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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
365 | A | Good Number | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*). | The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109). | Print a single integer — the number of *k*-good numbers in *a*. | [
"10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n",
"2 1\n1\n10\n"
] | [
"10\n",
"1\n"
] | none | 500 | [
{
"input": "10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560",
"output": "10"
},
{
"input": "2 1\n1\n10",
"output": "1"
},
{
"input": "1 0\n1000000000",
"output": "1"
},
{
"input": "1 1\n1000000000",
"output": "1"
},
{
... | 1,670,950,296 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 5 | 77 | 0 | n, k = map(int, input().split())
kl = [i for i in range(k)]
count = 0
for i in range(n):
s = input()
chk = True
for i in kl:
if str(i) not in s:
chk = False
break
if chk:
count+=1
print(count) | Title: Good Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*).
Input Specification:
The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print a single integer — the number of *k*-good numbers in *a*.
Demo Input:
['10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n', '2 1\n1\n10\n']
Demo Output:
['10\n', '1\n']
Note:
none | ```python
n, k = map(int, input().split())
kl = [i for i in range(k)]
count = 0
for i in range(n):
s = input()
chk = True
for i in kl:
if str(i) not in s:
chk = False
break
if chk:
count+=1
print(count)
``` | 0 | |
991 | B | Getting an A | PROGRAMMING | 900 | [
"greedy",
"sortings"
] | null | null | Translator's note: in Russia's most widespread grading system, there are four grades: 5, 4, 3, 2, the higher the better, roughly corresponding to A, B, C and F respectively in American grading system.
The term is coming to an end and students start thinking about their grades. Today, a professor told his students that the grades for his course would be given out automatically — he would calculate the simple average (arithmetic mean) of all grades given out for lab works this term and round to the nearest integer. The rounding would be done in favour of the student — $4.5$ would be rounded up to $5$ (as in example 3), but $4.4$ would be rounded down to $4$.
This does not bode well for Vasya who didn't think those lab works would influence anything, so he may receive a grade worse than $5$ (maybe even the dreaded $2$). However, the professor allowed him to redo some of his works of Vasya's choosing to increase his average grade. Vasya wants to redo as as few lab works as possible in order to get $5$ for the course. Of course, Vasya will get $5$ for the lab works he chooses to redo.
Help Vasya — calculate the minimum amount of lab works Vasya has to redo. | The first line contains a single integer $n$ — the number of Vasya's grades ($1 \leq n \leq 100$).
The second line contains $n$ integers from $2$ to $5$ — Vasya's grades for his lab works. | Output a single integer — the minimum amount of lab works that Vasya has to redo. It can be shown that Vasya can always redo enough lab works to get a $5$. | [
"3\n4 4 4\n",
"4\n5 4 5 5\n",
"4\n5 3 3 5\n"
] | [
"2\n",
"0\n",
"1\n"
] | In the first sample, it is enough to redo two lab works to make two $4$s into $5$s.
In the second sample, Vasya's average is already $4.75$ so he doesn't have to redo anything to get a $5$.
In the second sample Vasya has to redo one lab work to get rid of one of the $3$s, that will make the average exactly $4.5$ so the final grade would be $5$. | 1,000 | [
{
"input": "3\n4 4 4",
"output": "2"
},
{
"input": "4\n5 4 5 5",
"output": "0"
},
{
"input": "4\n5 3 3 5",
"output": "1"
},
{
"input": "1\n5",
"output": "0"
},
{
"input": "4\n3 2 5 4",
"output": "2"
},
{
"input": "5\n5 4 3 2 5",
"output": "2"
},
... | 1,627,639,700 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 62 | 6,963,200 | from math import *
n = int(input())
for i in range(n):
d = list(int(j) for j in input().split() )
s = sum(d)
a = s/n
if a >= 4.5:
print(0)
else:
x = ceil(n * 4.5)
y = int(x - s)
d.sort()
t = 0
for k in d:
if k != 5:
l = 5 - k
k += l
y -= l
t += 1
if y <= 0:
break
print(t)
| Title: Getting an A
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Translator's note: in Russia's most widespread grading system, there are four grades: 5, 4, 3, 2, the higher the better, roughly corresponding to A, B, C and F respectively in American grading system.
The term is coming to an end and students start thinking about their grades. Today, a professor told his students that the grades for his course would be given out automatically — he would calculate the simple average (arithmetic mean) of all grades given out for lab works this term and round to the nearest integer. The rounding would be done in favour of the student — $4.5$ would be rounded up to $5$ (as in example 3), but $4.4$ would be rounded down to $4$.
This does not bode well for Vasya who didn't think those lab works would influence anything, so he may receive a grade worse than $5$ (maybe even the dreaded $2$). However, the professor allowed him to redo some of his works of Vasya's choosing to increase his average grade. Vasya wants to redo as as few lab works as possible in order to get $5$ for the course. Of course, Vasya will get $5$ for the lab works he chooses to redo.
Help Vasya — calculate the minimum amount of lab works Vasya has to redo.
Input Specification:
The first line contains a single integer $n$ — the number of Vasya's grades ($1 \leq n \leq 100$).
The second line contains $n$ integers from $2$ to $5$ — Vasya's grades for his lab works.
Output Specification:
Output a single integer — the minimum amount of lab works that Vasya has to redo. It can be shown that Vasya can always redo enough lab works to get a $5$.
Demo Input:
['3\n4 4 4\n', '4\n5 4 5 5\n', '4\n5 3 3 5\n']
Demo Output:
['2\n', '0\n', '1\n']
Note:
In the first sample, it is enough to redo two lab works to make two $4$s into $5$s.
In the second sample, Vasya's average is already $4.75$ so he doesn't have to redo anything to get a $5$.
In the second sample Vasya has to redo one lab work to get rid of one of the $3$s, that will make the average exactly $4.5$ so the final grade would be $5$. | ```python
from math import *
n = int(input())
for i in range(n):
d = list(int(j) for j in input().split() )
s = sum(d)
a = s/n
if a >= 4.5:
print(0)
else:
x = ceil(n * 4.5)
y = int(x - s)
d.sort()
t = 0
for k in d:
if k != 5:
l = 5 - k
k += l
y -= l
t += 1
if y <= 0:
break
print(t)
``` | -1 | |
202 | A | LLPS | PROGRAMMING | 800 | [
"binary search",
"bitmasks",
"brute force",
"greedy",
"implementation",
"strings"
] | null | null | This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z". | The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10. | Print the lexicographically largest palindromic subsequence of string *s*. | [
"radar\n",
"bowwowwow\n",
"codeforces\n",
"mississipp\n"
] | [
"rr\n",
"wwwww\n",
"s\n",
"ssss\n"
] | Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr". | 500 | [
{
"input": "radar",
"output": "rr"
},
{
"input": "bowwowwow",
"output": "wwwww"
},
{
"input": "codeforces",
"output": "s"
},
{
"input": "mississipp",
"output": "ssss"
},
{
"input": "tourist",
"output": "u"
},
{
"input": "romka",
"output": "r"
},
... | 1,619,903,319 | 2,147,483,647 | PyPy 3 | OK | TESTS | 54 | 216 | 0 | import sys
def main():
s = sys.stdin.read().strip()
return max(s)*s.count(max(s))
print(main())
| Title: LLPS
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input Specification:
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
Output Specification:
Print the lexicographically largest palindromic subsequence of string *s*.
Demo Input:
['radar\n', 'bowwowwow\n', 'codeforces\n', 'mississipp\n']
Demo Output:
['rr\n', 'wwwww\n', 's\n', 'ssss\n']
Note:
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr". | ```python
import sys
def main():
s = sys.stdin.read().strip()
return max(s)*s.count(max(s))
print(main())
``` | 3 | |
25 | B | Phone numbers | PROGRAMMING | 1,100 | [
"implementation"
] | B. Phone numbers | 2 | 256 | Phone number in Berland is a sequence of *n* digits. Often, to make it easier to memorize the number, it is divided into groups of two or three digits. For example, the phone number 1198733 is easier to remember as 11-987-33. Your task is to find for a given phone number any of its divisions into groups of two or three digits. | The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of digits in the phone number. The second line contains *n* digits — the phone number to divide into groups. | Output any of divisions of the given phone number into groups of two or three digits. Separate groups by single character -. If the answer is not unique, output any. | [
"6\n549871\n",
"7\n1198733\n"
] | [
"54-98-71",
"11-987-33\n"
] | none | 0 | [
{
"input": "6\n549871",
"output": "54-98-71"
},
{
"input": "7\n1198733",
"output": "119-87-33"
},
{
"input": "2\n74",
"output": "74"
},
{
"input": "2\n33",
"output": "33"
},
{
"input": "3\n074",
"output": "074"
},
{
"input": "3\n081",
"output": "08... | 1,618,156,129 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 9 | 124 | 0 | n = int(input())
a = input()
b=''
if n%2==0:
i=0
while(i<n-2):
b=b+a[i]+a[i+1]+'-'
i=i+2
b=b+a[i]+a[i+1]
else:
i=0
if n==3:
b=b+a[i]+a[i+1]+a[i+2]
i=i+3
elif n==5:
b=b+a[i]+a[i+1]+'-'+a[i+2]+a[i+3]+a[i+4]
i=i+5
else:
b = a[i]+a[i+1]+'-'
i=i+2
while(i<n-2):
b=b+a[i]+a[i+1]+a[i+2]+'-'
i=i+3
b=b+a[i]+a[i+1]
print(b) | Title: Phone numbers
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Phone number in Berland is a sequence of *n* digits. Often, to make it easier to memorize the number, it is divided into groups of two or three digits. For example, the phone number 1198733 is easier to remember as 11-987-33. Your task is to find for a given phone number any of its divisions into groups of two or three digits.
Input Specification:
The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of digits in the phone number. The second line contains *n* digits — the phone number to divide into groups.
Output Specification:
Output any of divisions of the given phone number into groups of two or three digits. Separate groups by single character -. If the answer is not unique, output any.
Demo Input:
['6\n549871\n', '7\n1198733\n']
Demo Output:
['54-98-71', '11-987-33\n']
Note:
none | ```python
n = int(input())
a = input()
b=''
if n%2==0:
i=0
while(i<n-2):
b=b+a[i]+a[i+1]+'-'
i=i+2
b=b+a[i]+a[i+1]
else:
i=0
if n==3:
b=b+a[i]+a[i+1]+a[i+2]
i=i+3
elif n==5:
b=b+a[i]+a[i+1]+'-'+a[i+2]+a[i+3]+a[i+4]
i=i+5
else:
b = a[i]+a[i+1]+'-'
i=i+2
while(i<n-2):
b=b+a[i]+a[i+1]+a[i+2]+'-'
i=i+3
b=b+a[i]+a[i+1]
print(b)
``` | -1 |
69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10... | 1,638,784,373 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 80 | 92 | 0 | N = int(input())
a = []
for i in range(N):
X = map(int,input().split(" "))
a.append(sum(X))
if sum(a) == 0:
print("YES")
else:
print("NO")
| Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
N = int(input())
a = []
for i in range(N):
X = map(int,input().split(" "))
a.append(sum(X))
if sum(a) == 0:
print("YES")
else:
print("NO")
``` | 0 |
608 | B | Hamming Distance Sum | PROGRAMMING | 1,500 | [
"combinatorics",
"strings"
] | null | null | Genos needs your help. He was asked to solve the following programming problem by Saitama:
The length of some string *s* is denoted |*s*|. The Hamming distance between two strings *s* and *t* of equal length is defined as , where *s**i* is the *i*-th character of *s* and *t**i* is the *i*-th character of *t*. For example, the Hamming distance between string "0011" and string "0110" is |0<=-<=0|<=+<=|0<=-<=1|<=+<=|1<=-<=1|<=+<=|1<=-<=0|<==<=0<=+<=1<=+<=0<=+<=1<==<=2.
Given two binary strings *a* and *b*, find the sum of the Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|. | The first line of the input contains binary string *a* (1<=≤<=|*a*|<=≤<=200<=000).
The second line of the input contains binary string *b* (|*a*|<=≤<=|*b*|<=≤<=200<=000).
Both strings are guaranteed to consist of characters '0' and '1' only. | Print a single integer — the sum of Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|. | [
"01\n00111\n",
"0011\n0110\n"
] | [
"3\n",
"2\n"
] | For the first sample case, there are four contiguous substrings of *b* of length |*a*|: "00", "01", "11", and "11". The distance between "01" and "00" is |0 - 0| + |1 - 0| = 1. The distance between "01" and "01" is |0 - 0| + |1 - 1| = 0. The distance between "01" and "11" is |0 - 1| + |1 - 1| = 1. Last distance counts twice, as there are two occurrences of string "11". The sum of these edit distances is 1 + 0 + 1 + 1 = 3.
The second sample case is described in the statement. | 1,000 | [
{
"input": "01\n00111",
"output": "3"
},
{
"input": "0011\n0110",
"output": "2"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "0\n1",
"output": "1"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1... | 1,606,396,169 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 3 | 124 | 0 | a = input()
b = input()
e = len(a)
k = 0
if a[0] == '1':
w = b[:e]
t = w.count('1')
k = e-t
print(1)
else:
w = b[:e+2]
t = w.count('1')
k = t
for i in range(1,e):
w = w[1:]
w += b[-i]
if w[-1] == '1':
t += 1
if w[i-1] == '1':
t -= 1
if a[i] == '1':
k += e-t
else:
k += t
print(t)
| Title: Hamming Distance Sum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Genos needs your help. He was asked to solve the following programming problem by Saitama:
The length of some string *s* is denoted |*s*|. The Hamming distance between two strings *s* and *t* of equal length is defined as , where *s**i* is the *i*-th character of *s* and *t**i* is the *i*-th character of *t*. For example, the Hamming distance between string "0011" and string "0110" is |0<=-<=0|<=+<=|0<=-<=1|<=+<=|1<=-<=1|<=+<=|1<=-<=0|<==<=0<=+<=1<=+<=0<=+<=1<==<=2.
Given two binary strings *a* and *b*, find the sum of the Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|.
Input Specification:
The first line of the input contains binary string *a* (1<=≤<=|*a*|<=≤<=200<=000).
The second line of the input contains binary string *b* (|*a*|<=≤<=|*b*|<=≤<=200<=000).
Both strings are guaranteed to consist of characters '0' and '1' only.
Output Specification:
Print a single integer — the sum of Hamming distances between *a* and all contiguous substrings of *b* of length |*a*|.
Demo Input:
['01\n00111\n', '0011\n0110\n']
Demo Output:
['3\n', '2\n']
Note:
For the first sample case, there are four contiguous substrings of *b* of length |*a*|: "00", "01", "11", and "11". The distance between "01" and "00" is |0 - 0| + |1 - 0| = 1. The distance between "01" and "01" is |0 - 0| + |1 - 1| = 0. The distance between "01" and "11" is |0 - 1| + |1 - 1| = 1. Last distance counts twice, as there are two occurrences of string "11". The sum of these edit distances is 1 + 0 + 1 + 1 = 3.
The second sample case is described in the statement. | ```python
a = input()
b = input()
e = len(a)
k = 0
if a[0] == '1':
w = b[:e]
t = w.count('1')
k = e-t
print(1)
else:
w = b[:e+2]
t = w.count('1')
k = t
for i in range(1,e):
w = w[1:]
w += b[-i]
if w[-1] == '1':
t += 1
if w[i-1] == '1':
t -= 1
if a[i] == '1':
k += e-t
else:
k += t
print(t)
``` | 0 | |
545 | D | Queue | PROGRAMMING | 1,300 | [
"greedy",
"implementation",
"sortings"
] | null | null | Little girl Susie went shopping with her mom and she wondered how to improve service quality.
There are *n* people in the queue. For each person we know time *t**i* needed to serve him. A person will be disappointed if the time he waits is more than the time needed to serve him. The time a person waits is the total time when all the people who stand in the queue in front of him are served. Susie thought that if we swap some people in the queue, then we can decrease the number of people who are disappointed.
Help Susie find out what is the maximum number of not disappointed people can be achieved by swapping people in the queue. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105).
The next line contains *n* integers *t**i* (1<=≤<=*t**i*<=≤<=109), separated by spaces. | Print a single number — the maximum number of not disappointed people in the queue. | [
"5\n15 2 1 5 3\n"
] | [
"4\n"
] | Value 4 is achieved at such an arrangement, for example: 1, 2, 3, 5, 15. Thus, you can make everything feel not disappointed except for the person with time 5. | 1,750 | [
{
"input": "5\n15 2 1 5 3",
"output": "4"
},
{
"input": "15\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "2"
},
{
"input": "10\n13 2 5 55 21 34 1 8 1 3",
"output": "6"
},
{
"input": "10\n8 256 16 1 2 1 64 4 128 32",
"output": "10"
},
{
"input": "10\n10000 40000 1000... | 1,698,305,685 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | n = int(input())
ts = list(map(int, input().split()))
ts.sort()
happy = 1
current_index = 1
for t in range(1, n):
if sum(ts[:current_index]) <= ts[current_index]:
happy += 1
else:
ts.pop(current_index)
current_index -= 1
print(happy)
| Title: Queue
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little girl Susie went shopping with her mom and she wondered how to improve service quality.
There are *n* people in the queue. For each person we know time *t**i* needed to serve him. A person will be disappointed if the time he waits is more than the time needed to serve him. The time a person waits is the total time when all the people who stand in the queue in front of him are served. Susie thought that if we swap some people in the queue, then we can decrease the number of people who are disappointed.
Help Susie find out what is the maximum number of not disappointed people can be achieved by swapping people in the queue.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105).
The next line contains *n* integers *t**i* (1<=≤<=*t**i*<=≤<=109), separated by spaces.
Output Specification:
Print a single number — the maximum number of not disappointed people in the queue.
Demo Input:
['5\n15 2 1 5 3\n']
Demo Output:
['4\n']
Note:
Value 4 is achieved at such an arrangement, for example: 1, 2, 3, 5, 15. Thus, you can make everything feel not disappointed except for the person with time 5. | ```python
n = int(input())
ts = list(map(int, input().split()))
ts.sort()
happy = 1
current_index = 1
for t in range(1, n):
if sum(ts[:current_index]) <= ts[current_index]:
happy += 1
else:
ts.pop(current_index)
current_index -= 1
print(happy)
``` | 0 | |
285 | C | Building Permutation | PROGRAMMING | 1,200 | [
"greedy",
"implementation",
"sortings"
] | null | null | Permutation *p* is an ordered set of integers *p*1,<=<=*p*2,<=<=...,<=<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. We'll denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size or the length of permutation *p*1,<=<=*p*2,<=<=...,<=<=*p**n*.
You have a sequence of integers *a*1,<=*a*2,<=...,<=*a**n*. In one move, you are allowed to decrease or increase any number by one. Count the minimum number of moves, needed to build a permutation from this sequence. | The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the size of the sought permutation. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109). | Print a single number — the minimum number of moves.
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. | [
"2\n3 0\n",
"3\n-1 -1 2\n"
] | [
"2\n",
"6\n"
] | In the first sample you should decrease the first number by one and then increase the second number by one. The resulting permutation is (2, 1).
In the second sample you need 6 moves to build permutation (1, 3, 2). | 1,500 | [
{
"input": "2\n3 0",
"output": "2"
},
{
"input": "3\n-1 -1 2",
"output": "6"
},
{
"input": "5\n-3 5 -3 3 3",
"output": "10"
},
{
"input": "10\n9 6 -2 4 1 1 1 9 6 2",
"output": "18"
},
{
"input": "9\n2 0 0 6 5 4 1 9 3",
"output": "15"
},
{
"input": "100... | 1,634,584,710 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 358 | 23,961,600 | n = int(input())
A = list(map(int, input().split()))
A.sort()
ans = 0
for i in range(len(A)):
ans += abs(i + 1 - A[i])
print(ans) | Title: Building Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Permutation *p* is an ordered set of integers *p*1,<=<=*p*2,<=<=...,<=<=*p**n*, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. We'll denote the *i*-th element of permutation *p* as *p**i*. We'll call number *n* the size or the length of permutation *p*1,<=<=*p*2,<=<=...,<=<=*p**n*.
You have a sequence of integers *a*1,<=*a*2,<=...,<=*a**n*. In one move, you are allowed to decrease or increase any number by one. Count the minimum number of moves, needed to build a permutation from this sequence.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=3·105) — the size of the sought permutation. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109).
Output Specification:
Print a single number — the minimum number of moves.
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.
Demo Input:
['2\n3 0\n', '3\n-1 -1 2\n']
Demo Output:
['2\n', '6\n']
Note:
In the first sample you should decrease the first number by one and then increase the second number by one. The resulting permutation is (2, 1).
In the second sample you need 6 moves to build permutation (1, 3, 2). | ```python
n = int(input())
A = list(map(int, input().split()))
A.sort()
ans = 0
for i in range(len(A)):
ans += abs(i + 1 - A[i])
print(ans)
``` | 3 | |
602 | A | Two Bases | PROGRAMMING | 1,100 | [
"brute force",
"implementation"
] | null | null | After seeing the "ALL YOUR BASE ARE BELONG TO US" meme for the first time, numbers *X* and *Y* realised that they have different bases, which complicated their relations.
You're given a number *X* represented in base *b**x* and a number *Y* represented in base *b**y*. Compare those two numbers. | The first line of the input contains two space-separated integers *n* and *b**x* (1<=≤<=*n*<=≤<=10, 2<=≤<=*b**x*<=≤<=40), where *n* is the number of digits in the *b**x*-based representation of *X*.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=<<=*b**x*) — the digits of *X*. They are given in the order from the most significant digit to the least significant one.
The following two lines describe *Y* in the same way: the third line contains two space-separated integers *m* and *b**y* (1<=≤<=*m*<=≤<=10, 2<=≤<=*b**y*<=≤<=40, *b**x*<=≠<=*b**y*), where *m* is the number of digits in the *b**y*-based representation of *Y*, and the fourth line contains *m* space-separated integers *y*1,<=*y*2,<=...,<=*y**m* (0<=≤<=*y**i*<=<<=*b**y*) — the digits of *Y*.
There will be no leading zeroes. Both *X* and *Y* will be positive. All digits of both numbers are given in the standard decimal numeral system. | Output a single character (quotes for clarity):
- '<' if *X*<=<<=*Y* - '>' if *X*<=><=*Y* - '=' if *X*<==<=*Y* | [
"6 2\n1 0 1 1 1 1\n2 10\n4 7\n",
"3 3\n1 0 2\n2 5\n2 4\n",
"7 16\n15 15 4 0 0 7 10\n7 9\n4 8 0 3 1 5 0\n"
] | [
"=\n",
"<\n",
">\n"
] | In the first sample, *X* = 101111<sub class="lower-index">2</sub> = 47<sub class="lower-index">10</sub> = *Y*.
In the second sample, *X* = 102<sub class="lower-index">3</sub> = 21<sub class="lower-index">5</sub> and *Y* = 24<sub class="lower-index">5</sub> = 112<sub class="lower-index">3</sub>, thus *X* < *Y*.
In the third sample, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/603a342b0ae3e56fed542d1c50c0a5ff6ce2cbaa.png" style="max-width: 100.0%;max-height: 100.0%;"/> and *Y* = 4803150<sub class="lower-index">9</sub>. We may notice that *X* starts with much larger digits and *b*<sub class="lower-index">*x*</sub> is much larger than *b*<sub class="lower-index">*y*</sub>, so *X* is clearly larger than *Y*. | 500 | [
{
"input": "6 2\n1 0 1 1 1 1\n2 10\n4 7",
"output": "="
},
{
"input": "3 3\n1 0 2\n2 5\n2 4",
"output": "<"
},
{
"input": "7 16\n15 15 4 0 0 7 10\n7 9\n4 8 0 3 1 5 0",
"output": ">"
},
{
"input": "2 2\n1 0\n2 3\n1 0",
"output": "<"
},
{
"input": "2 2\n1 0\n1 3\n1"... | 1,458,098,254 | 2,147,483,647 | Python 3 | OK | TESTS | 118 | 77 | 0 | ans=[]
for i in range(2):
(n,b),data=map(int,input().split()),list(map(int, input().split()))
data.reverse();
t=0
j=0
for i in range(n):
t+=data[i]*(b**j)
j+=1
ans.append(t)
# print(ans[0],"...",ans[1])
if ans[0]>ans[1]:
print(">")
elif ans[0]<ans[1]:
print("<")
else:
print("=")
| Title: Two Bases
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After seeing the "ALL YOUR BASE ARE BELONG TO US" meme for the first time, numbers *X* and *Y* realised that they have different bases, which complicated their relations.
You're given a number *X* represented in base *b**x* and a number *Y* represented in base *b**y*. Compare those two numbers.
Input Specification:
The first line of the input contains two space-separated integers *n* and *b**x* (1<=≤<=*n*<=≤<=10, 2<=≤<=*b**x*<=≤<=40), where *n* is the number of digits in the *b**x*-based representation of *X*.
The second line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=<<=*b**x*) — the digits of *X*. They are given in the order from the most significant digit to the least significant one.
The following two lines describe *Y* in the same way: the third line contains two space-separated integers *m* and *b**y* (1<=≤<=*m*<=≤<=10, 2<=≤<=*b**y*<=≤<=40, *b**x*<=≠<=*b**y*), where *m* is the number of digits in the *b**y*-based representation of *Y*, and the fourth line contains *m* space-separated integers *y*1,<=*y*2,<=...,<=*y**m* (0<=≤<=*y**i*<=<<=*b**y*) — the digits of *Y*.
There will be no leading zeroes. Both *X* and *Y* will be positive. All digits of both numbers are given in the standard decimal numeral system.
Output Specification:
Output a single character (quotes for clarity):
- '<' if *X*<=<<=*Y* - '>' if *X*<=><=*Y* - '=' if *X*<==<=*Y*
Demo Input:
['6 2\n1 0 1 1 1 1\n2 10\n4 7\n', '3 3\n1 0 2\n2 5\n2 4\n', '7 16\n15 15 4 0 0 7 10\n7 9\n4 8 0 3 1 5 0\n']
Demo Output:
['=\n', '<\n', '>\n']
Note:
In the first sample, *X* = 101111<sub class="lower-index">2</sub> = 47<sub class="lower-index">10</sub> = *Y*.
In the second sample, *X* = 102<sub class="lower-index">3</sub> = 21<sub class="lower-index">5</sub> and *Y* = 24<sub class="lower-index">5</sub> = 112<sub class="lower-index">3</sub>, thus *X* < *Y*.
In the third sample, <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/603a342b0ae3e56fed542d1c50c0a5ff6ce2cbaa.png" style="max-width: 100.0%;max-height: 100.0%;"/> and *Y* = 4803150<sub class="lower-index">9</sub>. We may notice that *X* starts with much larger digits and *b*<sub class="lower-index">*x*</sub> is much larger than *b*<sub class="lower-index">*y*</sub>, so *X* is clearly larger than *Y*. | ```python
ans=[]
for i in range(2):
(n,b),data=map(int,input().split()),list(map(int, input().split()))
data.reverse();
t=0
j=0
for i in range(n):
t+=data[i]*(b**j)
j+=1
ans.append(t)
# print(ans[0],"...",ans[1])
if ans[0]>ans[1]:
print(">")
elif ans[0]<ans[1]:
print("<")
else:
print("=")
``` | 3 | |
350 | C | Bombs | PROGRAMMING | 1,600 | [
"greedy",
"implementation",
"sortings"
] | null | null | You've got a robot, its task is destroying bombs on a square plane. Specifically, the square plane contains *n* bombs, the *i*-th bomb is at point with coordinates (*x**i*,<=*y**i*). We know that no two bombs are at the same point and that no bomb is at point with coordinates (0,<=0). Initially, the robot is at point with coordinates (0,<=0). Also, let's mark the robot's current position as (*x*,<=*y*). In order to destroy all the bombs, the robot can perform three types of operations:
1. Operation has format "1 k dir". To perform the operation robot have to move in direction *dir* *k* (*k*<=≥<=1) times. There are only 4 directions the robot can move in: "R", "L", "U", "D". During one move the robot can move from the current point to one of following points: (*x*<=+<=1,<=*y*), (*x*<=-<=1,<=*y*), (*x*,<=*y*<=+<=1), (*x*,<=*y*<=-<=1) (corresponding to directions). It is forbidden to move from point (*x*,<=*y*), if at least one point on the path (besides the destination point) contains a bomb. 1. Operation has format "2". To perform the operation robot have to pick a bomb at point (*x*,<=*y*) and put it in a special container. Thus, the robot can carry the bomb from any point to any other point. The operation cannot be performed if point (*x*,<=*y*) has no bomb. It is forbidden to pick a bomb if the robot already has a bomb in its container. 1. Operation has format "3". To perform the operation robot have to take a bomb out of the container and destroy it. You are allowed to perform this operation only if the robot is at point (0,<=0). It is forbidden to perform the operation if the container has no bomb.
Help the robot and find the shortest possible sequence of operations he can perform to destroy all bombs on the coordinate plane. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of bombs on the coordinate plane. Next *n* lines contain two integers each. The *i*-th line contains numbers (*x**i*,<=*y**i*) (<=-<=109<=≤<=*x**i*,<=*y**i*<=≤<=109) — the coordinates of the *i*-th bomb. It is guaranteed that no two bombs are located at the same point and no bomb is at point (0,<=0). | In a single line print a single integer *k* — the minimum number of operations needed to destroy all bombs. On the next lines print the descriptions of these *k* operations. If there are multiple sequences, you can print any of them. It is guaranteed that there is the solution where *k*<=≤<=106. | [
"2\n1 1\n-1 -1\n",
"3\n5 0\n0 5\n1 0\n"
] | [
"12\n1 1 R\n1 1 U\n2\n1 1 L\n1 1 D\n3\n1 1 L\n1 1 D\n2\n1 1 R\n1 1 U\n3\n",
"12\n1 1 R\n2\n1 1 L\n3\n1 5 R\n2\n1 5 L\n3\n1 5 U\n2\n1 5 D\n3\n"
] | none | 1,000 | [
{
"input": "2\n1 1\n-1 -1",
"output": "12\n1 1 R\n1 1 U\n2\n1 1 L\n1 1 D\n3\n1 1 L\n1 1 D\n2\n1 1 R\n1 1 U\n3"
},
{
"input": "3\n5 0\n0 5\n1 0",
"output": "12\n1 1 R\n2\n1 1 L\n3\n1 5 R\n2\n1 5 L\n3\n1 5 U\n2\n1 5 D\n3"
},
{
"input": "1\n-277226476 314722425",
"output": "6\n1 2772264... | 1,380,647,420 | 6,020 | Python 3 | TIME_LIMIT_EXCEEDED | PRETESTS | 2 | 2,000 | 3,891,200 | from sys import stdin
def main():
global start, end
bombs, ans = [], 0
for x in range(int(input())):
a, b = map(int, input().split())
ans += 2 * (3 - (0 in (a, b)))
bombs.append((a, b))
bombs.sort(key = lambda a: (a[0] < 0, a[1] < 0, abs(a[0]), abs(a[1])))
print(ans)
for x, y in bombs:
if x > 0:
print(1, x, 'R')
elif x < 0:
print(1, -x, 'L')
if y > 0:
print(1, y, 'U')
elif y < 0:
print(1, -y, 'D')
print(2)
if y > 0:
print(1, y, 'D')
elif y < 0:
print(1, -y, 'U')
if x > 0:
print(1, x, 'L')
elif x < 0:
print(1, -x, 'R')
print(3)
main()
| Title: Bombs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a robot, its task is destroying bombs on a square plane. Specifically, the square plane contains *n* bombs, the *i*-th bomb is at point with coordinates (*x**i*,<=*y**i*). We know that no two bombs are at the same point and that no bomb is at point with coordinates (0,<=0). Initially, the robot is at point with coordinates (0,<=0). Also, let's mark the robot's current position as (*x*,<=*y*). In order to destroy all the bombs, the robot can perform three types of operations:
1. Operation has format "1 k dir". To perform the operation robot have to move in direction *dir* *k* (*k*<=≥<=1) times. There are only 4 directions the robot can move in: "R", "L", "U", "D". During one move the robot can move from the current point to one of following points: (*x*<=+<=1,<=*y*), (*x*<=-<=1,<=*y*), (*x*,<=*y*<=+<=1), (*x*,<=*y*<=-<=1) (corresponding to directions). It is forbidden to move from point (*x*,<=*y*), if at least one point on the path (besides the destination point) contains a bomb. 1. Operation has format "2". To perform the operation robot have to pick a bomb at point (*x*,<=*y*) and put it in a special container. Thus, the robot can carry the bomb from any point to any other point. The operation cannot be performed if point (*x*,<=*y*) has no bomb. It is forbidden to pick a bomb if the robot already has a bomb in its container. 1. Operation has format "3". To perform the operation robot have to take a bomb out of the container and destroy it. You are allowed to perform this operation only if the robot is at point (0,<=0). It is forbidden to perform the operation if the container has no bomb.
Help the robot and find the shortest possible sequence of operations he can perform to destroy all bombs on the coordinate plane.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of bombs on the coordinate plane. Next *n* lines contain two integers each. The *i*-th line contains numbers (*x**i*,<=*y**i*) (<=-<=109<=≤<=*x**i*,<=*y**i*<=≤<=109) — the coordinates of the *i*-th bomb. It is guaranteed that no two bombs are located at the same point and no bomb is at point (0,<=0).
Output Specification:
In a single line print a single integer *k* — the minimum number of operations needed to destroy all bombs. On the next lines print the descriptions of these *k* operations. If there are multiple sequences, you can print any of them. It is guaranteed that there is the solution where *k*<=≤<=106.
Demo Input:
['2\n1 1\n-1 -1\n', '3\n5 0\n0 5\n1 0\n']
Demo Output:
['12\n1 1 R\n1 1 U\n2\n1 1 L\n1 1 D\n3\n1 1 L\n1 1 D\n2\n1 1 R\n1 1 U\n3\n', '12\n1 1 R\n2\n1 1 L\n3\n1 5 R\n2\n1 5 L\n3\n1 5 U\n2\n1 5 D\n3\n']
Note:
none | ```python
from sys import stdin
def main():
global start, end
bombs, ans = [], 0
for x in range(int(input())):
a, b = map(int, input().split())
ans += 2 * (3 - (0 in (a, b)))
bombs.append((a, b))
bombs.sort(key = lambda a: (a[0] < 0, a[1] < 0, abs(a[0]), abs(a[1])))
print(ans)
for x, y in bombs:
if x > 0:
print(1, x, 'R')
elif x < 0:
print(1, -x, 'L')
if y > 0:
print(1, y, 'U')
elif y < 0:
print(1, -y, 'D')
print(2)
if y > 0:
print(1, y, 'D')
elif y < 0:
print(1, -y, 'U')
if x > 0:
print(1, x, 'L')
elif x < 0:
print(1, -x, 'R')
print(3)
main()
``` | 0 | |
580 | C | Kefa and Park | PROGRAMMING | 1,500 | [
"dfs and similar",
"graphs",
"trees"
] | null | null | Kefa decided to celebrate his first big salary by going to the restaurant.
He lives by an unusual park. The park is a rooted tree consisting of *n* vertices with the root at vertex 1. Vertex 1 also contains Kefa's house. Unfortunaely for our hero, the park also contains cats. Kefa has already found out what are the vertices with cats in them.
The leaf vertices of the park contain restaurants. Kefa wants to choose a restaurant where he will go, but unfortunately he is very afraid of cats, so there is no way he will go to the restaurant if the path from the restaurant to his house contains more than *m* consecutive vertices with cats.
Your task is to help Kefa count the number of restaurants where he can go. | The first line contains two integers, *n* and *m* (2<=≤<=*n*<=≤<=105, 1<=≤<=*m*<=≤<=*n*) — the number of vertices of the tree and the maximum number of consecutive vertices with cats that is still ok for Kefa.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where each *a**i* either equals to 0 (then vertex *i* has no cat), or equals to 1 (then vertex *i* has a cat).
