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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
32 | B | Borze | PROGRAMMING | 800 | [
"expression parsing",
"implementation"
] | B. Borze | 2 | 256 | Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet. | The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes). | Output the decoded ternary number. It can have leading zeroes. | [
".-.--\n",
"--.\n",
"-..-.--\n"
] | [
"012",
"20",
"1012"
] | none | 1,000 | [
{
"input": ".-.--",
"output": "012"
},
{
"input": "--.",
"output": "20"
},
{
"input": "-..-.--",
"output": "1012"
},
{
"input": "---..",
"output": "210"
},
{
"input": "..--.---..",
"output": "0020210"
},
{
"input": "-.....----.",
"output": "1000022... | 1,643,132,342 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <iostream>
#include<string>
using namespace std;
int main() {
string str1,str2;
cin>>str1;
for(int j=0;j<str1.length();j++)
{
if(str1[j]=='.'){
str2+='0';
}
if(str1[j]=='-'&&str1[j+1]=='.'){
str2+='1';
++j;
}
if(str1[j]=='-'&&str1[j+1]=='-'){
str2+='2';
++j;
}
}
cout<<str2;
} | Title: Borze
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.
Input Specification:
The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).
Output Specification:
Output the decoded ternary number. It can have leading zeroes.
Demo Input:
['.-.--\n', '--.\n', '-..-.--\n']
Demo Output:
['012', '20', '1012']
Note:
none | ```python
#include <iostream>
#include<string>
using namespace std;
int main() {
string str1,str2;
cin>>str1;
for(int j=0;j<str1.length();j++)
{
if(str1[j]=='.'){
str2+='0';
}
if(str1[j]=='-'&&str1[j+1]=='.'){
str2+='1';
++j;
}
if(str1[j]=='-'&&str1[j+1]=='-'){
str2+='2';
++j;
}
}
cout<<str2;
}
``` | -1 |
96 | A | Football | PROGRAMMING | 900 | [
"implementation",
"strings"
] | A. Football | 2 | 256 | Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not. | The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field. | Print "YES" if the situation is dangerous. Otherwise, print "NO". | [
"001001\n",
"1000000001\n"
] | [
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "001001",
"output": "NO"
},
{
"input": "1000000001",
"output": "YES"
},
{
"input": "00100110111111101",
"output": "YES"
},
{
"input": "11110111111111111",
"output": "YES"
},
{
"input": "01",
"output": "NO"
},
{
"input": "10100101",
"outp... | 1,691,151,375 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 40 | 92 | 0 | def Football():
playerPos = input()
playerPosArray = list(playerPos)
i = 1
no1Player = 0
no0Player = 0
answerFinal = False
while (i < len(playerPosArray)):
if(no1Player < 6 and no0Player < 6):
if (playerPosArray[i - 1] == '0' and playerPosArray[i] == '0'):
no0Player += 1
elif (playerPosArray[i - 1] == '1' and playerPosArray[i] == '1'):
no1Player += 1
else:
no1Player = 0
no0Player = 0
else:
answerFinal = True
break
i += 1
if (answerFinal == True):
print("YES")
else:
print("NO")
Football() | Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
Input Specification:
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
Output Specification:
Print "YES" if the situation is dangerous. Otherwise, print "NO".
Demo Input:
['001001\n', '1000000001\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none | ```python
def Football():
playerPos = input()
playerPosArray = list(playerPos)
i = 1
no1Player = 0
no0Player = 0
answerFinal = False
while (i < len(playerPosArray)):
if(no1Player < 6 and no0Player < 6):
if (playerPosArray[i - 1] == '0' and playerPosArray[i] == '0'):
no0Player += 1
elif (playerPosArray[i - 1] == '1' and playerPosArray[i] == '1'):
no1Player += 1
else:
no1Player = 0
no0Player = 0
else:
answerFinal = True
break
i += 1
if (answerFinal == True):
print("YES")
else:
print("NO")
Football()
``` | 0 |
863 | A | Quasi-palindrome | PROGRAMMING | 900 | [
"brute force",
"implementation"
] | null | null | Let quasi-palindromic number be such number that adding some leading zeros (possible none) to it produces a palindromic string.
String *t* is called a palindrome, if it reads the same from left to right and from right to left.
For example, numbers 131 and 2010200 are quasi-palindromic, they can be transformed to strings "131" and "002010200", respectively, which are palindromes.
You are given some integer number *x*. Check if it's a quasi-palindromic number. | The first line contains one integer number *x* (1<=≤<=*x*<=≤<=109). This number is given without any leading zeroes. | Print "YES" if number *x* is quasi-palindromic. Otherwise, print "NO" (without quotes). | [
"131\n",
"320\n",
"2010200\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | none | 0 | [
{
"input": "131",
"output": "YES"
},
{
"input": "320",
"output": "NO"
},
{
"input": "2010200",
"output": "YES"
},
{
"input": "1",
"output": "YES"
},
{
"input": "1000000000",
"output": "YES"
},
{
"input": "999999999",
"output": "YES"
},
{
"i... | 1,511,876,411 | 2,147,483,647 | Python 3 | OK | TESTS | 44 | 62 | 6,348,800 | s=input()
a=[]
b=[]
t=0
m=0
for i in range(len(s)):
if s[i]=='0':
t+=1
else:
break
for j in range(len(s)-1,-1,-1):
if s[j]=='0':
m+=1
else:
break
if m>t:
s=(m-t)*'0'+s
elif m<t:
s=s+'0'*(t-m)
for i in range(len(s)):
a.append(s[i])
b.append(s[len(s)-1-i])
if a==b:
print("YES")
else:
print("NO")
| Title: Quasi-palindrome
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let quasi-palindromic number be such number that adding some leading zeros (possible none) to it produces a palindromic string.
String *t* is called a palindrome, if it reads the same from left to right and from right to left.
For example, numbers 131 and 2010200 are quasi-palindromic, they can be transformed to strings "131" and "002010200", respectively, which are palindromes.
You are given some integer number *x*. Check if it's a quasi-palindromic number.
Input Specification:
The first line contains one integer number *x* (1<=≤<=*x*<=≤<=109). This number is given without any leading zeroes.
Output Specification:
Print "YES" if number *x* is quasi-palindromic. Otherwise, print "NO" (without quotes).
Demo Input:
['131\n', '320\n', '2010200\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
none | ```python
s=input()
a=[]
b=[]
t=0
m=0
for i in range(len(s)):
if s[i]=='0':
t+=1
else:
break
for j in range(len(s)-1,-1,-1):
if s[j]=='0':
m+=1
else:
break
if m>t:
s=(m-t)*'0'+s
elif m<t:
s=s+'0'*(t-m)
for i in range(len(s)):
a.append(s[i])
b.append(s[len(s)-1-i])
if a==b:
print("YES")
else:
print("NO")
``` | 3 | |
723 | A | The New Year: Meeting Friends | PROGRAMMING | 800 | [
"implementation",
"math",
"sortings"
] | null | null | There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer. | The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively. | Print one integer — the minimum total distance the friends need to travel in order to meet together. | [
"7 1 4\n",
"30 20 10\n"
] | [
"6\n",
"20\n"
] | In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4. | 500 | [
{
"input": "7 1 4",
"output": "6"
},
{
"input": "30 20 10",
"output": "20"
},
{
"input": "1 4 100",
"output": "99"
},
{
"input": "100 1 91",
"output": "99"
},
{
"input": "1 45 100",
"output": "99"
},
{
"input": "1 2 3",
"output": "2"
},
{
"... | 1,679,508,569 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | # A. The New Year: Meeting Friends
# Eles tem que se encontrar no ponto médio para os três
lista = [int(x) for x in input().split(' ')]
a = lista[0]
b = lista[1]
c = lista[2]
print(int((a+b+c)/3)) | Title: The New Year: Meeting Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
Input Specification:
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
Output Specification:
Print one integer — the minimum total distance the friends need to travel in order to meet together.
Demo Input:
['7 1 4\n', '30 20 10\n']
Demo Output:
['6\n', '20\n']
Note:
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4. | ```python
# A. The New Year: Meeting Friends
# Eles tem que se encontrar no ponto médio para os três
lista = [int(x) for x in input().split(' ')]
a = lista[0]
b = lista[1]
c = lista[2]
print(int((a+b+c)/3))
``` | 0 | |
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 ... | 1,687,211,870 | 2,147,483,647 | Python 3 | OK | TESTS | 63 | 46 | 9,113,600 | import math
def main():
_, events = input(), [int(el) for el in input().split()]
man_power = 0
crimes_untreated = 0
for e in events:
if e > 0:
man_power += e
else:
if man_power > 0:
man_power -= 1
else:
crimes_untreated += 1
print(crimes_untreated)
if __name__ == '__main__':
main() | 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
import math
def main():
_, events = input(), [int(el) for el in input().split()]
man_power = 0
crimes_untreated = 0
for e in events:
if e > 0:
man_power += e
else:
if man_power > 0:
man_power -= 1
else:
crimes_untreated += 1
print(crimes_untreated)
if __name__ == '__main__':
main()
``` | 3 | |
467 | A | George and Accommodation | PROGRAMMING | 800 | [
"implementation"
] | null | null | George has recently entered the BSUCP (Berland State University for Cool Programmers). George has a friend Alex who has also entered the university. Now they are moving into a dormitory.
George and Alex want to live in the same room. The dormitory has *n* rooms in total. At the moment the *i*-th room has *p**i* people living in it and the room can accommodate *q**i* people in total (*p**i*<=≤<=*q**i*). Your task is to count how many rooms has free place for both George and Alex. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of rooms.
The *i*-th of the next *n* lines contains two integers *p**i* and *q**i* (0<=≤<=*p**i*<=≤<=*q**i*<=≤<=100) — the number of people who already live in the *i*-th room and the room's capacity. | Print a single integer — the number of rooms where George and Alex can move in. | [
"3\n1 1\n2 2\n3 3\n",
"3\n1 10\n0 10\n10 10\n"
] | [
"0\n",
"2\n"
] | none | 500 | [
{
"input": "3\n1 1\n2 2\n3 3",
"output": "0"
},
{
"input": "3\n1 10\n0 10\n10 10",
"output": "2"
},
{
"input": "2\n36 67\n61 69",
"output": "2"
},
{
"input": "3\n21 71\n10 88\n43 62",
"output": "3"
},
{
"input": "3\n1 2\n2 3\n3 4",
"output": "0"
},
{
"... | 1,689,151,756 | 2,147,483,647 | Python 3 | OK | TESTS | 27 | 46 | 0 | ans=0
for i in range(int(input())):
a,b=map(int,input().split())
if b-a>=2:
ans+=1
print(ans) | Title: George and Accommodation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
George has recently entered the BSUCP (Berland State University for Cool Programmers). George has a friend Alex who has also entered the university. Now they are moving into a dormitory.
George and Alex want to live in the same room. The dormitory has *n* rooms in total. At the moment the *i*-th room has *p**i* people living in it and the room can accommodate *q**i* people in total (*p**i*<=≤<=*q**i*). Your task is to count how many rooms has free place for both George and Alex.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of rooms.
The *i*-th of the next *n* lines contains two integers *p**i* and *q**i* (0<=≤<=*p**i*<=≤<=*q**i*<=≤<=100) — the number of people who already live in the *i*-th room and the room's capacity.
Output Specification:
Print a single integer — the number of rooms where George and Alex can move in.
Demo Input:
['3\n1 1\n2 2\n3 3\n', '3\n1 10\n0 10\n10 10\n']
Demo Output:
['0\n', '2\n']
Note:
none | ```python
ans=0
for i in range(int(input())):
a,b=map(int,input().split())
if b-a>=2:
ans+=1
print(ans)
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | You are given several queries. In the *i*-th query you are given a single positive integer *n**i*. You are to represent *n**i* as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings.
An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself. | The first line contains single integer *q* (1<=≤<=*q*<=≤<=105) — the number of queries.
*q* lines follow. The (*i*<=+<=1)-th line contains single integer *n**i* (1<=≤<=*n**i*<=≤<=109) — the *i*-th query. | For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings. | [
"1\n12\n",
"2\n6\n8\n",
"3\n1\n2\n3\n"
] | [
"3\n",
"1\n2\n",
"-1\n-1\n-1\n"
] | 12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands.
8 = 4 + 4, 6 can't be split into several composite summands.
1, 2, 3 are less than any composite number, so they do not have valid splittings. | 0 | [
{
"input": "1\n12",
"output": "3"
},
{
"input": "2\n6\n8",
"output": "1\n2"
},
{
"input": "3\n1\n2\n3",
"output": "-1\n-1\n-1"
},
{
"input": "6\n1\n2\n3\n5\n7\n11",
"output": "-1\n-1\n-1\n-1\n-1\n-1"
},
{
"input": "3\n4\n6\n9",
"output": "1\n1\n1"
},
{
... | 1,595,086,385 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | def F(x):
s = 0
if x % 4 == 2:
s = 1
if x >= 4:
print(x // 4 + s)
if x < 4
print(-1)
n = int(input())
f = []
for i in range(n)
x = int(input())
f.append(P(x))
for j in f:
print(j)
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given several queries. In the *i*-th query you are given a single positive integer *n**i*. You are to represent *n**i* as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings.
An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself.
Input Specification:
The first line contains single integer *q* (1<=≤<=*q*<=≤<=105) — the number of queries.
*q* lines follow. The (*i*<=+<=1)-th line contains single integer *n**i* (1<=≤<=*n**i*<=≤<=109) — the *i*-th query.
Output Specification:
For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings.
Demo Input:
['1\n12\n', '2\n6\n8\n', '3\n1\n2\n3\n']
Demo Output:
['3\n', '1\n2\n', '-1\n-1\n-1\n']
Note:
12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands.
8 = 4 + 4, 6 can't be split into several composite summands.
1, 2, 3 are less than any composite number, so they do not have valid splittings. | ```python
def F(x):
s = 0
if x % 4 == 2:
s = 1
if x >= 4:
print(x // 4 + s)
if x < 4
print(-1)
n = int(input())
f = []
for i in range(n)
x = int(input())
f.append(P(x))
for j in f:
print(j)
``` | -1 | |
59 | B | Fortune Telling | PROGRAMMING | 1,200 | [
"implementation",
"number theory"
] | B. Fortune Telling | 2 | 256 | Marina loves Sasha. But she keeps wondering whether Sasha loves her. Of course, the best way to know it is fortune telling. There are many ways of telling fortune, but Marina has picked the easiest one. She takes in her hand one or several camomiles and tears off the petals one by one. After each petal she pronounces alternatively "Loves" and "Doesn't love", at that Marina always starts with "Loves". There are *n* camomiles growing in the field, possessing the numbers of petals equal to *a*1,<=*a*2,<=... *a**n*. Marina wants to pick a bouquet with the maximal possible total number of petals so that the result would still be "Loves". Help her do that; find the maximal number of petals possible in the bouquet. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100), which is the number of flowers growing in the field. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) which represent the number of petals on a given *i*-th camomile. | Print a single number which is the maximal number of petals in the bouquet, the fortune telling on which would result in "Loves". If there are no such bouquet, print 0 instead. The bouquet may consist of a single flower. | [
"1\n1\n",
"1\n2\n",
"3\n5 6 7\n"
] | [
"1\n",
"0\n",
"13\n"
] | none | 1,000 | [
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "3\n5 6 7",
"output": "13"
},
{
"input": "2\n5 7",
"output": "7"
},
{
"input": "3\n1 2 3",
"output": "5"
},
{
"input": "4\n4 3 1 2",
"output": "9"
},
{
"inpu... | 1,694,019,721 | 2,147,483,647 | Python 3 | OK | TESTS | 43 | 92 | 0 |
n = int(input())
a = list(map(int, input().split()))
odd_numbers_list = []
even_numbers_list = []
for petal in a:
if petal % 2 == 0:
even_numbers_list.append(petal)
else:
odd_numbers_list.append(petal)
odd_numbers_list.sort(reverse=True)
even_numbers_list.sort(reverse=True)
s = sum(a)
largest_odd = max(odd_numbers_list) if odd_numbers_list else 0
largest_even = max(even_numbers_list) if even_numbers_list else 0
len1 = len(even_numbers_list)
len2 = len(odd_numbers_list)
if n == 1:
if largest_even > 0:
print(0)
else:
print(largest_odd)
else:
if s % 2 == 1:
print(s)
else:
if (len1 !=0 and s - min(even_numbers_list)) % 2 == 1:
print(s - min(even_numbers_list))
elif (len1 == 0):
print(s - min(odd_numbers_list))
else:
if ( len2 !=0 and s - min(odd_numbers_list)) % 2 == 1:
print(s - min(odd_numbers_list))
elif (len2 == 0):
print(0) | Title: Fortune Telling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Marina loves Sasha. But she keeps wondering whether Sasha loves her. Of course, the best way to know it is fortune telling. There are many ways of telling fortune, but Marina has picked the easiest one. She takes in her hand one or several camomiles and tears off the petals one by one. After each petal she pronounces alternatively "Loves" and "Doesn't love", at that Marina always starts with "Loves". There are *n* camomiles growing in the field, possessing the numbers of petals equal to *a*1,<=*a*2,<=... *a**n*. Marina wants to pick a bouquet with the maximal possible total number of petals so that the result would still be "Loves". Help her do that; find the maximal number of petals possible in the bouquet.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100), which is the number of flowers growing in the field. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) which represent the number of petals on a given *i*-th camomile.
Output Specification:
Print a single number which is the maximal number of petals in the bouquet, the fortune telling on which would result in "Loves". If there are no such bouquet, print 0 instead. The bouquet may consist of a single flower.
Demo Input:
['1\n1\n', '1\n2\n', '3\n5 6 7\n']
Demo Output:
['1\n', '0\n', '13\n']
Note:
none | ```python
n = int(input())
a = list(map(int, input().split()))
odd_numbers_list = []
even_numbers_list = []
for petal in a:
if petal % 2 == 0:
even_numbers_list.append(petal)
else:
odd_numbers_list.append(petal)
odd_numbers_list.sort(reverse=True)
even_numbers_list.sort(reverse=True)
s = sum(a)
largest_odd = max(odd_numbers_list) if odd_numbers_list else 0
largest_even = max(even_numbers_list) if even_numbers_list else 0
len1 = len(even_numbers_list)
len2 = len(odd_numbers_list)
if n == 1:
if largest_even > 0:
print(0)
else:
print(largest_odd)
else:
if s % 2 == 1:
print(s)
else:
if (len1 !=0 and s - min(even_numbers_list)) % 2 == 1:
print(s - min(even_numbers_list))
elif (len1 == 0):
print(s - min(odd_numbers_list))
else:
if ( len2 !=0 and s - min(odd_numbers_list)) % 2 == 1:
print(s - min(odd_numbers_list))
elif (len2 == 0):
print(0)
``` | 3.977 |
222 | A | Shooshuns and Sequence | PROGRAMMING | 1,200 | [
"brute force",
"implementation"
] | null | null | One day shooshuns found a sequence of *n* integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes *k*-th in the current sequence and add the same number to the end of the sequence; 1. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same. | The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the sequence that the shooshuns found. | Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1. | [
"3 2\n3 1 1\n",
"3 1\n3 1 1\n"
] | [
"1\n",
"-1\n"
] | In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1. | 500 | [
{
"input": "3 2\n3 1 1",
"output": "1"
},
{
"input": "3 1\n3 1 1",
"output": "-1"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "0"
},
{
"input": "2 1\n2 1",
"output": "-1"
},
{
"input": "4 4\n1 2 3 4",
"output": "3"
}... | 1,693,772,104 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 60 | 0 | line=input()
n=int(line.split()[0])
k=int(line.split()[1])
out=0
num = list(map(int,input().split()))
for i in range(n):
num.append(num[k-1])
num.remove(num[0])
if len(set(num))== 1:
out=i+1
break
if not out:
print(-1)
else:
print(out) | Title: Shooshuns and Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day shooshuns found a sequence of *n* integers, written on a blackboard. The shooshuns can perform one operation with it, the operation consists of two steps:
1. Find the number that goes *k*-th in the current sequence and add the same number to the end of the sequence; 1. Delete the first number of the current sequence.
The shooshuns wonder after how many operations all numbers on the board will be the same and whether all numbers will ever be the same.
Input Specification:
The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105).
The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the sequence that the shooshuns found.
Output Specification:
Print the minimum number of operations, required for all numbers on the blackboard to become the same. If it is impossible to achieve, print -1.
Demo Input:
['3 2\n3 1 1\n', '3 1\n3 1 1\n']
Demo Output:
['1\n', '-1\n']
Note:
In the first test case after the first operation the blackboard will have sequence [1, 1, 1]. So, one operation is enough to make all numbers the same. Thus, the answer equals one.
In the second test case the sequence will never consist of the same numbers. It will always contain at least two distinct numbers 3 and 1. Thus, the answer equals -1. | ```python
line=input()
n=int(line.split()[0])
k=int(line.split()[1])
out=0
num = list(map(int,input().split()))
for i in range(n):
num.append(num[k-1])
num.remove(num[0])
if len(set(num))== 1:
out=i+1
break
if not out:
print(-1)
else:
print(out)
``` | 0 | |
625 | A | Guest From the Past | PROGRAMMING | 1,700 | [
"implementation",
"math"
] | null | null | Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs *a* rubles, or in glass liter bottle, that costs *b* rubles. Also, you may return empty glass bottle and get *c* (*c*<=<<=*b*) rubles back, but you cannot return plastic bottles.
Kolya has *n* rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help. | First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers *a*, *b* and *c* (1<=≤<=*a*<=≤<=1018, 1<=≤<=*c*<=<<=*b*<=≤<=1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively. | Print the only integer — maximum number of liters of kefir, that Kolya can drink. | [
"10\n11\n9\n8\n",
"10\n5\n6\n1\n"
] | [
"2\n",
"2\n"
] | In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir. | 750 | [
{
"input": "10\n11\n9\n8",
"output": "2"
},
{
"input": "10\n5\n6\n1",
"output": "2"
},
{
"input": "2\n2\n2\n1",
"output": "1"
},
{
"input": "10\n3\n3\n1",
"output": "4"
},
{
"input": "10\n1\n2\n1",
"output": "10"
},
{
"input": "10\n2\n3\n1",
"outpu... | 1,454,842,474 | 6,574 | Python 3 | WRONG_ANSWER | PRETESTS | 10 | 61 | 0 | n = int(input())
help = n
a = int(input())
b = int(input())
c = int(input())
answer = 0
if b - c < a:
answer = (n - b) // (b - c) + 1
else:
answer = n // a
print(answer) | Title: Guest From the Past
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the details of buying this delicious drink. One day, as you probably know, he found himself in year 2084, and buying kefir there is much more complicated.
Kolya is hungry, so he went to the nearest milk shop. In 2084 you may buy kefir in a plastic liter bottle, that costs *a* rubles, or in glass liter bottle, that costs *b* rubles. Also, you may return empty glass bottle and get *c* (*c*<=<<=*b*) rubles back, but you cannot return plastic bottles.
Kolya has *n* rubles and he is really hungry, so he wants to drink as much kefir as possible. There were no plastic bottles in his 1984, so Kolya doesn't know how to act optimally and asks for your help.
Input Specification:
First line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1018) — the number of rubles Kolya has at the beginning.
Then follow three lines containing integers *a*, *b* and *c* (1<=≤<=*a*<=≤<=1018, 1<=≤<=*c*<=<<=*b*<=≤<=1018) — the cost of one plastic liter bottle, the cost of one glass liter bottle and the money one can get back by returning an empty glass bottle, respectively.
Output Specification:
Print the only integer — maximum number of liters of kefir, that Kolya can drink.
Demo Input:
['10\n11\n9\n8\n', '10\n5\n6\n1\n']
Demo Output:
['2\n', '2\n']
Note:
In the first sample, Kolya can buy one glass bottle, then return it and buy one more glass bottle. Thus he will drink 2 liters of kefir.
In the second sample, Kolya can buy two plastic bottle and get two liters of kefir, or he can buy one liter glass bottle, then return it and buy one plastic bottle. In both cases he will drink two liters of kefir. | ```python
n = int(input())
help = n
a = int(input())
b = int(input())
c = int(input())
answer = 0
if b - c < a:
answer = (n - b) // (b - c) + 1
else:
answer = n // a
print(answer)
``` | 0 | |
870 | A | Search for Pretty Integers | PROGRAMMING | 900 | [
"brute force",
"implementation"
] | null | null | You are given two lists of non-zero digits.
Let's call an integer pretty if its (base 10) representation has at least one digit from the first list and at least one digit from the second list. What is the smallest positive pretty integer? | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=9) — the lengths of the first and the second lists, respectively.
The second line contains *n* distinct digits *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=9) — the elements of the first list.
The third line contains *m* distinct digits *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=9) — the elements of the second list. | Print the smallest pretty integer. | [
"2 3\n4 2\n5 7 6\n",
"8 8\n1 2 3 4 5 6 7 8\n8 7 6 5 4 3 2 1\n"
] | [
"25\n",
"1\n"
] | In the first example 25, 46, 24567 are pretty, as well as many other integers. The smallest among them is 25. 42 and 24 are not pretty because they don't have digits from the second list.
In the second example all integers that have at least one digit different from 9 are pretty. It's obvious that the smallest among them is 1, because it's the smallest positive integer. | 500 | [
{
"input": "2 3\n4 2\n5 7 6",
"output": "25"
},
{
"input": "8 8\n1 2 3 4 5 6 7 8\n8 7 6 5 4 3 2 1",
"output": "1"
},
{
"input": "1 1\n9\n1",
"output": "19"
},
{
"input": "9 1\n5 4 2 3 6 1 7 9 8\n9",
"output": "9"
},
{
"input": "5 3\n7 2 5 8 6\n3 1 9",
"output"... | 1,522,780,521 | 2,147,483,647 | Python 3 | OK | TESTS | 52 | 93 | 7,065,600 | n, m = map(int, input().split())
N = list(map(int, input().split()))
M = list(map(int, input().split()))
U = set(N) & set(M)
if (len(U)!=0):
print(min(U))
else:
q, w = min(N), min(M)
if (q>w):
q, w = w, q
print (q, w, sep="") | Title: Search for Pretty Integers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two lists of non-zero digits.
Let's call an integer pretty if its (base 10) representation has at least one digit from the first list and at least one digit from the second list. What is the smallest positive pretty integer?
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=9) — the lengths of the first and the second lists, respectively.
The second line contains *n* distinct digits *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=9) — the elements of the first list.
The third line contains *m* distinct digits *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=9) — the elements of the second list.
Output Specification:
Print the smallest pretty integer.
Demo Input:
['2 3\n4 2\n5 7 6\n', '8 8\n1 2 3 4 5 6 7 8\n8 7 6 5 4 3 2 1\n']
Demo Output:
['25\n', '1\n']
Note:
In the first example 25, 46, 24567 are pretty, as well as many other integers. The smallest among them is 25. 42 and 24 are not pretty because they don't have digits from the second list.
In the second example all integers that have at least one digit different from 9 are pretty. It's obvious that the smallest among them is 1, because it's the smallest positive integer. | ```python
n, m = map(int, input().split())
N = list(map(int, input().split()))
M = list(map(int, input().split()))
U = set(N) & set(M)
if (len(U)!=0):
print(min(U))
else:
q, w = min(N), min(M)
if (q>w):
q, w = w, q
print (q, w, sep="")
``` | 3 | |
982 | A | Row | PROGRAMMING | 1,200 | [
"brute force",
"constructive algorithms"
] | null | null | You're given a row with $n$ chairs. We call a seating of people "maximal" if the two following conditions hold:
1. There are no neighbors adjacent to anyone seated. 1. It's impossible to seat one more person without violating the first rule.
The seating is given as a string consisting of zeros and ones ($0$ means that the corresponding seat is empty, $1$ — occupied). The goal is to determine whether this seating is "maximal".
Note that the first and last seats are not adjacent (if $n \ne 2$). | The first line contains a single integer $n$ ($1 \leq n \leq 1000$) — the number of chairs.
The next line contains a string of $n$ characters, each of them is either zero or one, describing the seating. | Output "Yes" (without quotation marks) if the seating is "maximal". Otherwise print "No".
You are allowed to print letters in whatever case you'd like (uppercase or lowercase). | [
"3\n101\n",
"4\n1011\n",
"5\n10001\n"
] | [
"Yes\n",
"No\n",
"No\n"
] | In sample case one the given seating is maximal.
In sample case two the person at chair three has a neighbour to the right.
In sample case three it is possible to seat yet another person into chair three. | 500 | [
{
"input": "3\n101",
"output": "Yes"
},
{
"input": "4\n1011",
"output": "No"
},
{
"input": "5\n10001",
"output": "No"
},
{
"input": "1\n0",
"output": "No"
},
{
"input": "1\n1",
"output": "Yes"
},
{
"input": "100\n010100101010100101001001010100101010010... | 1,606,716,209 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 77 | 0 | val=input()
s=input()
if s[0]=='1' and s[-1]=='1':
print('NO')
elif '11' in s or '000' in s:
print('NO')
else:
print('YES')
| Title: Row
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You're given a row with $n$ chairs. We call a seating of people "maximal" if the two following conditions hold:
1. There are no neighbors adjacent to anyone seated. 1. It's impossible to seat one more person without violating the first rule.
The seating is given as a string consisting of zeros and ones ($0$ means that the corresponding seat is empty, $1$ — occupied). The goal is to determine whether this seating is "maximal".
Note that the first and last seats are not adjacent (if $n \ne 2$).
Input Specification:
The first line contains a single integer $n$ ($1 \leq n \leq 1000$) — the number of chairs.
The next line contains a string of $n$ characters, each of them is either zero or one, describing the seating.
Output Specification:
Output "Yes" (without quotation marks) if the seating is "maximal". Otherwise print "No".
You are allowed to print letters in whatever case you'd like (uppercase or lowercase).
Demo Input:
['3\n101\n', '4\n1011\n', '5\n10001\n']
Demo Output:
['Yes\n', 'No\n', 'No\n']
Note:
In sample case one the given seating is maximal.
In sample case two the person at chair three has a neighbour to the right.
In sample case three it is possible to seat yet another person into chair three. | ```python
val=input()
s=input()
if s[0]=='1' and s[-1]=='1':
print('NO')
elif '11' in s or '000' in s:
print('NO')
else:
print('YES')
``` | 0 | |
20 | C | Dijkstra? | PROGRAMMING | 1,900 | [
"graphs",
"shortest paths"
] | C. Dijkstra? | 1 | 64 | You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*. | The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices. | Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them. | [
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n",
"5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n"
] | [
"1 4 3 5 ",
"1 4 3 5 "
] | none | 1,500 | [
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1",
"output": "1 4 3 5 "
},
{
"input": "2 1\n1 2 1",
"output": "1 2 "
},
{
"input": "3 1\n1 2 1",
"output": "-1"
},
{
"input... | 1,644,086,301 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 46 | 512,000 | n,m=map(int,input().split())
lis=[[] for ____ in range(n)]
val=[0]+[10**9+5 for _____ in range(n-1)]
si=[0 for __ in range(n)]
"""
lis stores the graph
val stores the minimum value taken to reach the vertex i+1
si is the parent list retracing which you get the path
heap is a list of current elements that we havent tried yet.
"""
for _ in range(m):
a,b,w=map(int,input().split())
a-=1
b-=1
lis[a].append((w,b))
lis[b].append((w,a))
heap=[(0,0)]
while heap:
s,i=heap.pop(0)
for j in lis[i]:
x,y=j
if val[y]>s+x:
val[y]=s+x
si[y]=i
heap.append((s+x,y))
heap.sort()
if not si[n-1]:
print(-1)
else:
ans=[n]
k=n-1
while k!=0:
k=si[k]
ans.append(k+1)
ans.reverse()
print(*ans)
| Title: Dijkstra?
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
You are given a weighted undirected graph. The vertices are enumerated from 1 to *n*. Your task is to find the shortest path between the vertex 1 and the vertex *n*.
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=105,<=0<=≤<=*m*<=≤<=105), where *n* is the number of vertices and *m* is the number of edges. Following *m* lines contain one edge each in form *a**i*, *b**i* and *w**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*,<=1<=≤<=*w**i*<=≤<=106), where *a**i*,<=*b**i* are edge endpoints and *w**i* is the length of the edge.
It is possible that the graph has loops and multiple edges between pair of vertices.
Output Specification:
Write the only integer -1 in case of no path. Write the shortest path in opposite case. If there are many solutions, print any of them.
Demo Input:
['5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n', '5 6\n1 2 2\n2 5 5\n2 3 4\n1 4 1\n4 3 3\n3 5 1\n']
Demo Output:
['1 4 3 5 ', '1 4 3 5 ']
Note:
none | ```python
n,m=map(int,input().split())
lis=[[] for ____ in range(n)]
val=[0]+[10**9+5 for _____ in range(n-1)]
si=[0 for __ in range(n)]
"""
lis stores the graph
val stores the minimum value taken to reach the vertex i+1
si is the parent list retracing which you get the path
heap is a list of current elements that we havent tried yet.
"""
for _ in range(m):
a,b,w=map(int,input().split())
a-=1
b-=1
lis[a].append((w,b))
lis[b].append((w,a))
heap=[(0,0)]
while heap:
s,i=heap.pop(0)
for j in lis[i]:
x,y=j
if val[y]>s+x:
val[y]=s+x
si[y]=i
heap.append((s+x,y))
heap.sort()
if not si[n-1]:
print(-1)
else:
ans=[n]
k=n-1
while k!=0:
k=si[k]
ans.append(k+1)
ans.reverse()
print(*ans)
``` | 0 |
471 | A | MUH and Sticks | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Two polar bears Menshykov and Uslada from the St.Petersburg zoo and elephant Horace from the Kiev zoo got six sticks to play with and assess the animals' creativity. Menshykov, Uslada and Horace decided to make either an elephant or a bear from those sticks. They can make an animal from sticks in the following way:
- Four sticks represent the animal's legs, these sticks should have the same length. - Two remaining sticks represent the animal's head and body. The bear's head stick must be shorter than the body stick. The elephant, however, has a long trunk, so his head stick must be as long as the body stick. Note that there are no limits on the relations between the leg sticks and the head and body sticks.
Your task is to find out which animal can be made from the given stick set. The zoo keeper wants the sticks back after the game, so they must never be broken, even bears understand it. | The single line contains six space-separated integers *l**i* (1<=≤<=*l**i*<=≤<=9) — the lengths of the six sticks. It is guaranteed that the input is such that you cannot make both animals from the sticks. | If you can make a bear from the given set, print string "Bear" (without the quotes). If you can make an elephant, print string "Elephant" (wıthout the quotes). If you can make neither a bear nor an elephant, print string "Alien" (without the quotes). | [
"4 2 5 4 4 4\n",
"4 4 5 4 4 5\n",
"1 2 3 4 5 6\n"
] | [
"Bear",
"Elephant",
"Alien"
] | If you're out of creative ideas, see instructions below which show how to make a bear and an elephant in the first two samples. The stick of length 2 is in red, the sticks of length 4 are in green, the sticks of length 5 are in blue. | 500 | [
{
"input": "4 2 5 4 4 4",
"output": "Bear"
},
{
"input": "4 4 5 4 4 5",
"output": "Elephant"
},
{
"input": "1 2 3 4 5 6",
"output": "Alien"
},
{
"input": "5 5 5 5 5 5",
"output": "Elephant"
},
{
"input": "1 1 1 2 3 5",
"output": "Alien"
},
{
"input": "... | 1,639,492,724 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | from collections import Counter
l = list(map(int, input().split()))
dict = Counter(l)
if '4' in dict.values():
if len(dict) == 3:
print("Bear")
elif len(dict) == 2:
print("Elephant")
else:
print("Alien")
else:
print("Alien") | Title: MUH and Sticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two polar bears Menshykov and Uslada from the St.Petersburg zoo and elephant Horace from the Kiev zoo got six sticks to play with and assess the animals' creativity. Menshykov, Uslada and Horace decided to make either an elephant or a bear from those sticks. They can make an animal from sticks in the following way:
- Four sticks represent the animal's legs, these sticks should have the same length. - Two remaining sticks represent the animal's head and body. The bear's head stick must be shorter than the body stick. The elephant, however, has a long trunk, so his head stick must be as long as the body stick. Note that there are no limits on the relations between the leg sticks and the head and body sticks.
Your task is to find out which animal can be made from the given stick set. The zoo keeper wants the sticks back after the game, so they must never be broken, even bears understand it.
Input Specification:
The single line contains six space-separated integers *l**i* (1<=≤<=*l**i*<=≤<=9) — the lengths of the six sticks. It is guaranteed that the input is such that you cannot make both animals from the sticks.
Output Specification:
If you can make a bear from the given set, print string "Bear" (without the quotes). If you can make an elephant, print string "Elephant" (wıthout the quotes). If you can make neither a bear nor an elephant, print string "Alien" (without the quotes).
