]
for i in range(n):
l1.append(dt1[s[i]]**2)
l2.append(dt2[s[i]]**3)
l3.append(dt3[s[i]])
if i>0:
cl1.append(l1[-1]+cl1[-1])
cl2.append(l2[-1]+cl2[-1])
cl3.append(l3[-1]+cl3[-1])
s1=''
for i in range(n):
s1+=s[-1-i]
rl1,rl2,rl3=[],[],[]
rcl1,rcl2,rcl3=[dt1[s1[0]]**2+0],[dt2[s1[0]]**3+0],[dt3[s1[0]]+0]
for i in range(n):
rl1.append(dt1[s1[i]]**2)
rl2.append(dt2[s1[i]]**3)
rl3.append(dt3[s1[i]])
if i>0:
rcl1.append(rl1[-1]+rcl1[-1])
rcl2.append(rl2[-1]+rcl2[-1])
rcl3.append(rl3[-1]+rcl3[-1])
s1=''
j=0
for i in range(n):
if (cl1[i]==rcl1[i] and cl2[i]==rcl2[i]
and cl3[i]==rcl3[i]):
s1=cl1[i]
s2,s3=cl2[i],cl3[i]
j=i+1
break
ans=''
for i in range(1,n-j-1):
if (cl1[i+j]-cl1[i]==s1 and cl2[i+j]-cl2[i]==s2
and cl3[i+j]-cl3[i]==s3):
ans=s[:j]
break
if ans=='':
ans='just a legend'
print(ans)
``` | 0 | |
447 | B | DZY Loves Strings | PROGRAMMING | 1,000 | [
"greedy",
"implementation"
] | null | null | DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter *c* DZY knows its value *w**c*. For each special string *s*<==<=*s*1*s*2... *s*|*s*| (|*s*| is the length of the string) he represents its value with a function *f*(*s*), where
Now DZY has a string *s*. He wants to insert *k* lowercase letters into this string in order to get the largest possible value of the resulting string. Can you help him calculate the largest possible value he could get? | The first line contains a single string *s* (1<=≤<=|*s*|<=≤<=103).
The second line contains a single integer *k* (0<=≤<=*k*<=≤<=103).
The third line contains twenty-six integers from *w**a* to *w**z*. Each such number is non-negative and doesn't exceed 1000. | Print a single integer — the largest possible value of the resulting string DZY could get. | [
"abc\n3\n1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n"
] | [
"41\n"
] | In the test sample DZY can obtain "abcbbc", *value* = 1·1 + 2·2 + 3·2 + 4·2 + 5·2 + 6·2 = 41. | 1,000 | [
{
"input": "abc\n3\n1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "41"
},
{
"input": "mmzhr\n3\n443 497 867 471 195 670 453 413 579 466 553 881 847 642 269 996 666 702 487 209 257 741 974 133 519 453",
"output": "29978"
},
{
"input": "ajeeseerqnpaujubmajpibxrccazaawetyw... | 1,405,259,264 | 2,864 | Python 3 | OK | TESTS | 24 | 77 | 0 | st=input()
k=int(input())
w=list(map(int,input().split()))
w2={chr(ord('a')+i):w[i] for i in range(26)}
m=max(w)
print(sum([w2[st[i]]*(i+1) for i in range(len(st))])+sum([i*m for i in range(len(st)+1,len(st)+1+k)])) | Title: DZY Loves Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter *c* DZY knows its value *w**c*. For each special string *s*<==<=*s*1*s*2... *s*|*s*| (|*s*| is the length of the string) he represents its value with a function *f*(*s*), where
Now DZY has a string *s*. He wants to insert *k* lowercase letters into this string in order to get the largest possible value of the resulting string. Can you help him calculate the largest possible value he could get?
Input Specification:
The first line contains a single string *s* (1<=≤<=|*s*|<=≤<=103).
The second line contains a single integer *k* (0<=≤<=*k*<=≤<=103).
The third line contains twenty-six integers from *w**a* to *w**z*. Each such number is non-negative and doesn't exceed 1000.
Output Specification:
Print a single integer — the largest possible value of the resulting string DZY could get.
Demo Input:
['abc\n3\n1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n']
Demo Output:
['41\n']
Note:
In the test sample DZY can obtain "abcbbc", *value* = 1·1 + 2·2 + 3·2 + 4·2 + 5·2 + 6·2 = 41. | ```python
st=input()
k=int(input())
w=list(map(int,input().split()))
w2={chr(ord('a')+i):w[i] for i in range(26)}
m=max(w)
print(sum([w2[st[i]]*(i+1) for i in range(len(st))])+sum([i*m for i in range(len(st)+1,len(st)+1+k)]))
``` | 3 | |
514 | A | Chewbaсca and Number | PROGRAMMING | 1,200 | [
"greedy",
"implementation"
] | null | null | Luke Skywalker gave Chewbacca an integer number *x*. Chewbacca isn't good at numbers but he loves inverting digits in them. Inverting digit *t* means replacing it with digit 9<=-<=*t*.
Help Chewbacca to transform the initial number *x* to the minimum possible positive number by inverting some (possibly, zero) digits. The decimal representation of the final number shouldn't start with a zero. | The first line contains a single integer *x* (1<=≤<=*x*<=≤<=1018) — the number that Luke Skywalker gave to Chewbacca. | Print the minimum possible positive number that Chewbacca can obtain after inverting some digits. The number shouldn't contain leading zeroes. | [
"27\n",
"4545\n"
] | [
"22\n",
"4444\n"
] | none | 500 | [
{
"input": "27",
"output": "22"
},
{
"input": "4545",
"output": "4444"
},
{
"input": "1",
"output": "1"
},
{
"input": "9",
"output": "9"
},
{
"input": "8772",
"output": "1222"
},
{
"input": "81",
"output": "11"
},
{
"input": "71723447",
... | 1,699,808,370 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 31 | 0 | n=list(input())
index=0
for i in n:
if i>='5':
if index:n[index]=str(9-int(i))
index+=1
print(*n,sep='')
| Title: Chewbaсca and Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luke Skywalker gave Chewbacca an integer number *x*. Chewbacca isn't good at numbers but he loves inverting digits in them. Inverting digit *t* means replacing it with digit 9<=-<=*t*.
Help Chewbacca to transform the initial number *x* to the minimum possible positive number by inverting some (possibly, zero) digits. The decimal representation of the final number shouldn't start with a zero.
Input Specification:
The first line contains a single integer *x* (1<=≤<=*x*<=≤<=1018) — the number that Luke Skywalker gave to Chewbacca.
Output Specification:
Print the minimum possible positive number that Chewbacca can obtain after inverting some digits. The number shouldn't contain leading zeroes.
Demo Input:
['27\n', '4545\n']
Demo Output:
['22\n', '4444\n']
Note:
none | ```python
n=list(input())
index=0
for i in n:
if i>='5':
if index:n[index]=str(9-int(i))
index+=1
print(*n,sep='')
``` | 0 | |
478 | C | Table Decorations | PROGRAMMING | 1,800 | [
"greedy"
] | null | null | You have *r* red, *g* green and *b* blue balloons. To decorate a single table for the banquet you need exactly three balloons. Three balloons attached to some table shouldn't have the same color. What maximum number *t* of tables can be decorated if we know number of balloons of each color?
Your task is to write a program that for given values *r*, *g* and *b* will find the maximum number *t* of tables, that can be decorated in the required manner. | The single line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=2·109) — the number of red, green and blue baloons respectively. The numbers are separated by exactly one space. | Print a single integer *t* — the maximum number of tables that can be decorated in the required manner. | [
"5 4 3\n",
"1 1 1\n",
"2 3 3\n"
] | [
"4\n",
"1\n",
"2\n"
] | In the first sample you can decorate the tables with the following balloon sets: "rgg", "gbb", "brr", "rrg", where "r", "g" and "b" represent the red, green and blue balls, respectively. | 1,500 | [
{
"input": "5 4 3",
"output": "4"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "2 3 3",
"output": "2"
},
{
"input": "0 1 0",
"output": "0"
},
{
"input": "0 3 3",
"output": "2"
},
{
"input": "4 0 4",
"output": "2"
},
{
"input": "100000... | 1,668,599,366 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 42 | 77 | 0 | r,g,b = map(int,input().split())
print(min(r+g,r+b,b+g,(r+b+g)//3))
| Title: Table Decorations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have *r* red, *g* green and *b* blue balloons. To decorate a single table for the banquet you need exactly three balloons. Three balloons attached to some table shouldn't have the same color. What maximum number *t* of tables can be decorated if we know number of balloons of each color?
Your task is to write a program that for given values *r*, *g* and *b* will find the maximum number *t* of tables, that can be decorated in the required manner.
Input Specification:
The single line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=2·109) — the number of red, green and blue baloons respectively. The numbers are separated by exactly one space.
Output Specification:
Print a single integer *t* — the maximum number of tables that can be decorated in the required manner.
Demo Input:
['5 4 3\n', '1 1 1\n', '2 3 3\n']
Demo Output:
['4\n', '1\n', '2\n']
Note:
In the first sample you can decorate the tables with the following balloon sets: "rgg", "gbb", "brr", "rrg", where "r", "g" and "b" represent the red, green and blue balls, respectively. | ```python
r,g,b = map(int,input().split())
print(min(r+g,r+b,b+g,(r+b+g)//3))
``` | 3 | |
879 | A | Borya's Diagnosis | PROGRAMMING | 900 | [
"implementation"
] | null | null | It seems that Borya is seriously sick. He is going visit *n* doctors to find out the exact diagnosis. Each of the doctors needs the information about all previous visits, so Borya has to visit them in the prescribed order (i.e. Borya should first visit doctor 1, then doctor 2, then doctor 3 and so on). Borya will get the information about his health from the last doctor.
Doctors have a strange working schedule. The doctor *i* goes to work on the *s**i*-th day and works every *d**i* day. So, he works on days *s**i*,<=*s**i*<=+<=*d**i*,<=*s**i*<=+<=2*d**i*,<=....
The doctor's appointment takes quite a long time, so Borya can not see more than one doctor per day. What is the minimum time he needs to visit all doctors? | First line contains an integer *n* — number of doctors (1<=≤<=*n*<=≤<=1000).
Next *n* lines contain two numbers *s**i* and *d**i* (1<=≤<=*s**i*,<=*d**i*<=≤<=1000). | Output a single integer — the minimum day at which Borya can visit the last doctor. | [
"3\n2 2\n1 2\n2 2\n",
"2\n10 1\n6 5\n"
] | [
"4\n",
"11\n"
] | In the first sample case, Borya can visit all doctors on days 2, 3 and 4.
In the second sample case, Borya can visit all doctors on days 10 and 11. | 500 | [
{
"input": "3\n2 2\n1 2\n2 2",
"output": "4"
},
{
"input": "2\n10 1\n6 5",
"output": "11"
},
{
"input": "3\n6 10\n3 3\n8 2",
"output": "10"
},
{
"input": "4\n4 8\n10 10\n4 2\n8 2",
"output": "14"
},
{
"input": "5\n7 1\n5 1\n6 1\n1 6\n6 8",
"output": "14"
},
... | 1,548,411,327 | 2,147,483,647 | Python 3 | OK | TESTS | 51 | 124 | 0 | n = int(input())
cur = 0
for i in range(n):
s, d = map(int, input().split())
q = max(0, cur - s + d) // d
cur = q * d + s
print(cur)
| Title: Borya's Diagnosis
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It seems that Borya is seriously sick. He is going visit *n* doctors to find out the exact diagnosis. Each of the doctors needs the information about all previous visits, so Borya has to visit them in the prescribed order (i.e. Borya should first visit doctor 1, then doctor 2, then doctor 3 and so on). Borya will get the information about his health from the last doctor.
Doctors have a strange working schedule. The doctor *i* goes to work on the *s**i*-th day and works every *d**i* day. So, he works on days *s**i*,<=*s**i*<=+<=*d**i*,<=*s**i*<=+<=2*d**i*,<=....
The doctor's appointment takes quite a long time, so Borya can not see more than one doctor per day. What is the minimum time he needs to visit all doctors?
Input Specification:
First line contains an integer *n* — number of doctors (1<=≤<=*n*<=≤<=1000).
Next *n* lines contain two numbers *s**i* and *d**i* (1<=≤<=*s**i*,<=*d**i*<=≤<=1000).
Output Specification:
Output a single integer — the minimum day at which Borya can visit the last doctor.
Demo Input:
['3\n2 2\n1 2\n2 2\n', '2\n10 1\n6 5\n']
Demo Output:
['4\n', '11\n']
Note:
In the first sample case, Borya can visit all doctors on days 2, 3 and 4.
In the second sample case, Borya can visit all doctors on days 10 and 11. | ```python
n = int(input())
cur = 0
for i in range(n):
s, d = map(int, input().split())
q = max(0, cur - s + d) // d
cur = q * d + s
print(cur)
``` | 3 | |
8 | B | Obsession with Robots | PROGRAMMING | 1,400 | [
"constructive algorithms",
"graphs",
"implementation"
] | B. Obsession with Robots | 2 | 64 | The whole world got obsessed with robots,and to keep pace with the progress, great Berland's programmer Draude decided to build his own robot. He was working hard at the robot. He taught it to walk the shortest path from one point to another, to record all its movements, but like in many Draude's programs, there was a bug — the robot didn't always walk the shortest path. Fortunately, the robot recorded its own movements correctly. Now Draude wants to find out when his robot functions wrong. Heh, if Draude only remembered the map of the field, where he tested the robot, he would easily say if the robot walked in the right direction or not. But the field map was lost never to be found, that's why he asks you to find out if there exist at least one map, where the path recorded by the robot is the shortest.
The map is an infinite checkered field, where each square is either empty, or contains an obstruction. It is also known that the robot never tries to run into the obstruction. By the recorded robot's movements find out if there exist at least one such map, that it is possible to choose for the robot a starting square (the starting square should be empty) such that when the robot moves from this square its movements coincide with the recorded ones (the robot doesn't run into anything, moving along empty squares only), and the path from the starting square to the end one is the shortest.
In one movement the robot can move into the square (providing there are no obstrutions in this square) that has common sides with the square the robot is currently in. | The first line of the input file contains the recording of the robot's movements. This recording is a non-empty string, consisting of uppercase Latin letters L, R, U and D, standing for movements left, right, up and down respectively. The length of the string does not exceed 100. | In the first line output the only word OK (if the above described map exists), or BUG (if such a map does not exist). | [
"LLUUUR\n",
"RRUULLDD\n"
] | [
"OK\n",
"BUG\n"
] | none | 0 | [
{
"input": "LLUUUR",
"output": "OK"
},
{
"input": "RRUULLDD",
"output": "BUG"
},
{
"input": "L",
"output": "OK"
},
{
"input": "R",
"output": "OK"
},
{
"input": "R",
"output": "OK"
},
{
"input": "RR",
"output": "OK"
},
{
"input": "DL",
"... | 1,680,434,813 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 62 | 0 | def robots(s):
visited = set()
p = [0, 0]
for i in s:
if i == 'R': p[0] += 1
elif i == 'L': p[0] -= 1
elif i == 'U': p[1] += 1
elif i == 'D': p[1] -= 1
if str(p) in visited: return ("BUG")
visited.add(str([p[0]-1, p[1]]))
visited.add(str([p[0]+1, p[1]]))
visited.add(str([p[0], p[1]-1]))
visited.add(str([p[0], p[1]+1]))
return ("OK")
s = input()
print(robots(s)) | Title: Obsession with Robots
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
The whole world got obsessed with robots,and to keep pace with the progress, great Berland's programmer Draude decided to build his own robot. He was working hard at the robot. He taught it to walk the shortest path from one point to another, to record all its movements, but like in many Draude's programs, there was a bug — the robot didn't always walk the shortest path. Fortunately, the robot recorded its own movements correctly. Now Draude wants to find out when his robot functions wrong. Heh, if Draude only remembered the map of the field, where he tested the robot, he would easily say if the robot walked in the right direction or not. But the field map was lost never to be found, that's why he asks you to find out if there exist at least one map, where the path recorded by the robot is the shortest.
The map is an infinite checkered field, where each square is either empty, or contains an obstruction. It is also known that the robot never tries to run into the obstruction. By the recorded robot's movements find out if there exist at least one such map, that it is possible to choose for the robot a starting square (the starting square should be empty) such that when the robot moves from this square its movements coincide with the recorded ones (the robot doesn't run into anything, moving along empty squares only), and the path from the starting square to the end one is the shortest.
In one movement the robot can move into the square (providing there are no obstrutions in this square) that has common sides with the square the robot is currently in.
Input Specification:
The first line of the input file contains the recording of the robot's movements. This recording is a non-empty string, consisting of uppercase Latin letters L, R, U and D, standing for movements left, right, up and down respectively. The length of the string does not exceed 100.
Output Specification:
In the first line output the only word OK (if the above described map exists), or BUG (if such a map does not exist).
Demo Input:
['LLUUUR\n', 'RRUULLDD\n']
Demo Output:
['OK\n', 'BUG\n']
Note:
none | ```python
def robots(s):
visited = set()
p = [0, 0]
for i in s:
if i == 'R': p[0] += 1
elif i == 'L': p[0] -= 1
elif i == 'U': p[1] += 1
elif i == 'D': p[1] -= 1
if str(p) in visited: return ("BUG")
visited.add(str([p[0]-1, p[1]]))
visited.add(str([p[0]+1, p[1]]))
visited.add(str([p[0], p[1]-1]))
visited.add(str([p[0], p[1]+1]))
return ("OK")
s = input()
print(robots(s))
``` | 0 |
681 | A | A Good Contest | PROGRAMMING | 800 | [
"implementation"
] | null | null | Codeforces user' handle color depends on his rating — it is red if his rating is greater or equal to 2400; it is orange if his rating is less than 2400 but greater or equal to 2200, etc. Each time participant takes part in a rated contest, his rating is changed depending on his performance.
Anton wants the color of his handle to become red. He considers his performance in the rated contest to be good if he outscored some participant, whose handle was colored red before the contest and his rating has increased after it.
Anton has written a program that analyses contest results and determines whether he performed good or not. Are you able to do the same? | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants Anton has outscored in this contest .
The next *n* lines describe participants results: the *i*-th of them consists of a participant handle *name**i* and two integers *before**i* and *after**i* (<=-<=4000<=≤<=*before**i*,<=*after**i*<=≤<=4000) — participant's rating before and after the contest, respectively. Each handle is a non-empty string, consisting of no more than 10 characters, which might be lowercase and uppercase English letters, digits, characters «_» and «-» characters.
It is guaranteed that all handles are distinct. | Print «YES» (quotes for clarity), if Anton has performed good in the contest and «NO» (quotes for clarity) otherwise. | [
"3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749\n",
"3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450\n"
] | [
"YES",
"NO"
] | In the first sample, Anton has outscored user with handle Burunduk1, whose handle was colored red before the contest and his rating has increased after the contest.