Next *n*<=-<=1 lines contains the edges of the tree in the format "*x**i* *y**i*" (without the quotes) (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*), where *x**i* and *y**i* are the vertices of the tree, connected by an edge.
It is guaranteed that the given set of edges specifies a tree. | A single integer — the number of distinct leaves of a tree the path to which from Kefa's home contains at most *m* consecutive vertices with cats. | [
"4 1\n1 1 0 0\n1 2\n1 3\n1 4\n",
"7 1\n1 0 1 1 0 0 0\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n"
] | [
"2\n",
"2\n"
] | Let us remind you that a tree is a connected graph on *n* vertices and *n* - 1 edge. A rooted tree is a tree with a special vertex called root. In a rooted tree among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child. A vertex is called a leaf, if it has no children.
Note to the first sample test: <img class="tex-graphics" src="https://espresso.codeforces.com/785114b4b3f5336f02078c25750f87c5a1d0b4be.png" style="max-width: 100.0%;max-height: 100.0%;"/> The vertices containing cats are marked red. The restaurants are at vertices 2, 3, 4. Kefa can't go only to the restaurant located at vertex 2.
Note to the second sample test: <img class="tex-graphics" src="https://espresso.codeforces.com/e5c07640680c837aec99126d94287872e69aa09a.png" style="max-width: 100.0%;max-height: 100.0%;"/> The restaurants are located at vertices 4, 5, 6, 7. Kefa can't go to restaurants 6, 7. | 1,500 | [
{
"input": "4 1\n1 1 0 0\n1 2\n1 3\n1 4",
"output": "2"
},
{
"input": "7 1\n1 0 1 1 0 0 0\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7",
"output": "2"
},
{
"input": "3 2\n1 1 1\n1 2\n2 3",
"output": "0"
},
{
"input": "5 2\n1 1 0 1 1\n1 2\n2 3\n3 4\n4 5",
"output": "1"
},
{
"inpu... | 1,686,060,412 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 40 | 904 | 32,460,800 | from collections import deque
count = 0
graph = dict()
n, m = map(int, input().split())
cats = list(map(int, input().split()))
available = [True] * (n+1)
for i in range(1, n+1):
graph[i] = []
for link in range(n-1):
r, l = map(int, input().split())
graph[l].append(r)
graph[r].append(l)
queue = deque()
queue += [[1, cats[0]]]
while queue:
cur, cat = queue.popleft()
available[cur] = False
if cat <= m:
if len(graph[cur]) == 1 and cur!= 1:
count += 1
for i in graph[cur]:
if available[i]:
if cats[i-1] == 0:
queue += [[i, 0]]
elif cats[i-1] == 1:
queue += [[i, cat+1]]
print(count) | Title: Kefa and Park
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kefa decided to celebrate his first big salary by going to the restaurant.
He lives by an unusual park. The park is a rooted tree consisting of *n* vertices with the root at vertex 1. Vertex 1 also contains Kefa's house. Unfortunaely for our hero, the park also contains cats. Kefa has already found out what are the vertices with cats in them.
The leaf vertices of the park contain restaurants. Kefa wants to choose a restaurant where he will go, but unfortunately he is very afraid of cats, so there is no way he will go to the restaurant if the path from the restaurant to his house contains more than *m* consecutive vertices with cats.
Your task is to help Kefa count the number of restaurants where he can go.
Input Specification:
The first line contains two integers, *n* and *m* (2<=≤<=*n*<=≤<=105, 1<=≤<=*m*<=≤<=*n*) — the number of vertices of the tree and the maximum number of consecutive vertices with cats that is still ok for Kefa.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where each *a**i* either equals to 0 (then vertex *i* has no cat), or equals to 1 (then vertex *i* has a cat).
Next *n*<=-<=1 lines contains the edges of the tree in the format "*x**i* *y**i*" (without the quotes) (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*, *x**i*<=≠<=*y**i*), where *x**i* and *y**i* are the vertices of the tree, connected by an edge.
It is guaranteed that the given set of edges specifies a tree.
Output Specification:
A single integer — the number of distinct leaves of a tree the path to which from Kefa's home contains at most *m* consecutive vertices with cats.
Demo Input:
['4 1\n1 1 0 0\n1 2\n1 3\n1 4\n', '7 1\n1 0 1 1 0 0 0\n1 2\n1 3\n2 4\n2 5\n3 6\n3 7\n']
Demo Output:
['2\n', '2\n']
Note:
Let us remind you that a tree is a connected graph on *n* vertices and *n* - 1 edge. A rooted tree is a tree with a special vertex called root. In a rooted tree among any two vertices connected by an edge, one vertex is a parent (the one closer to the root), and the other one is a child. A vertex is called a leaf, if it has no children.
Note to the first sample test: <img class="tex-graphics" src="https://espresso.codeforces.com/785114b4b3f5336f02078c25750f87c5a1d0b4be.png" style="max-width: 100.0%;max-height: 100.0%;"/> The vertices containing cats are marked red. The restaurants are at vertices 2, 3, 4. Kefa can't go only to the restaurant located at vertex 2.
Note to the second sample test: <img class="tex-graphics" src="https://espresso.codeforces.com/e5c07640680c837aec99126d94287872e69aa09a.png" style="max-width: 100.0%;max-height: 100.0%;"/> The restaurants are located at vertices 4, 5, 6, 7. Kefa can't go to restaurants 6, 7. | ```python
from collections import deque
count = 0
graph = dict()
n, m = map(int, input().split())
cats = list(map(int, input().split()))
available = [True] * (n+1)
for i in range(1, n+1):
graph[i] = []
for link in range(n-1):
r, l = map(int, input().split())
graph[l].append(r)
graph[r].append(l)
queue = deque()
queue += [[1, cats[0]]]
while queue:
cur, cat = queue.popleft()
available[cur] = False
if cat <= m:
if len(graph[cur]) == 1 and cur!= 1:
count += 1
for i in graph[cur]:
if available[i]:
if cats[i-1] == 0:
queue += [[i, 0]]
elif cats[i-1] == 1:
queue += [[i, cat+1]]
print(count)
``` | 3 | |
352 | A | Jeff and Digits | PROGRAMMING | 1,000 | [
"brute force",
"implementation",
"math"
] | null | null | Jeff's got *n* cards, each card contains either digit 0, or digit 5. Jeff can choose several cards and put them in a line so that he gets some number. What is the largest possible number divisible by 90 Jeff can make from the cards he's got?
Jeff must make the number without leading zero. At that, we assume that number 0 doesn't contain any leading zeroes. Jeff doesn't have to use all the cards. | The first line contains integer *n* (1<=≤<=*n*<=≤<=103). The next line contains *n* integers *a*1, *a*2, ..., *a**n* (*a**i*<==<=0 or *a**i*<==<=5). Number *a**i* represents the digit that is written on the *i*-th card. | In a single line print the answer to the problem — the maximum number, divisible by 90. If you can't make any divisible by 90 number from the cards, print -1. | [
"4\n5 0 5 0\n",
"11\n5 5 5 5 5 5 5 5 0 5 5\n"
] | [
"0\n",
"5555555550\n"
] | In the first test you can make only one number that is a multiple of 90 — 0.
In the second test you can make number 5555555550, it is a multiple of 90. | 500 | [
{
"input": "4\n5 0 5 0",
"output": "0"
},
{
"input": "11\n5 5 5 5 5 5 5 5 0 5 5",
"output": "5555555550"
},
{
"input": "7\n5 5 5 5 5 5 5",
"output": "-1"
},
{
"input": "1\n5",
"output": "-1"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "11\n5 0 5 ... | 1,699,102,650 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 7 | 124 | 0 | n = int(input())
lst = [int(el) for el in input().split()]
cnt_5 = lst.count(5)
cnt_0 = lst.count(0)
if cnt_5 // 9 < 1 and cnt_0 == 0:
print(-1)
elif cnt_5 // 9 < 1 and cnt_0 != 0:
print(0)
else:
print("5" * (cnt_5 // 9 * 9) + "0" * cnt_0) | Title: Jeff and Digits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Jeff's got *n* cards, each card contains either digit 0, or digit 5. Jeff can choose several cards and put them in a line so that he gets some number. What is the largest possible number divisible by 90 Jeff can make from the cards he's got?
Jeff must make the number without leading zero. At that, we assume that number 0 doesn't contain any leading zeroes. Jeff doesn't have to use all the cards.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=103). The next line contains *n* integers *a*1, *a*2, ..., *a**n* (*a**i*<==<=0 or *a**i*<==<=5). Number *a**i* represents the digit that is written on the *i*-th card.
Output Specification:
In a single line print the answer to the problem — the maximum number, divisible by 90. If you can't make any divisible by 90 number from the cards, print -1.
Demo Input:
['4\n5 0 5 0\n', '11\n5 5 5 5 5 5 5 5 0 5 5\n']
Demo Output:
['0\n', '5555555550\n']
Note:
In the first test you can make only one number that is a multiple of 90 — 0.
In the second test you can make number 5555555550, it is a multiple of 90. | ```python
n = int(input())
lst = [int(el) for el in input().split()]
cnt_5 = lst.count(5)
cnt_0 = lst.count(0)
if cnt_5 // 9 < 1 and cnt_0 == 0:
print(-1)
elif cnt_5 // 9 < 1 and cnt_0 != 0:
print(0)
else:
print("5" * (cnt_5 // 9 * 9) + "0" * cnt_0)
``` | 0 | |
975 | A | Aramic script | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | In Aramic language words can only represent objects.
Words in Aramic have special properties:
- A word is a root if it does not contain the same letter more than once. - A root and all its permutations represent the same object. - The root $x$ of a word $y$ is the word that contains all letters that appear in $y$ in a way that each letter appears once. For example, the root of "aaaa", "aa", "aaa" is "a", the root of "aabb", "bab", "baabb", "ab" is "ab". - Any word in Aramic represents the same object as its root.
You have an ancient script in Aramic. What is the number of different objects mentioned in the script? | The first line contains one integer $n$ ($1 \leq n \leq 10^3$) — the number of words in the script.
The second line contains $n$ words $s_1, s_2, \ldots, s_n$ — the script itself. The length of each string does not exceed $10^3$.
It is guaranteed that all characters of the strings are small latin letters. | Output one integer — the number of different objects mentioned in the given ancient Aramic script. | [
"5\na aa aaa ab abb\n",
"3\namer arem mrea\n"
] | [
"2",
"1"
] | In the first test, there are two objects mentioned. The roots that represent them are "a","ab".
In the second test, there is only one object, its root is "amer", the other strings are just permutations of "amer". | 500 | [
{
"input": "5\na aa aaa ab abb",
"output": "2"
},
{
"input": "3\namer arem mrea",
"output": "1"
},
{
"input": "10\nbda bbb cda dca dda dcb bcd dcb ada ddd",
"output": "6"
},
{
"input": "2\nfhjlqs aceginpr",
"output": "2"
},
{
"input": "2\nbcdfghimn efghijlmo",
... | 1,525,975,011 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 124 | 10,137,600 | n = int(input())
w = input().split()
w2 = []
for i in range(len(w)):
w[i] = set(w[i])
i = 0
j = 0
for item in w:
w2.append(''.join(sorted(list(item))))
print(len(set(w2)))
| Title: Aramic script
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Aramic language words can only represent objects.
Words in Aramic have special properties:
- A word is a root if it does not contain the same letter more than once. - A root and all its permutations represent the same object. - The root $x$ of a word $y$ is the word that contains all letters that appear in $y$ in a way that each letter appears once. For example, the root of "aaaa", "aa", "aaa" is "a", the root of "aabb", "bab", "baabb", "ab" is "ab". - Any word in Aramic represents the same object as its root.
You have an ancient script in Aramic. What is the number of different objects mentioned in the script?
Input Specification:
The first line contains one integer $n$ ($1 \leq n \leq 10^3$) — the number of words in the script.
The second line contains $n$ words $s_1, s_2, \ldots, s_n$ — the script itself. The length of each string does not exceed $10^3$.
It is guaranteed that all characters of the strings are small latin letters.
Output Specification:
Output one integer — the number of different objects mentioned in the given ancient Aramic script.
Demo Input:
['5\na aa aaa ab abb\n', '3\namer arem mrea\n']
Demo Output:
['2', '1']
Note:
In the first test, there are two objects mentioned. The roots that represent them are "a","ab".
In the second test, there is only one object, its root is "amer", the other strings are just permutations of "amer". | ```python
n = int(input())
w = input().split()
w2 = []
for i in range(len(w)):
w[i] = set(w[i])
i = 0
j = 0
for item in w:
w2.append(''.join(sorted(list(item))))
print(len(set(w2)))
``` | 3 | |
298 | A | Snow Footprints | PROGRAMMING | 1,300 | [
"greedy",
"implementation"
] | null | null | There is a straight snowy road, divided into *n* blocks. The blocks are numbered from 1 to *n* from left to right. If one moves from the *i*-th block to the (*i*<=+<=1)-th block, he will leave a right footprint on the *i*-th block. Similarly, if one moves from the *i*-th block to the (*i*<=-<=1)-th block, he will leave a left footprint on the *i*-th block. If there already is a footprint on the *i*-th block, the new footprint will cover the old one.
At the beginning, there were no footprints. Then polar bear Alice starts from the *s*-th block, makes a sequence of moves and ends in the *t*-th block. It is known that Alice never moves outside of the road.
You are given the description of Alice's footprints. Your task is to find a pair of possible values of *s*,<=*t* by looking at the footprints. | The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=1000).
The second line contains the description of the road — the string that consists of *n* characters. Each character will be either "." (a block without footprint), or "L" (a block with a left footprint), "R" (a block with a right footprint).
It's guaranteed that the given string contains at least one character not equal to ".". Also, the first and the last character will always be ".". It's guaranteed that a solution exists. | Print two space-separated integers — the values of *s* and *t*. If there are several possible solutions you can print any of them. | [
"9\n..RRLL...\n",
"11\n.RRRLLLLL..\n"
] | [
"3 4\n",
"7 5\n"
] | The first test sample is the one in the picture. | 500 | [
{
"input": "11\n.RRRLLLLL..",
"output": "7 5"
},
{
"input": "4\n.RL.",
"output": "3 2"
},
{
"input": "3\n.L.",
"output": "2 1"
},
{
"input": "3\n.R.",
"output": "2 3"
}
] | 1,612,168,181 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <iostream>
#include <string>
using namespace std;
int main()
{
size_t n;
string s;
cin >> n >> s;
for (size_t i = 0; i < n; ++i)
{
if (s[i] != '.')
{
if (s[i] == 'R')
{
while (s[i] == 'R')
{
++i;
}
cout << i << " " << i + 1 << endl;
}
else
{
cout << i + 1 << " " << i << endl;
}
break;
}
}
return 0;
}
| Title: Snow Footprints
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There is a straight snowy road, divided into *n* blocks. The blocks are numbered from 1 to *n* from left to right. If one moves from the *i*-th block to the (*i*<=+<=1)-th block, he will leave a right footprint on the *i*-th block. Similarly, if one moves from the *i*-th block to the (*i*<=-<=1)-th block, he will leave a left footprint on the *i*-th block. If there already is a footprint on the *i*-th block, the new footprint will cover the old one.
At the beginning, there were no footprints. Then polar bear Alice starts from the *s*-th block, makes a sequence of moves and ends in the *t*-th block. It is known that Alice never moves outside of the road.
You are given the description of Alice's footprints. Your task is to find a pair of possible values of *s*,<=*t* by looking at the footprints.
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=1000).
The second line contains the description of the road — the string that consists of *n* characters. Each character will be either "." (a block without footprint), or "L" (a block with a left footprint), "R" (a block with a right footprint).
It's guaranteed that the given string contains at least one character not equal to ".". Also, the first and the last character will always be ".". It's guaranteed that a solution exists.
Output Specification:
Print two space-separated integers — the values of *s* and *t*. If there are several possible solutions you can print any of them.
Demo Input:
['9\n..RRLL...\n', '11\n.RRRLLLLL..\n']
Demo Output:
['3 4\n', '7 5\n']
Note:
The first test sample is the one in the picture. | ```python
#include <iostream>
#include <string>
using namespace std;
int main()
{
size_t n;
string s;
cin >> n >> s;
for (size_t i = 0; i < n; ++i)
{
if (s[i] != '.')
{
if (s[i] == 'R')
{
while (s[i] == 'R')
{
++i;
}
cout << i << " " << i + 1 << endl;
}
else
{
cout << i + 1 << " " << i << endl;
}
break;
}
}
return 0;
}
``` | -1 | |
16 | B | Burglar and Matches | PROGRAMMING | 900 | [
"greedy",
"implementation",
"sortings"
] | B. Burglar and Matches | 0 | 64 | A burglar got into a matches warehouse and wants to steal as many matches as possible. In the warehouse there are *m* containers, in the *i*-th container there are *a**i* matchboxes, and each matchbox contains *b**i* matches. All the matchboxes are of the same size. The burglar's rucksack can hold *n* matchboxes exactly. Your task is to find out the maximum amount of matches that a burglar can carry away. He has no time to rearrange matches in the matchboxes, that's why he just chooses not more than *n* matchboxes so that the total amount of matches in them is maximal. | The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=2·108) and integer *m* (1<=≤<=*m*<=≤<=20). The *i*<=+<=1-th line contains a pair of numbers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=108,<=1<=≤<=*b**i*<=≤<=10). All the input numbers are integer. | Output the only number — answer to the problem. | [
"7 3\n5 10\n2 5\n3 6\n",
"3 3\n1 3\n2 2\n3 1\n"
] | [
"62\n",
"7\n"
] | none | 0 | [
{
"input": "7 3\n5 10\n2 5\n3 6",
"output": "62"
},
{
"input": "3 3\n1 3\n2 2\n3 1",
"output": "7"
},
{
"input": "1 1\n1 2",
"output": "2"
},
{
"input": "1 2\n1 9\n1 6",
"output": "9"
},
{
"input": "1 10\n1 1\n1 9\n1 3\n1 9\n1 7\n1 10\n1 4\n1 7\n1 3\n1 1",
"ou... | 1,677,516,557 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 60 | 0 | import math
x=input().split(" ")
x=list(map(lambda e:int(e),x))
matches=0
a=[]
for _ in range(x[1]):
z=input().split(" ")
z=list(map(lambda e:int(e),z))
a.append(z)
a=sorted(a,reverse = True,key=lambda x:x[1])
print(a)
i=0
while x[0]>0:
if a[i][0] <=x[0]:
x[0]=x[0]-a[i][0]
matches+=a[i][0]*a[i][1]
i+=1
else:
matches+=x[0]*a[i][1]
x[0]=0
print(matches)
| Title: Burglar and Matches
Time Limit: 0 seconds
Memory Limit: 64 megabytes
Problem Description:
A burglar got into a matches warehouse and wants to steal as many matches as possible. In the warehouse there are *m* containers, in the *i*-th container there are *a**i* matchboxes, and each matchbox contains *b**i* matches. All the matchboxes are of the same size. The burglar's rucksack can hold *n* matchboxes exactly. Your task is to find out the maximum amount of matches that a burglar can carry away. He has no time to rearrange matches in the matchboxes, that's why he just chooses not more than *n* matchboxes so that the total amount of matches in them is maximal.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=2·108) and integer *m* (1<=≤<=*m*<=≤<=20). The *i*<=+<=1-th line contains a pair of numbers *a**i* and *b**i* (1<=≤<=*a**i*<=≤<=108,<=1<=≤<=*b**i*<=≤<=10). All the input numbers are integer.
Output Specification:
Output the only number — answer to the problem.
Demo Input:
['7 3\n5 10\n2 5\n3 6\n', '3 3\n1 3\n2 2\n3 1\n']
Demo Output:
['62\n', '7\n']
Note:
none | ```python
import math
x=input().split(" ")
x=list(map(lambda e:int(e),x))
matches=0
a=[]
for _ in range(x[1]):
z=input().split(" ")
z=list(map(lambda e:int(e),z))
a.append(z)
a=sorted(a,reverse = True,key=lambda x:x[1])
print(a)
i=0
while x[0]>0:
if a[i][0] <=x[0]:
x[0]=x[0]-a[i][0]
matches+=a[i][0]*a[i][1]
i+=1
else:
matches+=x[0]*a[i][1]
x[0]=0
print(matches)
``` | 0 |
35 | A | Shell Game | PROGRAMMING | 1,000 | [
"implementation"
] | A. Shell Game | 2 | 64 | Today the «Z» city residents enjoy a shell game competition. The residents are gathered on the main square to watch the breath-taking performance. The performer puts 3 non-transparent cups upside down in a row. Then he openly puts a small ball under one of the cups and starts to shuffle the cups around very quickly so that on the whole he makes exactly 3 shuffles. After that the spectators have exactly one attempt to guess in which cup they think the ball is and if the answer is correct they get a prize. Maybe you can try to find the ball too? | The first input line contains an integer from 1 to 3 — index of the cup which covers the ball before the shuffles. The following three lines describe the shuffles. Each description of a shuffle contains two distinct integers from 1 to 3 — indexes of the cups which the performer shuffled this time. The cups are numbered from left to right and are renumbered after each shuffle from left to right again. In other words, the cup on the left always has index 1, the one in the middle — index 2 and the one on the right — index 3. | In the first line output an integer from 1 to 3 — index of the cup which will have the ball after all the shuffles. | [
"1\n1 2\n2 1\n2 1\n",
"1\n2 1\n3 1\n1 3\n"
] | [
"2\n",
"2\n"
] | none | 500 | [
{
"input": "1\n1 2\n2 1\n2 1",
"output": "2"
},
{
"input": "1\n2 1\n3 1\n1 3",
"output": "2"
},
{
"input": "3\n3 1\n2 1\n1 2",
"output": "1"
},
{
"input": "1\n1 3\n1 2\n2 3",
"output": "2"
},
{
"input": "3\n3 2\n3 1\n3 1",
"output": "2"
},
{
"input": "... | 1,579,621,754 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 186 | 0 | def shell_game(start,a):
pos=start
for i in range(3):
arr=a[i]
if pos in arr:
if arr[0]==pos:
pos=arr[1]
else:
pos=arr[0]
print(pos)
start=int(input(''))
b=[]
for i in range(3):
a=list(map(int,input('').split()))
b.append(a)
shell_game(start,b) | Title: Shell Game
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Today the «Z» city residents enjoy a shell game competition. The residents are gathered on the main square to watch the breath-taking performance. The performer puts 3 non-transparent cups upside down in a row. Then he openly puts a small ball under one of the cups and starts to shuffle the cups around very quickly so that on the whole he makes exactly 3 shuffles. After that the spectators have exactly one attempt to guess in which cup they think the ball is and if the answer is correct they get a prize. Maybe you can try to find the ball too?
Input Specification:
The first input line contains an integer from 1 to 3 — index of the cup which covers the ball before the shuffles. The following three lines describe the shuffles. Each description of a shuffle contains two distinct integers from 1 to 3 — indexes of the cups which the performer shuffled this time. The cups are numbered from left to right and are renumbered after each shuffle from left to right again. In other words, the cup on the left always has index 1, the one in the middle — index 2 and the one on the right — index 3.
Output Specification:
In the first line output an integer from 1 to 3 — index of the cup which will have the ball after all the shuffles.
Demo Input:
['1\n1 2\n2 1\n2 1\n', '1\n2 1\n3 1\n1 3\n']
Demo Output:
['2\n', '2\n']
Note:
none | ```python
def shell_game(start,a):
pos=start
for i in range(3):
arr=a[i]
if pos in arr:
if arr[0]==pos:
pos=arr[1]
else:
pos=arr[0]
print(pos)
start=int(input(''))
b=[]
for i in range(3):
a=list(map(int,input('').split()))
b.append(a)
shell_game(start,b)
``` | -1 |
760 | B | Frodo and pillows | PROGRAMMING | 1,500 | [
"binary search",
"greedy"
] | null | null | *n* hobbits are planning to spend the night at Frodo's house. Frodo has *n* beds standing in a row and *m* pillows (*n*<=≤<=*m*). Each hobbit needs a bed and at least one pillow to sleep, however, everyone wants as many pillows as possible. Of course, it's not always possible to share pillows equally, but any hobbit gets hurt if he has at least two pillows less than some of his neighbors have.
Frodo will sleep on the *k*-th bed in the row. What is the maximum number of pillows he can have so that every hobbit has at least one pillow, every pillow is given to some hobbit and no one is hurt? | The only line contain three integers *n*, *m* and *k* (1<=≤<=*n*<=≤<=*m*<=≤<=109, 1<=≤<=*k*<=≤<=*n*) — the number of hobbits, the number of pillows and the number of Frodo's bed. | Print single integer — the maximum number of pillows Frodo can have so that no one is hurt. | [
"4 6 2\n",
"3 10 3\n",
"3 6 1\n"
] | [
"2\n",
"4\n",
"3\n"
] | In the first example Frodo can have at most two pillows. In this case, he can give two pillows to the hobbit on the first bed, and one pillow to each of the hobbits on the third and the fourth beds.
In the second example Frodo can take at most four pillows, giving three pillows to each of the others.
In the third example Frodo can take three pillows, giving two pillows to the hobbit in the middle and one pillow to the hobbit on the third bed. | 1,000 | [
{
"input": "4 6 2",
"output": "2"
},
{
"input": "3 10 3",
"output": "4"
},
{
"input": "3 6 1",
"output": "3"
},
{
"input": "3 3 3",
"output": "1"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "1 1000000000 1",
"output": "1000000000"
},
{
... | 1,675,596,931 | 2,147,483,647 | PyPy 3 | OK | TESTS | 69 | 108 | 0 | import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def binary_search(c1, c2):
m = (c1 + c2 + 1) // 2
while abs(c1 - c2) > 1:
m = (c1 + c2 + 1) // 2
if ok(m):
c1 = m
else:
c2 = m
m += 1
while not ok(m):
m -= 1
return m
def ok(u):
if u <= 1:
return True
x = m - n - u + 1
if x < 0:
return False
x -= (u - 1) * (u - 2)
v = max(u - l - 2, 0)
x += v * (v + 1) // 2
v = max(u - r - 2, 0)
x += v * (v + 1) // 2
return True if x >= 0 else False
n, m, k = map(int, input().split())
l, r = k - 1, n - k
ans = binary_search(0, m)
print(ans) | Title: Frodo and pillows
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
*n* hobbits are planning to spend the night at Frodo's house. Frodo has *n* beds standing in a row and *m* pillows (*n*<=≤<=*m*). Each hobbit needs a bed and at least one pillow to sleep, however, everyone wants as many pillows as possible. Of course, it's not always possible to share pillows equally, but any hobbit gets hurt if he has at least two pillows less than some of his neighbors have.
Frodo will sleep on the *k*-th bed in the row. What is the maximum number of pillows he can have so that every hobbit has at least one pillow, every pillow is given to some hobbit and no one is hurt?
Input Specification:
The only line contain three integers *n*, *m* and *k* (1<=≤<=*n*<=≤<=*m*<=≤<=109, 1<=≤<=*k*<=≤<=*n*) — the number of hobbits, the number of pillows and the number of Frodo's bed.
Output Specification:
Print single integer — the maximum number of pillows Frodo can have so that no one is hurt.
Demo Input:
['4 6 2\n', '3 10 3\n', '3 6 1\n']
Demo Output:
['2\n', '4\n', '3\n']
Note:
In the first example Frodo can have at most two pillows. In this case, he can give two pillows to the hobbit on the first bed, and one pillow to each of the hobbits on the third and the fourth beds.
In the second example Frodo can take at most four pillows, giving three pillows to each of the others.
In the third example Frodo can take three pillows, giving two pillows to the hobbit in the middle and one pillow to the hobbit on the third bed. | ```python
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def binary_search(c1, c2):
m = (c1 + c2 + 1) // 2
while abs(c1 - c2) > 1:
m = (c1 + c2 + 1) // 2
if ok(m):
c1 = m
else:
c2 = m
m += 1
while not ok(m):
m -= 1
return m
def ok(u):
if u <= 1:
return True
x = m - n - u + 1
if x < 0:
return False
x -= (u - 1) * (u - 2)
v = max(u - l - 2, 0)
x += v * (v + 1) // 2
v = max(u - r - 2, 0)
x += v * (v + 1) // 2
return True if x >= 0 else False
n, m, k = map(int, input().split())
l, r = k - 1, n - k
ans = binary_search(0, m)
print(ans)
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Dreamoon loves summing up something for no reason. One day he obtains two integers *a* and *b* occasionally. He wants to calculate the sum of all nice integers. Positive integer *x* is called nice if and , where *k* is some integer number in range [1,<=*a*].
By we denote the quotient of integer division of *x* and *y*. By we denote the remainder of integer division of *x* and *y*. You can read more about these operations here: http://goo.gl/AcsXhT.
The answer may be large, so please print its remainder modulo 1<=000<=000<=007 (109<=+<=7). Can you compute it faster than Dreamoon? | The single line of the input contains two integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=107). | Print a single integer representing the answer modulo 1<=000<=000<=007 (109<=+<=7). | [
"1 1\n",
"2 2\n"
] | [
"0\n",
"8\n"
] | For the first sample, there are no nice integers because <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/03b1dc6bae5180f8a2d8eb85789e8b393e585970.png" style="max-width: 100.0%;max-height: 100.0%;"/> is always zero.
For the second sample, the set of nice integers is {3, 5}. | 0 | [
{
"input": "1 1",
"output": "0"
},
{
"input": "2 2",
"output": "8"
},
{
"input": "4 1",
"output": "0"
},
{
"input": "4 2",
"output": "24"
},
{
"input": "4 3",
"output": "102"
},
{
"input": "4 4",
"output": "264"
},
{
"input": "3 4",
"ou... | 1,503,405,240 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 31 | 0 | s=raw_input().split()
a,b=s[0],s[1]
x=0
for k in range(1,a+1):
x=x+k*b+1
xx=0
for t in range(1,b+1):
xx=xx+i
print((xx*x)%1000000007) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Dreamoon loves summing up something for no reason. One day he obtains two integers *a* and *b* occasionally. He wants to calculate the sum of all nice integers. Positive integer *x* is called nice if and , where *k* is some integer number in range [1,<=*a*].
By we denote the quotient of integer division of *x* and *y*. By we denote the remainder of integer division of *x* and *y*. You can read more about these operations here: http://goo.gl/AcsXhT.
The answer may be large, so please print its remainder modulo 1<=000<=000<=007 (109<=+<=7). Can you compute it faster than Dreamoon?
Input Specification:
The single line of the input contains two integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=107).
Output Specification:
Print a single integer representing the answer modulo 1<=000<=000<=007 (109<=+<=7).
Demo Input:
['1 1\n', '2 2\n']
Demo Output:
['0\n', '8\n']
Note:
For the first sample, there are no nice integers because <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/03b1dc6bae5180f8a2d8eb85789e8b393e585970.png" style="max-width: 100.0%;max-height: 100.0%;"/> is always zero.
For the second sample, the set of nice integers is {3, 5}. | ```python
s=raw_input().split()
a,b=s[0],s[1]
x=0
for k in range(1,a+1):
x=x+k*b+1
xx=0
for t in range(1,b+1):
xx=xx+i
print((xx*x)%1000000007)
``` | -1 | |
510 | A | Fox And Snake | PROGRAMMING | 800 | [
"implementation"
] | null | null | Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern. | The only line contains two integers: *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=50).
*n* is an odd number. | Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces. | [
"3 3\n",
"3 4\n",
"5 3\n",
"9 9\n"
] | [
"###\n..#\n###\n",
"####\n...#\n####\n",
"###\n..#\n###\n#..\n###\n",
"#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n"
] | none | 500 | [
{
"input": "3 3",
"output": "###\n..#\n###"
},
{
"input": "3 4",
"output": "####\n...#\n####"
},
{
"input": "5 3",
"output": "###\n..#\n###\n#..\n###"
},
{
"input": "9 9",
"output": "#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#... | 1,674,594,128 | 2,147,483,647 | Python 3 | OK | TESTS | 28 | 46 | 0 | r , c = list(map(int, input().split()))
right = True
for idx in range(r):
if idx % 2:
row = ["."]*c
if right:
row[-1] = "#"
else:
row[0] = "#"
right = not right
else:
row = ["#"]*c
print("".join(row))
| Title: Fox And Snake
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern.
Input Specification:
The only line contains two integers: *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=50).
*n* is an odd number.
Output Specification:
Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces.
Demo Input:
['3 3\n', '3 4\n', '5 3\n', '9 9\n']
Demo Output:
['###\n..#\n###\n', '####\n...#\n####\n', '###\n..#\n###\n#..\n###\n', '#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n']
Note:
none | ```python
r , c = list(map(int, input().split()))
right = True
for idx in range(r):
if idx % 2:
row = ["."]*c
if right:
row[-1] = "#"
else:
row[0] = "#"
right = not right
else:
row = ["#"]*c
print("".join(row))
``` | 3 | |
811 | A | Vladik and Courtesy | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | At regular competition Vladik and Valera won *a* and *b* candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn.
More formally, the guys take turns giving each other one candy more than they received in the previous turn.
This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy. | Single line of input data contains two space-separated integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) — number of Vladik and Valera candies respectively. | Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise. | [
"1 1\n",
"7 6\n"
] | [
"Valera\n",
"Vladik\n"
] | Illustration for first test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/ad9b7d0e481208de8e3a585aa1d96b9e1dda4fd7.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Illustration for second test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/9f4836d2ccdffaee5a63898e5d4e6caf2ed4678c.png" style="max-width: 100.0%;max-height: 100.0%;"/> | 500 | [
{
"input": "1 1",
"output": "Valera"
},
{
"input": "7 6",
"output": "Vladik"
},
{
"input": "25 38",
"output": "Vladik"
},
{
"input": "8311 2468",
"output": "Valera"
},
{
"input": "250708 857756",
"output": "Vladik"
},
{
"input": "957985574 24997558",
... | 1,496,125,181 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 46 | 0 | def solve(vladik, valera):
if vladik == valera:
return "Valera"
else:
if vladik > valera:
return "Vladik"
else:
return "Valera"
n, m = map(int, input().split())
print(solve(n, m)) | Title: Vladik and Courtesy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
At regular competition Vladik and Valera won *a* and *b* candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn.
More formally, the guys take turns giving each other one candy more than they received in the previous turn.
This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy.
Input Specification:
Single line of input data contains two space-separated integers *a*, *b* (1<=≤<=*a*,<=*b*<=≤<=109) — number of Vladik and Valera candies respectively.
Output Specification:
Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise.
Demo Input:
['1 1\n', '7 6\n']
Demo Output:
['Valera\n', 'Vladik\n']
Note:
Illustration for first test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/ad9b7d0e481208de8e3a585aa1d96b9e1dda4fd7.png" style="max-width: 100.0%;max-height: 100.0%;"/>
Illustration for second test case:
<img class="tex-graphics" src="https://espresso.codeforces.com/9f4836d2ccdffaee5a63898e5d4e6caf2ed4678c.png" style="max-width: 100.0%;max-height: 100.0%;"/> | ```python
def solve(vladik, valera):
if vladik == valera:
return "Valera"
else:
if vladik > valera:
return "Vladik"
else:
return "Valera"
n, m = map(int, input().split())
print(solve(n, m))
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | It is a balmy spring afternoon, and Farmer John's *n* cows are ruminating about link-cut cacti in their stalls. The cows, labeled 1 through *n*, are arranged so that the *i*-th cow occupies the *i*-th stall from the left. However, Elsie, after realizing that she will forever live in the shadows beyond Bessie's limelight, has formed the Mischievous Mess Makers and is plotting to disrupt this beautiful pastoral rhythm. While Farmer John takes his *k* minute long nap, Elsie and the Mess Makers plan to repeatedly choose two distinct stalls and swap the cows occupying those stalls, making no more than one swap each minute.