Demo Input:
['4 2 5 4 4 4\n', '4 4 5 4 4 5\n', '1 2 3 4 5 6\n']
Demo Output:
['Bear', 'Elephant', 'Alien']
Note:
If you're out of creative ideas, see instructions below which show how to make a bear and an elephant in the first two samples. The stick of length 2 is in red, the sticks of length 4 are in green, the sticks of length 5 are in blue. | ```python
from collections import Counter
l = list(map(int, input().split()))
dict = Counter(l)
if '4' in dict.values():
if len(dict) == 3:
print("Bear")
elif len(dict) == 2:
print("Elephant")
else:
print("Alien")
else:
print("Alien")
``` | 0 | |
318 | A | Even Odds | PROGRAMMING | 900 | [
"math"
] | null | null | Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first *n*. He writes down the following sequence of numbers: firstly all odd integers from 1 to *n* (in ascending order), then all even integers from 1 to *n* (also in ascending order). Help our hero to find out which number will stand at the position number *k*. | The only line of input contains integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier. | Print the number that will stand at the position number *k* after Volodya's manipulations. | [
"10 3\n",
"7 7\n"
] | [
"5",
"6"
] | In the first sample Volodya's sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5. | 500 | [
{
"input": "10 3",
"output": "5"
},
{
"input": "7 7",
"output": "6"
},
{
"input": "7 1",
"output": "1"
},
{
"input": "7 2",
"output": "3"
},
{
"input": "8 5",
"output": "2"
},
{
"input": "8 3",
"output": "5"
},
{
"input": "8 4",
"output... | 1,696,590,254 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | # -*- coding: utf-8 -*-
"""
Created on Fri Oct 6 16:29:33 2023
@author: risha
"""
n,k=map(int,input().split())
l=[]
for i in range(k):
if i%2==1:
l.append(i)
for i in range(k):
if i%2==0:
l.append(i)
print(l[k-1]) | Title: Even Odds
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural number is positive integer number). He is determined to rearrange them. But there are too many natural numbers, so Volodya decided to start with the first *n*. He writes down the following sequence of numbers: firstly all odd integers from 1 to *n* (in ascending order), then all even integers from 1 to *n* (also in ascending order). Help our hero to find out which number will stand at the position number *k*.
Input Specification:
The only line of input contains integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print the number that will stand at the position number *k* after Volodya's manipulations.
Demo Input:
['10 3\n', '7 7\n']
Demo Output:
['5', '6']
Note:
In the first sample Volodya's sequence will look like this: {1, 3, 5, 7, 9, 2, 4, 6, 8, 10}. The third place in the sequence is therefore occupied by the number 5. | ```python
# -*- coding: utf-8 -*-
"""
Created on Fri Oct 6 16:29:33 2023
@author: risha
"""
n,k=map(int,input().split())
l=[]
for i in range(k):
if i%2==1:
l.append(i)
for i in range(k):
if i%2==0:
l.append(i)
print(l[k-1])
``` | 0 | |
357 | B | Flag Day | PROGRAMMING | 1,400 | [
"constructive algorithms",
"implementation"
] | null | null | In Berland, there is the national holiday coming — the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
- overall, there must be *m* dances;- exactly three people must take part in each dance;- each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has *n* dancers, and their number can be less than 3*m*. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the *m* dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the *n* dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances. | The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=105) and *m* (1<=≤<=*m*<=≤<=105) — the number of dancers and the number of dances, correspondingly. Then *m* lines follow, describing the dances in the order of dancing them. The *i*-th line contains three distinct integers — the numbers of the dancers that take part in the *i*-th dance. The dancers are numbered from 1 to *n*. Each dancer takes part in at least one dance. | Print *n* space-separated integers: the *i*-th number must represent the color of the *i*-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists. | [
"7 3\n1 2 3\n1 4 5\n4 6 7\n",
"9 3\n3 6 9\n2 5 8\n1 4 7\n",
"5 2\n4 1 5\n3 1 2\n"
] | [
"1 2 3 3 2 2 1 \n",
"1 1 1 2 2 2 3 3 3 \n",
"2 3 1 1 3 \n"
] | none | 1,000 | [
{
"input": "7 3\n1 2 3\n1 4 5\n4 6 7",
"output": "1 2 3 3 2 2 1 "
},
{
"input": "9 3\n3 6 9\n2 5 8\n1 4 7",
"output": "1 1 1 2 2 2 3 3 3 "
},
{
"input": "5 2\n4 1 5\n3 1 2",
"output": "2 3 1 1 3 "
},
{
"input": "14 5\n1 5 3\n13 10 11\n6 3 8\n14 9 2\n7 4 12",
"output": "1 ... | 1,678,001,028 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 61 | 0 | import sys
from os import path
from math import gcd,floor,sqrt,log,ceil
from bisect import bisect_left, bisect_right
import heapq
############ ---- Input Functions ---- ####################
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
def inmatrix(n,m):
arr=[]
for i in range(n):
temp=inlt()
arr.append(temp)
return arr
############################################################
def cuberoot(x):
if 0<=x: return x**(1./2.)
return -(-x)**(1./2.)
def is_anagram(str1, str2):
return sorted(str1) == sorted(str2)
############################################################
mod=1000000007
############################################################
if(path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
sys.stderr = open("error.txt", "w")
else:
input = sys.stdin.readline
############################################################
def solve():
n,m=invr()
a=[]
for i in range(m):
t=inlt()
a.append(t)
p=[0 for i in range(n+1)]
for i in range(m):
ss=set()
for j in range(3):
if p[a[i][j]]==0:
if 1 not in ss:
ss.add(1)
p[a[i][j]]=1
elif 2 not in ss:
ss.add(2)
p[a[i][j]]=2
elif 3 not in ss:
ss.add(3)
p[a[i][j]]=3
else:
ss.add(p[a[i][j]])
for i in range(1,n+1):
if p[i]:
print(p[i],end=" ")
p[i]=0
def main():
# T=inp()
T=1
for i in range(1,T+1):
#print("Case #{}:".format(i),end=" ")
solve()
main() | Title: Flag Day
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland, there is the national holiday coming — the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
- overall, there must be *m* dances;- exactly three people must take part in each dance;- each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has *n* dancers, and their number can be less than 3*m*. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the *m* dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the *n* dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances.
Input Specification:
The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=105) and *m* (1<=≤<=*m*<=≤<=105) — the number of dancers and the number of dances, correspondingly. Then *m* lines follow, describing the dances in the order of dancing them. The *i*-th line contains three distinct integers — the numbers of the dancers that take part in the *i*-th dance. The dancers are numbered from 1 to *n*. Each dancer takes part in at least one dance.
Output Specification:
Print *n* space-separated integers: the *i*-th number must represent the color of the *i*-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists.
Demo Input:
['7 3\n1 2 3\n1 4 5\n4 6 7\n', '9 3\n3 6 9\n2 5 8\n1 4 7\n', '5 2\n4 1 5\n3 1 2\n']
Demo Output:
['1 2 3 3 2 2 1 \n', '1 1 1 2 2 2 3 3 3 \n', '2 3 1 1 3 \n']
Note:
none | ```python
import sys
from os import path
from math import gcd,floor,sqrt,log,ceil
from bisect import bisect_left, bisect_right
import heapq
############ ---- Input Functions ---- ####################
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
def inmatrix(n,m):
arr=[]
for i in range(n):
temp=inlt()
arr.append(temp)
return arr
############################################################
def cuberoot(x):
if 0<=x: return x**(1./2.)
return -(-x)**(1./2.)
def is_anagram(str1, str2):
return sorted(str1) == sorted(str2)
############################################################
mod=1000000007
############################################################
if(path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
sys.stderr = open("error.txt", "w")
else:
input = sys.stdin.readline
############################################################
def solve():
n,m=invr()
a=[]
for i in range(m):
t=inlt()
a.append(t)
p=[0 for i in range(n+1)]
for i in range(m):
ss=set()
for j in range(3):
if p[a[i][j]]==0:
if 1 not in ss:
ss.add(1)
p[a[i][j]]=1
elif 2 not in ss:
ss.add(2)
p[a[i][j]]=2
elif 3 not in ss:
ss.add(3)
p[a[i][j]]=3
else:
ss.add(p[a[i][j]])
for i in range(1,n+1):
if p[i]:
print(p[i],end=" ")
p[i]=0
def main():
# T=inp()
T=1
for i in range(1,T+1):
#print("Case #{}:".format(i),end=" ")
solve()
main()
``` | 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"... | 1,693,315,161 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | w = input(int())
if w %2 == 1 or w == 2:
print('NO')
else:
print('YES') | 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
w = input(int())
if w %2 == 1 or w == 2:
print('NO')
else:
print('YES')
``` | -1 |
248 | A | Cupboards | PROGRAMMING | 800 | [
"implementation"
] | null | null | One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*. | The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces. | In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs. | [
"5\n0 1\n1 0\n0 1\n1 1\n0 1\n"
] | [
"3\n"
] | none | 500 | [
{
"input": "5\n0 1\n1 0\n0 1\n1 1\n0 1",
"output": "3"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1\n1 1\n1 1",
"output": "1"
},
{
"input": "8\n0 1\n1 0\n0 1\n1 1\n0 1\n1 0\n0 1\n1 0",
"output": "7"
},
{
"input": "8\n1 0\n1 0\n1 0\n0 1\n0 1\n1 ... | 1,618,061,035 | 2,147,483,647 | Python 3 | OK | TESTS | 57 | 154 | 0 | n=int(input())
L=0
R=0
for i in range(n):
l,r=map(int,input().split())
L+=l
R+=r
print(min(L,n-L)+min(R,n-R)) | Title: Cupboards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One foggy Stockholm morning, Karlsson decided to snack on some jam in his friend Lillebror Svantenson's house. Fortunately for Karlsson, there wasn't anybody in his friend's house. Karlsson was not going to be hungry any longer, so he decided to get some food in the house.
Karlsson's gaze immediately fell on *n* wooden cupboards, standing in the kitchen. He immediately realized that these cupboards have hidden jam stocks. Karlsson began to fly greedily around the kitchen, opening and closing the cupboards' doors, grab and empty all the jars of jam that he could find.
And now all jars of jam are empty, Karlsson has had enough and does not want to leave traces of his stay, so as not to let down his friend. Each of the cupboards has two doors: the left one and the right one. Karlsson remembers that when he rushed to the kitchen, all the cupboards' left doors were in the same position (open or closed), similarly, all the cupboards' right doors were in the same position (open or closed). Karlsson wants the doors to meet this condition as well by the time the family returns. Karlsson does not remember the position of all the left doors, also, he cannot remember the position of all the right doors. Therefore, it does not matter to him in what position will be all left or right doors. It is important to leave all the left doors in the same position, and all the right doors in the same position. For example, all the left doors may be closed, and all the right ones may be open.
Karlsson needs one second to open or close a door of a cupboard. He understands that he has very little time before the family returns, so he wants to know the minimum number of seconds *t*, in which he is able to bring all the cupboard doors in the required position.
Your task is to write a program that will determine the required number of seconds *t*.
Input Specification:
The first input line contains a single integer *n* — the number of cupboards in the kitchen (2<=≤<=*n*<=≤<=104). Then follow *n* lines, each containing two integers *l**i* and *r**i* (0<=≤<=*l**i*,<=*r**i*<=≤<=1). Number *l**i* equals one, if the left door of the *i*-th cupboard is opened, otherwise number *l**i* equals zero. Similarly, number *r**i* equals one, if the right door of the *i*-th cupboard is opened, otherwise number *r**i* equals zero.
The numbers in the lines are separated by single spaces.
Output Specification:
In the only output line print a single integer *t* — the minimum number of seconds Karlsson needs to change the doors of all cupboards to the position he needs.
Demo Input:
['5\n0 1\n1 0\n0 1\n1 1\n0 1\n']
Demo Output:
['3\n']
Note:
none | ```python
n=int(input())
L=0
R=0
for i in range(n):
l,r=map(int,input().split())
L+=l
R+=r
print(min(L,n-L)+min(R,n-R))
``` | 3 | |
992 | A | Nastya and an Array | PROGRAMMING | 800 | [
"implementation",
"sortings"
] | null | null | Nastya owns too many arrays now, so she wants to delete the least important of them. However, she discovered that this array is magic! Nastya now knows that the array has the following properties:
- In one second we can add an arbitrary (possibly negative) integer to all elements of the array that are not equal to zero. - When all elements of the array become equal to zero, the array explodes.
Nastya is always busy, so she wants to explode the array as fast as possible. Compute the minimum time in which the array can be exploded. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=105<=≤<=*a**i*<=≤<=105) — the elements of the array. | Print a single integer — the minimum number of seconds needed to make all elements of the array equal to zero. | [
"5\n1 1 1 1 1\n",
"3\n2 0 -1\n",
"4\n5 -6 -5 1\n"
] | [
"1\n",
"2\n",
"4\n"
] | In the first example you can add - 1 to all non-zero elements in one second and make them equal to zero.
In the second example you can add - 2 on the first second, then the array becomes equal to [0, 0, - 3]. On the second second you can add 3 to the third (the only non-zero) element. | 500 | [
{
"input": "5\n1 1 1 1 1",
"output": "1"
},
{
"input": "3\n2 0 -1",
"output": "2"
},
{
"input": "4\n5 -6 -5 1",
"output": "4"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "2\n21794 -79194",
"output": "2"
},
{
"input": "3\n-63526 95085 -5239",
... | 1,640,939,273 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 79 | 186 | 17,305,600 | n=int(input())
l=list(map(int,input().split()))
s=set(l)
x=len(s)
if 0 in s:
print(x-1)
else:
print(x) | Title: Nastya and an Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Nastya owns too many arrays now, so she wants to delete the least important of them. However, she discovered that this array is magic! Nastya now knows that the array has the following properties:
- In one second we can add an arbitrary (possibly negative) integer to all elements of the array that are not equal to zero. - When all elements of the array become equal to zero, the array explodes.
Nastya is always busy, so she wants to explode the array as fast as possible. Compute the minimum time in which the array can be exploded.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the size of the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=105<=≤<=*a**i*<=≤<=105) — the elements of the array.
Output Specification:
Print a single integer — the minimum number of seconds needed to make all elements of the array equal to zero.
Demo Input:
['5\n1 1 1 1 1\n', '3\n2 0 -1\n', '4\n5 -6 -5 1\n']
Demo Output:
['1\n', '2\n', '4\n']
Note:
In the first example you can add - 1 to all non-zero elements in one second and make them equal to zero.
In the second example you can add - 2 on the first second, then the array becomes equal to [0, 0, - 3]. On the second second you can add 3 to the third (the only non-zero) element. | ```python
n=int(input())
l=list(map(int,input().split()))
s=set(l)
x=len(s)
if 0 in s:
print(x-1)
else:
print(x)
``` | 3 | |
519 | B | A and B and Compilation Errors | PROGRAMMING | 1,100 | [
"data structures",
"implementation",
"sortings"
] | null | null | A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected? | The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one. | Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively. | [
"5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n",
"6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n"
] | [
"8\n123\n",
"1\n3\n"
] | In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | 1,000 | [
{
"input": "5\n1 5 8 123 7\n123 7 5 1\n5 1 7",
"output": "8\n123"
},
{
"input": "6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5",
"output": "1\n3"
},
{
"input": "3\n1 2 3\n3 2\n2",
"output": "1\n3"
},
{
"input": "10\n460626451 802090732 277246428 661369649 388684428 784303821 376287098 6... | 1,640,709,733 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 265 | 13,926,400 | n = int(input())
a = [int(x)for x in input().split()][:n]
a.sort()
a1 = [int(x)for x in input().split()][:n-1]
a1.sort()
a2 = [int(x)for x in input().split()][:n-2]
a2.sort()
for i in range(n-1):
if a[i]!= a1[i]:
print(a[i])
break
else:
print(a[-1])
for i in range(n-2):
if a1[i]!= a2[i]:
print(a1[i])
break
else:
print(a1[-1]) | Title: A and B and Compilation Errors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output Specification:
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Demo Input:
['5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n', '6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n']
Demo Output:
['8\n123\n', '1\n3\n']
Note:
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step. | ```python
n = int(input())
a = [int(x)for x in input().split()][:n]
a.sort()
a1 = [int(x)for x in input().split()][:n-1]
a1.sort()
a2 = [int(x)for x in input().split()][:n-2]
a2.sort()
for i in range(n-1):
if a[i]!= a1[i]:
print(a[i])
break
else:
print(a[-1])
for i in range(n-2):
if a1[i]!= a2[i]:
print(a1[i])
break
else:
print(a1[-1])
``` | 3 | |
939 | A | Love Triangle | PROGRAMMING | 800 | [
"graphs"
] | null | null | As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are *n* planes on Earth, numbered from 1 to *n*, and the plane with number *i* likes the plane with number *f**i*, where 1<=≤<=*f**i*<=≤<=*n* and *f**i*<=≠<=*i*.
We call a love triangle a situation in which plane *A* likes plane *B*, plane *B* likes plane *C* and plane *C* likes plane *A*. Find out if there is any love triangle on Earth. | The first line contains a single integer *n* (2<=≤<=*n*<=≤<=5000) — the number of planes.
The second line contains *n* integers *f*1,<=*f*2,<=...,<=*f**n* (1<=≤<=*f**i*<=≤<=*n*, *f**i*<=≠<=*i*), meaning that the *i*-th plane likes the *f**i*-th. | Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case. | [
"5\n2 4 5 1 3\n",
"5\n5 5 5 5 1\n"
] | [
"YES\n",
"NO\n"
] | In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles. | 500 | [
{
"input": "5\n2 4 5 1 3",
"output": "YES"
},
{
"input": "5\n5 5 5 5 1",
"output": "NO"
},
{
"input": "3\n3 1 2",
"output": "YES"
},
{
"input": "10\n4 10 9 5 3 1 5 10 6 4",
"output": "NO"
},
{
"input": "10\n5 5 4 9 10 9 9 5 3 1",
"output": "YES"
},
{
"... | 1,613,341,941 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 1 | 62 | 307,200 | #collaborated with Bhumi Patel
n=int(input())
string_input=input()
string_input=string_input.split()
array=[]
for i in range(n):
if int(string_input[i])==n:
continue
array.append(int(string_input[i]))
if(len(array)<3):
print("n")
counter=0
for i in range(len(array)):
temp1=array[i]
temp2=array[i+1]
temp3=array[i+2]
if temp3-1 == i:
print("YES")
counter=1
break
if counter==0:
print("NO")
| Title: Love Triangle
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are *n* planes on Earth, numbered from 1 to *n*, and the plane with number *i* likes the plane with number *f**i*, where 1<=≤<=*f**i*<=≤<=*n* and *f**i*<=≠<=*i*.
We call a love triangle a situation in which plane *A* likes plane *B*, plane *B* likes plane *C* and plane *C* likes plane *A*. Find out if there is any love triangle on Earth.
Input Specification:
The first line contains a single integer *n* (2<=≤<=*n*<=≤<=5000) — the number of planes.
The second line contains *n* integers *f*1,<=*f*2,<=...,<=*f**n* (1<=≤<=*f**i*<=≤<=*n*, *f**i*<=≠<=*i*), meaning that the *i*-th plane likes the *f**i*-th.
Output Specification:
Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».
You can output any letter in lower case or in upper case.
Demo Input:
['5\n2 4 5 1 3\n', '5\n5 5 5 5 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In first example plane 2 likes plane 4, plane 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.
In second example there are no love triangles. | ```python
#collaborated with Bhumi Patel
n=int(input())
string_input=input()
string_input=string_input.split()
array=[]
for i in range(n):
if int(string_input[i])==n:
continue
array.append(int(string_input[i]))
if(len(array)<3):
print("n")
counter=0
for i in range(len(array)):
temp1=array[i]
temp2=array[i+1]
temp3=array[i+2]
if temp3-1 == i:
print("YES")
counter=1
break
if counter==0:
print("NO")
``` | -1 | |
370 | B | Berland Bingo | PROGRAMMING | 1,300 | [
"implementation"
] | null | null | Lately, a national version of a bingo game has become very popular in Berland. There are *n* players playing the game, each player has a card with numbers. The numbers on each card are distinct, but distinct cards can have equal numbers. The card of the *i*-th player contains *m**i* numbers.
During the game the host takes numbered balls one by one from a bag. He reads the number aloud in a high and clear voice and then puts the ball away. All participants cross out the number if it occurs on their cards. The person who crosses out all numbers from his card first, wins. If multiple people cross out all numbers from their cards at the same time, there are no winners in the game. At the beginning of the game the bag contains 100 balls numbered 1 through 100, the numbers of all balls are distinct.
You are given the cards for each player. Write a program that determines whether a player can win the game at the most favorable for him scenario or not. | The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of the players. Then follow *n* lines, each line describes a player's card. The line that describes a card starts from integer *m**i* (1<=≤<=*m**i*<=≤<=100) that shows how many numbers the *i*-th player's card has. Then follows a sequence of integers *a**i*,<=1,<=*a**i*,<=2,<=...,<=*a**i*,<=*m**i* (1<=≤<=*a**i*,<=*k*<=≤<=100) — the numbers on the *i*-th player's card. The numbers in the lines are separated by single spaces.
It is guaranteed that all the numbers on each card are distinct. | Print *n* lines, the *i*-th line must contain word "YES" (without the quotes), if the *i*-th player can win, and "NO" (without the quotes) otherwise. | [
"3\n1 1\n3 2 4 1\n2 10 11\n",
"2\n1 1\n1 1\n"
] | [
"YES\nNO\nYES\n",
"NO\nNO\n"
] | none | 500 | [
{
"input": "3\n1 1\n3 2 4 1\n2 10 11",
"output": "YES\nNO\nYES"
},
{
"input": "2\n1 1\n1 1",
"output": "NO\nNO"
},
{
"input": "1\n1 1",
"output": "YES"
},
{
"input": "2\n1 2\n1 3",
"output": "YES\nYES"
},
{
"input": "2\n1 1\n2 1 2",
"output": "YES\nNO"
},
... | 1,386,439,393 | 2,147,483,647 | Python 3 | OK | TESTS | 24 | 78 | 307,200 | def contiene(l, s, n, m):
j=0
for i in range(n):
x = l[i]
while j<m and not l[i]==s[j]:
j+=1
if j==m and not i==n:
return False
return True
if __name__=='__main__':
n = int(input())
l = []
m = []
for i in range(n):
s = [int(x) for x in input().split()]
m.append( s[0] )
l.append( s[1:] )
l[i].sort()
for i in range(n):
canWin = True
for j in range(n):
if not j==i:
if m[i]<m[j]:
continue
if contiene( l[j], l[i], m[j], m[i] ):
canWin = False
break
if canWin:
print('YES')
else:
print('NO')
| Title: Berland Bingo
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Lately, a national version of a bingo game has become very popular in Berland. There are *n* players playing the game, each player has a card with numbers. The numbers on each card are distinct, but distinct cards can have equal numbers. The card of the *i*-th player contains *m**i* numbers.
During the game the host takes numbered balls one by one from a bag. He reads the number aloud in a high and clear voice and then puts the ball away. All participants cross out the number if it occurs on their cards. The person who crosses out all numbers from his card first, wins. If multiple people cross out all numbers from their cards at the same time, there are no winners in the game. At the beginning of the game the bag contains 100 balls numbered 1 through 100, the numbers of all balls are distinct.
You are given the cards for each player. Write a program that determines whether a player can win the game at the most favorable for him scenario or not.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of the players. Then follow *n* lines, each line describes a player's card. The line that describes a card starts from integer *m**i* (1<=≤<=*m**i*<=≤<=100) that shows how many numbers the *i*-th player's card has. Then follows a sequence of integers *a**i*,<=1,<=*a**i*,<=2,<=...,<=*a**i*,<=*m**i* (1<=≤<=*a**i*,<=*k*<=≤<=100) — the numbers on the *i*-th player's card. The numbers in the lines are separated by single spaces.
It is guaranteed that all the numbers on each card are distinct.
Output Specification:
Print *n* lines, the *i*-th line must contain word "YES" (without the quotes), if the *i*-th player can win, and "NO" (without the quotes) otherwise.
Demo Input:
['3\n1 1\n3 2 4 1\n2 10 11\n', '2\n1 1\n1 1\n']
Demo Output:
['YES\nNO\nYES\n', 'NO\nNO\n']
Note:
none | ```python
def contiene(l, s, n, m):
j=0
for i in range(n):
x = l[i]
while j<m and not l[i]==s[j]:
j+=1
if j==m and not i==n:
return False
return True
if __name__=='__main__':
n = int(input())
l = []
m = []
for i in range(n):
s = [int(x) for x in input().split()]
m.append( s[0] )
l.append( s[1:] )
l[i].sort()
for i in range(n):
canWin = True
for j in range(n):
if not j==i:
if m[i]<m[j]:
continue
if contiene( l[j], l[i], m[j], m[i] ):
canWin = False
break
if canWin:
print('YES')
else:
print('NO')
``` | 3 | |
854 | A | Fraction | PROGRAMMING | 800 | [
"brute force",
"constructive algorithms",
"math"
] | null | null | Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem. | In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction. | Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum. | [
"3\n",
"4\n",
"12\n"
] | [
"1 2\n",
"1 3\n",
"5 7\n"
] | none | 500 | [
{
"input": "3",
"output": "1 2"
},
{
"input": "4",
"output": "1 3"
},
{
"input": "12",
"output": "5 7"
},
{
"input": "34",
"output": "15 19"
},
{
"input": "13",
"output": "6 7"
},
{
"input": "11",
"output": "5 6"
},
{
"input": "24",
"ou... | 1,540,473,660 | 2,147,483,647 | Python 3 | OK | TESTS | 26 | 124 | 0 | from math import gcd
def inp():
return int(input())
def linp():
return list(map(int, input().split(' ')))
def minp():
return map(int, input().split(' '))
ls = []
n = inp()
for i in range(n//2+1):
k = n-i
if gcd(k,i)==1 :
ls.append(max(k,i))
ls = sorted(ls)
print((n-ls[0]), end=" ")
print(ls[0])
| Title: Fraction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (*a*<=<<=*b*) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).
During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button (<=+<=) instead of division button (÷) and got sum of numerator and denominator that was equal to *n* instead of the expected decimal notation.
Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That's why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals *n*. Help Petya deal with this problem.
Input Specification:
In the only line of input there is an integer *n* (3<=≤<=*n*<=≤<=1000), the sum of numerator and denominator of the fraction.
Output Specification:
Output two space-separated positive integers *a* and *b*, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.
Demo Input:
['3\n', '4\n', '12\n']
Demo Output:
['1 2\n', '1 3\n', '5 7\n']
Note:
none | ```python
from math import gcd
def inp():
return int(input())
def linp():
return list(map(int, input().split(' ')))
def minp():
return map(int, input().split(' '))
ls = []
n = inp()
for i in range(n//2+1):
k = n-i
if gcd(k,i)==1 :
ls.append(max(k,i))
ls = sorted(ls)
print((n-ls[0]), end=" ")
print(ls[0])
``` | 3 | |
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... | 1,694,151,039 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | import java.awt.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class Main {
static int mod = (int) 1e9 + 7;
static boolean[] vis;
public static void main(String[] args) throws Exception {
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
Point[] p=new Point[n];
for(int i=0;i<n;i++)p[i]=new Point(sc.nextInt(),sc.nextInt());
vis=new boolean[n];
int count=0;
for(int i=0;i<n;i++){
if(!vis[i]){
dfs(i,n,p);
count++;
}
}
System.out.println(count-1);
}
static void dfs(int i,int n,Point[] p){
vis[i]=true;
for(int j=0;j<n;j++){
if(!vis[j]){
if(p[i].x==p[j].x || p[i].y==p[j].y){
dfs(j,n,p);
}
}
}
}
} | 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
import java.awt.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class Main {
static int mod = (int) 1e9 + 7;
static boolean[] vis;
public static void main(String[] args) throws Exception {
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
Point[] p=new Point[n];
for(int i=0;i<n;i++)p[i]=new Point(sc.nextInt(),sc.nextInt());
vis=new boolean[n];
int count=0;
for(int i=0;i<n;i++){
if(!vis[i]){
dfs(i,n,p);
count++;
}
}
System.out.println(count-1);
}
static void dfs(int i,int n,Point[] p){
vis[i]=true;
for(int j=0;j<n;j++){
if(!vis[j]){
if(p[i].x==p[j].x || p[i].y==p[j].y){
dfs(j,n,p);
}
}
}
}
}
``` | -1 | |
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"... | 1,658,078,374 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 46 | 0 | import math
n,m,a=list(map(int,input().split(" ")))
area=n*m
print(math.ceil(m/a)*math.ceil(n/a)) | 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
import math
n,m,a=list(map(int,input().split(" ")))
area=n*m
print(math.ceil(m/a)*math.ceil(n/a))
``` | 3.977 |
9 | A | Die Roll | PROGRAMMING | 800 | [
"math",
"probabilities"
] | A. Die Roll | 1 | 64 | Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win. | The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls. | Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1». | [
"4 2\n"
] | [
"1/2\n"
] | Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | 0 | [
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
... | 1,627,115,801 | 2,147,483,647 | Python 3 | OK | TESTS | 36 | 124 | 6,758,400 | def gcd(a, b):
while b != 0:
a, b = b, a % b
return a
a, b = map(int, input().split())
ans = 6 - max(a, b) + 1
g = gcd(ans, 6)
print(ans // g, '/', 6 // g, sep='')
# Sat Jul 24 2021 11:36:40 GMT+0300 (Москва, стандартное время)
| Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | ```python
def gcd(a, b):
while b != 0:
a, b = b, a % b
return a
a, b = map(int, input().split())
ans = 6 - max(a, b) + 1
g = gcd(ans, 6)
print(ans // g, '/', 6 // g, sep='')
# Sat Jul 24 2021 11:36:40 GMT+0300 (Москва, стандартное время)
``` | 3.887646 |
777 | C | Alyona and Spreadsheet | PROGRAMMING | 1,600 | [
"binary search",
"data structures",
"dp",
"greedy",
"implementation",
"two pointers"
] | null | null | During the lesson small girl Alyona works with one famous spreadsheet computer program and learns how to edit tables.
Now she has a table filled with integers. The table consists of *n* rows and *m* columns. By *a**i*,<=*j* we will denote the integer located at the *i*-th row and the *j*-th column. We say that the table is sorted in non-decreasing order in the column *j* if *a**i*,<=*j*<=≤<=*a**i*<=+<=1,<=*j* for all *i* from 1 to *n*<=-<=1.
Teacher gave Alyona *k* tasks. For each of the tasks two integers *l* and *r* are given and Alyona has to answer the following question: if one keeps the rows from *l* to *r* inclusive and deletes all others, will the table be sorted in non-decreasing order in at least one column? Formally, does there exist such *j* that *a**i*,<=*j*<=≤<=*a**i*<=+<=1,<=*j* for all *i* from *l* to *r*<=-<=1 inclusive.
Alyona is too small to deal with this task and asks you to help! | The first line of the input contains two positive integers *n* and *m* (1<=≤<=*n*·*m*<=≤<=100<=000) — the number of rows and the number of columns in the table respectively. Note that your are given a constraint that bound the product of these two integers, i.e. the number of elements in the table.
Each of the following *n* lines contains *m* integers. The *j*-th integers in the *i* of these lines stands for *a**i*,<=*j* (1<=≤<=*a**i*,<=*j*<=≤<=109).
The next line of the input contains an integer *k* (1<=≤<=*k*<=≤<=100<=000) — the number of task that teacher gave to Alyona.
The *i*-th of the next *k* lines contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). | Print "Yes" to the *i*-th line of the output if the table consisting of rows from *l**i* to *r**i* inclusive is sorted in non-decreasing order in at least one column. Otherwise, print "No". | [
"5 4\n1 2 3 5\n3 1 3 2\n4 5 2 3\n5 5 3 2\n4 4 3 4\n6\n1 1\n2 5\n4 5\n3 5\n1 3\n1 5\n"
] | [
"Yes\nNo\nYes\nYes\nYes\nNo\n"
] | In the sample, the whole table is not sorted in any column. However, rows 1–3 are sorted in column 1, while rows 4–5 are sorted in column 3. | 1,500 | [
{
"input": "5 4\n1 2 3 5\n3 1 3 2\n4 5 2 3\n5 5 3 2\n4 4 3 4\n6\n1 1\n2 5\n4 5\n3 5\n1 3\n1 5",
"output": "Yes\nNo\nYes\nYes\nYes\nNo"
},
{
"input": "1 1\n1\n1\n1 1",
"output": "Yes"
},
{
"input": "10 1\n523130301\n127101624\n15573616\n703140639\n628818570\n957494759\n161270109\n38686565... | 1,599,924,454 | 6,154 | Python 3 | WRONG_ANSWER | TESTS | 0 | 93 | 307,200 | n,m=map(int,input().split())
t=[]
a=[]
b=0
c=0
for i in range(n):
t.append(list(map(int,input().split())))
t.append([0 for i in range(m)])
k=int(input())
for i in range(k):
l,r=map(int,input().split())
for j in range(m):
for y in range(l+1,r+1):
if t[y][j]<t[y-1][j]:
b=1
break
if b==0:
a.append("Yes")
a.append('\n')
c=1
break
b=0
if c==0:
a.append("No")
a.append('\n')
c=0
print(*a) | Title: Alyona and Spreadsheet
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
During the lesson small girl Alyona works with one famous spreadsheet computer program and learns how to edit tables.
Now she has a table filled with integers. The table consists of *n* rows and *m* columns. By *a**i*,<=*j* we will denote the integer located at the *i*-th row and the *j*-th column. We say that the table is sorted in non-decreasing order in the column *j* if *a**i*,<=*j*<=≤<=*a**i*<=+<=1,<=*j* for all *i* from 1 to *n*<=-<=1.
Teacher gave Alyona *k* tasks. For each of the tasks two integers *l* and *r* are given and Alyona has to answer the following question: if one keeps the rows from *l* to *r* inclusive and deletes all others, will the table be sorted in non-decreasing order in at least one column? Formally, does there exist such *j* that *a**i*,<=*j*<=≤<=*a**i*<=+<=1,<=*j* for all *i* from *l* to *r*<=-<=1 inclusive.
Alyona is too small to deal with this task and asks you to help!
Input Specification:
The first line of the input contains two positive integers *n* and *m* (1<=≤<=*n*·*m*<=≤<=100<=000) — the number of rows and the number of columns in the table respectively. Note that your are given a constraint that bound the product of these two integers, i.e. the number of elements in the table.
Each of the following *n* lines contains *m* integers. The *j*-th integers in the *i* of these lines stands for *a**i*,<=*j* (1<=≤<=*a**i*,<=*j*<=≤<=109).
The next line of the input contains an integer *k* (1<=≤<=*k*<=≤<=100<=000) — the number of task that teacher gave to Alyona.
The *i*-th of the next *k* lines contains two integers *l**i* and *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*).
Output Specification:
Print "Yes" to the *i*-th line of the output if the table consisting of rows from *l**i* to *r**i* inclusive is sorted in non-decreasing order in at least one column. Otherwise, print "No".
Demo Input:
['5 4\n1 2 3 5\n3 1 3 2\n4 5 2 3\n5 5 3 2\n4 4 3 4\n6\n1 1\n2 5\n4 5\n3 5\n1 3\n1 5\n']
Demo Output:
['Yes\nNo\nYes\nYes\nYes\nNo\n']
Note:
In the sample, the whole table is not sorted in any column. However, rows 1–3 are sorted in column 1, while rows 4–5 are sorted in column 3. | ```python
n,m=map(int,input().split())
t=[]
a=[]
b=0
c=0
for i in range(n):
t.append(list(map(int,input().split())))
t.append([0 for i in range(m)])
k=int(input())
for i in range(k):
l,r=map(int,input().split())
for j in range(m):
for y in range(l+1,r+1):
if t[y][j]<t[y-1][j]:
b=1
break
if b==0:
a.append("Yes")
a.append('\n')
c=1
break
b=0
if c==0:
a.append("No")
a.append('\n')
c=0
print(*a)
``` | 0 | |
501 | A | Contest | PROGRAMMING | 900 | [
"implementation"
] | null | null | Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points.
Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth. | The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180).
It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round). | Output on a single line:
"Misha" (without the quotes), if Misha got more points than Vasya.
"Vasya" (without the quotes), if Vasya got more points than Misha.
"Tie" (without the quotes), if both of them got the same number of points. | [
"500 1000 20 30\n",
"1000 1000 1 1\n",
"1500 1000 176 177\n"
] | [
"Vasya\n",
"Tie\n",
"Misha\n"
] | none | 500 | [
{
"input": "500 1000 20 30",
"output": "Vasya"
},
{
"input": "1000 1000 1 1",
"output": "Tie"
},
{
"input": "1500 1000 176 177",
"output": "Misha"
},
{
"input": "1500 1000 74 177",
"output": "Misha"
},
{
"input": "750 2500 175 178",
"output": "Vasya"
},
{
... | 1,655,313,067 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 46 | 0 | l1= list(map(int, input().split()))
a=l1[0]
b=l1[1]
c=l1[2]
d=l1[3]
m=max((3*a/10),a-(a*c/250))
n=max((3*b/10),b-(b*d/250))
if n>m:
print("Vasya")
elif m>n:
print("Misha")
else:
print("Tie")
| Title: Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points.
Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth.
Input Specification:
The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180).
It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round).
Output Specification:
Output on a single line:
"Misha" (without the quotes), if Misha got more points than Vasya.
"Vasya" (without the quotes), if Vasya got more points than Misha.
"Tie" (without the quotes), if both of them got the same number of points.
Demo Input:
['500 1000 20 30\n', '1000 1000 1 1\n', '1500 1000 176 177\n']
Demo Output:
['Vasya\n', 'Tie\n', 'Misha\n']
Note:
none | ```python
l1= list(map(int, input().split()))
a=l1[0]
b=l1[1]
c=l1[2]
d=l1[3]
m=max((3*a/10),a-(a*c/250))
n=max((3*b/10),b-(b*d/250))
if n>m:
print("Vasya")
elif m>n:
print("Misha")
else:
print("Tie")
``` | 3 | |
279 | B | Books | PROGRAMMING | 1,400 | [
"binary search",
"brute force",
"implementation",
"two pointers"
] | null | null | When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read. | The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book. | Print a single integer — the maximum number of books Valera can read. | [
"4 5\n3 1 2 1\n",
"3 3\n2 2 3\n"
] | [
"3\n",
"1\n"
] | none | 1,000 | [
{
"input": "4 5\n3 1 2 1",
"output": "3"
},
{
"input": "3 3\n2 2 3",
"output": "1"
},
{
"input": "1 3\n5",
"output": "0"
},
{
"input": "1 10\n4",
"output": "1"
},
{
"input": "2 10\n6 4",
"output": "2"
},
{
"input": "6 10\n2 3 4 2 1 1",
"output": "4... | 1,686,926,866 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 92 | 0 | n,i = map(int, input().split())
numbers = list(map(int, input().split()))
sffix_sum = [0]
c=[]
co=1
for x in range(n-1,-1,-1):
if numbers[x]>i:
co=1
sffix_sum.insert(0, 0)
continue
sffix_sum.insert(0,sffix_sum[0] + numbers[x])
c.insert(0,co)
co+=1
print(sffix_sum,c)
for x in range(n):
if sffix_sum[x]<=i:
print(c[x])
break
| Title: Books
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
When Valera has got some free time, he goes to the library to read some books. Today he's got *t* free minutes to read. That's why Valera took *n* books in the library and for each book he estimated the time he is going to need to read it. Let's number the books by integers from 1 to *n*. Valera needs *a**i* minutes to read the *i*-th book.
Valera decided to choose an arbitrary book with number *i* and read the books one by one, starting from this book. In other words, he will first read book number *i*, then book number *i*<=+<=1, then book number *i*<=+<=2 and so on. He continues the process until he either runs out of the free time or finishes reading the *n*-th book. Valera reads each book up to the end, that is, he doesn't start reading the book if he doesn't have enough free time to finish reading it.
Print the maximum number of books Valera can read.
Input Specification:
The first line contains two integers *n* and *t* (1<=≤<=*n*<=≤<=105; 1<=≤<=*t*<=≤<=109) — the number of books and the number of free minutes Valera's got. The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=104), where number *a**i* shows the number of minutes that the boy needs to read the *i*-th book.
Output Specification:
Print a single integer — the maximum number of books Valera can read.
Demo Input:
['4 5\n3 1 2 1\n', '3 3\n2 2 3\n']
Demo Output:
['3\n', '1\n']
Note:
none | ```python
n,i = map(int, input().split())
numbers = list(map(int, input().split()))
sffix_sum = [0]
c=[]
co=1
for x in range(n-1,-1,-1):
if numbers[x]>i:
co=1
sffix_sum.insert(0, 0)
continue
sffix_sum.insert(0,sffix_sum[0] + numbers[x])
c.insert(0,co)
co+=1
print(sffix_sum,c)
for x in range(n):
if sffix_sum[x]<=i:
print(c[x])
break
``` | 0 | |
868 | D | Huge Strings | PROGRAMMING | 2,200 | [
"bitmasks",
"brute force",
"dp",
"implementation",
"strings"
] | null | null | You are given *n* strings *s*1,<=*s*2,<=...,<=*s**n* consisting of characters 0 and 1. *m* operations are performed, on each of them you concatenate two existing strings into a new one. On the *i*-th operation the concatenation *s**a**i**s**b**i* is saved into a new string *s**n*<=+<=*i* (the operations are numbered starting from 1). After each operation you need to find the maximum positive integer *k* such that all possible strings consisting of 0 and 1 of length *k* (there are 2*k* such strings) are substrings of the new string. If there is no such *k*, print 0. | The first line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of strings. The next *n* lines contain strings *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=|*s**i*|<=≤<=100), one per line. The total length of strings is not greater than 100.
The next line contains single integer *m* (1<=≤<=*m*<=≤<=100) — the number of operations. *m* lines follow, each of them contains two integers *a**i* abd *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*<=+<=*i*<=-<=1) — the number of strings that are concatenated to form *s**n*<=+<=*i*. | Print *m* lines, each should contain one integer — the answer to the question after the corresponding operation. | [
"5\n01\n10\n101\n11111\n0\n3\n1 2\n6 5\n4 4\n"
] | [
"1\n2\n0\n"
] | On the first operation, a new string "0110" is created. For *k* = 1 the two possible binary strings of length *k* are "0" and "1", they are substrings of the new string. For *k* = 2 and greater there exist strings of length *k* that do not appear in this string (for *k* = 2 such string is "00"). So the answer is 1.
On the second operation the string "01100" is created. Now all strings of length *k* = 2 are present.
On the third operation the string "1111111111" is created. There is no zero, so the answer is 0. | 1,500 | [
{
"input": "5\n01\n10\n101\n11111\n0\n3\n1 2\n6 5\n4 4",
"output": "1\n2\n0"
},
{
"input": "5\n01\n1\n0011\n0\n01\n6\n5 5\n3 2\n4 2\n6 7\n5 1\n9 7",
"output": "1\n1\n1\n2\n1\n2"
},
{
"input": "5\n111101000111100011100110000100\n000111001\n01101000\n000011010010001001100100000001010010011... | 1,510,843,810 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 93 | 819,200 | # -*- coding: utf-8 -*-
import math
import collections
import bisect
import heapq
import time
import random
"""
created by shhuan at 2017/10/5 15:00
"""
N = int(input())
S = ['']
head = [""]
tail = [""]
for i in range(N):
S.append(input())
head.append(S[i][:10])
tail.append(S[i][:-10])
M = int(input())
A = [[set() for _ in range(10)] for _ in range(M+N+1)]
C = collections.defaultdict(tuple)
D = collections.defaultdict(int)
for i in range(1, N+1):
for j in range(1, 10):
s = S[i]
if j > len(s):
break
for k in range(len(s)-j+1):
A[i][j].add(int(s[k:k+j], 2))
if all(v in A[i][j] for v in range(2**j)):
D[i] = j
for i in range(M):
a, b = map(int, input().split())
s = S[a] + S[b]
S.append(s)
head.append(s[:10])
tail.append(s[:-10])
ai = i+N+1
C[ai] = (a, b)
d = max(D[a], D[b]) + 1
for dv in range(d, 10):
if len(S[i]) + len(S[b]) < dv:
break
A[ai][dv] = {int(S[a][-i:] + S[b][:dv-i], 2) for i in range(1, dv+1)}
q = [(ai)]
oz = set()
while q:
x = q.pop()
if x in C:
q.append(C[x][0])
q.append(C[x][1])
oz |= A[x][d]
if all(v in oz for v in range(2**d)):
print(d)
D[ai] = d
else:
print(d-1)
D[ai] = d-1
| Title: Huge Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given *n* strings *s*1,<=*s*2,<=...,<=*s**n* consisting of characters 0 and 1. *m* operations are performed, on each of them you concatenate two existing strings into a new one. On the *i*-th operation the concatenation *s**a**i**s**b**i* is saved into a new string *s**n*<=+<=*i* (the operations are numbered starting from 1). After each operation you need to find the maximum positive integer *k* such that all possible strings consisting of 0 and 1 of length *k* (there are 2*k* such strings) are substrings of the new string. If there is no such *k*, print 0.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of strings. The next *n* lines contain strings *s*1,<=*s*2,<=...,<=*s**n* (1<=≤<=|*s**i*|<=≤<=100), one per line. The total length of strings is not greater than 100.
The next line contains single integer *m* (1<=≤<=*m*<=≤<=100) — the number of operations. *m* lines follow, each of them contains two integers *a**i* abd *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*<=+<=*i*<=-<=1) — the number of strings that are concatenated to form *s**n*<=+<=*i*.
Output Specification:
Print *m* lines, each should contain one integer — the answer to the question after the corresponding operation.
Demo Input:
['5\n01\n10\n101\n11111\n0\n3\n1 2\n6 5\n4 4\n']
Demo Output:
['1\n2\n0\n']
Note:
On the first operation, a new string "0110" is created. For *k* = 1 the two possible binary strings of length *k* are "0" and "1", they are substrings of the new string. For *k* = 2 and greater there exist strings of length *k* that do not appear in this string (for *k* = 2 such string is "00"). So the answer is 1.
On the second operation the string "01100" is created. Now all strings of length *k* = 2 are present.
On the third operation the string "1111111111" is created. There is no zero, so the answer is 0. | ```python
# -*- coding: utf-8 -*-
import math
import collections
import bisect
import heapq
import time
import random
"""
created by shhuan at 2017/10/5 15:00
"""
N = int(input())
S = ['']
head = [""]
tail = [""]
for i in range(N):
S.append(input())
head.append(S[i][:10])
tail.append(S[i][:-10])
M = int(input())
A = [[set() for _ in range(10)] for _ in range(M+N+1)]
C = collections.defaultdict(tuple)
D = collections.defaultdict(int)
for i in range(1, N+1):
for j in range(1, 10):
s = S[i]
if j > len(s):
break
for k in range(len(s)-j+1):
A[i][j].add(int(s[k:k+j], 2))
if all(v in A[i][j] for v in range(2**j)):
D[i] = j
for i in range(M):
a, b = map(int, input().split())
s = S[a] + S[b]
S.append(s)
head.append(s[:10])
tail.append(s[:-10])
ai = i+N+1
C[ai] = (a, b)
d = max(D[a], D[b]) + 1
for dv in range(d, 10):
if len(S[i]) + len(S[b]) < dv:
break
A[ai][dv] = {int(S[a][-i:] + S[b][:dv-i], 2) for i in range(1, dv+1)}
q = [(ai)]
oz = set()
while q:
x = q.pop()
if x in C:
q.append(C[x][0])
q.append(C[x][1])
oz |= A[x][d]
if all(v in oz for v in range(2**d)):
print(d)
D[ai] = d
else:
print(d-1)
D[ai] = d-1
``` | 0 | |
80 | A | Panoramix's Prediction | PROGRAMMING | 800 | [
"brute force"
] | A. Panoramix's Prediction | 2 | 256 | A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix? | The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4. | Print YES, if *m* is the next prime number after *n*, or NO otherwise. | [
"3 5\n",
"7 11\n",
"7 9\n"
] | [
"YES",
"YES",
"NO"
] | none | 500 | [
{
"input": "3 5",
"output": "YES"
},
{
"input": "7 11",
"output": "YES"
},
{
"input": "7 9",
"output": "NO"
},
{
"input": "2 3",
"output": "YES"
},
{
"input": "2 4",
"output": "NO"
},
{
"input": "3 4",
"output": "NO"
},
{
"input": "3 5",
... | 1,673,856,856 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 154 | 2,150,400 | x,y=map(int,input().split())
t=y
while(t>0):
if(y%t==0):
c=c+1
t=t-1
if(c>2):
print("NO")
else:
print("YES") | Title: Panoramix's Prediction
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A prime number is a number which has exactly two distinct divisors: one and itself. For example, numbers 2, 7, 3 are prime, and 1, 6, 4 are not.
The next prime number after *x* is the smallest prime number greater than *x*. For example, the next prime number after 2 is 3, and the next prime number after 3 is 5. Note that there is exactly one next prime number after each number. So 5 is not the next prime number for 2.
One cold April morning Panoramix predicted that soon Kakofonix will break free from his straitjacket, and this will be a black day for the residents of the Gallic countryside.
Panoramix's prophecy tells that if some day Asterix and Obelix beat exactly *x* Roman soldiers, where *x* is a prime number, and next day they beat exactly *y* Roman soldiers, where *y* is the next prime number after *x*, then it's time to wait for Armageddon, for nothing can shut Kakofonix up while he sings his infernal song.
Yesterday the Gauls beat *n* Roman soldiers and it turned out that the number *n* was prime! Today their victims were a troop of *m* Romans (*m*<=><=*n*). Determine whether the Gauls should wait for the black day after today's victory of Asterix and Obelix?
Input Specification:
The first and only input line contains two positive integers — *n* and *m* (2<=≤<=*n*<=<<=*m*<=≤<=50). It is guaranteed that *n* is prime.
Pretests contain all the cases with restrictions 2<=≤<=*n*<=<<=*m*<=≤<=4.
Output Specification:
Print YES, if *m* is the next prime number after *n*, or NO otherwise.
Demo Input:
['3 5\n', '7 11\n', '7 9\n']
Demo Output:
['YES', 'YES', 'NO']
Note:
none | ```python
x,y=map(int,input().split())
t=y
while(t>0):
if(y%t==0):
c=c+1
t=t-1
if(c>2):
print("NO")
else:
print("YES")
``` | -1 |
476 | A | Dreamoon and Stairs | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Dreamoon wants to climb up a stair of *n* steps. He can climb 1 or 2 steps at each move. Dreamoon wants the number of moves to be a multiple of an integer *m*.
What is the minimal number of moves making him climb to the top of the stairs that satisfies his condition? | The single line contains two space separated integers *n*, *m* (0<=<<=*n*<=≤<=10000,<=1<=<<=*m*<=≤<=10). | Print a single integer — the minimal number of moves being a multiple of *m*. If there is no way he can climb satisfying condition print <=-<=1 instead. | [
"10 2\n",
"3 5\n"
] | [
"6\n",
"-1\n"
] | For the first sample, Dreamoon could climb in 6 moves with following sequence of steps: {2, 2, 2, 2, 1, 1}.
For the second sample, there are only three valid sequence of steps {2, 1}, {1, 2}, {1, 1, 1} with 2, 2, and 3 steps respectively. All these numbers are not multiples of 5. | 500 | [
{
"input": "10 2",
"output": "6"
},
{
"input": "3 5",
"output": "-1"
},
{
"input": "29 7",
"output": "21"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "1 2",
"output": "-1"
},
{
"input": "10000 2",
"output": "5000"
},
{
"input": "10000 ... | 1,694,408,303 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 46 | 0 | def main():
n, m = [int(i) for i in input().split()]
if n < m:
print(-1)
else:
minsteps = n // 2 + n % 2
if minsteps % m == 0:
print(minsteps)
else:
print(minsteps + m - (minsteps % m))
if __name__ == "__main__":
main()
| Title: Dreamoon and Stairs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dreamoon wants to climb up a stair of *n* steps. He can climb 1 or 2 steps at each move. Dreamoon wants the number of moves to be a multiple of an integer *m*.
What is the minimal number of moves making him climb to the top of the stairs that satisfies his condition?
Input Specification:
The single line contains two space separated integers *n*, *m* (0<=<<=*n*<=≤<=10000,<=1<=<<=*m*<=≤<=10).
Output Specification:
Print a single integer — the minimal number of moves being a multiple of *m*. If there is no way he can climb satisfying condition print <=-<=1 instead.
Demo Input:
['10 2\n', '3 5\n']
Demo Output:
['6\n', '-1\n']
Note:
For the first sample, Dreamoon could climb in 6 moves with following sequence of steps: {2, 2, 2, 2, 1, 1}.
For the second sample, there are only three valid sequence of steps {2, 1}, {1, 2}, {1, 1, 1} with 2, 2, and 3 steps respectively. All these numbers are not multiples of 5. | ```python
def main():
n, m = [int(i) for i in input().split()]
if n < m:
print(-1)
else:
minsteps = n // 2 + n % 2
if minsteps % m == 0:
print(minsteps)
else:
print(minsteps + m - (minsteps % m))
if __name__ == "__main__":
main()
``` | 3 | |
59 | A | Word | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Word | 2 | 256 | Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. | The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. | Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. | [
"HoUse\n",
"ViP\n",
"maTRIx\n"
] | [
"house\n",
"VIP\n",
"matrix\n"
] | none | 500 | [
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chn... | 1,577,017,842 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 218 | 307,200 | s=input()
l=0
u=0
for i in s:
if(i==i.lower()):
l+=1
elif(i==i.upper()):
u+=1
if(l>=u):
print(s.lower())
elif(u>l):
print(s.upper())
| Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none | ```python
s=input()
l=0
u=0
for i in s:
if(i==i.lower()):
l+=1
elif(i==i.upper()):
u+=1
if(l>=u):
print(s.lower())
elif(u>l):
print(s.upper())
``` | 3.944928 |
817 | B | Makes And The Product | PROGRAMMING | 1,500 | [
"combinatorics",
"implementation",
"math",
"sortings"
] | null | null | After returning from the army Makes received a gift — an array *a* consisting of *n* positive integer numbers. He hadn't been solving problems for a long time, so he became interested to answer a particular question: how many triples of indices (*i*,<= *j*,<= *k*) (*i*<=<<=*j*<=<<=*k*), such that *a**i*·*a**j*·*a**k* is minimum possible, are there in the array? Help him with it! | The first line of input contains a positive integer number *n* (3<=≤<=*n*<=≤<=105) — the number of elements in array *a*. The second line contains *n* positive integer numbers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of a given array. | Print one number — the quantity of triples (*i*,<= *j*,<= *k*) such that *i*,<= *j* and *k* are pairwise distinct and *a**i*·*a**j*·*a**k* is minimum possible. | [
"4\n1 1 1 1\n",
"5\n1 3 2 3 4\n",
"6\n1 3 3 1 3 2\n"
] | [
"4\n",
"2\n",
"1\n"
] | In the first example Makes always chooses three ones out of four, and the number of ways to choose them is 4.
In the second example a triple of numbers (1, 2, 3) is chosen (numbers, not indices). Since there are two ways to choose an element 3, then the answer is 2.
In the third example a triple of numbers (1, 1, 2) is chosen, and there's only one way to choose indices. | 0 | [
{
"input": "4\n1 1 1 1",
"output": "4"
},
{
"input": "5\n1 3 2 3 4",
"output": "2"
},
{
"input": "6\n1 3 3 1 3 2",
"output": "1"
},
{
"input": "3\n1000000000 1000000000 1000000000",
"output": "1"
},
{
"input": "4\n1 1 2 2",
"output": "2"
},
{
"input": ... | 1,604,841,435 | 2,147,483,647 | PyPy 3 | OK | TESTS | 80 | 265 | 10,956,800 | def ncr(n, r):
res = 1
for i in range(1, r + 1):
res = (res * (n + 1 - i) // i)
return res
n = int(input())
arr = [*map(int, input().split())]
a, b, c = sorted(arr)[:3]
ca, cb, cc = arr.count(a), arr.count(b), arr.count(c)
if a == b == c:
print(ncr(ca, 3))
elif a == b:
print(ncr(ca, 2) * cc)
elif b == c:
print(ncr(cb, 2) * ca)
else:
print(ca * cb * cc) | Title: Makes And The Product
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After returning from the army Makes received a gift — an array *a* consisting of *n* positive integer numbers. He hadn't been solving problems for a long time, so he became interested to answer a particular question: how many triples of indices (*i*,<= *j*,<= *k*) (*i*<=<<=*j*<=<<=*k*), such that *a**i*·*a**j*·*a**k* is minimum possible, are there in the array? Help him with it!
Input Specification:
The first line of input contains a positive integer number *n* (3<=≤<=*n*<=≤<=105) — the number of elements in array *a*. The second line contains *n* positive integer numbers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of a given array.
Output Specification:
Print one number — the quantity of triples (*i*,<= *j*,<= *k*) such that *i*,<= *j* and *k* are pairwise distinct and *a**i*·*a**j*·*a**k* is minimum possible.
Demo Input:
['4\n1 1 1 1\n', '5\n1 3 2 3 4\n', '6\n1 3 3 1 3 2\n']
Demo Output:
['4\n', '2\n', '1\n']
Note:
In the first example Makes always chooses three ones out of four, and the number of ways to choose them is 4.
In the second example a triple of numbers (1, 2, 3) is chosen (numbers, not indices). Since there are two ways to choose an element 3, then the answer is 2.
In the third example a triple of numbers (1, 1, 2) is chosen, and there's only one way to choose indices. | ```python
def ncr(n, r):
res = 1
for i in range(1, r + 1):
res = (res * (n + 1 - i) // i)
return res
n = int(input())
arr = [*map(int, input().split())]
a, b, c = sorted(arr)[:3]
ca, cb, cc = arr.count(a), arr.count(b), arr.count(c)
if a == b == c:
print(ncr(ca, 3))
elif a == b:
print(ncr(ca, 2) * cc)
elif b == c:
print(ncr(cb, 2) * ca)
else:
print(ca * cb * cc)
``` | 3 | |
560 | A | Currency System in Geraldion | PROGRAMMING | 1,000 | [
"implementation",
"sortings"
] | null | null | A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the system is not perfect and sometimes it happens that Geraldionians cannot express a certain sum of money with any set of banknotes. Of course, they can use any number of banknotes of each value. Such sum is called unfortunate. Gerald wondered: what is the minimum unfortunate sum? | The first line contains number *n* (1<=≤<=*n*<=≤<=1000) — the number of values of the banknotes that used in Geraldion.
The second line contains *n* distinct space-separated numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the values of the banknotes. | Print a single line — the minimum unfortunate sum. If there are no unfortunate sums, print <=-<=1. | [
"5\n1 2 3 4 5\n"
] | [
"-1\n"
] | none | 500 | [
{
"input": "5\n1 2 3 4 5",
"output": "-1"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "10\n371054 506438 397130 1 766759 208409 769264 549213 641270 771837",
"output": "-1"
},
{
"input": "10\n635370 154890 909382 220996 276501 716105 538714 140162 171960 271264",
"o... | 1,655,971,017 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 16 | 62 | 0 | n = int(input())
a = input().split()
if str(1) in a: print(-1)
else:print(1) | Title: Currency System in Geraldion
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the system is not perfect and sometimes it happens that Geraldionians cannot express a certain sum of money with any set of banknotes. Of course, they can use any number of banknotes of each value. Such sum is called unfortunate. Gerald wondered: what is the minimum unfortunate sum?
Input Specification:
The first line contains number *n* (1<=≤<=*n*<=≤<=1000) — the number of values of the banknotes that used in Geraldion.
The second line contains *n* distinct space-separated numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the values of the banknotes.
Output Specification:
Print a single line — the minimum unfortunate sum. If there are no unfortunate sums, print <=-<=1.
Demo Input:
['5\n1 2 3 4 5\n']
Demo Output:
['-1\n']
Note:
none | ```python
n = int(input())
a = input().split()
if str(1) in a: print(-1)
else:print(1)
``` | 3 | |
192 | A | Funky Numbers | PROGRAMMING | 1,300 | [
"binary search",
"brute force",
"implementation"
] | null | null | As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as , where *k* is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.
A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number *n*, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)! | The first input line contains an integer *n* (1<=≤<=*n*<=≤<=109). | Print "YES" (without the quotes), if *n* can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes). | [
"256\n",
"512\n"
] | [
"YES\n",
"NO\n"
] | In the first sample number <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/92095692c6ea93e9e3b837a0408ba7543549d5b2.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample number 512 can not be represented as a sum of two triangular numbers. | 500 | [
{
"input": "256",
"output": "YES"
},
{
"input": "512",
"output": "NO"
},
{
"input": "80",
"output": "NO"
},
{
"input": "828",
"output": "YES"
},
{
"input": "6035",
"output": "NO"
},
{
"input": "39210",
"output": "YES"
},
{
"input": "79712",... | 1,572,654,476 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 9 | 372 | 307,200 | import sys
import math
"Con esta función tarda mucho más que con binary"
def checkSum(array, x):
flag = False
for i in range(0, len(array)):
for j in range(0, len(array)):
if(lista[i]+lista[j] == x):
flag = True
break
return flag
def binarySearch(array, x):
low = 0
high = len(array)-1
mid = low + (high+low)/2
while(low <= high):
mid = low + math.floor((high-low)/2)
if x > lista[mid]:
low = mid+1
elif x < lista[mid]:
high = mid-1
else:
return True
return False
x = int(input())
if x == 1:
print("NO")
sys.exit()
lista = []
lista.append(1)
m = 2
for i in range(1, 10010):
lista.append(lista[i-1] + m)
m = m + 1
flag = False
for i in range(1, 10010):
if binarySearch(lista,x-lista[i]):
print("YES")
sys.exit()
if flag == False:
print("NO")
| Title: Funky Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as , where *k* is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.
A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number *n*, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!
Input Specification:
The first input line contains an integer *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
Print "YES" (without the quotes), if *n* can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).
Demo Input:
['256\n', '512\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample number <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/92095692c6ea93e9e3b837a0408ba7543549d5b2.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second sample number 512 can not be represented as a sum of two triangular numbers. | ```python
import sys
import math
"Con esta función tarda mucho más que con binary"
def checkSum(array, x):
flag = False
for i in range(0, len(array)):
for j in range(0, len(array)):
if(lista[i]+lista[j] == x):
flag = True
break
return flag
def binarySearch(array, x):
low = 0
high = len(array)-1
mid = low + (high+low)/2
while(low <= high):
mid = low + math.floor((high-low)/2)
if x > lista[mid]:
low = mid+1
elif x < lista[mid]:
high = mid-1
else:
return True
return False
x = int(input())
if x == 1:
print("NO")
sys.exit()
lista = []
lista.append(1)
m = 2
for i in range(1, 10010):
lista.append(lista[i-1] + m)
m = m + 1
flag = False
for i in range(1, 10010):
if binarySearch(lista,x-lista[i]):
print("YES")
sys.exit()
if flag == False:
print("NO")
``` | 0 | |
656 | A | Da Vinci Powers | PROGRAMMING | 1,900 | [
"*special"
] | null | null | The input contains a single integer *a* (0<=≤<=*a*<=≤<=35).
Output a single integer. | The input contains a single integer *a* (0<=≤<=*a*<=≤<=35). | Output a single integer. | [
"3\n",
"10\n"
] | [
"8\n",
"1024\n"
] | none | 0 | [
{
"input": "3",
"output": "8"
},
{
"input": "10",
"output": "1024"
},
{
"input": "35",
"output": "33940307968"
},
{
"input": "0",
"output": "1"
},
{
"input": "1",
"output": "2"
},
{
"input": "2",
"output": "4"
},
{
"input": "4",
"output... | 1,463,943,871 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 62 | 4,608,000 | n=input(n)
n=long(n)
stepen=long(2**n)
print(stepen) | Title: Da Vinci Powers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The input contains a single integer *a* (0<=≤<=*a*<=≤<=35).
Output a single integer.
Input Specification:
The input contains a single integer *a* (0<=≤<=*a*<=≤<=35).
Output Specification:
Output a single integer.
Demo Input:
['3\n', '10\n']
Demo Output:
['8\n', '1024\n']
Note:
none | ```python
n=input(n)
n=long(n)
stepen=long(2**n)
print(stepen)
``` | -1 | |
987 | C | Three displays | PROGRAMMING | 1,400 | [
"brute force",
"dp",
"implementation"
] | null | null | It is the middle of 2018 and Maria Stepanovna, who lives outside Krasnokamensk (a town in Zabaikalsky region), wants to rent three displays to highlight an important problem.
There are $n$ displays placed along a road, and the $i$-th of them can display a text with font size $s_i$ only. Maria Stepanovna wants to rent such three displays with indices $i < j < k$ that the font size increases if you move along the road in a particular direction. Namely, the condition $s_i < s_j < s_k$ should be held.
The rent cost is for the $i$-th display is $c_i$. Please determine the smallest cost Maria Stepanovna should pay. | The first line contains a single integer $n$ ($3 \le n \le 3\,000$) — the number of displays.
The second line contains $n$ integers $s_1, s_2, \ldots, s_n$ ($1 \le s_i \le 10^9$) — the font sizes on the displays in the order they stand along the road.
The third line contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \le c_i \le 10^8$) — the rent costs for each display. | If there are no three displays that satisfy the criteria, print -1. Otherwise print a single integer — the minimum total rent cost of three displays with indices $i < j < k$ such that $s_i < s_j < s_k$. | [
"5\n2 4 5 4 10\n40 30 20 10 40\n",
"3\n100 101 100\n2 4 5\n",
"10\n1 2 3 4 5 6 7 8 9 10\n10 13 11 14 15 12 13 13 18 13\n"
] | [
"90\n",
"-1\n",
"33\n"
] | In the first example you can, for example, choose displays $1$, $4$ and $5$, because $s_1 < s_4 < s_5$ ($2 < 4 < 10$), and the rent cost is $40 + 10 + 40 = 90$.
In the second example you can't select a valid triple of indices, so the answer is -1. | 1,250 | [
{
"input": "5\n2 4 5 4 10\n40 30 20 10 40",
"output": "90"
},
{
"input": "3\n100 101 100\n2 4 5",
"output": "-1"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10\n10 13 11 14 15 12 13 13 18 13",
"output": "33"
},
{
"input": "3\n1 2 3\n100000000 100000000 100000000",
"output": "300... | 1,687,783,829 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 5 | 62 | 0 | if __name__ == '__main__':
n = int(input())
s = list(map(int, input().strip().split()))
c = list(map(int, input().strip().split()))
INF = float("inf")
dp = [[INF] * 3 for _ in range(n)]; dp[0][0] = c[0]
res = INF
for i in range(1, n):
for j in range(i):
if s[j] >= s[i]:
continue
for k in range(1, 3):
dp[i][k] = min(dp[i][k], dp[j][k - 1] + c[i])
res = min(res, dp[i][2])
res = -1 if res == INF else res
print(res) | Title: Three displays
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It is the middle of 2018 and Maria Stepanovna, who lives outside Krasnokamensk (a town in Zabaikalsky region), wants to rent three displays to highlight an important problem.
There are $n$ displays placed along a road, and the $i$-th of them can display a text with font size $s_i$ only. Maria Stepanovna wants to rent such three displays with indices $i < j < k$ that the font size increases if you move along the road in a particular direction. Namely, the condition $s_i < s_j < s_k$ should be held.
The rent cost is for the $i$-th display is $c_i$. Please determine the smallest cost Maria Stepanovna should pay.
Input Specification:
The first line contains a single integer $n$ ($3 \le n \le 3\,000$) — the number of displays.
The second line contains $n$ integers $s_1, s_2, \ldots, s_n$ ($1 \le s_i \le 10^9$) — the font sizes on the displays in the order they stand along the road.
The third line contains $n$ integers $c_1, c_2, \ldots, c_n$ ($1 \le c_i \le 10^8$) — the rent costs for each display.
Output Specification:
If there are no three displays that satisfy the criteria, print -1. Otherwise print a single integer — the minimum total rent cost of three displays with indices $i < j < k$ such that $s_i < s_j < s_k$.
Demo Input:
['5\n2 4 5 4 10\n40 30 20 10 40\n', '3\n100 101 100\n2 4 5\n', '10\n1 2 3 4 5 6 7 8 9 10\n10 13 11 14 15 12 13 13 18 13\n']
Demo Output:
['90\n', '-1\n', '33\n']
Note:
In the first example you can, for example, choose displays $1$, $4$ and $5$, because $s_1 < s_4 < s_5$ ($2 < 4 < 10$), and the rent cost is $40 + 10 + 40 = 90$.
In the second example you can't select a valid triple of indices, so the answer is -1. | ```python
if __name__ == '__main__':
n = int(input())
s = list(map(int, input().strip().split()))
c = list(map(int, input().strip().split()))
INF = float("inf")
dp = [[INF] * 3 for _ in range(n)]; dp[0][0] = c[0]
res = INF
for i in range(1, n):
for j in range(i):
if s[j] >= s[i]:
continue
for k in range(1, 3):
dp[i][k] = min(dp[i][k], dp[j][k - 1] + c[i])
res = min(res, dp[i][2])
res = -1 if res == INF else res
print(res)
``` | 0 | |
246 | D | Colorful Graph | PROGRAMMING | 1,600 | [
"brute force",
"dfs and similar",
"graphs"
] | null | null | You've got an undirected graph, consisting of *n* vertices and *m* edges. We will consider the graph's vertices numbered with integers from 1 to *n*. Each vertex of the graph has a color. The color of the *i*-th vertex is an integer *c**i*.
Let's consider all vertices of the graph, that are painted some color *k*. Let's denote a set of such as *V*(*k*). Let's denote the value of the neighbouring color diversity for color *k* as the cardinality of the set *Q*(*k*)<==<={*c**u* :<= *c**u*<=≠<=*k* and there is vertex *v* belonging to set *V*(*k*) such that nodes *v* and *u* are connected by an edge of the graph}.
Your task is to find such color *k*, which makes the cardinality of set *Q*(*k*) maximum. In other words, you want to find the color that has the most diverse neighbours. Please note, that you want to find such color *k*, that the graph has at least one vertex with such color. | The first line contains two space-separated integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of vertices end edges of the graph, correspondingly. The second line contains a sequence of integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=105) — the colors of the graph vertices. The numbers on the line are separated by spaces.
Next *m* lines contain the description of the edges: the *i*-th line contains two space-separated integers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*) — the numbers of the vertices, connected by the *i*-th edge.
It is guaranteed that the given graph has no self-loops or multiple edges. | Print the number of the color which has the set of neighbours with the maximum cardinality. It there are multiple optimal colors, print the color with the minimum number. Please note, that you want to find such color, that the graph has at least one vertex with such color. | [
"6 6\n1 1 2 3 5 8\n1 2\n3 2\n1 4\n4 3\n4 5\n4 6\n",
"5 6\n4 2 5 2 4\n1 2\n2 3\n3 1\n5 3\n5 4\n3 4\n"
] | [
"3\n",
"2\n"
] | none | 2,000 | [
{
"input": "6 6\n1 1 2 3 5 8\n1 2\n3 2\n1 4\n4 3\n4 5\n4 6",
"output": "3"
},
{
"input": "5 6\n4 2 5 2 4\n1 2\n2 3\n3 1\n5 3\n5 4\n3 4",
"output": "2"
},
{
"input": "3 1\n13 13 4\n1 2",
"output": "4"
},
{
"input": "2 1\n500 300\n1 2",
"output": "300"
},
{
"input":... | 1,682,686,950 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 60 | 0 | n, m = list(map(int, input().split()))
color = list(map(int, input().split()))
graph = [[] for i in range(n)]
for i in range(m):
a, b = list(map(int, input().split()))
graph[a - 1].append(b - 1)
graph[b - 1].append(a - 1)
ans = [color[0], 0]
for vertex in range(n):
count = set()
for adjvertex in graph[vertex]:
count.add(color[adjvertex])
if ans[1] < len(count):
ans = [color[vertex], len(count)]
elif ans[1] == len(count):
ans = [min(color[vertex], ans[0]), len(count)]
print(ans[0]) | Title: Colorful Graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got an undirected graph, consisting of *n* vertices and *m* edges. We will consider the graph's vertices numbered with integers from 1 to *n*. Each vertex of the graph has a color. The color of the *i*-th vertex is an integer *c**i*.
Let's consider all vertices of the graph, that are painted some color *k*. Let's denote a set of such as *V*(*k*). Let's denote the value of the neighbouring color diversity for color *k* as the cardinality of the set *Q*(*k*)<==<={*c**u* :<= *c**u*<=≠<=*k* and there is vertex *v* belonging to set *V*(*k*) such that nodes *v* and *u* are connected by an edge of the graph}.
Your task is to find such color *k*, which makes the cardinality of set *Q*(*k*) maximum. In other words, you want to find the color that has the most diverse neighbours. Please note, that you want to find such color *k*, that the graph has at least one vertex with such color.
Input Specification:
The first line contains two space-separated integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of vertices end edges of the graph, correspondingly. The second line contains a sequence of integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=105) — the colors of the graph vertices. The numbers on the line are separated by spaces.
Next *m* lines contain the description of the edges: the *i*-th line contains two space-separated integers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*) — the numbers of the vertices, connected by the *i*-th edge.
It is guaranteed that the given graph has no self-loops or multiple edges.
Output Specification:
Print the number of the color which has the set of neighbours with the maximum cardinality. It there are multiple optimal colors, print the color with the minimum number. Please note, that you want to find such color, that the graph has at least one vertex with such color.