In the second sample, Applejack's rating has not increased after the contest, while both Fluttershy's and Pinkie_Pie's handles were not colored red before the contest. | 500 | [
{
"input": "3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749",
"output": "YES"
},
{
"input": "3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450",
"output": "NO"
},
{
"input": "1\nDb -3373 3591",
"output": "NO"
},
{
"input": "5\nQ2bz 960 2342... | 1,561,213,214 | 2,147,483,647 | Python 3 | OK | TESTS | 60 | 109 | 0 | n = int(input())
flag = 'NO'
for i in range(n):
text, f, s = input().split()
if int(s) > int(f) >= 2400:
flag = 'YES'
print(flag) | Title: A Good Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Codeforces user' handle color depends on his rating — it is red if his rating is greater or equal to 2400; it is orange if his rating is less than 2400 but greater or equal to 2200, etc. Each time participant takes part in a rated contest, his rating is changed depending on his performance.
Anton wants the color of his handle to become red. He considers his performance in the rated contest to be good if he outscored some participant, whose handle was colored red before the contest and his rating has increased after it.
Anton has written a program that analyses contest results and determines whether he performed good or not. Are you able to do the same?
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants Anton has outscored in this contest .
The next *n* lines describe participants results: the *i*-th of them consists of a participant handle *name**i* and two integers *before**i* and *after**i* (<=-<=4000<=≤<=*before**i*,<=*after**i*<=≤<=4000) — participant's rating before and after the contest, respectively. Each handle is a non-empty string, consisting of no more than 10 characters, which might be lowercase and uppercase English letters, digits, characters «_» and «-» characters.
It is guaranteed that all handles are distinct.
Output Specification:
Print «YES» (quotes for clarity), if Anton has performed good in the contest and «NO» (quotes for clarity) otherwise.
Demo Input:
['3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749\n', '3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450\n']
Demo Output:
['YES', 'NO']
Note:
In the first sample, Anton has outscored user with handle Burunduk1, whose handle was colored red before the contest and his rating has increased after the contest.
In the second sample, Applejack's rating has not increased after the contest, while both Fluttershy's and Pinkie_Pie's handles were not colored red before the contest. | ```python
n = int(input())
flag = 'NO'
for i in range(n):
text, f, s = input().split()
if int(s) > int(f) >= 2400:
flag = 'YES'
print(flag)
``` | 3 | |
443 | A | Anton and Letters | PROGRAMMING | 800 | [
"constructive algorithms",
"implementation"
] | null | null | Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set. | The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space. | Print a single number — the number of distinct letters in Anton's set. | [
"{a, b, c}\n",
"{b, a, b, a}\n",
"{}\n"
] | [
"3\n",
"2\n",
"0\n"
] | none | 500 | [
{
"input": "{a, b, c}",
"output": "3"
},
{
"input": "{b, a, b, a}",
"output": "2"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{a, a, c, b, b, b, c, c, c, c}",
"output": "3"
},
{
"input": "{a, c, b, b}",
"output": "3"
},
{
"input": "{a, b}",
"o... | 1,693,950,888 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | print(len(set(list(input().split())[1:-1]))) | Title: Anton and Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
Input Specification:
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
Output Specification:
Print a single number — the number of distinct letters in Anton's set.
Demo Input:
['{a, b, c}\n', '{b, a, b, a}\n', '{}\n']
Demo Output:
['3\n', '2\n', '0\n']
Note:
none | ```python
print(len(set(list(input().split())[1:-1])))
``` | 0 | |
691 | A | Fashion in Berland | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | According to rules of the Berland fashion, a jacket should be fastened by all the buttons except only one, but not necessarily it should be the last one. Also if the jacket has only one button, it should be fastened, so the jacket will not swinging open.
You are given a jacket with *n* buttons. Determine if it is fastened in a right way. | The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of buttons on the jacket.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1). The number *a**i*<==<=0 if the *i*-th button is not fastened. Otherwise *a**i*<==<=1. | In the only line print the word "YES" if the jacket is fastened in a right way. Otherwise print the word "NO". | [
"3\n1 0 1\n",
"3\n1 0 0\n"
] | [
"YES\n",
"NO\n"
] | none | 0 | [
{
"input": "3\n1 0 1",
"output": "YES"
},
{
"input": "3\n1 0 0",
"output": "NO"
},
{
"input": "3\n1 1 0",
"output": "YES"
},
{
"input": "3\n1 1 1",
"output": "NO"
},
{
"input": "3\n0 0 1",
"output": "NO"
},
{
"input": "3\n0 0 0",
"output": "NO"
}... | 1,523,693,238 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 77 | 7,065,600 | n = input()
list = [int(x) for x in input().split()]
print(list)
result = True
k = list[0]
for i in range(1, len(list)):
if list[i] == list[i-1]:
result = False
break
if result:
print('YES')
else:
print('NO') | Title: Fashion in Berland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
According to rules of the Berland fashion, a jacket should be fastened by all the buttons except only one, but not necessarily it should be the last one. Also if the jacket has only one button, it should be fastened, so the jacket will not swinging open.
You are given a jacket with *n* buttons. Determine if it is fastened in a right way.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of buttons on the jacket.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1). The number *a**i*<==<=0 if the *i*-th button is not fastened. Otherwise *a**i*<==<=1.
Output Specification:
In the only line print the word "YES" if the jacket is fastened in a right way. Otherwise print the word "NO".
Demo Input:
['3\n1 0 1\n', '3\n1 0 0\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
n = input()
list = [int(x) for x in input().split()]
print(list)
result = True
k = list[0]
for i in range(1, len(list)):
if list[i] == list[i-1]:
result = False
break
if result:
print('YES')
else:
print('NO')
``` | 0 | |
798 | A | Mike and palindrome | PROGRAMMING | 1,000 | [
"brute force",
"constructive algorithms",
"strings"
] | null | null | Mike has a string *s* consisting of only lowercase English letters. He wants to change exactly one character from the string so that the resulting one is a palindrome.
A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not. | The first and single line contains string *s* (1<=≤<=|*s*|<=≤<=15). | Print "YES" (without quotes) if Mike can change exactly one character so that the resulting string is palindrome or "NO" (without quotes) otherwise. | [
"abccaa\n",
"abbcca\n",
"abcda\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "abccaa",
"output": "YES"
},
{
"input": "abbcca",
"output": "NO"
},
{
"input": "abcda",
"output": "YES"
},
{
"input": "kyw",
"output": "YES"
},
{
"input": "fccf",
"output": "NO"
},
{
"input": "mnlm",
"output": "YES"
},
{
"input":... | 1,637,732,392 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 7 | 31 | 0 | n=input()
l=len(n)
s=0
for i in range(l):
if n[i]!=n[l-i-1]:
s+=1
if s<=2 and s>=1:
print("YES")
else:
print("NO") | Title: Mike and palindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mike has a string *s* consisting of only lowercase English letters. He wants to change exactly one character from the string so that the resulting one is a palindrome.
A palindrome is a string that reads the same backward as forward, for example strings "z", "aaa", "aba", "abccba" are palindromes, but strings "codeforces", "reality", "ab" are not.
Input Specification:
The first and single line contains string *s* (1<=≤<=|*s*|<=≤<=15).
Output Specification:
Print "YES" (without quotes) if Mike can change exactly one character so that the resulting string is palindrome or "NO" (without quotes) otherwise.
Demo Input:
['abccaa\n', 'abbcca\n', 'abcda\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none | ```python
n=input()
l=len(n)
s=0
for i in range(l):
if n[i]!=n[l-i-1]:
s+=1
if s<=2 and s>=1:
print("YES")
else:
print("NO")
``` | 0 | |
629 | B | Far Relative’s Problem | PROGRAMMING | 1,100 | [
"brute force"
] | null | null | Famil Door wants to celebrate his birthday with his friends from Far Far Away. He has *n* friends and each of them can come to the party in a specific range of days of the year from *a**i* to *b**i*. Of course, Famil Door wants to have as many friends celebrating together with him as possible.
Far cars are as weird as Far Far Away citizens, so they can only carry two people of opposite gender, that is exactly one male and one female. However, Far is so far from here that no other transportation may be used to get to the party.
Famil Door should select some day of the year and invite some of his friends, such that they all are available at this moment and the number of male friends invited is equal to the number of female friends invited. Find the maximum number of friends that may present at the party. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=5000) — then number of Famil Door's friends.
Then follow *n* lines, that describe the friends. Each line starts with a capital letter 'F' for female friends and with a capital letter 'M' for male friends. Then follow two integers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=*b**i*<=≤<=366), providing that the *i*-th friend can come to the party from day *a**i* to day *b**i* inclusive. | Print the maximum number of people that may come to Famil Door's party. | [
"4\nM 151 307\nF 343 352\nF 117 145\nM 24 128\n",
"6\nM 128 130\nF 128 131\nF 131 140\nF 131 141\nM 131 200\nM 140 200\n"
] | [
"2\n",
"4\n"
] | In the first sample, friends 3 and 4 can come on any day in range [117, 128].
In the second sample, friends with indices 3, 4, 5 and 6 can come on day 140. | 1,000 | [
{
"input": "4\nM 151 307\nF 343 352\nF 117 145\nM 24 128",
"output": "2"
},
{
"input": "6\nM 128 130\nF 128 131\nF 131 140\nF 131 141\nM 131 200\nM 140 200",
"output": "4"
},
{
"input": "1\nF 68 307",
"output": "0"
},
{
"input": "40\nM 55 363\nF 117 252\nM 157 282\nF 322 345\... | 1,489,481,431 | 2,147,483,647 | Python 3 | OK | TESTS | 76 | 546 | 5,017,600 | n = int(input())
gender = []
fromdate = []
todate = []
for k in range(n):
a, b, c = input().split()
gender.append(a)
fromdate.append(int(b))
todate.append(int(c))
minimum = min(fromdate)
maximum = max(todate)
#print("min = {} max = {}".format(minimum,maximum))
ans = 0
for i in range(minimum, maximum + 1):
male = 0
women = 0
for j in range(n):
if fromdate[j] <= i and todate[j] >= i:
if gender[j] == 'M':
male += 1
else:
women += 1
count = min(male, women) * 2
#print(count)
if count > ans:
ans = count
print(ans)
| Title: Far Relative’s Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Famil Door wants to celebrate his birthday with his friends from Far Far Away. He has *n* friends and each of them can come to the party in a specific range of days of the year from *a**i* to *b**i*. Of course, Famil Door wants to have as many friends celebrating together with him as possible.
Far cars are as weird as Far Far Away citizens, so they can only carry two people of opposite gender, that is exactly one male and one female. However, Far is so far from here that no other transportation may be used to get to the party.
Famil Door should select some day of the year and invite some of his friends, such that they all are available at this moment and the number of male friends invited is equal to the number of female friends invited. Find the maximum number of friends that may present at the party.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=5000) — then number of Famil Door's friends.
Then follow *n* lines, that describe the friends. Each line starts with a capital letter 'F' for female friends and with a capital letter 'M' for male friends. Then follow two integers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=*b**i*<=≤<=366), providing that the *i*-th friend can come to the party from day *a**i* to day *b**i* inclusive.
Output Specification:
Print the maximum number of people that may come to Famil Door's party.
Demo Input:
['4\nM 151 307\nF 343 352\nF 117 145\nM 24 128\n', '6\nM 128 130\nF 128 131\nF 131 140\nF 131 141\nM 131 200\nM 140 200\n']
Demo Output:
['2\n', '4\n']
Note:
In the first sample, friends 3 and 4 can come on any day in range [117, 128].
In the second sample, friends with indices 3, 4, 5 and 6 can come on day 140. | ```python
n = int(input())
gender = []
fromdate = []
todate = []
for k in range(n):
a, b, c = input().split()
gender.append(a)
fromdate.append(int(b))
todate.append(int(c))
minimum = min(fromdate)
maximum = max(todate)
#print("min = {} max = {}".format(minimum,maximum))
ans = 0
for i in range(minimum, maximum + 1):
male = 0
women = 0
for j in range(n):
if fromdate[j] <= i and todate[j] >= i:
if gender[j] == 'M':
male += 1
else:
women += 1
count = min(male, women) * 2
#print(count)
if count > ans:
ans = count
print(ans)
``` | 3 | |
771 | A | Bear and Friendship Condition | PROGRAMMING | 1,500 | [
"dfs and similar",
"dsu",
"graphs"
] | null | null | Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).
There are *n* members, numbered 1 through *n*. *m* pairs of members are friends. Of course, a member can't be a friend with themselves.
Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.
For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.
Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes. | The first line of the input contain two integers *n* and *m* (3<=≤<=*n*<=≤<=150<=000, ) — the number of members and the number of pairs of members that are friends.
The *i*-th of the next *m* lines contains two distinct integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*). Members *a**i* and *b**i* are friends with each other. No pair of members will appear more than once in the input. | If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes). | [
"4 3\n1 3\n3 4\n1 4\n",
"4 4\n3 1\n2 3\n3 4\n1 2\n",
"10 4\n4 3\n5 10\n8 9\n1 2\n",
"3 2\n1 2\n2 3\n"
] | [
"YES\n",
"NO\n",
"YES\n",
"NO\n"
] | The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not. | 250 | [
{
"input": "4 3\n1 3\n3 4\n1 4",
"output": "YES"
},
{
"input": "4 4\n3 1\n2 3\n3 4\n1 2",
"output": "NO"
},
{
"input": "10 4\n4 3\n5 10\n8 9\n1 2",
"output": "YES"
},
{
"input": "3 2\n1 2\n2 3",
"output": "NO"
},
{
"input": "3 0",
"output": "YES"
},
{
... | 1,630,293,735 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 9 | 1,000 | 110,182,400 | from sys import stdin
from collections import deque
def main():
input = stdin.readline
n,m = [int(i) for i in input().split()]
graph = {i+1:set() for i in range(n)}
for _ in range(m):
a,b = [int(i) for i in input().split()]
graph[a].add(b)
graph[b].add(a)
INF = 10**10
seen = set()
for i in range(1,n+1):
que = deque([(i, INF, INF)])
seen.add((i, INF, INF))
while que:
cur, prev, pprev = que.popleft()
# print(cur,prev,pprev)
if pprev != INF and pprev not in graph[cur]:
print("NO")
exit()
for nxt in graph[cur]:
t = (nxt, cur, prev)
if t in seen or nxt == prev:
continue
seen.add(t)
que.append(t)
print("YES")
if __name__ == '__main__':
main()
| Title: Bear and Friendship Condition
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).
There are *n* members, numbered 1 through *n*. *m* pairs of members are friends. Of course, a member can't be a friend with themselves.
Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.
For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.
Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes.
Input Specification:
The first line of the input contain two integers *n* and *m* (3<=≤<=*n*<=≤<=150<=000, ) — the number of members and the number of pairs of members that are friends.
The *i*-th of the next *m* lines contains two distinct integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=*a**i*<=≠<=*b**i*). Members *a**i* and *b**i* are friends with each other. No pair of members will appear more than once in the input.
Output Specification:
If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes).
Demo Input:
['4 3\n1 3\n3 4\n1 4\n', '4 4\n3 1\n2 3\n3 4\n1 2\n', '10 4\n4 3\n5 10\n8 9\n1 2\n', '3 2\n1 2\n2 3\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'NO\n']
Note:
The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not. | ```python
from sys import stdin
from collections import deque
def main():
input = stdin.readline
n,m = [int(i) for i in input().split()]
graph = {i+1:set() for i in range(n)}
for _ in range(m):
a,b = [int(i) for i in input().split()]
graph[a].add(b)
graph[b].add(a)
INF = 10**10
seen = set()
for i in range(1,n+1):
que = deque([(i, INF, INF)])
seen.add((i, INF, INF))
while que:
cur, prev, pprev = que.popleft()
# print(cur,prev,pprev)
if pprev != INF and pprev not in graph[cur]:
print("NO")
exit()
for nxt in graph[cur]:
t = (nxt, cur, prev)
if t in seen or nxt == prev:
continue
seen.add(t)
que.append(t)
print("YES")
if __name__ == '__main__':
main()
``` | 0 | |
601 | A | The Two Routes | PROGRAMMING | 1,600 | [
"graphs",
"shortest paths"
] | null | null | In Absurdistan, there are *n* towns (numbered 1 through *n*) and *m* bidirectional railways. There is also an absurdly simple road network — for each pair of different towns *x* and *y*, there is a bidirectional road between towns *x* and *y* if and only if there is no railway between them. Travelling to a different town using one railway or one road always takes exactly one hour.
A train and a bus leave town 1 at the same time. They both have the same destination, town *n*, and don't make any stops on the way (but they can wait in town *n*). The train can move only along railways and the bus can move only along roads.
You've been asked to plan out routes for the vehicles; each route can use any road/railway multiple times. One of the most important aspects to consider is safety — in order to avoid accidents at railway crossings, the train and the bus must not arrive at the same town (except town *n*) simultaneously.
Under these constraints, what is the minimum number of hours needed for both vehicles to reach town *n* (the maximum of arrival times of the bus and the train)? Note, that bus and train are not required to arrive to the town *n* at the same moment of time, but are allowed to do so. | The first line of the input contains two integers *n* and *m* (2<=≤<=*n*<=≤<=400, 0<=≤<=*m*<=≤<=*n*(*n*<=-<=1)<=/<=2) — the number of towns and the number of railways respectively.
Each of the next *m* lines contains two integers *u* and *v*, denoting a railway between towns *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*).
You may assume that there is at most one railway connecting any two towns. | Output one integer — the smallest possible time of the later vehicle's arrival in town *n*. If it's impossible for at least one of the vehicles to reach town *n*, output <=-<=1. | [
"4 2\n1 3\n3 4\n",
"4 6\n1 2\n1 3\n1 4\n2 3\n2 4\n3 4\n",
"5 5\n4 2\n3 5\n4 5\n5 1\n1 2\n"
] | [
"2\n",
"-1\n",
"3\n"
] | In the first sample, the train can take the route <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7c0aa60a06309ef607b7159fd7f3687ea0d943ce.png" style="max-width: 100.0%;max-height: 100.0%;"/> and the bus can take the route <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/a26c2f3e93c9d9be6c21cb5d2bd6ac1f99f4ff55.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Note that they can arrive at town 4 at the same time.
In the second sample, Absurdistan is ruled by railwaymen. There are no roads, so there's no way for the bus to reach town 4. | 500 | [
{
"input": "4 2\n1 3\n3 4",
"output": "2"
},
{
"input": "4 6\n1 2\n1 3\n1 4\n2 3\n2 4\n3 4",
"output": "-1"
},
{
"input": "5 5\n4 2\n3 5\n4 5\n5 1\n1 2",
"output": "3"
},
{
"input": "5 4\n1 2\n3 2\n3 4\n5 4",
"output": "4"
},
{
"input": "3 1\n1 2",
"output": "... | 1,690,044,953 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 11 | 46 | 0 | import math
def main():
n, m = map(int, input().split())
graph = [[False] * (n + 1) for _ in range(n + 1)]
for _ in range(m):
a, b = map(int, input().split())
graph[a][b] = True
graph[b][a] = True
dp = [float('inf')] * (n + 1)
dp[1] = 0
for i in range(2, n + 1):
for j in range(1, i):
if graph[i][j]:
dp[i] = min(dp[i], dp[j] + 1)
dp[j] = min(dp[j], dp[i] + 1)
train_time = dp[-1]
if math.isinf(train_time):
print(-1)
return
dp = [float('inf')] * (n + 1)
dp[1] = 0
for i in range(2, n + 1):
for j in range(1, i):
if not graph[i][j]:
dp[i] = min(dp[i], dp[j] + 1)
dp[j] = min(dp[j], dp[i] + 1)
bus_time = dp[-1]
if math.isinf(bus_time):
print(-1)
return
print(max(train_time, bus_time))
main()
| Title: The Two Routes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Absurdistan, there are *n* towns (numbered 1 through *n*) and *m* bidirectional railways. There is also an absurdly simple road network — for each pair of different towns *x* and *y*, there is a bidirectional road between towns *x* and *y* if and only if there is no railway between them. Travelling to a different town using one railway or one road always takes exactly one hour.