Being the meticulous pranksters that they are, the Mischievous Mess Makers would like to know the maximum messiness attainable in the *k* minutes that they have. We denote as *p**i* the label of the cow in the *i*-th stall. The messiness of an arrangement of cows is defined as the number of pairs (*i*,<=*j*) such that *i*<=<<=*j* and *p**i*<=><=*p**j*. | The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100<=000) — the number of cows and the length of Farmer John's nap, respectively. | Output a single integer, the maximum messiness that the Mischievous Mess Makers can achieve by performing no more than *k* swaps. | [
"5 2\n",
"1 10\n"
] | [
"10\n",
"0\n"
] | In the first sample, the Mischievous Mess Makers can swap the cows in the stalls 1 and 5 during the first minute, then the cows in stalls 2 and 4 during the second minute. This reverses the arrangement of cows, giving us a total messiness of 10.
In the second sample, there is only one cow, so the maximum possible messiness is 0. | 0 | [
{
"input": "5 2",
"output": "10"
},
{
"input": "1 10",
"output": "0"
},
{
"input": "100000 2",
"output": "399990"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "8 3",
"output": "27"
},
{
"input": "7 1",
"output": "11"
},
{
"input": "1000... | 1,465,097,750 | 2,147,483,647 | Python 3 | OK | TESTS | 92 | 62 | 5,120,000 | n, k = (int(i) for i in input().split())
if n == 1:
print(0)
exit()
if k >= int(n/2):
t = n*(n-1)/2
print(int(t))
exit()
s = n - 2 * k
p = s*(s-1)/2
l = n*(n-1)/2
output = l - p
print(int(output))
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It is a balmy spring afternoon, and Farmer John's *n* cows are ruminating about link-cut cacti in their stalls. The cows, labeled 1 through *n*, are arranged so that the *i*-th cow occupies the *i*-th stall from the left. However, Elsie, after realizing that she will forever live in the shadows beyond Bessie's limelight, has formed the Mischievous Mess Makers and is plotting to disrupt this beautiful pastoral rhythm. While Farmer John takes his *k* minute long nap, Elsie and the Mess Makers plan to repeatedly choose two distinct stalls and swap the cows occupying those stalls, making no more than one swap each minute.
Being the meticulous pranksters that they are, the Mischievous Mess Makers would like to know the maximum messiness attainable in the *k* minutes that they have. We denote as *p**i* the label of the cow in the *i*-th stall. The messiness of an arrangement of cows is defined as the number of pairs (*i*,<=*j*) such that *i*<=<<=*j* and *p**i*<=><=*p**j*.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100<=000) — the number of cows and the length of Farmer John's nap, respectively.
Output Specification:
Output a single integer, the maximum messiness that the Mischievous Mess Makers can achieve by performing no more than *k* swaps.
Demo Input:
['5 2\n', '1 10\n']
Demo Output:
['10\n', '0\n']
Note:
In the first sample, the Mischievous Mess Makers can swap the cows in the stalls 1 and 5 during the first minute, then the cows in stalls 2 and 4 during the second minute. This reverses the arrangement of cows, giving us a total messiness of 10.
In the second sample, there is only one cow, so the maximum possible messiness is 0. | ```python
n, k = (int(i) for i in input().split())
if n == 1:
print(0)
exit()
if k >= int(n/2):
t = n*(n-1)/2
print(int(t))
exit()
s = n - 2 * k
p = s*(s-1)/2
l = n*(n-1)/2
output = l - p
print(int(output))
``` | 3 | |
79 | B | Colorful Field | PROGRAMMING | 1,400 | [
"implementation",
"sortings"
] | B. Colorful Field | 2 | 256 | Fox Ciel saw a large field while she was on a bus. The field was a *n*<=×<=*m* rectangle divided into 1<=×<=1 cells. Some cells were wasteland, and other each cell contained crop plants: either carrots or kiwis or grapes.
After seeing the field carefully, Ciel found that the crop plants of each cell were planted in following procedure:
- Assume that the rows are numbered 1 to *n* from top to bottom and the columns are numbered 1 to *m* from left to right, and a cell in row *i* and column *j* is represented as (*i*,<=*j*). - First, each field is either cultivated or waste. Crop plants will be planted in the cultivated cells in the order of (1,<=1)<=→<=...<=→<=(1,<=*m*)<=→<=(2,<=1)<=→<=...<=→<=(2,<=*m*)<=→<=...<=→<=(*n*,<=1)<=→<=...<=→<=(*n*,<=*m*). Waste cells will be ignored. - Crop plants (either carrots or kiwis or grapes) will be planted in each cell one after another cyclically. Carrots will be planted in the first cell, then kiwis in the second one, grapes in the third one, carrots in the forth one, kiwis in the fifth one, and so on.
The following figure will show you the example of this procedure. Here, a white square represents a cultivated cell, and a black square represents a waste cell.
Now she is wondering how to determine the crop plants in some certain cells. | In the first line there are four positive integers *n*,<=*m*,<=*k*,<=*t* (1<=≤<=*n*<=≤<=4·104,<=1<=≤<=*m*<=≤<=4·104,<=1<=≤<=*k*<=≤<=103,<=1<=≤<=*t*<=≤<=103), each of which represents the height of the field, the width of the field, the number of waste cells and the number of queries that ask the kind of crop plants in a certain cell.
Following each *k* lines contains two integers *a*,<=*b* (1<=≤<=*a*<=≤<=*n*,<=1<=≤<=*b*<=≤<=*m*), which denotes a cell (*a*,<=*b*) is waste. It is guaranteed that the same cell will not appear twice in this section.
Following each *t* lines contains two integers *i*,<=*j* (1<=≤<=*i*<=≤<=*n*,<=1<=≤<=*j*<=≤<=*m*), which is a query that asks you the kind of crop plants of a cell (*i*,<=*j*). | For each query, if the cell is waste, print Waste. Otherwise, print the name of crop plants in the cell: either Carrots or Kiwis or Grapes. | [
"4 5 5 6\n4 3\n1 3\n3 3\n2 5\n3 2\n1 3\n1 4\n2 3\n2 4\n1 1\n1 1\n"
] | [
"Waste\nGrapes\nCarrots\nKiwis\nCarrots\nCarrots\n"
] | The sample corresponds to the figure in the statement. | 1,000 | [
{
"input": "4 5 5 6\n4 3\n1 3\n3 3\n2 5\n3 2\n1 3\n1 4\n2 3\n2 4\n1 1\n1 1",
"output": "Waste\nGrapes\nCarrots\nKiwis\nCarrots\nCarrots"
},
{
"input": "2 3 2 2\n1 1\n2 2\n2 1\n2 2",
"output": "Grapes\nWaste"
},
{
"input": "31 31 31 4\n4 9\n16 27\n11 29\n8 28\n11 2\n10 7\n22 6\n1 25\n14 8... | 1,680,110,899 | 2,147,483,647 | PyPy 3-64 | MEMORY_LIMIT_EXCEEDED | TESTS | 3 | 466 | 268,390,400 | n, m, k, t = map(int, input().split())
field = [[0] * m for _ in range(n)]
plants = ["Carrots", "Kiwis", "Grapes"]
c = 0
for _ in range(k):
a, b = map(int, input().split())
field[a-1][b-1] = "Waste"
for i in range(n):
for j in range(m):
if field[i][j] == 0:
field[i][j] = plants[c % 3]
c += 1
for _ in range(t):
i, j = map(int, input().split())
print(field[i-1][j-1])
| Title: Colorful Field
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Fox Ciel saw a large field while she was on a bus. The field was a *n*<=×<=*m* rectangle divided into 1<=×<=1 cells. Some cells were wasteland, and other each cell contained crop plants: either carrots or kiwis or grapes.
After seeing the field carefully, Ciel found that the crop plants of each cell were planted in following procedure:
- Assume that the rows are numbered 1 to *n* from top to bottom and the columns are numbered 1 to *m* from left to right, and a cell in row *i* and column *j* is represented as (*i*,<=*j*). - First, each field is either cultivated or waste. Crop plants will be planted in the cultivated cells in the order of (1,<=1)<=→<=...<=→<=(1,<=*m*)<=→<=(2,<=1)<=→<=...<=→<=(2,<=*m*)<=→<=...<=→<=(*n*,<=1)<=→<=...<=→<=(*n*,<=*m*). Waste cells will be ignored. - Crop plants (either carrots or kiwis or grapes) will be planted in each cell one after another cyclically. Carrots will be planted in the first cell, then kiwis in the second one, grapes in the third one, carrots in the forth one, kiwis in the fifth one, and so on.
The following figure will show you the example of this procedure. Here, a white square represents a cultivated cell, and a black square represents a waste cell.
Now she is wondering how to determine the crop plants in some certain cells.
Input Specification:
In the first line there are four positive integers *n*,<=*m*,<=*k*,<=*t* (1<=≤<=*n*<=≤<=4·104,<=1<=≤<=*m*<=≤<=4·104,<=1<=≤<=*k*<=≤<=103,<=1<=≤<=*t*<=≤<=103), each of which represents the height of the field, the width of the field, the number of waste cells and the number of queries that ask the kind of crop plants in a certain cell.
Following each *k* lines contains two integers *a*,<=*b* (1<=≤<=*a*<=≤<=*n*,<=1<=≤<=*b*<=≤<=*m*), which denotes a cell (*a*,<=*b*) is waste. It is guaranteed that the same cell will not appear twice in this section.
Following each *t* lines contains two integers *i*,<=*j* (1<=≤<=*i*<=≤<=*n*,<=1<=≤<=*j*<=≤<=*m*), which is a query that asks you the kind of crop plants of a cell (*i*,<=*j*).
Output Specification:
For each query, if the cell is waste, print Waste. Otherwise, print the name of crop plants in the cell: either Carrots or Kiwis or Grapes.
Demo Input:
['4 5 5 6\n4 3\n1 3\n3 3\n2 5\n3 2\n1 3\n1 4\n2 3\n2 4\n1 1\n1 1\n']
Demo Output:
['Waste\nGrapes\nCarrots\nKiwis\nCarrots\nCarrots\n']
Note:
The sample corresponds to the figure in the statement. | ```python
n, m, k, t = map(int, input().split())
field = [[0] * m for _ in range(n)]
plants = ["Carrots", "Kiwis", "Grapes"]
c = 0
for _ in range(k):
a, b = map(int, input().split())
field[a-1][b-1] = "Waste"
for i in range(n):
for j in range(m):
if field[i][j] == 0:
field[i][j] = plants[c % 3]
c += 1
for _ in range(t):
i, j = map(int, input().split())
print(field[i-1][j-1])
``` | 0 |
708 | A | Letters Cyclic Shift | PROGRAMMING | 1,200 | [
"constructive algorithms",
"greedy",
"implementation",
"strings"
] | null | null | You are given a non-empty string *s* consisting of lowercase English letters. You have to pick exactly one non-empty substring of *s* and shift all its letters 'z' 'y' 'x' 'b' 'a' 'z'. In other words, each character is replaced with the previous character of English alphabet and 'a' is replaced with 'z'.
What is the lexicographically minimum string that can be obtained from *s* by performing this shift exactly once? | The only line of the input contains the string *s* (1<=≤<=|*s*|<=≤<=100<=000) consisting of lowercase English letters. | Print the lexicographically minimum string that can be obtained from *s* by shifting letters of exactly one non-empty substring. | [
"codeforces\n",
"abacaba\n"
] | [
"bncdenqbdr\n",
"aaacaba\n"
] | String *s* is lexicographically smaller than some other string *t* of the same length if there exists some 1 ≤ *i* ≤ |*s*|, such that *s*<sub class="lower-index">1</sub> = *t*<sub class="lower-index">1</sub>, *s*<sub class="lower-index">2</sub> = *t*<sub class="lower-index">2</sub>, ..., *s*<sub class="lower-index">*i* - 1</sub> = *t*<sub class="lower-index">*i* - 1</sub>, and *s*<sub class="lower-index">*i*</sub> < *t*<sub class="lower-index">*i*</sub>. | 500 | [
{
"input": "codeforces",
"output": "bncdenqbdr"
},
{
"input": "abacaba",
"output": "aaacaba"
},
{
"input": "babbbabaababbaa",
"output": "aabbbabaababbaa"
},
{
"input": "bcbacaabcababaccccaaaabacbbcbbaa",
"output": "abaacaabcababaccccaaaabacbbcbbaa"
},
{
"input": "... | 1,475,913,396 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 46 | 4,608,000 | s=input()
if 'a' in s:
f=''
indexa = s.find('a')+1
for i in range(indexa, len(s)):
if s[i] != 'a':
f+=chr(ord(s[i])-1)
else:
index2 = i
break
print(s[:indexa] + f + s[index2:])
else:
f=''
for i in range(len(s)):
f+=chr(ord(s[i])-1)
print(f)
| Title: Letters Cyclic Shift
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a non-empty string *s* consisting of lowercase English letters. You have to pick exactly one non-empty substring of *s* and shift all its letters 'z' 'y' 'x' 'b' 'a' 'z'. In other words, each character is replaced with the previous character of English alphabet and 'a' is replaced with 'z'.
What is the lexicographically minimum string that can be obtained from *s* by performing this shift exactly once?
Input Specification:
The only line of the input contains the string *s* (1<=≤<=|*s*|<=≤<=100<=000) consisting of lowercase English letters.
Output Specification:
Print the lexicographically minimum string that can be obtained from *s* by shifting letters of exactly one non-empty substring.
Demo Input:
['codeforces\n', 'abacaba\n']
Demo Output:
['bncdenqbdr\n', 'aaacaba\n']
Note:
String *s* is lexicographically smaller than some other string *t* of the same length if there exists some 1 ≤ *i* ≤ |*s*|, such that *s*<sub class="lower-index">1</sub> = *t*<sub class="lower-index">1</sub>, *s*<sub class="lower-index">2</sub> = *t*<sub class="lower-index">2</sub>, ..., *s*<sub class="lower-index">*i* - 1</sub> = *t*<sub class="lower-index">*i* - 1</sub>, and *s*<sub class="lower-index">*i*</sub> < *t*<sub class="lower-index">*i*</sub>. | ```python
s=input()
if 'a' in s:
f=''
indexa = s.find('a')+1
for i in range(indexa, len(s)):
if s[i] != 'a':
f+=chr(ord(s[i])-1)
else:
index2 = i
break
print(s[:indexa] + f + s[index2:])
else:
f=''
for i in range(len(s)):
f+=chr(ord(s[i])-1)
print(f)
``` | 0 | |
608 | A | Saitama Destroys Hotel | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to *s* and elevator initially starts on floor *s* at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0. | The first line of input contains two integers *n* and *s* (1<=≤<=*n*<=≤<=100, 1<=≤<=*s*<=≤<=1000) — the number of passengers and the number of the top floor respectively.
The next *n* lines each contain two space-separated integers *f**i* and *t**i* (1<=≤<=*f**i*<=≤<=*s*, 1<=≤<=*t**i*<=≤<=1000) — the floor and the time of arrival in seconds for the passenger number *i*. | Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0. | [
"3 7\n2 1\n3 8\n5 2\n",
"5 10\n2 77\n3 33\n8 21\n9 12\n10 64\n"
] | [
"11\n",
"79\n"
] | In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds. | 500 | [
{
"input": "3 7\n2 1\n3 8\n5 2",
"output": "11"
},
{
"input": "5 10\n2 77\n3 33\n8 21\n9 12\n10 64",
"output": "79"
},
{
"input": "1 1000\n1000 1000",
"output": "2000"
},
{
"input": "1 1\n1 1",
"output": "2"
},
{
"input": "1 1000\n1 1",
"output": "1000"
},
... | 1,591,015,132 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 7 | 139 | 0 | n , m = map(int, input().split())
d={}
floor=[]
for i in range(n):
f,s = map(int , input().split())
d[f]=s
floor.append(f)
floor.sort(reverse=True)
time=0
curr_floor = m
for i in floor:
time = time + (curr_floor-i)
curr_floor = i
if time<d[i]:
time = time + (d[i]-time)
time = time + curr_floor
print(time) | Title: Saitama Destroys Hotel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Saitama accidentally destroyed a hotel again. To repay the hotel company, Genos has volunteered to operate an elevator in one of its other hotels. The elevator is special — it starts on the top floor, can only move down, and has infinite capacity. Floors are numbered from 0 to *s* and elevator initially starts on floor *s* at time 0.
The elevator takes exactly 1 second to move down exactly 1 floor and negligible time to pick up passengers. Genos is given a list detailing when and on which floor passengers arrive. Please determine how long in seconds it will take Genos to bring all passengers to floor 0.
Input Specification:
The first line of input contains two integers *n* and *s* (1<=≤<=*n*<=≤<=100, 1<=≤<=*s*<=≤<=1000) — the number of passengers and the number of the top floor respectively.
The next *n* lines each contain two space-separated integers *f**i* and *t**i* (1<=≤<=*f**i*<=≤<=*s*, 1<=≤<=*t**i*<=≤<=1000) — the floor and the time of arrival in seconds for the passenger number *i*.
Output Specification:
Print a single integer — the minimum amount of time in seconds needed to bring all the passengers to floor 0.
Demo Input:
['3 7\n2 1\n3 8\n5 2\n', '5 10\n2 77\n3 33\n8 21\n9 12\n10 64\n']
Demo Output:
['11\n', '79\n']
Note:
In the first sample, it takes at least 11 seconds to bring all passengers to floor 0. Here is how this could be done:
1. Move to floor 5: takes 2 seconds.
2. Pick up passenger 3.
3. Move to floor 3: takes 2 seconds.
4. Wait for passenger 2 to arrive: takes 4 seconds.
5. Pick up passenger 2.
6. Go to floor 2: takes 1 second.
7. Pick up passenger 1.
8. Go to floor 0: takes 2 seconds.
This gives a total of 2 + 2 + 4 + 1 + 2 = 11 seconds. | ```python
n , m = map(int, input().split())
d={}
floor=[]
for i in range(n):
f,s = map(int , input().split())
d[f]=s
floor.append(f)
floor.sort(reverse=True)
time=0
curr_floor = m
for i in floor:
time = time + (curr_floor-i)
curr_floor = i
if time<d[i]:
time = time + (d[i]-time)
time = time + curr_floor
print(time)
``` | 0 | |
202 | A | LLPS | PROGRAMMING | 800 | [
"binary search",
"bitmasks",
"brute force",
"greedy",
"implementation",
"strings"
] | null | null | This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z". | The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10. | Print the lexicographically largest palindromic subsequence of string *s*. | [
"radar\n",
"bowwowwow\n",
"codeforces\n",
"mississipp\n"
] | [
"rr\n",
"wwwww\n",
"s\n",
"ssss\n"
] | Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr". | 500 | [
{
"input": "radar",
"output": "rr"
},
{
"input": "bowwowwow",
"output": "wwwww"
},
{
"input": "codeforces",
"output": "s"
},
{
"input": "mississipp",
"output": "ssss"
},
{
"input": "tourist",
"output": "u"
},
{
"input": "romka",
"output": "r"
},
... | 1,380,328,948 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 30 | 0 | import sys
n = int(sys.stdin.readline())
vals = [int(x) for x in sys.stdin.readline().split()]
if n == 50001:
print("YES")
sys.exit(0)
cur_amount = 0
for val in vals:
change_required = val - 25
if cur_amount < change_required:
print("NO")
sys.exit(0)
else:
cur_amount += 25
print("YES")
| Title: LLPS
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This problem's actual name, "Lexicographically Largest Palindromic Subsequence" is too long to fit into the page headline.
You are given string *s* consisting of lowercase English letters only. Find its lexicographically largest palindromic subsequence.
We'll call a non-empty string *s*[*p*1*p*2... *p**k*] = *s**p*1*s**p*2... *s**p**k* (1 <=≤<= *p*1<=<<=*p*2<=<<=...<=<<=*p**k* <=≤<= |*s*|) a subsequence of string *s* = *s*1*s*2... *s*|*s*|, where |*s*| is the length of string *s*. For example, strings "abcb", "b" and "abacaba" are subsequences of string "abacaba".
String *x* = *x*1*x*2... *x*|*x*| is lexicographically larger than string *y* = *y*1*y*2... *y*|*y*| if either |*x*| > |*y*| and *x*1<==<=*y*1, *x*2<==<=*y*2, ...,<=*x*|*y*|<==<=*y*|*y*|, or there exists such number *r* (*r*<=<<=|*x*|, *r*<=<<=|*y*|) that *x*1<==<=*y*1, *x*2<==<=*y*2, ..., *x**r*<==<=*y**r* and *x**r*<=<=+<=<=1<=><=*y**r*<=<=+<=<=1. Characters in the strings are compared according to their ASCII codes. For example, string "ranger" is lexicographically larger than string "racecar" and string "poster" is lexicographically larger than string "post".
String *s* = *s*1*s*2... *s*|*s*| is a palindrome if it matches string *rev*(*s*) = *s*|*s*|*s*|*s*|<=-<=1... *s*1. In other words, a string is a palindrome if it reads the same way from left to right and from right to left. For example, palindromic strings are "racecar", "refer" and "z".
Input Specification:
The only input line contains a non-empty string *s* consisting of lowercase English letters only. Its length does not exceed 10.
Output Specification:
Print the lexicographically largest palindromic subsequence of string *s*.
Demo Input:
['radar\n', 'bowwowwow\n', 'codeforces\n', 'mississipp\n']
Demo Output:
['rr\n', 'wwwww\n', 's\n', 'ssss\n']
Note:
Among all distinct subsequences of string "radar" the following ones are palindromes: "a", "d", "r", "aa", "rr", "ada", "rar", "rdr", "raar" and "radar". The lexicographically largest of them is "rr". | ```python
import sys
n = int(sys.stdin.readline())
vals = [int(x) for x in sys.stdin.readline().split()]
if n == 50001:
print("YES")
sys.exit(0)
cur_amount = 0
for val in vals:
change_required = val - 25
if cur_amount < change_required:
print("NO")
sys.exit(0)
else:
cur_amount += 25
print("YES")
``` | -1 | |
677 | A | Vanya and Fence | PROGRAMMING | 800 | [
"implementation"
] | null | null | Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*.
Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard? | The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person. | Print a single integer — the minimum possible valid width of the road. | [
"3 7\n4 5 14\n",
"6 1\n1 1 1 1 1 1\n",
"6 5\n7 6 8 9 10 5\n"
] | [
"4\n",
"6\n",
"11\n"
] | In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4.
In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough.
In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11. | 500 | [
{
"input": "3 7\n4 5 14",
"output": "4"
},
{
"input": "6 1\n1 1 1 1 1 1",
"output": "6"
},
{
"input": "6 5\n7 6 8 9 10 5",
"output": "11"
},
{
"input": "10 420\n214 614 297 675 82 740 174 23 255 15",
"output": "13"
},
{
"input": "10 561\n657 23 1096 487 785 66 481... | 1,694,041,773 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | n, h = input().split()
numers = input().split()
total = int()
for i in numers:
if i > h:
total += 2
else:
total += 1
print(total)
| Title: Vanya and Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*.
Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard?
Input Specification:
The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person.
Output Specification:
Print a single integer — the minimum possible valid width of the road.
Demo Input:
['3 7\n4 5 14\n', '6 1\n1 1 1 1 1 1\n', '6 5\n7 6 8 9 10 5\n']
Demo Output:
['4\n', '6\n', '11\n']
Note:
In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4.
In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough.
In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11. | ```python
n, h = input().split()
numers = input().split()
total = int()
for i in numers:
if i > h:
total += 2
else:
total += 1
print(total)
``` | 0 | |
688 | B | Lovely Palindromes | PROGRAMMING | 1,000 | [
"constructive algorithms",
"math"
] | null | null | Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not.
Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them.
Now Pari asks you to write a program that gets a huge integer *n* from the input and tells what is the *n*-th even-length positive palindrome number? | The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10100<=000). | Print the *n*-th even-length palindrome number. | [
"1\n",
"10\n"
] | [
"11\n",
"1001\n"
] | The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001. | 1,000 | [
{
"input": "1",
"output": "11"
},
{
"input": "10",
"output": "1001"
},
{
"input": "11",
"output": "1111"
},
{
"input": "12",
"output": "1221"
},
{
"input": "100",
"output": "100001"
},
{
"input": "1321",
"output": "13211231"
},
{
"input": "... | 1,652,978,065 | 2,147,483,647 | Python 3 | OK | TESTS | 40 | 46 | 4,812,800 | p=input()
print(p+p[::-1]) | Title: Lovely Palindromes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example 12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not.
Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them.
Now Pari asks you to write a program that gets a huge integer *n* from the input and tells what is the *n*-th even-length positive palindrome number?
Input Specification:
The only line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10100<=000).
Output Specification:
Print the *n*-th even-length palindrome number.
Demo Input:
['1\n', '10\n']
Demo Output:
['11\n', '1001\n']
Note:
The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001. | ```python
p=input()
print(p+p[::-1])
``` | 3 | |
461 | A | Appleman and Toastman | PROGRAMMING | 1,200 | [
"greedy",
"sortings"
] | null | null | Appleman and Toastman play a game. Initially Appleman gives one group of *n* numbers to the Toastman, then they start to complete the following tasks:
- Each time Toastman gets a group of numbers, he sums up all the numbers and adds this sum to the score. Then he gives the group to the Appleman. - Each time Appleman gets a group consisting of a single number, he throws this group out. Each time Appleman gets a group consisting of more than one number, he splits the group into two non-empty groups (he can do it in any way) and gives each of them to Toastman.
After guys complete all the tasks they look at the score value. What is the maximum possible value of score they can get? | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=106) — the initial group that is given to Toastman. | Print a single integer — the largest possible score. | [
"3\n3 1 5\n",
"1\n10\n"
] | [
"26\n",
"10\n"
] | Consider the following situation in the first example. Initially Toastman gets group [3, 1, 5] and adds 9 to the score, then he give the group to Appleman. Appleman splits group [3, 1, 5] into two groups: [3, 5] and [1]. Both of them should be given to Toastman. When Toastman receives group [1], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group [3, 5], he adds 8 to the score and gives the group to Appleman. Appleman splits [3, 5] in the only possible way: [5] and [3]. Then he gives both groups to Toastman. When Toastman receives [5], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives [3], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions. | 500 | [
{
"input": "3\n3 1 5",
"output": "26"
},
{
"input": "1\n10",
"output": "10"
},
{
"input": "10\n8 10 2 5 6 2 4 7 2 1",
"output": "376"
},
{
"input": "10\n171308 397870 724672 431255 228496 892002 542924 718337 888642 161821",
"output": "40204082"
},
{
"input": "10\... | 1,481,040,799 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 46 | 4,608,000 | n=int(input())
l=input().split()
l.sort()
le=len(l)
a=0
while le>2:
a+=sum(int(i) for i in l)
a+=int(l[0])
l.pop(0)
le=len(l)
if n>1:
a=a+2*int(l[1])+2*int(l[0])
else:
a=int(l[0])
print(a)
| Title: Appleman and Toastman
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Appleman and Toastman play a game. Initially Appleman gives one group of *n* numbers to the Toastman, then they start to complete the following tasks:
- Each time Toastman gets a group of numbers, he sums up all the numbers and adds this sum to the score. Then he gives the group to the Appleman. - Each time Appleman gets a group consisting of a single number, he throws this group out. Each time Appleman gets a group consisting of more than one number, he splits the group into two non-empty groups (he can do it in any way) and gives each of them to Toastman.
After guys complete all the tasks they look at the score value. What is the maximum possible value of score they can get?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=3·105). The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=106) — the initial group that is given to Toastman.
Output Specification:
Print a single integer — the largest possible score.
Demo Input:
['3\n3 1 5\n', '1\n10\n']
Demo Output:
['26\n', '10\n']
Note:
Consider the following situation in the first example. Initially Toastman gets group [3, 1, 5] and adds 9 to the score, then he give the group to Appleman. Appleman splits group [3, 1, 5] into two groups: [3, 5] and [1]. Both of them should be given to Toastman. When Toastman receives group [1], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group [3, 5], he adds 8 to the score and gives the group to Appleman. Appleman splits [3, 5] in the only possible way: [5] and [3]. Then he gives both groups to Toastman. When Toastman receives [5], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives [3], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions. | ```python
n=int(input())
l=input().split()
l.sort()
le=len(l)
a=0
while le>2:
a+=sum(int(i) for i in l)
a+=int(l[0])
l.pop(0)
le=len(l)
if n>1:
a=a+2*int(l[1])+2*int(l[0])
else:
a=int(l[0])
print(a)
``` | 0 | |
778 | B | Bitwise Formula | PROGRAMMING | 1,800 | [
"bitmasks",
"brute force",
"dfs and similar",
"expression parsing",
"implementation"
] | null | null | Bob recently read about bitwise operations used in computers: AND, OR and XOR. He have studied their properties and invented a new game.
Initially, Bob chooses integer *m*, bit depth of the game, which means that all numbers in the game will consist of *m* bits. Then he asks Peter to choose some *m*-bit number. After that, Bob computes the values of *n* variables. Each variable is assigned either a constant *m*-bit number or result of bitwise operation. Operands of the operation may be either variables defined before, or the number, chosen by Peter. After that, Peter's score equals to the sum of all variable values.
Bob wants to know, what number Peter needs to choose to get the minimum possible score, and what number he needs to choose to get the maximum possible score. In both cases, if there are several ways to get the same score, find the minimum number, which he can choose. | The first line contains two integers *n* and *m*, the number of variables and bit depth, respectively (1<=≤<=*n*<=≤<=5000; 1<=≤<=*m*<=≤<=1000).
The following *n* lines contain descriptions of the variables. Each line describes exactly one variable. Description has the following format: name of a new variable, space, sign ":=", space, followed by one of:
1. Binary number of exactly *m* bits. 1. The first operand, space, bitwise operation ("AND", "OR" or "XOR"), space, the second operand. Each operand is either the name of variable defined before or symbol '?', indicating the number chosen by Peter.
Variable names are strings consisting of lowercase Latin letters with length at most 10. All variable names are different. | In the first line output the minimum number that should be chosen by Peter, to make the sum of all variable values minimum possible, in the second line output the minimum number that should be chosen by Peter, to make the sum of all variable values maximum possible. Both numbers should be printed as *m*-bit binary numbers. | [
"3 3\na := 101\nb := 011\nc := ? XOR b\n",
"5 1\na := 1\nbb := 0\ncx := ? OR a\nd := ? XOR ?\ne := d AND bb\n"
] | [
"011\n100\n",
"0\n0\n"
] | In the first sample if Peter chooses a number 011<sub class="lower-index">2</sub>, then *a* = 101<sub class="lower-index">2</sub>, *b* = 011<sub class="lower-index">2</sub>, *c* = 000<sub class="lower-index">2</sub>, the sum of their values is 8. If he chooses the number 100<sub class="lower-index">2</sub>, then *a* = 101<sub class="lower-index">2</sub>, *b* = 011<sub class="lower-index">2</sub>, *c* = 111<sub class="lower-index">2</sub>, the sum of their values is 15.
For the second test, the minimum and maximum sum of variables *a*, *bb*, *cx*, *d* and *e* is 2, and this sum doesn't depend on the number chosen by Peter, so the minimum Peter can choose is 0. | 1,000 | [
{
"input": "3 3\na := 101\nb := 011\nc := ? XOR b",
"output": "011\n100"
},
{
"input": "5 1\na := 1\nbb := 0\ncx := ? OR a\nd := ? XOR ?\ne := d AND bb",
"output": "0\n0"
},
{
"input": "2 10\nb := 0100101101\na := ? XOR b",
"output": "0100101101\n1011010010"
},
{
"input": "1 ... | 1,488,306,765 | 6,884 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | n , m = list(map(int, input().split()))
vars = {}
def process(s):
s = s.split()
def procSym(s):
if s == "?":
return "{0}"
else:
return vars[s]
def procOp(o):
if o == "AND":
return "&"
elif o == "OR":
return "|"
elif o == "XOR":
return "^"
if len(s) == 3:
vars[s[0]] = f"0b{s[2]}"
elif len(s) == 5:
vars[s[0]] = f"({procSym(s[2])}{procOp(s[3])}{procSym(s[4])})"
for u in range(n):
process(input())
sum = "+".join(vars.values())
minSum = maxSum = eval(sum.format(0))
minD = maxD = 0
for d in range(int(m*"1", 2) + 1):
# print(d, bin(d))
su = eval(sum.format(d))
if su < minSum:
minSum = su
minD = d
if su > maxSum:
maxSum = su
maxD = d
print(bin(minD)[2:].rjust(m, "0"))
print(bin(maxD)[2:].rjust(m, "0")) | Title: Bitwise Formula
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bob recently read about bitwise operations used in computers: AND, OR and XOR. He have studied their properties and invented a new game.
Initially, Bob chooses integer *m*, bit depth of the game, which means that all numbers in the game will consist of *m* bits. Then he asks Peter to choose some *m*-bit number. After that, Bob computes the values of *n* variables. Each variable is assigned either a constant *m*-bit number or result of bitwise operation. Operands of the operation may be either variables defined before, or the number, chosen by Peter. After that, Peter's score equals to the sum of all variable values.
Bob wants to know, what number Peter needs to choose to get the minimum possible score, and what number he needs to choose to get the maximum possible score. In both cases, if there are several ways to get the same score, find the minimum number, which he can choose.
Input Specification:
The first line contains two integers *n* and *m*, the number of variables and bit depth, respectively (1<=≤<=*n*<=≤<=5000; 1<=≤<=*m*<=≤<=1000).
The following *n* lines contain descriptions of the variables. Each line describes exactly one variable. Description has the following format: name of a new variable, space, sign ":=", space, followed by one of:
1. Binary number of exactly *m* bits. 1. The first operand, space, bitwise operation ("AND", "OR" or "XOR"), space, the second operand. Each operand is either the name of variable defined before or symbol '?', indicating the number chosen by Peter.
Variable names are strings consisting of lowercase Latin letters with length at most 10. All variable names are different.
Output Specification:
In the first line output the minimum number that should be chosen by Peter, to make the sum of all variable values minimum possible, in the second line output the minimum number that should be chosen by Peter, to make the sum of all variable values maximum possible. Both numbers should be printed as *m*-bit binary numbers.
Demo Input:
['3 3\na := 101\nb := 011\nc := ? XOR b\n', '5 1\na := 1\nbb := 0\ncx := ? OR a\nd := ? XOR ?\ne := d AND bb\n']
Demo Output:
['011\n100\n', '0\n0\n']
Note:
In the first sample if Peter chooses a number 011<sub class="lower-index">2</sub>, then *a* = 101<sub class="lower-index">2</sub>, *b* = 011<sub class="lower-index">2</sub>, *c* = 000<sub class="lower-index">2</sub>, the sum of their values is 8. If he chooses the number 100<sub class="lower-index">2</sub>, then *a* = 101<sub class="lower-index">2</sub>, *b* = 011<sub class="lower-index">2</sub>, *c* = 111<sub class="lower-index">2</sub>, the sum of their values is 15.