Demo Input:
['6 6\n1 1 2 3 5 8\n1 2\n3 2\n1 4\n4 3\n4 5\n4 6\n', '5 6\n4 2 5 2 4\n1 2\n2 3\n3 1\n5 3\n5 4\n3 4\n']
Demo Output:
['3\n', '2\n']
Note:
none | ```python
n, m = list(map(int, input().split()))
color = list(map(int, input().split()))
graph = [[] for i in range(n)]
for i in range(m):
a, b = list(map(int, input().split()))
graph[a - 1].append(b - 1)
graph[b - 1].append(a - 1)
ans = [color[0], 0]
for vertex in range(n):
count = set()
for adjvertex in graph[vertex]:
count.add(color[adjvertex])
if ans[1] < len(count):
ans = [color[vertex], len(count)]
elif ans[1] == len(count):
ans = [min(color[vertex], ans[0]), len(count)]
print(ans[0])
``` | 0 | |
957 | A | Tritonic Iridescence | PROGRAMMING | 1,300 | [
"implementation"
] | null | null | Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into *n* consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them. | The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the canvas.
The second line contains a string *s* of *n* characters, the *i*-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one). | If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower). | [
"5\nCY??Y\n",
"5\nC?C?Y\n",
"5\n?CYC?\n",
"5\nC??MM\n",
"3\nMMY\n"
] | [
"Yes\n",
"Yes\n",
"Yes\n",
"No\n",
"No\n"
] | For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example. | 500 | [
{
"input": "5\nCY??Y",
"output": "Yes"
},
{
"input": "5\nC?C?Y",
"output": "Yes"
},
{
"input": "5\n?CYC?",
"output": "Yes"
},
{
"input": "5\nC??MM",
"output": "No"
},
{
"input": "3\nMMY",
"output": "No"
},
{
"input": "15\n??YYYYYY??YYYY?",
"output"... | 1,541,671,735 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 6 | 109 | 0 | n = int (input ())
a = list (input ())
b = 0; t = 1
for I in range(n-1):
if a[I]==a[I+1] and a [I]!="?": t = 0; break
if a.count("?") > 1 and t: print ("Yes")
else : print ("No") | Title: Tritonic Iridescence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Overlooking the captivating blend of myriads of vernal hues, Arkady the painter lays out a long, long canvas.
Arkady has a sufficiently large amount of paint of three colours: cyan, magenta, and yellow. On the one-dimensional canvas split into *n* consecutive segments, each segment needs to be painted in one of the colours.
Arkady has already painted some (possibly none or all) segments and passes the paintbrush to you. You are to determine whether there are at least two ways of colouring all the unpainted segments so that no two adjacent segments are of the same colour. Two ways are considered different if and only if a segment is painted in different colours in them.
Input Specification:
The first line contains a single positive integer *n* (1<=≤<=*n*<=≤<=100) — the length of the canvas.
The second line contains a string *s* of *n* characters, the *i*-th of which is either 'C' (denoting a segment painted in cyan), 'M' (denoting one painted in magenta), 'Y' (one painted in yellow), or '?' (an unpainted one).
Output Specification:
If there are at least two different ways of painting, output "Yes"; otherwise output "No" (both without quotes).
You can print each character in any case (upper or lower).
Demo Input:
['5\nCY??Y\n', '5\nC?C?Y\n', '5\n?CYC?\n', '5\nC??MM\n', '3\nMMY\n']
Demo Output:
['Yes\n', 'Yes\n', 'Yes\n', 'No\n', 'No\n']
Note:
For the first example, there are exactly two different ways of colouring: CYCMY and CYMCY.
For the second example, there are also exactly two different ways of colouring: CMCMY and CYCMY.
For the third example, there are four ways of colouring: MCYCM, MCYCY, YCYCM, and YCYCY.
For the fourth example, no matter how the unpainted segments are coloured, the existing magenta segments will prevent the painting from satisfying the requirements. The similar is true for the fifth example. | ```python
n = int (input ())
a = list (input ())
b = 0; t = 1
for I in range(n-1):
if a[I]==a[I+1] and a [I]!="?": t = 0; break
if a.count("?") > 1 and t: print ("Yes")
else : print ("No")
``` | 0 | |
569 | B | Inventory | PROGRAMMING | 1,200 | [
"greedy",
"math"
] | null | null | Companies always have a lot of equipment, furniture and other things. All of them should be tracked. To do this, there is an inventory number assigned with each item. It is much easier to create a database by using those numbers and keep the track of everything.
During an audit, you were surprised to find out that the items are not numbered sequentially, and some items even share the same inventory number! There is an urgent need to fix it. You have chosen to make the numbers of the items sequential, starting with 1. Changing a number is quite a time-consuming process, and you would like to make maximum use of the current numbering.
You have been given information on current inventory numbers for *n* items in the company. Renumber items so that their inventory numbers form a permutation of numbers from 1 to *n* by changing the number of as few items as possible. Let us remind you that a set of *n* numbers forms a permutation if all the numbers are in the range from 1 to *n*, and no two numbers are equal. | The first line contains a single integer *n* — the number of items (1<=≤<=*n*<=≤<=105).
The second line contains *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the initial inventory numbers of the items. | Print *n* numbers — the final inventory numbers of the items in the order they occur in the input. If there are multiple possible answers, you may print any of them. | [
"3\n1 3 2\n",
"4\n2 2 3 3\n",
"1\n2\n"
] | [
"1 3 2 \n",
"2 1 3 4 \n",
"1 \n"
] | In the first test the numeration is already a permutation, so there is no need to change anything.
In the second test there are two pairs of equal numbers, in each pair you need to replace one number.
In the third test you need to replace 2 by 1, as the numbering should start from one. | 1,000 | [
{
"input": "3\n1 3 2",
"output": "1 3 2 "
},
{
"input": "4\n2 2 3 3",
"output": "2 1 3 4 "
},
{
"input": "1\n2",
"output": "1 "
},
{
"input": "3\n3 3 1",
"output": "3 2 1 "
},
{
"input": "5\n1 1 1 1 1",
"output": "1 2 3 4 5 "
},
{
"input": "5\n5 3 4 4 ... | 1,439,226,691 | 2,491 | Python 3 | OK | TESTS | 29 | 467 | 7,168,000 | n = int(input())
a = [int(i) for i in input().split()]
c = [0] * (n+1)
for e in a:
if e <= n:
c[e] = 1
l = [e for e in range(1,n+1) if c[e] == 0]
s = 0
c = [0] * (n+1)
for i in range(n):
if a[i] <= n and c[a[i]] < 1:
c[a[i]] = 1
else:
a[i] = l[s]
s += 1
for e in a:
print(e, end = ' ')
| Title: Inventory
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Companies always have a lot of equipment, furniture and other things. All of them should be tracked. To do this, there is an inventory number assigned with each item. It is much easier to create a database by using those numbers and keep the track of everything.
During an audit, you were surprised to find out that the items are not numbered sequentially, and some items even share the same inventory number! There is an urgent need to fix it. You have chosen to make the numbers of the items sequential, starting with 1. Changing a number is quite a time-consuming process, and you would like to make maximum use of the current numbering.
You have been given information on current inventory numbers for *n* items in the company. Renumber items so that their inventory numbers form a permutation of numbers from 1 to *n* by changing the number of as few items as possible. Let us remind you that a set of *n* numbers forms a permutation if all the numbers are in the range from 1 to *n*, and no two numbers are equal.
Input Specification:
The first line contains a single integer *n* — the number of items (1<=≤<=*n*<=≤<=105).
The second line contains *n* numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=105) — the initial inventory numbers of the items.
Output Specification:
Print *n* numbers — the final inventory numbers of the items in the order they occur in the input. If there are multiple possible answers, you may print any of them.
Demo Input:
['3\n1 3 2\n', '4\n2 2 3 3\n', '1\n2\n']
Demo Output:
['1 3 2 \n', '2 1 3 4 \n', '1 \n']
Note:
In the first test the numeration is already a permutation, so there is no need to change anything.
In the second test there are two pairs of equal numbers, in each pair you need to replace one number.
In the third test you need to replace 2 by 1, as the numbering should start from one. | ```python
n = int(input())
a = [int(i) for i in input().split()]
c = [0] * (n+1)
for e in a:
if e <= n:
c[e] = 1
l = [e for e in range(1,n+1) if c[e] == 0]
s = 0
c = [0] * (n+1)
for i in range(n):
if a[i] <= n and c[a[i]] < 1:
c[a[i]] = 1
else:
a[i] = l[s]
s += 1
for e in a:
print(e, end = ' ')
``` | 3 | |
38 | A | Army | PROGRAMMING | 800 | [
"implementation"
] | A. Army | 2 | 256 | The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank.
One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible.
Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream. | The first input line contains an integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=≤<=*d**i*<=≤<=100). The third input line contains two integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=*n*). The numbers on the lines are space-separated. | Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*. | [
"3\n5 6\n1 2\n",
"3\n5 6\n1 3\n"
] | [
"5\n",
"11\n"
] | none | 0 | [
{
"input": "3\n5 6\n1 2",
"output": "5"
},
{
"input": "3\n5 6\n1 3",
"output": "11"
},
{
"input": "2\n55\n1 2",
"output": "55"
},
{
"input": "3\n85 78\n1 3",
"output": "163"
},
{
"input": "4\n63 4 49\n2 3",
"output": "4"
},
{
"input": "5\n93 83 42 56\n... | 1,389,877,526 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 92 | 307,200 | n=int(input())
d=[]
for i in range(n-1):
d.append(0)
d[i]=int(input())
a=int(input())
b=int(input())
c=0
y=0
while a!=b:
a=a+1
y=y+d[c]
c=c+1
print(y)
| Title: Army
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The Berland Armed Forces System consists of *n* ranks that are numbered using natural numbers from 1 to *n*, where 1 is the lowest rank and *n* is the highest rank.
One needs exactly *d**i* years to rise from rank *i* to rank *i*<=+<=1. Reaching a certain rank *i* having not reached all the previous *i*<=-<=1 ranks is impossible.
Vasya has just reached a new rank of *a*, but he dreams of holding the rank of *b*. Find for how many more years Vasya should serve in the army until he can finally realize his dream.
Input Specification:
The first input line contains an integer *n* (2<=≤<=*n*<=≤<=100). The second line contains *n*<=-<=1 integers *d**i* (1<=≤<=*d**i*<=≤<=100). The third input line contains two integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=*n*). The numbers on the lines are space-separated.
Output Specification:
Print the single number which is the number of years that Vasya needs to rise from rank *a* to rank *b*.
Demo Input:
['3\n5 6\n1 2\n', '3\n5 6\n1 3\n']
Demo Output:
['5\n', '11\n']
Note:
none | ```python
n=int(input())
d=[]
for i in range(n-1):
d.append(0)
d[i]=int(input())
a=int(input())
b=int(input())
c=0
y=0
while a!=b:
a=a+1
y=y+d[c]
c=c+1
print(y)
``` | -1 |
873 | B | Balanced Substring | PROGRAMMING | 1,500 | [
"dp",
"implementation"
] | null | null | You are given a string *s* consisting only of characters 0 and 1. A substring [*l*,<=*r*] of *s* is a string *s**l**s**l*<=+<=1*s**l*<=+<=2... *s**r*, and its length equals to *r*<=-<=*l*<=+<=1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of *s*. | The first line contains *n* (1<=≤<=*n*<=≤<=100000) — the number of characters in *s*.
The second line contains a string *s* consisting of exactly *n* characters. Only characters 0 and 1 can appear in *s*. | If there is no non-empty balanced substring in *s*, print 0. Otherwise, print the length of the longest balanced substring. | [
"8\n11010111\n",
"3\n111\n"
] | [
"4\n",
"0\n"
] | In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring. | 0 | [
{
"input": "8\n11010111",
"output": "4"
},
{
"input": "3\n111",
"output": "0"
},
{
"input": "11\n00001000100",
"output": "2"
},
{
"input": "10\n0100000000",
"output": "2"
},
{
"input": "13\n0001000011010",
"output": "6"
},
{
"input": "14\n0000010010101... | 1,694,006,215 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | n = int(input())
string = input()
pref_balanced = [0] * (n + 1)
for i, _ in enumerate(string):
pref_balanced[i + 1] = pref_balanced[i] + (1 if (string[i] == '1') else -1)
indexes = {}
max_lenght = 0
for i, _ in enumerate(pref_balanced):
if pref_balanced[i] in indexes:
lenght = i - pref_balanced[i]
max_lenght = max(lenght, max_lenght)
else:
indexes[pref_balanced[i]] = i
| Title: Balanced Substring
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string *s* consisting only of characters 0 and 1. A substring [*l*,<=*r*] of *s* is a string *s**l**s**l*<=+<=1*s**l*<=+<=2... *s**r*, and its length equals to *r*<=-<=*l*<=+<=1. A substring is called balanced if the number of zeroes (0) equals to the number of ones in this substring.
You have to determine the length of the longest balanced substring of *s*.
Input Specification:
The first line contains *n* (1<=≤<=*n*<=≤<=100000) — the number of characters in *s*.
The second line contains a string *s* consisting of exactly *n* characters. Only characters 0 and 1 can appear in *s*.
Output Specification:
If there is no non-empty balanced substring in *s*, print 0. Otherwise, print the length of the longest balanced substring.
Demo Input:
['8\n11010111\n', '3\n111\n']
Demo Output:
['4\n', '0\n']
Note:
In the first example you can choose the substring [3, 6]. It is balanced, and its length is 4. Choosing the substring [2, 5] is also possible.
In the second example it's impossible to find a non-empty balanced substring. | ```python
n = int(input())
string = input()
pref_balanced = [0] * (n + 1)
for i, _ in enumerate(string):
pref_balanced[i + 1] = pref_balanced[i] + (1 if (string[i] == '1') else -1)
indexes = {}
max_lenght = 0
for i, _ in enumerate(pref_balanced):
if pref_balanced[i] in indexes:
lenght = i - pref_balanced[i]
max_lenght = max(lenght, max_lenght)
else:
indexes[pref_balanced[i]] = i
``` | 0 | |
340 | B | Maximal Area Quadrilateral | PROGRAMMING | 2,100 | [
"brute force",
"geometry"
] | null | null | Iahub has drawn a set of *n* points in the cartesian plane which he calls "special points". A quadrilateral is a simple polygon without self-intersections with four sides (also called edges) and four vertices (also called corners). Please note that a quadrilateral doesn't have to be convex. A special quadrilateral is one which has all four vertices in the set of special points. Given the set of special points, please calculate the maximal area of a special quadrilateral. | The first line contains integer *n* (4<=≤<=*n*<=≤<=300). Each of the next *n* lines contains two integers: *x**i*, *y**i* (<=-<=1000<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the cartesian coordinates of *i*th special point. It is guaranteed that no three points are on the same line. It is guaranteed that no two points coincide. | Output a single real number — the maximal area of a special quadrilateral. The answer will be considered correct if its absolute or relative error does't exceed 10<=-<=9. | [
"5\n0 0\n0 4\n4 0\n4 4\n2 3\n"
] | [
"16.000000"
] | In the test example we can choose first 4 points to be the vertices of the quadrilateral. They form a square by side 4, so the area is 4·4 = 16. | 3,000 | [
{
"input": "5\n0 0\n0 4\n4 0\n4 4\n2 3",
"output": "16.000000"
},
{
"input": "10\n-6 -4\n-7 5\n-7 -7\n5 -7\n4 -9\n-6 7\n2 9\n-4 -6\n2 10\n-10 -4",
"output": "166.000000"
},
{
"input": "4\n-3 3\n0 3\n-2 -1\n2 2",
"output": "11.000000"
},
{
"input": "5\n-4 -3\n-3 -2\n3 3\n-1 2\... | 1,389,723,775 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 92 | 307,200 | c, n = 0, int(input())
t = tuple(complex(map(int, input().split())) for i in range(n))
for x, i in enumerate(t, 1):
for j in t[x: ]:
a = b = 0
for k in t:
d = (i.real - k.real) * (j.imag - k.imag) - (i.imag - k.imag) * (j.real - k.real)
a, b = min(d, a), max(d, b)
c = max(c, b - a)
print(c / 2) | Title: Maximal Area Quadrilateral
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub has drawn a set of *n* points in the cartesian plane which he calls "special points". A quadrilateral is a simple polygon without self-intersections with four sides (also called edges) and four vertices (also called corners). Please note that a quadrilateral doesn't have to be convex. A special quadrilateral is one which has all four vertices in the set of special points. Given the set of special points, please calculate the maximal area of a special quadrilateral.
Input Specification:
The first line contains integer *n* (4<=≤<=*n*<=≤<=300). Each of the next *n* lines contains two integers: *x**i*, *y**i* (<=-<=1000<=≤<=*x**i*,<=*y**i*<=≤<=1000) — the cartesian coordinates of *i*th special point. It is guaranteed that no three points are on the same line. It is guaranteed that no two points coincide.
Output Specification:
Output a single real number — the maximal area of a special quadrilateral. The answer will be considered correct if its absolute or relative error does't exceed 10<=-<=9.
Demo Input:
['5\n0 0\n0 4\n4 0\n4 4\n2 3\n']
Demo Output:
['16.000000']
Note:
In the test example we can choose first 4 points to be the vertices of the quadrilateral. They form a square by side 4, so the area is 4·4 = 16. | ```python
c, n = 0, int(input())
t = tuple(complex(map(int, input().split())) for i in range(n))
for x, i in enumerate(t, 1):
for j in t[x: ]:
a = b = 0
for k in t:
d = (i.real - k.real) * (j.imag - k.imag) - (i.imag - k.imag) * (j.real - k.real)
a, b = min(d, a), max(d, b)
c = max(c, b - a)
print(c / 2)
``` | -1 | |
59 | A | Word | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Word | 2 | 256 | Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. | The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. | Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. | [
"HoUse\n",
"ViP\n",
"maTRIx\n"
] | [
"house\n",
"VIP\n",
"matrix\n"
] | none | 500 | [
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chn... | 1,596,643,760 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 218 | 6,758,400 | s = input()
count_Up = 0
count_Lower = 0
for i in s:
if i.islower():
count_Lower += 1
elif i.isupper():
count_Up += 1
if count_Lower > count_Up:
s = s.lower()
elif count_Lower < count_Up:
s = s.upper()
else:
s = s.lower()
print(s) | Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none | ```python
s = input()
count_Up = 0
count_Lower = 0
for i in s:
if i.islower():
count_Lower += 1
elif i.isupper():
count_Up += 1
if count_Lower > count_Up:
s = s.lower()
elif count_Lower < count_Up:
s = s.upper()
else:
s = s.lower()
print(s)
``` | 3.932911 |
479 | A | Expression | PROGRAMMING | 1,000 | [
"brute force",
"math"
] | null | null | Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get. | The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10). | Print the maximum value of the expression that you can obtain. | [
"1\n2\n3\n",
"2\n10\n3\n"
] | [
"9\n",
"60\n"
] | none | 500 | [
{
"input": "1\n2\n3",
"output": "9"
},
{
"input": "2\n10\n3",
"output": "60"
},
{
"input": "1\n1\n1",
"output": "3"
},
{
"input": "1\n2\n1",
"output": "4"
},
{
"input": "10\n10\n10",
"output": "1000"
},
{
"input": "5\n1\n3",
"output": "20"
},
{... | 1,691,685,247 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 5 | 46 | 0 | import sys
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return int(input())
def inlt():
return list(map(int, input().split()))
def insr():
s = input()
return list(s[:len(s) - 1])
def invr():
return map(int, input().split())
n1 = inp()
n2 = inp()
n3 = inp()
maxy = 0
if n1 == 1 and n3 == 1:
maxy = n1 + n2 + n3 # Here, the sum of all three numbers is used
elif n1 == 1:
maxy = (n1 + n2) * n3
elif n3 == 1:
maxy = max(n1 * (n2 + n3), (n1 + n2) * n3)
else:
maxy = n1 * n2 * n3
print(maxy)
| Title: Expression
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya studies in a school and he adores Maths. His class has been studying arithmetic expressions. On the last class the teacher wrote three positive integers *a*, *b*, *c* on the blackboard. The task was to insert signs of operations '+' and '*', and probably brackets between the numbers so that the value of the resulting expression is as large as possible. Let's consider an example: assume that the teacher wrote numbers 1, 2 and 3 on the blackboard. Here are some ways of placing signs and brackets:
- 1+2*3=7 - 1*(2+3)=5 - 1*2*3=6 - (1+2)*3=9
Note that you can insert operation signs only between *a* and *b*, and between *b* and *c*, that is, you cannot swap integers. For instance, in the given sample you cannot get expression (1+3)*2.
It's easy to see that the maximum value that you can obtain is 9.
Your task is: given *a*, *b* and *c* print the maximum value that you can get.
Input Specification:
The input contains three integers *a*, *b* and *c*, each on a single line (1<=≤<=*a*,<=*b*,<=*c*<=≤<=10).
Output Specification:
Print the maximum value of the expression that you can obtain.
Demo Input:
['1\n2\n3\n', '2\n10\n3\n']
Demo Output:
['9\n', '60\n']
Note:
none | ```python
import sys
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return int(input())
def inlt():
return list(map(int, input().split()))
def insr():
s = input()
return list(s[:len(s) - 1])
def invr():
return map(int, input().split())
n1 = inp()
n2 = inp()
n3 = inp()
maxy = 0
if n1 == 1 and n3 == 1:
maxy = n1 + n2 + n3 # Here, the sum of all three numbers is used
elif n1 == 1:
maxy = (n1 + n2) * n3
elif n3 == 1:
maxy = max(n1 * (n2 + n3), (n1 + n2) * n3)
else:
maxy = n1 * n2 * n3
print(maxy)
``` | 0 | |
538 | B | Quasi Binary | PROGRAMMING | 1,400 | [
"constructive algorithms",
"dp",
"greedy",
"implementation"
] | null | null | A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 — are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer *n*. Represent it as a sum of minimum number of quasibinary numbers. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=106). | In the first line print a single integer *k* — the minimum number of numbers in the representation of number *n* as a sum of quasibinary numbers.
In the second line print *k* numbers — the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal *n*. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them. | [
"9\n",
"32\n"
] | [
"9\n1 1 1 1 1 1 1 1 1 \n",
"3\n10 11 11 \n"
] | none | 1,000 | [
{
"input": "9",
"output": "9\n1 1 1 1 1 1 1 1 1 "
},
{
"input": "32",
"output": "3\n10 11 11 "
},
{
"input": "1",
"output": "1\n1 "
},
{
"input": "415",
"output": "5\n1 101 101 101 111 "
},
{
"input": "10011",
"output": "1\n10011 "
},
{
"input": "10201... | 1,473,289,429 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 61 | 0 | d=set()
def f(n):
if n>10**6:return
d.add(n);f(n*10);f(n*10+1)
f(1)
r=[]
n=int(input())
for i in sorted(d)[::-1]:
while n>=i:n-=i;r+=[i]
print(len(r),'\n'+' '.join(map(str,r))) | Title: Quasi Binary
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A number is called quasibinary if its decimal representation contains only digits 0 or 1. For example, numbers 0, 1, 101, 110011 — are quasibinary and numbers 2, 12, 900 are not.
You are given a positive integer *n*. Represent it as a sum of minimum number of quasibinary numbers.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=106).
Output Specification:
In the first line print a single integer *k* — the minimum number of numbers in the representation of number *n* as a sum of quasibinary numbers.
In the second line print *k* numbers — the elements of the sum. All these numbers should be quasibinary according to the definition above, their sum should equal *n*. Do not have to print the leading zeroes in the numbers. The order of numbers doesn't matter. If there are multiple possible representations, you are allowed to print any of them.
Demo Input:
['9\n', '32\n']
Demo Output:
['9\n1 1 1 1 1 1 1 1 1 \n', '3\n10 11 11 \n']
Note:
none | ```python
d=set()
def f(n):
if n>10**6:return
d.add(n);f(n*10);f(n*10+1)
f(1)
r=[]
n=int(input())
for i in sorted(d)[::-1]:
while n>=i:n-=i;r+=[i]
print(len(r),'\n'+' '.join(map(str,r)))
``` | 0 | |
492 | B | Vanya and Lanterns | PROGRAMMING | 1,200 | [
"binary search",
"implementation",
"math",
"sortings"
] | null | null | Vanya walks late at night along a straight street of length *l*, lit by *n* lanterns. Consider the coordinate system with the beginning of the street corresponding to the point 0, and its end corresponding to the point *l*. Then the *i*-th lantern is at the point *a**i*. The lantern lights all points of the street that are at the distance of at most *d* from it, where *d* is some positive number, common for all lanterns.
Vanya wonders: what is the minimum light radius *d* should the lanterns have to light the whole street? | The first line contains two integers *n*, *l* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*l*<=≤<=109) — the number of lanterns and the length of the street respectively.
The next line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=*l*). Multiple lanterns can be located at the same point. The lanterns may be located at the ends of the street. | Print the minimum light radius *d*, needed to light the whole street. The answer will be considered correct if its absolute or relative error doesn't exceed 10<=-<=9. | [
"7 15\n15 5 3 7 9 14 0\n",
"2 5\n2 5\n"
] | [
"2.5000000000\n",
"2.0000000000\n"
] | Consider the second sample. At *d* = 2 the first lantern will light the segment [0, 4] of the street, and the second lantern will light segment [3, 5]. Thus, the whole street will be lit. | 1,000 | [
{
"input": "7 15\n15 5 3 7 9 14 0",
"output": "2.5000000000"
},
{
"input": "2 5\n2 5",
"output": "2.0000000000"
},
{
"input": "46 615683844\n431749087 271781274 274974690 324606253 480870261 401650581 13285442 478090364 266585394 425024433 588791449 492057200 391293435 563090494 317950 1... | 1,695,109,484 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 31 | 0 | a=input().split()
l=list(map(int,input().split()))
l.sort()
L=[0]
for x in range(len(l)-1):
L.append(l[x+1]-l[x])
L.append(a[1])
print(0.5*max(L)) | Title: Vanya and Lanterns
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya walks late at night along a straight street of length *l*, lit by *n* lanterns. Consider the coordinate system with the beginning of the street corresponding to the point 0, and its end corresponding to the point *l*. Then the *i*-th lantern is at the point *a**i*. The lantern lights all points of the street that are at the distance of at most *d* from it, where *d* is some positive number, common for all lanterns.
Vanya wonders: what is the minimum light radius *d* should the lanterns have to light the whole street?
Input Specification:
The first line contains two integers *n*, *l* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*l*<=≤<=109) — the number of lanterns and the length of the street respectively.
The next line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=*l*). Multiple lanterns can be located at the same point. The lanterns may be located at the ends of the street.
Output Specification:
Print the minimum light radius *d*, needed to light the whole street. The answer will be considered correct if its absolute or relative error doesn't exceed 10<=-<=9.
Demo Input:
['7 15\n15 5 3 7 9 14 0\n', '2 5\n2 5\n']
Demo Output:
['2.5000000000\n', '2.0000000000\n']
Note:
Consider the second sample. At *d* = 2 the first lantern will light the segment [0, 4] of the street, and the second lantern will light segment [3, 5]. Thus, the whole street will be lit. | ```python
a=input().split()
l=list(map(int,input().split()))
l.sort()
L=[0]
for x in range(len(l)-1):
L.append(l[x+1]-l[x])
L.append(a[1])
print(0.5*max(L))
``` | -1 | |
228 | D | Zigzag | PROGRAMMING | 2,100 | [
"data structures"
] | null | null | The court wizard Zigzag wants to become a famous mathematician. For that, he needs his own theorem, like the Cauchy theorem, or his sum, like the Minkowski sum. But most of all he wants to have his sequence, like the Fibonacci sequence, and his function, like the Euler's totient function.
The Zigag's sequence with the zigzag factor z is an infinite sequence *S**i**z* (*i*<=≥<=1; *z*<=≥<=2), that is determined as follows:
- *S**i**z*<==<=2, when ; - , when ; - , when .
Operation means taking the remainder from dividing number *x* by number *y*. For example, the beginning of sequence *S**i*3 (zigzag factor 3) looks as follows: 1, 2, 3, 2, 1, 2, 3, 2, 1.
Let's assume that we are given an array *a*, consisting of *n* integers. Let's define element number *i* (1<=≤<=*i*<=≤<=*n*) of the array as *a**i*. The Zigzag function is function , where *l*,<=*r*,<=*z* satisfy the inequalities 1<=≤<=*l*<=≤<=*r*<=≤<=*n*, *z*<=≥<=2.
To become better acquainted with the Zigzag sequence and the Zigzag function, the wizard offers you to implement the following operations on the given array *a*.
1. The assignment operation. The operation parameters are (*p*,<=*v*). The operation denotes assigning value *v* to the *p*-th array element. After the operation is applied, the value of the array element *a**p* equals *v*. 1. The Zigzag operation. The operation parameters are (*l*,<=*r*,<=*z*). The operation denotes calculating the Zigzag function *Z*(*l*,<=*r*,<=*z*).
Explore the magical powers of zigzags, implement the described operations. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — The number of elements in array *a*. The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of operations. Next *m* lines contain the operations' descriptions. An operation's description starts with integer *t**i* (1<=≤<=*t**i*<=≤<=2) — the operation type.
- If *t**i*<==<=1 (assignment operation), then on the line follow two space-separated integers: *p**i*,<=*v**i* (1<=≤<=*p**i*<=≤<=*n*; 1<=≤<=*v**i*<=≤<=109) — the parameters of the assigning operation. - If *t**i*<==<=2 (Zigzag operation), then on the line follow three space-separated integers: *l**i*,<=*r**i*,<=*z**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*; 2<=≤<=*z**i*<=≤<=6) — the parameters of the Zigzag operation.
You should execute the operations in the order, in which they are given in the input. | For each Zigzag operation print the calculated value of the Zigzag function on a single line. Print the values for Zigzag functions in the order, in which they are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier. | [
"5\n2 3 1 5 5\n4\n2 2 3 2\n2 1 5 3\n1 3 5\n2 1 5 3\n"
] | [
"5\n26\n38\n"
] | Explanation of the sample test:
- Result of the first operation is *Z*(2, 3, 2) = 3·1 + 1·2 = 5. - Result of the second operation is *Z*(1, 5, 3) = 2·1 + 3·2 + 1·3 + 5·2 + 5·1 = 26. - After the third operation array *a* is equal to 2, 3, 5, 5, 5. - Result of the forth operation is *Z*(1, 5, 3) = 2·1 + 3·2 + 5·3 + 5·2 + 5·1 = 38. | 2,000 | [] | 1,691,442,542 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 62 | 0 | print("_RANDOM_GUESS_1691442542.3355112")# 1691442542.3355286 | Title: Zigzag
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The court wizard Zigzag wants to become a famous mathematician. For that, he needs his own theorem, like the Cauchy theorem, or his sum, like the Minkowski sum. But most of all he wants to have his sequence, like the Fibonacci sequence, and his function, like the Euler's totient function.
The Zigag's sequence with the zigzag factor z is an infinite sequence *S**i**z* (*i*<=≥<=1; *z*<=≥<=2), that is determined as follows:
- *S**i**z*<==<=2, when ; - , when ; - , when .
Operation means taking the remainder from dividing number *x* by number *y*. For example, the beginning of sequence *S**i*3 (zigzag factor 3) looks as follows: 1, 2, 3, 2, 1, 2, 3, 2, 1.
Let's assume that we are given an array *a*, consisting of *n* integers. Let's define element number *i* (1<=≤<=*i*<=≤<=*n*) of the array as *a**i*. The Zigzag function is function , where *l*,<=*r*,<=*z* satisfy the inequalities 1<=≤<=*l*<=≤<=*r*<=≤<=*n*, *z*<=≥<=2.
To become better acquainted with the Zigzag sequence and the Zigzag function, the wizard offers you to implement the following operations on the given array *a*.
1. The assignment operation. The operation parameters are (*p*,<=*v*). The operation denotes assigning value *v* to the *p*-th array element. After the operation is applied, the value of the array element *a**p* equals *v*. 1. The Zigzag operation. The operation parameters are (*l*,<=*r*,<=*z*). The operation denotes calculating the Zigzag function *Z*(*l*,<=*r*,<=*z*).
Explore the magical powers of zigzags, implement the described operations.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — The number of elements in array *a*. The second line contains *n* space-separated integers: *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of operations. Next *m* lines contain the operations' descriptions. An operation's description starts with integer *t**i* (1<=≤<=*t**i*<=≤<=2) — the operation type.
- If *t**i*<==<=1 (assignment operation), then on the line follow two space-separated integers: *p**i*,<=*v**i* (1<=≤<=*p**i*<=≤<=*n*; 1<=≤<=*v**i*<=≤<=109) — the parameters of the assigning operation. - If *t**i*<==<=2 (Zigzag operation), then on the line follow three space-separated integers: *l**i*,<=*r**i*,<=*z**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*; 2<=≤<=*z**i*<=≤<=6) — the parameters of the Zigzag operation.
You should execute the operations in the order, in which they are given in the input.
Output Specification:
For each Zigzag operation print the calculated value of the Zigzag function on a single line. Print the values for Zigzag functions in the order, in which they are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Demo Input:
['5\n2 3 1 5 5\n4\n2 2 3 2\n2 1 5 3\n1 3 5\n2 1 5 3\n']
Demo Output:
['5\n26\n38\n']
Note:
Explanation of the sample test:
- Result of the first operation is *Z*(2, 3, 2) = 3·1 + 1·2 = 5. - Result of the second operation is *Z*(1, 5, 3) = 2·1 + 3·2 + 1·3 + 5·2 + 5·1 = 26. - After the third operation array *a* is equal to 2, 3, 5, 5, 5. - Result of the forth operation is *Z*(1, 5, 3) = 2·1 + 3·2 + 5·3 + 5·2 + 5·1 = 38. | ```python
print("_RANDOM_GUESS_1691442542.3355112")# 1691442542.3355286
``` | 0 | |
821 | B | Okabe and Banana Trees | PROGRAMMING | 1,300 | [
"brute force",
"math"
] | null | null | Okabe needs bananas for one of his experiments for some strange reason. So he decides to go to the forest and cut banana trees.
Consider the point (*x*,<=*y*) in the 2D plane such that *x* and *y* are integers and 0<=≤<=*x*,<=*y*. There is a tree in such a point, and it has *x*<=+<=*y* bananas. There are no trees nor bananas in other points. Now, Okabe draws a line with equation . Okabe can select a single rectangle with axis aligned sides with all points on or under the line and cut all the trees in all points that are inside or on the border of this rectangle and take their bananas. Okabe's rectangle can be degenerate; that is, it can be a line segment or even a point.
Help Okabe and find the maximum number of bananas he can get if he chooses the rectangle wisely.
Okabe is sure that the answer does not exceed 1018. You can trust him. | The first line of input contains two space-separated integers *m* and *b* (1<=≤<=*m*<=≤<=1000, 1<=≤<=*b*<=≤<=10000). | Print the maximum number of bananas Okabe can get from the trees he cuts. | [
"1 5\n",
"2 3\n"
] | [
"30\n",
"25\n"
] | The graph above corresponds to sample test 1. The optimal rectangle is shown in red and has 30 bananas. | 1,000 | [
{
"input": "1 5",
"output": "30"
},
{
"input": "2 3",
"output": "25"
},
{
"input": "4 6",
"output": "459"
},
{
"input": "6 3",
"output": "171"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "10 1",
"output": "55"
},
{
"input": "20 10",
... | 1,515,657,751 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 5,632,000 | m,b= map(int, input().split(" "))
mx=0
for j in range(0,b):
x=(b-j)*m
s =int(((j+x)*(x+1)*(j+1)))
mx = max(mx, s)
print(int(mx))
print((mx//2))
| Title: Okabe and Banana Trees
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Okabe needs bananas for one of his experiments for some strange reason. So he decides to go to the forest and cut banana trees.
Consider the point (*x*,<=*y*) in the 2D plane such that *x* and *y* are integers and 0<=≤<=*x*,<=*y*. There is a tree in such a point, and it has *x*<=+<=*y* bananas. There are no trees nor bananas in other points. Now, Okabe draws a line with equation . Okabe can select a single rectangle with axis aligned sides with all points on or under the line and cut all the trees in all points that are inside or on the border of this rectangle and take their bananas. Okabe's rectangle can be degenerate; that is, it can be a line segment or even a point.
Help Okabe and find the maximum number of bananas he can get if he chooses the rectangle wisely.
Okabe is sure that the answer does not exceed 1018. You can trust him.
Input Specification:
The first line of input contains two space-separated integers *m* and *b* (1<=≤<=*m*<=≤<=1000, 1<=≤<=*b*<=≤<=10000).
Output Specification:
Print the maximum number of bananas Okabe can get from the trees he cuts.