A train and a bus leave town 1 at the same time. They both have the same destination, town *n*, and don't make any stops on the way (but they can wait in town *n*). The train can move only along railways and the bus can move only along roads.
You've been asked to plan out routes for the vehicles; each route can use any road/railway multiple times. One of the most important aspects to consider is safety — in order to avoid accidents at railway crossings, the train and the bus must not arrive at the same town (except town *n*) simultaneously.
Under these constraints, what is the minimum number of hours needed for both vehicles to reach town *n* (the maximum of arrival times of the bus and the train)? Note, that bus and train are not required to arrive to the town *n* at the same moment of time, but are allowed to do so.
Input Specification:
The first line of the input contains two integers *n* and *m* (2<=≤<=*n*<=≤<=400, 0<=≤<=*m*<=≤<=*n*(*n*<=-<=1)<=/<=2) — the number of towns and the number of railways respectively.
Each of the next *m* lines contains two integers *u* and *v*, denoting a railway between towns *u* and *v* (1<=≤<=*u*,<=*v*<=≤<=*n*, *u*<=≠<=*v*).
You may assume that there is at most one railway connecting any two towns.
Output Specification:
Output one integer — the smallest possible time of the later vehicle's arrival in town *n*. If it's impossible for at least one of the vehicles to reach town *n*, output <=-<=1.
Demo Input:
['4 2\n1 3\n3 4\n', '4 6\n1 2\n1 3\n1 4\n2 3\n2 4\n3 4\n', '5 5\n4 2\n3 5\n4 5\n5 1\n1 2\n']
Demo Output:
['2\n', '-1\n', '3\n']
Note:
In the first sample, the train can take the route <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7c0aa60a06309ef607b7159fd7f3687ea0d943ce.png" style="max-width: 100.0%;max-height: 100.0%;"/> and the bus can take the route <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/a26c2f3e93c9d9be6c21cb5d2bd6ac1f99f4ff55.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Note that they can arrive at town 4 at the same time.
In the second sample, Absurdistan is ruled by railwaymen. There are no roads, so there's no way for the bus to reach town 4. | ```python
import math
def main():
n, m = map(int, input().split())
graph = [[False] * (n + 1) for _ in range(n + 1)]
for _ in range(m):
a, b = map(int, input().split())
graph[a][b] = True
graph[b][a] = True
dp = [float('inf')] * (n + 1)
dp[1] = 0
for i in range(2, n + 1):
for j in range(1, i):
if graph[i][j]:
dp[i] = min(dp[i], dp[j] + 1)
dp[j] = min(dp[j], dp[i] + 1)
train_time = dp[-1]
if math.isinf(train_time):
print(-1)
return
dp = [float('inf')] * (n + 1)
dp[1] = 0
for i in range(2, n + 1):
for j in range(1, i):
if not graph[i][j]:
dp[i] = min(dp[i], dp[j] + 1)
dp[j] = min(dp[j], dp[i] + 1)
bus_time = dp[-1]
if math.isinf(bus_time):
print(-1)
return
print(max(train_time, bus_time))
main()
``` | 0 | |
678 | D | Iterated Linear Function | PROGRAMMING | 1,700 | [
"math",
"number theory"
] | null | null | Consider a linear function *f*(*x*)<==<=*Ax*<=+<=*B*. Let's define *g*(0)(*x*)<==<=*x* and *g*(*n*)(*x*)<==<=*f*(*g*(*n*<=-<=1)(*x*)) for *n*<=><=0. For the given integer values *A*, *B*, *n* and *x* find the value of *g*(*n*)(*x*) modulo 109<=+<=7. | The only line contains four integers *A*, *B*, *n* and *x* (1<=≤<=*A*,<=*B*,<=*x*<=≤<=109,<=1<=≤<=*n*<=≤<=1018) — the parameters from the problem statement.
Note that the given value *n* can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type. | Print the only integer *s* — the value *g*(*n*)(*x*) modulo 109<=+<=7. | [
"3 4 1 1\n",
"3 4 2 1\n",
"3 4 3 1\n"
] | [
"7\n",
"25\n",
"79\n"
] | none | 0 | [
{
"input": "3 4 1 1",
"output": "7"
},
{
"input": "3 4 2 1",
"output": "25"
},
{
"input": "3 4 3 1",
"output": "79"
},
{
"input": "1 1 1 1",
"output": "2"
},
{
"input": "3 10 723 6",
"output": "443623217"
},
{
"input": "14 81 51 82",
"output": "908... | 1,615,326,218 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<utility>
#include<tuple>
#include<queue>
#include<stack>
#include<deque>
#include<map>
#include<set>
#include<string.h>
#include<assert.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
const ll oo = 1e18;
const ll MOD = 1e9 + 7;
ll a, b, n, x;
ll binpow(ll a, ll b) {
a %= MOD;
ll res = 1;
while (b > 0) {
if (b & 1)
res = res * a % MOD;
a = a * a % MOD;
b >>= 1;
}
return res;
}
int main(){
cin >> a >> b >> n >> x;
}
| Title: Iterated Linear Function
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Consider a linear function *f*(*x*)<==<=*Ax*<=+<=*B*. Let's define *g*(0)(*x*)<==<=*x* and *g*(*n*)(*x*)<==<=*f*(*g*(*n*<=-<=1)(*x*)) for *n*<=><=0. For the given integer values *A*, *B*, *n* and *x* find the value of *g*(*n*)(*x*) modulo 109<=+<=7.
Input Specification:
The only line contains four integers *A*, *B*, *n* and *x* (1<=≤<=*A*,<=*B*,<=*x*<=≤<=109,<=1<=≤<=*n*<=≤<=1018) — the parameters from the problem statement.
Note that the given value *n* can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Output Specification:
Print the only integer *s* — the value *g*(*n*)(*x*) modulo 109<=+<=7.
Demo Input:
['3 4 1 1\n', '3 4 2 1\n', '3 4 3 1\n']
Demo Output:
['7\n', '25\n', '79\n']
Note:
none | ```python
#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<utility>
#include<tuple>
#include<queue>
#include<stack>
#include<deque>
#include<map>
#include<set>
#include<string.h>
#include<assert.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
const ll oo = 1e18;
const ll MOD = 1e9 + 7;
ll a, b, n, x;
ll binpow(ll a, ll b) {
a %= MOD;
ll res = 1;
while (b > 0) {
if (b & 1)
res = res * a % MOD;
a = a * a % MOD;
b >>= 1;
}
return res;
}
int main(){
cin >> a >> b >> n >> x;
}
``` | -1 | |
769 | B | News About Credit | PROGRAMMING | 1,200 | [
"*special",
"greedy",
"two pointers"
] | null | null | Polycarp studies at the university in the group which consists of *n* students (including himself). All they are registrated in the social net "TheContacnt!".
Not all students are equally sociable. About each student you know the value *a**i* — the maximum number of messages which the *i*-th student is agree to send per day. The student can't send messages to himself.
In early morning Polycarp knew important news that the programming credit will be tomorrow. For this reason it is necessary to urgently inform all groupmates about this news using private messages.
Your task is to make a plan of using private messages, so that:
- the student *i* sends no more than *a**i* messages (for all *i* from 1 to *n*); - all students knew the news about the credit (initially only Polycarp knew it); - the student can inform the other student only if he knows it himself.
Let's consider that all students are numerated by distinct numbers from 1 to *n*, and Polycarp always has the number 1.
In that task you shouldn't minimize the number of messages, the moment of time, when all knew about credit or some other parameters. Find any way how to use private messages which satisfies requirements above. | The first line contains the positive integer *n* (2<=≤<=*n*<=≤<=100) — the number of students.
The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* equals to the maximum number of messages which can the *i*-th student agree to send. Consider that Polycarp always has the number 1. | Print -1 to the first line if it is impossible to inform all students about credit.
Otherwise, in the first line print the integer *k* — the number of messages which will be sent. In each of the next *k* lines print two distinct integers *f* and *t*, meaning that the student number *f* sent the message with news to the student number *t*. All messages should be printed in chronological order. It means that the student, who is sending the message, must already know this news. It is assumed that students can receive repeated messages with news of the credit.
If there are several answers, it is acceptable to print any of them. | [
"4\n1 2 1 0\n",
"6\n2 0 1 3 2 0\n",
"3\n0 2 2\n"
] | [
"3\n1 2\n2 4\n2 3\n",
"6\n1 3\n3 4\n1 2\n4 5\n5 6\n4 6\n",
"-1\n"
] | In the first test Polycarp (the student number 1) can send the message to the student number 2, who after that can send the message to students number 3 and 4. Thus, all students knew about the credit. | 1,000 | [
{
"input": "4\n1 2 1 0",
"output": "3\n1 2\n2 3\n2 4"
},
{
"input": "6\n2 0 1 3 2 0",
"output": "5\n1 4\n1 5\n4 3\n4 2\n4 6"
},
{
"input": "3\n0 2 2",
"output": "-1"
},
{
"input": "2\n0 0",
"output": "-1"
},
{
"input": "2\n1 0",
"output": "1\n1 2"
},
{
... | 1,546,334,404 | 2,147,483,647 | PyPy 3 | OK | TESTS | 60 | 140 | 0 | n = int(input())
call = list(map(int, input().split()))
cp = []
ans = []
for i in range(1, n):
x = call[i]
cp.append([x, i + 1])
cp.sort(reverse=True)
l = [[call[0], 1]]
j = 0
while len(cp) > 0 and len(l) > 0 and j < len(cp):
s = l[0]
x = s[0]
p = s[1]
for i in range(j, min(j + x, len(cp))):
ans.append([str(p), str(cp[i][1])])
l.append(cp[i])
j = min(j + x, len(cp))
l.remove(l[0])
if j < len(cp):
print(-1)
else:
print(len(ans))
for i in range(len(ans)):
print(' '.join(ans[i])) | Title: News About Credit
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp studies at the university in the group which consists of *n* students (including himself). All they are registrated in the social net "TheContacnt!".
Not all students are equally sociable. About each student you know the value *a**i* — the maximum number of messages which the *i*-th student is agree to send per day. The student can't send messages to himself.
In early morning Polycarp knew important news that the programming credit will be tomorrow. For this reason it is necessary to urgently inform all groupmates about this news using private messages.
Your task is to make a plan of using private messages, so that:
- the student *i* sends no more than *a**i* messages (for all *i* from 1 to *n*); - all students knew the news about the credit (initially only Polycarp knew it); - the student can inform the other student only if he knows it himself.
Let's consider that all students are numerated by distinct numbers from 1 to *n*, and Polycarp always has the number 1.
In that task you shouldn't minimize the number of messages, the moment of time, when all knew about credit or some other parameters. Find any way how to use private messages which satisfies requirements above.
Input Specification:
The first line contains the positive integer *n* (2<=≤<=*n*<=≤<=100) — the number of students.
The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* equals to the maximum number of messages which can the *i*-th student agree to send. Consider that Polycarp always has the number 1.
Output Specification:
Print -1 to the first line if it is impossible to inform all students about credit.
Otherwise, in the first line print the integer *k* — the number of messages which will be sent. In each of the next *k* lines print two distinct integers *f* and *t*, meaning that the student number *f* sent the message with news to the student number *t*. All messages should be printed in chronological order. It means that the student, who is sending the message, must already know this news. It is assumed that students can receive repeated messages with news of the credit.
If there are several answers, it is acceptable to print any of them.
Demo Input:
['4\n1 2 1 0\n', '6\n2 0 1 3 2 0\n', '3\n0 2 2\n']
Demo Output:
['3\n1 2\n2 4\n2 3\n', '6\n1 3\n3 4\n1 2\n4 5\n5 6\n4 6\n', '-1\n']
Note:
In the first test Polycarp (the student number 1) can send the message to the student number 2, who after that can send the message to students number 3 and 4. Thus, all students knew about the credit. | ```python
n = int(input())
call = list(map(int, input().split()))
cp = []
ans = []
for i in range(1, n):
x = call[i]
cp.append([x, i + 1])
cp.sort(reverse=True)
l = [[call[0], 1]]
j = 0
while len(cp) > 0 and len(l) > 0 and j < len(cp):
s = l[0]
x = s[0]
p = s[1]
for i in range(j, min(j + x, len(cp))):
ans.append([str(p), str(cp[i][1])])
l.append(cp[i])
j = min(j + x, len(cp))
l.remove(l[0])
if j < len(cp):
print(-1)
else:
print(len(ans))
for i in range(len(ans)):
print(' '.join(ans[i]))
``` | 3 | |
1,005 | A | Tanya and Stairways | PROGRAMMING | 800 | [
"implementation"
] | null | null | Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairway, she starts counting steps from $1$ to the number of steps in this stairway. She speaks every number aloud. For example, if she climbs two stairways, the first of which contains $3$ steps, and the second contains $4$ steps, she will pronounce the numbers $1, 2, 3, 1, 2, 3, 4$.
You are given all the numbers pronounced by Tanya. How many stairways did she climb? Also, output the number of steps in each stairway.
The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways. | The first line contains $n$ ($1 \le n \le 1000$) — the total number of numbers pronounced by Tanya.
The second line contains integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — all the numbers Tanya pronounced while climbing the stairs, in order from the first to the last pronounced number. Passing a stairway with $x$ steps, she will pronounce the numbers $1, 2, \dots, x$ in that order.
The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways. | In the first line, output $t$ — the number of stairways that Tanya climbed. In the second line, output $t$ numbers — the number of steps in each stairway she climbed. Write the numbers in the correct order of passage of the stairways. | [
"7\n1 2 3 1 2 3 4\n",
"4\n1 1 1 1\n",
"5\n1 2 3 4 5\n",
"5\n1 2 1 2 1\n"
] | [
"2\n3 4 ",
"4\n1 1 1 1 ",
"1\n5 ",
"3\n2 2 1 "
] | none | 0 | [
{
"input": "7\n1 2 3 1 2 3 4",
"output": "2\n3 4 "
},
{
"input": "4\n1 1 1 1",
"output": "4\n1 1 1 1 "
},
{
"input": "5\n1 2 3 4 5",
"output": "1\n5 "
},
{
"input": "5\n1 2 1 2 1",
"output": "3\n2 2 1 "
},
{
"input": "1\n1",
"output": "1\n1 "
},
{
"inp... | 1,637,832,748 | 2,147,483,647 | Python 3 | OK | TESTS | 16 | 62 | 0 | n = int(input())
li = list(map(int, input().split()))
count = 1
newli = []
for i in range(1,n):
if li[i]<li[i-1]:
count +=1
newli.append(li[i-1])
if li[i]==li[i-1]:
count +=1
newli.append(li[i-1])
newli.append(li[-1])
print(count)
print(" ".join([str(x) for x in newli])) | Title: Tanya and Stairways
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little girl Tanya climbs the stairs inside a multi-storey building. Every time Tanya climbs a stairway, she starts counting steps from $1$ to the number of steps in this stairway. She speaks every number aloud. For example, if she climbs two stairways, the first of which contains $3$ steps, and the second contains $4$ steps, she will pronounce the numbers $1, 2, 3, 1, 2, 3, 4$.
You are given all the numbers pronounced by Tanya. How many stairways did she climb? Also, output the number of steps in each stairway.
The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.
Input Specification:
The first line contains $n$ ($1 \le n \le 1000$) — the total number of numbers pronounced by Tanya.
The second line contains integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — all the numbers Tanya pronounced while climbing the stairs, in order from the first to the last pronounced number. Passing a stairway with $x$ steps, she will pronounce the numbers $1, 2, \dots, x$ in that order.
The given sequence will be a valid sequence that Tanya could have pronounced when climbing one or more stairways.
Output Specification:
In the first line, output $t$ — the number of stairways that Tanya climbed. In the second line, output $t$ numbers — the number of steps in each stairway she climbed. Write the numbers in the correct order of passage of the stairways.
Demo Input:
['7\n1 2 3 1 2 3 4\n', '4\n1 1 1 1\n', '5\n1 2 3 4 5\n', '5\n1 2 1 2 1\n']
Demo Output:
['2\n3 4 ', '4\n1 1 1 1 ', '1\n5 ', '3\n2 2 1 ']
Note:
none | ```python
n = int(input())
li = list(map(int, input().split()))
count = 1
newli = []
for i in range(1,n):
if li[i]<li[i-1]:
count +=1
newli.append(li[i-1])
if li[i]==li[i-1]:
count +=1
newli.append(li[i-1])
newli.append(li[-1])
print(count)
print(" ".join([str(x) for x in newli]))
``` | 3 | |
631 | B | Print Check | PROGRAMMING | 1,200 | [
"constructive algorithms",
"implementation"
] | null | null | Kris works in a large company "Blake Technologies". As a best engineer of the company he was assigned a task to develop a printer that will be able to print horizontal and vertical strips. First prototype is already built and Kris wants to tests it. He wants you to implement the program that checks the result of the printing.
Printer works with a rectangular sheet of paper of size *n*<=×<=*m*. Consider the list as a table consisting of *n* rows and *m* columns. Rows are numbered from top to bottom with integers from 1 to *n*, while columns are numbered from left to right with integers from 1 to *m*. Initially, all cells are painted in color 0.
Your program has to support two operations:
1. Paint all cells in row *r**i* in color *a**i*; 1. Paint all cells in column *c**i* in color *a**i*.
If during some operation *i* there is a cell that have already been painted, the color of this cell also changes to *a**i*.
Your program has to print the resulting table after *k* operation. | The first line of the input contains three integers *n*, *m* and *k* (1<=<=≤<=<=*n*,<=<=*m*<=<=≤<=5000, *n*·*m*<=≤<=100<=000, 1<=≤<=*k*<=≤<=100<=000) — the dimensions of the sheet and the number of operations, respectively.