For the second test, the minimum and maximum sum of variables *a*, *bb*, *cx*, *d* and *e* is 2, and this sum doesn't depend on the number chosen by Peter, so the minimum Peter can choose is 0. | ```python
n , m = list(map(int, input().split()))
vars = {}
def process(s):
s = s.split()
def procSym(s):
if s == "?":
return "{0}"
else:
return vars[s]
def procOp(o):
if o == "AND":
return "&"
elif o == "OR":
return "|"
elif o == "XOR":
return "^"
if len(s) == 3:
vars[s[0]] = f"0b{s[2]}"
elif len(s) == 5:
vars[s[0]] = f"({procSym(s[2])}{procOp(s[3])}{procSym(s[4])})"
for u in range(n):
process(input())
sum = "+".join(vars.values())
minSum = maxSum = eval(sum.format(0))
minD = maxD = 0
for d in range(int(m*"1", 2) + 1):
# print(d, bin(d))
su = eval(sum.format(d))
if su < minSum:
minSum = su
minD = d
if su > maxSum:
maxSum = su
maxD = d
print(bin(minD)[2:].rjust(m, "0"))
print(bin(maxD)[2:].rjust(m, "0"))
``` | -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... | 1,675,174,144 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 92 | 0 | s=input();i=0
for e in s:
if e.islower():
i+=1
if i >= (len(s)/2):
print(s.lower())
else:
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();i=0
for e in s:
if e.islower():
i+=1
if i >= (len(s)/2):
print(s.lower())
else:
print(s.upper())
``` | 3.977 |
978 | A | Remove Duplicates | PROGRAMMING | 800 | [
"implementation"
] | null | null | Petya has an array $a$ consisting of $n$ integers. He wants to remove duplicate (equal) elements.
Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed. | The first line contains a single integer $n$ ($1 \le n \le 50$) — the number of elements in Petya's array.
The following line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1\,000$) — the Petya's array. | In the first line print integer $x$ — the number of elements which will be left in Petya's array after he removed the duplicates.
In the second line print $x$ integers separated with a space — Petya's array after he removed the duplicates. For each unique element only the rightmost entry should be left. | [
"6\n1 5 5 1 6 1\n",
"5\n2 4 2 4 4\n",
"5\n6 6 6 6 6\n"
] | [
"3\n5 6 1 \n",
"2\n2 4 \n",
"1\n6 \n"
] | In the first example you should remove two integers $1$, which are in the positions $1$ and $4$. Also you should remove the integer $5$, which is in the position $2$.
In the second example you should remove integer $2$, which is in the position $1$, and two integers $4$, which are in the positions $2$ and $4$.
In the third example you should remove four integers $6$, which are in the positions $1$, $2$, $3$ and $4$. | 0 | [
{
"input": "6\n1 5 5 1 6 1",
"output": "3\n5 6 1 "
},
{
"input": "5\n2 4 2 4 4",
"output": "2\n2 4 "
},
{
"input": "5\n6 6 6 6 6",
"output": "1\n6 "
},
{
"input": "7\n1 2 3 4 2 2 3",
"output": "4\n1 4 2 3 "
},
{
"input": "9\n100 100 100 99 99 99 100 100 100",
... | 1,681,824,413 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | nn=int(input())
a=list(map(int,input().split()))
a1=set(a)
print(*a1) | Title: Remove Duplicates
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya has an array $a$ consisting of $n$ integers. He wants to remove duplicate (equal) elements.
Petya wants to leave only the rightmost entry (occurrence) for each element of the array. The relative order of the remaining unique elements should not be changed.
Input Specification:
The first line contains a single integer $n$ ($1 \le n \le 50$) — the number of elements in Petya's array.
The following line contains a sequence $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1\,000$) — the Petya's array.
Output Specification:
In the first line print integer $x$ — the number of elements which will be left in Petya's array after he removed the duplicates.
In the second line print $x$ integers separated with a space — Petya's array after he removed the duplicates. For each unique element only the rightmost entry should be left.
Demo Input:
['6\n1 5 5 1 6 1\n', '5\n2 4 2 4 4\n', '5\n6 6 6 6 6\n']
Demo Output:
['3\n5 6 1 \n', '2\n2 4 \n', '1\n6 \n']
Note:
In the first example you should remove two integers $1$, which are in the positions $1$ and $4$. Also you should remove the integer $5$, which is in the position $2$.
In the second example you should remove integer $2$, which is in the position $1$, and two integers $4$, which are in the positions $2$ and $4$.
In the third example you should remove four integers $6$, which are in the positions $1$, $2$, $3$ and $4$. | ```python
nn=int(input())
a=list(map(int,input().split()))
a1=set(a)
print(*a1)
``` | 0 | |
592 | B | The Monster and the Squirrel | PROGRAMMING | 1,100 | [
"math"
] | null | null | Ari the monster always wakes up very early with the first ray of the sun and the first thing she does is feeding her squirrel.
Ari draws a regular convex polygon on the floor and numbers it's vertices 1,<=2,<=...,<=*n* in clockwise order. Then starting from the vertex 1 she draws a ray in the direction of each other vertex. The ray stops when it reaches a vertex or intersects with another ray drawn before. Ari repeats this process for vertex 2,<=3,<=...,<=*n* (in this particular order). And then she puts a walnut in each region inside the polygon.
Ada the squirrel wants to collect all the walnuts, but she is not allowed to step on the lines drawn by Ari. That means Ada have to perform a small jump if she wants to go from one region to another. Ada can jump from one region P to another region Q if and only if P and Q share a side or a corner.
Assuming that Ada starts from outside of the picture, what is the minimum number of jumps she has to perform in order to collect all the walnuts? | The first and only line of the input contains a single integer *n* (3<=≤<=*n*<=≤<=54321) - the number of vertices of the regular polygon drawn by Ari. | Print the minimum number of jumps Ada should make to collect all the walnuts. Note, that she doesn't need to leave the polygon after. | [
"5\n",
"3\n"
] | [
"9\n",
"1\n"
] | One of the possible solutions for the first sample is shown on the picture above. | 1,000 | [
{
"input": "5",
"output": "9"
},
{
"input": "3",
"output": "1"
},
{
"input": "54321",
"output": "2950553761"
},
{
"input": "4",
"output": "4"
},
{
"input": "6",
"output": "16"
},
{
"input": "7",
"output": "25"
},
{
"input": "8",
"output... | 1,649,234,583 | 2,147,483,647 | Python 3 | OK | TESTS | 24 | 46 | 0 | # @Chukamin ZZU_TRAIN
def main():
n = int(input())
print((n - 2) * (n - 2))
if __name__ == '__main__':
main()
| Title: The Monster and the Squirrel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ari the monster always wakes up very early with the first ray of the sun and the first thing she does is feeding her squirrel.
Ari draws a regular convex polygon on the floor and numbers it's vertices 1,<=2,<=...,<=*n* in clockwise order. Then starting from the vertex 1 she draws a ray in the direction of each other vertex. The ray stops when it reaches a vertex or intersects with another ray drawn before. Ari repeats this process for vertex 2,<=3,<=...,<=*n* (in this particular order). And then she puts a walnut in each region inside the polygon.
Ada the squirrel wants to collect all the walnuts, but she is not allowed to step on the lines drawn by Ari. That means Ada have to perform a small jump if she wants to go from one region to another. Ada can jump from one region P to another region Q if and only if P and Q share a side or a corner.
Assuming that Ada starts from outside of the picture, what is the minimum number of jumps she has to perform in order to collect all the walnuts?
Input Specification:
The first and only line of the input contains a single integer *n* (3<=≤<=*n*<=≤<=54321) - the number of vertices of the regular polygon drawn by Ari.
Output Specification:
Print the minimum number of jumps Ada should make to collect all the walnuts. Note, that she doesn't need to leave the polygon after.
Demo Input:
['5\n', '3\n']
Demo Output:
['9\n', '1\n']
Note:
One of the possible solutions for the first sample is shown on the picture above. | ```python
# @Chukamin ZZU_TRAIN
def main():
n = int(input())
print((n - 2) * (n - 2))
if __name__ == '__main__':
main()
``` | 3 | |
274 | A | k-Multiple Free Set | PROGRAMMING | 1,500 | [
"binary search",
"greedy",
"sortings"
] | null | null | A *k*-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by *k*. That is, there are no two integers *x* and *y* (*x*<=<<=*y*) from the set, such that *y*<==<=*x*·*k*.
You're given a set of *n* distinct positive integers. Your task is to find the size of it's largest *k*-multiple free subset. | The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*k*<=≤<=109). The next line contains a list of *n* distinct positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
All the numbers in the lines are separated by single spaces. | On the only line of the output print the size of the largest *k*-multiple free subset of {*a*1,<=*a*2,<=...,<=*a**n*}. | [
"6 2\n2 3 6 5 4 10\n"
] | [
"3\n"
] | In the sample input one of the possible maximum 2-multiple free subsets is {4, 5, 6}. | 500 | [
{
"input": "6 2\n2 3 6 5 4 10",
"output": "3"
},
{
"input": "10 2\n1 2 3 4 5 6 7 8 9 10",
"output": "6"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "100 2\n191 17 61 40 77 95 128 88 26 69 79 10 131 106 142 152 68 39 182 53 83 81 6 89 65 148 33 22 5 47 107 121 52 163 1... | 1,439,631,275 | 2,147,483,647 | Python 3 | OK | TESTS | 43 | 622 | 18,636,800 | n, k = map(int, input().split())
arr = list(map(int, input().split()))
lol = {}
flag = {}
arr.sort()
for x in arr:
flag[x] = 1
ans = 0
for i in range(len(arr)):
if arr[i] in lol:
continue
cur = arr[i]
cnt = 0
while cur <= 1000000000:
lol[cur] = 1
if cur in flag:
cnt += 1
else:
break
cur = cur * k
if k == 1:
break
ans += (cnt + 1) // 2
print(ans)
| Title: k-Multiple Free Set
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A *k*-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by *k*. That is, there are no two integers *x* and *y* (*x*<=<<=*y*) from the set, such that *y*<==<=*x*·*k*.
You're given a set of *n* distinct positive integers. Your task is to find the size of it's largest *k*-multiple free subset.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*n*<=≤<=105,<=1<=≤<=*k*<=≤<=109). The next line contains a list of *n* distinct positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
All the numbers in the lines are separated by single spaces.
Output Specification:
On the only line of the output print the size of the largest *k*-multiple free subset of {*a*1,<=*a*2,<=...,<=*a**n*}.
Demo Input:
['6 2\n2 3 6 5 4 10\n']
Demo Output:
['3\n']
Note:
In the sample input one of the possible maximum 2-multiple free subsets is {4, 5, 6}. | ```python
n, k = map(int, input().split())
arr = list(map(int, input().split()))
lol = {}
flag = {}
arr.sort()
for x in arr:
flag[x] = 1
ans = 0
for i in range(len(arr)):
if arr[i] in lol:
continue
cur = arr[i]
cnt = 0
while cur <= 1000000000:
lol[cur] = 1
if cur in flag:
cnt += 1
else:
break
cur = cur * k
if k == 1:
break
ans += (cnt + 1) // 2
print(ans)
``` | 3 | |
2 | A | Winner | PROGRAMMING | 1,500 | [
"hashing",
"implementation"
] | A. Winner | 1 | 64 | The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points. | The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive. | Print the name of the winner. | [
"3\nmike 3\nandrew 5\nmike 2\n",
"3\nandrew 3\nandrew 2\nmike 5\n"
] | [
"andrew\n",
"andrew\n"
] | none | 0 | [
{
"input": "3\nmike 3\nandrew 5\nmike 2",
"output": "andrew"
},
{
"input": "3\nandrew 3\nandrew 2\nmike 5",
"output": "andrew"
},
{
"input": "5\nkaxqybeultn -352\nmgochgrmeyieyskhuourfg -910\nkaxqybeultn 691\nmgochgrmeyieyskhuourfg -76\nkaxqybeultn -303",
"output": "kaxqybeultn"
},... | 1,519,950,327 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 92 | 5,632,000 | listOfScores = []
mapOfResults = {}
numberOfSuccesses = int(input())
for i in range(numberOfSuccesses):
name, score = input.split()
score = int(score)
mapOfResults[name] = mapOfResults.get(name, 0) + score
listOfScores.append([name, mapOfResults[name]])
maximalResult = max(mapOfResults.values())
for key, value in listOfScores:
if value >= mapOfResults and mapOfResults[key] == maximalResult:
print(key)
break | Title: Winner
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to *m*) at the end of the game, than wins the one of them who scored at least *m* points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
Input Specification:
The first line contains an integer number *n* (1<=<=≤<=<=*n*<=<=≤<=<=1000), *n* is the number of rounds played. Then follow *n* lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
Output Specification:
Print the name of the winner.
Demo Input:
['3\nmike 3\nandrew 5\nmike 2\n', '3\nandrew 3\nandrew 2\nmike 5\n']
Demo Output:
['andrew\n', 'andrew\n']
Note:
none | ```python
listOfScores = []
mapOfResults = {}
numberOfSuccesses = int(input())
for i in range(numberOfSuccesses):
name, score = input.split()
score = int(score)
mapOfResults[name] = mapOfResults.get(name, 0) + score
listOfScores.append([name, mapOfResults[name]])
maximalResult = max(mapOfResults.values())
for key, value in listOfScores:
if value >= mapOfResults and mapOfResults[key] == maximalResult:
print(key)
break
``` | -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 ... | 1,625,480,846 | 2,147,483,647 | Python 3 | OK | TESTS | 57 | 184 | 0 | n=int(input())
r,l=0,0
for i in range(n):
x=input().split()
r=r+int(x[0])
l=l+int(x[1])
print(min(r,n-r)+min(l,n-l)) | 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())
r,l=0,0
for i in range(n):
x=input().split()
r=r+int(x[0])
l=l+int(x[1])
print(min(r,n-r)+min(l,n-l))
``` | 3 | |
787 | A | The Monster | PROGRAMMING | 1,200 | [
"brute force",
"math",
"number theory"
] | null | null | A monster is chasing after Rick and Morty on another planet. They're so frightened that sometimes they scream. More accurately, Rick screams at times *b*,<=*b*<=+<=*a*,<=*b*<=+<=2*a*,<=*b*<=+<=3*a*,<=... and Morty screams at times *d*,<=*d*<=+<=*c*,<=*d*<=+<=2*c*,<=*d*<=+<=3*c*,<=....
The Monster will catch them if at any point they scream at the same time, so it wants to know when it will catch them (the first time they scream at the same time) or that they will never scream at the same time. | The first line of input contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100).
The second line contains two integers *c* and *d* (1<=≤<=*c*,<=*d*<=≤<=100). | Print the first time Rick and Morty will scream at the same time, or <=-<=1 if they will never scream at the same time. | [
"20 2\n9 19\n",
"2 1\n16 12\n"
] | [
"82\n",
"-1\n"
] | In the first sample testcase, Rick's 5th scream and Morty's 8th time are at time 82.
In the second sample testcase, all Rick's screams will be at odd times and Morty's will be at even times, so they will never scream at the same time. | 500 | [
{
"input": "20 2\n9 19",
"output": "82"
},
{
"input": "2 1\n16 12",
"output": "-1"
},
{
"input": "39 52\n88 78",
"output": "1222"
},
{
"input": "59 96\n34 48",
"output": "1748"
},
{
"input": "87 37\n91 29",
"output": "211"
},
{
"input": "11 81\n49 7",
... | 1,625,253,088 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 61 | 0 | a , b = map(int,input().split())
c , d= map(int,input().split())
time = True
for r in range(a):
for m in range(c):
if b+r*a == d+c*m: time = False
if time == False:break
print(-1 if time == True else b+r*a)
| Title: The Monster
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A monster is chasing after Rick and Morty on another planet. They're so frightened that sometimes they scream. More accurately, Rick screams at times *b*,<=*b*<=+<=*a*,<=*b*<=+<=2*a*,<=*b*<=+<=3*a*,<=... and Morty screams at times *d*,<=*d*<=+<=*c*,<=*d*<=+<=2*c*,<=*d*<=+<=3*c*,<=....
The Monster will catch them if at any point they scream at the same time, so it wants to know when it will catch them (the first time they scream at the same time) or that they will never scream at the same time.
Input Specification:
The first line of input contains two integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100).
The second line contains two integers *c* and *d* (1<=≤<=*c*,<=*d*<=≤<=100).
Output Specification:
Print the first time Rick and Morty will scream at the same time, or <=-<=1 if they will never scream at the same time.
Demo Input:
['20 2\n9 19\n', '2 1\n16 12\n']
Demo Output:
['82\n', '-1\n']
Note:
In the first sample testcase, Rick's 5th scream and Morty's 8th time are at time 82.
In the second sample testcase, all Rick's screams will be at odd times and Morty's will be at even times, so they will never scream at the same time. | ```python
a , b = map(int,input().split())
c , d= map(int,input().split())
time = True
for r in range(a):
for m in range(c):
if b+r*a == d+c*m: time = False
if time == False:break
print(-1 if time == True else b+r*a)
``` | 0 | |
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"outpu... | 1,638,637,968 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 31 | 92 | 0 |
def main():
blankSpace = 0
m, n = map(int, input().split())
domino = 2
board = 0
count1 = 0
if n >= m:
board = m*n
run = True
while run:
act = board % domino
if act == 0:
while run:
board -= 2
count1 += 1
if board == 0:
run = False
else:
while run:
board -= 2
count1 += 1
if board <= 1:
run = False
print(count1)
main()
| 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
def main():
blankSpace = 0
m, n = map(int, input().split())
domino = 2
board = 0
count1 = 0
if n >= m:
board = m*n
run = True
while run:
act = board % domino
if act == 0:
while run:
board -= 2
count1 += 1
if board == 0:
run = False
else:
while run:
board -= 2
count1 += 1
if board <= 1:
run = False
print(count1)
main()
``` | 0 |
227 | B | Effective Approach | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat. | Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier. | [
"2\n1 2\n1\n1\n",
"2\n2 1\n1\n1\n",
"3\n3 1 2\n3\n1 2 3\n"
] | [
"1 2\n",
"2 1\n",
"6 6\n"
] | In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | 1,000 | [
{
"input": "2\n1 2\n1\n1",
"output": "1 2"
},
{
"input": "2\n2 1\n1\n1",
"output": "2 1"
},
{
"input": "3\n3 1 2\n3\n1 2 3",
"output": "6 6"
},
{
"input": "9\n2 9 3 1 6 4 7 8 5\n9\n5 1 5 2 8 4 4 4 5",
"output": "58 32"
},
{
"input": "10\n3 10 9 2 7 6 5 8 4 1\n1\n4... | 1,633,615,225 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 62 | 6,963,200 | n=int(input())
arr=list(map(int,input().split()))
m=int(input())
q=list(map(int,input().split()))
v=0
p=0
s=set(q)
for i in s:
k=arr.index(i)
c=arr.count(i)
v+=c*(k+1)
p+=c*(n-k)
print(str(v)+' '+str(p)) | Title: Effective Approach
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
Output Specification:
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Demo Input:
['2\n1 2\n1\n1\n', '2\n2 1\n1\n1\n', '3\n3 1 2\n3\n1 2 3\n']
Demo Output:
['1 2\n', '2 1\n', '6 6\n']
Note:
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | ```python
n=int(input())
arr=list(map(int,input().split()))
m=int(input())
q=list(map(int,input().split()))
v=0
p=0
s=set(q)
for i in s:
k=arr.index(i)
c=arr.count(i)
v+=c*(k+1)
p+=c*(n-k)
print(str(v)+' '+str(p))
``` | 0 | |
629 | A | Far Relative’s Birthday Cake | PROGRAMMING | 800 | [
"brute force",
"combinatorics",
"constructive algorithms",
"implementation"
] | null | null | Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird!
The cake is a *n*<=×<=*n* square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be?
Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column. | In the first line of the input, you are given a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the side of the cake.
Then follow *n* lines, each containing *n* characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'. | Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column. | [
"3\n.CC\nC..\nC.C\n",
"4\nCC..\nC..C\n.CC.\n.CC.\n"
] | [
"4\n",
"9\n"
] | If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are:
1. (1, 2) and (1, 3) 1. (3, 1) and (3, 3) 1. (2, 1) and (3, 1) 1. (1, 3) and (3, 3) | 500 | [
{
"input": "3\n.CC\nC..\nC.C",
"output": "4"
},
{
"input": "4\nCC..\nC..C\n.CC.\n.CC.",
"output": "9"
},
{
"input": "5\n.CCCC\nCCCCC\n.CCC.\nCC...\n.CC.C",
"output": "46"
},
{
"input": "7\n.CC..CC\nCC.C..C\nC.C..C.\nC...C.C\nCCC.CCC\n.CC...C\n.C.CCC.",
"output": "84"
},... | 1,699,280,244 | 2,147,483,647 | Python 3 | OK | TESTS | 48 | 46 | 0 | import math
n = int(input())
a = []
for i in range(n):
x = input()
a.append(x)
# row count
r = 0
for i in range(n):
x = a[i].count('C')
r+=math.comb(x,2)
# column count
c = 0
for i in range(n):
x = 0
for j in range(n):
if a[j][i]=='C':
x+=1
c+=math.comb(x,2)
print(r+c)
| Title: Far Relative’s Birthday Cake
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird!
The cake is a *n*<=×<=*n* square consisting of equal squares with side length 1. Each square is either empty or consists of a single chocolate. They bought the cake and randomly started to put the chocolates on the cake. The value of Famil Door's happiness will be equal to the number of pairs of cells with chocolates that are in the same row or in the same column of the cake. Famil Doors's family is wondering what is the amount of happiness of Famil going to be?
Please, note that any pair can be counted no more than once, as two different cells can't share both the same row and the same column.
Input Specification:
In the first line of the input, you are given a single integer *n* (1<=≤<=*n*<=≤<=100) — the length of the side of the cake.
Then follow *n* lines, each containing *n* characters. Empty cells are denoted with '.', while cells that contain chocolates are denoted by 'C'.
Output Specification:
Print the value of Famil Door's happiness, i.e. the number of pairs of chocolate pieces that share the same row or the same column.
Demo Input:
['3\n.CC\nC..\nC.C\n', '4\nCC..\nC..C\n.CC.\n.CC.\n']
Demo Output:
['4\n', '9\n']
Note:
If we number rows from top to bottom and columns from left to right, then, pieces that share the same row in the first sample are:
1. (1, 2) and (1, 3) 1. (3, 1) and (3, 3) 1. (2, 1) and (3, 1) 1. (1, 3) and (3, 3) | ```python
import math
n = int(input())
a = []
for i in range(n):
x = input()
a.append(x)
# row count
r = 0
for i in range(n):
x = a[i].count('C')
r+=math.comb(x,2)
# column count
c = 0
for i in range(n):
x = 0
for j in range(n):
if a[j][i]=='C':
x+=1
c+=math.comb(x,2)
print(r+c)
``` | 3 | |
216 | B | Forming Teams | PROGRAMMING | 1,700 | [
"dfs and similar",
"implementation"
] | null | null | One day *n* students come to the stadium. They want to play football, and for that they need to split into teams, the teams must have an equal number of people.
We know that this group of people has archenemies. Each student has at most two archenemies. Besides, if student *A* is an archenemy to student *B*, then student *B* is an archenemy to student *A*.
The students want to split so as no two archenemies were in one team. If splitting in the required manner is impossible, some students will have to sit on the bench.
Determine the minimum number of students you will have to send to the bench in order to form the two teams in the described manner and begin the game at last. | The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=100, 1<=≤<=*m*<=≤<=100) — the number of students and the number of pairs of archenemies correspondingly.
Next *m* lines describe enmity between students. Each enmity is described as two numbers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*) — the indexes of the students who are enemies to each other. Each enmity occurs in the list exactly once. It is guaranteed that each student has no more than two archenemies.
You can consider the students indexed in some manner with distinct integers from 1 to *n*. | Print a single integer — the minimum number of students you will have to send to the bench in order to start the game. | [
"5 4\n1 2\n2 4\n5 3\n1 4\n",
"6 2\n1 4\n3 4\n",
"6 6\n1 2\n2 3\n3 1\n4 5\n5 6\n6 4\n"
] | [
"1",
"0",
"2"
] | none | 1,500 | [
{
"input": "5 4\n1 2\n2 4\n5 3\n1 4",
"output": "1"
},
{
"input": "6 2\n1 4\n3 4",
"output": "0"
},
{
"input": "6 6\n1 2\n2 3\n3 1\n4 5\n5 6\n6 4",
"output": "2"
},
{
"input": "5 1\n1 2",
"output": "1"
},
{
"input": "8 8\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 1",
... | 1,538,063,665 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 23 | 278 | 0 | n,m = map(int,input().split())
L,b,c = [[] for _ in ' '*(n+1)],[1 for _ in ' '*(n+1)],0
for _ in ' '*m:
i,j = map(int,input().split());L[i].append(j);L[j].append(i)
for i in range(1,n+1):
if b[i] ==1:
for j in L[i]:
if b[j] == 1:
for k in L[i]:
if k in L[j]:b[i] = b[j] = b[k] = 0 ; c+=1
print(c if (n-c)%2 == 0 else c+1) | Title: Forming Teams
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day *n* students come to the stadium. They want to play football, and for that they need to split into teams, the teams must have an equal number of people.
We know that this group of people has archenemies. Each student has at most two archenemies. Besides, if student *A* is an archenemy to student *B*, then student *B* is an archenemy to student *A*.
The students want to split so as no two archenemies were in one team. If splitting in the required manner is impossible, some students will have to sit on the bench.
Determine the minimum number of students you will have to send to the bench in order to form the two teams in the described manner and begin the game at last.
Input Specification:
The first line contains two integers *n* and *m* (2<=≤<=*n*<=≤<=100, 1<=≤<=*m*<=≤<=100) — the number of students and the number of pairs of archenemies correspondingly.
Next *m* lines describe enmity between students. Each enmity is described as two numbers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*) — the indexes of the students who are enemies to each other. Each enmity occurs in the list exactly once. It is guaranteed that each student has no more than two archenemies.
You can consider the students indexed in some manner with distinct integers from 1 to *n*.
Output Specification:
Print a single integer — the minimum number of students you will have to send to the bench in order to start the game.
Demo Input:
['5 4\n1 2\n2 4\n5 3\n1 4\n', '6 2\n1 4\n3 4\n', '6 6\n1 2\n2 3\n3 1\n4 5\n5 6\n6 4\n']
Demo Output:
['1', '0', '2']
Note:
none | ```python
n,m = map(int,input().split())
L,b,c = [[] for _ in ' '*(n+1)],[1 for _ in ' '*(n+1)],0
for _ in ' '*m:
i,j = map(int,input().split());L[i].append(j);L[j].append(i)
for i in range(1,n+1):
if b[i] ==1:
for j in L[i]:
if b[j] == 1:
for k in L[i]:
if k in L[j]:b[i] = b[j] = b[k] = 0 ; c+=1
print(c if (n-c)%2 == 0 else c+1)
``` | 0 | |
8 | C | Looking for Order | PROGRAMMING | 2,000 | [
"bitmasks",
"dp"
] | C. Looking for Order | 4 | 512 | Girl Lena likes it when everything is in order, and looks for order everywhere. Once she was getting ready for the University and noticed that the room was in a mess — all the objects from her handbag were thrown about the room. Of course, she wanted to put them back into her handbag. The problem is that the girl cannot carry more than two objects at a time, and cannot move the handbag. Also, if he has taken an object, she cannot put it anywhere except her handbag — her inherent sense of order does not let her do so.
You are given the coordinates of the handbag and the coordinates of the objects in some Сartesian coordinate system. It is known that the girl covers the distance between any two objects in the time equal to the squared length of the segment between the points of the objects. It is also known that initially the coordinates of the girl and the handbag are the same. You are asked to find such an order of actions, that the girl can put all the objects back into her handbag in a minimum time period. | The first line of the input file contains the handbag's coordinates *x**s*,<=*y**s*. The second line contains number *n* (1<=≤<=*n*<=≤<=24) — the amount of objects the girl has. The following *n* lines contain the objects' coordinates. All the coordinates do not exceed 100 in absolute value. All the given positions are different. All the numbers are integer. | In the first line output the only number — the minimum time the girl needs to put the objects into her handbag.
In the second line output the possible optimum way for Lena. Each object in the input is described by its index number (from 1 to *n*), the handbag's point is described by number 0. The path should start and end in the handbag's point. If there are several optimal paths, print any of them. | [
"0 0\n2\n1 1\n-1 1\n",
"1 1\n3\n4 3\n3 4\n0 0\n"
] | [
"8\n0 1 2 0 \n",
"32\n0 1 2 0 3 0 \n"
] | none | 0 | [
{
"input": "0 0\n2\n1 1\n-1 1",
"output": "8\n0 1 2 0 "
},
{
"input": "1 1\n3\n4 3\n3 4\n0 0",
"output": "32\n0 1 2 0 3 0 "
},
{
"input": "-3 4\n1\n2 2",
"output": "58\n0 1 0 "
},
{
"input": "7 -7\n2\n3 1\n-3 8",
"output": "490\n0 1 2 0 "
},
{
"input": "3 -9\n3\n0... | 1,464,294,231 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 46 | 5,120,000 | #!/usr/bin/env python
"""
CodeForces: 8C. Looking for Order
Strategy
The optimum way to pick up the pair of objects is:
dd = distance between two objects
d1 = distance between the handbag and one of object
d2 = distance between the handbag and other object
if dd < d1 + d2:
pick up the two objects at once
optimum time = d1 + dd + d2
else:
pick up the two objects one by one
optimum time = (d1 + d2) * 2
The optimum pair can be calculated by:
optimum pair = min(min(dd, d1 + d2) for all possible pairs)
"""
def dist2(dx, dy):
return (dx * dx) + (dy * dy)
def main():
x0, y0 = map(int, input().split())
n = int(input())
# pre-calculate the optimum way for all possible pairs
xyd = []
pair = []
for o1 in range(n):
x1, y1 = map(int, input().split())
d1 = dist2(x1 - x0, y1 - y0)
xyd.append((x1, y1, d1))
other = []
for o2 in range(o1):
x2, y2, d2 = xyd[o2]
d0 = d1 + d2
dd = dist2(x2 -x1, y2 - y1)
other.append((min(dd, d0), o2, d0))
other.sort(reverse=True)
pair.append(other)
# calculate the optimum order by finding the optimum (n/2) pairs
order = [0]
total = 0
for _ in range(n // 2):
best_d = 1000000
best_o = 0
for o1 in range(1, n):
if xyd[o1] is None:
continue
other = pair[o1]
while other:
dd, o2, _ = other[-1]
if xyd[o2] is not None:
if dd < best_d:
best_o = o1
best_d = dd
break
other.pop()
_, o2, d0 = pair[best_o].pop()
total += best_d + d0
order.append(o2 + 1)
if best_d == d0:
order.append(0)
order.append(best_o + 1)
order.append(0)
xyd[best_o] = xyd[o2] = None
# pick up the last object
if n & 1:
for o1 in range(n):
if xyd[o1] is not None:
total += xyd[o1][2] * 2
order.append(o1 + 1)
order.append(0)
break
print(total)
print(" ".join(map(str, order)))
if __name__ == '__main__':
main()
| Title: Looking for Order
Time Limit: 4 seconds
Memory Limit: 512 megabytes
Problem Description:
Girl Lena likes it when everything is in order, and looks for order everywhere. Once she was getting ready for the University and noticed that the room was in a mess — all the objects from her handbag were thrown about the room. Of course, she wanted to put them back into her handbag. The problem is that the girl cannot carry more than two objects at a time, and cannot move the handbag. Also, if he has taken an object, she cannot put it anywhere except her handbag — her inherent sense of order does not let her do so.
You are given the coordinates of the handbag and the coordinates of the objects in some Сartesian coordinate system. It is known that the girl covers the distance between any two objects in the time equal to the squared length of the segment between the points of the objects. It is also known that initially the coordinates of the girl and the handbag are the same. You are asked to find such an order of actions, that the girl can put all the objects back into her handbag in a minimum time period.
Input Specification:
The first line of the input file contains the handbag's coordinates *x**s*,<=*y**s*. The second line contains number *n* (1<=≤<=*n*<=≤<=24) — the amount of objects the girl has. The following *n* lines contain the objects' coordinates. All the coordinates do not exceed 100 in absolute value. All the given positions are different. All the numbers are integer.
Output Specification:
In the first line output the only number — the minimum time the girl needs to put the objects into her handbag.
In the second line output the possible optimum way for Lena. Each object in the input is described by its index number (from 1 to *n*), the handbag's point is described by number 0. The path should start and end in the handbag's point. If there are several optimal paths, print any of them.
Demo Input:
['0 0\n2\n1 1\n-1 1\n', '1 1\n3\n4 3\n3 4\n0 0\n']
Demo Output:
['8\n0 1 2 0 \n', '32\n0 1 2 0 3 0 \n']
Note:
none | ```python
#!/usr/bin/env python
"""
CodeForces: 8C. Looking for Order
Strategy
The optimum way to pick up the pair of objects is:
dd = distance between two objects
d1 = distance between the handbag and one of object
d2 = distance between the handbag and other object
if dd < d1 + d2:
pick up the two objects at once
optimum time = d1 + dd + d2
else:
pick up the two objects one by one
optimum time = (d1 + d2) * 2
The optimum pair can be calculated by:
optimum pair = min(min(dd, d1 + d2) for all possible pairs)
"""
def dist2(dx, dy):
return (dx * dx) + (dy * dy)
def main():
x0, y0 = map(int, input().split())
n = int(input())
# pre-calculate the optimum way for all possible pairs
xyd = []
pair = []
for o1 in range(n):
x1, y1 = map(int, input().split())
d1 = dist2(x1 - x0, y1 - y0)
xyd.append((x1, y1, d1))
other = []
for o2 in range(o1):
x2, y2, d2 = xyd[o2]
d0 = d1 + d2
dd = dist2(x2 -x1, y2 - y1)
other.append((min(dd, d0), o2, d0))
other.sort(reverse=True)
pair.append(other)
# calculate the optimum order by finding the optimum (n/2) pairs
order = [0]
total = 0
for _ in range(n // 2):
best_d = 1000000
best_o = 0
for o1 in range(1, n):
if xyd[o1] is None:
continue
other = pair[o1]
while other:
dd, o2, _ = other[-1]
if xyd[o2] is not None:
if dd < best_d:
best_o = o1
best_d = dd
break
other.pop()
_, o2, d0 = pair[best_o].pop()
total += best_d + d0
order.append(o2 + 1)
if best_d == d0:
order.append(0)
order.append(best_o + 1)
order.append(0)
xyd[best_o] = xyd[o2] = None
# pick up the last object
if n & 1:
for o1 in range(n):
if xyd[o1] is not None:
total += xyd[o1][2] * 2
order.append(o1 + 1)
order.append(0)
break
print(total)
print(" ".join(map(str, order)))
if __name__ == '__main__':
main()
``` | 0 |
725 | A | Jumping Ball | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | In a new version of the famous Pinball game, one of the most important parts of the game field is a sequence of *n* bumpers. The bumpers are numbered with integers from 1 to *n* from left to right. There are two types of bumpers. They are denoted by the characters '<' and '>'. When the ball hits the bumper at position *i* it goes one position to the right (to the position *i*<=+<=1) if the type of this bumper is '>', or one position to the left (to *i*<=-<=1) if the type of the bumper at position *i* is '<'. If there is no such position, in other words if *i*<=-<=1<=<<=1 or *i*<=+<=1<=><=*n*, the ball falls from the game field.
Depending on the ball's starting position, the ball may eventually fall from the game field or it may stay there forever. You are given a string representing the bumpers' types. Calculate the number of positions such that the ball will eventually fall from the game field if it starts at that position. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the sequence of bumpers. The second line contains the string, which consists of the characters '<' and '>'. The character at the *i*-th position of this string corresponds to the type of the *i*-th bumper. | Print one integer — the number of positions in the sequence such that the ball will eventually fall from the game field if it starts at that position. | [
"4\n<<><\n",
"5\n>>>>>\n",
"4\n>><<\n"
] | [
"2",
"5",
"0"
] | In the first sample, the ball will fall from the field if starts at position 1 or position 2.