Demo Input:
['1 5\n', '2 3\n']
Demo Output:
['30\n', '25\n']
Note:
The graph above corresponds to sample test 1. The optimal rectangle is shown in red and has 30 bananas. | ```python
m,b= map(int, input().split(" "))
mx=0
for j in range(0,b):
x=(b-j)*m
s =int(((j+x)*(x+1)*(j+1)))
mx = max(mx, s)
print(int(mx))
print((mx//2))
``` | 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... | 1,527,517,199 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 61 | 0 | #-*- coding:utf-8 -*-
tmpstr = input().strip()
tmpvec = tmpstr.split(' ')
n = int(tmpvec[0])
k = int(tmpvec[1])
l = int(tmpvec[2])
tmpstr = input().strip()
tmpvec = tmpstr.split(' ')
vec_len = len(tmpvec)
data = map(int, tmpvec)
dlist = list(data)
minivalue = min(dlist)
dlist.sort()
volume = 0
index = 0
for i in range(n):
if dlist[index] > dlist[0] + l:
print(0)
exit(0)
volunme = volume + dlist[index]
index = index +4
print(volunme) | 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
#-*- coding:utf-8 -*-
tmpstr = input().strip()
tmpvec = tmpstr.split(' ')
n = int(tmpvec[0])
k = int(tmpvec[1])
l = int(tmpvec[2])
tmpstr = input().strip()
tmpvec = tmpstr.split(' ')
vec_len = len(tmpvec)
data = map(int, tmpvec)
dlist = list(data)
minivalue = min(dlist)
dlist.sort()
volume = 0
index = 0
for i in range(n):
if dlist[index] > dlist[0] + l:
print(0)
exit(0)
volunme = volume + dlist[index]
index = index +4
print(volunme)
``` | -1 | |
723 | D | Lakes in Berland | PROGRAMMING | 1,600 | [
"dfs and similar",
"dsu",
"graphs",
"greedy",
"implementation"
] | null | null | The map of Berland is a rectangle of the size *n*<=×<=*m*, which consists of cells of size 1<=×<=1. Each cell is either land or water. The map is surrounded by the ocean.
Lakes are the maximal regions of water cells, connected by sides, which are not connected with the ocean. Formally, lake is a set of water cells, such that it's possible to get from any cell of the set to any other without leaving the set and moving only to cells adjacent by the side, none of them is located on the border of the rectangle, and it's impossible to add one more water cell to the set such that it will be connected with any other cell.
You task is to fill up with the earth the minimum number of water cells so that there will be exactly *k* lakes in Berland. Note that the initial number of lakes on the map is not less than *k*. | The first line of the input contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=50, 0<=≤<=*k*<=≤<=50) — the sizes of the map and the number of lakes which should be left on the map.
The next *n* lines contain *m* characters each — the description of the map. Each of the characters is either '.' (it means that the corresponding cell is water) or '*' (it means that the corresponding cell is land).
It is guaranteed that the map contain at least *k* lakes. | In the first line print the minimum number of cells which should be transformed from water to land.
In the next *n* lines print *m* symbols — the map after the changes. The format must strictly follow the format of the map in the input data (there is no need to print the size of the map). If there are several answers, print any of them.
It is guaranteed that the answer exists on the given data. | [
"5 4 1\n****\n*..*\n****\n**.*\n..**\n",
"3 3 0\n***\n*.*\n***\n"
] | [
"1\n****\n*..*\n****\n****\n..**\n",
"1\n***\n***\n***\n"
] | In the first example there are only two lakes — the first consists of the cells (2, 2) and (2, 3), the second consists of the cell (4, 3). It is profitable to cover the second lake because it is smaller. Pay attention that the area of water in the lower left corner is not a lake because this area share a border with the ocean. | 2,000 | [
{
"input": "5 4 1\n****\n*..*\n****\n**.*\n..**",
"output": "1\n****\n*..*\n****\n****\n..**"
},
{
"input": "3 3 0\n***\n*.*\n***",
"output": "1\n***\n***\n***"
},
{
"input": "3 5 1\n.**.*\n*.*.*\n***..",
"output": "0\n.**.*\n*.*.*\n***.."
},
{
"input": "3 5 0\n.**.*\n*.*.*\n... | 1,674,064,623 | 2,147,483,647 | Python 3 | OK | TESTS | 26 | 46 | 0 | n,m,k=map(int,input().split())
visited=[[False]*(52) for _ in range(54)]
l=[]
matrix=[]
for _ in range(n):
matrix.append(list(el for el in input()))
def dfs(i,j):
visited[i][j]=True
q=[(i,j)]
t=[]
f=False
while q:
x,y=q.pop()
t.append((x,y))
if x==0 or x==n-1 or y==0 or y==m-1:
f=True
for dx,dy in [(1,0),(0,1),(-1,0),(0,-1)]:
nx,ny=dx+x,dy+y
if 0<=nx<n and 0<=ny<m and matrix[nx][ny]=="." and visited[nx][ny]==False:
q.append((nx,ny))
visited[nx][ny]=True
if not f:
l.append(t)
for i in range(n):
for j in range(m):
if visited[i][j]==False and matrix[i][j]==".":
dfs(i,j)
l.sort(key=lambda x:(len(x)))
ans=0
for i in range(len(l)-k):
ans+=len(l[i])
for x,y in l[i]:
matrix[x][y]="*"
print(ans)
for el in matrix:
print("".join(el))
| Title: Lakes in Berland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The map of Berland is a rectangle of the size *n*<=×<=*m*, which consists of cells of size 1<=×<=1. Each cell is either land or water. The map is surrounded by the ocean.
Lakes are the maximal regions of water cells, connected by sides, which are not connected with the ocean. Formally, lake is a set of water cells, such that it's possible to get from any cell of the set to any other without leaving the set and moving only to cells adjacent by the side, none of them is located on the border of the rectangle, and it's impossible to add one more water cell to the set such that it will be connected with any other cell.
You task is to fill up with the earth the minimum number of water cells so that there will be exactly *k* lakes in Berland. Note that the initial number of lakes on the map is not less than *k*.
Input Specification:
The first line of the input contains three integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=50, 0<=≤<=*k*<=≤<=50) — the sizes of the map and the number of lakes which should be left on the map.
The next *n* lines contain *m* characters each — the description of the map. Each of the characters is either '.' (it means that the corresponding cell is water) or '*' (it means that the corresponding cell is land).
It is guaranteed that the map contain at least *k* lakes.
Output Specification:
In the first line print the minimum number of cells which should be transformed from water to land.
In the next *n* lines print *m* symbols — the map after the changes. The format must strictly follow the format of the map in the input data (there is no need to print the size of the map). If there are several answers, print any of them.
It is guaranteed that the answer exists on the given data.
Demo Input:
['5 4 1\n****\n*..*\n****\n**.*\n..**\n', '3 3 0\n***\n*.*\n***\n']
Demo Output:
['1\n****\n*..*\n****\n****\n..**\n', '1\n***\n***\n***\n']
Note:
In the first example there are only two lakes — the first consists of the cells (2, 2) and (2, 3), the second consists of the cell (4, 3). It is profitable to cover the second lake because it is smaller. Pay attention that the area of water in the lower left corner is not a lake because this area share a border with the ocean. | ```python
n,m,k=map(int,input().split())
visited=[[False]*(52) for _ in range(54)]
l=[]
matrix=[]
for _ in range(n):
matrix.append(list(el for el in input()))
def dfs(i,j):
visited[i][j]=True
q=[(i,j)]
t=[]
f=False
while q:
x,y=q.pop()
t.append((x,y))
if x==0 or x==n-1 or y==0 or y==m-1:
f=True
for dx,dy in [(1,0),(0,1),(-1,0),(0,-1)]:
nx,ny=dx+x,dy+y
if 0<=nx<n and 0<=ny<m and matrix[nx][ny]=="." and visited[nx][ny]==False:
q.append((nx,ny))
visited[nx][ny]=True
if not f:
l.append(t)
for i in range(n):
for j in range(m):
if visited[i][j]==False and matrix[i][j]==".":
dfs(i,j)
l.sort(key=lambda x:(len(x)))
ans=0
for i in range(len(l)-k):
ans+=len(l[i])
for x,y in l[i]:
matrix[x][y]="*"
print(ans)
for el in matrix:
print("".join(el))
``` | 3 | |
96 | A | Football | PROGRAMMING | 900 | [
"implementation",
"strings"
] | A. Football | 2 | 256 | Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not. | The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field. | Print "YES" if the situation is dangerous. Otherwise, print "NO". | [
"001001\n",
"1000000001\n"
] | [
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "001001",
"output": "NO"
},
{
"input": "1000000001",
"output": "YES"
},
{
"input": "00100110111111101",
"output": "YES"
},
{
"input": "11110111111111111",
"output": "YES"
},
{
"input": "01",
"output": "NO"
},
{
"input": "10100101",
"outp... | 1,686,373,204 | 2,147,483,647 | Python 3 | OK | TESTS | 44 | 62 | 0 | g = input()
if '0' * 7 in g or '1' * 7 in g:
print('YES')
else:
print('NO')
| Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
Input Specification:
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
Output Specification:
Print "YES" if the situation is dangerous. Otherwise, print "NO".
Demo Input:
['001001\n', '1000000001\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none | ```python
g = input()
if '0' * 7 in g or '1' * 7 in g:
print('YES')
else:
print('NO')
``` | 3.9845 |
520 | A | Pangram | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.
You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string.
The second line contains the string. The string consists only of uppercase and lowercase Latin letters. | Output "YES", if the string is a pangram and "NO" otherwise. | [
"12\ntoosmallword\n",
"35\nTheQuickBrownFoxJumpsOverTheLazyDog\n"
] | [
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "12\ntoosmallword",
"output": "NO"
},
{
"input": "35\nTheQuickBrownFoxJumpsOverTheLazyDog",
"output": "YES"
},
{
"input": "1\na",
"output": "NO"
},
{
"input": "26\nqwertyuiopasdfghjklzxcvbnm",
"output": "YES"
},
{
"input": "26\nABCDEFGHIJKLMNOPQRSTUVWXY... | 1,694,189,699 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 921,600 | import re
inp = input()
reg = r'([a-zA-Z]).*?\1'
if re.search(reg, inp):
print("No")
else:
print("Yes")
| Title: Pangram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.
You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string.
The second line contains the string. The string consists only of uppercase and lowercase Latin letters.
Output Specification:
Output "YES", if the string is a pangram and "NO" otherwise.
Demo Input:
['12\ntoosmallword\n', '35\nTheQuickBrownFoxJumpsOverTheLazyDog\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none | ```python
import re
inp = input()
reg = r'([a-zA-Z]).*?\1'
if re.search(reg, inp):
print("No")
else:
print("Yes")
``` | 0 | |
961 | A | Tetris | PROGRAMMING | 900 | [
"implementation"
] | null | null | You are given a following process.
There is a platform with $n$ columns. $1 \times 1$ squares are appearing one after another in some columns on this platform. If there are no squares in the column, a square will occupy the bottom row. Otherwise a square will appear at the top of the highest square of this column.
When all of the $n$ columns have at least one square in them, the bottom row is being removed. You will receive $1$ point for this, and all the squares left will fall down one row.
You task is to calculate the amount of points you will receive. | The first line of input contain 2 integer numbers $n$ and $m$ ($1 \le n, m \le 1000$) — the length of the platform and the number of the squares.
The next line contain $m$ integer numbers $c_1, c_2, \dots, c_m$ ($1 \le c_i \le n$) — column in which $i$-th square will appear. | Print one integer — the amount of points you will receive. | [
"3 9\n1 1 2 2 2 3 1 2 3\n"
] | [
"2\n"
] | In the sample case the answer will be equal to $2$ because after the appearing of $6$-th square will be removed one row (counts of the squares on the platform will look like $[2~ 3~ 1]$, and after removing one row will be $[1~ 2~ 0]$).
After the appearing of $9$-th square counts will be $[2~ 3~ 1]$, and after removing one row it will look like $[1~ 2~ 0]$.
So the answer will be equal to $2$. | 0 | [
{
"input": "3 9\n1 1 2 2 2 3 1 2 3",
"output": "2"
},
{
"input": "1 7\n1 1 1 1 1 1 1",
"output": "7"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "3 5\n1 1 1 2 3",
"output": "1"
},
{
"input": "4 6\n4 4 4 4 4 4",
"output": "0"
},
{
"input": "4 6\... | 1,544,713,051 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 93 | 0 | a,b=map(int,input().split())
c=list(map(int,input().split()))
d=0
e=0
for i in range(1,len(c)):
if c[i]==c[i-1]:
if i==1:
d+=2
else:
d+=1
elif c[i]!=c[i-1]:
if d>=2:
e+=(a//2)
d=0
print(e) | Title: Tetris
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a following process.
There is a platform with $n$ columns. $1 \times 1$ squares are appearing one after another in some columns on this platform. If there are no squares in the column, a square will occupy the bottom row. Otherwise a square will appear at the top of the highest square of this column.
When all of the $n$ columns have at least one square in them, the bottom row is being removed. You will receive $1$ point for this, and all the squares left will fall down one row.
You task is to calculate the amount of points you will receive.
Input Specification:
The first line of input contain 2 integer numbers $n$ and $m$ ($1 \le n, m \le 1000$) — the length of the platform and the number of the squares.
The next line contain $m$ integer numbers $c_1, c_2, \dots, c_m$ ($1 \le c_i \le n$) — column in which $i$-th square will appear.
Output Specification:
Print one integer — the amount of points you will receive.
Demo Input:
['3 9\n1 1 2 2 2 3 1 2 3\n']
Demo Output:
['2\n']
Note:
In the sample case the answer will be equal to $2$ because after the appearing of $6$-th square will be removed one row (counts of the squares on the platform will look like $[2~ 3~ 1]$, and after removing one row will be $[1~ 2~ 0]$).
After the appearing of $9$-th square counts will be $[2~ 3~ 1]$, and after removing one row it will look like $[1~ 2~ 0]$.
So the answer will be equal to $2$. | ```python
a,b=map(int,input().split())
c=list(map(int,input().split()))
d=0
e=0
for i in range(1,len(c)):
if c[i]==c[i-1]:
if i==1:
d+=2
else:
d+=1
elif c[i]!=c[i-1]:
if d>=2:
e+=(a//2)
d=0
print(e)
``` | 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"... | 1,694,824,432 | 2,147,483,647 | Python 3 | OK | TESTS | 20 | 92 | 0 | x = int(input())
print("NO" if x % 2 == 1 or x == 2 else "YES") | 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
x = int(input())
print("NO" if x % 2 == 1 or x == 2 else "YES")
``` | 3.954 |
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 ... | 1,652,070,114 | 2,147,483,647 | Python 3 | OK | TESTS | 41 | 77 | 2,048,000 | n=int(input())
count=0
list0=[]
while(1):
if n==2 or n==3:
print(count+1)
list0.append(n)
print(" ".join(str(x) for x in list0))
break
else:
n-=2
count+=1
list0.append(2)
| 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())
count=0
list0=[]
while(1):
if n==2 or n==3:
print(count+1)
list0.append(n)
print(" ".join(str(x) for x in list0))
break
else:
n-=2
count+=1
list0.append(2)
``` | 3 | |
1,008 | A | Romaji | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.
In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.
Help Vitya find out if a word $s$ is Berlanese. | The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters. | Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".
You can print each letter in any case (upper or lower). | [
"sumimasen\n",
"ninja\n",
"codeforces\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.
In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese. | 500 | [
{
"input": "sumimasen",
"output": "YES"
},
{
"input": "ninja",
"output": "YES"
},
{
"input": "codeforces",
"output": "NO"
},
{
"input": "auuaoonntanonnuewannnnpuuinniwoonennyolonnnvienonpoujinndinunnenannmuveoiuuhikucuziuhunnnmunzancenen",
"output": "YES"
},
{
"in... | 1,640,796,403 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | str1=input()
n=len(str1)
str2='aeioun'
if str1[n-1] not in str2:
re1='NO'
else:
re1='YES'
for i in range(0,n-1):
print(i)
#if str1[i] not in str2 and str1[i+1] not in str2:
if str1[i] not in str2:
re2='NO'
if str1[i+1] in str2:
re2='YES'
if re1!=re2:
print('NO')
else:
print('YES') | Title: Romaji
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.
In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.
Help Vitya find out if a word $s$ is Berlanese.
Input Specification:
The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters.
Output Specification:
Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".
You can print each letter in any case (upper or lower).
Demo Input:
['sumimasen\n', 'ninja\n', 'codeforces\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.
In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese. | ```python
str1=input()
n=len(str1)
str2='aeioun'
if str1[n-1] not in str2:
re1='NO'
else:
re1='YES'
for i in range(0,n-1):
print(i)
#if str1[i] not in str2 and str1[i+1] not in str2:
if str1[i] not in str2:
re2='NO'
if str1[i+1] in str2:
re2='YES'
if re1!=re2:
print('NO')
else:
print('YES')
``` | 0 | |
912 | A | Tricky Alchemy | PROGRAMMING | 800 | [
"implementation"
] | null | null | During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are *A* yellow and *B* blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls. | The first line features two integers *A* and *B* (0<=≤<=*A*,<=*B*<=≤<=109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=109) — the respective amounts of yellow, green and blue balls to be obtained. | Print a single integer — the minimum number of crystals that Grisha should acquire in addition. | [
"4 3\n2 1 1\n",
"3 9\n1 1 3\n",
"12345678 87654321\n43043751 1000000000 53798715\n"
] | [
"2\n",
"1\n",
"2147483648\n"
] | In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue. | 500 | [
{
"input": "4 3\n2 1 1",
"output": "2"
},
{
"input": "3 9\n1 1 3",
"output": "1"
},
{
"input": "12345678 87654321\n43043751 1000000000 53798715",
"output": "2147483648"
},
{
"input": "12 12\n3 5 2",
"output": "0"
},
{
"input": "770 1390\n170 442 311",
"output"... | 1,620,193,764 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 62 | 0 | a, b = map(int, input().split())
x, y, z = map(int, input().split())
blue = z * 3 + y
yellow = x * 2 + y
x = blue + yellow
if x >= a + b:
print(x - a - b)
else:
print(0)
| Title: Tricky Alchemy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are *A* yellow and *B* blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input Specification:
The first line features two integers *A* and *B* (0<=≤<=*A*,<=*B*<=≤<=109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=109) — the respective amounts of yellow, green and blue balls to be obtained.
Output Specification:
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Demo Input:
['4 3\n2 1 1\n', '3 9\n1 1 3\n', '12345678 87654321\n43043751 1000000000 53798715\n']
Demo Output:
['2\n', '1\n', '2147483648\n']
Note:
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue. | ```python
a, b = map(int, input().split())
x, y, z = map(int, input().split())
blue = z * 3 + y
yellow = x * 2 + y
x = blue + yellow
if x >= a + b:
print(x - a - b)
else:
print(0)
``` | 0 | |
25 | A | IQ test | PROGRAMMING | 1,300 | [
"brute force"
] | A. IQ test | 2 | 256 | Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness. | The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness. | Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order. | [
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] | [
"3\n",
"2\n"
] | none | 0 | [
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output"... | 1,593,540,321 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 218 | 0 | n = int(input())
a = list(map(int,input().split()))
a = list(map(lambda x : x % 2, a))
ev = a.count(0)
od = a.count(1)
if ev > od:
print(a.index(1)+1)
else:
print(a.index(0)+1) | Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none | ```python
n = int(input())
a = list(map(int,input().split()))
a = list(map(lambda x : x % 2, a))
ev = a.count(0)
od = a.count(1)
if ev > od:
print(a.index(1)+1)
else:
print(a.index(0)+1)
``` | 3.9455 |
687 | B | Remainders Game | PROGRAMMING | 1,800 | [
"chinese remainder theorem",
"math",
"number theory"
] | null | null | Today Pari and Arya are playing a game called Remainders.
Pari chooses two positive integer *x* and *k*, and tells Arya *k* but not *x*. Arya have to find the value . There are *n* ancient numbers *c*1,<=*c*2,<=...,<=*c**n* and Pari has to tell Arya if Arya wants. Given *k* and the ancient values, tell us if Arya has a winning strategy independent of value of *x* or not. Formally, is it true that Arya can understand the value for any positive integer *x*?
Note, that means the remainder of *x* after dividing it by *y*. | The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<= *k*<=≤<=1<=000<=000) — the number of ancient integers and value *k* that is chosen by Pari.
The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=1<=000<=000). | Print "Yes" (without quotes) if Arya has a winning strategy independent of value of *x*, or "No" (without quotes) otherwise. | [
"4 5\n2 3 5 12\n",
"2 7\n2 3\n"
] | [
"Yes\n",
"No\n"
] | In the first sample, Arya can understand <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/d170efffcde0907ee6bcf32de21051bce0677a2c.png" style="max-width: 100.0%;max-height: 100.0%;"/> because 5 is one of the ancient numbers.
In the second sample, Arya can't be sure what <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/57b5f6a96f5db073270dd3ed4266c69299ec701d.png" style="max-width: 100.0%;max-height: 100.0%;"/> is. For example 1 and 7 have the same remainders after dividing by 2 and 3, but they differ in remainders after dividing by 7. | 1,000 | [
{
"input": "4 5\n2 3 5 12",
"output": "Yes"
},
{
"input": "2 7\n2 3",
"output": "No"
},
{
"input": "1 6\n8",
"output": "No"
},
{
"input": "2 3\n9 4",
"output": "Yes"
},
{
"input": "4 16\n19 16 13 9",
"output": "Yes"
},
{
"input": "5 10\n5 16 19 9 17",
... | 1,589,288,686 | 2,147,483,647 | PyPy 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | def gcd(a,b):
if b==0:
return a
return gcd(b,a%b)
def lcm(a,b):
return (a*b)/gcd(a,b)
n,k=map(int,raw_input().split())
l=1
a=map(int,raw_input().split())
for i in a:
g=gcd(i,k)
l=lcm(l,g)
if l==k:
print "Yes"
else:
print "No" | Title: Remainders Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today Pari and Arya are playing a game called Remainders.
Pari chooses two positive integer *x* and *k*, and tells Arya *k* but not *x*. Arya have to find the value . There are *n* ancient numbers *c*1,<=*c*2,<=...,<=*c**n* and Pari has to tell Arya if Arya wants. Given *k* and the ancient values, tell us if Arya has a winning strategy independent of value of *x* or not. Formally, is it true that Arya can understand the value for any positive integer *x*?
Note, that means the remainder of *x* after dividing it by *y*.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<= *k*<=≤<=1<=000<=000) — the number of ancient integers and value *k* that is chosen by Pari.
The second line contains *n* integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=1<=000<=000).
Output Specification:
Print "Yes" (without quotes) if Arya has a winning strategy independent of value of *x*, or "No" (without quotes) otherwise.
Demo Input:
['4 5\n2 3 5 12\n', '2 7\n2 3\n']
Demo Output:
['Yes\n', 'No\n']
Note:
In the first sample, Arya can understand <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/d170efffcde0907ee6bcf32de21051bce0677a2c.png" style="max-width: 100.0%;max-height: 100.0%;"/> because 5 is one of the ancient numbers.
In the second sample, Arya can't be sure what <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/57b5f6a96f5db073270dd3ed4266c69299ec701d.png" style="max-width: 100.0%;max-height: 100.0%;"/> is. For example 1 and 7 have the same remainders after dividing by 2 and 3, but they differ in remainders after dividing by 7. | ```python
def gcd(a,b):
if b==0:
return a
return gcd(b,a%b)
def lcm(a,b):
return (a*b)/gcd(a,b)
n,k=map(int,raw_input().split())
l=1
a=map(int,raw_input().split())
for i in a:
g=gcd(i,k)
l=lcm(l,g)
if l==k:
print "Yes"
else:
print "No"
``` | -1 | |
721 | A | One-dimensional Japanese Crossword | PROGRAMMING | 800 | [
"implementation"
] | null | null | Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword.
Help Adaltik find the numbers encrypting the row he drew. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew). | The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right. | [
"3\nBBW\n",
"5\nBWBWB\n",
"4\nWWWW\n",
"4\nBBBB\n",
"13\nWBBBBWWBWBBBW\n"
] | [
"1\n2 ",
"3\n1 1 1 ",
"0\n",
"1\n4 ",
"3\n4 1 3 "
] | The last sample case correspond to the picture in the statement. | 500 | [
{
"input": "3\nBBW",
"output": "1\n2 "
},
{
"input": "5\nBWBWB",
"output": "3\n1 1 1 "
},
{
"input": "4\nWWWW",
"output": "0"
},
{
"input": "4\nBBBB",
"output": "1\n4 "
},
{
"input": "13\nWBBBBWWBWBBBW",
"output": "3\n4 1 3 "
},
{
"input": "1\nB",
... | 1,586,512,026 | 2,147,483,647 | Python 3 | OK | TESTS | 54 | 109 | 307,200 | n = int(input())
s = input()
count = 0
i = 0
enc = []
while (i < n):
temp = 0
if (s[i] == 'B'):
count += 1
while(i < n and s[i] == 'B'):
i += 1
temp += 1
if (temp > 0):
enc.append(temp)
i -= 1
i += 1
print(count)
for i in enc:
print(i, end = ' ') | Title: One-dimensional Japanese Crossword
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently Adaltik discovered japanese crosswords. Japanese crossword is a picture, represented as a table sized *a*<=×<=*b* squares, and each square is colored white or black. There are integers to the left of the rows and to the top of the columns, encrypting the corresponding row or column. The number of integers represents how many groups of black squares there are in corresponding row or column, and the integers themselves represents the number of consecutive black squares in corresponding group (you can find more detailed explanation in Wikipedia [https://en.wikipedia.org/wiki/Japanese_crossword](https://en.wikipedia.org/wiki/Japanese_crossword)).
Adaltik decided that the general case of japanese crossword is too complicated and drew a row consisting of *n* squares (e.g. japanese crossword sized 1<=×<=*n*), which he wants to encrypt in the same way as in japanese crossword.
Help Adaltik find the numbers encrypting the row he drew.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the row. The second line of the input contains a single string consisting of *n* characters 'B' or 'W', ('B' corresponds to black square, 'W' — to white square in the row that Adaltik drew).
Output Specification:
The first line should contain a single integer *k* — the number of integers encrypting the row, e.g. the number of groups of black squares in the row.
The second line should contain *k* integers, encrypting the row, e.g. corresponding to sizes of groups of consecutive black squares in the order from left to right.
Demo Input:
['3\nBBW\n', '5\nBWBWB\n', '4\nWWWW\n', '4\nBBBB\n', '13\nWBBBBWWBWBBBW\n']
Demo Output:
['1\n2 ', '3\n1 1 1 ', '0\n', '1\n4 ', '3\n4 1 3 ']
Note:
The last sample case correspond to the picture in the statement. | ```python
n = int(input())
s = input()
count = 0
i = 0
enc = []
while (i < n):
temp = 0
if (s[i] == 'B'):
count += 1
while(i < n and s[i] == 'B'):
i += 1
temp += 1
if (temp > 0):
enc.append(temp)
i -= 1
i += 1
print(count)
for i in enc:
print(i, end = ' ')
``` | 3 | |
246 | D | Colorful Graph | PROGRAMMING | 1,600 | [
"brute force",
"dfs and similar",
"graphs"
] | null | null | You've got an undirected graph, consisting of *n* vertices and *m* edges. We will consider the graph's vertices numbered with integers from 1 to *n*. Each vertex of the graph has a color. The color of the *i*-th vertex is an integer *c**i*.
Let's consider all vertices of the graph, that are painted some color *k*. Let's denote a set of such as *V*(*k*). Let's denote the value of the neighbouring color diversity for color *k* as the cardinality of the set *Q*(*k*)<==<={*c**u* :<= *c**u*<=≠<=*k* and there is vertex *v* belonging to set *V*(*k*) such that nodes *v* and *u* are connected by an edge of the graph}.
Your task is to find such color *k*, which makes the cardinality of set *Q*(*k*) maximum. In other words, you want to find the color that has the most diverse neighbours. Please note, that you want to find such color *k*, that the graph has at least one vertex with such color. | The first line contains two space-separated integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of vertices end edges of the graph, correspondingly. The second line contains a sequence of integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=105) — the colors of the graph vertices. The numbers on the line are separated by spaces.
Next *m* lines contain the description of the edges: the *i*-th line contains two space-separated integers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*) — the numbers of the vertices, connected by the *i*-th edge.
It is guaranteed that the given graph has no self-loops or multiple edges. | Print the number of the color which has the set of neighbours with the maximum cardinality. It there are multiple optimal colors, print the color with the minimum number. Please note, that you want to find such color, that the graph has at least one vertex with such color. | [
"6 6\n1 1 2 3 5 8\n1 2\n3 2\n1 4\n4 3\n4 5\n4 6\n",
"5 6\n4 2 5 2 4\n1 2\n2 3\n3 1\n5 3\n5 4\n3 4\n"
] | [
"3\n",
"2\n"
] | none | 2,000 | [
{
"input": "6 6\n1 1 2 3 5 8\n1 2\n3 2\n1 4\n4 3\n4 5\n4 6",
"output": "3"
},
{
"input": "5 6\n4 2 5 2 4\n1 2\n2 3\n3 1\n5 3\n5 4\n3 4",
"output": "2"
},
{
"input": "3 1\n13 13 4\n1 2",
"output": "4"
},
{
"input": "2 1\n500 300\n1 2",
"output": "300"
},
{
"input":... | 1,590,386,065 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 10 | 716 | 1,228,800 | from collections import defaultdict
n,m = map(int,input().split())
g = defaultdict(list)
c = list(map(int,input().split()))
for _ in range(m):
x,y = map(int,input().split())
if(c[x-1]!=c[y-1]):g[c[x-1]].append(c[y-1])
if(c[y-1]!=c[x-1]):g[c[y-1]].append(c[x-1])
ma = 0
co = 10**5 + 1
f = 0
for i in g:
ma = max(ma,len(g[i]))
for i in g:
if(len(g[i])==ma):
f+=1
co = min(co,i)
if(f==0):
co = min(c)
print(co) | Title: Colorful Graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got an undirected graph, consisting of *n* vertices and *m* edges. We will consider the graph's vertices numbered with integers from 1 to *n*. Each vertex of the graph has a color. The color of the *i*-th vertex is an integer *c**i*.
Let's consider all vertices of the graph, that are painted some color *k*. Let's denote a set of such as *V*(*k*). Let's denote the value of the neighbouring color diversity for color *k* as the cardinality of the set *Q*(*k*)<==<={*c**u* :<= *c**u*<=≠<=*k* and there is vertex *v* belonging to set *V*(*k*) such that nodes *v* and *u* are connected by an edge of the graph}.
Your task is to find such color *k*, which makes the cardinality of set *Q*(*k*) maximum. In other words, you want to find the color that has the most diverse neighbours. Please note, that you want to find such color *k*, that the graph has at least one vertex with such color.
Input Specification:
The first line contains two space-separated integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of vertices end edges of the graph, correspondingly. The second line contains a sequence of integers *c*1,<=*c*2,<=...,<=*c**n* (1<=≤<=*c**i*<=≤<=105) — the colors of the graph vertices. The numbers on the line are separated by spaces.
Next *m* lines contain the description of the edges: the *i*-th line contains two space-separated integers *a**i*,<=*b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*; *a**i*<=≠<=*b**i*) — the numbers of the vertices, connected by the *i*-th edge.
It is guaranteed that the given graph has no self-loops or multiple edges.
Output Specification:
Print the number of the color which has the set of neighbours with the maximum cardinality. It there are multiple optimal colors, print the color with the minimum number. Please note, that you want to find such color, that the graph has at least one vertex with such color.
Demo Input:
['6 6\n1 1 2 3 5 8\n1 2\n3 2\n1 4\n4 3\n4 5\n4 6\n', '5 6\n4 2 5 2 4\n1 2\n2 3\n3 1\n5 3\n5 4\n3 4\n']
Demo Output:
['3\n', '2\n']
Note:
none | ```python
from collections import defaultdict
n,m = map(int,input().split())
g = defaultdict(list)
c = list(map(int,input().split()))
for _ in range(m):
x,y = map(int,input().split())
if(c[x-1]!=c[y-1]):g[c[x-1]].append(c[y-1])
if(c[y-1]!=c[x-1]):g[c[y-1]].append(c[x-1])
ma = 0
co = 10**5 + 1
f = 0
for i in g:
ma = max(ma,len(g[i]))
for i in g:
if(len(g[i])==ma):
f+=1
co = min(co,i)
if(f==0):
co = min(c)
print(co)
``` | 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... | 1,610,811,679 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 62 | 307,200 | # -*- coding: utf-8 -*-
if __name__ == "__main__":
A, B, n, X = list(map(int, input().strip().split()))
mod_tmp = 10 ** 9 + 7
if n >= mod_tmp:
n = n % mod_tmp
cache_B = -1
if A == 1:
cache_B = n
else:
cache_B = int(1 * (1 - pow(A, n, mod_tmp))/(1-A))
# print(cache_B)
item_B = (B % mod_tmp * (cache_B % mod_tmp)) % mod_tmp
# result = ((A ** n) % mod_tmp) * (X % mod_tmp) + (item_B) % mod_tmp
result = pow(A, n, mod_tmp) * (X % mod_tmp) + (item_B) % mod_tmp
result = result % mod_tmp
print(int(result))
| 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
# -*- coding: utf-8 -*-
if __name__ == "__main__":
A, B, n, X = list(map(int, input().strip().split()))
mod_tmp = 10 ** 9 + 7
if n >= mod_tmp:
n = n % mod_tmp
cache_B = -1
if A == 1:
cache_B = n
else:
cache_B = int(1 * (1 - pow(A, n, mod_tmp))/(1-A))
# print(cache_B)
item_B = (B % mod_tmp * (cache_B % mod_tmp)) % mod_tmp
# result = ((A ** n) % mod_tmp) * (X % mod_tmp) + (item_B) % mod_tmp
result = pow(A, n, mod_tmp) * (X % mod_tmp) + (item_B) % mod_tmp
result = result % mod_tmp
print(int(result))
``` | 0 | |
10 | A | Power Consumption Calculation | PROGRAMMING | 900 | [
"implementation"
] | A. Power Consumption Calculation | 1 | 256 | Tom is interested in power consumption of his favourite laptop. His laptop has three modes. In normal mode laptop consumes *P*1 watt per minute. *T*1 minutes after Tom moved the mouse or touched the keyboard for the last time, a screensaver starts and power consumption changes to *P*2 watt per minute. Finally, after *T*2 minutes from the start of the screensaver, laptop switches to the "sleep" mode and consumes *P*3 watt per minute. If Tom moves the mouse or touches the keyboard when the laptop is in the second or in the third mode, it switches to the first (normal) mode. Tom's work with the laptop can be divided into *n* time periods [*l*1,<=*r*1],<=[*l*2,<=*r*2],<=...,<=[*l**n*,<=*r**n*]. During each interval Tom continuously moves the mouse and presses buttons on the keyboard. Between the periods Tom stays away from the laptop. Find out the total amount of power consumed by the laptop during the period [*l*1,<=*r**n*]. | The first line contains 6 integer numbers *n*, *P*1, *P*2, *P*3, *T*1, *T*2 (1<=≤<=*n*<=≤<=100,<=0<=≤<=*P*1,<=*P*2,<=*P*3<=≤<=100,<=1<=≤<=*T*1,<=*T*2<=≤<=60). The following *n* lines contain description of Tom's work. Each *i*-th of these lines contains two space-separated integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=1440, *r**i*<=<<=*l**i*<=+<=1 for *i*<=<<=*n*), which stand for the start and the end of the *i*-th period of work. | Output the answer to the problem. | [
"1 3 2 1 5 10\n0 10\n",
"2 8 4 2 5 10\n20 30\n50 100\n"
] | [
"30",
"570"
] | none | 0 | [
{
"input": "1 3 2 1 5 10\n0 10",
"output": "30"
},
{
"input": "2 8 4 2 5 10\n20 30\n50 100",
"output": "570"
},
{
"input": "3 15 9 95 39 19\n873 989\n1003 1137\n1172 1436",
"output": "8445"
},
{
"input": "4 73 2 53 58 16\n51 52\n209 242\n281 407\n904 945",
"output": "5287... | 1,688,482,949 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 92 | 0 | n, p1, p2, p3, t1, t2 = map(int, input().split())
times = [list(map(int, input().split())) for _ in range(n)]
power = 0
end = times[0][0]
for i in times:
start = i[0]
for j in range(start - end):
if j < t1:
power += p1
elif j < t2 + t1:
power += p2
else:
power += p3
end = i[1]
power += (end - start) * p1
print(power)
| Title: Power Consumption Calculation
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Tom is interested in power consumption of his favourite laptop. His laptop has three modes. In normal mode laptop consumes *P*1 watt per minute. *T*1 minutes after Tom moved the mouse or touched the keyboard for the last time, a screensaver starts and power consumption changes to *P*2 watt per minute. Finally, after *T*2 minutes from the start of the screensaver, laptop switches to the "sleep" mode and consumes *P*3 watt per minute. If Tom moves the mouse or touches the keyboard when the laptop is in the second or in the third mode, it switches to the first (normal) mode. Tom's work with the laptop can be divided into *n* time periods [*l*1,<=*r*1],<=[*l*2,<=*r*2],<=...,<=[*l**n*,<=*r**n*]. During each interval Tom continuously moves the mouse and presses buttons on the keyboard. Between the periods Tom stays away from the laptop. Find out the total amount of power consumed by the laptop during the period [*l*1,<=*r**n*].
Input Specification:
The first line contains 6 integer numbers *n*, *P*1, *P*2, *P*3, *T*1, *T*2 (1<=≤<=*n*<=≤<=100,<=0<=≤<=*P*1,<=*P*2,<=*P*3<=≤<=100,<=1<=≤<=*T*1,<=*T*2<=≤<=60). The following *n* lines contain description of Tom's work. Each *i*-th of these lines contains two space-separated integers *l**i* and *r**i* (0<=≤<=*l**i*<=<<=*r**i*<=≤<=1440, *r**i*<=<<=*l**i*<=+<=1 for *i*<=<<=*n*), which stand for the start and the end of the *i*-th period of work.