Each of the next *k* lines contains the description of exactly one query:
- 1 *r**i* *a**i* (1<=≤<=*r**i*<=≤<=*n*, 1<=≤<=*a**i*<=≤<=109), means that row *r**i* is painted in color *a**i*; - 2 *c**i* *a**i* (1<=≤<=*c**i*<=≤<=*m*, 1<=≤<=*a**i*<=≤<=109), means that column *c**i* is painted in color *a**i*. | Print *n* lines containing *m* integers each — the resulting table after all operations are applied. | [
"3 3 3\n1 1 3\n2 2 1\n1 2 2\n",
"5 3 5\n1 1 1\n1 3 1\n1 5 1\n2 1 1\n2 3 1\n"
] | [
"3 1 3 \n2 2 2 \n0 1 0 \n",
"1 1 1 \n1 0 1 \n1 1 1 \n1 0 1 \n1 1 1 \n"
] | The figure below shows all three operations for the first sample step by step. The cells that were painted on the corresponding step are marked gray. | 1,000 | [
{
"input": "3 3 3\n1 1 3\n2 2 1\n1 2 2",
"output": "3 1 3 \n2 2 2 \n0 1 0 "
},
{
"input": "5 3 5\n1 1 1\n1 3 1\n1 5 1\n2 1 1\n2 3 1",
"output": "1 1 1 \n1 0 1 \n1 1 1 \n1 0 1 \n1 1 1 "
},
{
"input": "5 5 4\n1 2 1\n1 4 1\n2 2 1\n2 4 1",
"output": "0 1 0 1 0 \n1 1 1 1 1 \n0 1 0 1 0 \n1... | 1,582,505,857 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 10 | 1,000 | 1,126,400 | n, m, k = map(int, input().split())
ary = [[0 for i in range(m)]for j in range(n)]
for i in range(k):
x, r, a = map(int, input().split())
if x == 1:
for j in range(m):
ary[r-1][j] = a
else:
for j in range(n):
ary[j][r-1] = a
for i in range(n):
iar = map(str, ary[i])
print (" ".join(iar))
| Title: Print Check
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kris works in a large company "Blake Technologies". As a best engineer of the company he was assigned a task to develop a printer that will be able to print horizontal and vertical strips. First prototype is already built and Kris wants to tests it. He wants you to implement the program that checks the result of the printing.
Printer works with a rectangular sheet of paper of size *n*<=×<=*m*. Consider the list as a table consisting of *n* rows and *m* columns. Rows are numbered from top to bottom with integers from 1 to *n*, while columns are numbered from left to right with integers from 1 to *m*. Initially, all cells are painted in color 0.
Your program has to support two operations:
1. Paint all cells in row *r**i* in color *a**i*; 1. Paint all cells in column *c**i* in color *a**i*.
If during some operation *i* there is a cell that have already been painted, the color of this cell also changes to *a**i*.
Your program has to print the resulting table after *k* operation.
Input Specification:
The first line of the input contains three integers *n*, *m* and *k* (1<=<=≤<=<=*n*,<=<=*m*<=<=≤<=5000, *n*·*m*<=≤<=100<=000, 1<=≤<=*k*<=≤<=100<=000) — the dimensions of the sheet and the number of operations, respectively.
Each of the next *k* lines contains the description of exactly one query:
- 1 *r**i* *a**i* (1<=≤<=*r**i*<=≤<=*n*, 1<=≤<=*a**i*<=≤<=109), means that row *r**i* is painted in color *a**i*; - 2 *c**i* *a**i* (1<=≤<=*c**i*<=≤<=*m*, 1<=≤<=*a**i*<=≤<=109), means that column *c**i* is painted in color *a**i*.
Output Specification:
Print *n* lines containing *m* integers each — the resulting table after all operations are applied.
Demo Input:
['3 3 3\n1 1 3\n2 2 1\n1 2 2\n', '5 3 5\n1 1 1\n1 3 1\n1 5 1\n2 1 1\n2 3 1\n']
Demo Output:
['3 1 3 \n2 2 2 \n0 1 0 \n', '1 1 1 \n1 0 1 \n1 1 1 \n1 0 1 \n1 1 1 \n']
Note:
The figure below shows all three operations for the first sample step by step. The cells that were painted on the corresponding step are marked gray. | ```python
n, m, k = map(int, input().split())
ary = [[0 for i in range(m)]for j in range(n)]
for i in range(k):
x, r, a = map(int, input().split())
if x == 1:
for j in range(m):
ary[r-1][j] = a
else:
for j in range(n):
ary[j][r-1] = a
for i in range(n):
iar = map(str, ary[i])
print (" ".join(iar))
``` | 0 | |
255 | A | Greg's Workout | PROGRAMMING | 800 | [
"implementation"
] | null | null | Greg is a beginner bodybuilder. Today the gym coach gave him the training plan. All it had was *n* integers *a*1,<=*a*2,<=...,<=*a**n*. These numbers mean that Greg needs to do exactly *n* exercises today. Besides, Greg should repeat the *i*-th in order exercise *a**i* times.
Greg now only does three types of exercises: "chest" exercises, "biceps" exercises and "back" exercises. Besides, his training is cyclic, that is, the first exercise he does is a "chest" one, the second one is "biceps", the third one is "back", the fourth one is "chest", the fifth one is "biceps", and so on to the *n*-th exercise.
Now Greg wonders, which muscle will get the most exercise during his training. We know that the exercise Greg repeats the maximum number of times, trains the corresponding muscle the most. Help Greg, determine which muscle will get the most training. | The first line contains integer *n* (1<=≤<=*n*<=≤<=20). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=25) — the number of times Greg repeats the exercises. | Print word "chest" (without the quotes), if the chest gets the most exercise, "biceps" (without the quotes), if the biceps gets the most exercise and print "back" (without the quotes) if the back gets the most exercise.
It is guaranteed that the input is such that the answer to the problem is unambiguous. | [
"2\n2 8\n",
"3\n5 1 10\n",
"7\n3 3 2 7 9 6 8\n"
] | [
"biceps\n",
"back\n",
"chest\n"
] | In the first sample Greg does 2 chest, 8 biceps and zero back exercises, so the biceps gets the most exercises.
In the second sample Greg does 5 chest, 1 biceps and 10 back exercises, so the back gets the most exercises.
In the third sample Greg does 18 chest, 12 biceps and 8 back exercises, so the chest gets the most exercise. | 500 | [
{
"input": "2\n2 8",
"output": "biceps"
},
{
"input": "3\n5 1 10",
"output": "back"
},
{
"input": "7\n3 3 2 7 9 6 8",
"output": "chest"
},
{
"input": "4\n5 6 6 2",
"output": "chest"
},
{
"input": "5\n8 2 2 6 3",
"output": "chest"
},
{
"input": "6\n8 7 ... | 1,649,615,359 | 2,147,483,647 | Python 3 | OK | TESTS | 61 | 92 | 0 | n=int(input(''))
l=list(map(int,input().split()))
c,bc,bk=[],[],[]
for i in range(0,n,3):
c.append(l[i])
for j in range(1,n,3):
bc.append(l[j])
for k in range(2,n,3):
bk.append(l[k])
t=[sum(c),sum(bc),sum(bk)]
m=t.index(max(t))
if m==0:
print('chest')
elif m==1:
print('biceps')
else:
print('back') | Title: Greg's Workout
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greg is a beginner bodybuilder. Today the gym coach gave him the training plan. All it had was *n* integers *a*1,<=*a*2,<=...,<=*a**n*. These numbers mean that Greg needs to do exactly *n* exercises today. Besides, Greg should repeat the *i*-th in order exercise *a**i* times.
Greg now only does three types of exercises: "chest" exercises, "biceps" exercises and "back" exercises. Besides, his training is cyclic, that is, the first exercise he does is a "chest" one, the second one is "biceps", the third one is "back", the fourth one is "chest", the fifth one is "biceps", and so on to the *n*-th exercise.
Now Greg wonders, which muscle will get the most exercise during his training. We know that the exercise Greg repeats the maximum number of times, trains the corresponding muscle the most. Help Greg, determine which muscle will get the most training.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=20). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=25) — the number of times Greg repeats the exercises.
Output Specification:
Print word "chest" (without the quotes), if the chest gets the most exercise, "biceps" (without the quotes), if the biceps gets the most exercise and print "back" (without the quotes) if the back gets the most exercise.
It is guaranteed that the input is such that the answer to the problem is unambiguous.
Demo Input:
['2\n2 8\n', '3\n5 1 10\n', '7\n3 3 2 7 9 6 8\n']
Demo Output:
['biceps\n', 'back\n', 'chest\n']
Note:
In the first sample Greg does 2 chest, 8 biceps and zero back exercises, so the biceps gets the most exercises.
In the second sample Greg does 5 chest, 1 biceps and 10 back exercises, so the back gets the most exercises.
In the third sample Greg does 18 chest, 12 biceps and 8 back exercises, so the chest gets the most exercise. | ```python
n=int(input(''))
l=list(map(int,input().split()))
c,bc,bk=[],[],[]
for i in range(0,n,3):
c.append(l[i])
for j in range(1,n,3):
bc.append(l[j])
for k in range(2,n,3):
bk.append(l[k])
t=[sum(c),sum(bc),sum(bk)]
m=t.index(max(t))
if m==0:
print('chest')
elif m==1:
print('biceps')
else:
print('back')
``` | 3 | |
313 | B | Ilya and Queries | PROGRAMMING | 1,100 | [
"dp",
"implementation"
] | null | null | Ilya the Lion wants to help all his friends with passing exams. They need to solve the following problem to pass the IT exam.
You've got string *s*<==<=*s*1*s*2... *s**n* (*n* is the length of the string), consisting only of characters "." and "#" and *m* queries. Each query is described by a pair of integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=*n*). The answer to the query *l**i*,<=*r**i* is the number of such integers *i* (*l**i*<=≤<=*i*<=<<=*r**i*), that *s**i*<==<=*s**i*<=+<=1.
Ilya the Lion wants to help his friends but is there anyone to help him? Help Ilya, solve the problem. | The first line contains string *s* of length *n* (2<=≤<=*n*<=≤<=105). It is guaranteed that the given string only consists of characters "." and "#".
The next line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. Each of the next *m* lines contains the description of the corresponding query. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=*n*). | Print *m* integers — the answers to the queries in the order in which they are given in the input. | [
"......\n4\n3 4\n2 3\n1 6\n2 6\n",
"#..###\n5\n1 3\n5 6\n1 5\n3 6\n3 4\n"
] | [
"1\n1\n5\n4\n",
"1\n1\n2\n2\n0\n"
] | none | 1,000 | [
{
"input": "......\n4\n3 4\n2 3\n1 6\n2 6",
"output": "1\n1\n5\n4"
},
{
"input": "#..###\n5\n1 3\n5 6\n1 5\n3 6\n3 4",
"output": "1\n1\n2\n2\n0"
},
{
"input": ".#...#..\n6\n1 5\n2 3\n6 7\n2 4\n2 5\n1 3",
"output": "2\n0\n0\n1\n2\n0"
},
{
"input": "#.#.#..\n5\n3 4\n4 5\n5 7\n5... | 1,695,285,387 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 1 | 124 | 2,048,000 | s = input()
out = 0
dp = [0]
for i in range(1,len(s)):
if s[i] == s[i-1]:
out +=1
dp.append(out)
m=int(input())
for i in range(m):
left,right = map(int, input().split())
print(dp[right-1] - dp[left-1]) | Title: Ilya and Queries
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ilya the Lion wants to help all his friends with passing exams. They need to solve the following problem to pass the IT exam.
You've got string *s*<==<=*s*1*s*2... *s**n* (*n* is the length of the string), consisting only of characters "." and "#" and *m* queries. Each query is described by a pair of integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=*n*). The answer to the query *l**i*,<=*r**i* is the number of such integers *i* (*l**i*<=≤<=*i*<=<<=*r**i*), that *s**i*<==<=*s**i*<=+<=1.
Ilya the Lion wants to help his friends but is there anyone to help him? Help Ilya, solve the problem.
Input Specification:
The first line contains string *s* of length *n* (2<=≤<=*n*<=≤<=105). It is guaranteed that the given string only consists of characters "." and "#".
The next line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. Each of the next *m* lines contains the description of the corresponding query. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=<<=*r**i*<=≤<=*n*).
Output Specification:
Print *m* integers — the answers to the queries in the order in which they are given in the input.
Demo Input:
['......\n4\n3 4\n2 3\n1 6\n2 6\n', '#..###\n5\n1 3\n5 6\n1 5\n3 6\n3 4\n']
Demo Output:
['1\n1\n5\n4\n', '1\n1\n2\n2\n0\n']
Note:
none | ```python
s = input()
out = 0
dp = [0]
for i in range(1,len(s)):
if s[i] == s[i-1]:
out +=1
dp.append(out)
m=int(input())
for i in range(m):
left,right = map(int, input().split())
print(dp[right-1] - dp[left-1])
``` | -1 | |
901 | B | GCD of Polynomials | PROGRAMMING | 2,200 | [
"constructive algorithms",
"math"
] | null | null | Suppose you have two polynomials and . Then polynomial can be uniquely represented in the following way:
This can be done using [long division](https://en.wikipedia.org/wiki/Polynomial_long_division). Here, denotes the degree of polynomial *P*(*x*). is called the remainder of division of polynomial by polynomial , it is also denoted as .
Since there is a way to divide polynomials with remainder, we can define Euclid's algorithm of finding the greatest common divisor of two polynomials. The algorithm takes two polynomials . If the polynomial is zero, the result is , otherwise the result is the value the algorithm returns for pair . On each step the degree of the second argument decreases, so the algorithm works in finite number of steps. But how large that number could be? You are to answer this question.
You are given an integer *n*. You have to build two polynomials with degrees not greater than *n*, such that their coefficients are integers not exceeding 1 by their absolute value, the leading coefficients (ones with the greatest power of *x*) are equal to one, and the described Euclid's algorithm performs exactly *n* steps finding their greatest common divisor. Moreover, the degree of the first polynomial should be greater than the degree of the second. By a step of the algorithm we mean the transition from pair to pair . | You are given a single integer *n* (1<=≤<=*n*<=≤<=150) — the number of steps of the algorithm you need to reach. | Print two polynomials in the following format.
In the first line print a single integer *m* (0<=≤<=*m*<=≤<=*n*) — the degree of the polynomial.
In the second line print *m*<=+<=1 integers between <=-<=1 and 1 — the coefficients of the polynomial, from constant to leading.
The degree of the first polynomial should be greater than the degree of the second polynomial, the leading coefficients should be equal to 1. Euclid's algorithm should perform exactly *n* steps when called using these polynomials.
If there is no answer for the given *n*, print -1.
If there are multiple answer, print any of them. | [
"1\n",
"2\n"
] | [
"1\n0 1\n0\n1\n",
"2\n-1 0 1\n1\n0 1\n"
] | In the second example you can print polynomials *x*<sup class="upper-index">2</sup> - 1 and *x*. The sequence of transitions is
There are two steps in it. | 1,000 | [
{
"input": "1",
"output": "1\n0 1\n0\n1"
},
{
"input": "2",
"output": "2\n-1 0 1\n1\n0 1"
},
{
"input": "3",
"output": "3\n0 0 0 1\n2\n-1 0 1"
},
{
"input": "4",
"output": "4\n1 0 -1 0 1\n3\n0 0 0 1"
},
{
"input": "5",
"output": "5\n0 1 0 0 0 1\n4\n1 0 -1 0 1"... | 1,513,703,401 | 5,701 | Python 3 | COMPILATION_ERROR | PRETESTS | 0 | 0 | 0 | n = input()
a = [0, 1]
b = [1, 0]
for x in xrange(n-1):
c = [0, 0]
c[0] = a[0] + b[0]
c[1] = a[1] + b[1]
a = b
b = c
print 1
print b[1], b[0]
print 1
print a[1], a[0]
| Title: GCD of Polynomials
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Suppose you have two polynomials and . Then polynomial can be uniquely represented in the following way:
This can be done using [long division](https://en.wikipedia.org/wiki/Polynomial_long_division). Here, denotes the degree of polynomial *P*(*x*). is called the remainder of division of polynomial by polynomial , it is also denoted as .
Since there is a way to divide polynomials with remainder, we can define Euclid's algorithm of finding the greatest common divisor of two polynomials. The algorithm takes two polynomials . If the polynomial is zero, the result is , otherwise the result is the value the algorithm returns for pair . On each step the degree of the second argument decreases, so the algorithm works in finite number of steps. But how large that number could be? You are to answer this question.
You are given an integer *n*. You have to build two polynomials with degrees not greater than *n*, such that their coefficients are integers not exceeding 1 by their absolute value, the leading coefficients (ones with the greatest power of *x*) are equal to one, and the described Euclid's algorithm performs exactly *n* steps finding their greatest common divisor. Moreover, the degree of the first polynomial should be greater than the degree of the second. By a step of the algorithm we mean the transition from pair to pair .
Input Specification:
You are given a single integer *n* (1<=≤<=*n*<=≤<=150) — the number of steps of the algorithm you need to reach.
Output Specification:
Print two polynomials in the following format.
In the first line print a single integer *m* (0<=≤<=*m*<=≤<=*n*) — the degree of the polynomial.
In the second line print *m*<=+<=1 integers between <=-<=1 and 1 — the coefficients of the polynomial, from constant to leading.
The degree of the first polynomial should be greater than the degree of the second polynomial, the leading coefficients should be equal to 1. Euclid's algorithm should perform exactly *n* steps when called using these polynomials.
If there is no answer for the given *n*, print -1.
If there are multiple answer, print any of them.
Demo Input:
['1\n', '2\n']
Demo Output:
['1\n0 1\n0\n1\n', '2\n-1 0 1\n1\n0 1\n']
Note:
In the second example you can print polynomials *x*<sup class="upper-index">2</sup> - 1 and *x*. The sequence of transitions is
There are two steps in it. | ```python
n = input()
a = [0, 1]
b = [1, 0]
for x in xrange(n-1):
c = [0, 0]
c[0] = a[0] + b[0]
c[1] = a[1] + b[1]
a = b
b = c
print 1
print b[1], b[0]
print 1
print a[1], a[0]
``` | -1 | |
221 | A | Little Elephant and Function | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | The Little Elephant enjoys recursive functions.
This time he enjoys the sorting function. Let *a* is a permutation of an integers from 1 to *n*, inclusive, and *a**i* denotes the *i*-th element of the permutation. The Little Elephant's recursive function *f*(*x*), that sorts the first *x* permutation's elements, works as follows:
- If *x*<==<=1, exit the function. - Otherwise, call *f*(*x*<=-<=1), and then make *swap*(*a**x*<=-<=1,<=*a**x*) (swap the *x*-th and (*x*<=-<=1)-th elements of *a*).
The Little Elephant's teacher believes that this function does not work correctly. But that-be do not get an F, the Little Elephant wants to show the performance of its function. Help him, find a permutation of numbers from 1 to *n*, such that after performing the Little Elephant's function (that is call *f*(*n*)), the permutation will be sorted in ascending order. | A single line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the size of permutation. | In a single line print *n* distinct integers from 1 to *n* — the required permutation. Numbers in a line should be separated by spaces.
It is guaranteed that the answer exists. | [
"1\n",
"2\n"
] | [
"1 ",
"2 1 "
] | none | 500 | [
{
"input": "1",
"output": "1 "
},
{
"input": "2",
"output": "2 1 "
},
{
"input": "3",
"output": "3 1 2 "
},
{
"input": "4",
"output": "4 1 2 3 "
},
{
"input": "5",
"output": "5 1 2 3 4 "
},
{
"input": "6",
"output": "6 1 2 3 4 5 "
},
{
"inp... | 1,621,970,804 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 154 | 0 | x = int(input())
def solve(x):
if x == 1:
print(1)
elif x == 2:
print("2 1")
else:
x = [z for z in range(1, x)]
print(x, *x)
solve(x) | Title: Little Elephant and Function
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant enjoys recursive functions.
This time he enjoys the sorting function. Let *a* is a permutation of an integers from 1 to *n*, inclusive, and *a**i* denotes the *i*-th element of the permutation. The Little Elephant's recursive function *f*(*x*), that sorts the first *x* permutation's elements, works as follows:
- If *x*<==<=1, exit the function. - Otherwise, call *f*(*x*<=-<=1), and then make *swap*(*a**x*<=-<=1,<=*a**x*) (swap the *x*-th and (*x*<=-<=1)-th elements of *a*).