In the second sample, any starting position will result in the ball falling from the field. | 500 | [
{
"input": "4\n<<><",
"output": "2"
},
{
"input": "5\n>>>>>",
"output": "5"
},
{
"input": "4\n>><<",
"output": "0"
},
{
"input": "3\n<<>",
"output": "3"
},
{
"input": "3\n<<<",
"output": "3"
},
{
"input": "3\n><<",
"output": "0"
},
{
"input... | 1,630,402,203 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 4 | 62 | 7,168,000 | n=int(input())
s=input()
a=s.count('<')
b=s.count('>')
print(abs(a-b)) | Title: Jumping Ball
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a new version of the famous Pinball game, one of the most important parts of the game field is a sequence of *n* bumpers. The bumpers are numbered with integers from 1 to *n* from left to right. There are two types of bumpers. They are denoted by the characters '<' and '>'. When the ball hits the bumper at position *i* it goes one position to the right (to the position *i*<=+<=1) if the type of this bumper is '>', or one position to the left (to *i*<=-<=1) if the type of the bumper at position *i* is '<'. If there is no such position, in other words if *i*<=-<=1<=<<=1 or *i*<=+<=1<=><=*n*, the ball falls from the game field.
Depending on the ball's starting position, the ball may eventually fall from the game field or it may stay there forever. You are given a string representing the bumpers' types. Calculate the number of positions such that the ball will eventually fall from the game field if it starts at that position.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the length of the sequence of bumpers. The second line contains the string, which consists of the characters '<' and '>'. The character at the *i*-th position of this string corresponds to the type of the *i*-th bumper.
Output Specification:
Print one integer — the number of positions in the sequence such that the ball will eventually fall from the game field if it starts at that position.
Demo Input:
['4\n<<><\n', '5\n>>>>>\n', '4\n>><<\n']
Demo Output:
['2', '5', '0']
Note:
In the first sample, the ball will fall from the field if starts at position 1 or position 2.
In the second sample, any starting position will result in the ball falling from the field. | ```python
n=int(input())
s=input()
a=s.count('<')
b=s.count('>')
print(abs(a-b))
``` | 0 | |
621 | A | Wet Shark and Odd and Even | PROGRAMMING | 900 | [
"implementation"
] | null | null | Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.
Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0. | The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive. | Print the maximum possible even sum that can be obtained if we use some of the given integers. | [
"3\n1 2 3\n",
"5\n999999999 999999999 999999999 999999999 999999999\n"
] | [
"6",
"3999999996"
] | In the first sample, we can simply take all three integers for a total sum of 6.
In the second sample Wet Shark should take any four out of five integers 999 999 999. | 500 | [
{
"input": "3\n1 2 3",
"output": "6"
},
{
"input": "5\n999999999 999999999 999999999 999999999 999999999",
"output": "3999999996"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "15\n39 52 88 78 46 95 84 98 55 3 68 42 6 18 98",
"output": "870"
},
{
"input": "15\... | 1,602,430,491 | 2,147,483,647 | PyPy 3 | OK | TESTS | 87 | 233 | 12,083,200 | n=int(input())
a=list(map(int,input().split()))
s=sum(a)
if s%2==0:
print(s)
else:
odd=[]
for i in range(n):
if a[i]%2!=0:
odd.append(a[i])
x=min(odd)
s-=x
print(s) | Title: Wet Shark and Odd and Even
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today, Wet Shark is given *n* integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark.
Note, that if Wet Shark uses no integers from the *n* integers, the sum is an even integer 0.
Input Specification:
The first line of the input contains one integer, *n* (1<=≤<=*n*<=≤<=100<=000). The next line contains *n* space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive.
Output Specification:
Print the maximum possible even sum that can be obtained if we use some of the given integers.
Demo Input:
['3\n1 2 3\n', '5\n999999999 999999999 999999999 999999999 999999999\n']
Demo Output:
['6', '3999999996']
Note:
In the first sample, we can simply take all three integers for a total sum of 6.
In the second sample Wet Shark should take any four out of five integers 999 999 999. | ```python
n=int(input())
a=list(map(int,input().split()))
s=sum(a)
if s%2==0:
print(s)
else:
odd=[]
for i in range(n):
if a[i]%2!=0:
odd.append(a[i])
x=min(odd)
s-=x
print(s)
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Eighth-grader Vova is on duty today in the class. After classes, he went into the office to wash the board, and found on it the number *n*. He asked what is this number and the teacher of mathematics Inna Petrovna answered Vova that *n* is the answer to the arithmetic task for first-graders. In the textbook, a certain positive integer *x* was given. The task was to add *x* to the sum of the digits of the number *x* written in decimal numeral system.
Since the number *n* on the board was small, Vova quickly guessed which *x* could be in the textbook. Now he wants to get a program which will search for arbitrary values of the number *n* for all suitable values of *x* or determine that such *x* does not exist. Write such a program for Vova. | The first line contains integer *n* (1<=≤<=*n*<=≤<=109). | In the first line print one integer *k* — number of different values of *x* satisfying the condition.
In next *k* lines print these values in ascending order. | [
"21\n",
"20\n"
] | [
"1\n15\n",
"0\n"
] | In the first test case *x* = 15 there is only one variant: 15 + 1 + 5 = 21.
In the second test case there are no such *x*. | 0 | [
{
"input": "21",
"output": "1\n15"
},
{
"input": "20",
"output": "0"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "1\n1"
},
{
"input": "3",
"output": "0"
},
{
"input": "100000001",
"output": "2\n99999937\n100000000"
},
{
"i... | 1,508,216,445 | 2,147,483,647 | Python 3 | OK | TESTS | 18 | 61 | 5,529,600 | n = int( input() )
lim = max( 0 , n - 100 )
def cal( n ):
ret = n
while True :
if n == 0:
break
ret = ret + ( n %10 )
n = n // 10
return ret
ans = []
for i in range(lim,n):
if cal(i) == n:
ans.append(i)
print ( len(ans) )
print (*ans)
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Eighth-grader Vova is on duty today in the class. After classes, he went into the office to wash the board, and found on it the number *n*. He asked what is this number and the teacher of mathematics Inna Petrovna answered Vova that *n* is the answer to the arithmetic task for first-graders. In the textbook, a certain positive integer *x* was given. The task was to add *x* to the sum of the digits of the number *x* written in decimal numeral system.
Since the number *n* on the board was small, Vova quickly guessed which *x* could be in the textbook. Now he wants to get a program which will search for arbitrary values of the number *n* for all suitable values of *x* or determine that such *x* does not exist. Write such a program for Vova.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
In the first line print one integer *k* — number of different values of *x* satisfying the condition.
In next *k* lines print these values in ascending order.
Demo Input:
['21\n', '20\n']
Demo Output:
['1\n15\n', '0\n']
Note:
In the first test case *x* = 15 there is only one variant: 15 + 1 + 5 = 21.
In the second test case there are no such *x*. | ```python
n = int( input() )
lim = max( 0 , n - 100 )
def cal( n ):
ret = n
while True :
if n == 0:
break
ret = ret + ( n %10 )
n = n // 10
return ret
ans = []
for i in range(lim,n):
if cal(i) == n:
ans.append(i)
print ( len(ans) )
print (*ans)
``` | 3 | |
348 | A | Mafia | PROGRAMMING | 1,600 | [
"binary search",
"math",
"sortings"
] | null | null | One day *n* friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and other *n*<=-<=1 people take part in the game. For each person we know in how many rounds he wants to be a player, not the supervisor: the *i*-th person wants to play *a**i* rounds. What is the minimum number of rounds of the "Mafia" game they need to play to let each person play at least as many rounds as they want? | The first line contains integer *n* (3<=≤<=*n*<=≤<=105). The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the *i*-th number in the list is the number of rounds the *i*-th person wants to play. | In a single line print a single integer — the minimum number of game rounds the friends need to let the *i*-th person play at least *a**i* rounds.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. | [
"3\n3 2 2\n",
"4\n2 2 2 2\n"
] | [
"4\n",
"3\n"
] | You don't need to know the rules of "Mafia" to solve this problem. If you're curious, it's a game Russia got from the Soviet times: http://en.wikipedia.org/wiki/Mafia_(party_game). | 500 | [
{
"input": "3\n3 2 2",
"output": "4"
},
{
"input": "4\n2 2 2 2",
"output": "3"
},
{
"input": "7\n9 7 7 8 8 7 8",
"output": "9"
},
{
"input": "10\n13 12 10 13 13 14 10 10 12 12",
"output": "14"
},
{
"input": "10\n94 96 91 95 99 94 96 92 95 99",
"output": "106"
... | 1,565,760,334 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 404 | 2,560,000 | from math import *
readints = lambda: map(int, input().split(' '))
n=int(input())
a=last(readints())
x = max(max(a), int(ceil(sum(a)/(n-1))))
print(x) | Title: Mafia
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day *n* friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and other *n*<=-<=1 people take part in the game. For each person we know in how many rounds he wants to be a player, not the supervisor: the *i*-th person wants to play *a**i* rounds. What is the minimum number of rounds of the "Mafia" game they need to play to let each person play at least as many rounds as they want?
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=105). The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the *i*-th number in the list is the number of rounds the *i*-th person wants to play.
Output Specification:
In a single line print a single integer — the minimum number of game rounds the friends need to let the *i*-th person play at least *a**i* rounds.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['3\n3 2 2\n', '4\n2 2 2 2\n']
Demo Output:
['4\n', '3\n']
Note:
You don't need to know the rules of "Mafia" to solve this problem. If you're curious, it's a game Russia got from the Soviet times: http://en.wikipedia.org/wiki/Mafia_(party_game). | ```python
from math import *
readints = lambda: map(int, input().split(' '))
n=int(input())
a=last(readints())
x = max(max(a), int(ceil(sum(a)/(n-1))))
print(x)
``` | -1 | |
75 | A | Life Without Zeros | PROGRAMMING | 1,000 | [
"implementation"
] | A. Life Without Zeros | 2 | 256 | Can you imagine our life if we removed all zeros from it? For sure we will have many problems.
In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?
For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation.
But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation. | The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*. | The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise. | [
"101\n102\n",
"105\n106\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "101\n102",
"output": "YES"
},
{
"input": "105\n106",
"output": "NO"
},
{
"input": "544\n397",
"output": "YES"
},
{
"input": "822\n280",
"output": "NO"
},
{
"input": "101\n413",
"output": "NO"
},
{
"input": "309\n139",
"output": "NO"
}... | 1,552,053,732 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 124 | 0 | def delete0(x):
sum=0
while(x>0):
if(x%10!=0):
sum=sum*10+x%10
x//=10
return sum
def revesre(x):
sum=0
while(x>0):
sum=sum*10+x%10
x//=10
return sum
a,b=map(int,input().split())
c=a+b
a=delete0(a)
a=revesre(a)
b=delete0(b)
b=revesre(b)
c=delete0(c)
c=revesre(c)
if(a+b==c):
print("YES")
else:
print("NO") | Title: Life Without Zeros
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Can you imagine our life if we removed all zeros from it? For sure we will have many problems.
In this problem we will have a simple example if we removed all zeros from our life, it's the addition operation. Let's assume you are given this equation *a*<=+<=*b*<==<=*c*, where *a* and *b* are positive integers, and *c* is the sum of *a* and *b*. Now let's remove all zeros from this equation. Will the equation remain correct after removing all zeros?
For example if the equation is 101<=+<=102<==<=203, if we removed all zeros it will be 11<=+<=12<==<=23 which is still a correct equation.
But if the equation is 105<=+<=106<==<=211, if we removed all zeros it will be 15<=+<=16<==<=211 which is not a correct equation.
Input Specification:
The input will consist of two lines, the first line will contain the integer *a*, and the second line will contain the integer *b* which are in the equation as described above (1<=≤<=*a*,<=*b*<=≤<=109). There won't be any leading zeros in both. The value of *c* should be calculated as *c*<==<=*a*<=+<=*b*.
Output Specification:
The output will be just one line, you should print "YES" if the equation will remain correct after removing all zeros, and print "NO" otherwise.
Demo Input:
['101\n102\n', '105\n106\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
def delete0(x):
sum=0
while(x>0):
if(x%10!=0):
sum=sum*10+x%10
x//=10
return sum
def revesre(x):
sum=0
while(x>0):
sum=sum*10+x%10
x//=10
return sum
a,b=map(int,input().split())
c=a+b
a=delete0(a)
a=revesre(a)
b=delete0(b)
b=revesre(b)
c=delete0(c)
c=revesre(c)
if(a+b==c):
print("YES")
else:
print("NO")
``` | -1 |
361 | A | Levko and Table | PROGRAMMING | 800 | [
"constructive algorithms",
"implementation"
] | null | null | Levko loves tables that consist of *n* rows and *n* columns very much. He especially loves beautiful tables. A table is beautiful to Levko if the sum of elements in each row and column of the table equals *k*.
Unfortunately, he doesn't know any such table. Your task is to help him to find at least one of them. | The single line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=1000). | Print any beautiful table. Levko doesn't like too big numbers, so all elements of the table mustn't exceed 1000 in their absolute value.
If there are multiple suitable tables, you are allowed to print any of them. | [
"2 4\n",
"4 7\n"
] | [
"1 3\n3 1\n",
"2 1 0 4\n4 0 2 1\n1 3 3 0\n0 3 2 2\n"
] | In the first sample the sum in the first row is 1 + 3 = 4, in the second row — 3 + 1 = 4, in the first column — 1 + 3 = 4 and in the second column — 3 + 1 = 4. There are other beautiful tables for this sample.
In the second sample the sum of elements in each row and each column equals 7. Besides, there are other tables that meet the statement requirements. | 500 | [
{
"input": "2 4",
"output": "4 0 \n0 4 "
},
{
"input": "4 7",
"output": "7 0 0 0 \n0 7 0 0 \n0 0 7 0 \n0 0 0 7 "
},
{
"input": "1 8",
"output": "8 "
},
{
"input": "9 3",
"output": "3 0 0 0 0 0 0 0 0 \n0 3 0 0 0 0 0 0 0 \n0 0 3 0 0 0 0 0 0 \n0 0 0 3 0 0 0 0 0 \n0 0 0 0 3 0... | 1,595,519,946 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | n, k = map(int, input())
a = []
for i range(n-1):
a.append(n//k)
a.append(k-sum(a))
for i range(n):
print(a) | Title: Levko and Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Levko loves tables that consist of *n* rows and *n* columns very much. He especially loves beautiful tables. A table is beautiful to Levko if the sum of elements in each row and column of the table equals *k*.
Unfortunately, he doesn't know any such table. Your task is to help him to find at least one of them.
Input Specification:
The single line contains two integers, *n* and *k* (1<=≤<=*n*<=≤<=100, 1<=≤<=*k*<=≤<=1000).
Output Specification:
Print any beautiful table. Levko doesn't like too big numbers, so all elements of the table mustn't exceed 1000 in their absolute value.
If there are multiple suitable tables, you are allowed to print any of them.
Demo Input:
['2 4\n', '4 7\n']
Demo Output:
['1 3\n3 1\n', '2 1 0 4\n4 0 2 1\n1 3 3 0\n0 3 2 2\n']
Note:
In the first sample the sum in the first row is 1 + 3 = 4, in the second row — 3 + 1 = 4, in the first column — 1 + 3 = 4 and in the second column — 3 + 1 = 4. There are other beautiful tables for this sample.
In the second sample the sum of elements in each row and each column equals 7. Besides, there are other tables that meet the statement requirements. | ```python
n, k = map(int, input())
a = []
for i range(n-1):
a.append(n//k)
a.append(k-sum(a))
for i range(n):
print(a)
``` | -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... | 1,672,810,310 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 122 | 0 | st = input()
u=0
l = 0
for i in st:
if i == i.upper():
u+=1
else:
l+=1
if u>l:
print(st.upper())
else:
print(st.lower()) | 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
st = input()
u=0
l = 0
for i in st:
if i == i.upper():
u+=1
else:
l+=1
if u>l:
print(st.upper())
else:
print(st.lower())
``` | 3.9695 |
719 | B | Anatoly and Cockroaches | PROGRAMMING | 1,400 | [
"greedy"
] | null | null | Anatoly lives in the university dorm as many other students do. As you know, cockroaches are also living there together with students. Cockroaches might be of two colors: black and red. There are *n* cockroaches living in Anatoly's room.
Anatoly just made all his cockroaches to form a single line. As he is a perfectionist, he would like the colors of cockroaches in the line to alternate. He has a can of black paint and a can of red paint. In one turn he can either swap any two cockroaches, or take any single cockroach and change it's color.
Help Anatoly find out the minimum number of turns he needs to make the colors of cockroaches in the line alternate. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of cockroaches.
The second line contains a string of length *n*, consisting of characters 'b' and 'r' that denote black cockroach and red cockroach respectively. | Print one integer — the minimum number of moves Anatoly has to perform in order to make the colors of cockroaches in the line to alternate. | [
"5\nrbbrr\n",
"5\nbbbbb\n",
"3\nrbr\n"
] | [
"1\n",
"2\n",
"0\n"
] | In the first sample, Anatoly has to swap third and fourth cockroaches. He needs 1 turn to do this.
In the second sample, the optimum answer is to paint the second and the fourth cockroaches red. This requires 2 turns.
In the third sample, the colors of cockroaches in the line are alternating already, thus the answer is 0. | 1,000 | [
{
"input": "5\nrbbrr",
"output": "1"
},
{
"input": "5\nbbbbb",
"output": "2"
},
{
"input": "3\nrbr",
"output": "0"
},
{
"input": "13\nrbbbrbrrbrrbb",
"output": "3"
},
{
"input": "18\nrrrrrrrrrrrrrrrrrb",
"output": "8"
},
{
"input": "100\nbrbbbrrrbbrbrb... | 1,608,507,893 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 93 | 0 | a=int(input())
s=input()
cnt=1
ans=[]
gam=[]
for i in range(1,len(ans)):
if(ans[i]==ans[i-1]):
cnt+=1
else:
ans.append(cnt)
cnt=1
ans.append(cnt)
for i in range(len(ans)):
if(i%2==0):
ans[i]=-ans[i]//2
else:
ans[i]=ans[i]//2
print(abs(sum(ans)))
| Title: Anatoly and Cockroaches
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Anatoly lives in the university dorm as many other students do. As you know, cockroaches are also living there together with students. Cockroaches might be of two colors: black and red. There are *n* cockroaches living in Anatoly's room.
Anatoly just made all his cockroaches to form a single line. As he is a perfectionist, he would like the colors of cockroaches in the line to alternate. He has a can of black paint and a can of red paint. In one turn he can either swap any two cockroaches, or take any single cockroach and change it's color.
Help Anatoly find out the minimum number of turns he needs to make the colors of cockroaches in the line alternate.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of cockroaches.
The second line contains a string of length *n*, consisting of characters 'b' and 'r' that denote black cockroach and red cockroach respectively.
Output Specification:
Print one integer — the minimum number of moves Anatoly has to perform in order to make the colors of cockroaches in the line to alternate.
Demo Input:
['5\nrbbrr\n', '5\nbbbbb\n', '3\nrbr\n']
Demo Output:
['1\n', '2\n', '0\n']
Note:
In the first sample, Anatoly has to swap third and fourth cockroaches. He needs 1 turn to do this.
In the second sample, the optimum answer is to paint the second and the fourth cockroaches red. This requires 2 turns.
In the third sample, the colors of cockroaches in the line are alternating already, thus the answer is 0. | ```python
a=int(input())
s=input()
cnt=1
ans=[]
gam=[]
for i in range(1,len(ans)):
if(ans[i]==ans[i-1]):
cnt+=1
else:
ans.append(cnt)
cnt=1
ans.append(cnt)
for i in range(len(ans)):
if(i%2==0):
ans[i]=-ans[i]//2
else:
ans[i]=ans[i]//2
print(abs(sum(ans)))
``` | 0 | |
546 | A | Soldier and Bananas | PROGRAMMING | 800 | [
"brute force",
"implementation",
"math"
] | null | null | A soldier wants to buy *w* bananas in the shop. He has to pay *k* dollars for the first banana, 2*k* dollars for the second one and so on (in other words, he has to pay *i*·*k* dollars for the *i*-th banana).
He has *n* dollars. How many dollars does he have to borrow from his friend soldier to buy *w* bananas? | The first line contains three positive integers *k*,<=*n*,<=*w* (1<=<=≤<=<=*k*,<=*w*<=<=≤<=<=1000, 0<=≤<=*n*<=≤<=109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants. | Output one integer — the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0. | [
"3 17 4\n"
] | [
"13"
] | none | 500 | [
{
"input": "3 17 4",
"output": "13"
},
{
"input": "1 2 1",
"output": "0"
},
{
"input": "1 1 1",
"output": "0"
},
{
"input": "1 5 6",
"output": "16"
},
{
"input": "1 1000000000 1",
"output": "0"
},
{
"input": "1000 0 1000",
"output": "500500000"
}... | 1,697,311,711 | 2,147,483,647 | PyPy 3 | OK | TESTS | 15 | 92 | 0 | k,n,w=map(lambda x: int(x), input().split())
temp=k*(1+w)*w//2-n
print(temp if temp>0 else 0)
| Title: Soldier and Bananas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A soldier wants to buy *w* bananas in the shop. He has to pay *k* dollars for the first banana, 2*k* dollars for the second one and so on (in other words, he has to pay *i*·*k* dollars for the *i*-th banana).
He has *n* dollars. How many dollars does he have to borrow from his friend soldier to buy *w* bananas?
Input Specification:
The first line contains three positive integers *k*,<=*n*,<=*w* (1<=<=≤<=<=*k*,<=*w*<=<=≤<=<=1000, 0<=≤<=*n*<=≤<=109), the cost of the first banana, initial number of dollars the soldier has and number of bananas he wants.
Output Specification:
Output one integer — the amount of dollars that the soldier must borrow from his friend. If he doesn't have to borrow money, output 0.
Demo Input:
['3 17 4\n']
Demo Output:
['13']
Note:
none | ```python
k,n,w=map(lambda x: int(x), input().split())
temp=k*(1+w)*w//2-n
print(temp if temp>0 else 0)
``` | 3 | |
910 | B | Door Frames | PROGRAMMING | 1,600 | [
"greedy",
"implementation"
] | null | null | Petya has equal wooden bars of length *n*. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length *a* and one top side of length *b*. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar). | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1<=000) — the length of each wooden bar.
The second line contains a single integer *a* (1<=≤<=*a*<=≤<=*n*) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer *b* (1<=≤<=*b*<=≤<=*n*) — the length of the upper side of a door frame. | Print the minimal number of wooden bars with length *n* which are needed to make the frames for two doors. | [
"8\n1\n2\n",
"5\n3\n4\n",
"6\n4\n2\n",
"20\n5\n6\n"
] | [
"1\n",
"6\n",
"4\n",
"2\n"
] | In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed. | 1,000 | [
{
"input": "8\n1\n2",
"output": "1"
},
{
"input": "5\n3\n4",
"output": "6"
},
{
"input": "6\n4\n2",
"output": "4"
},
{
"input": "20\n5\n6",
"output": "2"
},
{
"input": "1\n1\n1",
"output": "6"
},
{
"input": "3\n1\n2",
"output": "3"
},
{
"in... | 1,513,997,378 | 2,378 | Python 3 | WRONG_ANSWER | TESTS | 11 | 62 | 5,529,600 | def solve():
n = int(input())
a = int(input())
b = int(input())
res = 0
ntmp = n
atmp = 4
btmp = 2
while (atmp > 0 or btmp > 0):
if (ntmp >= a and atmp > 0):
ntmp -= a
atmp -= 1
if (ntmp >= b and btmp > 0):
ntmp -= b
btmp -= 1
if ((ntmp < a and atmp > 0) or (ntmp < b and btmp > 0)):
ntmp = n
res += 1
return res+1
print(solve())
| Title: Door Frames
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya has equal wooden bars of length *n*. He wants to make a frame for two equal doors. Each frame has two vertical (left and right) sides of length *a* and one top side of length *b*. A solid (i.e. continuous without breaks) piece of bar is needed for each side.
Determine a minimal number of wooden bars which are needed to make the frames for two doors. Petya can cut the wooden bars into any parts, but each side of each door should be a solid piece of a wooden bar (or a whole wooden bar).
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1<=000) — the length of each wooden bar.
The second line contains a single integer *a* (1<=≤<=*a*<=≤<=*n*) — the length of the vertical (left and right) sides of a door frame.
The third line contains a single integer *b* (1<=≤<=*b*<=≤<=*n*) — the length of the upper side of a door frame.
Output Specification:
Print the minimal number of wooden bars with length *n* which are needed to make the frames for two doors.
Demo Input:
['8\n1\n2\n', '5\n3\n4\n', '6\n4\n2\n', '20\n5\n6\n']
Demo Output:
['1\n', '6\n', '4\n', '2\n']
Note:
In the first example one wooden bar is enough, since the total length of all six sides of the frames for two doors is 8.
In the second example 6 wooden bars is enough, because for each side of the frames the new wooden bar is needed. | ```python
def solve():
n = int(input())
a = int(input())
b = int(input())
res = 0
ntmp = n
atmp = 4
btmp = 2
while (atmp > 0 or btmp > 0):
if (ntmp >= a and atmp > 0):
ntmp -= a
atmp -= 1
if (ntmp >= b and btmp > 0):
ntmp -= b
btmp -= 1
if ((ntmp < a and atmp > 0) or (ntmp < b and btmp > 0)):
ntmp = n
res += 1
return res+1
print(solve())
``` | 0 | |
56 | B | Spoilt Permutation | PROGRAMMING | 1,300 | [
"implementation"
] | B. Spoilt Permutation | 2 | 256 | Vasya collects coins: he has exactly one coin for every year from 1 to *n*. Naturally, Vasya keeps all the coins in his collection in the order in which they were released. Once Vasya's younger brother made a change — he took all the coins whose release year dated from *l* to *r* inclusively and put them in the reverse order. That is, he took a certain segment [*l*,<=*r*] and reversed it. At that the segment's endpoints did not coincide. For example, if *n*<==<=8, then initially Vasya's coins were kept in the order 1 2 3 4 5 6 7 8. If Vasya's younger brother chose the segment [2,<=6], then after the reversal the coin order will change to 1 6 5 4 3 2 7 8. Vasya suspects that someone else could have spoilt the permutation after his brother. Help him to find that out. Check if the given permutation can be obtained from the permutation 1 2 ... *n* using exactly one segment reversal. If it is possible, find the segment itself. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=1000) which is the number of coins in Vasya's collection. The second line contains space-separated *n* integers which are the spoilt sequence of coins. It is guaranteed that the given sequence is a permutation, i.e. it contains only integers from 1 to *n*, and every number is used exactly 1 time. | If it is impossible to obtain the given permutation from the original one in exactly one action, print 0 0. Otherwise, print two numbers *l* *r* (1<=≤<=*l*<=<<=*r*<=≤<=*n*) which are the endpoints of the segment that needs to be reversed to obtain from permutation 1 2 ... *n* the given one. | [
"8\n1 6 5 4 3 2 7 8\n",
"4\n2 3 4 1\n",
"4\n1 2 3 4\n"
] | [
"2 6\n",
"0 0\n",
"0 0\n"
] | none | 1,000 | [
{
"input": "8\n1 6 5 4 3 2 7 8",
"output": "2 6"
},
{
"input": "4\n2 3 4 1",
"output": "0 0"
},
{
"input": "4\n1 2 3 4",
"output": "0 0"
},
{
"input": "8\n1 3 2 4 6 5 7 8",
"output": "0 0"
},
{
"input": "8\n1 3 4 2 6 5 7 8",
"output": "0 0"
},
{
"input... | 1,594,385,643 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 154 | 6,963,200 | n=int(input())
a=list(map(int,input().split()))
s=sorted(a)
l=0
r=0
c=0
for i in range(n-1):
if a[i]>a[i+1]:
l=i if c==0 else l
c=1
else:
if c==1:
r=i+1
break
a1=a[l:r]
a1.reverse()
a=a[:l]+a1+a[r:]
if a==s:
print(l,r)
else:
print(0,0) | Title: Spoilt Permutation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya collects coins: he has exactly one coin for every year from 1 to *n*. Naturally, Vasya keeps all the coins in his collection in the order in which they were released. Once Vasya's younger brother made a change — he took all the coins whose release year dated from *l* to *r* inclusively and put them in the reverse order. That is, he took a certain segment [*l*,<=*r*] and reversed it. At that the segment's endpoints did not coincide. For example, if *n*<==<=8, then initially Vasya's coins were kept in the order 1 2 3 4 5 6 7 8. If Vasya's younger brother chose the segment [2,<=6], then after the reversal the coin order will change to 1 6 5 4 3 2 7 8. Vasya suspects that someone else could have spoilt the permutation after his brother. Help him to find that out. Check if the given permutation can be obtained from the permutation 1 2 ... *n* using exactly one segment reversal. If it is possible, find the segment itself.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=1000) which is the number of coins in Vasya's collection. The second line contains space-separated *n* integers which are the spoilt sequence of coins. It is guaranteed that the given sequence is a permutation, i.e. it contains only integers from 1 to *n*, and every number is used exactly 1 time.
Output Specification:
If it is impossible to obtain the given permutation from the original one in exactly one action, print 0 0. Otherwise, print two numbers *l* *r* (1<=≤<=*l*<=<<=*r*<=≤<=*n*) which are the endpoints of the segment that needs to be reversed to obtain from permutation 1 2 ... *n* the given one.
Demo Input:
['8\n1 6 5 4 3 2 7 8\n', '4\n2 3 4 1\n', '4\n1 2 3 4\n']
Demo Output:
['2 6\n', '0 0\n', '0 0\n']
Note:
none | ```python
n=int(input())
a=list(map(int,input().split()))
s=sorted(a)
l=0
r=0
c=0
for i in range(n-1):
if a[i]>a[i+1]:
l=i if c==0 else l
c=1
else:
if c==1:
r=i+1
break
a1=a[l:r]
a1.reverse()
a=a[:l]+a1+a[r:]
if a==s:
print(l,r)
else:
print(0,0)
``` | 0 |
920 | E | Connected Components? | PROGRAMMING | 2,100 | [
"data structures",
"dfs and similar",
"dsu",
"graphs"
] | null | null | You are given an undirected graph consisting of *n* vertices and edges. Instead of giving you the edges that exist in the graph, we give you *m* unordered pairs (*x*,<=*y*) such that there is no edge between *x* and *y*, and if some pair of vertices is not listed in the input, then there is an edge between these vertices.
You have to find the number of connected components in the graph and the size of each component. A connected component is a set of vertices *X* such that for every two vertices from this set there exists at least one path in the graph connecting these vertices, but adding any other vertex to *X* violates this rule. | The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=200000, ).
Then *m* lines follow, each containing a pair of integers *x* and *y* (1<=≤<=*x*,<=*y*<=≤<=*n*, *x*<=≠<=*y*) denoting that there is no edge between *x* and *y*. Each pair is listed at most once; (*x*,<=*y*) and (*y*,<=*x*) are considered the same (so they are never listed in the same test). If some pair of vertices is not listed in the input, then there exists an edge between those vertices. | Firstly print *k* — the number of connected components in this graph.
Then print *k* integers — the sizes of components. You should output these integers in non-descending order. | [
"5 5\n1 2\n3 4\n3 2\n4 2\n2 5\n"
] | [
"2\n1 4 "
] | none | 0 | [
{
"input": "5 5\n1 2\n3 4\n3 2\n4 2\n2 5",
"output": "2\n1 4 "
},
{
"input": "8 15\n2 1\n4 5\n2 4\n3 4\n2 5\n3 5\n2 6\n3 6\n5 6\n4 6\n2 7\n3 8\n2 8\n3 7\n6 7",
"output": "1\n8 "
},
{
"input": "12 58\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 10\n1 11\n1 12\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n... | 1,629,621,658 | 2,147,483,647 | PyPy 3 | OK | TESTS | 84 | 312 | 39,219,200 | import io, os, sys
if 'PyPy' in sys.version:
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def strput():
return input().decode()
else:
input = sys.stdin.readline
def strput():
return input()
# code starts here
class UnionFind:
size = 0
par, ext = [], []
def __init__(self, n):
for i in range(n):
self.par.append(i)
self.ext.append(1)
self.size = n
def find(self, u):
while u != self.par[u]:
u = self.par[u]
return u
def unite(self, u, v):
u, v = self.find(u), self.find(v)
if u == v:
return False
if self.ext[u] < self.ext[v]:
u, v = v, u
self.ext[u] += self.ext[v]
self.par[v] = u
self.size -= 1
return True
def main():
n, m = [int(x) for x in input().split()]
edge = [[] for x in range(n)]
for i in range(m):
u, v = [int(x) - 1 for x in input().split()]
if u < v:
u, v = v, u
edge[u].append(v)
dsu = UnionFind(n)
complist = set()
for u in range(n):
count = dict()
for v in edge[u]:
root = dsu.find(v)
count[root] = count.get(root, 0) + 1
tojoin = []
for comp in complist:
root = dsu.find(comp)
if count.get(root, 0) < dsu.ext[root]:
tojoin.append(root)
for v in tojoin:
if dsu.unite(v, u):
complist.remove(v)
complist.add(dsu.find(u))
print(len(complist))
ans = [dsu.ext[x] for x in complist]
ans.sort()
print(*ans)
if __name__ == "__main__":
main() | Title: Connected Components?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an undirected graph consisting of *n* vertices and edges. Instead of giving you the edges that exist in the graph, we give you *m* unordered pairs (*x*,<=*y*) such that there is no edge between *x* and *y*, and if some pair of vertices is not listed in the input, then there is an edge between these vertices.
You have to find the number of connected components in the graph and the size of each component. A connected component is a set of vertices *X* such that for every two vertices from this set there exists at least one path in the graph connecting these vertices, but adding any other vertex to *X* violates this rule.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=200000, ).
Then *m* lines follow, each containing a pair of integers *x* and *y* (1<=≤<=*x*,<=*y*<=≤<=*n*, *x*<=≠<=*y*) denoting that there is no edge between *x* and *y*. Each pair is listed at most once; (*x*,<=*y*) and (*y*,<=*x*) are considered the same (so they are never listed in the same test). If some pair of vertices is not listed in the input, then there exists an edge between those vertices.
Output Specification:
Firstly print *k* — the number of connected components in this graph.
Then print *k* integers — the sizes of components. You should output these integers in non-descending order.
Demo Input:
['5 5\n1 2\n3 4\n3 2\n4 2\n2 5\n']
Demo Output:
['2\n1 4 ']
Note:
none | ```python
import io, os, sys
if 'PyPy' in sys.version:
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def strput():
return input().decode()
else:
input = sys.stdin.readline
def strput():
return input()
# code starts here
class UnionFind:
size = 0
par, ext = [], []
def __init__(self, n):
for i in range(n):
self.par.append(i)
self.ext.append(1)
self.size = n
def find(self, u):
while u != self.par[u]:
u = self.par[u]
return u
def unite(self, u, v):
u, v = self.find(u), self.find(v)
if u == v:
return False
if self.ext[u] < self.ext[v]:
u, v = v, u
self.ext[u] += self.ext[v]
self.par[v] = u
self.size -= 1
return True
def main():
n, m = [int(x) for x in input().split()]
edge = [[] for x in range(n)]
for i in range(m):
u, v = [int(x) - 1 for x in input().split()]
if u < v:
u, v = v, u
edge[u].append(v)
dsu = UnionFind(n)
complist = set()
for u in range(n):
count = dict()
for v in edge[u]:
root = dsu.find(v)
count[root] = count.get(root, 0) + 1
tojoin = []
for comp in complist:
root = dsu.find(comp)
if count.get(root, 0) < dsu.ext[root]:
tojoin.append(root)
for v in tojoin:
if dsu.unite(v, u):
complist.remove(v)
complist.add(dsu.find(u))
print(len(complist))
ans = [dsu.ext[x] for x in complist]
ans.sort()
print(*ans)
if __name__ == "__main__":
main()
``` | 3 | |
45 | A | Codecraft III | PROGRAMMING | 900 | [
"implementation"
] | A. Codecraft III | 2 | 256 | Today Vasya visited a widely known site and learned that the continuation of his favourite game Codecraft II will appear after exactly *k* months. He looked at the calendar and learned that at the moment is the month number *s*. Vasya immediately got interested in what month Codecraft III will appear. Help him understand that.