Output Specification:
Output the answer to the problem.
Demo Input:
['1 3 2 1 5 10\n0 10\n', '2 8 4 2 5 10\n20 30\n50 100\n']
Demo Output:
['30', '570']
Note:
none | ```python
n, p1, p2, p3, t1, t2 = map(int, input().split())
times = [list(map(int, input().split())) for _ in range(n)]
power = 0
end = times[0][0]
for i in times:
start = i[0]
for j in range(start - end):
if j < t1:
power += p1
elif j < t2 + t1:
power += p2
else:
power += p3
end = i[1]
power += (end - start) * p1
print(power)
``` | 3.954 |
505 | B | Mr. Kitayuta's Colorful Graph | PROGRAMMING | 1,400 | [
"dfs and similar",
"dp",
"dsu",
"graphs"
] | null | null | Mr. Kitayuta has just bought an undirected graph consisting of *n* vertices and *m* edges. The vertices of the graph are numbered from 1 to *n*. Each edge, namely edge *i*, has a color *c**i*, connecting vertex *a**i* and *b**i*.
Mr. Kitayuta wants you to process the following *q* queries.
In the *i*-th query, he gives you two integers — *u**i* and *v**i*.
Find the number of the colors that satisfy the following condition: the edges of that color connect vertex *u**i* and vertex *v**i* directly or indirectly. | The first line of the input contains space-separated two integers — *n* and *m* (2<=≤<=*n*<=≤<=100,<=1<=≤<=*m*<=≤<=100), denoting the number of the vertices and the number of the edges, respectively.
The next *m* lines contain space-separated three integers — *a**i*, *b**i* (1<=≤<=*a**i*<=<<=*b**i*<=≤<=*n*) and *c**i* (1<=≤<=*c**i*<=≤<=*m*). Note that there can be multiple edges between two vertices. However, there are no multiple edges of the same color between two vertices, that is, if *i*<=≠<=*j*, (*a**i*,<=*b**i*,<=*c**i*)<=≠<=(*a**j*,<=*b**j*,<=*c**j*).
The next line contains a integer — *q* (1<=≤<=*q*<=≤<=100), denoting the number of the queries.
Then follows *q* lines, containing space-separated two integers — *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*). It is guaranteed that *u**i*<=≠<=*v**i*. | For each query, print the answer in a separate line. | [
"4 5\n1 2 1\n1 2 2\n2 3 1\n2 3 3\n2 4 3\n3\n1 2\n3 4\n1 4\n",
"5 7\n1 5 1\n2 5 1\n3 5 1\n4 5 1\n1 2 2\n2 3 2\n3 4 2\n5\n1 5\n5 1\n2 5\n1 5\n1 4\n"
] | [
"2\n1\n0\n",
"1\n1\n1\n1\n2\n"
] | Let's consider the first sample.
- Vertex 1 and vertex 2 are connected by color 1 and 2. - Vertex 3 and vertex 4 are connected by color 3. - Vertex 1 and vertex 4 are not connected by any single color. | 1,000 | [
{
"input": "4 5\n1 2 1\n1 2 2\n2 3 1\n2 3 3\n2 4 3\n3\n1 2\n3 4\n1 4",
"output": "2\n1\n0"
},
{
"input": "5 7\n1 5 1\n2 5 1\n3 5 1\n4 5 1\n1 2 2\n2 3 2\n3 4 2\n5\n1 5\n5 1\n2 5\n1 5\n1 4",
"output": "1\n1\n1\n1\n2"
},
{
"input": "2 1\n1 2 1\n1\n1 2",
"output": "1"
},
{
"input... | 1,585,485,666 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 124 | 409,600 | from collections import defaultdict
class Graph:
def __init__(self,n,m):
self.colors = defaultdict(list)
self.vertices = n
self.maxColor = m
for i in range(1,m+1):
self.colors[i] = [-1]*(n+1)
def addColor(self,e1,e2,col):
x = self.colors[col][e1]
y = self.colors[col][e2]
if(x==-1 and y == -1): #Both are Fresh
self.colors[col][e1] = e2
elif(x==-1 and y!=-1):
self.colors[col][e1] = self.colors[col][e2]
else:
self.colors[col][e2] = self.colors[col][e1]
def findParent(self,e,i):
if(self.colors[i][e]==-1):
return e
else:
return self.findParent(self.colors[i][e],i)
def solve(self,e1,e2):
count = 0
for i in range(1,self.maxColor+1):
if(self.findParent(e1,i)==self.findParent(e2,i)):
count+=1
return count
n,m = [int(x) for x in input().split()]
G = Graph(n,m)
for _ in range(m):
e1,e2,col = [int(x) for x in input().split()]
G.addColor(e1,e2,col)
Q = int(input())
for __ in range(Q):
e1,e2 = map(int,input().split())
print(G.solve(e1,e2)) | Title: Mr. Kitayuta's Colorful Graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mr. Kitayuta has just bought an undirected graph consisting of *n* vertices and *m* edges. The vertices of the graph are numbered from 1 to *n*. Each edge, namely edge *i*, has a color *c**i*, connecting vertex *a**i* and *b**i*.
Mr. Kitayuta wants you to process the following *q* queries.
In the *i*-th query, he gives you two integers — *u**i* and *v**i*.
Find the number of the colors that satisfy the following condition: the edges of that color connect vertex *u**i* and vertex *v**i* directly or indirectly.
Input Specification:
The first line of the input contains space-separated two integers — *n* and *m* (2<=≤<=*n*<=≤<=100,<=1<=≤<=*m*<=≤<=100), denoting the number of the vertices and the number of the edges, respectively.
The next *m* lines contain space-separated three integers — *a**i*, *b**i* (1<=≤<=*a**i*<=<<=*b**i*<=≤<=*n*) and *c**i* (1<=≤<=*c**i*<=≤<=*m*). Note that there can be multiple edges between two vertices. However, there are no multiple edges of the same color between two vertices, that is, if *i*<=≠<=*j*, (*a**i*,<=*b**i*,<=*c**i*)<=≠<=(*a**j*,<=*b**j*,<=*c**j*).
The next line contains a integer — *q* (1<=≤<=*q*<=≤<=100), denoting the number of the queries.
Then follows *q* lines, containing space-separated two integers — *u**i* and *v**i* (1<=≤<=*u**i*,<=*v**i*<=≤<=*n*). It is guaranteed that *u**i*<=≠<=*v**i*.
Output Specification:
For each query, print the answer in a separate line.
Demo Input:
['4 5\n1 2 1\n1 2 2\n2 3 1\n2 3 3\n2 4 3\n3\n1 2\n3 4\n1 4\n', '5 7\n1 5 1\n2 5 1\n3 5 1\n4 5 1\n1 2 2\n2 3 2\n3 4 2\n5\n1 5\n5 1\n2 5\n1 5\n1 4\n']
Demo Output:
['2\n1\n0\n', '1\n1\n1\n1\n2\n']
Note:
Let's consider the first sample.
- Vertex 1 and vertex 2 are connected by color 1 and 2. - Vertex 3 and vertex 4 are connected by color 3. - Vertex 1 and vertex 4 are not connected by any single color. | ```python
from collections import defaultdict
class Graph:
def __init__(self,n,m):
self.colors = defaultdict(list)
self.vertices = n
self.maxColor = m
for i in range(1,m+1):
self.colors[i] = [-1]*(n+1)
def addColor(self,e1,e2,col):
x = self.colors[col][e1]
y = self.colors[col][e2]
if(x==-1 and y == -1): #Both are Fresh
self.colors[col][e1] = e2
elif(x==-1 and y!=-1):
self.colors[col][e1] = self.colors[col][e2]
else:
self.colors[col][e2] = self.colors[col][e1]
def findParent(self,e,i):
if(self.colors[i][e]==-1):
return e
else:
return self.findParent(self.colors[i][e],i)
def solve(self,e1,e2):
count = 0
for i in range(1,self.maxColor+1):
if(self.findParent(e1,i)==self.findParent(e2,i)):
count+=1
return count
n,m = [int(x) for x in input().split()]
G = Graph(n,m)
for _ in range(m):
e1,e2,col = [int(x) for x in input().split()]
G.addColor(e1,e2,col)
Q = int(input())
for __ in range(Q):
e1,e2 = map(int,input().split())
print(G.solve(e1,e2))
``` | 0 | |
180 | C | Letter | PROGRAMMING | 1,400 | [
"dp"
] | null | null | Patrick has just finished writing a message to his sweetheart Stacey when he noticed that the message didn't look fancy. Patrick was nervous while writing the message, so some of the letters there were lowercase and some of them were uppercase.
Patrick believes that a message is fancy if any uppercase letter stands to the left of any lowercase one. In other words, this rule describes the strings where first go zero or more uppercase letters, and then — zero or more lowercase letters.
To make the message fancy, Patrick can erase some letter and add the same letter in the same place in the opposite case (that is, he can replace an uppercase letter with the lowercase one and vice versa). Patrick got interested in the following question: what minimum number of actions do we need to make a message fancy? Changing a letter's case in the message counts as one action. Patrick cannot perform any other actions. | The only line of the input contains a non-empty string consisting of uppercase and lowercase letters. The string's length does not exceed 105. | Print a single number — the least number of actions needed to make the message fancy. | [
"PRuvetSTAaYA\n",
"OYPROSTIYAOPECHATALSYAPRIVETSTASYA\n",
"helloworld\n"
] | [
"5\n",
"0\n",
"0\n"
] | none | 0 | [
{
"input": "PRuvetSTAaYA",
"output": "5"
},
{
"input": "OYPROSTIYAOPECHATALSYAPRIVETSTASYA",
"output": "0"
},
{
"input": "helloworld",
"output": "0"
},
{
"input": "P",
"output": "0"
},
{
"input": "t",
"output": "0"
},
{
"input": "XdJ",
"output": "1... | 1,650,482,878 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 10 | 92 | 0 | s=input()
v=[]
t=[]
for i in range(len(s)):
if s[i].isupper()==True:
v.append(i)
else:
t.append(i)
if len(v)==0:
print(0)
else:
j=max(v)
c=0
for i in range(len(t)):
if t[i]<j:
c+=1
print(c)
| Title: Letter
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Patrick has just finished writing a message to his sweetheart Stacey when he noticed that the message didn't look fancy. Patrick was nervous while writing the message, so some of the letters there were lowercase and some of them were uppercase.
Patrick believes that a message is fancy if any uppercase letter stands to the left of any lowercase one. In other words, this rule describes the strings where first go zero or more uppercase letters, and then — zero or more lowercase letters.
To make the message fancy, Patrick can erase some letter and add the same letter in the same place in the opposite case (that is, he can replace an uppercase letter with the lowercase one and vice versa). Patrick got interested in the following question: what minimum number of actions do we need to make a message fancy? Changing a letter's case in the message counts as one action. Patrick cannot perform any other actions.
Input Specification:
The only line of the input contains a non-empty string consisting of uppercase and lowercase letters. The string's length does not exceed 105.
Output Specification:
Print a single number — the least number of actions needed to make the message fancy.
Demo Input:
['PRuvetSTAaYA\n', 'OYPROSTIYAOPECHATALSYAPRIVETSTASYA\n', 'helloworld\n']
Demo Output:
['5\n', '0\n', '0\n']
Note:
none | ```python
s=input()
v=[]
t=[]
for i in range(len(s)):
if s[i].isupper()==True:
v.append(i)
else:
t.append(i)
if len(v)==0:
print(0)
else:
j=max(v)
c=0
for i in range(len(t)):
if t[i]<j:
c+=1
print(c)
``` | 0 | |
178 | D3 | Magic Squares | PROGRAMMING | 2,100 | [] | null | null | The Smart Beaver from ABBYY loves puzzles. One of his favorite puzzles is the magic square. He has recently had an idea to automate the solution of this puzzle. The Beaver decided to offer this challenge to the ABBYY Cup contestants.
The magic square is a matrix of size *n*<=×<=*n*. The elements of this matrix are integers. The sum of numbers in each row of the matrix is equal to some number *s*. The sum of numbers in each column of the matrix is also equal to *s*. In addition, the sum of the elements on the main diagonal is equal to *s* and the sum of elements on the secondary diagonal is equal to *s*. Examples of magic squares are given in the following figure:
You are given a set of *n*2 integers *a**i*. It is required to place these numbers into a square matrix of size *n*<=×<=*n* so that they form a magic square. Note that each number must occur in the matrix exactly the same number of times as it occurs in the original set.
It is guaranteed that a solution exists! | The first input line contains a single integer *n*. The next line contains *n*2 integers *a**i* (<=-<=108<=≤<=*a**i*<=≤<=108), separated by single spaces.
The input limitations for getting 20 points are:
- 1<=≤<=*n*<=≤<=3
The input limitations for getting 50 points are:
- 1<=≤<=*n*<=≤<=4 - It is guaranteed that there are no more than 9 distinct numbers among *a**i*.
The input limitations for getting 100 points are:
- 1<=≤<=*n*<=≤<=4 | The first line of the output should contain a single integer *s*. In each of the following *n* lines print *n* integers, separated by spaces and describing the resulting magic square. In the resulting magic square the sums in the rows, columns and diagonals must be equal to *s*. If there are multiple solutions, you are allowed to print any of them. | [
"3\n1 2 3 4 5 6 7 8 9\n",
"3\n1 0 -1 0 2 -1 -2 0 1\n",
"2\n5 5 5 5\n"
] | [
"15\n2 7 6\n9 5 1\n4 3 8\n",
"0\n1 0 -1\n-2 0 2\n1 0 -1\n",
"10\n5 5\n5 5\n"
] | none | 50 | [
{
"input": "3\n1 2 3 4 5 6 7 8 9",
"output": "15\n2 7 6\n9 5 1\n4 3 8"
},
{
"input": "3\n1 0 -1 0 2 -1 -2 0 1",
"output": "0\n1 0 -1\n-2 0 2\n1 0 -1"
},
{
"input": "2\n5 5 5 5",
"output": "10\n5 5\n5 5"
},
{
"input": "2\n-1 -1 -1 -1",
"output": "-2\n-1 -1\n-1 -1"
},
{... | 1,617,454,169 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS3 | 15 | 2,000 | 14,540,800 | import sys, random
def f(a):
if len(a) == 4:
rez = 0
s = [0] * 6
s[0] = a[0] + a[1]
s[1] = a[2] + a[3]
s[2] = a[0] + a[2]
s[3] = a[1] + a[3]
s[4] = a[0] + a[3]
s[5] = a[2] + a[1]
for i in s:
rez += abs(su - i)
return rez
elif len(a) == 9:
s = [0] * 8
rez = 0
s[0] = a[0] + a[1] + a[2]
s[1] = a[3] + a[4] + a[5]
s[2] = a[6] + a[7] + a[8]
s[3] = a[0] + a[3] + a[6]
s[4] = a[1] + a[4] + a[7]
s[5] = a[2] + a[5] + a[8]
s[6] = a[0] + a[4] + a[8]
s[7] = a[2] + a[4] + a[6]
for i in s:
rez += abs(su - i)
return rez
elif len(a) == 16:
s = [0] * 10
rez = 0
s[0] = a[0] + a[1] + a[2] + a[3]
s[1] = a[4] + a[5] + a[6] + a[7]
s[2] = a[8] + a[9] + a[10] + a[11]
s[3] = a[12] + a[13] + a[14] + a[15]
s[4] = a[0] + a[1] + a[2] + a[3]
s[5] = a[4] + a[5] + a[6] + a[7]
s[6] = a[8] + a[9] + a[10] + a[11]
s[7] = a[12] + a[13] + a[14] + a[15]
s[8] = a[0] + a[5] + a[10] + a[15]
s[9] = a[3] + a[6] + a[9] + a[12]
for i in s:
rez += abs(su - i)
return rez
# sys.stdin = open("input.txt", 'r')
input = sys.stdin.readline
n = int(input())
d = list(map(int, input().split()))
su = sum(d) // n
if len(d) == 1:
print(*d)
print(*d)
exit(0)
p = f(d)
while p:
random.shuffle(d)
for k in range(500):
p = f(d)
i = random.randint(0, n)
j = random.randint(0, n)
while i == j:
j = random.randint(0, n)
d[i], d[j] = d[j], d[i]
q = f(d)
if q < p:
p = q
else:
d[i], d[j] = d[j], d[i]
if p == 0:
print(su)
b = 0
c = len(d) // n
for i in range(len(d) // n):
print(*d[b:c])
b += n
c += n
exit(0)
if p == 0:
print(su)
b = 0
c = len(d) // n
print(su)
for i in range(len(d) // n):
print(*d[b:c])
b += n
c += n
exit(0)
| Title: Magic Squares
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Smart Beaver from ABBYY loves puzzles. One of his favorite puzzles is the magic square. He has recently had an idea to automate the solution of this puzzle. The Beaver decided to offer this challenge to the ABBYY Cup contestants.
The magic square is a matrix of size *n*<=×<=*n*. The elements of this matrix are integers. The sum of numbers in each row of the matrix is equal to some number *s*. The sum of numbers in each column of the matrix is also equal to *s*. In addition, the sum of the elements on the main diagonal is equal to *s* and the sum of elements on the secondary diagonal is equal to *s*. Examples of magic squares are given in the following figure:
You are given a set of *n*2 integers *a**i*. It is required to place these numbers into a square matrix of size *n*<=×<=*n* so that they form a magic square. Note that each number must occur in the matrix exactly the same number of times as it occurs in the original set.
It is guaranteed that a solution exists!
Input Specification:
The first input line contains a single integer *n*. The next line contains *n*2 integers *a**i* (<=-<=108<=≤<=*a**i*<=≤<=108), separated by single spaces.
The input limitations for getting 20 points are:
- 1<=≤<=*n*<=≤<=3
The input limitations for getting 50 points are:
- 1<=≤<=*n*<=≤<=4 - It is guaranteed that there are no more than 9 distinct numbers among *a**i*.
The input limitations for getting 100 points are:
- 1<=≤<=*n*<=≤<=4
Output Specification:
The first line of the output should contain a single integer *s*. In each of the following *n* lines print *n* integers, separated by spaces and describing the resulting magic square. In the resulting magic square the sums in the rows, columns and diagonals must be equal to *s*. If there are multiple solutions, you are allowed to print any of them.
Demo Input:
['3\n1 2 3 4 5 6 7 8 9\n', '3\n1 0 -1 0 2 -1 -2 0 1\n', '2\n5 5 5 5\n']
Demo Output:
['15\n2 7 6\n9 5 1\n4 3 8\n', '0\n1 0 -1\n-2 0 2\n1 0 -1\n', '10\n5 5\n5 5\n']
Note:
none | ```python
import sys, random
def f(a):
if len(a) == 4:
rez = 0
s = [0] * 6
s[0] = a[0] + a[1]
s[1] = a[2] + a[3]
s[2] = a[0] + a[2]
s[3] = a[1] + a[3]
s[4] = a[0] + a[3]
s[5] = a[2] + a[1]
for i in s:
rez += abs(su - i)
return rez
elif len(a) == 9:
s = [0] * 8
rez = 0
s[0] = a[0] + a[1] + a[2]
s[1] = a[3] + a[4] + a[5]
s[2] = a[6] + a[7] + a[8]
s[3] = a[0] + a[3] + a[6]
s[4] = a[1] + a[4] + a[7]
s[5] = a[2] + a[5] + a[8]
s[6] = a[0] + a[4] + a[8]
s[7] = a[2] + a[4] + a[6]
for i in s:
rez += abs(su - i)
return rez
elif len(a) == 16:
s = [0] * 10
rez = 0
s[0] = a[0] + a[1] + a[2] + a[3]
s[1] = a[4] + a[5] + a[6] + a[7]
s[2] = a[8] + a[9] + a[10] + a[11]
s[3] = a[12] + a[13] + a[14] + a[15]
s[4] = a[0] + a[1] + a[2] + a[3]
s[5] = a[4] + a[5] + a[6] + a[7]
s[6] = a[8] + a[9] + a[10] + a[11]
s[7] = a[12] + a[13] + a[14] + a[15]
s[8] = a[0] + a[5] + a[10] + a[15]
s[9] = a[3] + a[6] + a[9] + a[12]
for i in s:
rez += abs(su - i)
return rez
# sys.stdin = open("input.txt", 'r')
input = sys.stdin.readline
n = int(input())
d = list(map(int, input().split()))
su = sum(d) // n
if len(d) == 1:
print(*d)
print(*d)
exit(0)
p = f(d)
while p:
random.shuffle(d)
for k in range(500):
p = f(d)
i = random.randint(0, n)
j = random.randint(0, n)
while i == j:
j = random.randint(0, n)
d[i], d[j] = d[j], d[i]
q = f(d)
if q < p:
p = q
else:
d[i], d[j] = d[j], d[i]
if p == 0:
print(su)
b = 0
c = len(d) // n
for i in range(len(d) // n):
print(*d[b:c])
b += n
c += n
exit(0)
if p == 0:
print(su)
b = 0
c = len(d) // n
print(su)
for i in range(len(d) // n):
print(*d[b:c])
b += n
c += n
exit(0)
``` | 0 | |
535 | B | Tavas and SaDDas | PROGRAMMING | 1,100 | [
"bitmasks",
"brute force",
"combinatorics",
"implementation"
] | null | null | Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you."
The problem is:
You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
If we sort all lucky numbers in increasing order, what's the 1-based index of *n*?
Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back. | The first and only line of input contains a lucky number *n* (1<=≤<=*n*<=≤<=109). | Print the index of *n* among all lucky numbers. | [
"4\n",
"7\n",
"77\n"
] | [
"1\n",
"2\n",
"6\n"
] | none | 1,000 | [
{
"input": "4",
"output": "1"
},
{
"input": "7",
"output": "2"
},
{
"input": "77",
"output": "6"
},
{
"input": "4",
"output": "1"
},
{
"input": "474744",
"output": "83"
},
{
"input": "777774",
"output": "125"
},
{
"input": "447",
"outpu... | 1,699,582,728 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 46 | 307,200 | from collections import deque
n = input()
q = deque([""])
res = 0
while True:
cur = q.pop()
if cur == n:
break
q.appendleft(cur + "4")
q.appendleft(cur + "7")
res += 1
print(res)
| Title: Tavas and SaDDas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you."
The problem is:
You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
If we sort all lucky numbers in increasing order, what's the 1-based index of *n*?
Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back.
Input Specification:
The first and only line of input contains a lucky number *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
Print the index of *n* among all lucky numbers.
Demo Input:
['4\n', '7\n', '77\n']
Demo Output:
['1\n', '2\n', '6\n']
Note:
none | ```python
from collections import deque
n = input()
q = deque([""])
res = 0
while True:
cur = q.pop()
if cur == n:
break
q.appendleft(cur + "4")
q.appendleft(cur + "7")
res += 1
print(res)
``` | 3 | |
9 | A | Die Roll | PROGRAMMING | 800 | [
"math",
"probabilities"
] | A. Die Roll | 1 | 64 | Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win. | The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls. | Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1». | [
"4 2\n"
] | [
"1/2\n"
] | Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | 0 | [
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
... | 1,693,979,482 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 62 | 0 | import sys
numb = sys.stdin.readline().strip().split()
need = 6 - max(int(numb[0]), int(numb[1])) + 1
if need == 6:
print('1/1')
else:
print(str(need) +'/' + '6')
| Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points. | ```python
import sys
numb = sys.stdin.readline().strip().split()
need = 6 - max(int(numb[0]), int(numb[1])) + 1
if need == 6:
print('1/1')
else:
print(str(need) +'/' + '6')
``` | 0 |
61 | A | Ultra-Fast Mathematician | PROGRAMMING | 800 | [
"implementation"
] | A. Ultra-Fast Mathematician | 2 | 256 | Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate. | There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. | Write one line — the corresponding answer. Do not omit the leading 0s. | [
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] | [
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] | none | 500 | [
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "... | 1,617,911,181 | 2,147,483,647 | Python 3 | OK | TESTS | 102 | 77 | 0 | a = input()
b = input()
def maths(a,b):
r = ''
for n in range(len(a)):
if a[n] == b[n]:
r += '0'
else:
r += '1'
return r
print(maths(a,b)) | Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none | ```python
a = input()
b = input()
def maths(a,b):
r = ''
for n in range(len(a)):
if a[n] == b[n]:
r += '0'
else:
r += '1'
return r
print(maths(a,b))
``` | 3.98075 |
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": ... | 1,681,242,401 | 2,147,483,647 | Python 3 | OK | TESTS | 106 | 872 | 0 | #22mces08
n=int(input())
l=[-10]
for i in range(n):
r=int(input())
l.append(r)
ma=0
for j in range(1,n+1):
p=0
while l[j]!=-1:
p+=1
j=l[j]
ma=max(p,ma)
print(ma+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
#22mces08
n=int(input())
l=[-10]
for i in range(n):
r=int(input())
l.append(r)
ma=0
for j in range(1,n+1):
p=0
while l[j]!=-1:
p+=1
j=l[j]
ma=max(p,ma)
print(ma+1)
``` | 3 | |
103 | B | Cthulhu | PROGRAMMING | 1,500 | [
"dfs and similar",
"dsu",
"graphs"
] | B. Cthulhu | 2 | 256 | ...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with *n* vertices and *m* edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops. | The first line contains two integers — the number of vertices *n* and the number of edges *m* of the graph (1<=≤<=*n*<=≤<=100, 0<=≤<=*m*<=≤<=).
Each of the following *m* lines contains a pair of integers *x* and *y*, that show that an edge exists between vertices *x* and *y* (1<=≤<=*x*,<=*y*<=≤<=*n*,<=*x*<=≠<=*y*). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself. | Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is. | [
"6 6\n6 3\n6 4\n5 1\n2 5\n1 4\n5 4\n",
"6 5\n5 6\n4 6\n3 1\n5 1\n1 2\n"
] | [
"FHTAGN!",
"NO"
] | Let us denote as a simple cycle a set of *v* vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., *v* - 1 and *v*, *v* and 1.
A tree is a connected undirected graph consisting of *n* vertices and *n* - 1 edges (*n* > 0).
A rooted tree is a tree where one vertex is selected to be the root. | 1,000 | [
{
"input": "6 6\n6 3\n6 4\n5 1\n2 5\n1 4\n5 4",
"output": "FHTAGN!"
},
{
"input": "6 5\n5 6\n4 6\n3 1\n5 1\n1 2",
"output": "NO"
},
{
"input": "10 10\n4 10\n8 5\n2 8\n4 9\n9 3\n2 7\n10 6\n10 2\n9 8\n1 8",
"output": "FHTAGN!"
},
{
"input": "5 4\n1 5\n1 3\n1 4\n3 2",
"outpu... | 1,649,941,057 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 92 | 512,000 | class Graph:
def __init__(self, edges, a):
# Adjacency list representation
self.adjacencyList = [[] for _ in range(a)]
for (source, destination) in edges:
print(source,destination)
self.adjacencyList[source].append(destination)
self.adjacencyList[destination].append(source)
# Function for DFS_Traversal traversal
def DFS_Traversal(graph, v, visited, parent_node=-1):
# assign current node as
visited[v] = True
# loop for every edge (v, u)
for u in graph.adjacencyList[v]:
# if `u` is not visited
if not visited[u]:
if DFS_Traversal(graph, u, visited, v):
return True
# if `u` is visited, and `u` is not a parent_node
elif u != parent_node:
# found a back-edge
return True
# No back-edges were found
return False
n,m=map(int,input().split())
"""
n = 6
m = 6
data = [
"6 3",
"6 4",
"5 1",
"2 5",
"1 4",
"5 4",
]
"""
edges = []
for i in range(m):
a,b=map(int,input().split())
#a,b=map(int,data[i].split(' '))
#g.addEdge(a, b)
edges.append((a-1,b-1))
constructed_graph = Graph(edges, n)
# note the visited and unvisited nodes
visited = [False] * n
if DFS_Traversal(constructed_graph, 0, visited):
print("FHTAGN!")
else:
print("NO")
| Title: Cthulhu
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
...Once upon a time a man came to the sea. The sea was stormy and dark. The man started to call for the little mermaid to appear but alas, he only woke up Cthulhu...
Whereas on the other end of the world Pentagon is actively collecting information trying to predict the monster's behavior and preparing the secret super weapon. Due to high seismic activity and poor weather conditions the satellites haven't yet been able to make clear shots of the monster. The analysis of the first shot resulted in an undirected graph with *n* vertices and *m* edges. Now the world's best minds are about to determine whether this graph can be regarded as Cthulhu or not.
To add simplicity, let's suppose that Cthulhu looks from the space like some spherical body with tentacles attached to it. Formally, we shall regard as Cthulhu such an undirected graph that can be represented as a set of three or more rooted trees, whose roots are connected by a simple cycle.
It is guaranteed that the graph contains no multiple edges and self-loops.
Input Specification:
The first line contains two integers — the number of vertices *n* and the number of edges *m* of the graph (1<=≤<=*n*<=≤<=100, 0<=≤<=*m*<=≤<=).
Each of the following *m* lines contains a pair of integers *x* and *y*, that show that an edge exists between vertices *x* and *y* (1<=≤<=*x*,<=*y*<=≤<=*n*,<=*x*<=≠<=*y*). For each pair of vertices there will be at most one edge between them, no edge connects a vertex to itself.
Output Specification:
Print "NO", if the graph is not Cthulhu and "FHTAGN!" if it is.
Demo Input:
['6 6\n6 3\n6 4\n5 1\n2 5\n1 4\n5 4\n', '6 5\n5 6\n4 6\n3 1\n5 1\n1 2\n']
Demo Output:
['FHTAGN!', 'NO']
Note:
Let us denote as a simple cycle a set of *v* vertices that can be numbered so that the edges will only exist between vertices number 1 and 2, 2 and 3, ..., *v* - 1 and *v*, *v* and 1.
A tree is a connected undirected graph consisting of *n* vertices and *n* - 1 edges (*n* > 0).
A rooted tree is a tree where one vertex is selected to be the root. | ```python
class Graph:
def __init__(self, edges, a):
# Adjacency list representation
self.adjacencyList = [[] for _ in range(a)]
for (source, destination) in edges:
print(source,destination)
self.adjacencyList[source].append(destination)
self.adjacencyList[destination].append(source)
# Function for DFS_Traversal traversal
def DFS_Traversal(graph, v, visited, parent_node=-1):
# assign current node as
visited[v] = True
# loop for every edge (v, u)
for u in graph.adjacencyList[v]:
# if `u` is not visited
if not visited[u]:
if DFS_Traversal(graph, u, visited, v):
return True
# if `u` is visited, and `u` is not a parent_node
elif u != parent_node:
# found a back-edge
return True
# No back-edges were found
return False
n,m=map(int,input().split())
"""
n = 6
m = 6
data = [
"6 3",
"6 4",
"5 1",
"2 5",
"1 4",
"5 4",
]
"""
edges = []
for i in range(m):
a,b=map(int,input().split())
#a,b=map(int,data[i].split(' '))
#g.addEdge(a, b)
edges.append((a-1,b-1))
constructed_graph = Graph(edges, n)
# note the visited and unvisited nodes
visited = [False] * n
if DFS_Traversal(constructed_graph, 0, visited):
print("FHTAGN!")
else:
print("NO")
``` | 0 |
490 | A | Team Olympiad | PROGRAMMING | 800 | [
"greedy",
"implementation",
"sortings"
] | null | null | The School №0 of the capital of Berland has *n* children studying in it. All the children in this school are gifted: some of them are good at programming, some are good at maths, others are good at PE (Physical Education). Hence, for each child we know value *t**i*:
- *t**i*<==<=1, if the *i*-th child is good at programming, - *t**i*<==<=2, if the *i*-th child is good at maths, - *t**i*<==<=3, if the *i*-th child is good at PE
Each child happens to be good at exactly one of these three subjects.
The Team Scientific Decathlon Olympias requires teams of three students. The school teachers decided that the teams will be composed of three children that are good at different subjects. That is, each team must have one mathematician, one programmer and one sportsman. Of course, each child can be a member of no more than one team.
What is the maximum number of teams that the school will be able to present at the Olympiad? How should the teams be formed for that? | The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of children in the school. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=3), where *t**i* describes the skill of the *i*-th child. | In the first line output integer *w* — the largest possible number of teams.
Then print *w* lines, containing three numbers in each line. Each triple represents the indexes of the children forming the team. You can print both the teams, and the numbers in the triplets in any order. The children are numbered from 1 to *n* in the order of their appearance in the input. Each child must participate in no more than one team. If there are several solutions, print any of them.
If no teams can be compiled, print the only line with value *w* equal to 0. | [
"7\n1 3 1 3 2 1 2\n",
"4\n2 1 1 2\n"
] | [
"2\n3 5 2\n6 7 4\n",
"0\n"
] | none | 500 | [
{
"input": "7\n1 3 1 3 2 1 2",
"output": "2\n3 5 2\n6 7 4"
},
{
"input": "4\n2 1 1 2",
"output": "0"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "2\n3 1",
"output": "0"
},
{
"input": "3\n2 1 2",
"output": "0"
},
{
"input": "3\n1 2 3",
"output... | 1,682,175,089 | 2,147,483,647 | Python 3 | OK | TESTS | 41 | 77 | 0 | n = int(input())
t = list(map(int, input().split()))
p = []
m = []
s = []
for i in range(n):
if t[i] == 1:
p.append(i+1)
elif t[i] == 2:
m.append(i+1)
else:
s.append(i+1)
num_teams = min(len(p), len(m), len(s))
print(num_teams)
if num_teams > 0:
for i in range(num_teams):
print(p[i], m[i], s[i])
| Title: Team Olympiad
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The School №0 of the capital of Berland has *n* children studying in it. All the children in this school are gifted: some of them are good at programming, some are good at maths, others are good at PE (Physical Education). Hence, for each child we know value *t**i*:
- *t**i*<==<=1, if the *i*-th child is good at programming, - *t**i*<==<=2, if the *i*-th child is good at maths, - *t**i*<==<=3, if the *i*-th child is good at PE
Each child happens to be good at exactly one of these three subjects.
The Team Scientific Decathlon Olympias requires teams of three students. The school teachers decided that the teams will be composed of three children that are good at different subjects. That is, each team must have one mathematician, one programmer and one sportsman. Of course, each child can be a member of no more than one team.
What is the maximum number of teams that the school will be able to present at the Olympiad? How should the teams be formed for that?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=5000) — the number of children in the school. The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=3), where *t**i* describes the skill of the *i*-th child.
Output Specification:
In the first line output integer *w* — the largest possible number of teams.
Then print *w* lines, containing three numbers in each line. Each triple represents the indexes of the children forming the team. You can print both the teams, and the numbers in the triplets in any order. The children are numbered from 1 to *n* in the order of their appearance in the input. Each child must participate in no more than one team. If there are several solutions, print any of them.
If no teams can be compiled, print the only line with value *w* equal to 0.
Demo Input:
['7\n1 3 1 3 2 1 2\n', '4\n2 1 1 2\n']
Demo Output:
['2\n3 5 2\n6 7 4\n', '0\n']
Note:
none | ```python
n = int(input())
t = list(map(int, input().split()))
p = []
m = []
s = []
for i in range(n):
if t[i] == 1:
p.append(i+1)
elif t[i] == 2:
m.append(i+1)
else:
s.append(i+1)
num_teams = min(len(p), len(m), len(s))
print(num_teams)
if num_teams > 0:
for i in range(num_teams):
print(p[i], m[i], s[i])
``` | 3 | |
572 | A | Arrays | PROGRAMMING | 900 | [
"sortings"
] | null | null | You are given two arrays *A* and *B* consisting of integers, sorted in non-decreasing order. Check whether it is possible to choose *k* numbers in array *A* and choose *m* numbers in array *B* so that any number chosen in the first array is strictly less than any number chosen in the second array. | The first line contains two integers *n**A*,<=*n**B* (1<=≤<=*n**A*,<=*n**B*<=≤<=105), separated by a space — the sizes of arrays *A* and *B*, correspondingly.
The second line contains two integers *k* and *m* (1<=≤<=*k*<=≤<=*n**A*,<=1<=≤<=*m*<=≤<=*n**B*), separated by a space.
The third line contains *n**A* numbers *a*1,<=*a*2,<=... *a**n**A* (<=-<=109<=≤<=*a*1<=≤<=*a*2<=≤<=...<=≤<=*a**n**A*<=≤<=109), separated by spaces — elements of array *A*.
The fourth line contains *n**B* integers *b*1,<=*b*2,<=... *b**n**B* (<=-<=109<=≤<=*b*1<=≤<=*b*2<=≤<=...<=≤<=*b**n**B*<=≤<=109), separated by spaces — elements of array *B*. | Print "YES" (without the quotes), if you can choose *k* numbers in array *A* and *m* numbers in array *B* so that any number chosen in array *A* was strictly less than any number chosen in array *B*. Otherwise, print "NO" (without the quotes). | [
"3 3\n2 1\n1 2 3\n3 4 5\n",
"3 3\n3 3\n1 2 3\n3 4 5\n",
"5 2\n3 1\n1 1 1 1 1\n2 2\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | In the first sample test you can, for example, choose numbers 1 and 2 from array *A* and number 3 from array *B* (1 < 3 and 2 < 3).