The Little Elephant's teacher believes that this function does not work correctly. But that-be do not get an F, the Little Elephant wants to show the performance of its function. Help him, find a permutation of numbers from 1 to *n*, such that after performing the Little Elephant's function (that is call *f*(*n*)), the permutation will be sorted in ascending order.
Input Specification:
A single line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the size of permutation.
Output Specification:
In a single line print *n* distinct integers from 1 to *n* — the required permutation. Numbers in a line should be separated by spaces.
It is guaranteed that the answer exists.
Demo Input:
['1\n', '2\n']
Demo Output:
['1 ', '2 1 ']
Note:
none | ```python
x = int(input())
def solve(x):
if x == 1:
print(1)
elif x == 2:
print("2 1")
else:
x = [z for z in range(1, x)]
print(x, *x)
solve(x)
``` | 0 | |
4 | A | Watermelon | PROGRAMMING | 800 | [
"brute force",
"math"
] | A. Watermelon | 1 | 64 | One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight. | The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys. | Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case. | [
"8\n"
] | [
"YES\n"
] | For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos). | 0 | [
{
"input": "8",
"output": "YES"
},
{
"input": "5",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "3",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "1",
"output": "NO"
},
{
"input": "7",
"output": "NO"... | 1,692,190,731 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | number = int(input())
if number > 2 and number % 2 == 0:
return "YES"
else:
return "NO"
| Title: Watermelon
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
One hot summer day Pete and his friend Billy decided to buy a watermelon. They chose the biggest and the ripest one, in their opinion. After that the watermelon was weighed, and the scales showed *w* kilos. They rushed home, dying of thirst, and decided to divide the berry, however they faced a hard problem.
Pete and Billy are great fans of even numbers, that's why they want to divide the watermelon in such a way that each of the two parts weighs even number of kilos, at the same time it is not obligatory that the parts are equal. The boys are extremely tired and want to start their meal as soon as possible, that's why you should help them and find out, if they can divide the watermelon in the way they want. For sure, each of them should get a part of positive weight.
Input Specification:
The first (and the only) input line contains integer number *w* (1<=≤<=*w*<=≤<=100) — the weight of the watermelon bought by the boys.
Output Specification:
Print YES, if the boys can divide the watermelon into two parts, each of them weighing even number of kilos; and NO in the opposite case.
Demo Input:
['8\n']
Demo Output:
['YES\n']
Note:
For example, the boys can divide the watermelon into two parts of 2 and 6 kilos respectively (another variant — two parts of 4 and 4 kilos). | ```python
number = int(input())
if number > 2 and number % 2 == 0:
return "YES"
else:
return "NO"
``` | -1 |
115 | A | Party | PROGRAMMING | 900 | [
"dfs and similar",
"graphs",
"trees"
] | null | null | A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true:
- Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*.
What is the minimum number of groups that must be formed? | The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees.
The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles. | Print a single integer denoting the minimum number of groups that will be formed in the party. | [
"5\n-1\n1\n2\n1\n-1\n"
] | [
"3\n"
] | For the first example, three groups are sufficient, for example:
- Employee 1 - Employees 2 and 4 - Employees 3 and 5 | 500 | [
{
"input": "5\n-1\n1\n2\n1\n-1",
"output": "3"
},
{
"input": "4\n-1\n1\n2\n3",
"output": "4"
},
{
"input": "12\n-1\n1\n2\n3\n-1\n5\n6\n7\n-1\n9\n10\n11",
"output": "4"
},
{
"input": "6\n-1\n-1\n2\n3\n1\n1",
"output": "3"
},
{
"input": "3\n-1\n1\n1",
"output": ... | 1,698,417,265 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 11 | 186 | 4,096,000 | test=int(input())
res=[]
for i in range (test):
n=int(input())
if n==-1: res.append(0)
else: res.append(n)
for i in range (test):
if res[i]!=0: res[i]=res[res[i]-1]+1
print(max(res)+1) | Title: Party
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A company has *n* employees numbered from 1 to *n*. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee *A* is said to be the superior of another employee *B* if at least one of the following is true:
- Employee *A* is the immediate manager of employee *B* - Employee *B* has an immediate manager employee *C* such that employee *A* is the superior of employee *C*.
The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.
Today the company is going to arrange a party. This involves dividing all *n* employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees *A* and *B* such that *A* is the superior of *B*.
What is the minimum number of groups that must be formed?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=2000) — the number of employees.
The next *n* lines contain the integers *p**i* (1<=≤<=*p**i*<=≤<=*n* or *p**i*<==<=-1). Every *p**i* denotes the immediate manager for the *i*-th employee. If *p**i* is -1, that means that the *i*-th employee does not have an immediate manager.
It is guaranteed, that no employee will be the immediate manager of him/herself (*p**i*<=≠<=*i*). Also, there will be no managerial cycles.
Output Specification:
Print a single integer denoting the minimum number of groups that will be formed in the party.
Demo Input:
['5\n-1\n1\n2\n1\n-1\n']
Demo Output:
['3\n']
Note:
For the first example, three groups are sufficient, for example:
- Employee 1 - Employees 2 and 4 - Employees 3 and 5 | ```python
test=int(input())
res=[]
for i in range (test):
n=int(input())
if n==-1: res.append(0)
else: res.append(n)
for i in range (test):
if res[i]!=0: res[i]=res[res[i]-1]+1
print(max(res)+1)
``` | 0 | |
821 | C | Okabe and Boxes | PROGRAMMING | 1,500 | [
"data structures",
"greedy",
"trees"
] | null | null | Okabe and Super Hacker Daru are stacking and removing boxes. There are *n* boxes numbered from 1 to *n*. Initially there are no boxes on the stack.
Okabe, being a control freak, gives Daru 2*n* commands: *n* of which are to add a box to the top of the stack, and *n* of which are to remove a box from the top of the stack and throw it in the trash. Okabe wants Daru to throw away the boxes in the order from 1 to *n*. Of course, this means that it might be impossible for Daru to perform some of Okabe's remove commands, because the required box is not on the top of the stack.
That's why Daru can decide to wait until Okabe looks away and then reorder the boxes in the stack in any way he wants. He can do it at any point of time between Okabe's commands, but he can't add or remove boxes while he does it.
Tell Daru the minimum number of times he needs to reorder the boxes so that he can successfully complete all of Okabe's commands. It is guaranteed that every box is added before it is required to be removed. | The first line of input contains the integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of boxes.
Each of the next 2*n* lines of input starts with a string "add" or "remove". If the line starts with the "add", an integer *x* (1<=≤<=*x*<=≤<=*n*) follows, indicating that Daru should add the box with number *x* to the top of the stack.
It is guaranteed that exactly *n* lines contain "add" operations, all the boxes added are distinct, and *n* lines contain "remove" operations. It is also guaranteed that a box is always added before it is required to be removed. | Print the minimum number of times Daru needs to reorder the boxes to successfully complete all of Okabe's commands. | [
"3\nadd 1\nremove\nadd 2\nadd 3\nremove\nremove\n",
"7\nadd 3\nadd 2\nadd 1\nremove\nadd 4\nremove\nremove\nremove\nadd 6\nadd 7\nadd 5\nremove\nremove\nremove\n"
] | [
"1\n",
"2\n"
] | In the first sample, Daru should reorder the boxes after adding box 3 to the stack.
In the second sample, Daru should reorder the boxes after adding box 4 and box 7 to the stack. | 1,500 | [
{
"input": "3\nadd 1\nremove\nadd 2\nadd 3\nremove\nremove",
"output": "1"
},
{
"input": "7\nadd 3\nadd 2\nadd 1\nremove\nadd 4\nremove\nremove\nremove\nadd 6\nadd 7\nadd 5\nremove\nremove\nremove",
"output": "2"
},
{
"input": "4\nadd 1\nadd 3\nremove\nadd 4\nadd 2\nremove\nremove\nremov... | 1,611,993,529 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 8 | 3,000 | 7,680,000 | t = int(input())
ans = 0
curr = 1
lst = []
for i in range(2*t):
s = input().split()
if s[0]=="add":
n = int(s[1])
lst.append(n)
else:
if lst[-1]==curr:
curr += 1
lst.pop()
else:
lst.sort(reverse = True)
lst.pop()
ans += 1
curr += 1
print(ans) | Title: Okabe and Boxes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Okabe and Super Hacker Daru are stacking and removing boxes. There are *n* boxes numbered from 1 to *n*. Initially there are no boxes on the stack.
Okabe, being a control freak, gives Daru 2*n* commands: *n* of which are to add a box to the top of the stack, and *n* of which are to remove a box from the top of the stack and throw it in the trash. Okabe wants Daru to throw away the boxes in the order from 1 to *n*. Of course, this means that it might be impossible for Daru to perform some of Okabe's remove commands, because the required box is not on the top of the stack.
That's why Daru can decide to wait until Okabe looks away and then reorder the boxes in the stack in any way he wants. He can do it at any point of time between Okabe's commands, but he can't add or remove boxes while he does it.
Tell Daru the minimum number of times he needs to reorder the boxes so that he can successfully complete all of Okabe's commands. It is guaranteed that every box is added before it is required to be removed.
Input Specification:
The first line of input contains the integer *n* (1<=≤<=*n*<=≤<=3·105) — the number of boxes.
Each of the next 2*n* lines of input starts with a string "add" or "remove". If the line starts with the "add", an integer *x* (1<=≤<=*x*<=≤<=*n*) follows, indicating that Daru should add the box with number *x* to the top of the stack.
It is guaranteed that exactly *n* lines contain "add" operations, all the boxes added are distinct, and *n* lines contain "remove" operations. It is also guaranteed that a box is always added before it is required to be removed.
Output Specification:
Print the minimum number of times Daru needs to reorder the boxes to successfully complete all of Okabe's commands.
Demo Input:
['3\nadd 1\nremove\nadd 2\nadd 3\nremove\nremove\n', '7\nadd 3\nadd 2\nadd 1\nremove\nadd 4\nremove\nremove\nremove\nadd 6\nadd 7\nadd 5\nremove\nremove\nremove\n']
Demo Output:
['1\n', '2\n']
Note:
In the first sample, Daru should reorder the boxes after adding box 3 to the stack.
In the second sample, Daru should reorder the boxes after adding box 4 and box 7 to the stack. | ```python
t = int(input())
ans = 0
curr = 1
lst = []
for i in range(2*t):
s = input().split()
if s[0]=="add":
n = int(s[1])
lst.append(n)
else:
if lst[-1]==curr:
curr += 1
lst.pop()
else:
lst.sort(reverse = True)
lst.pop()
ans += 1
curr += 1
print(ans)
``` | 0 | |
749 | A | Bachgold Problem | PROGRAMMING | 800 | [
"greedy",
"implementation",
"math",
"number theory"
] | null | null | Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*. | The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000). | The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them. | [
"5\n",
"6\n"
] | [
"2\n2 3\n",
"3\n2 2 2\n"
] | none | 500 | [
{
"input": "5",
"output": "2\n2 3"
},
{
"input": "6",
"output": "3\n2 2 2"
},
{
"input": "2",
"output": "1\n2"
},
{
"input": "3",
"output": "1\n3"
},
{
"input": "99999",
"output": "49999\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ... | 1,641,485,037 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | n=int(input())
if n%2==0 : s='2 '*(n//2)
else : s='2 '*((n-3)//2)+'3'
print(s) | Title: Bachgold Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*.
Input Specification:
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000).
Output Specification:
The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them.
Demo Input:
['5\n', '6\n']
Demo Output:
['2\n2 3\n', '3\n2 2 2\n']
Note:
none | ```python
n=int(input())
if n%2==0 : s='2 '*(n//2)
else : s='2 '*((n-3)//2)+'3'
print(s)
``` | 0 | |
735 | B | Urbanization | PROGRAMMING | 1,100 | [
"greedy",
"number theory",
"sortings"
] | null | null | Local authorities have heard a lot about combinatorial abilities of Ostap Bender so they decided to ask his help in the question of urbanization. There are *n* people who plan to move to the cities. The wealth of the *i* of them is equal to *a**i*. Authorities plan to build two cities, first for *n*1 people and second for *n*2 people. Of course, each of *n* candidates can settle in only one of the cities. Thus, first some subset of candidates of size *n*1 settle in the first city and then some subset of size *n*2 is chosen among the remaining candidates and the move to the second city. All other candidates receive an official refuse and go back home.
To make the statistic of local region look better in the eyes of their bosses, local authorities decided to pick subsets of candidates in such a way that the sum of arithmetic mean of wealth of people in each of the cities is as large as possible. Arithmetic mean of wealth in one city is the sum of wealth *a**i* among all its residents divided by the number of them (*n*1 or *n*2 depending on the city). The division should be done in real numbers without any rounding.
Please, help authorities find the optimal way to pick residents for two cities. | The first line of the input contains three integers *n*, *n*1 and *n*2 (1<=≤<=*n*,<=*n*1,<=*n*2<=≤<=100<=000, *n*1<=+<=*n*2<=≤<=*n*) — the number of candidates who want to move to the cities, the planned number of residents of the first city and the planned number of residents of the second city.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000), the *i*-th of them is equal to the wealth of the *i*-th candidate. | Print one real value — the maximum possible sum of arithmetic means of wealth of cities' residents. You answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if . | [
"2 1 1\n1 5\n",
"4 2 1\n1 4 2 3\n"
] | [
"6.00000000\n",
"6.50000000\n"
] | In the first sample, one of the optimal solutions is to move candidate 1 to the first city and candidate 2 to the second.
In the second sample, the optimal solution is to pick candidates 3 and 4 for the first city, and candidate 2 for the second one. Thus we obtain (*a*<sub class="lower-index">3</sub> + *a*<sub class="lower-index">4</sub>) / 2 + *a*<sub class="lower-index">2</sub> = (3 + 2) / 2 + 4 = 6.5 | 1,000 | [
{
"input": "2 1 1\n1 5",
"output": "6.00000000"
},
{
"input": "4 2 1\n1 4 2 3",
"output": "6.50000000"
},
{
"input": "3 1 2\n1 2 3",
"output": "4.50000000"
},
{
"input": "10 4 6\n3 5 7 9 12 25 67 69 83 96",
"output": "88.91666667"
},
{
"input": "19 7 12\n1 2 4 8 1... | 1,632,338,841 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 77 | 6,963,200 | # بسم الله الرحمن الرحيم
def main():
n, n1, n2 = [int(i) for i in input().split()]
wealthes = [int(i) for i in input().split()]
sorted_wealthes = sorted(wealthes)
# print(sorted_wealthes)
max_mean1 = 0
max_mean2 = 0
if n1 <= n2:
# start fill n1 with people hav maxis wealthes
sum = 0
for i in range(n-1, n-n1-1, -1):
sum += sorted_wealthes[i]
max_mean1 = sum / n1
sum = 0
for i in range(n-n1-1, n-n2-1-1, -1):
sum += sorted_wealthes[i]
max_mean2 = sum / n2
else:
# start fill n2 with people hav maxis wealthes
sum = 0
for i in range(n-1, n-n2-1, -1): # [4]
sum += sorted_wealthes[i]
max_mean2 = sum / n2
# then continue fill n1 with people hav maxis wealthes after those on n2
sum = 0
for i in range(n-n2-1, n-n1-1-1, -1): # [2, 1]
sum += sorted_wealthes[i]
max_mean1 = sum / n1
print(max_mean1+max_mean2)
if __name__=='__main__':
main()
| Title: Urbanization
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Local authorities have heard a lot about combinatorial abilities of Ostap Bender so they decided to ask his help in the question of urbanization. There are *n* people who plan to move to the cities. The wealth of the *i* of them is equal to *a**i*. Authorities plan to build two cities, first for *n*1 people and second for *n*2 people. Of course, each of *n* candidates can settle in only one of the cities. Thus, first some subset of candidates of size *n*1 settle in the first city and then some subset of size *n*2 is chosen among the remaining candidates and the move to the second city. All other candidates receive an official refuse and go back home.
To make the statistic of local region look better in the eyes of their bosses, local authorities decided to pick subsets of candidates in such a way that the sum of arithmetic mean of wealth of people in each of the cities is as large as possible. Arithmetic mean of wealth in one city is the sum of wealth *a**i* among all its residents divided by the number of them (*n*1 or *n*2 depending on the city). The division should be done in real numbers without any rounding.
Please, help authorities find the optimal way to pick residents for two cities.
Input Specification:
The first line of the input contains three integers *n*, *n*1 and *n*2 (1<=≤<=*n*,<=*n*1,<=*n*2<=≤<=100<=000, *n*1<=+<=*n*2<=≤<=*n*) — the number of candidates who want to move to the cities, the planned number of residents of the first city and the planned number of residents of the second city.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000), the *i*-th of them is equal to the wealth of the *i*-th candidate.
Output Specification:
Print one real value — the maximum possible sum of arithmetic means of wealth of cities' residents. You answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
Demo Input:
['2 1 1\n1 5\n', '4 2 1\n1 4 2 3\n']
Demo Output:
['6.00000000\n', '6.50000000\n']
Note:
In the first sample, one of the optimal solutions is to move candidate 1 to the first city and candidate 2 to the second.
In the second sample, the optimal solution is to pick candidates 3 and 4 for the first city, and candidate 2 for the second one. Thus we obtain (*a*<sub class="lower-index">3</sub> + *a*<sub class="lower-index">4</sub>) / 2 + *a*<sub class="lower-index">2</sub> = (3 + 2) / 2 + 4 = 6.5 | ```python
# بسم الله الرحمن الرحيم
def main():
n, n1, n2 = [int(i) for i in input().split()]
wealthes = [int(i) for i in input().split()]
sorted_wealthes = sorted(wealthes)
# print(sorted_wealthes)
max_mean1 = 0
max_mean2 = 0
if n1 <= n2:
# start fill n1 with people hav maxis wealthes
sum = 0
for i in range(n-1, n-n1-1, -1):
sum += sorted_wealthes[i]
max_mean1 = sum / n1
sum = 0
for i in range(n-n1-1, n-n2-1-1, -1):
sum += sorted_wealthes[i]
max_mean2 = sum / n2
else:
# start fill n2 with people hav maxis wealthes
sum = 0
for i in range(n-1, n-n2-1, -1): # [4]
sum += sorted_wealthes[i]
max_mean2 = sum / n2
# then continue fill n1 with people hav maxis wealthes after those on n2
sum = 0
for i in range(n-n2-1, n-n1-1-1, -1): # [2, 1]
sum += sorted_wealthes[i]
max_mean1 = sum / n1
print(max_mean1+max_mean2)
if __name__=='__main__':
main()
``` | 0 | |
448 | A | Rewards | PROGRAMMING | 800 | [
"implementation"
] | null | null | Bizon the Champion is called the Champion for a reason.
Bizon the Champion has recently got a present — a new glass cupboard with *n* shelves and he decided to put all his presents there. All the presents can be divided into two types: medals and cups. Bizon the Champion has *a*1 first prize cups, *a*2 second prize cups and *a*3 third prize cups. Besides, he has *b*1 first prize medals, *b*2 second prize medals and *b*3 third prize medals.