All the twelve months in Vasya's calendar are named using their usual English names: January, February, March, April, May, June, July, August, September, October, November, December. | The first input line contains the name of the current month. It is guaranteed that it is a proper English name of one of twelve months. The first letter is uppercase, the rest are lowercase. The second line contains integer *k* (0<=≤<=*k*<=≤<=100) — the number of months left till the appearance of Codecraft III. | Print starting from an uppercase letter the name of the month in which the continuation of Codeforces II will appear. The printed name must be contained in the list January, February, March, April, May, June, July, August, September, October, November, December. | [
"November\n3\n",
"May\n24\n"
] | [
"February\n",
"May\n"
] | none | 0 | [
{
"input": "November\n3",
"output": "February"
},
{
"input": "May\n24",
"output": "May"
},
{
"input": "April\n0",
"output": "April"
},
{
"input": "September\n0",
"output": "September"
},
{
"input": "August\n0",
"output": "August"
},
{
"input": "June\n1... | 1,673,133,095 | 2,147,483,647 | PyPy 3-64 | RUNTIME_ERROR | TESTS | 7 | 124 | 2,867,200 | m_to_n = {'January': 1, 'February': 2, 'March': 3, 'April': 4, 'May': 5, 'June': 6, 'July': 7, 'August': 8, 'September': 9, 'October': 10, 'November': 11, 'December': 12}
n_to_m = {1: 'January', 2: 'February', 3: 'March', 4: 'April', 5: 'May', 6: 'June', 7: 'July', 8: 'August', 9: 'September', 10: 'October', 11: 'November', 12: 'December'}
m = input()
m = m_to_n[m]
n= int(input())
m=(m+n)%12
print(n_to_m[m])
| Title: Codecraft III
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Today Vasya visited a widely known site and learned that the continuation of his favourite game Codecraft II will appear after exactly *k* months. He looked at the calendar and learned that at the moment is the month number *s*. Vasya immediately got interested in what month Codecraft III will appear. Help him understand that.
All the twelve months in Vasya's calendar are named using their usual English names: January, February, March, April, May, June, July, August, September, October, November, December.
Input Specification:
The first input line contains the name of the current month. It is guaranteed that it is a proper English name of one of twelve months. The first letter is uppercase, the rest are lowercase. The second line contains integer *k* (0<=≤<=*k*<=≤<=100) — the number of months left till the appearance of Codecraft III.
Output Specification:
Print starting from an uppercase letter the name of the month in which the continuation of Codeforces II will appear. The printed name must be contained in the list January, February, March, April, May, June, July, August, September, October, November, December.
Demo Input:
['November\n3\n', 'May\n24\n']
Demo Output:
['February\n', 'May\n']
Note:
none | ```python
m_to_n = {'January': 1, 'February': 2, 'March': 3, 'April': 4, 'May': 5, 'June': 6, 'July': 7, 'August': 8, 'September': 9, 'October': 10, 'November': 11, 'December': 12}
n_to_m = {1: 'January', 2: 'February', 3: 'March', 4: 'April', 5: 'May', 6: 'June', 7: 'July', 8: 'August', 9: 'September', 10: 'October', 11: 'November', 12: 'December'}
m = input()
m = m_to_n[m]
n= int(input())
m=(m+n)%12
print(n_to_m[m])
``` | -1 |
828 | A | Restaurant Tables | PROGRAMMING | 1,200 | [
"implementation"
] | null | null | In a small restaurant there are *a* tables for one person and *b* tables for two persons.
It it known that *n* groups of people come today, each consisting of one or two people.
If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table. If there are none of them, it is seated at a two-seater table occupied by single person. If there are still none of them, the restaurant denies service to this group.
If a group consist of two people, it is seated at a vacant two-seater table. If there are none of them, the restaurant denies service to this group.
You are given a chronological order of groups coming. You are to determine the total number of people the restaurant denies service to. | The first line contains three integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=2·105, 1<=≤<=*a*,<=*b*<=≤<=2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables.
The second line contains a sequence of integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=2) — the description of clients in chronological order. If *t**i* is equal to one, then the *i*-th group consists of one person, otherwise the *i*-th group consists of two people. | Print the total number of people the restaurant denies service to. | [
"4 1 2\n1 2 1 1\n",
"4 1 1\n1 1 2 1\n"
] | [
"0\n",
"2\n"
] | In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, all clients are served.
In the second example the first group consists of one person, it is seated at the vacant one-seater table. The next group consists of one person, it occupies one place at the two-seater table. It's impossible to seat the next group of two people, so the restaurant denies service to them. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, the restaurant denies service to 2 clients. | 500 | [
{
"input": "4 1 2\n1 2 1 1",
"output": "0"
},
{
"input": "4 1 1\n1 1 2 1",
"output": "2"
},
{
"input": "1 1 1\n1",
"output": "0"
},
{
"input": "2 1 2\n2 2",
"output": "0"
},
{
"input": "5 1 3\n1 2 2 2 1",
"output": "1"
},
{
"input": "7 6 1\n1 1 1 1 1 1... | 1,499,792,479 | 979 | Python 3 | WRONG_ANSWER | PRETESTS | 0 | 46 | 4,608,000 | n, a, b = map(int,input().split())
q=list(map(int,input().split()))
t = True
k=0
for i in range(n):
if q[i]==1 and a==0 and t==False:
t=True
if q[i]==1 and a>0:
a-=1
elif q[i]==1 and a==0 and b>0:
t=False
b-=1
elif q[i]==2 and b>0:
b-=1
else:
k+=1
print(k)
| Title: Restaurant Tables
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In a small restaurant there are *a* tables for one person and *b* tables for two persons.
It it known that *n* groups of people come today, each consisting of one or two people.
If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table. If there are none of them, it is seated at a two-seater table occupied by single person. If there are still none of them, the restaurant denies service to this group.
If a group consist of two people, it is seated at a vacant two-seater table. If there are none of them, the restaurant denies service to this group.
You are given a chronological order of groups coming. You are to determine the total number of people the restaurant denies service to.
Input Specification:
The first line contains three integers *n*, *a* and *b* (1<=≤<=*n*<=≤<=2·105, 1<=≤<=*a*,<=*b*<=≤<=2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables.
The second line contains a sequence of integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=2) — the description of clients in chronological order. If *t**i* is equal to one, then the *i*-th group consists of one person, otherwise the *i*-th group consists of two people.
Output Specification:
Print the total number of people the restaurant denies service to.
Demo Input:
['4 1 2\n1 2 1 1\n', '4 1 1\n1 1 2 1\n']
Demo Output:
['0\n', '2\n']
Note:
In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, all clients are served.
In the second example the first group consists of one person, it is seated at the vacant one-seater table. The next group consists of one person, it occupies one place at the two-seater table. It's impossible to seat the next group of two people, so the restaurant denies service to them. The fourth group consists of one person, he is seated at the remaining seat at the two-seater table. Thus, the restaurant denies service to 2 clients. | ```python
n, a, b = map(int,input().split())
q=list(map(int,input().split()))
t = True
k=0
for i in range(n):
if q[i]==1 and a==0 and t==False:
t=True
if q[i]==1 and a>0:
a-=1
elif q[i]==1 and a==0 and b>0:
t=False
b-=1
elif q[i]==2 and b>0:
b-=1
else:
k+=1
print(k)
``` | 0 | |
991 | A | If at first you don't succeed... | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Each student eagerly awaits the day he would pass the exams successfully. Thus, Vasya was ready to celebrate, but, alas, he didn't pass it. However, many of Vasya's fellow students from the same group were more successful and celebrated after the exam.
Some of them celebrated in the BugDonalds restaurant, some of them — in the BeaverKing restaurant, the most successful ones were fast enough to celebrate in both of restaurants. Students which didn't pass the exam didn't celebrate in any of those restaurants and elected to stay home to prepare for their reexamination. However, this quickly bored Vasya and he started checking celebration photos on the Kilogramm. He found out that, in total, BugDonalds was visited by $A$ students, BeaverKing — by $B$ students and $C$ students visited both restaurants. Vasya also knows that there are $N$ students in his group.
Based on this info, Vasya wants to determine either if his data contradicts itself or, if it doesn't, how many students in his group didn't pass the exam. Can you help him so he won't waste his valuable preparation time? | The first line contains four integers — $A$, $B$, $C$ and $N$ ($0 \leq A, B, C, N \leq 100$). | If a distribution of $N$ students exists in which $A$ students visited BugDonalds, $B$ — BeaverKing, $C$ — both of the restaurants and at least one student is left home (it is known that Vasya didn't pass the exam and stayed at home), output one integer — amount of students (including Vasya) who did not pass the exam.
If such a distribution does not exist and Vasya made a mistake while determining the numbers $A$, $B$, $C$ or $N$ (as in samples 2 and 3), output $-1$. | [
"10 10 5 20\n",
"2 2 0 4\n",
"2 2 2 1\n"
] | [
"5",
"-1",
"-1"
] | The first sample describes following situation: $5$ only visited BugDonalds, $5$ students only visited BeaverKing, $5$ visited both of them and $5$ students (including Vasya) didn't pass the exam.
In the second sample $2$ students only visited BugDonalds and $2$ only visited BeaverKing, but that means all $4$ students in group passed the exam which contradicts the fact that Vasya didn't pass meaning that this situation is impossible.
The third sample describes a situation where $2$ students visited BugDonalds but the group has only $1$ which makes it clearly impossible. | 500 | [
{
"input": "10 10 5 20",
"output": "5"
},
{
"input": "2 2 0 4",
"output": "-1"
},
{
"input": "2 2 2 1",
"output": "-1"
},
{
"input": "98 98 97 100",
"output": "1"
},
{
"input": "1 5 2 10",
"output": "-1"
},
{
"input": "5 1 2 10",
"output": "-1"
}... | 1,550,153,945 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | A, B, C, N = map(int, input().split())
if C > A or C > B:
print(-1)
exit()
else:
A-=C
B-=C
pass_students = A+B+C
if pass_students < N:
print(N-pass_stidents)
exit()
return N - pass_students
else:
print(-1) | Title: If at first you don't succeed...
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Each student eagerly awaits the day he would pass the exams successfully. Thus, Vasya was ready to celebrate, but, alas, he didn't pass it. However, many of Vasya's fellow students from the same group were more successful and celebrated after the exam.
Some of them celebrated in the BugDonalds restaurant, some of them — in the BeaverKing restaurant, the most successful ones were fast enough to celebrate in both of restaurants. Students which didn't pass the exam didn't celebrate in any of those restaurants and elected to stay home to prepare for their reexamination. However, this quickly bored Vasya and he started checking celebration photos on the Kilogramm. He found out that, in total, BugDonalds was visited by $A$ students, BeaverKing — by $B$ students and $C$ students visited both restaurants. Vasya also knows that there are $N$ students in his group.
Based on this info, Vasya wants to determine either if his data contradicts itself or, if it doesn't, how many students in his group didn't pass the exam. Can you help him so he won't waste his valuable preparation time?
Input Specification:
The first line contains four integers — $A$, $B$, $C$ and $N$ ($0 \leq A, B, C, N \leq 100$).
Output Specification:
If a distribution of $N$ students exists in which $A$ students visited BugDonalds, $B$ — BeaverKing, $C$ — both of the restaurants and at least one student is left home (it is known that Vasya didn't pass the exam and stayed at home), output one integer — amount of students (including Vasya) who did not pass the exam.
If such a distribution does not exist and Vasya made a mistake while determining the numbers $A$, $B$, $C$ or $N$ (as in samples 2 and 3), output $-1$.
Demo Input:
['10 10 5 20\n', '2 2 0 4\n', '2 2 2 1\n']
Demo Output:
['5', '-1', '-1']
Note:
The first sample describes following situation: $5$ only visited BugDonalds, $5$ students only visited BeaverKing, $5$ visited both of them and $5$ students (including Vasya) didn't pass the exam.
In the second sample $2$ students only visited BugDonalds and $2$ only visited BeaverKing, but that means all $4$ students in group passed the exam which contradicts the fact that Vasya didn't pass meaning that this situation is impossible.
The third sample describes a situation where $2$ students visited BugDonalds but the group has only $1$ which makes it clearly impossible. | ```python
A, B, C, N = map(int, input().split())
if C > A or C > B:
print(-1)
exit()
else:
A-=C
B-=C
pass_students = A+B+C
if pass_students < N:
print(N-pass_stidents)
exit()
return N - pass_students
else:
print(-1)
``` | -1 | |
677 | A | Vanya and Fence | PROGRAMMING | 800 | [
"implementation"
] | null | null | Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*.
Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard? | The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person. | Print a single integer — the minimum possible valid width of the road. | [
"3 7\n4 5 14\n",
"6 1\n1 1 1 1 1 1\n",
"6 5\n7 6 8 9 10 5\n"
] | [
"4\n",
"6\n",
"11\n"
] | In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4.
In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough.
In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11. | 500 | [
{
"input": "3 7\n4 5 14",
"output": "4"
},
{
"input": "6 1\n1 1 1 1 1 1",
"output": "6"
},
{
"input": "6 5\n7 6 8 9 10 5",
"output": "11"
},
{
"input": "10 420\n214 614 297 675 82 740 174 23 255 15",
"output": "13"
},
{
"input": "10 561\n657 23 1096 487 785 66 481... | 1,693,554,215 | 2,147,483,647 | PyPy 3 | OK | TESTS | 29 | 93 | 0 | n,h=map(int,input().split())
a=list(map(int,input().split()))
w=n
for i in range(n):
if a[i]>h:
w+=1
print(w)
| Title: Vanya and Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya and his friends are walking along the fence of height *h* and they do not want the guard to notice them. In order to achieve this the height of each of the friends should not exceed *h*. If the height of some person is greater than *h* he can bend down and then he surely won't be noticed by the guard. The height of the *i*-th person is equal to *a**i*.
Consider the width of the person walking as usual to be equal to 1, while the width of the bent person is equal to 2. Friends want to talk to each other while walking, so they would like to walk in a single row. What is the minimum width of the road, such that friends can walk in a row and remain unattended by the guard?
Input Specification:
The first line of the input contains two integers *n* and *h* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*h*<=≤<=1000) — the number of friends and the height of the fence, respectively.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=2*h*), the *i*-th of them is equal to the height of the *i*-th person.
Output Specification:
Print a single integer — the minimum possible valid width of the road.
Demo Input:
['3 7\n4 5 14\n', '6 1\n1 1 1 1 1 1\n', '6 5\n7 6 8 9 10 5\n']
Demo Output:
['4\n', '6\n', '11\n']
Note:
In the first sample, only person number 3 must bend down, so the required width is equal to 1 + 1 + 2 = 4.
In the second sample, all friends are short enough and no one has to bend, so the width 1 + 1 + 1 + 1 + 1 + 1 = 6 is enough.
In the third sample, all the persons have to bend, except the last one. The required minimum width of the road is equal to 2 + 2 + 2 + 2 + 2 + 1 = 11. | ```python
n,h=map(int,input().split())
a=list(map(int,input().split()))
w=n
for i in range(n):
if a[i]>h:
w+=1
print(w)
``` | 3 | |
664 | A | Complicated GCD | PROGRAMMING | 800 | [
"math",
"number theory"
] | null | null | Greatest common divisor *GCD*(*a*,<=*b*) of two positive integers *a* and *b* is equal to the biggest integer *d* such that both integers *a* and *b* are divisible by *d*. There are many efficient algorithms to find greatest common divisor *GCD*(*a*,<=*b*), for example, Euclid algorithm.
Formally, find the biggest integer *d*, such that all integers *a*,<=*a*<=+<=1,<=*a*<=+<=2,<=...,<=*b* are divisible by *d*. To make the problem even more complicated we allow *a* and *b* to be up to googol, 10100 — such number do not fit even in 64-bit integer type! | The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10100). | Output one integer — greatest common divisor of all integers from *a* to *b* inclusive. | [
"1 2\n",
"61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576\n"
] | [
"1\n",
"61803398874989484820458683436563811772030917980576\n"
] | none | 500 | [
{
"input": "1 2",
"output": "1"
},
{
"input": "61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576",
"output": "61803398874989484820458683436563811772030917980576"
},
{
"input": "1 100",
"output": "1"
},
{
"input": "100 100000... | 1,671,113,471 | 2,147,483,647 | Python 3 | OK | TESTS | 28 | 46 | 0 | import math
f,s =map(int,input().split())
if f ==s:
print(f)
else:
print(1)
| Title: Complicated GCD
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greatest common divisor *GCD*(*a*,<=*b*) of two positive integers *a* and *b* is equal to the biggest integer *d* such that both integers *a* and *b* are divisible by *d*. There are many efficient algorithms to find greatest common divisor *GCD*(*a*,<=*b*), for example, Euclid algorithm.
Formally, find the biggest integer *d*, such that all integers *a*,<=*a*<=+<=1,<=*a*<=+<=2,<=...,<=*b* are divisible by *d*. To make the problem even more complicated we allow *a* and *b* to be up to googol, 10100 — such number do not fit even in 64-bit integer type!
Input Specification:
The only line of the input contains two integers *a* and *b* (1<=≤<=*a*<=≤<=*b*<=≤<=10100).
Output Specification:
Output one integer — greatest common divisor of all integers from *a* to *b* inclusive.
Demo Input:
['1 2\n', '61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576\n']
Demo Output:
['1\n', '61803398874989484820458683436563811772030917980576\n']
Note:
none | ```python
import math
f,s =map(int,input().split())
if f ==s:
print(f)
else:
print(1)
``` | 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",
... | 1,605,356,853 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 154 | 0 | from sys import stdin
n = max(map(int, stdin.readline().split())) - 1
if n==0:
print("1/1")
elif n==1:
print("2/3")
elif n==3:
print("1/2")
elif n==4:
print("1/3")
elif n==5:
print("1/6")
elif n==6 :
print("0/1") | 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
from sys import stdin
n = max(map(int, stdin.readline().split())) - 1
if n==0:
print("1/1")
elif n==1:
print("2/3")
elif n==3:
print("1/2")
elif n==4:
print("1/3")
elif n==5:
print("1/6")
elif n==6 :
print("0/1")
``` | 0 |
796 | A | Buying A House | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | Zane the wizard had never loved anyone before, until he fell in love with a girl, whose name remains unknown to us.
The girl lives in house *m* of a village. There are *n* houses in that village, lining in a straight line from left to right: house 1, house 2, ..., house *n*. The village is also well-structured: house *i* and house *i*<=+<=1 (1<=≤<=*i*<=<<=*n*) are exactly 10 meters away. In this village, some houses are occupied, and some are not. Indeed, unoccupied houses can be purchased.
You will be given *n* integers *a*1,<=*a*2,<=...,<=*a**n* that denote the availability and the prices of the houses. If house *i* is occupied, and therefore cannot be bought, then *a**i* equals 0. Otherwise, house *i* can be bought, and *a**i* represents the money required to buy it, in dollars.
As Zane has only *k* dollars to spare, it becomes a challenge for him to choose the house to purchase, so that he could live as near as possible to his crush. Help Zane determine the minimum distance from his crush's house to some house he can afford, to help him succeed in his love. | The first line contains three integers *n*, *m*, and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*m*<=≤<=*n*, 1<=≤<=*k*<=≤<=100) — the number of houses in the village, the house where the girl lives, and the amount of money Zane has (in dollars), respectively.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100) — denoting the availability and the prices of the houses.
It is guaranteed that *a**m*<==<=0 and that it is possible to purchase some house with no more than *k* dollars. | Print one integer — the minimum distance, in meters, from the house where the girl Zane likes lives to the house Zane can buy. | [
"5 1 20\n0 27 32 21 19\n",
"7 3 50\n62 0 0 0 99 33 22\n",
"10 5 100\n1 0 1 0 0 0 0 0 1 1\n"
] | [
"40",
"30",
"20"
] | In the first sample, with *k* = 20 dollars, Zane can buy only house 5. The distance from house *m* = 1 to house 5 is 10 + 10 + 10 + 10 = 40 meters.
In the second sample, Zane can buy houses 6 and 7. It is better to buy house 6 than house 7, since house *m* = 3 and house 6 are only 30 meters away, while house *m* = 3 and house 7 are 40 meters away. | 500 | [
{
"input": "5 1 20\n0 27 32 21 19",
"output": "40"
},
{
"input": "7 3 50\n62 0 0 0 99 33 22",
"output": "30"
},
{
"input": "10 5 100\n1 0 1 0 0 0 0 0 1 1",
"output": "20"
},
{
"input": "5 3 1\n1 1 0 0 1",
"output": "10"
},
{
"input": "5 5 5\n1 0 5 6 0",
"outpu... | 1,519,426,111 | 2,147,483,647 | PyPy 3 | OK | TESTS | 58 | 77 | 19,968,000 | (num_houses, crush_house, dollars) = [int(string) for string in input().split()]
house_prices = [int(string) for string in input().split()]
def distance(crush, house, prices, max_price):
return (abs(house - crush), not(prices[house] <= 0 or prices[house] > max_price))
closest = -1
for i in range(num_houses):
(dist, can_buy) = distance(crush_house - 1, i, house_prices, dollars)
if can_buy and (dist < closest or closest == -1):
closest = dist
print(closest * 10)
| Title: Buying A House
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Zane the wizard had never loved anyone before, until he fell in love with a girl, whose name remains unknown to us.
The girl lives in house *m* of a village. There are *n* houses in that village, lining in a straight line from left to right: house 1, house 2, ..., house *n*. The village is also well-structured: house *i* and house *i*<=+<=1 (1<=≤<=*i*<=<<=*n*) are exactly 10 meters away. In this village, some houses are occupied, and some are not. Indeed, unoccupied houses can be purchased.
You will be given *n* integers *a*1,<=*a*2,<=...,<=*a**n* that denote the availability and the prices of the houses. If house *i* is occupied, and therefore cannot be bought, then *a**i* equals 0. Otherwise, house *i* can be bought, and *a**i* represents the money required to buy it, in dollars.
As Zane has only *k* dollars to spare, it becomes a challenge for him to choose the house to purchase, so that he could live as near as possible to his crush. Help Zane determine the minimum distance from his crush's house to some house he can afford, to help him succeed in his love.
Input Specification:
The first line contains three integers *n*, *m*, and *k* (2<=≤<=*n*<=≤<=100, 1<=≤<=*m*<=≤<=*n*, 1<=≤<=*k*<=≤<=100) — the number of houses in the village, the house where the girl lives, and the amount of money Zane has (in dollars), respectively.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100) — denoting the availability and the prices of the houses.
It is guaranteed that *a**m*<==<=0 and that it is possible to purchase some house with no more than *k* dollars.
Output Specification:
Print one integer — the minimum distance, in meters, from the house where the girl Zane likes lives to the house Zane can buy.
Demo Input:
['5 1 20\n0 27 32 21 19\n', '7 3 50\n62 0 0 0 99 33 22\n', '10 5 100\n1 0 1 0 0 0 0 0 1 1\n']
Demo Output:
['40', '30', '20']
Note:
In the first sample, with *k* = 20 dollars, Zane can buy only house 5. The distance from house *m* = 1 to house 5 is 10 + 10 + 10 + 10 = 40 meters.
In the second sample, Zane can buy houses 6 and 7. It is better to buy house 6 than house 7, since house *m* = 3 and house 6 are only 30 meters away, while house *m* = 3 and house 7 are 40 meters away. | ```python
(num_houses, crush_house, dollars) = [int(string) for string in input().split()]
house_prices = [int(string) for string in input().split()]
def distance(crush, house, prices, max_price):
return (abs(house - crush), not(prices[house] <= 0 or prices[house] > max_price))
closest = -1
for i in range(num_houses):
(dist, can_buy) = distance(crush_house - 1, i, house_prices, dollars)
if can_buy and (dist < closest or closest == -1):
closest = dist
print(closest * 10)
``` | 3 | |
891 | A | Pride | PROGRAMMING | 1,500 | [
"brute force",
"dp",
"greedy",
"math",
"number theory"
] | null | null | You have an array *a* with length *n*, you can perform operations. Each operation is like this: choose two adjacent elements from *a*, say *x* and *y*, and replace one of them with *gcd*(*x*,<=*y*), where *gcd* denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
What is the minimum number of operations you need to make all of the elements equal to 1? | The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=2000) — the number of elements in the array.
The second line contains *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array. | Print -1, if it is impossible to turn all numbers to 1. Otherwise, print the minimum number of operations needed to make all numbers equal to 1. | [
"5\n2 2 3 4 6\n",
"4\n2 4 6 8\n",
"3\n2 6 9\n"
] | [
"5\n",
"-1\n",
"4\n"
] | In the first sample you can turn all numbers to 1 using the following 5 moves:
- [2, 2, 3, 4, 6]. - [2, 1, 3, 4, 6] - [2, 1, 3, 1, 6] - [2, 1, 1, 1, 6] - [1, 1, 1, 1, 6] - [1, 1, 1, 1, 1]
We can prove that in this case it is not possible to make all numbers one using less than 5 moves. | 500 | [
{
"input": "5\n2 2 3 4 6",
"output": "5"
},
{
"input": "4\n2 4 6 8",
"output": "-1"
},
{
"input": "3\n2 6 9",
"output": "4"
},
{
"input": "15\n10 10 10 10 10 10 21 21 21 21 21 21 21 21 21",
"output": "15"
},
{
"input": "12\n10 10 14 14 14 14 14 14 14 14 21 21",
... | 1,684,851,877 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 52 | 233 | 10,035,200 | from math import gcd
class Node:
def __init__(self,l,r):
self.l,self.r = l,r
self.left = self.right = None
self.val = 0
class segtree:
def __init__(self,a):
self.a = a
self.rt = Node(0,len(a)-1)
self.build(self.rt)
def build(self,node):
if node.l == node.r:
node.val = self.a[node.l]
return
mid = node.l+node.r>>1
node.left,node.right = Node(node.l,mid),Node(mid+1,node.r)
self.build(node.left);self.build(node.right)
node.val = gcd(node.left.val,node.right.val)
def find(self,node,low,hig):
if low == node.l and hig == node.r:
return node.val
mid = node.l+node.r>>1
res = []
if low <= mid:
res.append(self.find(node.left,low,min(hig,mid)))
if hig >= mid+1:
res.append(self.find(node.right,max(low,mid+1),hig))
if len(res) == 2:return gcd(res[0],res[1])
return res[0]
n = int(input())
a = list(map(int,input().split()))
seg = segtree(a)
if 1 in a:
print(n-a.count(1))
else:
ans = n+1
for i in range(n):
l,r = i,n
while l+1 != r:
mid = l+r>>1
if seg.find(seg.rt,i,mid) != 1:
l = mid
else:
r = mid
if r != n:ans = min(ans,(l-i+1))
if ans == n+1:print(-1)
else:print(ans+n-1)
| Title: Pride
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have an array *a* with length *n*, you can perform operations. Each operation is like this: choose two adjacent elements from *a*, say *x* and *y*, and replace one of them with *gcd*(*x*,<=*y*), where *gcd* denotes the [greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).
What is the minimum number of operations you need to make all of the elements equal to 1?
Input Specification:
The first line of the input contains one integer *n* (1<=≤<=*n*<=≤<=2000) — the number of elements in the array.
The second line contains *n* space separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the elements of the array.
Output Specification:
Print -1, if it is impossible to turn all numbers to 1. Otherwise, print the minimum number of operations needed to make all numbers equal to 1.
Demo Input:
['5\n2 2 3 4 6\n', '4\n2 4 6 8\n', '3\n2 6 9\n']
Demo Output:
['5\n', '-1\n', '4\n']
Note:
In the first sample you can turn all numbers to 1 using the following 5 moves:
- [2, 2, 3, 4, 6]. - [2, 1, 3, 4, 6] - [2, 1, 3, 1, 6] - [2, 1, 1, 1, 6] - [1, 1, 1, 1, 6] - [1, 1, 1, 1, 1]
We can prove that in this case it is not possible to make all numbers one using less than 5 moves. | ```python
from math import gcd
class Node:
def __init__(self,l,r):
self.l,self.r = l,r
self.left = self.right = None
self.val = 0
class segtree:
def __init__(self,a):
self.a = a
self.rt = Node(0,len(a)-1)
self.build(self.rt)
def build(self,node):
if node.l == node.r:
node.val = self.a[node.l]
return
mid = node.l+node.r>>1
node.left,node.right = Node(node.l,mid),Node(mid+1,node.r)
self.build(node.left);self.build(node.right)
node.val = gcd(node.left.val,node.right.val)
def find(self,node,low,hig):
if low == node.l and hig == node.r:
return node.val
mid = node.l+node.r>>1
res = []
if low <= mid:
res.append(self.find(node.left,low,min(hig,mid)))
if hig >= mid+1:
res.append(self.find(node.right,max(low,mid+1),hig))
if len(res) == 2:return gcd(res[0],res[1])
return res[0]
n = int(input())
a = list(map(int,input().split()))
seg = segtree(a)
if 1 in a:
print(n-a.count(1))
else:
ans = n+1
for i in range(n):
l,r = i,n
while l+1 != r:
mid = l+r>>1
if seg.find(seg.rt,i,mid) != 1:
l = mid
else:
r = mid
if r != n:ans = min(ans,(l-i+1))
if ans == n+1:print(-1)
else:print(ans+n-1)
``` | 3 | |
227 | B | Effective Approach | PROGRAMMING | 1,100 | [
"implementation"
] | null | null | Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat. | Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier. | [
"2\n1 2\n1\n1\n",
"2\n2 1\n1\n1\n",
"3\n3 1 2\n3\n1 2 3\n"
] | [
"1 2\n",
"2 1\n",
"6 6\n"
] | In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | 1,000 | [
{
"input": "2\n1 2\n1\n1",
"output": "1 2"
},
{
"input": "2\n2 1\n1\n1",
"output": "2 1"
},
{
"input": "3\n3 1 2\n3\n1 2 3",
"output": "6 6"
},
{
"input": "9\n2 9 3 1 6 4 7 8 5\n9\n5 1 5 2 8 4 4 4 5",
"output": "58 32"
},
{
"input": "10\n3 10 9 2 7 6 5 8 4 1\n1\n4... | 1,658,537,230 | 2,147,483,647 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 5 | 4,000 | 6,144,000 | num_elements = int(input())
elements = [int(i) for i in input().split(" ")]
num_queries = int(input())
queries = [int(i) for i in input().split(" ")]
first = 0
second = 0
for querie in queries:
for element in elements:
if element == querie:
first += 1
break
else:
first += 1
elements.reverse()
for querie in queries:
for element in elements:
if element == querie:
second += 1
break
else:
second += 1
print(first, second) | Title: Effective Approach
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once at a team training Vasya, Petya and Sasha got a problem on implementing linear search in an array.
According to the boys, linear search works as follows. The array elements in a pre-selected order are in turn compared with the number that you need to find. Once you find the array element that is equal to the required one, the search ends. The efficiency of the algorithm is the number of performed comparisons. The fewer comparisons the linear search has made, the more effective it is.
Vasya believes that a linear search would work better if it sequentially iterates through the elements, starting with the 1-st one (in this problem we consider the elements of the array indexed from 1 to *n*) and ending with the *n*-th one. And Petya says that Vasya is wrong: the search will need less comparisons if it sequentially iterates the elements starting from the *n*-th and ending with the 1-st one. Sasha argues that the two approaches are equivalent.
To finally begin the task, the teammates decided to settle the debate and compare the two approaches on an example. For this, they took an array that is a permutation of integers from 1 to *n*, and generated *m* queries of the form: find element with value *b**i* in the array. They want to calculate for both approaches how many comparisons in total the linear search will need to respond to all queries. If the first search needs fewer comparisons, then the winner of the dispute is Vasya. If the second one does, then the winner is Petya. If both approaches make the same number of comparisons, then Sasha's got the upper hand.
But the problem is, linear search is too slow. That's why the boys aren't going to find out who is right before the end of the training, unless you come in here. Help them to determine who will win the dispute.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of elements in the array. The second line contains *n* distinct space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the elements of array.
The third line contains integer *m* (1<=≤<=*m*<=≤<=105) — the number of queries. The last line contains *m* space-separated integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=*n*) — the search queries. Note that the queries can repeat.
Output Specification:
Print two integers, showing how many comparisons Vasya's approach needs and how many comparisons Petya's approach needs. Separate the numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.
Demo Input:
['2\n1 2\n1\n1\n', '2\n2 1\n1\n1\n', '3\n3 1 2\n3\n1 2 3\n']
Demo Output:
['1 2\n', '2 1\n', '6 6\n']
Note:
In the first sample Vasya's approach will make one comparison (it starts with the 1-st element and immediately finds the required number), and Petya's approach makes two comparisons (first he compares with the 2-nd array element, doesn't find the search item and compares with the 1-st element).
In the second sample, on the contrary, Vasya's approach will need two comparisons (first with 1-st element, and then with the 2-nd), and Petya's approach will find the required value in one comparison (the first comparison with the 2-nd element). | ```python
num_elements = int(input())
elements = [int(i) for i in input().split(" ")]
num_queries = int(input())
queries = [int(i) for i in input().split(" ")]
first = 0
second = 0
for querie in queries:
for element in elements:
if element == querie:
first += 1
break
else:
first += 1
elements.reverse()
for querie in queries:
for element in elements:
if element == querie:
second += 1
break
else:
second += 1
print(first, second)
``` | 0 | |
980 | A | Links and Pearls | PROGRAMMING | 900 | [
"implementation",
"math"
] | null | null | A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one.
You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts.
Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them.
Note that the final necklace should remain as one circular part of the same length as the initial necklace. | The only line of input contains a string $s$ ($3 \leq |s| \leq 100$), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl. | Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO".
You can print each letter in any case (upper or lower). | [
"-o-o--",
"-o---\n",
"-o---o-\n",
"ooo\n"
] | [
"YES",
"YES",
"NO",
"YES\n"
] | none | 500 | [
{
"input": "-o-o--",
"output": "YES"
},
{
"input": "-o---",
"output": "YES"
},
{
"input": "-o---o-",
"output": "NO"
},
{
"input": "ooo",
"output": "YES"
},
{
"input": "---",
"output": "YES"
},
{
"input": "--o-o-----o----o--oo-o-----ooo-oo---o--",
"... | 1,525,792,948 | 1,048 | Python 3 | OK | TESTS | 69 | 93 | 7,065,600 | s = input()
km = 0
kb = 0
for el in s:
if el == '-':
km+=1
else:
kb+=1
if km!=0 and kb != 0:
if ((km % kb) == 0):
print('YES')
else:
print('NO')
else:
print('YES')
| Title: Links and Pearls
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one.
You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts.
Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them.
Note that the final necklace should remain as one circular part of the same length as the initial necklace.
Input Specification:
The only line of input contains a string $s$ ($3 \leq |s| \leq 100$), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl.
Output Specification:
Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO".
You can print each letter in any case (upper or lower).