In the second sample test the only way to choose *k* elements in the first array and *m* elements in the second one is to choose all numbers in both arrays, but then not all the numbers chosen in *A* will be less than all the numbers chosen in *B*: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7280148ed5eab0a7d418d4f92b32061243a8ca58.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | 500 | [
{
"input": "3 3\n2 1\n1 2 3\n3 4 5",
"output": "YES"
},
{
"input": "3 3\n3 3\n1 2 3\n3 4 5",
"output": "NO"
},
{
"input": "5 2\n3 1\n1 1 1 1 1\n2 2",
"output": "YES"
},
{
"input": "3 5\n1 1\n5 5 5\n5 5 5 5 5",
"output": "NO"
},
{
"input": "1 1\n1 1\n1\n1",
"ou... | 1,664,390,063 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 8 | 2,000 | 12,800,000 | x,y=map(int,input().split())
n,m=map(int,input().split())
l1=list(map(int,input().split()))
l2=list(map(int,input().split()))
l=[]
k=[]
for i in range(n):
l.append(min(l1))
l1.remove(min(l1))
for j in range(m):
k.append(max(l2))
l2.remove(max(l2))
if max(l)<min(k):
print('YES')
else:
print('NO') | Title: Arrays
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two arrays *A* and *B* consisting of integers, sorted in non-decreasing order. Check whether it is possible to choose *k* numbers in array *A* and choose *m* numbers in array *B* so that any number chosen in the first array is strictly less than any number chosen in the second array.
Input Specification:
The first line contains two integers *n**A*,<=*n**B* (1<=≤<=*n**A*,<=*n**B*<=≤<=105), separated by a space — the sizes of arrays *A* and *B*, correspondingly.
The second line contains two integers *k* and *m* (1<=≤<=*k*<=≤<=*n**A*,<=1<=≤<=*m*<=≤<=*n**B*), separated by a space.
The third line contains *n**A* numbers *a*1,<=*a*2,<=... *a**n**A* (<=-<=109<=≤<=*a*1<=≤<=*a*2<=≤<=...<=≤<=*a**n**A*<=≤<=109), separated by spaces — elements of array *A*.
The fourth line contains *n**B* integers *b*1,<=*b*2,<=... *b**n**B* (<=-<=109<=≤<=*b*1<=≤<=*b*2<=≤<=...<=≤<=*b**n**B*<=≤<=109), separated by spaces — elements of array *B*.
Output Specification:
Print "YES" (without the quotes), if you can choose *k* numbers in array *A* and *m* numbers in array *B* so that any number chosen in array *A* was strictly less than any number chosen in array *B*. Otherwise, print "NO" (without the quotes).
Demo Input:
['3 3\n2 1\n1 2 3\n3 4 5\n', '3 3\n3 3\n1 2 3\n3 4 5\n', '5 2\n3 1\n1 1 1 1 1\n2 2\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first sample test you can, for example, choose numbers 1 and 2 from array *A* and number 3 from array *B* (1 < 3 and 2 < 3).
In the second sample test the only way to choose *k* elements in the first array and *m* elements in the second one is to choose all numbers in both arrays, but then not all the numbers chosen in *A* will be less than all the numbers chosen in *B*: <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/7280148ed5eab0a7d418d4f92b32061243a8ca58.png" style="max-width: 100.0%;max-height: 100.0%;"/>. | ```python
x,y=map(int,input().split())
n,m=map(int,input().split())
l1=list(map(int,input().split()))
l2=list(map(int,input().split()))
l=[]
k=[]
for i in range(n):
l.append(min(l1))
l1.remove(min(l1))
for j in range(m):
k.append(max(l2))
l2.remove(max(l2))
if max(l)<min(k):
print('YES')
else:
print('NO')
``` | 0 | |
501 | A | Contest | PROGRAMMING | 900 | [
"implementation"
] | null | null | Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points.
Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth. | The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180).
It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round). | Output on a single line:
"Misha" (without the quotes), if Misha got more points than Vasya.
"Vasya" (without the quotes), if Vasya got more points than Misha.
"Tie" (without the quotes), if both of them got the same number of points. | [
"500 1000 20 30\n",
"1000 1000 1 1\n",
"1500 1000 176 177\n"
] | [
"Vasya\n",
"Tie\n",
"Misha\n"
] | none | 500 | [
{
"input": "500 1000 20 30",
"output": "Vasya"
},
{
"input": "1000 1000 1 1",
"output": "Tie"
},
{
"input": "1500 1000 176 177",
"output": "Misha"
},
{
"input": "1500 1000 74 177",
"output": "Misha"
},
{
"input": "750 2500 175 178",
"output": "Vasya"
},
{
... | 1,603,780,173 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 108 | 0 | a,b,c,d=map(int,input().split())
s1=max(3*a//10,a-a*c//250)
s2=max(3*b//10,b-b*d//250)
if s1>s2:
print('Misha')
elif s2>s1:
print('Vasya')
else:
print('Tie')
| Title: Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points.
Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth.
Input Specification:
The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180).
It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round).
Output Specification:
Output on a single line:
"Misha" (without the quotes), if Misha got more points than Vasya.
"Vasya" (without the quotes), if Vasya got more points than Misha.
"Tie" (without the quotes), if both of them got the same number of points.
Demo Input:
['500 1000 20 30\n', '1000 1000 1 1\n', '1500 1000 176 177\n']
Demo Output:
['Vasya\n', 'Tie\n', 'Misha\n']
Note:
none | ```python
a,b,c,d=map(int,input().split())
s1=max(3*a//10,a-a*c//250)
s2=max(3*b//10,b-b*d//250)
if s1>s2:
print('Misha')
elif s2>s1:
print('Vasya')
else:
print('Tie')
``` | 3 | |
275 | A | Lights Out | PROGRAMMING | 900 | [
"implementation"
] | null | null | Lenny is playing a game on a 3<=×<=3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. The goal of the game is to switch all the lights off. We consider the toggling as follows: if the light was switched on then it will be switched off, if it was switched off then it will be switched on.
Lenny has spent some time playing with the grid and by now he has pressed each light a certain number of times. Given the number of times each light is pressed, you have to print the current state of each light. | The input consists of three rows. Each row contains three integers each between 0 to 100 inclusive. The *j*-th number in the *i*-th row is the number of times the *j*-th light of the *i*-th row of the grid is pressed. | Print three lines, each containing three characters. The *j*-th character of the *i*-th line is "1" if and only if the corresponding light is switched on, otherwise it's "0". | [
"1 0 0\n0 0 0\n0 0 1\n",
"1 0 1\n8 8 8\n2 0 3\n"
] | [
"001\n010\n100\n",
"010\n011\n100\n"
] | none | 500 | [
{
"input": "1 0 0\n0 0 0\n0 0 1",
"output": "001\n010\n100"
},
{
"input": "1 0 1\n8 8 8\n2 0 3",
"output": "010\n011\n100"
},
{
"input": "13 85 77\n25 50 45\n65 79 9",
"output": "000\n010\n000"
},
{
"input": "96 95 5\n8 84 74\n67 31 61",
"output": "011\n011\n101"
},
{... | 1,642,098,607 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 0 | 140 | 10,342,400 | for t in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
s=a[0]
c=0
for i in range(1,n):
s += a[i]
if s < i*(i+1)//2:
c=1
print("no")
break
if c==0:
print("yes") | Title: Lights Out
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Lenny is playing a game on a 3<=×<=3 grid of lights. In the beginning of the game all lights are switched on. Pressing any of the lights will toggle it and all side-adjacent lights. The goal of the game is to switch all the lights off. We consider the toggling as follows: if the light was switched on then it will be switched off, if it was switched off then it will be switched on.
Lenny has spent some time playing with the grid and by now he has pressed each light a certain number of times. Given the number of times each light is pressed, you have to print the current state of each light.
Input Specification:
The input consists of three rows. Each row contains three integers each between 0 to 100 inclusive. The *j*-th number in the *i*-th row is the number of times the *j*-th light of the *i*-th row of the grid is pressed.
Output Specification:
Print three lines, each containing three characters. The *j*-th character of the *i*-th line is "1" if and only if the corresponding light is switched on, otherwise it's "0".
Demo Input:
['1 0 0\n0 0 0\n0 0 1\n', '1 0 1\n8 8 8\n2 0 3\n']
Demo Output:
['001\n010\n100\n', '010\n011\n100\n']
Note:
none | ```python
for t in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
s=a[0]
c=0
for i in range(1,n):
s += a[i]
if s < i*(i+1)//2:
c=1
print("no")
break
if c==0:
print("yes")
``` | -1 | |
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": "... | 1,600,692,710 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 3 | 280 | 0 | import math
target, chip = [int(x) for x in input().split()]
n = ((-1 + math.sqrt(1 + 8 * chip)) / 2)
while n >= target:
n = math.floor(n)
summation = int((n ** 2 + n) / 2)
chip -= summation
n = ((-1 + math.sqrt(1 + 8 * chip)) / 2)
n = math.floor(n)
summation = int((n ** 2 + n) / 2)
print(chip - summation) | 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
import math
target, chip = [int(x) for x in input().split()]
n = ((-1 + math.sqrt(1 + 8 * chip)) / 2)
while n >= target:
n = math.floor(n)
summation = int((n ** 2 + n) / 2)
chip -= summation
n = ((-1 + math.sqrt(1 + 8 * chip)) / 2)
n = math.floor(n)
summation = int((n ** 2 + n) / 2)
print(chip - summation)
``` | 0 |
450 | A | Jzzhu and Children | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies.
Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:
1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty.
Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order? | The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100). | Output a single integer, representing the number of the last child. | [
"5 2\n1 3 1 4 2\n",
"6 4\n1 1 2 2 3 3\n"
] | [
"4\n",
"6\n"
] | Let's consider the first sample.
Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home.
Child 4 is the last one who goes home. | 500 | [
{
"input": "5 2\n1 3 1 4 2",
"output": "4"
},
{
"input": "6 4\n1 1 2 2 3 3",
"output": "6"
},
{
"input": "7 3\n6 1 5 4 2 3 1",
"output": "4"
},
{
"input": "10 5\n2 7 3 6 2 5 1 3 4 5",
"output": "4"
},
{
"input": "100 1\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18... | 1,647,866,767 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 49 | 62 | 2,048,000 | x = input().split()
temp = input().split()
list1 = []
for i in temp:
list1.append(int(i))
n = int(x[0])
m = int(x[1])
b1 = False
start = 0
list2 = []
for i in range(1,n+1):
list2.append(i)
while b1==False:
if m >= list1[0]:
if len(list1) == 1:
break
list1.pop(0)
list2.pop(0)
continue
if list1[0] > m:
if len(list1) == 1:
break
list1[0] = list1[0] - m
temp1 = list1[0]
temp2 = list2[0]
list1.pop(0)
list1.append(temp1)
list2.pop(0)
list2.append(temp2)
continue
print(list2[-1])
| Title: Jzzhu and Children
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to *n*. The *i*-th child wants to get at least *a**i* candies.
Jzzhu asks children to line up. Initially, the *i*-th child stands at the *i*-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:
1. Give *m* candies to the first child of the line. 1. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home. 1. Repeat the first two steps while the line is not empty.
Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order?
Input Specification:
The first line contains two integers *n*,<=*m* (1<=≤<=*n*<=≤<=100; 1<=≤<=*m*<=≤<=100). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100).
Output Specification:
Output a single integer, representing the number of the last child.
Demo Input:
['5 2\n1 3 1 4 2\n', '6 4\n1 1 2 2 3 3\n']
Demo Output:
['4\n', '6\n']
Note:
Let's consider the first sample.
Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home.
Child 4 is the last one who goes home. | ```python
x = input().split()
temp = input().split()
list1 = []
for i in temp:
list1.append(int(i))
n = int(x[0])
m = int(x[1])
b1 = False
start = 0
list2 = []
for i in range(1,n+1):
list2.append(i)
while b1==False:
if m >= list1[0]:
if len(list1) == 1:
break
list1.pop(0)
list2.pop(0)
continue
if list1[0] > m:
if len(list1) == 1:
break
list1[0] = list1[0] - m
temp1 = list1[0]
temp2 = list2[0]
list1.pop(0)
list1.append(temp1)
list2.pop(0)
list2.append(temp2)
continue
print(list2[-1])
``` | 3 | |
446 | A | DZY Loves Sequences | PROGRAMMING | 1,600 | [
"dp",
"implementation",
"two pointers"
] | null | null | DZY has a sequence *a*, consisting of *n* integers.
We'll call a sequence *a**i*,<=*a**i*<=+<=1,<=...,<=*a**j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) a subsegment of the sequence *a*. The value (*j*<=-<=*i*<=+<=1) denotes the length of the subsegment.
Your task is to find the longest subsegment of *a*, such that it is possible to change at most one number (change one number to any integer you want) from the subsegment to make the subsegment strictly increasing.
You only need to output the length of the subsegment you find. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109). | In a single line print the answer to the problem — the maximum length of the required subsegment. | [
"6\n7 2 3 1 5 6\n"
] | [
"5\n"
] | You can choose subsegment *a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">3</sub>, *a*<sub class="lower-index">4</sub>, *a*<sub class="lower-index">5</sub>, *a*<sub class="lower-index">6</sub> and change its 3rd element (that is *a*<sub class="lower-index">4</sub>) to 4. | 500 | [
{
"input": "6\n7 2 3 1 5 6",
"output": "5"
},
{
"input": "10\n424238336 649760493 681692778 714636916 719885387 804289384 846930887 957747794 596516650 189641422",
"output": "9"
},
{
"input": "50\n804289384 846930887 681692778 714636916 957747794 424238336 719885387 649760493 596516650 1... | 1,600,670,659 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 108 | 0 | from sys import stdin
input = stdin.readline
n = int(input())
a = [-100] + [*map(int, input().split())] + [10**9 + 100]
c = 0
ans = []
for i, j in enumerate(a):
if 1 <= i <= n:
if a[i - 1] < a[i]:
c += 1
else:
ans.append(c)
c = 1
if a[i+1] - a[i-1] > 1 or a[i] - a[i-2] > 1:
ans.append(True)
else:
ans.append(False)
if i == n + 1:
ans.append(c)
k = 0
for i, j in enumerate(ans):
if type(j) == bool:
if j:
k = max(k, ans[i-1] + ans[i+1])
else:
k = max(k, ans[i-1], ans[i+1])
else:
k = max(k, j)
print(k)
| Title: DZY Loves Sequences
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY has a sequence *a*, consisting of *n* integers.
We'll call a sequence *a**i*,<=*a**i*<=+<=1,<=...,<=*a**j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) a subsegment of the sequence *a*. The value (*j*<=-<=*i*<=+<=1) denotes the length of the subsegment.
Your task is to find the longest subsegment of *a*, such that it is possible to change at most one number (change one number to any integer you want) from the subsegment to make the subsegment strictly increasing.
You only need to output the length of the subsegment you find.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
In a single line print the answer to the problem — the maximum length of the required subsegment.
Demo Input:
['6\n7 2 3 1 5 6\n']
Demo Output:
['5\n']
Note:
You can choose subsegment *a*<sub class="lower-index">2</sub>, *a*<sub class="lower-index">3</sub>, *a*<sub class="lower-index">4</sub>, *a*<sub class="lower-index">5</sub>, *a*<sub class="lower-index">6</sub> and change its 3rd element (that is *a*<sub class="lower-index">4</sub>) to 4. | ```python
from sys import stdin
input = stdin.readline
n = int(input())
a = [-100] + [*map(int, input().split())] + [10**9 + 100]
c = 0
ans = []
for i, j in enumerate(a):
if 1 <= i <= n:
if a[i - 1] < a[i]:
c += 1
else:
ans.append(c)
c = 1
if a[i+1] - a[i-1] > 1 or a[i] - a[i-2] > 1:
ans.append(True)
else:
ans.append(False)
if i == n + 1:
ans.append(c)
k = 0
for i, j in enumerate(ans):
if type(j) == bool:
if j:
k = max(k, ans[i-1] + ans[i+1])
else:
k = max(k, ans[i-1], ans[i+1])
else:
k = max(k, j)
print(k)
``` | 0 | |
967 | B | Watering System | PROGRAMMING | 1,000 | [
"math",
"sortings"
] | null | null | Arkady wants to water his only flower. Unfortunately, he has a very poor watering system that was designed for $n$ flowers and so it looks like a pipe with $n$ holes. Arkady can only use the water that flows from the first hole.
Arkady can block some of the holes, and then pour $A$ liters of water into the pipe. After that, the water will flow out from the non-blocked holes proportionally to their sizes $s_1, s_2, \ldots, s_n$. In other words, if the sum of sizes of non-blocked holes is $S$, and the $i$-th hole is not blocked, $\frac{s_i \cdot A}{S}$ liters of water will flow out of it.
What is the minimum number of holes Arkady should block to make at least $B$ liters of water flow out of the first hole? | The first line contains three integers $n$, $A$, $B$ ($1 \le n \le 100\,000$, $1 \le B \le A \le 10^4$) — the number of holes, the volume of water Arkady will pour into the system, and the volume he wants to get out of the first hole.
The second line contains $n$ integers $s_1, s_2, \ldots, s_n$ ($1 \le s_i \le 10^4$) — the sizes of the holes. | Print a single integer — the number of holes Arkady should block. | [
"4 10 3\n2 2 2 2\n",
"4 80 20\n3 2 1 4\n",
"5 10 10\n1000 1 1 1 1\n"
] | [
"1\n",
"0\n",
"4\n"
] | In the first example Arkady should block at least one hole. After that, $\frac{10 \cdot 2}{6} \approx 3.333$ liters of water will flow out of the first hole, and that suits Arkady.
In the second example even without blocking any hole, $\frac{80 \cdot 3}{10} = 24$ liters will flow out of the first hole, that is not less than $20$.
In the third example Arkady has to block all holes except the first to make all water flow out of the first hole. | 1,000 | [
{
"input": "4 10 3\n2 2 2 2",
"output": "1"
},
{
"input": "4 80 20\n3 2 1 4",
"output": "0"
},
{
"input": "5 10 10\n1000 1 1 1 1",
"output": "4"
},
{
"input": "10 300 100\n20 1 3 10 8 5 3 6 4 3",
"output": "1"
},
{
"input": "10 300 100\n20 25 68 40 60 37 44 85 23 ... | 1,587,582,518 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 3 | 140 | 0 | n,a,b=list(map(int,input().split()))
t=list(map(int,input().split()))
t.sort()
s= (t[n-1]*a)/sum(t)
if s>=b:
print(0)
else:
h=0
for k in range(n-1):
p = (t[n-1]*a)/(sum(t)-t[k])
if p>= b:
print(k+1)
h+=1
break
if h==0:
print(n-1)
| Title: Watering System
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Arkady wants to water his only flower. Unfortunately, he has a very poor watering system that was designed for $n$ flowers and so it looks like a pipe with $n$ holes. Arkady can only use the water that flows from the first hole.
Arkady can block some of the holes, and then pour $A$ liters of water into the pipe. After that, the water will flow out from the non-blocked holes proportionally to their sizes $s_1, s_2, \ldots, s_n$. In other words, if the sum of sizes of non-blocked holes is $S$, and the $i$-th hole is not blocked, $\frac{s_i \cdot A}{S}$ liters of water will flow out of it.
What is the minimum number of holes Arkady should block to make at least $B$ liters of water flow out of the first hole?
Input Specification:
The first line contains three integers $n$, $A$, $B$ ($1 \le n \le 100\,000$, $1 \le B \le A \le 10^4$) — the number of holes, the volume of water Arkady will pour into the system, and the volume he wants to get out of the first hole.
The second line contains $n$ integers $s_1, s_2, \ldots, s_n$ ($1 \le s_i \le 10^4$) — the sizes of the holes.
Output Specification:
Print a single integer — the number of holes Arkady should block.
Demo Input:
['4 10 3\n2 2 2 2\n', '4 80 20\n3 2 1 4\n', '5 10 10\n1000 1 1 1 1\n']
Demo Output:
['1\n', '0\n', '4\n']
Note:
In the first example Arkady should block at least one hole. After that, $\frac{10 \cdot 2}{6} \approx 3.333$ liters of water will flow out of the first hole, and that suits Arkady.
In the second example even without blocking any hole, $\frac{80 \cdot 3}{10} = 24$ liters will flow out of the first hole, that is not less than $20$.
In the third example Arkady has to block all holes except the first to make all water flow out of the first hole. | ```python
n,a,b=list(map(int,input().split()))
t=list(map(int,input().split()))
t.sort()
s= (t[n-1]*a)/sum(t)
if s>=b:
print(0)
else:
h=0
for k in range(n-1):
p = (t[n-1]*a)/(sum(t)-t[k])
if p>= b:
print(k+1)
h+=1
break
if h==0:
print(n-1)
``` | 0 | |
287 | A | IQ Test | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | In the city of Ultima Thule job applicants are often offered an IQ test.
The test is as follows: the person gets a piece of squared paper with a 4<=×<=4 square painted on it. Some of the square's cells are painted black and others are painted white. Your task is to repaint at most one cell the other color so that the picture has a 2<=×<=2 square, completely consisting of cells of the same color. If the initial picture already has such a square, the person should just say so and the test will be completed.
Your task is to write a program that determines whether it is possible to pass the test. You cannot pass the test if either repainting any cell or no action doesn't result in a 2<=×<=2 square, consisting of cells of the same color. | Four lines contain four characters each: the *j*-th character of the *i*-th line equals "." if the cell in the *i*-th row and the *j*-th column of the square is painted white, and "#", if the cell is black. | Print "YES" (without the quotes), if the test can be passed and "NO" (without the quotes) otherwise. | [
"####\n.#..\n####\n....\n",
"####\n....\n####\n....\n"
] | [
"YES\n",
"NO\n"
] | In the first test sample it is enough to repaint the first cell in the second row. After such repainting the required 2 × 2 square is on the intersection of the 1-st and 2-nd row with the 1-st and 2-nd column. | 500 | [
{
"input": "###.\n...#\n###.\n...#",
"output": "NO"
},
{
"input": ".##.\n#..#\n.##.\n#..#",
"output": "NO"
},
{
"input": ".#.#\n#.#.\n.#.#\n#.#.",
"output": "NO"
},
{
"input": "##..\n..##\n##..\n..##",
"output": "NO"
},
{
"input": "#.#.\n#.#.\n.#.#\n.#.#",
"ou... | 1,659,096,055 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 0 | 61 | 2,764,800 | square = []
for i in range(5):
square.append(input())
dot = '.'
hashtag = '#'
hash_count = 0
dot_count = 0
movement = 0
for i in square:
dot_nums = i.count(dot)
hash_nums = i.count(hashtag)
if dot_nums == 0 and hash_nums == 4:
hash_count += 1
elif dot_nums == 4 and hash_nums == 0:
dot_count += 1
elif dot_nums == 3 or hash_nums == 3:
movement += 1
if hash_count == dot_count:
print("NO")
elif movement >= 1:
print("YES")
| Title: IQ Test
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In the city of Ultima Thule job applicants are often offered an IQ test.
The test is as follows: the person gets a piece of squared paper with a 4<=×<=4 square painted on it. Some of the square's cells are painted black and others are painted white. Your task is to repaint at most one cell the other color so that the picture has a 2<=×<=2 square, completely consisting of cells of the same color. If the initial picture already has such a square, the person should just say so and the test will be completed.
Your task is to write a program that determines whether it is possible to pass the test. You cannot pass the test if either repainting any cell or no action doesn't result in a 2<=×<=2 square, consisting of cells of the same color.
Input Specification:
Four lines contain four characters each: the *j*-th character of the *i*-th line equals "." if the cell in the *i*-th row and the *j*-th column of the square is painted white, and "#", if the cell is black.
Output Specification:
Print "YES" (without the quotes), if the test can be passed and "NO" (without the quotes) otherwise.
Demo Input:
['####\n.#..\n####\n....\n', '####\n....\n####\n....\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first test sample it is enough to repaint the first cell in the second row. After such repainting the required 2 × 2 square is on the intersection of the 1-st and 2-nd row with the 1-st and 2-nd column. | ```python
square = []
for i in range(5):
square.append(input())
dot = '.'
hashtag = '#'
hash_count = 0
dot_count = 0
movement = 0
for i in square:
dot_nums = i.count(dot)
hash_nums = i.count(hashtag)
if dot_nums == 0 and hash_nums == 4:
hash_count += 1
elif dot_nums == 4 and hash_nums == 0:
dot_count += 1
elif dot_nums == 3 or hash_nums == 3:
movement += 1
if hash_count == dot_count:
print("NO")
elif movement >= 1:
print("YES")
``` | -1 | |
912 | A | Tricky Alchemy | PROGRAMMING | 800 | [
"implementation"
] | null | null | During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are *A* yellow and *B* blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls. | The first line features two integers *A* and *B* (0<=≤<=*A*,<=*B*<=≤<=109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=109) — the respective amounts of yellow, green and blue balls to be obtained. | Print a single integer — the minimum number of crystals that Grisha should acquire in addition. | [
"4 3\n2 1 1\n",
"3 9\n1 1 3\n",
"12345678 87654321\n43043751 1000000000 53798715\n"
] | [
"2\n",
"1\n",
"2147483648\n"
] | In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue. | 500 | [
{
"input": "4 3\n2 1 1",
"output": "2"
},
{
"input": "3 9\n1 1 3",
"output": "1"
},
{
"input": "12345678 87654321\n43043751 1000000000 53798715",
"output": "2147483648"
},
{
"input": "12 12\n3 5 2",
"output": "0"
},
{
"input": "770 1390\n170 442 311",
"output"... | 1,517,645,299 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 61 | 5,632,000 | y,b=input().split()
y=int(y)
b=int(b)
oy,og,ob=input().split()
oy=int(oy)
og=int(og)
ob=int(ob)
min=(oy*2 + og*2 + ob*3) - (y+b)
print(min) | Title: Tricky Alchemy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are *A* yellow and *B* blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input Specification:
The first line features two integers *A* and *B* (0<=≤<=*A*,<=*B*<=≤<=109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=109) — the respective amounts of yellow, green and blue balls to be obtained.
Output Specification:
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Demo Input:
['4 3\n2 1 1\n', '3 9\n1 1 3\n', '12345678 87654321\n43043751 1000000000 53798715\n']
Demo Output:
['2\n', '1\n', '2147483648\n']
Note:
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue. | ```python
y,b=input().split()
y=int(y)
b=int(b)
oy,og,ob=input().split()
oy=int(oy)
og=int(og)
ob=int(ob)
min=(oy*2 + og*2 + ob*3) - (y+b)
print(min)
``` | 0 | |
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":... | 1,683,276,455 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | ip = int(raw_input())
for i in range(ip):
ll = raw_input()
if len(ll) > 10:
print ll[0] + str(len(ll) - 2) + ll[len(ll) - 1]
else:
print ll | 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
ip = int(raw_input())
for i in range(ip):
ll = raw_input()
if len(ll) > 10:
print ll[0] + str(len(ll) - 2) + ll[len(ll) - 1]
else:
print ll
``` | -1 |
719 | A | Vitya in the Countryside | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Every summer Vitya comes to visit his grandmother in the countryside. This summer, he got a huge wart. Every grandma knows that one should treat warts when the moon goes down. Thus, Vitya has to catch the moment when the moon is down.
Moon cycle lasts 30 days. The size of the visible part of the moon (in Vitya's units) for each day is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, and then cycle repeats, thus after the second 1 again goes 0.
As there is no internet in the countryside, Vitya has been watching the moon for *n* consecutive days and for each of these days he wrote down the size of the visible part of the moon. Help him find out whether the moon will be up or down next day, or this cannot be determined by the data he has. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=92) — the number of consecutive days Vitya was watching the size of the visible part of the moon.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=15) — Vitya's records.
It's guaranteed that the input data is consistent. | If Vitya can be sure that the size of visible part of the moon on day *n*<=+<=1 will be less than the size of the visible part on day *n*, then print "DOWN" at the only line of the output. If he might be sure that the size of the visible part will increase, then print "UP". If it's impossible to determine what exactly will happen with the moon, print -1. | [
"5\n3 4 5 6 7\n",
"7\n12 13 14 15 14 13 12\n",
"1\n8\n"
] | [
"UP\n",
"DOWN\n",
"-1\n"
] | In the first sample, the size of the moon on the next day will be equal to 8, thus the answer is "UP".
In the second sample, the size of the moon on the next day will be 11, thus the answer is "DOWN".
In the third sample, there is no way to determine whether the size of the moon on the next day will be 7 or 9, thus the answer is -1. | 500 | [
{
"input": "5\n3 4 5 6 7",
"output": "UP"
},
{
"input": "7\n12 13 14 15 14 13 12",
"output": "DOWN"
},
{
"input": "1\n8",
"output": "-1"
},
{
"input": "44\n7 8 9 10 11 12 13 14 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 14 13 12 11 10",
"out... | 1,645,167,400 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 46 | 512,000 | def main():
n = int(input())
arr = list(map(int, input().split()))
if n == 1:
print(-1)
else:
is_decreasing = True
for i in range(n - 1):
if arr[i] < arr[i + 1]:
is_decreasing = False
break
is_increasing = True
for i in range(n - 1):
if arr[i] > arr[i + 1]:
is_increasing = False
break
if is_decreasing:
if arr[-1] == 0:
print("UP")
else:
print("DOWN")
elif is_increasing:
if arr[-1] == 15:
print("DOWN")
else:
print("UP")
else:
print("DOWN")
if __name__ == "__main__":
main()
| Title: Vitya in the Countryside
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Every summer Vitya comes to visit his grandmother in the countryside. This summer, he got a huge wart. Every grandma knows that one should treat warts when the moon goes down. Thus, Vitya has to catch the moment when the moon is down.
Moon cycle lasts 30 days. The size of the visible part of the moon (in Vitya's units) for each day is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, and then cycle repeats, thus after the second 1 again goes 0.
As there is no internet in the countryside, Vitya has been watching the moon for *n* consecutive days and for each of these days he wrote down the size of the visible part of the moon. Help him find out whether the moon will be up or down next day, or this cannot be determined by the data he has.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=92) — the number of consecutive days Vitya was watching the size of the visible part of the moon.
The second line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=15) — Vitya's records.
It's guaranteed that the input data is consistent.
Output Specification:
If Vitya can be sure that the size of visible part of the moon on day *n*<=+<=1 will be less than the size of the visible part on day *n*, then print "DOWN" at the only line of the output. If he might be sure that the size of the visible part will increase, then print "UP". If it's impossible to determine what exactly will happen with the moon, print -1.
Demo Input:
['5\n3 4 5 6 7\n', '7\n12 13 14 15 14 13 12\n', '1\n8\n']
Demo Output:
['UP\n', 'DOWN\n', '-1\n']
Note:
In the first sample, the size of the moon on the next day will be equal to 8, thus the answer is "UP".
In the second sample, the size of the moon on the next day will be 11, thus the answer is "DOWN".
In the third sample, there is no way to determine whether the size of the moon on the next day will be 7 or 9, thus the answer is -1. | ```python
def main():
n = int(input())
arr = list(map(int, input().split()))
if n == 1:
print(-1)
else:
is_decreasing = True
for i in range(n - 1):
if arr[i] < arr[i + 1]:
is_decreasing = False
break
is_increasing = True
for i in range(n - 1):
if arr[i] > arr[i + 1]:
is_increasing = False
break
if is_decreasing:
if arr[-1] == 0:
print("UP")
else:
print("DOWN")
elif is_increasing:
if arr[-1] == 15:
print("DOWN")
else:
print("UP")
else:
print("DOWN")
if __name__ == "__main__":
main()
``` | 0 | |
59 | A | Word | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Word | 2 | 256 | Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. | The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. | Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. | [
"HoUse\n",
"ViP\n",
"maTRIx\n"
] | [
"house\n",
"VIP\n",
"matrix\n"
] | none | 500 | [
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chn... | 1,672,425,452 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 0 | 154 | 3,276,800 | a=input()
count=0
for i in a:
if i.isupper():
count+=1
if count>len(a)//2:
print(a.upper())
else:
print(a.l0wer()) | Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none | ```python
a=input()
count=0
for i in a:
if i.isupper():
count+=1
if count>len(a)//2:
print(a.upper())
else:
print(a.l0wer())
``` | -1 |
372 | A | Counting Kangaroos is Fun | PROGRAMMING | 1,600 | [
"binary search",
"greedy",
"sortings",
"two pointers"
] | null | null | There are *n* kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.
Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.
The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible. | The first line contains a single integer — *n* (1<=≤<=*n*<=≤<=5·105). Each of the next *n* lines contains an integer *s**i* — the size of the *i*-th kangaroo (1<=≤<=*s**i*<=≤<=105). | Output a single integer — the optimal number of visible kangaroos. | [
"8\n2\n5\n7\n6\n9\n8\n4\n2\n",
"8\n9\n1\n6\n2\n6\n5\n8\n3\n"
] | [
"5\n",
"5\n"
] | none | 500 | [
{
"input": "8\n2\n5\n7\n6\n9\n8\n4\n2",
"output": "5"
},
{
"input": "8\n9\n1\n6\n2\n6\n5\n8\n3",
"output": "5"
},
{
"input": "12\n3\n99\n24\n46\n75\n63\n57\n55\n10\n62\n34\n52",
"output": "7"
},
{
"input": "12\n55\n75\n1\n98\n63\n64\n9\n39\n82\n18\n47\n9",
"output": "6"
... | 1,614,413,803 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 5 | 77 | 1,228,800 | n = int(input())
l=[]
for i in range(n):l.append(int(input()))
l.sort()
j=n-2
i=n-1
ans = -1
for i in range(n):
for j in range(i+1,n):
if l[i]*2<=l[j]:
ans = max(ans,i+1)
break
print(n-ans) | Title: Counting Kangaroos is Fun
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.
Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.
The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible.
Input Specification:
The first line contains a single integer — *n* (1<=≤<=*n*<=≤<=5·105). Each of the next *n* lines contains an integer *s**i* — the size of the *i*-th kangaroo (1<=≤<=*s**i*<=≤<=105).
Output Specification:
Output a single integer — the optimal number of visible kangaroos.
Demo Input:
['8\n2\n5\n7\n6\n9\n8\n4\n2\n', '8\n9\n1\n6\n2\n6\n5\n8\n3\n']
Demo Output:
['5\n', '5\n']
Note:
none | ```python
n = int(input())
l=[]
for i in range(n):l.append(int(input()))
l.sort()
j=n-2
i=n-1
ans = -1
for i in range(n):
for j in range(i+1,n):
if l[i]*2<=l[j]:
ans = max(ans,i+1)
break
print(n-ans)
``` | 0 | |
78 | A | Haiku | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Haiku | 2 | 256 | Haiku is a genre of Japanese traditional poetry.
A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words.
To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u".
Three phases from a certain poem are given. Determine whether it is haiku or not. | The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The *i*-th line contains the *i*-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification. | Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes). | [
"on codeforces \nbeta round is running\n a rustling of keys \n",
"how many gallons\nof edo s rain did you drink\n cuckoo\n"
] | [
"YES",
"NO"
] | none | 500 | [
{
"input": "on codeforces \nbeta round is running\n a rustling of keys ",
"output": "YES"
},
{
"input": "how many gallons\nof edo s rain did you drink\n cuckoo",
"output": "NO"
},
{
"input": " hatsu shigure\n saru mo komino wo\nhoshige nari",
... | 1,610,564,724 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 124 | 307,200 | # Codeforces A. Haiku
# Created by Abdulrahman Elsayed on 13/01/2021
l1 = input()
l2 = input()
l3 = input()
w1 = 0
w2 = 0
w3 = 0
for c in l1:
if ((c == 'a') or (c == 'e') or (c == 'i') or (c == 'o') or (c == 'u')):
w1 += 1
for c in l2:
if ((c == 'a') or (c == 'e') or (c == 'i') or (c == 'o') or (c == 'u')):
w2 += 1
for c in l3:
if ((c == 'a') or (c == 'e') or (c == 'i') or (c == 'o') or (c == 'u')):
w3 += 1
if ((w1 == 5) and (w2 == 7) and (w3 == 5)):
print('YES')
else:
print('NO') | Title: Haiku
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Haiku is a genre of Japanese traditional poetry.
A haiku poem consists of 17 syllables split into three phrases, containing 5, 7 and 5 syllables correspondingly (the first phrase should contain exactly 5 syllables, the second phrase should contain exactly 7 syllables, and the third phrase should contain exactly 5 syllables). A haiku masterpiece contains a description of a moment in those three phrases. Every word is important in a small poem, which is why haiku are rich with symbols. Each word has a special meaning, a special role. The main principle of haiku is to say much using a few words.
To simplify the matter, in the given problem we will consider that the number of syllable in the phrase is equal to the number of vowel letters there. Only the following letters are regarded as vowel letters: "a", "e", "i", "o" and "u".