Naturally, the rewards in the cupboard must look good, that's why Bizon the Champion decided to follow the rules:
- any shelf cannot contain both cups and medals at the same time; - no shelf can contain more than five cups; - no shelf can have more than ten medals.
Help Bizon the Champion find out if we can put all the rewards so that all the conditions are fulfilled. | The first line contains integers *a*1, *a*2 and *a*3 (0<=≤<=*a*1,<=*a*2,<=*a*3<=≤<=100). The second line contains integers *b*1, *b*2 and *b*3 (0<=≤<=*b*1,<=*b*2,<=*b*3<=≤<=100). The third line contains integer *n* (1<=≤<=*n*<=≤<=100).
The numbers in the lines are separated by single spaces. | Print "YES" (without the quotes) if all the rewards can be put on the shelves in the described manner. Otherwise, print "NO" (without the quotes). | [
"1 1 1\n1 1 1\n4\n",
"1 1 3\n2 3 4\n2\n",
"1 0 0\n1 0 0\n1\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "1 1 1\n1 1 1\n4",
"output": "YES"
},
{
"input": "1 1 3\n2 3 4\n2",
"output": "YES"
},
{
"input": "1 0 0\n1 0 0\n1",
"output": "NO"
},
{
"input": "0 0 0\n0 0 0\n1",
"output": "YES"
},
{
"input": "100 100 100\n100 100 100\n100",
"output": "YES"
},
... | 1,655,994,472 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 25 | 186 | 12,800,000 | import sys
import math
import collections
import heapq
import decimal
input=sys.stdin.readline
a=[int(i) for i in input().split()]
b=[int(i) for i in input().split()]
n=int(input())
if(math.ceil(sum(a)/5)+math.ceil(sum(b)/10)<=n):
print("YES")
else:
print("NO") | Title: Rewards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bizon the Champion is called the Champion for a reason.
Bizon the Champion has recently got a present — a new glass cupboard with *n* shelves and he decided to put all his presents there. All the presents can be divided into two types: medals and cups. Bizon the Champion has *a*1 first prize cups, *a*2 second prize cups and *a*3 third prize cups. Besides, he has *b*1 first prize medals, *b*2 second prize medals and *b*3 third prize medals.
Naturally, the rewards in the cupboard must look good, that's why Bizon the Champion decided to follow the rules:
- any shelf cannot contain both cups and medals at the same time; - no shelf can contain more than five cups; - no shelf can have more than ten medals.
Help Bizon the Champion find out if we can put all the rewards so that all the conditions are fulfilled.
Input Specification:
The first line contains integers *a*1, *a*2 and *a*3 (0<=≤<=*a*1,<=*a*2,<=*a*3<=≤<=100). The second line contains integers *b*1, *b*2 and *b*3 (0<=≤<=*b*1,<=*b*2,<=*b*3<=≤<=100). The third line contains integer *n* (1<=≤<=*n*<=≤<=100).
The numbers in the lines are separated by single spaces.
Output Specification:
Print "YES" (without the quotes) if all the rewards can be put on the shelves in the described manner. Otherwise, print "NO" (without the quotes).
Demo Input:
['1 1 1\n1 1 1\n4\n', '1 1 3\n2 3 4\n2\n', '1 0 0\n1 0 0\n1\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
none | ```python
import sys
import math
import collections
import heapq
import decimal
input=sys.stdin.readline
a=[int(i) for i in input().split()]
b=[int(i) for i in input().split()]
n=int(input())
if(math.ceil(sum(a)/5)+math.ceil(sum(b)/10)<=n):
print("YES")
else:
print("NO")
``` | 3 | |
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... | 1,685,903,866 | 2,147,483,647 | Python 3 | OK | TESTS | 51 | 109 | 15,769,600 | n, k = list(map(int, input().split()))
arr = list(map(int, input().split()))
from math import *
summ = 0
min_sum = inf
for right in range(k):
summ += arr[right]
min_sum = min(min_sum, summ)
idx = 0
for i in range(k, n):
summ = summ - arr[i - k] + arr[i]
if summ < min_sum:
min_sum = summ
idx = i - k + 1
print(idx + 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 = list(map(int, input().split()))
arr = list(map(int, input().split()))
from math import *
summ = 0
min_sum = inf
for right in range(k):
summ += arr[right]
min_sum = min(min_sum, summ)
idx = 0
for i in range(k, n):
summ = summ - arr[i - k] + arr[i]
if summ < min_sum:
min_sum = summ
idx = i - k + 1
print(idx + 1)
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | One social network developer recently suggested a new algorithm of choosing ads for users.
There are *n* slots which advertisers can buy. It is possible to buy a segment of consecutive slots at once. The more slots you own, the bigger are the chances your ad will be shown to users.
Every time it is needed to choose ads to show, some segment of slots is picked by a secret algorithm. Then some advertisers are chosen. The only restriction is that it should be guaranteed for advertisers which own at least *p*% of slots composing this segment that their ad will be shown.
From the other side, users don't like ads. So it was decided to show no more than ads at once. You are asked to develop a system to sell segments of slots and choose ads in accordance with the rules described above. | The first line of the input contains three integers *n*, *m* and *p* (1<=≤<=*n*,<=*m*<=≤<=150<=000,<=20<=≤<=*p*<=≤<=100) — the number of slots, the number of queries to your system and threshold for which display of the ad is guaranteed.
Next line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=150<=000), where the *i*-th number means id of advertiser who currently owns the *i*-th slot.
Next *m* lines contain queries descriptions. Each description is of one of the following forms:
- 1 l r id (1<=≤<=*l*<=≤<=*r*<=≤<=*n*,<=1<=≤<=*id*<=≤<=150 000) — advertiser *id* bought all slots in a range from *l* to *r* inclusive; - 2 l r (1<=≤<=*l*<=≤<=*r*) — you need to choose advertisers for segment [*l*,<=*r*]. | For each query of the second type answer should be printed in a separate line. First integer of the answer should be the number of advertisements that will be shown . Next *cnt* integers should be advertisers' ids.
It is allowed to print one advertiser more than once, but each advertiser that owns at least slots of the segment from *l* to *r* should be in your answer. | [
"5 9 33\n1 2 1 3 3\n2 1 5\n2 1 5\n2 1 3\n2 3 3\n1 2 4 5\n2 1 5\n2 3 5\n1 4 5 1\n2 1 5\n"
] | [
"3 1 2 3\n2 1 3\n2 2 1\n3 1 1000 1000\n1 5\n2 5 3\n2 1 5"
] | Samples demonstrate that you actually have quite a lot of freedom in choosing advertisers. | 0 | [] | 1,689,600,010 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | print("_RANDOM_GUESS_1689600009.992197")# 1689600009.9922159 | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One social network developer recently suggested a new algorithm of choosing ads for users.
There are *n* slots which advertisers can buy. It is possible to buy a segment of consecutive slots at once. The more slots you own, the bigger are the chances your ad will be shown to users.
Every time it is needed to choose ads to show, some segment of slots is picked by a secret algorithm. Then some advertisers are chosen. The only restriction is that it should be guaranteed for advertisers which own at least *p*% of slots composing this segment that their ad will be shown.
From the other side, users don't like ads. So it was decided to show no more than ads at once. You are asked to develop a system to sell segments of slots and choose ads in accordance with the rules described above.
Input Specification:
The first line of the input contains three integers *n*, *m* and *p* (1<=≤<=*n*,<=*m*<=≤<=150<=000,<=20<=≤<=*p*<=≤<=100) — the number of slots, the number of queries to your system and threshold for which display of the ad is guaranteed.
Next line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=150<=000), where the *i*-th number means id of advertiser who currently owns the *i*-th slot.
Next *m* lines contain queries descriptions. Each description is of one of the following forms:
- 1 l r id (1<=≤<=*l*<=≤<=*r*<=≤<=*n*,<=1<=≤<=*id*<=≤<=150 000) — advertiser *id* bought all slots in a range from *l* to *r* inclusive; - 2 l r (1<=≤<=*l*<=≤<=*r*) — you need to choose advertisers for segment [*l*,<=*r*].
Output Specification:
For each query of the second type answer should be printed in a separate line. First integer of the answer should be the number of advertisements that will be shown . Next *cnt* integers should be advertisers' ids.
It is allowed to print one advertiser more than once, but each advertiser that owns at least slots of the segment from *l* to *r* should be in your answer.
Demo Input:
['5 9 33\n1 2 1 3 3\n2 1 5\n2 1 5\n2 1 3\n2 3 3\n1 2 4 5\n2 1 5\n2 3 5\n1 4 5 1\n2 1 5\n']
Demo Output:
['3 1 2 3\n2 1 3\n2 2 1\n3 1 1000 1000\n1 5\n2 5 3\n2 1 5']
Note:
Samples demonstrate that you actually have quite a lot of freedom in choosing advertisers. | ```python
print("_RANDOM_GUESS_1689600009.992197")# 1689600009.9922159
``` | 0 | |
707 | A | Brain's Photos | PROGRAMMING | 800 | [
"implementation"
] | null | null | Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored. | The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'. | Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line. | [
"2 2\nC M\nY Y\n",
"3 2\nW W\nW W\nB B\n",
"1 1\nW\n"
] | [
"#Color",
"#Black&White",
"#Black&White"
] | none | 500 | [
{
"input": "2 2\nC M\nY Y",
"output": "#Color"
},
{
"input": "3 2\nW W\nW W\nB B",
"output": "#Black&White"
},
{
"input": "1 1\nW",
"output": "#Black&White"
},
{
"input": "2 3\nW W W\nB G Y",
"output": "#Color"
},
{
"input": "1 1\nW",
"output": "#Black&White"
... | 1,692,145,421 | 2,147,483,647 | Python 3 | OK | TESTS | 50 | 46 | 0 | A = []
def Color (A) :
for i in A:
for j in i:
if j!='B' and j!='W' and j!='G' :
return '#Color'
return '#Black&White'
a = list(map(int,input().split()))
for i in range(a[0]):
b = list(map(str, input().split()))
A.append(b)
print(Color(A))
| Title: Brain's Photos
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'.
Output Specification:
Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line.
Demo Input:
['2 2\nC M\nY Y\n', '3 2\nW W\nW W\nB B\n', '1 1\nW\n']
Demo Output:
['#Color', '#Black&White', '#Black&White']
Note:
none | ```python
A = []
def Color (A) :
for i in A:
for j in i:
if j!='B' and j!='W' and j!='G' :
return '#Color'
return '#Black&White'
a = list(map(int,input().split()))
for i in range(a[0]):
b = list(map(str, input().split()))
A.append(b)
print(Color(A))
``` | 3 | |
707 | C | Pythagorean Triples | PROGRAMMING | 1,500 | [
"math",
"number theory"
] | null | null | Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, that there are triples of positive integers such that you can construct a right triangle with segments of lengths corresponding to triple. Such triples are called Pythagorean triples.
For example, triples (3,<=4,<=5), (5,<=12,<=13) and (6,<=8,<=10) are Pythagorean triples.
Here Katya wondered if she can specify the length of some side of right triangle and find any Pythagorean triple corresponding to such length? Note that the side which length is specified can be a cathetus as well as hypotenuse.
Katya had no problems with completing this task. Will you do the same? | The only line of the input contains single integer *n* (1<=≤<=*n*<=≤<=109) — the length of some side of a right triangle. | Print two integers *m* and *k* (1<=≤<=*m*,<=*k*<=≤<=1018), such that *n*, *m* and *k* form a Pythagorean triple, in the only line.
In case if there is no any Pythagorean triple containing integer *n*, print <=-<=1 in the only line. If there are many answers, print any of them. | [
"3\n",
"6\n",
"1\n",
"17\n",
"67\n"
] | [
"4 5",
"8 10",
"-1",
"144 145",
"2244 2245"
] | Illustration for the first sample. | 1,500 | [
{
"input": "3",
"output": "4 5"
},
{
"input": "6",
"output": "8 10"
},
{
"input": "1",
"output": "-1"
},
{
"input": "17",
"output": "144 145"
},
{
"input": "67",
"output": "2244 2245"
},
{
"input": "10",
"output": "24 26"
},
{
"input": "14"... | 1,681,307,754 | 2,147,483,647 | Python 3 | OK | TESTS | 90 | 46 | 0 | n = int(input())
if n < 3:
print(-1)
elif n % 2 == 1:
k = 1
a = (n**2//k - k)//2
b = (n**2//k + k)//2
print(a, b)
else:
k = 2
a = (n ** 2 // k - k) // 2
b = (n ** 2 // k + k) // 2
print(a, b)
| Title: Pythagorean Triples
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, that there are triples of positive integers such that you can construct a right triangle with segments of lengths corresponding to triple. Such triples are called Pythagorean triples.
For example, triples (3,<=4,<=5), (5,<=12,<=13) and (6,<=8,<=10) are Pythagorean triples.
Here Katya wondered if she can specify the length of some side of right triangle and find any Pythagorean triple corresponding to such length? Note that the side which length is specified can be a cathetus as well as hypotenuse.
Katya had no problems with completing this task. Will you do the same?
Input Specification:
The only line of the input contains single integer *n* (1<=≤<=*n*<=≤<=109) — the length of some side of a right triangle.
Output Specification:
Print two integers *m* and *k* (1<=≤<=*m*,<=*k*<=≤<=1018), such that *n*, *m* and *k* form a Pythagorean triple, in the only line.
In case if there is no any Pythagorean triple containing integer *n*, print <=-<=1 in the only line. If there are many answers, print any of them.
Demo Input:
['3\n', '6\n', '1\n', '17\n', '67\n']
Demo Output:
['4 5', '8 10', '-1', '144 145', '2244 2245']
Note:
Illustration for the first sample. | ```python
n = int(input())
if n < 3:
print(-1)
elif n % 2 == 1:
k = 1
a = (n**2//k - k)//2
b = (n**2//k + k)//2
print(a, b)
else:
k = 2
a = (n ** 2 // k - k) // 2
b = (n ** 2 // k + k) // 2
print(a, b)
``` | 3 | |
14 | A | Letter | PROGRAMMING | 800 | [
"implementation"
] | A. Letter | 1 | 64 | A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sheet with *n* rows and *m* columns. Bob shaded some of the squares on the sheet. Having seen his masterpiece, he decided to share it with his elder brother, who lives in Flatland. Now Bob has to send his picture by post, but because of the world economic crisis and high oil prices, he wants to send his creation, but to spend as little money as possible. For each sent square of paper (no matter whether it is shaded or not) Bob has to pay 3.14 burles. Please, help Bob cut out of his masterpiece a rectangle of the minimum cost, that will contain all the shaded squares. The rectangle's sides should be parallel to the sheet's sides. | The first line of the input data contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50), *n* — amount of lines, and *m* — amount of columns on Bob's sheet. The following *n* lines contain *m* characters each. Character «.» stands for a non-shaded square on the sheet, and «*» — for a shaded square. It is guaranteed that Bob has shaded at least one square. | Output the required rectangle of the minimum cost. Study the output data in the sample tests to understand the output format better. | [
"6 7\n.......\n..***..\n..*....\n..***..\n..*....\n..***..\n",
"3 3\n***\n*.*\n***\n"
] | [
"***\n*..\n***\n*..\n***\n",
"***\n*.*\n***\n"
] | none | 0 | [
{
"input": "6 7\n.......\n..***..\n..*....\n..***..\n..*....\n..***..",
"output": "***\n*..\n***\n*..\n***"
},
{
"input": "3 3\n***\n*.*\n***",
"output": "***\n*.*\n***"
},
{
"input": "1 1\n*",
"output": "*"
},
{
"input": "2 1\n*\n*",
"output": "*\n*"
},
{
"input"... | 1,657,381,659 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 62 | 0 | n, m = input().split(" ")
matrix = []
imin, imax = int(n), -1
jmin, jmax = int(m), -1
for i in range(int(n)):
row = input()
matrix.append(row)
for j in range(len(row)):
if row[j] == '*':
if i < imin:
imin = i
if i > imax:
imax = i
if j < jmin:
jmin = j
if j > jmax:
jmax = j
for i in range(imin, imax+1, 1):
for j in range(jmin, jmax+1, 1):
print(matrix[i][j], end=" ")
print()
| Title: Letter
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
A boy Bob likes to draw. Not long ago he bought a rectangular graph (checked) sheet with *n* rows and *m* columns. Bob shaded some of the squares on the sheet. Having seen his masterpiece, he decided to share it with his elder brother, who lives in Flatland. Now Bob has to send his picture by post, but because of the world economic crisis and high oil prices, he wants to send his creation, but to spend as little money as possible. For each sent square of paper (no matter whether it is shaded or not) Bob has to pay 3.14 burles. Please, help Bob cut out of his masterpiece a rectangle of the minimum cost, that will contain all the shaded squares. The rectangle's sides should be parallel to the sheet's sides.
Input Specification:
The first line of the input data contains numbers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50), *n* — amount of lines, and *m* — amount of columns on Bob's sheet. The following *n* lines contain *m* characters each. Character «.» stands for a non-shaded square on the sheet, and «*» — for a shaded square. It is guaranteed that Bob has shaded at least one square.
Output Specification:
Output the required rectangle of the minimum cost. Study the output data in the sample tests to understand the output format better.
Demo Input:
['6 7\n.......\n..***..\n..*....\n..***..\n..*....\n..***..\n', '3 3\n***\n*.*\n***\n']
Demo Output:
['***\n*..\n***\n*..\n***\n', '***\n*.*\n***\n']
Note:
none | ```python
n, m = input().split(" ")
matrix = []
imin, imax = int(n), -1
jmin, jmax = int(m), -1
for i in range(int(n)):
row = input()
matrix.append(row)
for j in range(len(row)):
if row[j] == '*':
if i < imin:
imin = i
if i > imax:
imax = i
if j < jmin:
jmin = j
if j > jmax:
jmax = j
for i in range(imin, imax+1, 1):
for j in range(jmin, jmax+1, 1):
print(matrix[i][j], end=" ")
print()
``` | 0 |
660 | C | Hard Process | PROGRAMMING | 1,600 | [
"binary search",
"dp",
"two pointers"
] | null | null | You are given an array *a* with *n* elements. Each element of *a* is either 0 or 1.
Let's denote the length of the longest subsegment of consecutive elements in *a*, consisting of only numbers one, as *f*(*a*). You can change no more than *k* zeroes to ones to maximize *f*(*a*). | The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=3·105,<=0<=≤<=*k*<=≤<=*n*) — the number of elements in *a* and the parameter *k*.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1) — the elements of *a*. | On the first line print a non-negative integer *z* — the maximal value of *f*(*a*) after no more than *k* changes of zeroes to ones.
On the second line print *n* integers *a**j* — the elements of the array *a* after the changes.