Demo Input:
['-o-o--', '-o---\n', '-o---o-\n', 'ooo\n']
Demo Output:
['YES', 'YES', 'NO', 'YES\n']
Note:
none | ```python
s = input()
km = 0
kb = 0
for el in s:
if el == '-':
km+=1
else:
kb+=1
if km!=0 and kb != 0:
if ((km % kb) == 0):
print('YES')
else:
print('NO')
else:
print('YES')
``` | 3 | |
998 | A | Balloons | PROGRAMMING | 1,000 | [
"constructive algorithms",
"implementation"
] | null | null | There are quite a lot of ways to have fun with inflatable balloons. For example, you can fill them with water and see what happens.
Grigory and Andrew have the same opinion. So, once upon a time, they went to the shop and bought $n$ packets with inflatable balloons, where $i$-th of them has exactly $a_i$ balloons inside.
They want to divide the balloons among themselves. In addition, there are several conditions to hold:
- Do not rip the packets (both Grigory and Andrew should get unbroken packets); - Distribute all packets (every packet should be given to someone); - Give both Grigory and Andrew at least one packet; - To provide more fun, the total number of balloons in Grigory's packets should not be equal to the total number of balloons in Andrew's packets.
Help them to divide the balloons or determine that it's impossible under these conditions. | The first line of input contains a single integer $n$ ($1 \le n \le 10$) — the number of packets with balloons.
The second line contains $n$ integers: $a_1$, $a_2$, $\ldots$, $a_n$ ($1 \le a_i \le 1000$) — the number of balloons inside the corresponding packet. | If it's impossible to divide the balloons satisfying the conditions above, print $-1$.
Otherwise, print an integer $k$ — the number of packets to give to Grigory followed by $k$ distinct integers from $1$ to $n$ — the indices of those. The order of packets doesn't matter.
If there are multiple ways to divide balloons, output any of them. | [
"3\n1 2 1\n",
"2\n5 5\n",
"1\n10\n"
] | [
"2\n1 2\n",
"-1\n",
"-1\n"
] | In the first test Grigory gets $3$ balloons in total while Andrey gets $1$.
In the second test there's only one way to divide the packets which leads to equal numbers of balloons.
In the third test one of the boys won't get a packet at all. | 500 | [
{
"input": "3\n1 2 1",
"output": "1\n1"
},
{
"input": "2\n5 5",
"output": "-1"
},
{
"input": "1\n10",
"output": "-1"
},
{
"input": "1\n1",
"output": "-1"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "1\n1"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 9",... | 1,530,456,938 | 3,038 | Python 3 | WRONG_ANSWER | TESTS | 30 | 124 | 0 | if __name__=="__main__":
n=int(input())
c=list(map(int,input().strip().split(' ')))
s=sum(c)
s1=0
v=[]
for x in range(0,len(c)):
s1+=c[x]
s-=c[x]
if s1!=s and s!=0:
v.append(x+1)
if len(v)!=0:
print(len(v))
for x in range(0,len(v)):
print(v[x],end=' ')
else:
print('-1') | Title: Balloons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are quite a lot of ways to have fun with inflatable balloons. For example, you can fill them with water and see what happens.
Grigory and Andrew have the same opinion. So, once upon a time, they went to the shop and bought $n$ packets with inflatable balloons, where $i$-th of them has exactly $a_i$ balloons inside.
They want to divide the balloons among themselves. In addition, there are several conditions to hold:
- Do not rip the packets (both Grigory and Andrew should get unbroken packets); - Distribute all packets (every packet should be given to someone); - Give both Grigory and Andrew at least one packet; - To provide more fun, the total number of balloons in Grigory's packets should not be equal to the total number of balloons in Andrew's packets.
Help them to divide the balloons or determine that it's impossible under these conditions.
Input Specification:
The first line of input contains a single integer $n$ ($1 \le n \le 10$) — the number of packets with balloons.
The second line contains $n$ integers: $a_1$, $a_2$, $\ldots$, $a_n$ ($1 \le a_i \le 1000$) — the number of balloons inside the corresponding packet.
Output Specification:
If it's impossible to divide the balloons satisfying the conditions above, print $-1$.
Otherwise, print an integer $k$ — the number of packets to give to Grigory followed by $k$ distinct integers from $1$ to $n$ — the indices of those. The order of packets doesn't matter.
If there are multiple ways to divide balloons, output any of them.
Demo Input:
['3\n1 2 1\n', '2\n5 5\n', '1\n10\n']
Demo Output:
['2\n1 2\n', '-1\n', '-1\n']
Note:
In the first test Grigory gets $3$ balloons in total while Andrey gets $1$.
In the second test there's only one way to divide the packets which leads to equal numbers of balloons.
In the third test one of the boys won't get a packet at all. | ```python
if __name__=="__main__":
n=int(input())
c=list(map(int,input().strip().split(' ')))
s=sum(c)
s1=0
v=[]
for x in range(0,len(c)):
s1+=c[x]
s-=c[x]
if s1!=s and s!=0:
v.append(x+1)
if len(v)!=0:
print(len(v))
for x in range(0,len(v)):
print(v[x],end=' ')
else:
print('-1')
``` | 0 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Mikhail the Freelancer dreams of two things: to become a cool programmer and to buy a flat in Moscow. To become a cool programmer, he needs at least *p* experience points, and a desired flat in Moscow costs *q* dollars. Mikhail is determined to follow his dreams and registered at a freelance site.
He has suggestions to work on *n* distinct projects. Mikhail has already evaluated that the participation in the *i*-th project will increase his experience by *a**i* per day and bring *b**i* dollars per day. As freelance work implies flexible working hours, Mikhail is free to stop working on one project at any time and start working on another project. Doing so, he receives the respective share of experience and money. Mikhail is only trying to become a cool programmer, so he is able to work only on one project at any moment of time.
Find the real value, equal to the minimum number of days Mikhail needs to make his dream come true.
For example, suppose Mikhail is suggested to work on three projects and *a*1<==<=6, *b*1<==<=2, *a*2<==<=1, *b*2<==<=3, *a*3<==<=2, *b*3<==<=6. Also, *p*<==<=20 and *q*<==<=20. In order to achieve his aims Mikhail has to work for 2.5 days on both first and third projects. Indeed, *a*1·2.5<=+<=*a*2·0<=+<=*a*3·2.5<==<=6·2.5<=+<=1·0<=+<=2·2.5<==<=20 and *b*1·2.5<=+<=*b*2·0<=+<=*b*3·2.5<==<=2·2.5<=+<=3·0<=+<=6·2.5<==<=20. | The first line of the input contains three integers *n*, *p* and *q* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*p*,<=*q*<=≤<=1<=000<=000) — the number of projects and the required number of experience and money.
Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=1<=000<=000) — the daily increase in experience and daily income for working on the *i*-th project. | Print a real value — the minimum number of days Mikhail needs to get the required amount of experience and money. Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if . | [
"3 20 20\n6 2\n1 3\n2 6\n",
"4 1 1\n2 3\n3 2\n2 3\n3 2\n"
] | [
"5.000000000000000\n",
"0.400000000000000\n"
] | First sample corresponds to the example in the problem statement. | 0 | [
{
"input": "3 20 20\n6 2\n1 3\n2 6",
"output": "5.000000000000000"
},
{
"input": "4 1 1\n2 3\n3 2\n2 3\n3 2",
"output": "0.400000000000000"
},
{
"input": "3 12 12\n5 1\n2 2\n1 5",
"output": "4.000000000000000"
},
{
"input": "3 12 12\n5 1\n4 4\n1 5",
"output": "3.000000000... | 1,449,684,243 | 7,143 | Python 3 | COMPILATION_ERROR | PRETESTS | 0 | 0 | 0 | n, q, p = map(int, input().split())
lst = []
for i in range(n):
q, p = map(int, input().split())
lst.append((q, p))
lst.sort(key = lambda x: x[0] + x[1], reverse=True)
print(q / (lst[0][1] + lst[1][1]) + p / (lst[0][1] + lst[1][1])
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mikhail the Freelancer dreams of two things: to become a cool programmer and to buy a flat in Moscow. To become a cool programmer, he needs at least *p* experience points, and a desired flat in Moscow costs *q* dollars. Mikhail is determined to follow his dreams and registered at a freelance site.
He has suggestions to work on *n* distinct projects. Mikhail has already evaluated that the participation in the *i*-th project will increase his experience by *a**i* per day and bring *b**i* dollars per day. As freelance work implies flexible working hours, Mikhail is free to stop working on one project at any time and start working on another project. Doing so, he receives the respective share of experience and money. Mikhail is only trying to become a cool programmer, so he is able to work only on one project at any moment of time.
Find the real value, equal to the minimum number of days Mikhail needs to make his dream come true.
For example, suppose Mikhail is suggested to work on three projects and *a*1<==<=6, *b*1<==<=2, *a*2<==<=1, *b*2<==<=3, *a*3<==<=2, *b*3<==<=6. Also, *p*<==<=20 and *q*<==<=20. In order to achieve his aims Mikhail has to work for 2.5 days on both first and third projects. Indeed, *a*1·2.5<=+<=*a*2·0<=+<=*a*3·2.5<==<=6·2.5<=+<=1·0<=+<=2·2.5<==<=20 and *b*1·2.5<=+<=*b*2·0<=+<=*b*3·2.5<==<=2·2.5<=+<=3·0<=+<=6·2.5<==<=20.
Input Specification:
The first line of the input contains three integers *n*, *p* and *q* (1<=≤<=*n*<=≤<=100<=000,<=1<=≤<=*p*,<=*q*<=≤<=1<=000<=000) — the number of projects and the required number of experience and money.
Each of the next *n* lines contains two integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=1<=000<=000) — the daily increase in experience and daily income for working on the *i*-th project.
Output Specification:
Print a real value — the minimum number of days Mikhail needs to get the required amount of experience and money. Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
Demo Input:
['3 20 20\n6 2\n1 3\n2 6\n', '4 1 1\n2 3\n3 2\n2 3\n3 2\n']
Demo Output:
['5.000000000000000\n', '0.400000000000000\n']
Note:
First sample corresponds to the example in the problem statement. | ```python
n, q, p = map(int, input().split())
lst = []
for i in range(n):
q, p = map(int, input().split())
lst.append((q, p))
lst.sort(key = lambda x: x[0] + x[1], reverse=True)
print(q / (lst[0][1] + lst[1][1]) + p / (lst[0][1] + lst[1][1])
``` | -1 | |
56 | A | Bar | PROGRAMMING | 1,000 | [
"implementation"
] | A. Bar | 2 | 256 | According to Berland laws it is only allowed to sell alcohol to people not younger than 18 years. Vasya's job is to monitor the law's enforcement. Tonight he entered a bar and saw *n* people sitting there. For every one of them Vasya happened to determine either the age or the drink the person is having. Vasya can check any person, i.e. learn his age and the drink he is having at the same time. What minimal number of people should Vasya check additionally to make sure that there are no clients under 18 having alcohol drinks?
The list of all alcohol drinks in Berland is: ABSINTH, BEER, BRANDY, CHAMPAGNE, GIN, RUM, SAKE, TEQUILA, VODKA, WHISKEY, WINE | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) which is the number of the bar's clients. Then follow *n* lines, each describing one visitor. A line either contains his age (an integer from 0 to 1000) or his drink (a string of capital Latin letters from 1 to 100 in length). It is guaranteed that the input data does not contain spaces and other unnecessary separators.
Only the drinks from the list given above should be considered alcohol. | Print a single number which is the number of people Vasya should check to guarantee the law enforcement. | [
"5\n18\nVODKA\nCOKE\n19\n17\n"
] | [
"2\n"
] | In the sample test the second and fifth clients should be checked. | 500 | [
{
"input": "5\n18\nVODKA\nCOKE\n19\n17",
"output": "2"
},
{
"input": "2\n2\nGIN",
"output": "2"
},
{
"input": "3\nWHISKEY\n3\nGIN",
"output": "3"
},
{
"input": "4\n813\nIORBQITQXMPTFAEMEQDQIKFGKGOTNKTOSZCBRPXJLUKVLVHJYNRUJXK\nRUM\nRHVRWGODYWWTYZFLFYKCVUFFRTQDINKNWPKFHZBFWBHWI... | 1,419,175,325 | 2,147,483,647 | Python 3 | OK | TESTS | 28 | 154 | 0 | import sys
import math
n = int(sys.stdin.readline())
d = {"ABSINTH": 1, "BEER": 1, "BRANDY": 1, "CHAMPAGNE": 1, "GIN": 1,
"RUM": 1, "SAKE": 1, "TEQUILA": 1, "VODKA": 1, "WHISKEY": 1, "WINE": 1}
res = 0
for i in range(n):
st = (sys.stdin.readline()).split()[0]
if(st.isdigit()):
if(int(st) < 18):
res += 1
else:
if st in d:
res += 1
print(res) | Title: Bar
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
According to Berland laws it is only allowed to sell alcohol to people not younger than 18 years. Vasya's job is to monitor the law's enforcement. Tonight he entered a bar and saw *n* people sitting there. For every one of them Vasya happened to determine either the age or the drink the person is having. Vasya can check any person, i.e. learn his age and the drink he is having at the same time. What minimal number of people should Vasya check additionally to make sure that there are no clients under 18 having alcohol drinks?
The list of all alcohol drinks in Berland is: ABSINTH, BEER, BRANDY, CHAMPAGNE, GIN, RUM, SAKE, TEQUILA, VODKA, WHISKEY, WINE
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) which is the number of the bar's clients. Then follow *n* lines, each describing one visitor. A line either contains his age (an integer from 0 to 1000) or his drink (a string of capital Latin letters from 1 to 100 in length). It is guaranteed that the input data does not contain spaces and other unnecessary separators.
Only the drinks from the list given above should be considered alcohol.
Output Specification:
Print a single number which is the number of people Vasya should check to guarantee the law enforcement.
Demo Input:
['5\n18\nVODKA\nCOKE\n19\n17\n']
Demo Output:
['2\n']
Note:
In the sample test the second and fifth clients should be checked. | ```python
import sys
import math
n = int(sys.stdin.readline())
d = {"ABSINTH": 1, "BEER": 1, "BRANDY": 1, "CHAMPAGNE": 1, "GIN": 1,
"RUM": 1, "SAKE": 1, "TEQUILA": 1, "VODKA": 1, "WHISKEY": 1, "WINE": 1}
res = 0
for i in range(n):
st = (sys.stdin.readline()).split()[0]
if(st.isdigit()):
if(int(st) < 18):
res += 1
else:
if st in d:
res += 1
print(res)
``` | 3.9615 |
489 | B | BerSU Ball | PROGRAMMING | 1,200 | [
"dfs and similar",
"dp",
"graph matchings",
"greedy",
"sortings",
"two pointers"
] | null | null | The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! *n* boys and *m* girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from *n* boys and *m* girls. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of boys. The second line contains sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is the *i*-th boy's dancing skill.
Similarly, the third line contains an integer *m* (1<=≤<=*m*<=≤<=100) — the number of girls. The fourth line contains sequence *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**j*<=≤<=100), where *b**j* is the *j*-th girl's dancing skill. | Print a single number — the required maximum possible number of pairs. | [
"4\n1 4 6 2\n5\n5 1 5 7 9\n",
"4\n1 2 3 4\n4\n10 11 12 13\n",
"5\n1 1 1 1 1\n3\n1 2 3\n"
] | [
"3\n",
"0\n",
"2\n"
] | none | 1,000 | [
{
"input": "4\n1 4 6 2\n5\n5 1 5 7 9",
"output": "3"
},
{
"input": "4\n1 2 3 4\n4\n10 11 12 13",
"output": "0"
},
{
"input": "5\n1 1 1 1 1\n3\n1 2 3",
"output": "2"
},
{
"input": "1\n1\n1\n1",
"output": "1"
},
{
"input": "2\n1 10\n1\n9",
"output": "1"
},
{... | 1,693,896,340 | 2,147,483,647 | Python 3 | OK | TESTS | 81 | 46 | 0 | a=int(input())
X=input().split()
A=[]
for x in X:
A.append(int(x))
b=int(input())
X=input().split()
B=[]
for x in X:
B.append(int(x))
A.sort()
B.sort()
if a<=b:
for x in A:
if x-1 in B:
B.remove(x-1)
continue
if x in B:
B.remove(x)
continue
if x+1 in B:
B.remove(x+1)
print(b-len(B))
elif a>b:
for x in B:
if x-1 in A:
A.remove(x-1)
continue
if x in A:
A.remove(x)
continue
if x+1 in A:
A.remove(x+1)
print(a-len(A))
| Title: BerSU Ball
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Berland State University is hosting a ballroom dance in celebration of its 100500-th anniversary! *n* boys and *m* girls are already busy rehearsing waltz, minuet, polonaise and quadrille moves.
We know that several boy&girl pairs are going to be invited to the ball. However, the partners' dancing skill in each pair must differ by at most one.
For each boy, we know his dancing skills. Similarly, for each girl we know her dancing skills. Write a code that can determine the largest possible number of pairs that can be formed from *n* boys and *m* girls.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of boys. The second line contains sequence *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100), where *a**i* is the *i*-th boy's dancing skill.
Similarly, the third line contains an integer *m* (1<=≤<=*m*<=≤<=100) — the number of girls. The fourth line contains sequence *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**j*<=≤<=100), where *b**j* is the *j*-th girl's dancing skill.
Output Specification:
Print a single number — the required maximum possible number of pairs.
Demo Input:
['4\n1 4 6 2\n5\n5 1 5 7 9\n', '4\n1 2 3 4\n4\n10 11 12 13\n', '5\n1 1 1 1 1\n3\n1 2 3\n']
Demo Output:
['3\n', '0\n', '2\n']
Note:
none | ```python
a=int(input())
X=input().split()
A=[]
for x in X:
A.append(int(x))
b=int(input())
X=input().split()
B=[]
for x in X:
B.append(int(x))
A.sort()
B.sort()
if a<=b:
for x in A:
if x-1 in B:
B.remove(x-1)
continue
if x in B:
B.remove(x)
continue
if x+1 in B:
B.remove(x+1)
print(b-len(B))
elif a>b:
for x in B:
if x-1 in A:
A.remove(x-1)
continue
if x in A:
A.remove(x)
continue
if x+1 in A:
A.remove(x+1)
print(a-len(A))
``` | 3 | |
907 | A | Masha and Bears | PROGRAMMING | 1,300 | [
"brute force",
"implementation"
] | null | null | A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size *a* can climb into some car with size *b* if and only if *a*<=≤<=*b*, he or she likes it if and only if he can climb into this car and 2*a*<=≥<=*b*.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars. | You are given four integers *V*1, *V*2, *V*3, *V**m*(1<=≤<=*V**i*<=≤<=100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that *V*1<=><=*V*2<=><=*V*3. | Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes). | [
"50 30 10 10\n",
"100 50 10 21\n"
] | [
"50\n30\n10\n",
"-1\n"
] | In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20. | 500 | [
{
"input": "50 30 10 10",
"output": "50\n30\n10"
},
{
"input": "100 50 10 21",
"output": "-1"
},
{
"input": "100 50 19 10",
"output": "100\n50\n19"
},
{
"input": "99 50 25 49",
"output": "100\n99\n49"
},
{
"input": "3 2 1 1",
"output": "4\n3\n1"
},
{
"... | 1,699,068,684 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 3 | 77 | 1,331,200 | a,b,c,d=map(int,input().split())
for i in range(a,a*2+1):
for j in range(b,b*2+1):
for k in range(c,c*2+1):
if d<=i and d<=j and d<=k and d*2<i and d*2<j and d*2>=k:
print('%d\n%d\n%d'%(i,j,k))
exit()
print(-1) | Title: Masha and Bears
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size *a* can climb into some car with size *b* if and only if *a*<=≤<=*b*, he or she likes it if and only if he can climb into this car and 2*a*<=≥<=*b*.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input Specification:
You are given four integers *V*1, *V*2, *V*3, *V**m*(1<=≤<=*V**i*<=≤<=100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that *V*1<=><=*V*2<=><=*V*3.
Output Specification:
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Demo Input:
['50 30 10 10\n', '100 50 10 21\n']
Demo Output:
['50\n30\n10\n', '-1\n']
Note:
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20. | ```python
a,b,c,d=map(int,input().split())
for i in range(a,a*2+1):
for j in range(b,b*2+1):
for k in range(c,c*2+1):
if d<=i and d<=j and d<=k and d*2<i and d*2<j and d*2>=k:
print('%d\n%d\n%d'%(i,j,k))
exit()
print(-1)
``` | 0 | |
615 | D | Multipliers | PROGRAMMING | 2,000 | [
"math",
"number theory"
] | null | null | Ayrat has number *n*, represented as it's prime factorization *p**i* of size *m*, i.e. *n*<==<=*p*1·*p*2·...·*p**m*. Ayrat got secret information that that the product of all divisors of *n* taken modulo 109<=+<=7 is the password to the secret data base. Now he wants to calculate this value. | The first line of the input contains a single integer *m* (1<=≤<=*m*<=≤<=200<=000) — the number of primes in factorization of *n*.
The second line contains *m* primes numbers *p**i* (2<=≤<=*p**i*<=≤<=200<=000). | Print one integer — the product of all divisors of *n* modulo 109<=+<=7. | [
"2\n2 3\n",
"3\n2 3 2\n"
] | [
"36\n",
"1728\n"
] | In the first sample *n* = 2·3 = 6. The divisors of 6 are 1, 2, 3 and 6, their product is equal to 1·2·3·6 = 36.
In the second sample 2·3·2 = 12. The divisors of 12 are 1, 2, 3, 4, 6 and 12. 1·2·3·4·6·12 = 1728. | 2,000 | [
{
"input": "2\n2 3",
"output": "36"
},
{
"input": "3\n2 3 2",
"output": "1728"
},
{
"input": "1\n2017",
"output": "2017"
},
{
"input": "2\n63997 63997",
"output": "135893224"
},
{
"input": "5\n11 7 11 7 11",
"output": "750455957"
},
{
"input": "5\n2 2 ... | 1,619,568,813 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 92 | 2,048,000 | m = int(input())
f = {}
n = 1
MOD = int(1e9+7)
p = list(map(int,input().split()))
for i in range(m):
n *= p[i]
if p[i] in f:
f[p[i]]+=1
else: f[p[i]] = 1
d = 1
for k in f:
d *= f[k]+1
if (d & 1) == 1:
sq = 1
for k in f:
for i in range(f[k]//2):
sq *= k
print(pow(sq,d,MOD))
else:
print(pow(n,d/2,MOD))
| Title: Multipliers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ayrat has number *n*, represented as it's prime factorization *p**i* of size *m*, i.e. *n*<==<=*p*1·*p*2·...·*p**m*. Ayrat got secret information that that the product of all divisors of *n* taken modulo 109<=+<=7 is the password to the secret data base. Now he wants to calculate this value.
Input Specification:
The first line of the input contains a single integer *m* (1<=≤<=*m*<=≤<=200<=000) — the number of primes in factorization of *n*.
The second line contains *m* primes numbers *p**i* (2<=≤<=*p**i*<=≤<=200<=000).
Output Specification:
Print one integer — the product of all divisors of *n* modulo 109<=+<=7.
Demo Input:
['2\n2 3\n', '3\n2 3 2\n']
Demo Output:
['36\n', '1728\n']
Note:
In the first sample *n* = 2·3 = 6. The divisors of 6 are 1, 2, 3 and 6, their product is equal to 1·2·3·6 = 36.
In the second sample 2·3·2 = 12. The divisors of 12 are 1, 2, 3, 4, 6 and 12. 1·2·3·4·6·12 = 1728. | ```python
m = int(input())
f = {}
n = 1
MOD = int(1e9+7)
p = list(map(int,input().split()))
for i in range(m):
n *= p[i]
if p[i] in f:
f[p[i]]+=1
else: f[p[i]] = 1
d = 1
for k in f:
d *= f[k]+1
if (d & 1) == 1:
sq = 1
for k in f:
for i in range(f[k]//2):
sq *= k
print(pow(sq,d,MOD))
else:
print(pow(n,d/2,MOD))
``` | -1 | |
665 | C | Simple Strings | PROGRAMMING | 1,300 | [
"dp",
"greedy",
"strings"
] | null | null | zscoder loves simple strings! A string *t* is called simple if every pair of adjacent characters are distinct. For example ab, aba, zscoder are simple whereas aa, add are not simple.
zscoder is given a string *s*. He wants to change a minimum number of characters so that the string *s* becomes simple. Help him with this task! | The only line contains the string *s* (1<=≤<=|*s*|<=≤<=2·105) — the string given to zscoder. The string *s* consists of only lowercase English letters. | Print the simple string *s*' — the string *s* after the minimal number of changes. If there are multiple solutions, you may output any of them.
Note that the string *s*' should also consist of only lowercase English letters. | [
"aab\n",
"caaab\n",
"zscoder\n"
] | [
"bab\n",
"cabab\n",
"zscoder\n"
] | none | 0 | [
{
"input": "aab",
"output": "bab"
},
{
"input": "caaab",
"output": "cabab"
},
{
"input": "zscoder",
"output": "zscoder"
},
{
"input": "u",
"output": "u"
},
{
"input": "h",
"output": "h"
},
{
"input": "dtottttotd",
"output": "dtotataotd"
},
{
... | 1,641,663,024 | 2,147,483,647 | Python 3 | OK | TESTS | 27 | 187 | 2,764,800 | def inc(a):
return chr(97 + (ord(a)-96)%26)
l = [i for i in input()]
i = 1
while i<len(l):
if l[i] == l[i-1]:
l[i] = inc(l[i])
if i+1 < len(l) and l[i] == l[i+1]:
l[i] = inc(l[i])
i += 2
continue
i += 1
print("".join(l))
| Title: Simple Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
zscoder loves simple strings! A string *t* is called simple if every pair of adjacent characters are distinct. For example ab, aba, zscoder are simple whereas aa, add are not simple.
zscoder is given a string *s*. He wants to change a minimum number of characters so that the string *s* becomes simple. Help him with this task!
Input Specification:
The only line contains the string *s* (1<=≤<=|*s*|<=≤<=2·105) — the string given to zscoder. The string *s* consists of only lowercase English letters.
Output Specification:
Print the simple string *s*' — the string *s* after the minimal number of changes. If there are multiple solutions, you may output any of them.
Note that the string *s*' should also consist of only lowercase English letters.
Demo Input:
['aab\n', 'caaab\n', 'zscoder\n']
Demo Output:
['bab\n', 'cabab\n', 'zscoder\n']
Note:
none | ```python
def inc(a):
return chr(97 + (ord(a)-96)%26)
l = [i for i in input()]
i = 1
while i<len(l):
if l[i] == l[i-1]:
l[i] = inc(l[i])
if i+1 < len(l) and l[i] == l[i+1]:
l[i] = inc(l[i])
i += 2
continue
i += 1
print("".join(l))
``` | 3 | |
628 | A | Tennis Tournament | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | null | null | A tennis tournament with *n* participants is running. The participants are playing by an olympic system, so the winners move on and the losers drop out.
The tournament takes place in the following way (below, *m* is the number of the participants of the current round):
- let *k* be the maximal power of the number 2 such that *k*<=≤<=*m*, - *k* participants compete in the current round and a half of them passes to the next round, the other *m*<=-<=*k* participants pass to the next round directly, - when only one participant remains, the tournament finishes.
Each match requires *b* bottles of water for each participant and one bottle for the judge. Besides *p* towels are given to each participant for the whole tournament.
Find the number of bottles and towels needed for the tournament.
Note that it's a tennis tournament so in each match two participants compete (one of them will win and the other will lose). | The only line contains three integers *n*,<=*b*,<=*p* (1<=≤<=*n*,<=*b*,<=*p*<=≤<=500) — the number of participants and the parameters described in the problem statement. | Print two integers *x* and *y* — the number of bottles and towels need for the tournament. | [
"5 2 3\n",
"8 2 4\n"
] | [
"20 15\n",
"35 32\n"
] | In the first example will be three rounds:
1. in the first round will be two matches and for each match 5 bottles of water are needed (two for each of the participants and one for the judge), 1. in the second round will be only one match, so we need another 5 bottles of water, 1. in the third round will also be only one match, so we need another 5 bottles of water.
So in total we need 20 bottles of water.
In the second example no participant will move on to some round directly. | 0 | [
{
"input": "5 2 3",
"output": "20 15"
},
{
"input": "8 2 4",
"output": "35 32"
},
{
"input": "10 1 500",
"output": "27 5000"
},
{
"input": "20 500 1",
"output": "19019 20"
},
{
"input": "100 123 99",
"output": "24453 9900"
},
{
"input": "500 1 1",
... | 1,665,071,386 | 2,147,483,647 | Python 3 | OK | TESTS | 38 | 46 | 0 | # 628A Tennis Tournament
def tennis():
n, b, p = map(int, input().split())
participants = n
t = 0
while participants > 1:
t += participants // 2
participants = participants // 2 + participants % 2
print(f'{t + t * b * 2} {p*n}')
tennis() | Title: Tennis Tournament
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A tennis tournament with *n* participants is running. The participants are playing by an olympic system, so the winners move on and the losers drop out.
The tournament takes place in the following way (below, *m* is the number of the participants of the current round):
- let *k* be the maximal power of the number 2 such that *k*<=≤<=*m*, - *k* participants compete in the current round and a half of them passes to the next round, the other *m*<=-<=*k* participants pass to the next round directly, - when only one participant remains, the tournament finishes.
Each match requires *b* bottles of water for each participant and one bottle for the judge. Besides *p* towels are given to each participant for the whole tournament.
Find the number of bottles and towels needed for the tournament.
Note that it's a tennis tournament so in each match two participants compete (one of them will win and the other will lose).
Input Specification:
The only line contains three integers *n*,<=*b*,<=*p* (1<=≤<=*n*,<=*b*,<=*p*<=≤<=500) — the number of participants and the parameters described in the problem statement.
Output Specification:
Print two integers *x* and *y* — the number of bottles and towels need for the tournament.
Demo Input:
['5 2 3\n', '8 2 4\n']
Demo Output:
['20 15\n', '35 32\n']
Note:
In the first example will be three rounds:
1. in the first round will be two matches and for each match 5 bottles of water are needed (two for each of the participants and one for the judge), 1. in the second round will be only one match, so we need another 5 bottles of water, 1. in the third round will also be only one match, so we need another 5 bottles of water.
So in total we need 20 bottles of water.
In the second example no participant will move on to some round directly. | ```python
# 628A Tennis Tournament
def tennis():
n, b, p = map(int, input().split())
participants = n
t = 0
while participants > 1:
t += participants // 2
participants = participants // 2 + participants % 2
print(f'{t + t * b * 2} {p*n}')
tennis()
``` | 3 | |
584 | A | Olesya and Rodion | PROGRAMMING | 1,000 | [
"math"
] | null | null | Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them.
Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1. | The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by. | Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them. | [
"3 2\n"
] | [
"712"
] | none | 500 | [
{
"input": "3 2",
"output": "222"
},
{
"input": "2 2",
"output": "22"
},
{
"input": "4 3",
"output": "3333"
},
{
"input": "5 3",
"output": "33333"
},
{
"input": "10 7",
"output": "7777777777"
},
{
"input": "2 9",
"output": "99"
},
{
"input"... | 1,679,773,503 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 31 | 0 | n,t = map(int, intput().split())
if n ==1 and t==10:
print(-1)
if t==10:
t=1
for i in range(n):
print(0)
| Title: Olesya and Rodion
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Olesya loves numbers consisting of *n* digits, and Rodion only likes numbers that are divisible by *t*. Find some number that satisfies both of them.
Your task is: given the *n* and *t* print an integer strictly larger than zero consisting of *n* digits that is divisible by *t*. If such number doesn't exist, print <=-<=1.
Input Specification:
The single line contains two numbers, *n* and *t* (1<=≤<=*n*<=≤<=100, 2<=≤<=*t*<=≤<=10) — the length of the number and the number it should be divisible by.
Output Specification:
Print one such positive number without leading zeroes, — the answer to the problem, or <=-<=1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.
Demo Input:
['3 2\n']
Demo Output:
['712']
Note:
none | ```python
n,t = map(int, intput().split())
if n ==1 and t==10:
print(-1)
if t==10:
t=1
for i in range(n):
print(0)
``` | -1 | |
1,003 | B | Binary String Constructing | PROGRAMMING | 1,300 | [
"constructive algorithms"
] | null | null | You are given three integers $a$, $b$ and $x$. Your task is to construct a binary string $s$ of length $n = a + b$ such that there are exactly $a$ zeroes, exactly $b$ ones and exactly $x$ indices $i$ (where $1 \le i < n$) such that $s_i \ne s_{i + 1}$. It is guaranteed that the answer always exists.
For example, for the string "01010" there are four indices $i$ such that $1 \le i < n$ and $s_i \ne s_{i + 1}$ ($i = 1, 2, 3, 4$). For the string "111001" there are two such indices $i$ ($i = 3, 5$).
Recall that binary string is a non-empty sequence of characters where each character is either 0 or 1. | The first line of the input contains three integers $a$, $b$ and $x$ ($1 \le a, b \le 100, 1 \le x < a + b)$. | Print only one string $s$, where $s$ is any binary string satisfying conditions described above. It is guaranteed that the answer always exists. | [
"2 2 1\n",
"3 3 3\n",
"5 3 6\n"
] | [
"1100\n",
"101100\n",
"01010100\n"
] | All possible answers for the first example:
- 1100; - 0011.
All possible answers for the second example:
- 110100; - 101100; - 110010; - 100110; - 011001; - 001101; - 010011; - 001011. | 0 | [
{
"input": "2 2 1",
"output": "1100"
},
{
"input": "3 3 3",
"output": "101100"
},
{
"input": "5 3 6",
"output": "01010100"
},
{
"input": "100 1 2",
"output": "01000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
... | 1,641,494,356 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 4 | 61 | 0 | import sys
input = sys.stdin.readline
import math
a,b,x=map(int,input().split())
if a>b:
s='01'
else:
s='10'
repeat=x//2
string = s*repeat
if x%2!=0 and s=='10':
string=string+'1'*(a-repeat)+'0'*(b-repeat)
elif x%2!=0 and s=='01':
string=string+'0'*(b-repeat)+'1'*(a-repeat)
elif s=='01':
string=string+'1'*(b-repeat)+'0'*(a-repeat)
else:
string=string+'0'*(a-repeat)+'1'*(b-repeat)
print(string)
| Title: Binary String Constructing
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given three integers $a$, $b$ and $x$. Your task is to construct a binary string $s$ of length $n = a + b$ such that there are exactly $a$ zeroes, exactly $b$ ones and exactly $x$ indices $i$ (where $1 \le i < n$) such that $s_i \ne s_{i + 1}$. It is guaranteed that the answer always exists.
For example, for the string "01010" there are four indices $i$ such that $1 \le i < n$ and $s_i \ne s_{i + 1}$ ($i = 1, 2, 3, 4$). For the string "111001" there are two such indices $i$ ($i = 3, 5$).
Recall that binary string is a non-empty sequence of characters where each character is either 0 or 1.
Input Specification:
The first line of the input contains three integers $a$, $b$ and $x$ ($1 \le a, b \le 100, 1 \le x < a + b)$.
Output Specification:
Print only one string $s$, where $s$ is any binary string satisfying conditions described above. It is guaranteed that the answer always exists.
Demo Input:
['2 2 1\n', '3 3 3\n', '5 3 6\n']
Demo Output:
['1100\n', '101100\n', '01010100\n']
Note:
All possible answers for the first example:
- 1100; - 0011.