Three phases from a certain poem are given. Determine whether it is haiku or not.
Input Specification:
The input data consists of three lines. The length of each line is between 1 and 100, inclusive. The *i*-th line contains the *i*-th phrase of the poem. Each phrase consists of one or more words, which are separated by one or more spaces. A word is a non-empty sequence of lowercase Latin letters. Leading and/or trailing spaces in phrases are allowed. Every phrase has at least one non-space character. See the example for clarification.
Output Specification:
Print "YES" (without the quotes) if the poem is a haiku. Otherwise, print "NO" (also without the quotes).
Demo Input:
['on codeforces \nbeta round is running\n a rustling of keys \n', 'how many gallons\nof edo s rain did you drink\n cuckoo\n']
Demo Output:
['YES', 'NO']
Note:
none | ```python
# Codeforces A. Haiku
# Created by Abdulrahman Elsayed on 13/01/2021
l1 = input()
l2 = input()
l3 = input()
w1 = 0
w2 = 0
w3 = 0
for c in l1:
if ((c == 'a') or (c == 'e') or (c == 'i') or (c == 'o') or (c == 'u')):
w1 += 1
for c in l2:
if ((c == 'a') or (c == 'e') or (c == 'i') or (c == 'o') or (c == 'u')):
w2 += 1
for c in l3:
if ((c == 'a') or (c == 'e') or (c == 'i') or (c == 'o') or (c == 'u')):
w3 += 1
if ((w1 == 5) and (w2 == 7) and (w3 == 5)):
print('YES')
else:
print('NO')
``` | 3.968428 |
48 | D | Permutations | PROGRAMMING | 1,500 | [
"greedy"
] | D. Permutations | 1 | 256 | A permutation is a sequence of integers from 1 to *n* of length *n* containing each number exactly once. For example, (1), (4,<=3,<=5,<=1,<=2), (3,<=2,<=1) are permutations, and (1,<=1), (4,<=3,<=1), (2,<=3,<=4) are not.
There are many tasks on permutations. Today you are going to solve one of them. Let’s imagine that somebody took several permutations (perhaps, with a different number of elements), wrote them down consecutively as one array and then shuffled the resulting array. The task is to restore the initial permutations if it is possible. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The next line contains the mixed array of *n* integers, divided with a single space. The numbers in the array are from 1 to 105. | If this array can be split into several permutations so that every element of the array belongs to exactly one permutation, print in the first line the number of permutations. The second line should contain *n* numbers, corresponding to the elements of the given array. If the *i*-th element belongs to the first permutation, the *i*-th number should be 1, if it belongs to the second one, then its number should be 2 and so on. The order of the permutations’ numbering is free.
If several solutions are possible, print any one of them. If there’s no solution, print in the first line <=-<=1. | [
"9\n1 2 3 1 2 1 4 2 5\n",
"4\n4 3 2 1\n",
"4\n1 2 2 3\n"
] | [
"3\n3 1 2 1 2 2 2 3 2\n",
"1\n1 1 1 1 ",
"-1\n"
] | In the first sample test the array is split into three permutations: (2, 1), (3, 2, 1, 4, 5), (1, 2). The first permutation is formed by the second and the fourth elements of the array, the second one — by the third, the fifth, the sixth, the seventh and the ninth elements, the third one — by the first and the eigth elements. Clearly, there are other splitting variants possible. | 0 | [
{
"input": "9\n1 2 3 1 2 1 4 2 5",
"output": "3\n1 1 1 2 2 3 1 3 1 "
},
{
"input": "4\n4 3 2 1",
"output": "1\n1 1 1 1 "
},
{
"input": "4\n1 2 2 3",
"output": "-1"
},
{
"input": "1\n1",
"output": "1\n1 "
},
{
"input": "1\n2",
"output": "-1"
},
{
"input... | 1,584,789,965 | 2,147,483,647 | Python 3 | OK | TESTS | 60 | 343 | 7,680,000 | import sys
n=int(sys.stdin.readline())
a=list(map(int,sys.stdin.readline().split()))
l=[0]*(10**5)
for i in range(n):
l[a[i]-1]+=1
flag=True
for i in range(1,len(l)):
if l[i]>l[i-1]:
print(-1)
flag=False
break
if flag:
arr=[0]*(10**5+1)
ans=[0]*n
for i in range(n):
ans[i]=arr[a[i]]+1
arr[a[i]]+=1
print(max(ans))
print(*ans) | Title: Permutations
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
A permutation is a sequence of integers from 1 to *n* of length *n* containing each number exactly once. For example, (1), (4,<=3,<=5,<=1,<=2), (3,<=2,<=1) are permutations, and (1,<=1), (4,<=3,<=1), (2,<=3,<=4) are not.
There are many tasks on permutations. Today you are going to solve one of them. Let’s imagine that somebody took several permutations (perhaps, with a different number of elements), wrote them down consecutively as one array and then shuffled the resulting array. The task is to restore the initial permutations if it is possible.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105). The next line contains the mixed array of *n* integers, divided with a single space. The numbers in the array are from 1 to 105.
Output Specification:
If this array can be split into several permutations so that every element of the array belongs to exactly one permutation, print in the first line the number of permutations. The second line should contain *n* numbers, corresponding to the elements of the given array. If the *i*-th element belongs to the first permutation, the *i*-th number should be 1, if it belongs to the second one, then its number should be 2 and so on. The order of the permutations’ numbering is free.
If several solutions are possible, print any one of them. If there’s no solution, print in the first line <=-<=1.
Demo Input:
['9\n1 2 3 1 2 1 4 2 5\n', '4\n4 3 2 1\n', '4\n1 2 2 3\n']
Demo Output:
['3\n3 1 2 1 2 2 2 3 2\n', '1\n1 1 1 1 ', '-1\n']
Note:
In the first sample test the array is split into three permutations: (2, 1), (3, 2, 1, 4, 5), (1, 2). The first permutation is formed by the second and the fourth elements of the array, the second one — by the third, the fifth, the sixth, the seventh and the ninth elements, the third one — by the first and the eigth elements. Clearly, there are other splitting variants possible. | ```python
import sys
n=int(sys.stdin.readline())
a=list(map(int,sys.stdin.readline().split()))
l=[0]*(10**5)
for i in range(n):
l[a[i]-1]+=1
flag=True
for i in range(1,len(l)):
if l[i]>l[i-1]:
print(-1)
flag=False
break
if flag:
arr=[0]*(10**5+1)
ans=[0]*n
for i in range(n):
ans[i]=arr[a[i]]+1
arr[a[i]]+=1
print(max(ans))
print(*ans)
``` | 3.814195 |
237 | A | Free Cash | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately.
Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe.
Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), that is the number of cafe visitors.
Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=≤<=*h**i*<=≤<=23; 0<=≤<=*m**i*<=≤<=59), representing the time when the *i*-th person comes into the cafe.
Note that the time is given in the chronological order. All time is given within one 24-hour period. | Print a single integer — the minimum number of cashes, needed to serve all clients next day. | [
"4\n8 0\n8 10\n8 10\n8 45\n",
"3\n0 12\n10 11\n22 22\n"
] | [
"2\n",
"1\n"
] | In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away.
In the second sample all visitors will come in different times, so it will be enough one cash. | 500 | [
{
"input": "4\n8 0\n8 10\n8 10\n8 45",
"output": "2"
},
{
"input": "3\n0 12\n10 11\n22 22",
"output": "1"
},
{
"input": "5\n12 8\n15 27\n15 27\n16 2\n19 52",
"output": "2"
},
{
"input": "7\n5 6\n7 34\n7 34\n7 34\n12 29\n15 19\n20 23",
"output": "3"
},
{
"input": "... | 1,679,567,370 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 10 | 1,184 | 13,209,600 | n=int(input())
i=0
list1=[]
list2=[]
while i<n:
hi,mi=(map(int,input().split()))
list1.append(hi)
list2.append(mi)
i+=1
list_3=[]
count1=1
j=0
while j<n-1:
if (list1[j]==list1[j+1]) and (list2[j]==list2[j+1]):
count1+=1
else:
list_3.append(count1)
count1=1
j+=1
print(max(list_3))
| Title: Free Cash
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately.
Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe.
Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), that is the number of cafe visitors.
Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=≤<=*h**i*<=≤<=23; 0<=≤<=*m**i*<=≤<=59), representing the time when the *i*-th person comes into the cafe.
Note that the time is given in the chronological order. All time is given within one 24-hour period.
Output Specification:
Print a single integer — the minimum number of cashes, needed to serve all clients next day.
Demo Input:
['4\n8 0\n8 10\n8 10\n8 45\n', '3\n0 12\n10 11\n22 22\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away.
In the second sample all visitors will come in different times, so it will be enough one cash. | ```python
n=int(input())
i=0
list1=[]
list2=[]
while i<n:
hi,mi=(map(int,input().split()))
list1.append(hi)
list2.append(mi)
i+=1
list_3=[]
count1=1
j=0
while j<n-1:
if (list1[j]==list1[j+1]) and (list2[j]==list2[j+1]):
count1+=1
else:
list_3.append(count1)
count1=1
j+=1
print(max(list_3))
``` | -1 | |
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"outpu... | 1,685,551,637 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 19 | 92 | 0 | m,n=map(int,input().split())
if(m>1 and m%2==0):
print(n*(m//2))
elif(m>1 and m%2==1):
print(n+(n//2))
elif(m==1):
print(n//2)
| Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
m,n=map(int,input().split())
if(m>1 and m%2==0):
print(n*(m//2))
elif(m>1 and m%2==1):
print(n+(n//2))
elif(m==1):
print(n//2)
``` | 0 |
660 | A | Co-prime Array | PROGRAMMING | 1,200 | [
"greedy",
"implementation",
"math",
"number theory"
] | null | null | You are given an array of *n* elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers *a* and *b* are said to be co-prime if the only positive integer that divides both of them is 1. | The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the given array.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*. | Print integer *k* on the first line — the least number of elements needed to add to the array *a* to make it co-prime.
The second line should contain *n*<=+<=*k* integers *a**j* — the elements of the array *a* after adding *k* elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array *a* by adding *k* elements to it.
If there are multiple answers you can print any one of them. | [
"3\n2 7 28\n"
] | [
"1\n2 7 9 28\n"
] | none | 0 | [
{
"input": "3\n2 7 28",
"output": "1\n2 7 1 28"
},
{
"input": "1\n1",
"output": "0\n1"
},
{
"input": "1\n548",
"output": "0\n548"
},
{
"input": "1\n963837006",
"output": "0\n963837006"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "0\n1 1 1 1 1 1 1 1 1 1"... | 1,669,224,093 | 2,147,483,647 | Python 3 | OK | TESTS | 93 | 46 | 0 | def gcd(a, b):
if (b == 0):
return a
return gcd(b, a%b)
n = int(input().strip())
arr = list(map(int, input().strip().split()))
i, count = 0, 0
for i in range(n-1):
if gcd(arr[i], arr[i+1]) != 1:
count += 1
print(count)
print(arr[0], end=" ")
for i in range(1, n):
if gcd(arr[i], arr[i-1]) != 1:
print(1, end=" ")
print(arr[i], end=" ")
| Title: Co-prime Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an array of *n* elements, you must make it a co-prime array in as few moves as possible.
In each move you can insert any positive integral number you want not greater than 109 in any place in the array.
An array is co-prime if any two adjacent numbers of it are co-prime.
In the number theory, two integers *a* and *b* are said to be co-prime if the only positive integer that divides both of them is 1.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the given array.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*.
Output Specification:
Print integer *k* on the first line — the least number of elements needed to add to the array *a* to make it co-prime.
The second line should contain *n*<=+<=*k* integers *a**j* — the elements of the array *a* after adding *k* elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array *a* by adding *k* elements to it.
If there are multiple answers you can print any one of them.
Demo Input:
['3\n2 7 28\n']
Demo Output:
['1\n2 7 9 28\n']
Note:
none | ```python
def gcd(a, b):
if (b == 0):
return a
return gcd(b, a%b)
n = int(input().strip())
arr = list(map(int, input().strip().split()))
i, count = 0, 0
for i in range(n-1):
if gcd(arr[i], arr[i+1]) != 1:
count += 1
print(count)
print(arr[0], end=" ")
for i in range(1, n):
if gcd(arr[i], arr[i-1]) != 1:
print(1, end=" ")
print(arr[i], end=" ")
``` | 3 | |
669 | A | Little Artem and Presents | PROGRAMMING | 800 | [
"math"
] | null | null | Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that.
How many times can Artem give presents to Masha? | The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — number of stones Artem received on his birthday. | Print the maximum possible number of times Artem can give presents to Masha. | [
"1\n",
"2\n",
"3\n",
"4\n"
] | [
"1\n",
"1\n",
"2\n",
"3\n"
] | In the first sample, Artem can only give 1 stone to Masha.
In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times.
In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone.
In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again. | 500 | [
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "1"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "3"
},
{
"input": "100",
"output": "67"
},
{
"input": "101",
"output": "67"
},
{
"input": "102",
"output": "68"... | 1,648,307,772 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 31 | 0 | n=int(input())
print(n-1 if n!=1 else 1) | Title: Little Artem and Presents
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Artem got *n* stones on his birthday and now wants to give some of them to Masha. He knows that Masha cares more about the fact of receiving the present, rather than the value of that present, so he wants to give her stones as many times as possible. However, Masha remembers the last present she received, so Artem can't give her the same number of stones twice in a row. For example, he can give her 3 stones, then 1 stone, then again 3 stones, but he can't give her 3 stones and then again 3 stones right after that.
How many times can Artem give presents to Masha?
Input Specification:
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=109) — number of stones Artem received on his birthday.
Output Specification:
Print the maximum possible number of times Artem can give presents to Masha.
Demo Input:
['1\n', '2\n', '3\n', '4\n']
Demo Output:
['1\n', '1\n', '2\n', '3\n']
Note:
In the first sample, Artem can only give 1 stone to Masha.
In the second sample, Atrem can give Masha 1 or 2 stones, though he can't give her 1 stone two times.
In the third sample, Atrem can first give Masha 2 stones, a then 1 more stone.
In the fourth sample, Atrem can first give Masha 1 stone, then 2 stones, and finally 1 stone again. | ```python
n=int(input())
print(n-1 if n!=1 else 1)
``` | 0 | |
408 | B | Garland | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | Once little Vasya read an article in a magazine on how to make beautiful handmade garland from colored paper. Vasya immediately went to the store and bought *n* colored sheets of paper, the area of each sheet is 1 square meter.
The garland must consist of exactly *m* pieces of colored paper of arbitrary area, each piece should be of a certain color. To make the garland, Vasya can arbitrarily cut his existing colored sheets into pieces. Vasya is not obliged to use all the sheets to make the garland.
Vasya wants the garland to be as attractive as possible, so he wants to maximize the total area of *m* pieces of paper in the garland. Calculate what the maximum total area of the pieces of paper in the garland Vasya can get. | The first line contains a non-empty sequence of *n* (1<=≤<=*n*<=≤<=1000) small English letters ("a"..."z"). Each letter means that Vasya has a sheet of paper of the corresponding color.
The second line contains a non-empty sequence of *m* (1<=≤<=*m*<=≤<=1000) small English letters that correspond to the colors of the pieces of paper in the garland that Vasya wants to make. | Print an integer that is the maximum possible total area of the pieces of paper in the garland Vasya wants to get or -1, if it is impossible to make the garland from the sheets he's got. It is guaranteed that the answer is always an integer. | [
"aaabbac\naabbccac\n",
"a\nz\n"
] | [
"6\n",
"-1"
] | In the first test sample Vasya can make an garland of area 6: he can use both sheets of color *b*, three (but not four) sheets of color *a* and cut a single sheet of color *c* in three, for example, equal pieces. Vasya can use the resulting pieces to make a garland of area 6.
In the second test sample Vasya cannot make a garland at all — he doesn't have a sheet of color *z*. | 1,000 | [
{
"input": "aaabbac\naabbccac",
"output": "6"
},
{
"input": "a\nz",
"output": "-1"
},
{
"input": "r\nr",
"output": "1"
},
{
"input": "stnsdn\nndnndsn",
"output": "4"
},
{
"input": "yqfqfp\ntttwtqq",
"output": "-1"
},
{
"input": "zzbbrrtrtzr\ntbbtrrrzr"... | 1,560,520,781 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 109 | 0 | n = list(input())
m = list(input())
c = 0
for x in m:
if x in n:
n.remove(x)
c += 1
if not c:
print(-1)
else:
print(c)
| Title: Garland
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once little Vasya read an article in a magazine on how to make beautiful handmade garland from colored paper. Vasya immediately went to the store and bought *n* colored sheets of paper, the area of each sheet is 1 square meter.
The garland must consist of exactly *m* pieces of colored paper of arbitrary area, each piece should be of a certain color. To make the garland, Vasya can arbitrarily cut his existing colored sheets into pieces. Vasya is not obliged to use all the sheets to make the garland.
Vasya wants the garland to be as attractive as possible, so he wants to maximize the total area of *m* pieces of paper in the garland. Calculate what the maximum total area of the pieces of paper in the garland Vasya can get.
Input Specification:
The first line contains a non-empty sequence of *n* (1<=≤<=*n*<=≤<=1000) small English letters ("a"..."z"). Each letter means that Vasya has a sheet of paper of the corresponding color.
The second line contains a non-empty sequence of *m* (1<=≤<=*m*<=≤<=1000) small English letters that correspond to the colors of the pieces of paper in the garland that Vasya wants to make.
Output Specification:
Print an integer that is the maximum possible total area of the pieces of paper in the garland Vasya wants to get or -1, if it is impossible to make the garland from the sheets he's got. It is guaranteed that the answer is always an integer.
Demo Input:
['aaabbac\naabbccac\n', 'a\nz\n']
Demo Output:
['6\n', '-1']
Note:
In the first test sample Vasya can make an garland of area 6: he can use both sheets of color *b*, three (but not four) sheets of color *a* and cut a single sheet of color *c* in three, for example, equal pieces. Vasya can use the resulting pieces to make a garland of area 6.
In the second test sample Vasya cannot make a garland at all — he doesn't have a sheet of color *z*. | ```python
n = list(input())
m = list(input())
c = 0
for x in m:
if x in n:
n.remove(x)
c += 1
if not c:
print(-1)
else:
print(c)
``` | 0 | |
149 | A | Business trip | PROGRAMMING | 900 | [
"greedy",
"implementation",
"sortings"
] | null | null | What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" — thought Petya. He know for a fact that if he fulfills the parents' task in the *i*-th (1<=≤<=*i*<=≤<=12) month of the year, then the flower will grow by *a**i* centimeters, and if he doesn't water the flower in the *i*-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by *k* centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by *k* centimeters. | The first line contains exactly one integer *k* (0<=≤<=*k*<=≤<=100). The next line contains twelve space-separated integers: the *i*-th (1<=≤<=*i*<=≤<=12) number in the line represents *a**i* (0<=≤<=*a**i*<=≤<=100). | Print the only integer — the minimum number of months when Petya has to water the flower so that the flower grows no less than by *k* centimeters. If the flower can't grow by *k* centimeters in a year, print -1. | [
"5\n1 1 1 1 2 2 3 2 2 1 1 1\n",
"0\n0 0 0 0 0 0 0 1 1 2 3 0\n",
"11\n1 1 4 1 1 5 1 1 4 1 1 1\n"
] | [
"2\n",
"0\n",
"3\n"
] | Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (*k* = 0). So, it is possible for Petya not to water the flower at all. | 500 | [
{
"input": "5\n1 1 1 1 2 2 3 2 2 1 1 1",
"output": "2"
},
{
"input": "0\n0 0 0 0 0 0 0 1 1 2 3 0",
"output": "0"
},
{
"input": "11\n1 1 4 1 1 5 1 1 4 1 1 1",
"output": "3"
},
{
"input": "15\n20 1 1 1 1 2 2 1 2 2 1 1",
"output": "1"
},
{
"input": "7\n8 9 100 12 14 ... | 1,646,222,915 | 2,147,483,647 | Python 3 | OK | TESTS | 39 | 92 | 0 | k=int(input())
waters=[int(i) for i in input().split()]
waters=sorted(waters)
height=0;i=0
while(height<k and i<12):
height+=waters[11-i]
i+=1
if(height<k):
print(-1)
else:
print(i) | Title: Business trip
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
What joy! Petya's parents went on a business trip for the whole year and the playful kid is left all by himself. Petya got absolutely happy. He jumped on the bed and threw pillows all day long, until...
Today Petya opened the cupboard and found a scary note there. His parents had left him with duties: he should water their favourite flower all year, each day, in the morning, in the afternoon and in the evening. "Wait a second!" — thought Petya. He know for a fact that if he fulfills the parents' task in the *i*-th (1<=≤<=*i*<=≤<=12) month of the year, then the flower will grow by *a**i* centimeters, and if he doesn't water the flower in the *i*-th month, then the flower won't grow this month. Petya also knows that try as he might, his parents won't believe that he has been watering the flower if it grows strictly less than by *k* centimeters.
Help Petya choose the minimum number of months when he will water the flower, given that the flower should grow no less than by *k* centimeters.
Input Specification:
The first line contains exactly one integer *k* (0<=≤<=*k*<=≤<=100). The next line contains twelve space-separated integers: the *i*-th (1<=≤<=*i*<=≤<=12) number in the line represents *a**i* (0<=≤<=*a**i*<=≤<=100).
Output Specification:
Print the only integer — the minimum number of months when Petya has to water the flower so that the flower grows no less than by *k* centimeters. If the flower can't grow by *k* centimeters in a year, print -1.
Demo Input:
['5\n1 1 1 1 2 2 3 2 2 1 1 1\n', '0\n0 0 0 0 0 0 0 1 1 2 3 0\n', '11\n1 1 4 1 1 5 1 1 4 1 1 1\n']
Demo Output:
['2\n', '0\n', '3\n']
Note:
Let's consider the first sample test. There it is enough to water the flower during the seventh and the ninth month. Then the flower grows by exactly five centimeters.
In the second sample Petya's parents will believe him even if the flower doesn't grow at all (*k* = 0). So, it is possible for Petya not to water the flower at all. | ```python
k=int(input())
waters=[int(i) for i in input().split()]
waters=sorted(waters)
height=0;i=0
while(height<k and i<12):
height+=waters[11-i]
i+=1
if(height<k):
print(-1)
else:
print(i)
``` | 3 | |
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"
... | 1,697,512,281 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 15 | 0 | def is_black_and_white(grid):
for row in grid:
for c in row:
if c != 'W' and c != 'B':
return False
return True
def main():
n, m = map(int, input().split())
grid = [input() for _ in range(n)]
if is_black_and_white(grid):
print("#Black&White")
else:
print("#Color")
if __name__ == "__main__":
main()
| 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
def is_black_and_white(grid):
for row in grid:
for c in row:
if c != 'W' and c != 'B':
return False
return True
def main():
n, m = map(int, input().split())
grid = [input() for _ in range(n)]
if is_black_and_white(grid):
print("#Black&White")
else:
print("#Color")
if __name__ == "__main__":
main()
``` | 0 | |
760 | A | Petr and a calendar | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Petr wants to make a calendar for current month. For this purpose he draws a table in which columns correspond to weeks (a week is seven consequent days from Monday to Sunday), rows correspond to weekdays, and cells contain dates. For example, a calendar for January 2017 should look like on the picture:
Petr wants to know how many columns his table should have given the month and the weekday of the first date of that month? Assume that the year is non-leap. | The only line contain two integers *m* and *d* (1<=≤<=*m*<=≤<=12, 1<=≤<=*d*<=≤<=7) — the number of month (January is the first month, December is the twelfth) and the weekday of the first date of this month (1 is Monday, 7 is Sunday). | Print single integer: the number of columns the table should have. | [
"1 7\n",
"1 1\n",
"11 6\n"
] | [
"6\n",
"5\n",
"5\n"
] | The first example corresponds to the January 2017 shown on the picture in the statements.
In the second example 1-st January is Monday, so the whole month fits into 5 columns.
In the third example 1-st November is Saturday and 5 columns is enough. | 500 | [
{
"input": "1 7",
"output": "6"
},
{
"input": "1 1",
"output": "5"
},
{
"input": "11 6",
"output": "5"
},
{
"input": "2 7",
"output": "5"
},
{
"input": "2 1",
"output": "4"
},
{
"input": "8 6",
"output": "6"
},
{
"input": "1 1",
"output... | 1,485,111,328 | 2,428 | Python 3 | RUNTIME_ERROR | PRETESTS | 0 | 31 | 4,608,000 | import numpy as np
m, d = [int(x) for x in input().split()]
md = np.array ([31,28,31,30,31,30,31,31,30,31,30,31])
n = (md[(m - 1)] - (8-d) )// 7
if (md[(m - 1)] - (8-d) ) % 7 == 0:
print (n+1)
else:
print (n+2)
| Title: Petr and a calendar
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petr wants to make a calendar for current month. For this purpose he draws a table in which columns correspond to weeks (a week is seven consequent days from Monday to Sunday), rows correspond to weekdays, and cells contain dates. For example, a calendar for January 2017 should look like on the picture:
Petr wants to know how many columns his table should have given the month and the weekday of the first date of that month? Assume that the year is non-leap.
Input Specification:
The only line contain two integers *m* and *d* (1<=≤<=*m*<=≤<=12, 1<=≤<=*d*<=≤<=7) — the number of month (January is the first month, December is the twelfth) and the weekday of the first date of this month (1 is Monday, 7 is Sunday).
Output Specification:
Print single integer: the number of columns the table should have.
Demo Input:
['1 7\n', '1 1\n', '11 6\n']
Demo Output:
['6\n', '5\n', '5\n']
Note:
The first example corresponds to the January 2017 shown on the picture in the statements.
In the second example 1-st January is Monday, so the whole month fits into 5 columns.
In the third example 1-st November is Saturday and 5 columns is enough. | ```python
import numpy as np
m, d = [int(x) for x in input().split()]
md = np.array ([31,28,31,30,31,30,31,31,30,31,30,31])
n = (md[(m - 1)] - (8-d) )// 7
if (md[(m - 1)] - (8-d) ) % 7 == 0:
print (n+1)
else:
print (n+2)
``` | -1 | |
414 | A | Mashmokh and Numbers | PROGRAMMING | 1,500 | [
"constructive algorithms",
"number theory"
] | null | null | It's holiday. Mashmokh and his boss, Bimokh, are playing a game invented by Mashmokh.
In this game Mashmokh writes sequence of *n* distinct integers on the board. Then Bimokh makes several (possibly zero) moves. On the first move he removes the first and the second integer from from the board, on the second move he removes the first and the second integer of the remaining sequence from the board, and so on. Bimokh stops when the board contains less than two numbers. When Bimokh removes numbers *x* and *y* from the board, he gets *gcd*(*x*,<=*y*) points. At the beginning of the game Bimokh has zero points.
Mashmokh wants to win in the game. For this reason he wants his boss to get exactly *k* points in total. But the guy doesn't know how choose the initial sequence in the right way.
Please, help him. Find *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* such that his boss will score exactly *k* points. Also Mashmokh can't memorize too huge numbers. Therefore each of these integers must be at most 109. | The first line of input contains two space-separated integers *n*,<=*k* (1<=≤<=*n*<=≤<=105; 0<=≤<=*k*<=≤<=108). | If such sequence doesn't exist output -1 otherwise output *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109). | [
"5 2\n",
"5 3",
"7 2\n"
] | [
"1 2 3 4 5\n",
"2 4 3 7 1",
"-1\n"
] | *gcd*(*x*, *y*) is greatest common divisor of *x* and *y*. | 500 | [
{
"input": "5 2",
"output": "1 2 3 4 5"
},
{
"input": "5 3",
"output": "2 4 5 6 7"
},
{
"input": "7 2",
"output": "-1"
},
{
"input": "1 1",
"output": "-1"
},
{
"input": "2 0",
"output": "-1"
},
{
"input": "1 10",
"output": "-1"
},
{
"input"... | 1,652,545,219 | 2,147,483,647 | Python 3 | OK | TESTS | 84 | 78 | 8,396,800 | # 2022-05-15T00:50:46.010Z
def proc(n, k):
if n == 1:
return [1] if k == 0 else [-1]
if n // 2 > k:
return [-1]
num_of_pairs = n // 2
k -= (num_of_pairs - 1)
ans = [k, 2 * k]
base = 2 * k + 1
for i in range(n - 2):
ans.append(base)
base += 1
return ans
n, k = map(int, input().split())
ans = proc(n, k)
print(' '.join(map(str, ans)))
| Title: Mashmokh and Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's holiday. Mashmokh and his boss, Bimokh, are playing a game invented by Mashmokh.
In this game Mashmokh writes sequence of *n* distinct integers on the board. Then Bimokh makes several (possibly zero) moves. On the first move he removes the first and the second integer from from the board, on the second move he removes the first and the second integer of the remaining sequence from the board, and so on. Bimokh stops when the board contains less than two numbers. When Bimokh removes numbers *x* and *y* from the board, he gets *gcd*(*x*,<=*y*) points. At the beginning of the game Bimokh has zero points.
Mashmokh wants to win in the game. For this reason he wants his boss to get exactly *k* points in total. But the guy doesn't know how choose the initial sequence in the right way.
Please, help him. Find *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* such that his boss will score exactly *k* points. Also Mashmokh can't memorize too huge numbers. Therefore each of these integers must be at most 109.
Input Specification:
The first line of input contains two space-separated integers *n*,<=*k* (1<=≤<=*n*<=≤<=105; 0<=≤<=*k*<=≤<=108).
Output Specification:
If such sequence doesn't exist output -1 otherwise output *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Demo Input:
['5 2\n', '5 3', '7 2\n']
Demo Output:
['1 2 3 4 5\n', '2 4 3 7 1', '-1\n']
Note:
*gcd*(*x*, *y*) is greatest common divisor of *x* and *y*. | ```python
# 2022-05-15T00:50:46.010Z
def proc(n, k):
if n == 1:
return [1] if k == 0 else [-1]
if n // 2 > k:
return [-1]
num_of_pairs = n // 2
k -= (num_of_pairs - 1)
ans = [k, 2 * k]
base = 2 * k + 1
for i in range(n - 2):
ans.append(base)
base += 1
return ans
n, k = map(int, input().split())
ans = proc(n, k)
print(' '.join(map(str, ans)))
``` | 3 | |
802 | G | Fake News (easy) | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | As it's the first of April, Heidi is suspecting that the news she reads today are fake, and she does not want to look silly in front of all the contestants. She knows that a newspiece is fake if it contains heidi as a subsequence. Help Heidi assess whether the given piece is true, but please be discreet about it... | The first and only line of input contains a single nonempty string *s* of length at most 1000 composed of lowercase letters (a-z). | Output YES if the string *s* contains heidi as a subsequence and NO otherwise. | [
"abcheaibcdi\n",
"hiedi\n"
] | [
"YES",
"NO"
] | A string *s* contains another string *p* as a subsequence if it is possible to delete some characters from *s* and obtain *p*. | 0 | [
{
"input": "abcheaibcdi",
"output": "YES"
},
{
"input": "hiedi",
"output": "NO"
},
{
"input": "ihied",
"output": "NO"
},
{
"input": "diehi",
"output": "NO"
},
{
"input": "deiih",
"output": "NO"
},
{
"input": "iheid",
"output": "NO"
},
{
"in... | 1,610,480,082 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 62 | 0 | string= input()
if 'heidi' in string:
print('NO')
else:
print('YES')
| Title: Fake News (easy)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As it's the first of April, Heidi is suspecting that the news she reads today are fake, and she does not want to look silly in front of all the contestants. She knows that a newspiece is fake if it contains heidi as a subsequence. Help Heidi assess whether the given piece is true, but please be discreet about it...
Input Specification:
The first and only line of input contains a single nonempty string *s* of length at most 1000 composed of lowercase letters (a-z).
Output Specification:
Output YES if the string *s* contains heidi as a subsequence and NO otherwise.
Demo Input:
['abcheaibcdi\n', 'hiedi\n']
Demo Output:
['YES', 'NO']
Note:
A string *s* contains another string *p* as a subsequence if it is possible to delete some characters from *s* and obtain *p*. | ```python
string= input()
if 'heidi' in string:
print('NO')
else:
print('YES')
``` | 0 | |
244 | A | Dividing Orange | PROGRAMMING | 900 | [
"implementation"
] | null | null | One day Ms Swan bought an orange in a shop. The orange consisted of *n*·*k* segments, numbered with integers from 1 to *n*·*k*.
There were *k* children waiting for Ms Swan at home. The children have recently learned about the orange and they decided to divide it between them. For that each child took a piece of paper and wrote the number of the segment that he would like to get: the *i*-th (1<=≤<=*i*<=≤<=*k*) child wrote the number *a**i* (1<=≤<=*a**i*<=≤<=*n*·*k*). All numbers *a**i* accidentally turned out to be different.
Now the children wonder, how to divide the orange so as to meet these conditions:
- each child gets exactly *n* orange segments; - the *i*-th child gets the segment with number *a**i* for sure; - no segment goes to two children simultaneously.
Help the children, divide the orange and fulfill the requirements, described above. | The first line contains two integers *n*, *k* (1<=≤<=*n*,<=*k*<=≤<=30). The second line contains *k* space-separated integers *a*1,<=*a*2,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=*n*·*k*), where *a**i* is the number of the orange segment that the *i*-th child would like to get.
It is guaranteed that all numbers *a**i* are distinct. | Print exactly *n*·*k* distinct integers. The first *n* integers represent the indexes of the segments the first child will get, the second *n* integers represent the indexes of the segments the second child will get, and so on. Separate the printed numbers with whitespaces.
You can print a child's segment indexes in any order. It is guaranteed that the answer always exists. If there are multiple correct answers, print any of them. | [
"2 2\n4 1\n",
"3 1\n2\n"
] | [
"2 4 \n1 3 \n",
"3 2 1 \n"
] | none | 500 | [
{
"input": "2 2\n4 1",
"output": "2 4 \n1 3 "
},
{
"input": "3 1\n2",
"output": "3 2 1 "
},
{
"input": "5 5\n25 24 23 22 21",
"output": "2 3 1 25 4 \n7 6 8 5 24 \n10 12 9 23 11 \n13 15 14 16 22 \n19 21 20 17 18 "
},
{
"input": "1 30\n8 22 13 25 10 30 12 27 6 4 7 2 20 16 26 14... | 1,548,578,791 | 2,147,483,647 | Python 3 | OK | TESTS | 29 | 248 | 102,400 | n, k = [int(x) for x in input().split()]
likes = [int(x) for x in input().split()]
oranges = [ True ] * (n * k)
for i in likes:
oranges[i - 1] = False
oranges_cnt = 0
for l in likes:
print(l, end=" ")
for _ in range(n - 1):
while not oranges[oranges_cnt]:
oranges_cnt += 1
print(oranges_cnt + 1, end=" ")
oranges_cnt += 1
print()
| Title: Dividing Orange
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Ms Swan bought an orange in a shop. The orange consisted of *n*·*k* segments, numbered with integers from 1 to *n*·*k*.
There were *k* children waiting for Ms Swan at home. The children have recently learned about the orange and they decided to divide it between them. For that each child took a piece of paper and wrote the number of the segment that he would like to get: the *i*-th (1<=≤<=*i*<=≤<=*k*) child wrote the number *a**i* (1<=≤<=*a**i*<=≤<=*n*·*k*). All numbers *a**i* accidentally turned out to be different.
Now the children wonder, how to divide the orange so as to meet these conditions:
- each child gets exactly *n* orange segments; - the *i*-th child gets the segment with number *a**i* for sure; - no segment goes to two children simultaneously.
Help the children, divide the orange and fulfill the requirements, described above.
Input Specification:
The first line contains two integers *n*, *k* (1<=≤<=*n*,<=*k*<=≤<=30). The second line contains *k* space-separated integers *a*1,<=*a*2,<=...,<=*a**k* (1<=≤<=*a**i*<=≤<=*n*·*k*), where *a**i* is the number of the orange segment that the *i*-th child would like to get.
It is guaranteed that all numbers *a**i* are distinct.
Output Specification:
Print exactly *n*·*k* distinct integers. The first *n* integers represent the indexes of the segments the first child will get, the second *n* integers represent the indexes of the segments the second child will get, and so on. Separate the printed numbers with whitespaces.
You can print a child's segment indexes in any order. It is guaranteed that the answer always exists. If there are multiple correct answers, print any of them.
Demo Input:
['2 2\n4 1\n', '3 1\n2\n']
Demo Output:
['2 4 \n1 3 \n', '3 2 1 \n']
Note:
none | ```python
n, k = [int(x) for x in input().split()]
likes = [int(x) for x in input().split()]
oranges = [ True ] * (n * k)
for i in likes:
oranges[i - 1] = False
oranges_cnt = 0
for l in likes:
print(l, end=" ")
for _ in range(n - 1):
while not oranges[oranges_cnt]:
oranges_cnt += 1
print(oranges_cnt + 1, end=" ")
oranges_cnt += 1
print()
``` | 3 |
Subsets and Splits
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