If there are multiple answers, you can print any one of them. | [
"7 1\n1 0 0 1 1 0 1\n",
"10 2\n1 0 0 1 0 1 0 1 0 1\n"
] | [
"4\n1 0 0 1 1 1 1\n",
"5\n1 0 0 1 1 1 1 1 0 1\n"
] | none | 0 | [
{
"input": "7 1\n1 0 0 1 1 0 1",
"output": "4\n1 0 0 1 1 1 1"
},
{
"input": "10 2\n1 0 0 1 0 1 0 1 0 1",
"output": "5\n1 0 0 1 1 1 1 1 0 1"
},
{
"input": "1 0\n0",
"output": "0\n0"
},
{
"input": "1 0\n0",
"output": "0\n0"
},
{
"input": "7 0\n0 1 0 0 0 1 0",
"o... | 1,562,019,448 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 93 | 0 |
n, k = map(int, input().split())
A = [int(x) for x in input().split()]
a, b, c = 0, 0, 0
fa, fl = 0, 0
while b < len(A):
while b < len(A) and c <= k:
if A[b] == 0:
c += 1
b += 1
while c > k and a <= b:
if A[a] == 0:
c -= 1
a += 1
c = max([0, c])
if b - a > fl:
fl = b - a
fa = a
print(fl)
for i in range(len(A)):
print(1 if fa <= i and i < fa + fl else A[i], end=' ')
| Title: Hard Process
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array *a* with *n* elements. Each element of *a* is either 0 or 1.
Let's denote the length of the longest subsegment of consecutive elements in *a*, consisting of only numbers one, as *f*(*a*). You can change no more than *k* zeroes to ones to maximize *f*(*a*).
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=3·105,<=0<=≤<=*k*<=≤<=*n*) — the number of elements in *a* and the parameter *k*.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=1) — the elements of *a*.
Output Specification:
On the first line print a non-negative integer *z* — the maximal value of *f*(*a*) after no more than *k* changes of zeroes to ones.
On the second line print *n* integers *a**j* — the elements of the array *a* after the changes.
If there are multiple answers, you can print any one of them.
Demo Input:
['7 1\n1 0 0 1 1 0 1\n', '10 2\n1 0 0 1 0 1 0 1 0 1\n']
Demo Output:
['4\n1 0 0 1 1 1 1\n', '5\n1 0 0 1 1 1 1 1 0 1\n']
Note:
none | ```python
n, k = map(int, input().split())
A = [int(x) for x in input().split()]
a, b, c = 0, 0, 0
fa, fl = 0, 0
while b < len(A):
while b < len(A) and c <= k:
if A[b] == 0:
c += 1
b += 1
while c > k and a <= b:
if A[a] == 0:
c -= 1
a += 1
c = max([0, c])
if b - a > fl:
fl = b - a
fa = a
print(fl)
for i in range(len(A)):
print(1 if fa <= i and i < fa + fl else A[i], end=' ')
``` | 0 | |
217 | A | Ice Skating | PROGRAMMING | 1,200 | [
"brute force",
"dfs and similar",
"dsu",
"graphs"
] | null | null | Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.
We assume that Bajtek can only heap up snow drifts at integer coordinates. | The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the coordinates of the *i*-th snow drift.
Note that the north direction coinсides with the direction of *Oy* axis, so the east direction coinсides with the direction of the *Ox* axis. All snow drift's locations are distinct. | Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one. | [
"2\n2 1\n1 2\n",
"2\n2 1\n4 1\n"
] | [
"1\n",
"0\n"
] | none | 500 | [
{
"input": "2\n2 1\n1 2",
"output": "1"
},
{
"input": "2\n2 1\n4 1",
"output": "0"
},
{
"input": "24\n171 35\n261 20\n4 206\n501 446\n961 912\n581 748\n946 978\n463 514\n841 889\n341 466\n842 967\n54 102\n235 261\n925 889\n682 672\n623 636\n268 94\n635 710\n474 510\n697 794\n586 663\n182... | 1,614,400,099 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 11 | 218 | 307,200 | from collections import defaultdict
visitedr=defaultdict(lambda:False)
visitedc=defaultdict(lambda:False)
cnt=0
for _ in range(int(input())):
a,b=map(int,input().split())
if not visitedr:
visitedr[a]=True
visitedc[b]=True
continue
if visitedr[a] or visitedc[b]:
visitedr[a]=True
visitedc[b]=True
else:
visitedr[a]=True
visitedc[b]=True
cnt+=1
print(cnt)
| Title: Ice Skating
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bajtek is learning to skate on ice. He's a beginner, so his only mode of transportation is pushing off from a snow drift to the north, east, south or west and sliding until he lands in another snow drift. He has noticed that in this way it's impossible to get from some snow drifts to some other by any sequence of moves. He now wants to heap up some additional snow drifts, so that he can get from any snow drift to any other one. He asked you to find the minimal number of snow drifts that need to be created.
We assume that Bajtek can only heap up snow drifts at integer coordinates.
Input Specification:
The first line of input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of snow drifts. Each of the following *n* lines contains two integers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the coordinates of the *i*-th snow drift.
Note that the north direction coinсides with the direction of *Oy* axis, so the east direction coinсides with the direction of the *Ox* axis. All snow drift's locations are distinct.
Output Specification:
Output the minimal number of snow drifts that need to be created in order for Bajtek to be able to reach any snow drift from any other one.
Demo Input:
['2\n2 1\n1 2\n', '2\n2 1\n4 1\n']
Demo Output:
['1\n', '0\n']
Note:
none | ```python
from collections import defaultdict
visitedr=defaultdict(lambda:False)
visitedc=defaultdict(lambda:False)
cnt=0
for _ in range(int(input())):
a,b=map(int,input().split())
if not visitedr:
visitedr[a]=True
visitedc[b]=True
continue
if visitedr[a] or visitedc[b]:
visitedr[a]=True
visitedc[b]=True
else:
visitedr[a]=True
visitedc[b]=True
cnt+=1
print(cnt)
``` | 0 | |
274 | A | k-Multiple Free Set | PROGRAMMING | 1,500 | [
"binary search",
"greedy",
"sortings"
] | null | null | A *k*-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by *k*. That is, there are no two integers *x* and *y* (*x*<=<<=*y*) from the set, such that *y*<==<=*x*·*k*.
You're given a set of *n* distinct positive integers. Your task is to find the size of it's largest *k*-multiple free subset. | The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*k*<=≤<=109). The next line contains a list of *n* distinct positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
All the numbers in the lines are separated by single spaces. | On the only line of the output print the size of the largest *k*-multiple free subset of {*a*1,<=*a*2,<=...,<=*a**n*}. | [
"6 2\n2 3 6 5 4 10\n"
] | [
"3\n"
] | In the sample input one of the possible maximum 2-multiple free subsets is {4, 5, 6}. | 500 | [
{
"input": "6 2\n2 3 6 5 4 10",
"output": "3"
},
{
"input": "10 2\n1 2 3 4 5 6 7 8 9 10",
"output": "6"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "100 2\n191 17 61 40 77 95 128 88 26 69 79 10 131 106 142 152 68 39 182 53 83 81 6 89 65 148 33 22 5 47 107 121 52 163 1... | 1,687,614,473 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 92 | 1,024,000 | from collections import defaultdict, deque, Counter
from math import sqrt, ceil, log2, factorial, gcd
from itertools import combinations, permutations
from functools import lru_cache
from heapq import heappop, heappush, heapreplace, heapify
from bisect import bisect_left, bisect_right
from fractions import Fraction
from sys import stdin, stdout, setrecursionlimit
inf = float('inf')
setrecursionlimit(1 << 21)
write = lambda x='', end='\n': stdout.write(str(x) + end)
writes = lambda x, step=' ': stdout.write(step.join(map(str, x)) + '\n')
read = lambda x=int: x(stdin.readline().strip())
reads = lambda x=int: map(x, stdin.readline().strip().split())
n, k = reads()
s = set()
for i in sorted(reads()):
t = i // k
if t not in s:
s.add(i)
else:
s.remove(t)
s.add(i)
write(len(s))
| Title: k-Multiple Free Set
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A *k*-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by *k*. That is, there are no two integers *x* and *y* (*x*<=<<=*y*) from the set, such that *y*<==<=*x*·*k*.
You're given a set of *n* distinct positive integers. Your task is to find the size of it's largest *k*-multiple free subset.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*k*<=≤<=109). The next line contains a list of *n* distinct positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
All the numbers in the lines are separated by single spaces.
Output Specification:
On the only line of the output print the size of the largest *k*-multiple free subset of {*a*1,<=*a*2,<=...,<=*a**n*}.
Demo Input:
['6 2\n2 3 6 5 4 10\n']
Demo Output:
['3\n']
Note:
In the sample input one of the possible maximum 2-multiple free subsets is {4, 5, 6}. | ```python
from collections import defaultdict, deque, Counter
from math import sqrt, ceil, log2, factorial, gcd
from itertools import combinations, permutations
from functools import lru_cache
from heapq import heappop, heappush, heapreplace, heapify
from bisect import bisect_left, bisect_right
from fractions import Fraction
from sys import stdin, stdout, setrecursionlimit
inf = float('inf')
setrecursionlimit(1 << 21)
write = lambda x='', end='\n': stdout.write(str(x) + end)
writes = lambda x, step=' ': stdout.write(step.join(map(str, x)) + '\n')
read = lambda x=int: x(stdin.readline().strip())
reads = lambda x=int: map(x, stdin.readline().strip().split())
n, k = reads()
s = set()
for i in sorted(reads()):
t = i // k
if t not in s:
s.add(i)
else:
s.remove(t)
s.add(i)
write(len(s))
``` | 0 | |
52 | A | 123-sequence | PROGRAMMING | 900 | [
"implementation"
] | A. 123-sequence | 2 | 256 | There is a given sequence of integers *a*1,<=*a*2,<=...,<=*a**n*, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=106). The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3). | Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal. | [
"9\n1 3 2 2 2 1 1 2 3\n"
] | [
"5\n"
] | In the example all the numbers equal to 1 and 3 should be replaced by 2. | 500 | [
{
"input": "9\n1 3 2 2 2 1 1 2 3",
"output": "5"
},
{
"input": "6\n3 3 2 2 1 3",
"output": "3"
},
{
"input": "12\n3 1 3 1 2 1 3 2 2 1 2 1",
"output": "7"
},
{
"input": "15\n3 2 1 1 1 1 3 2 2 3 3 1 2 3 2",
"output": "10"
},
{
"input": "2\n2 1",
"output": "1"
... | 1,600,524,465 | 2,147,483,647 | Python 3 | OK | TESTS | 23 | 904 | 14,643,200 | n=int(input())
count=0
one=0
two=0
three=0
x=list(map(int,input().split()))
for i in x:
if i==1:
one=one+1
elif i==2:
two=two+1
elif i==3:
three=three+1
max_=max(one,two ,three)
answer=n-max_
print(answer) | Title: 123-sequence
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There is a given sequence of integers *a*1,<=*a*2,<=...,<=*a**n*, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=106). The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3).
Output Specification:
Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal.
Demo Input:
['9\n1 3 2 2 2 1 1 2 3\n']
Demo Output:
['5\n']
Note:
In the example all the numbers equal to 1 and 3 should be replaced by 2. | ```python
n=int(input())
count=0
one=0
two=0
three=0
x=list(map(int,input().split()))
for i in x:
if i==1:
one=one+1
elif i==2:
two=two+1
elif i==3:
three=three+1
max_=max(one,two ,three)
answer=n-max_
print(answer)
``` | 3.746725 |
71 | A | Way Too Long Words | PROGRAMMING | 800 | [
"strings"
] | A. Way Too Long Words | 1 | 256 | Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. | Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. | [
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] | [
"word\nl10n\ni18n\np43s\n"
] | none | 500 | [
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input":... | 1,694,370,480 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 1 | 62 | 0 | x = int(input())
arr = []
for i in range(x):
arr.append(input())
for i in range(x):
if len(arr[i])==x:
print(arr[i])
else:
s = arr[i][1:-1]
print(arr[i][0]+str(len(s))+arr[i][-1])
| Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none | ```python
x = int(input())
arr = []
for i in range(x):
arr.append(input())
for i in range(x):
if len(arr[i])==x:
print(arr[i])
else:
s = arr[i][1:-1]
print(arr[i][0]+str(len(s))+arr[i][-1])
``` | 0 |
343 | C | Read Time | PROGRAMMING | 1,900 | [
"binary search",
"greedy",
"two pointers"
] | null | null | Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but *n* different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the *i*-th reading head is above the track number *h**i*. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered *h*1, *h*2, ..., *h**n* have been read at the beginning of the operation.
Mike needs to read the data on *m* distinct tracks with numbers *p*1, *p*2, ..., *p**m*. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read. | The first line of the input contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of disk heads and the number of tracks to read, accordingly. The second line contains *n* distinct integers *h**i* in ascending order (1<=≤<=*h**i*<=≤<=1010, *h**i*<=<<=*h**i*<=+<=1) — the initial positions of the heads. The third line contains *m* distinct integers *p**i* in ascending order (1<=≤<=*p**i*<=≤<=1010, *p**i*<=<<=*p**i*<=+<=1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is recommended to use the cin, cout streams or the %I64d specifier. | Print a single number — the minimum time required, in seconds, to read all the needed tracks. | [
"3 4\n2 5 6\n1 3 6 8\n",
"3 3\n1 2 3\n1 2 3\n",
"1 2\n165\n142 200\n"
] | [
"2\n",
"0\n",
"81\n"
] | The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there; 1. move the second head to the left twice; 1. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track. | 1,500 | [
{
"input": "3 4\n2 5 6\n1 3 6 8",
"output": "2"
},
{
"input": "3 3\n1 2 3\n1 2 3",
"output": "0"
},
{
"input": "1 2\n165\n142 200",
"output": "81"
},
{
"input": "1 2\n5000000000\n1 10000000000",
"output": "14999999998"
},
{
"input": "2 4\n3 12\n1 7 8 14",
"out... | 1,689,194,924 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | print("_RANDOM_GUESS_1689194924.175988")# 1689194924.1760361 | Title: Read Time
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but *n* different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the *i*-th reading head is above the track number *h**i*. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered *h*1, *h*2, ..., *h**n* have been read at the beginning of the operation.
Mike needs to read the data on *m* distinct tracks with numbers *p*1, *p*2, ..., *p**m*. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input Specification:
The first line of the input contains two space-separated integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of disk heads and the number of tracks to read, accordingly. The second line contains *n* distinct integers *h**i* in ascending order (1<=≤<=*h**i*<=≤<=1010, *h**i*<=<<=*h**i*<=+<=1) — the initial positions of the heads. The third line contains *m* distinct integers *p**i* in ascending order (1<=≤<=*p**i*<=≤<=1010, *p**i*<=<<=*p**i*<=+<=1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print a single number — the minimum time required, in seconds, to read all the needed tracks.
Demo Input:
['3 4\n2 5 6\n1 3 6 8\n', '3 3\n1 2 3\n1 2 3\n', '1 2\n165\n142 200\n']
Demo Output:
['2\n', '0\n', '81\n']
Note:
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there; 1. move the second head to the left twice; 1. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track. | ```python
print("_RANDOM_GUESS_1689194924.175988")# 1689194924.1760361
``` | 0 | |
43 | A | Football | PROGRAMMING | 1,000 | [
"strings"
] | A. Football | 2 | 256 | One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams. | Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner. | [
"1\nABC\n",
"5\nA\nABA\nABA\nA\nA\n"
] | [
"ABC\n",
"A\n"
] | none | 500 | [
{
"input": "1\nABC",
"output": "ABC"
},
{
"input": "5\nA\nABA\nABA\nA\nA",
"output": "A"
},
{
"input": "2\nXTSJEP\nXTSJEP",
"output": "XTSJEP"
},
{
"input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ",
"output": "XZYDJAEDZ"
},
{
"input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD",
... | 1,651,472,400 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 2 | 62 | 0 | n = int(input())
l = list()
d = dict()
for i in range(n):
team=str(input())
if team not in d:
d[team]=1
l.append(team)
else:
d[team]+=1
if n==1:
print(l[0])
elif int(d[l[0]]) > int(d[l[1]]):
print(l[0])
else:
print(l[1])
| Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
Output Specification:
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
Demo Input:
['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n']
Demo Output:
['ABC\n', 'A\n']
Note:
none | ```python
n = int(input())
l = list()
d = dict()
for i in range(n):
team=str(input())
if team not in d:
d[team]=1
l.append(team)
else:
d[team]+=1
if n==1:
print(l[0])
elif int(d[l[0]]) > int(d[l[1]]):
print(l[0])
else:
print(l[1])
``` | -1 |
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",
... | 1,682,173,812 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 54 | 124 | 3,891,200 | import sys
input = sys.stdin.readline
def f(u):
return u*(u+1)//2
n, m = map(int, input().split())
x1, x2 = n//2, (n-1)//2
x, y, c, d = f(x1) + f(x2), f(n-1), 0, 0
for i in range(m):
a, b = map(int, input().split())
c += a
d += max(b*x, b*y)
print((c*n+d)/n) | 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
import sys
input = sys.stdin.readline
def f(u):
return u*(u+1)//2
n, m = map(int, input().split())
x1, x2 = n//2, (n-1)//2
x, y, c, d = f(x1) + f(x2), f(n-1), 0, 0
for i in range(m):
a, b = map(int, input().split())
c += a
d += max(b*x, b*y)
print((c*n+d)/n)
``` | 3 | |
92 | A | Chips | PROGRAMMING | 800 | [
"implementation",
"math"
] | A. Chips | 2 | 256 | There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end. | The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly. | Print the number of chips the presenter ended up with. | [
"4 11\n",
"17 107\n",
"3 8\n"
] | [
"0\n",
"2\n",
"1\n"
] | In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip. | 500 | [
{
"input": "4 11",
"output": "0"
},
{
"input": "17 107",
"output": "2"
},
{
"input": "3 8",
"output": "1"
},
{
"input": "46 7262",
"output": "35"
},
{
"input": "32 6864",
"output": "0"
},
{
"input": "36 6218",
"output": "14"
},
{
"input": "... | 1,507,366,718 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 124 | 0 | n,m = map(int,input().split())
s = int(n*(n+1)/2)
m = m%s
for i in range(1,n):
if(m<i):
break
m= m-i
print(m) | Title: Chips
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There are *n* walruses sitting in a circle. All of them are numbered in the clockwise order: the walrus number 2 sits to the left of the walrus number 1, the walrus number 3 sits to the left of the walrus number 2, ..., the walrus number 1 sits to the left of the walrus number *n*.
The presenter has *m* chips. The presenter stands in the middle of the circle and starts giving the chips to the walruses starting from walrus number 1 and moving clockwise. The walrus number *i* gets *i* chips. If the presenter can't give the current walrus the required number of chips, then the presenter takes the remaining chips and the process ends. Determine by the given *n* and *m* how many chips the presenter will get in the end.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=50, 1<=≤<=*m*<=≤<=104) — the number of walruses and the number of chips correspondingly.
Output Specification:
Print the number of chips the presenter ended up with.
Demo Input:
['4 11\n', '17 107\n', '3 8\n']
Demo Output:
['0\n', '2\n', '1\n']
Note:
In the first sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, four chips to the walrus number 4, then again one chip to the walrus number 1. After that the presenter runs out of chips. He can't give anything to the walrus number 2 and the process finishes.
In the third sample the presenter gives one chip to the walrus number 1, two chips to the walrus number 2, three chips to the walrus number 3, then again one chip to the walrus number 1. The presenter has one chip left and he can't give two chips to the walrus number 2, that's why the presenter takes the last chip. | ```python
n,m = map(int,input().split())
s = int(n*(n+1)/2)
m = m%s
for i in range(1,n):
if(m<i):
break
m= m-i
print(m)
``` | 3.969 |
774 | D | Lie or Truth | PROGRAMMING | 1,500 | [
"*special",
"constructive algorithms",
"implementation",
"sortings"
] | null | null | Vasya has a sequence of cubes and exactly one integer is written on each cube. Vasya exhibited all his cubes in a row. So the sequence of numbers written on the cubes in the order from the left to the right equals to *a*1,<=*a*2,<=...,<=*a**n*.
While Vasya was walking, his little brother Stepan played with Vasya's cubes and changed their order, so now the sequence of numbers written on the cubes became equal to *b*1,<=*b*2,<=...,<=*b**n*.
Stepan said that he swapped only cubes which where on the positions between *l* and *r*, inclusive, and did not remove or add any other cubes (i. e. he said that he reordered cubes between positions *l* and *r*, inclusive, in some way).
Your task is to determine if it is possible that Stepan said the truth, or it is guaranteed that Stepan deceived his brother. | The first line contains three integers *n*, *l*, *r* (1<=≤<=*n*<=≤<=105, 1<=≤<=*l*<=≤<=*r*<=≤<=*n*) — the number of Vasya's cubes and the positions told by Stepan.
The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the sequence of integers written on cubes in the Vasya's order.
The third line contains the sequence *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=*n*) — the sequence of integers written on cubes after Stepan rearranged their order.
It is guaranteed that Stepan did not remove or add other cubes, he only rearranged Vasya's cubes. | Print "LIE" (without quotes) if it is guaranteed that Stepan deceived his brother. In the other case, print "TRUTH" (without quotes). | [
"5 2 4\n3 4 2 3 1\n3 2 3 4 1\n",
"3 1 2\n1 2 3\n3 1 2\n",
"4 2 4\n1 1 1 1\n1 1 1 1\n"
] | [
"TRUTH\n",
"LIE\n",
"TRUTH\n"
] | In the first example there is a situation when Stepan said the truth. Initially the sequence of integers on the cubes was equal to [3, 4, 2, 3, 1]. Stepan could at first swap cubes on positions 2 and 3 (after that the sequence of integers on cubes became equal to [3, 2, 4, 3, 1]), and then swap cubes in positions 3 and 4 (after that the sequence of integers on cubes became equal to [3, 2, 3, 4, 1]).
In the second example it is not possible that Stepan said truth because he said that he swapped cubes only between positions 1 and 2, but we can see that it is guaranteed that he changed the position of the cube which was on the position 3 at first. So it is guaranteed that Stepan deceived his brother.
In the third example for any values *l* and *r* there is a situation when Stepan said the truth. | 0 | [
{
"input": "5 2 4\n3 4 2 3 1\n3 2 3 4 1",
"output": "TRUTH"
},
{
"input": "3 1 2\n1 2 3\n3 1 2",
"output": "LIE"
},
{
"input": "4 2 4\n1 1 1 1\n1 1 1 1",
"output": "TRUTH"
},
{
"input": "5 1 3\n2 2 2 1 2\n2 2 2 1 2",
"output": "TRUTH"
},
{
"input": "7 1 4\n2 5 5 5... | 1,491,407,788 | 1,288 | Python 3 | OK | TESTS | 52 | 140 | 15,155,200 | import sys
def main():
yes = True
array = sys.stdin.readline().strip().split(' ')
array1 = sys.stdin.readline().strip().split(' ')
array2 = sys.stdin.readline().strip().split(' ')
for i in range(len(array1)):
if i < int(array[1])-1 or i >= int(array[2]):
if array1[i] != array2[i]:
yes = False
break
if yes:
print("TRUTH")
else:
print("LIE")
return
main() | Title: Lie or Truth
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has a sequence of cubes and exactly one integer is written on each cube. Vasya exhibited all his cubes in a row. So the sequence of numbers written on the cubes in the order from the left to the right equals to *a*1,<=*a*2,<=...,<=*a**n*.
While Vasya was walking, his little brother Stepan played with Vasya's cubes and changed their order, so now the sequence of numbers written on the cubes became equal to *b*1,<=*b*2,<=...,<=*b**n*.
Stepan said that he swapped only cubes which where on the positions between *l* and *r*, inclusive, and did not remove or add any other cubes (i. e. he said that he reordered cubes between positions *l* and *r*, inclusive, in some way).
Your task is to determine if it is possible that Stepan said the truth, or it is guaranteed that Stepan deceived his brother.
Input Specification:
The first line contains three integers *n*, *l*, *r* (1<=≤<=*n*<=≤<=105, 1<=≤<=*l*<=≤<=*r*<=≤<=*n*) — the number of Vasya's cubes and the positions told by Stepan.
The second line contains the sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the sequence of integers written on cubes in the Vasya's order.
The third line contains the sequence *b*1,<=*b*2,<=...,<=*b**n* (1<=≤<=*b**i*<=≤<=*n*) — the sequence of integers written on cubes after Stepan rearranged their order.
It is guaranteed that Stepan did not remove or add other cubes, he only rearranged Vasya's cubes.
Output Specification:
Print "LIE" (without quotes) if it is guaranteed that Stepan deceived his brother. In the other case, print "TRUTH" (without quotes).
Demo Input:
['5 2 4\n3 4 2 3 1\n3 2 3 4 1\n', '3 1 2\n1 2 3\n3 1 2\n', '4 2 4\n1 1 1 1\n1 1 1 1\n']
Demo Output:
['TRUTH\n', 'LIE\n', 'TRUTH\n']
Note:
In the first example there is a situation when Stepan said the truth. Initially the sequence of integers on the cubes was equal to [3, 4, 2, 3, 1]. Stepan could at first swap cubes on positions 2 and 3 (after that the sequence of integers on cubes became equal to [3, 2, 4, 3, 1]), and then swap cubes in positions 3 and 4 (after that the sequence of integers on cubes became equal to [3, 2, 3, 4, 1]).
In the second example it is not possible that Stepan said truth because he said that he swapped cubes only between positions 1 and 2, but we can see that it is guaranteed that he changed the position of the cube which was on the position 3 at first. So it is guaranteed that Stepan deceived his brother.
In the third example for any values *l* and *r* there is a situation when Stepan said the truth. | ```python
import sys
def main():
yes = True
array = sys.stdin.readline().strip().split(' ')
array1 = sys.stdin.readline().strip().split(' ')
array2 = sys.stdin.readline().strip().split(' ')
for i in range(len(array1)):
if i < int(array[1])-1 or i >= int(array[2]):
if array1[i] != array2[i]:
yes = False
break
if yes:
print("TRUTH")
else:
print("LIE")
return
main()
``` | 3 | |
220 | A | Little Elephant and Problem | PROGRAMMING | 1,300 | [
"implementation",
"sortings"
] | null | null | The Little Elephant has got a problem — somebody has been touching his sorted by non-decreasing array *a* of length *n* and possibly swapped some elements of the array.
The Little Elephant doesn't want to call the police until he understands if he could have accidentally changed the array himself. He thinks that he could have accidentally changed array *a*, only if array *a* can be sorted in no more than one operation of swapping elements (not necessarily adjacent). That is, the Little Elephant could have accidentally swapped some two elements.
Help the Little Elephant, determine if he could have accidentally changed the array *a*, sorted by non-decreasing, himself. | The first line contains a single integer *n* (2<=≤<=*n*<=≤<=105) — the size of array *a*. The next line contains *n* positive integers, separated by single spaces and not exceeding 109, — array *a*.
Note that the elements of the array are not necessarily distinct numbers. | In a single line print "YES" (without the quotes) if the Little Elephant could have accidentally changed the array himself, and "NO" (without the quotes) otherwise. | [
"2\n1 2\n",
"3\n3 2 1\n",
"4\n4 3 2 1\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | In the first sample the array has already been sorted, so to sort it, we need 0 swap operations, that is not more than 1. Thus, the answer is "YES".
In the second sample we can sort the array if we swap elements 1 and 3, so we need 1 swap operation to sort the array. Thus, the answer is "YES".
In the third sample we can't sort the array in more than one swap operation, so the answer is "NO". | 500 | [
{
"input": "2\n1 2",
"output": "YES"
},
{
"input": "3\n3 2 1",
"output": "YES"
},
{
"input": "4\n4 3 2 1",
"output": "NO"
},
{
"input": "3\n1 3 2",
"output": "YES"
},
{
"input": "2\n2 1",
"output": "YES"
},
{
"input": "9\n7 7 8 8 10 10 10 10 1000000000... | 1,462,009,338 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 46 | 4,915,200 | '''
Created on Apr 30, 2016
Gmail : [email protected]
@author: Md. Rezwanul Haque
'''
n = int(input())
a = list(map(int,input().split()))
a.sort(key=None, reverse=False)
#for i in range(n):
#if(a[i+1])
if (len(a)%2==1 ):
print("YES")
else:
print("NO") | Title: Little Elephant and Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant has got a problem — somebody has been touching his sorted by non-decreasing array *a* of length *n* and possibly swapped some elements of the array.
The Little Elephant doesn't want to call the police until he understands if he could have accidentally changed the array himself. He thinks that he could have accidentally changed array *a*, only if array *a* can be sorted in no more than one operation of swapping elements (not necessarily adjacent). That is, the Little Elephant could have accidentally swapped some two elements.
Help the Little Elephant, determine if he could have accidentally changed the array *a*, sorted by non-decreasing, himself.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=105) — the size of array *a*. The next line contains *n* positive integers, separated by single spaces and not exceeding 109, — array *a*.
Note that the elements of the array are not necessarily distinct numbers.
Output Specification:
In a single line print "YES" (without the quotes) if the Little Elephant could have accidentally changed the array himself, and "NO" (without the quotes) otherwise.
Demo Input:
['2\n1 2\n', '3\n3 2 1\n', '4\n4 3 2 1\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first sample the array has already been sorted, so to sort it, we need 0 swap operations, that is not more than 1. Thus, the answer is "YES".
In the second sample we can sort the array if we swap elements 1 and 3, so we need 1 swap operation to sort the array. Thus, the answer is "YES".
In the third sample we can't sort the array in more than one swap operation, so the answer is "NO". | ```python
'''
Created on Apr 30, 2016
Gmail : [email protected]
@author: Md. Rezwanul Haque
'''
n = int(input())
a = list(map(int,input().split()))
a.sort(key=None, reverse=False)
#for i in range(n):
#if(a[i+1])
if (len(a)%2==1 ):
print("YES")
else:
print("NO")
``` | 0 | |
776 | B | Sherlock and his girlfriend | PROGRAMMING | 1,200 | [
"constructive algorithms",
"number theory"
] | null | null | Sherlock has a new girlfriend (so unlike him!). Valentine's day is coming and he wants to gift her some jewelry.
He bought *n* pieces of jewelry. The *i*-th piece has price equal to *i*<=+<=1, that is, the prices of the jewelry are 2,<=3,<=4,<=... *n*<=+<=1.
Watson gave Sherlock a challenge to color these jewelry pieces such that two pieces don't have the same color if the price of one piece is a prime divisor of the price of the other piece. Also, Watson asked him to minimize the number of different colors used.
Help Sherlock complete this trivial task. | The only line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the number of jewelry pieces. | The first line of output should contain a single integer *k*, the minimum number of colors that can be used to color the pieces of jewelry with the given constraints.
The next line should consist of *n* space-separated integers (between 1 and *k*) that specify the color of each piece in the order of increasing price.
If there are multiple ways to color the pieces using *k* colors, you can output any of them. | [
"3\n",
"4\n"
] | [
"2\n1 1 2 ",
"2\n2 1 1 2\n"
] | In the first input, the colors for first, second and third pieces of jewelry having respective prices 2, 3 and 4 are 1, 1 and 2 respectively.
In this case, as 2 is a prime divisor of 4, colors of jewelry having prices 2 and 4 must be distinct. | 1,000 | [
{
"input": "3",
"output": "2\n1 1 2 "
},
{
"input": "4",
"output": "2\n1 1 2 1 "
},
{
"input": "17",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 "
},
{
"input": "25",
"output": "2\n1 1 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 1 2 2 2 1 2 2 2 "
},
{
"input": "85",
"ou... | 1,521,092,554 | 2,147,483,647 | Python 3 | OK | TESTS | 53 | 623 | 6,758,400 | n=int(input())
a=[0]*(n+2)
for i in range(2,n+2):
for j in range(i,n+2,i):
a[j]+=1
if n>2:print(2)
else:print(1)
for i in range(2,n+2):
if a[i]==1:
print(1,end=' ')
else:
print(2,end=' ') | Title: Sherlock and his girlfriend
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sherlock has a new girlfriend (so unlike him!). Valentine's day is coming and he wants to gift her some jewelry.
He bought *n* pieces of jewelry. The *i*-th piece has price equal to *i*<=+<=1, that is, the prices of the jewelry are 2,<=3,<=4,<=... *n*<=+<=1.
Watson gave Sherlock a challenge to color these jewelry pieces such that two pieces don't have the same color if the price of one piece is a prime divisor of the price of the other piece. Also, Watson asked him to minimize the number of different colors used.
Help Sherlock complete this trivial task.
Input Specification:
The only line contains single integer *n* (1<=≤<=*n*<=≤<=100000) — the number of jewelry pieces.
Output Specification:
The first line of output should contain a single integer *k*, the minimum number of colors that can be used to color the pieces of jewelry with the given constraints.
The next line should consist of *n* space-separated integers (between 1 and *k*) that specify the color of each piece in the order of increasing price.
If there are multiple ways to color the pieces using *k* colors, you can output any of them.
Demo Input:
['3\n', '4\n']
Demo Output:
['2\n1 1 2 ', '2\n2 1 1 2\n']
Note:
In the first input, the colors for first, second and third pieces of jewelry having respective prices 2, 3 and 4 are 1, 1 and 2 respectively.
In this case, as 2 is a prime divisor of 4, colors of jewelry having prices 2 and 4 must be distinct. | ```python
n=int(input())
a=[0]*(n+2)
for i in range(2,n+2):
for j in range(i,n+2,i):
a[j]+=1
if n>2:print(2)
else:print(1)
for i in range(2,n+2):
if a[i]==1:
print(1,end=' ')
else:
print(2,end=' ')
``` | 3 | |
133 | A | HQ9+ | PROGRAMMING | 900 | [
"implementation"
] | null | null | HQ9+ is a joke programming language which has only four one-character instructions:
- "H" prints "Hello, World!",- "Q" prints the source code of the program itself,- "9" prints the lyrics of "99 Bottles of Beer" song, - "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output. | The input will consist of a single line *p* which will give a program in HQ9+. String *p* will contain between 1 and 100 characters, inclusive. ASCII-code of each character of *p* will be between 33 (exclamation mark) and 126 (tilde), inclusive. | Output "YES", if executing the program will produce any output, and "NO" otherwise. | [
"Hi!\n",
"Codeforces\n"
] | [
"YES\n",
"NO\n"
] | In the first case the program contains only one instruction — "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions. | 500 | [
{
"input": "Hi!",
"output": "YES"
},
{
"input": "Codeforces",
"output": "NO"
},
{
"input": "a+b=c",
"output": "NO"
},
{
"input": "hq-lowercase",
"output": "NO"
},
{
"input": "Q",
"output": "YES"
},
{
"input": "9",
"output": "YES"
},
{
"inpu... | 1,689,393,338 | 2,147,483,647 | Python 3 | OK | TESTS | 85 | 92 | 0 | print(["NO", "YES"][bool(set("HQ9") & set(input()))])
| Title: HQ9+
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
HQ9+ is a joke programming language which has only four one-character instructions:
- "H" prints "Hello, World!",- "Q" prints the source code of the program itself,- "9" prints the lyrics of "99 Bottles of Beer" song, - "+" increments the value stored in the internal accumulator.
Instructions "H" and "Q" are case-sensitive and must be uppercase. The characters of the program which are not instructions are ignored.
You are given a program written in HQ9+. You have to figure out whether executing this program will produce any output.
Input Specification:
The input will consist of a single line *p* which will give a program in HQ9+. String *p* will contain between 1 and 100 characters, inclusive. ASCII-code of each character of *p* will be between 33 (exclamation mark) and 126 (tilde), inclusive.
Output Specification:
Output "YES", if executing the program will produce any output, and "NO" otherwise.
Demo Input:
['Hi!\n', 'Codeforces\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first case the program contains only one instruction — "H", which prints "Hello, World!".
In the second case none of the program characters are language instructions. | ```python
print(["NO", "YES"][bool(set("HQ9") & set(input()))])
``` | 3 | |
678 | A | Johny Likes Numbers | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*. | The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109). | Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*. | [
"5 3\n",
"25 13\n",
"26 13\n"
] | [
"6\n",
"26\n",
"39\n"
] | none | 0 | [
{
"input": "5 3",
"output": "6"
},
{
"input": "25 13",
"output": "26"
},
{
"input": "26 13",
"output": "39"
},
{
"input": "1 1",
"output": "2"
},
{
"input": "8 8",
"output": "16"
},
{
"input": "14 15",
"output": "15"
},
{
"input": "197 894"... | 1,663,689,304 | 2,147,483,647 | Python 3 | OK | TESTS | 37 | 46 | 0 | n, k = input().split()
n = int(n)
k = int(k)
print(n-(n%k)+k) | Title: Johny Likes Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Johny likes numbers *n* and *k* very much. Now Johny wants to find the smallest integer *x* greater than *n*, so it is divisible by the number *k*.
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=109).
Output Specification:
Print the smallest integer *x*<=><=*n*, so it is divisible by the number *k*.
Demo Input:
['5 3\n', '25 13\n', '26 13\n']
Demo Output:
['6\n', '26\n', '39\n']
Note:
none | ```python
n, k = input().split()
n = int(n)
k = int(k)
print(n-(n%k)+k)
``` | 3 | |
22 | A | Second Order Statistics | PROGRAMMING | 800 | [
"brute force"
] | A. Second Order Statistics | 2 | 256 | Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem. | The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value. | If the given sequence has the second order statistics, output this order statistics, otherwise output NO. | [
"4\n1 2 2 -4\n",
"5\n1 2 3 1 1\n"
] | [
"1\n",
"2\n"
] | none | 0 | [
{
"input": "4\n1 2 2 -4",
"output": "1"
},
{
"input": "5\n1 2 3 1 1",
"output": "2"
},
{
"input": "1\n28",
"output": "NO"
},
{
"input": "2\n-28 12",
"output": "12"
},
{
"input": "3\n-83 40 -80",
"output": "-80"
},
{
"input": "8\n93 77 -92 26 21 -48 53 ... | 1,671,630,512 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 92 | 0 | n1=int(input())
lis=list(map(int,input().split()))
lis=list(set(lis))
lis.sort()
if len(lis)>1:
print(lis[1])
else:
print("NO")
| Title: Second Order Statistics
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem.
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value.
Output Specification:
If the given sequence has the second order statistics, output this order statistics, otherwise output NO.
Demo Input:
['4\n1 2 2 -4\n', '5\n1 2 3 1 1\n']
Demo Output:
['1\n', '2\n']
Note:
none | ```python
n1=int(input())
lis=list(map(int,input().split()))
lis=list(set(lis))
lis.sort()
if len(lis)>1:
print(lis[1])
else:
print("NO")
``` | 3.977 |
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