All possible answers for the second example:
- 110100; - 101100; - 110010; - 100110; - 011001; - 001101; - 010011; - 001011. | ```python
import sys
input = sys.stdin.readline
import math
a,b,x=map(int,input().split())
if a>b:
s='01'
else:
s='10'
repeat=x//2
string = s*repeat
if x%2!=0 and s=='10':
string=string+'1'*(a-repeat)+'0'*(b-repeat)
elif x%2!=0 and s=='01':
string=string+'0'*(b-repeat)+'1'*(a-repeat)
elif s=='01':
string=string+'1'*(b-repeat)+'0'*(a-repeat)
else:
string=string+'0'*(a-repeat)+'1'*(b-repeat)
print(string)
``` | 0 | |
670 | B | Game of Robots | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | In late autumn evening *n* robots gathered in the cheerful company of friends. Each robot has a unique identifier — an integer from 1 to 109.
At some moment, robots decided to play the game "Snowball". Below there are the rules of this game. First, all robots stand in a row. Then the first robot says his identifier. After that the second robot says the identifier of the first robot and then says his own identifier. Then the third robot says the identifier of the first robot, then says the identifier of the second robot and after that says his own. This process continues from left to right until the *n*-th robot says his identifier.
Your task is to determine the *k*-th identifier to be pronounced. | The first line contains two positive integers *n* and *k* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*k*<=≤<=*min*(2·109,<=*n*·(*n*<=+<=1)<=/<=2).
The second line contains the sequence *id*1,<=*id*2,<=...,<=*id**n* (1<=≤<=*id**i*<=≤<=109) — identifiers of roborts. It is guaranteed that all identifiers are different. | Print the *k*-th pronounced identifier (assume that the numeration starts from 1). | [
"2 2\n1 2\n",
"4 5\n10 4 18 3\n"
] | [
"1\n",
"4\n"
] | In the first sample identifiers of robots will be pronounced in the following order: 1, 1, 2. As *k* = 2, the answer equals to 1.
In the second test case identifiers of robots will be pronounced in the following order: 10, 10, 4, 10, 4, 18, 10, 4, 18, 3. As *k* = 5, the answer equals to 4. | 750 | [
{
"input": "2 2\n1 2",
"output": "1"
},
{
"input": "4 5\n10 4 18 3",
"output": "4"
},
{
"input": "1 1\n4",
"output": "4"
},
{
"input": "2 1\n5 1",
"output": "5"
},
{
"input": "2 2\n1 4",
"output": "1"
},
{
"input": "2 3\n6 7",
"output": "7"
},
... | 1,691,015,296 | 2,147,483,647 | PyPy 3-64 | TIME_LIMIT_EXCEEDED | TESTS | 12 | 1,000 | 11,264,000 | tamanho_numero = str(input()).split(" ")
tamanho = int(tamanho_numero[0])
num = int(tamanho_numero[1])
identificadores = str(input()).split(" ")
convertido = [int(x) for x in identificadores]
meio = tamanho//2
def area(base, altura):
area = (base*altura) // 2
return area
def busca(mid, numero):
if mid == 0:
return 1
if area(mid,mid+1) == numero:
return mid
elif area(mid, mid+1) < numero:
if(tamanho - mid) == 1:
mid += 1
return busca(mid, numero)
mid += (tamanho-mid) // 2
return busca(mid, numero)
else:
if area(mid, mid+1) - numero <= numero:
return mid
else:
mid -= 1
return busca(mid, numero)
def posicao(linha, numero):
quantia = area(linha-1, linha)
contador = 0
while quantia != numero:
quantia +=1
contador += 1
return contador
ans = posicao(busca(meio,num), num)
print(convertido[ans-1])
| Title: Game of Robots
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In late autumn evening *n* robots gathered in the cheerful company of friends. Each robot has a unique identifier — an integer from 1 to 109.
At some moment, robots decided to play the game "Snowball". Below there are the rules of this game. First, all robots stand in a row. Then the first robot says his identifier. After that the second robot says the identifier of the first robot and then says his own identifier. Then the third robot says the identifier of the first robot, then says the identifier of the second robot and after that says his own. This process continues from left to right until the *n*-th robot says his identifier.
Your task is to determine the *k*-th identifier to be pronounced.
Input Specification:
The first line contains two positive integers *n* and *k* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*k*<=≤<=*min*(2·109,<=*n*·(*n*<=+<=1)<=/<=2).
The second line contains the sequence *id*1,<=*id*2,<=...,<=*id**n* (1<=≤<=*id**i*<=≤<=109) — identifiers of roborts. It is guaranteed that all identifiers are different.
Output Specification:
Print the *k*-th pronounced identifier (assume that the numeration starts from 1).
Demo Input:
['2 2\n1 2\n', '4 5\n10 4 18 3\n']
Demo Output:
['1\n', '4\n']
Note:
In the first sample identifiers of robots will be pronounced in the following order: 1, 1, 2. As *k* = 2, the answer equals to 1.
In the second test case identifiers of robots will be pronounced in the following order: 10, 10, 4, 10, 4, 18, 10, 4, 18, 3. As *k* = 5, the answer equals to 4. | ```python
tamanho_numero = str(input()).split(" ")
tamanho = int(tamanho_numero[0])
num = int(tamanho_numero[1])
identificadores = str(input()).split(" ")
convertido = [int(x) for x in identificadores]
meio = tamanho//2
def area(base, altura):
area = (base*altura) // 2
return area
def busca(mid, numero):
if mid == 0:
return 1
if area(mid,mid+1) == numero:
return mid
elif area(mid, mid+1) < numero:
if(tamanho - mid) == 1:
mid += 1
return busca(mid, numero)
mid += (tamanho-mid) // 2
return busca(mid, numero)
else:
if area(mid, mid+1) - numero <= numero:
return mid
else:
mid -= 1
return busca(mid, numero)
def posicao(linha, numero):
quantia = area(linha-1, linha)
contador = 0
while quantia != numero:
quantia +=1
contador += 1
return contador
ans = posicao(busca(meio,num), num)
print(convertido[ans-1])
``` | 0 | |
370 | A | Rook, Bishop and King | PROGRAMMING | 1,100 | [
"graphs",
"math",
"shortest paths"
] | null | null | Little Petya is learning to play chess. He has already learned how to move a king, a rook and a bishop. Let us remind you the rules of moving chess pieces. A chessboard is 64 square fields organized into an 8<=×<=8 table. A field is represented by a pair of integers (*r*,<=*c*) — the number of the row and the number of the column (in a classical game the columns are traditionally indexed by letters). Each chess piece takes up exactly one field. To make a move is to move a chess piece, the pieces move by the following rules:
- A rook moves any number of fields horizontally or vertically. - A bishop moves any number of fields diagonally. - A king moves one field in any direction — horizontally, vertically or diagonally.
Petya is thinking about the following problem: what minimum number of moves is needed for each of these pieces to move from field (*r*1,<=*c*1) to field (*r*2,<=*c*2)? At that, we assume that there are no more pieces besides this one on the board. Help him solve this problem. | The input contains four integers *r*1,<=*c*1,<=*r*2,<=*c*2 (1<=≤<=*r*1,<=*c*1,<=*r*2,<=*c*2<=≤<=8) — the coordinates of the starting and the final field. The starting field doesn't coincide with the final one.
You can assume that the chessboard rows are numbered from top to bottom 1 through 8, and the columns are numbered from left to right 1 through 8. | Print three space-separated integers: the minimum number of moves the rook, the bishop and the king (in this order) is needed to move from field (*r*1,<=*c*1) to field (*r*2,<=*c*2). If a piece cannot make such a move, print a 0 instead of the corresponding number. | [
"4 3 1 6\n",
"5 5 5 6\n"
] | [
"2 1 3\n",
"1 0 1\n"
] | none | 500 | [
{
"input": "4 3 1 6",
"output": "2 1 3"
},
{
"input": "5 5 5 6",
"output": "1 0 1"
},
{
"input": "1 1 8 8",
"output": "2 1 7"
},
{
"input": "1 1 8 1",
"output": "1 0 7"
},
{
"input": "1 1 1 8",
"output": "1 0 7"
},
{
"input": "8 1 1 1",
"output": "... | 1,684,688,734 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 31 | 614,400 | from queue import deque
def rook_neighbours(pos):
i, j = pos
return [(i,k) for k in range(1,9) if k != j] + [(k,j) for k in range(1,9) if k!=i]
def bishop_neighbours(pos):
i, j = pos
return [(i+k,j+k) for k in range(-8,9) if 1<=i+k<=8 and 1<=j+k<=8 and k!= 0]+[(i+k, j-k) for k in range(-8,9) if 1<=i+k<=8 and 1<=j-k<=8 and k!= 0]
def king_neighbours(pos):
i,j = pos
l = []
if i > 1:
l.append((i-1,j))
if j > 1:
l.append((i-1,j-1))
if j < 8:
l.append((i-1, j+1))
if i < 8:
l.append((i+1,j))
if j > 1:
l.append((i+1,j-1))
if j < 8:
l.append((i+1,j+1))
if j > 1:
l.append((i,j-1))
if j < 8:
l.append((i,j+1))
return l
def bfs(pos1, pos2, neighbours):
seen = set(pos1)
d = deque()
d.appendleft((pos1,0))
while d:
pos1, n = d.pop()
seen.add(pos1)
if pos1 == pos2:
return n
nb = neighbours(pos1)
d.extendleft((x,n+1) for x in nb if x not in seen)
a1, b1, a2,b2 = tuple(map(int,input().split()))
pos1 = (a1,b1)
pos2 = (a2,b2)
nb_rook = bfs(pos1, pos2, rook_neighbours)
nb_king = bfs(pos1,pos2, king_neighbours)
if (pos1[0]+pos2[0]-(pos1[1]+pos2[1]))%2 == 0:
nb_bishop = bfs(pos1, pos2, bishop_neighbours)
else:
nb_bishop = 0
print(bishop_neighbours((4,3)))
print(nb_rook, nb_bishop, nb_king) | Title: Rook, Bishop and King
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Petya is learning to play chess. He has already learned how to move a king, a rook and a bishop. Let us remind you the rules of moving chess pieces. A chessboard is 64 square fields organized into an 8<=×<=8 table. A field is represented by a pair of integers (*r*,<=*c*) — the number of the row and the number of the column (in a classical game the columns are traditionally indexed by letters). Each chess piece takes up exactly one field. To make a move is to move a chess piece, the pieces move by the following rules:
- A rook moves any number of fields horizontally or vertically. - A bishop moves any number of fields diagonally. - A king moves one field in any direction — horizontally, vertically or diagonally.
Petya is thinking about the following problem: what minimum number of moves is needed for each of these pieces to move from field (*r*1,<=*c*1) to field (*r*2,<=*c*2)? At that, we assume that there are no more pieces besides this one on the board. Help him solve this problem.
Input Specification:
The input contains four integers *r*1,<=*c*1,<=*r*2,<=*c*2 (1<=≤<=*r*1,<=*c*1,<=*r*2,<=*c*2<=≤<=8) — the coordinates of the starting and the final field. The starting field doesn't coincide with the final one.
You can assume that the chessboard rows are numbered from top to bottom 1 through 8, and the columns are numbered from left to right 1 through 8.
Output Specification:
Print three space-separated integers: the minimum number of moves the rook, the bishop and the king (in this order) is needed to move from field (*r*1,<=*c*1) to field (*r*2,<=*c*2). If a piece cannot make such a move, print a 0 instead of the corresponding number.
Demo Input:
['4 3 1 6\n', '5 5 5 6\n']
Demo Output:
['2 1 3\n', '1 0 1\n']
Note:
none | ```python
from queue import deque
def rook_neighbours(pos):
i, j = pos
return [(i,k) for k in range(1,9) if k != j] + [(k,j) for k in range(1,9) if k!=i]
def bishop_neighbours(pos):
i, j = pos
return [(i+k,j+k) for k in range(-8,9) if 1<=i+k<=8 and 1<=j+k<=8 and k!= 0]+[(i+k, j-k) for k in range(-8,9) if 1<=i+k<=8 and 1<=j-k<=8 and k!= 0]
def king_neighbours(pos):
i,j = pos
l = []
if i > 1:
l.append((i-1,j))
if j > 1:
l.append((i-1,j-1))
if j < 8:
l.append((i-1, j+1))
if i < 8:
l.append((i+1,j))
if j > 1:
l.append((i+1,j-1))
if j < 8:
l.append((i+1,j+1))
if j > 1:
l.append((i,j-1))
if j < 8:
l.append((i,j+1))
return l
def bfs(pos1, pos2, neighbours):
seen = set(pos1)
d = deque()
d.appendleft((pos1,0))
while d:
pos1, n = d.pop()
seen.add(pos1)
if pos1 == pos2:
return n
nb = neighbours(pos1)
d.extendleft((x,n+1) for x in nb if x not in seen)
a1, b1, a2,b2 = tuple(map(int,input().split()))
pos1 = (a1,b1)
pos2 = (a2,b2)
nb_rook = bfs(pos1, pos2, rook_neighbours)
nb_king = bfs(pos1,pos2, king_neighbours)
if (pos1[0]+pos2[0]-(pos1[1]+pos2[1]))%2 == 0:
nb_bishop = bfs(pos1, pos2, bishop_neighbours)
else:
nb_bishop = 0
print(bishop_neighbours((4,3)))
print(nb_rook, nb_bishop, nb_king)
``` | 0 | |
270 | A | Fancy Fence | PROGRAMMING | 1,100 | [
"geometry",
"implementation",
"math"
] | null | null | Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*? | The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees. | For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible. | [
"3\n30\n60\n90\n"
] | [
"NO\nYES\nYES\n"
] | In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square. | 500 | [
{
"input": "3\n30\n60\n90",
"output": "NO\nYES\nYES"
},
{
"input": "6\n1\n2\n3\n170\n179\n25",
"output": "NO\nNO\nNO\nYES\nYES\nNO"
}
] | 1,666,950,063 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | =int(input())
a=0
b=0
for i in range(t):
i=int(input())
b=360//i
a=(b-2)*180//b
if a==i:
print('YES')
else:
print('NO')
| Title: Fancy Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
Input Specification:
The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees.
Output Specification:
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
Demo Input:
['3\n30\n60\n90\n']
Demo Output:
['NO\nYES\nYES\n']
Note:
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square. | ```python
=int(input())
a=0
b=0
for i in range(t):
i=int(input())
b=360//i
a=(b-2)*180//b
if a==i:
print('YES')
else:
print('NO')
``` | -1 | |
612 | E | Square Root of Permutation | PROGRAMMING | 2,200 | [
"combinatorics",
"constructive algorithms",
"dfs and similar",
"graphs",
"math"
] | null | null | A permutation of length *n* is an array containing each integer from 1 to *n* exactly once. For example, *q*<==<=[4,<=5,<=1,<=2,<=3] is a permutation. For the permutation *q* the square of permutation is the permutation *p* that *p*[*i*]<==<=*q*[*q*[*i*]] for each *i*<==<=1... *n*. For example, the square of *q*<==<=[4,<=5,<=1,<=2,<=3] is *p*<==<=*q*2<==<=[2,<=3,<=4,<=5,<=1].
This problem is about the inverse operation: given the permutation *p* you task is to find such permutation *q* that *q*2<==<=*p*. If there are several such *q* find any of them. | The first line contains integer *n* (1<=≤<=*n*<=≤<=106) — the number of elements in permutation *p*.
The second line contains *n* distinct integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — the elements of permutation *p*. | If there is no permutation *q* such that *q*2<==<=*p* print the number "-1".
If the answer exists print it. The only line should contain *n* different integers *q**i* (1<=≤<=*q**i*<=≤<=*n*) — the elements of the permutation *q*. If there are several solutions print any of them. | [
"4\n2 1 4 3\n",
"4\n2 1 3 4\n",
"5\n2 3 4 5 1\n"
] | [
"3 4 2 1\n",
"-1\n",
"4 5 1 2 3\n"
] | none | 0 | [
{
"input": "4\n2 1 4 3",
"output": "3 4 2 1"
},
{
"input": "4\n2 1 3 4",
"output": "-1"
},
{
"input": "5\n2 3 4 5 1",
"output": "4 5 1 2 3"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n8 2 10 3 4 6 1 7 9 5"... | 1,621,177,221 | 2,147,483,647 | PyPy 3 | OK | TESTS | 18 | 1,730 | 96,358,400 | import sys
import bisect
from bisect import bisect_left as lb
input_=lambda: sys.stdin.readline().strip("\r\n")
from math import log
from math import gcd
from math import atan2,acos
from random import randint
sa=lambda :input_()
sb=lambda:int(input_())
sc=lambda:input_().split()
sd=lambda:list(map(int,input_().split()))
se=lambda:float(input_())
sf=lambda:list(input_())
flsh=lambda: sys.stdout.flush()
#sys.setrecursionlimit(10**6)
mod=10**9+7
gp=[]
cost=[]
dp=[]
mx=[]
ans1=[]
ans2=[]
special=[]
specnode=[]
a=0
kthpar=[]
def dfs(root,par):
if par!=-1:
dp[root]=dp[par]+1
for i in range(1,20):
if kthpar[root][i-1]!=-1:
kthpar[root][i]=kthpar[kthpar[root][i-1]][i-1]
for child in gp[root]:
if child==par:continue
kthpar[child][0]=root
dfs(child,root)
def hnbhai():
n=sb()
a=[0]+sd()
ans=[-1]*(n+1)
d={}
visited=[0]*(n+1)
for i in range(1,n+1):
if visited[i]==0:
g=[]
abe=i
while not visited[abe]:
visited[abe]=1
g.append(abe)
abe=a[abe]
if len(g)%2:
mid=(len(g)+1)//2
for i in range(len(g)):
ans[g[i]]=g[(i+mid)%len(g)]
else:
if d.get(len(g)):
temp=d[len(g)]
for i in range(len(g)):
ans[g[i]]=temp[(i+1)%len(g)]
ans[temp[i]]=g[i]
del d[len(g)]
else:
d[len(g)]=g
if len(d):
print(-1)
return
print(*ans[1:])
for _ in range(1):
hnbhai()
| Title: Square Root of Permutation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A permutation of length *n* is an array containing each integer from 1 to *n* exactly once. For example, *q*<==<=[4,<=5,<=1,<=2,<=3] is a permutation. For the permutation *q* the square of permutation is the permutation *p* that *p*[*i*]<==<=*q*[*q*[*i*]] for each *i*<==<=1... *n*. For example, the square of *q*<==<=[4,<=5,<=1,<=2,<=3] is *p*<==<=*q*2<==<=[2,<=3,<=4,<=5,<=1].
This problem is about the inverse operation: given the permutation *p* you task is to find such permutation *q* that *q*2<==<=*p*. If there are several such *q* find any of them.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=106) — the number of elements in permutation *p*.
The second line contains *n* distinct integers *p*1,<=*p*2,<=...,<=*p**n* (1<=≤<=*p**i*<=≤<=*n*) — the elements of permutation *p*.
Output Specification:
If there is no permutation *q* such that *q*2<==<=*p* print the number "-1".
If the answer exists print it. The only line should contain *n* different integers *q**i* (1<=≤<=*q**i*<=≤<=*n*) — the elements of the permutation *q*. If there are several solutions print any of them.
Demo Input:
['4\n2 1 4 3\n', '4\n2 1 3 4\n', '5\n2 3 4 5 1\n']
Demo Output:
['3 4 2 1\n', '-1\n', '4 5 1 2 3\n']
Note:
none | ```python
import sys
import bisect
from bisect import bisect_left as lb
input_=lambda: sys.stdin.readline().strip("\r\n")
from math import log
from math import gcd
from math import atan2,acos
from random import randint
sa=lambda :input_()
sb=lambda:int(input_())
sc=lambda:input_().split()
sd=lambda:list(map(int,input_().split()))
se=lambda:float(input_())
sf=lambda:list(input_())
flsh=lambda: sys.stdout.flush()
#sys.setrecursionlimit(10**6)
mod=10**9+7
gp=[]
cost=[]
dp=[]
mx=[]
ans1=[]
ans2=[]
special=[]
specnode=[]
a=0
kthpar=[]
def dfs(root,par):
if par!=-1:
dp[root]=dp[par]+1
for i in range(1,20):
if kthpar[root][i-1]!=-1:
kthpar[root][i]=kthpar[kthpar[root][i-1]][i-1]
for child in gp[root]:
if child==par:continue
kthpar[child][0]=root
dfs(child,root)
def hnbhai():
n=sb()
a=[0]+sd()
ans=[-1]*(n+1)
d={}
visited=[0]*(n+1)
for i in range(1,n+1):
if visited[i]==0:
g=[]
abe=i
while not visited[abe]:
visited[abe]=1
g.append(abe)
abe=a[abe]
if len(g)%2:
mid=(len(g)+1)//2
for i in range(len(g)):
ans[g[i]]=g[(i+mid)%len(g)]
else:
if d.get(len(g)):
temp=d[len(g)]
for i in range(len(g)):
ans[g[i]]=temp[(i+1)%len(g)]
ans[temp[i]]=g[i]
del d[len(g)]
else:
d[len(g)]=g
if len(d):
print(-1)
return
print(*ans[1:])
for _ in range(1):
hnbhai()
``` | 3 | |
534 | C | Polycarpus' Dice | PROGRAMMING | 1,600 | [
"math"
] | null | null | Polycarp has *n* dice *d*1,<=*d*2,<=...,<=*d**n*. The *i*-th dice shows numbers from 1 to *d**i*. Polycarp rolled all the dice and the sum of numbers they showed is *A*. Agrippina didn't see which dice showed what number, she knows only the sum *A* and the values *d*1,<=*d*2,<=...,<=*d**n*. However, she finds it enough to make a series of statements of the following type: dice *i* couldn't show number *r*. For example, if Polycarp had two six-faced dice and the total sum is *A*<==<=11, then Agrippina can state that each of the two dice couldn't show a value less than five (otherwise, the remaining dice must have a value of at least seven, which is impossible).
For each dice find the number of values for which it can be guaranteed that the dice couldn't show these values if the sum of the shown values is *A*. | The first line contains two integers *n*,<=*A* (1<=≤<=*n*<=≤<=2·105,<=*n*<=≤<=*A*<=≤<=*s*) — the number of dice and the sum of shown values where *s*<==<=*d*1<=+<=*d*2<=+<=...<=+<=*d**n*.
The second line contains *n* integers *d*1,<=*d*2,<=...,<=*d**n* (1<=≤<=*d**i*<=≤<=106), where *d**i* is the maximum value that the *i*-th dice can show. | Print *n* integers *b*1,<=*b*2,<=...,<=*b**n*, where *b**i* is the number of values for which it is guaranteed that the *i*-th dice couldn't show them. | [
"2 8\n4 4\n",
"1 3\n5\n",
"2 3\n2 3\n"
] | [
"3 3 ",
"4 ",
"0 1 "
] | In the first sample from the statement *A* equal to 8 could be obtained in the only case when both the first and the second dice show 4. Correspondingly, both dice couldn't show values 1, 2 or 3.
In the second sample from the statement *A* equal to 3 could be obtained when the single dice shows 3. Correspondingly, it couldn't show 1, 2, 4 or 5.
In the third sample from the statement *A* equal to 3 could be obtained when one dice shows 1 and the other dice shows 2. That's why the first dice doesn't have any values it couldn't show and the second dice couldn't show 3. | 1,500 | [
{
"input": "2 8\n4 4",
"output": "3 3 "
},
{
"input": "1 3\n5",
"output": "4 "
},
{
"input": "2 3\n2 3",
"output": "0 1 "
},
{
"input": "1 1\n3",
"output": "2 "
},
{
"input": "1 2\n3",
"output": "2 "
},
{
"input": "2 2\n2 3",
"output": "1 2 "
},
... | 1,429,019,828 | 2,147,483,647 | PyPy 3 | OK | TESTS | 99 | 467 | 19,558,400 | #!/usr/bin/python3
import sys
n, A = list(map(int, sys.stdin.readline().split()))
d = list(map(int, sys.stdin.readline().split()))
s = sum(d)
for x in d:
z = min(A - n + 1, x) - max(1, A - s + x) + 1
if z < 0:
z = 0
sys.stdout.write("%d " % (x - z))
sys.stdout.write('\n')
| Title: Polycarpus' Dice
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has *n* dice *d*1,<=*d*2,<=...,<=*d**n*. The *i*-th dice shows numbers from 1 to *d**i*. Polycarp rolled all the dice and the sum of numbers they showed is *A*. Agrippina didn't see which dice showed what number, she knows only the sum *A* and the values *d*1,<=*d*2,<=...,<=*d**n*. However, she finds it enough to make a series of statements of the following type: dice *i* couldn't show number *r*. For example, if Polycarp had two six-faced dice and the total sum is *A*<==<=11, then Agrippina can state that each of the two dice couldn't show a value less than five (otherwise, the remaining dice must have a value of at least seven, which is impossible).
For each dice find the number of values for which it can be guaranteed that the dice couldn't show these values if the sum of the shown values is *A*.
Input Specification:
The first line contains two integers *n*,<=*A* (1<=≤<=*n*<=≤<=2·105,<=*n*<=≤<=*A*<=≤<=*s*) — the number of dice and the sum of shown values where *s*<==<=*d*1<=+<=*d*2<=+<=...<=+<=*d**n*.
The second line contains *n* integers *d*1,<=*d*2,<=...,<=*d**n* (1<=≤<=*d**i*<=≤<=106), where *d**i* is the maximum value that the *i*-th dice can show.
Output Specification:
Print *n* integers *b*1,<=*b*2,<=...,<=*b**n*, where *b**i* is the number of values for which it is guaranteed that the *i*-th dice couldn't show them.
Demo Input:
['2 8\n4 4\n', '1 3\n5\n', '2 3\n2 3\n']
Demo Output:
['3 3 ', '4 ', '0 1 ']
Note:
In the first sample from the statement *A* equal to 8 could be obtained in the only case when both the first and the second dice show 4. Correspondingly, both dice couldn't show values 1, 2 or 3.
In the second sample from the statement *A* equal to 3 could be obtained when the single dice shows 3. Correspondingly, it couldn't show 1, 2, 4 or 5.
In the third sample from the statement *A* equal to 3 could be obtained when one dice shows 1 and the other dice shows 2. That's why the first dice doesn't have any values it couldn't show and the second dice couldn't show 3. | ```python
#!/usr/bin/python3
import sys
n, A = list(map(int, sys.stdin.readline().split()))
d = list(map(int, sys.stdin.readline().split()))
s = sum(d)
for x in d:
z = min(A - n + 1, x) - max(1, A - s + x) + 1
if z < 0:
z = 0
sys.stdout.write("%d " % (x - z))
sys.stdout.write('\n')
``` | 3 | |
368 | A | Sereja and Coat Rack | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Sereja owns a restaurant for *n* people. The restaurant hall has a coat rack with *n* hooks. Each restaurant visitor can use a hook to hang his clothes on it. Using the *i*-th hook costs *a**i* rubles. Only one person can hang clothes on one hook.
Tonight Sereja expects *m* guests in the restaurant. Naturally, each guest wants to hang his clothes on an available hook with minimum price (if there are multiple such hooks, he chooses any of them). However if the moment a guest arrives the rack has no available hooks, Sereja must pay a *d* ruble fine to the guest.
Help Sereja find out the profit in rubles (possibly negative) that he will get tonight. You can assume that before the guests arrive, all hooks on the rack are available, all guests come at different time, nobody besides the *m* guests is visiting Sereja's restaurant tonight. | The first line contains two integers *n* and *d* (1<=≤<=*n*,<=*d*<=≤<=100). The next line contains integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100). The third line contains integer *m* (1<=≤<=*m*<=≤<=100). | In a single line print a single integer — the answer to the problem. | [
"2 1\n2 1\n2\n",
"2 1\n2 1\n10\n"
] | [
"3\n",
"-5\n"
] | In the first test both hooks will be used, so Sereja gets 1 + 2 = 3 rubles.
In the second test both hooks will be used but Sereja pays a fine 8 times, so the answer is 3 - 8 = - 5. | 500 | [
{
"input": "2 1\n2 1\n2",
"output": "3"
},
{
"input": "2 1\n2 1\n10",
"output": "-5"
},
{
"input": "1 1\n1\n2",
"output": "0"
},
{
"input": "3 96\n83 22 17\n19",
"output": "-1414"
},
{
"input": "8 4\n27 72 39 70 13 68 100 36\n95",
"output": "77"
},
{
"... | 1,679,053,928 | 2,147,483,647 | PyPy 3 | OK | TESTS | 35 | 78 | 0 | I=lambda:map(int,input().split())
n,d=I()
a=sorted(I())+[-d]*99
print(sum(a[:next(I())])) | Title: Sereja and Coat Rack
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja owns a restaurant for *n* people. The restaurant hall has a coat rack with *n* hooks. Each restaurant visitor can use a hook to hang his clothes on it. Using the *i*-th hook costs *a**i* rubles. Only one person can hang clothes on one hook.
Tonight Sereja expects *m* guests in the restaurant. Naturally, each guest wants to hang his clothes on an available hook with minimum price (if there are multiple such hooks, he chooses any of them). However if the moment a guest arrives the rack has no available hooks, Sereja must pay a *d* ruble fine to the guest.
Help Sereja find out the profit in rubles (possibly negative) that he will get tonight. You can assume that before the guests arrive, all hooks on the rack are available, all guests come at different time, nobody besides the *m* guests is visiting Sereja's restaurant tonight.
Input Specification:
The first line contains two integers *n* and *d* (1<=≤<=*n*,<=*d*<=≤<=100). The next line contains integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=100). The third line contains integer *m* (1<=≤<=*m*<=≤<=100).
Output Specification:
In a single line print a single integer — the answer to the problem.
Demo Input:
['2 1\n2 1\n2\n', '2 1\n2 1\n10\n']
Demo Output:
['3\n', '-5\n']
Note:
In the first test both hooks will be used, so Sereja gets 1 + 2 = 3 rubles.
In the second test both hooks will be used but Sereja pays a fine 8 times, so the answer is 3 - 8 = - 5. | ```python
I=lambda:map(int,input().split())
n,d=I()
a=sorted(I())+[-d]*99
print(sum(a[:next(I())]))
``` | 3 | |
1,011 | A | Stages | PROGRAMMING | 900 | [
"greedy",
"implementation",
"sortings"
] | null | null | Natasha is going to fly to Mars. She needs to build a rocket, which consists of several stages in some order. Each of the stages is defined by a lowercase Latin letter. This way, the rocket can be described by the string — concatenation of letters, which correspond to the stages.
There are $n$ stages available. The rocket must contain exactly $k$ of them. Stages in the rocket should be ordered by their weight. So, after the stage with some letter can go only stage with a letter, which is at least two positions after in the alphabet (skipping one letter in between, or even more). For example, after letter 'c' can't go letters 'a', 'b', 'c' and 'd', but can go letters 'e', 'f', ..., 'z'.
For the rocket to fly as far as possible, its weight should be minimal. The weight of the rocket is equal to the sum of the weights of its stages. The weight of the stage is the number of its letter in the alphabet. For example, the stage 'a 'weighs one ton,' b 'weighs two tons, and' z' — $26$ tons.
Build the rocket with the minimal weight or determine, that it is impossible to build a rocket at all. Each stage can be used at most once. | The first line of input contains two integers — $n$ and $k$ ($1 \le k \le n \le 50$) – the number of available stages and the number of stages to use in the rocket.
The second line contains string $s$, which consists of exactly $n$ lowercase Latin letters. Each letter defines a new stage, which can be used to build the rocket. Each stage can be used at most once. | Print a single integer — the minimal total weight of the rocket or -1, if it is impossible to build the rocket at all. | [
"5 3\nxyabd\n",
"7 4\nproblem\n",
"2 2\nab\n",
"12 1\nabaabbaaabbb\n"
] | [
"29",
"34",
"-1",
"1"
] | In the first example, the following rockets satisfy the condition:
- "adx" (weight is $1+4+24=29$);- "ady" (weight is $1+4+25=30$);- "bdx" (weight is $2+4+24=30$);- "bdy" (weight is $2+4+25=31$).
Rocket "adx" has the minimal weight, so the answer is $29$.
In the second example, target rocket is "belo". Its weight is $2+5+12+15=34$.
In the third example, $n=k=2$, so the rocket must have both stages: 'a' and 'b'. This rocket doesn't satisfy the condition, because these letters are adjacent in the alphabet. Answer is -1. | 500 | [
{
"input": "5 3\nxyabd",
"output": "29"
},
{
"input": "7 4\nproblem",
"output": "34"
},
{
"input": "2 2\nab",
"output": "-1"
},
{
"input": "12 1\nabaabbaaabbb",
"output": "1"
},
{
"input": "50 13\nqwertyuiopasdfghjklzxcvbnmaaaaaaaaaaaaaaaaaaaaaaaa",
"output": ... | 1,533,044,535 | 2,147,483,647 | Python 3 | OK | TESTS | 29 | 109 | 0 | n, k = map(int, input().split())
s = input()
alph = 'abcdefghijklmnopqrstuvwxyz'
s = sorted(s)
min_count = float('inf')
for i in range(n):
c = 1
j = 0
count = alph.find(s[i]) + 1
x = alph.find(s[i])
while c != k and j != n:
if i != j and alph.find(s[j]) > x + 1:
c += 1
count += alph.find(s[j]) + 1
x = alph.find(s[j])
j += 1
if c == k and count < min_count:
min_count = count
print(-1 if min_count == float('inf') else min_count) | Title: Stages
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Natasha is going to fly to Mars. She needs to build a rocket, which consists of several stages in some order. Each of the stages is defined by a lowercase Latin letter. This way, the rocket can be described by the string — concatenation of letters, which correspond to the stages.
There are $n$ stages available. The rocket must contain exactly $k$ of them. Stages in the rocket should be ordered by their weight. So, after the stage with some letter can go only stage with a letter, which is at least two positions after in the alphabet (skipping one letter in between, or even more). For example, after letter 'c' can't go letters 'a', 'b', 'c' and 'd', but can go letters 'e', 'f', ..., 'z'.
For the rocket to fly as far as possible, its weight should be minimal. The weight of the rocket is equal to the sum of the weights of its stages. The weight of the stage is the number of its letter in the alphabet. For example, the stage 'a 'weighs one ton,' b 'weighs two tons, and' z' — $26$ tons.
Build the rocket with the minimal weight or determine, that it is impossible to build a rocket at all. Each stage can be used at most once.
Input Specification:
The first line of input contains two integers — $n$ and $k$ ($1 \le k \le n \le 50$) – the number of available stages and the number of stages to use in the rocket.
The second line contains string $s$, which consists of exactly $n$ lowercase Latin letters. Each letter defines a new stage, which can be used to build the rocket. Each stage can be used at most once.
Output Specification:
Print a single integer — the minimal total weight of the rocket or -1, if it is impossible to build the rocket at all.
Demo Input:
['5 3\nxyabd\n', '7 4\nproblem\n', '2 2\nab\n', '12 1\nabaabbaaabbb\n']
Demo Output:
['29', '34', '-1', '1']
Note:
In the first example, the following rockets satisfy the condition:
- "adx" (weight is $1+4+24=29$);- "ady" (weight is $1+4+25=30$);- "bdx" (weight is $2+4+24=30$);- "bdy" (weight is $2+4+25=31$).
Rocket "adx" has the minimal weight, so the answer is $29$.
In the second example, target rocket is "belo". Its weight is $2+5+12+15=34$.
In the third example, $n=k=2$, so the rocket must have both stages: 'a' and 'b'. This rocket doesn't satisfy the condition, because these letters are adjacent in the alphabet. Answer is -1. | ```python
n, k = map(int, input().split())
s = input()
alph = 'abcdefghijklmnopqrstuvwxyz'
s = sorted(s)
min_count = float('inf')
for i in range(n):
c = 1
j = 0
count = alph.find(s[i]) + 1
x = alph.find(s[i])
while c != k and j != n:
if i != j and alph.find(s[j]) > x + 1:
c += 1
count += alph.find(s[j]) + 1
x = alph.find(s[j])
j += 1
if c == k and count < min_count:
min_count = count
print(-1 if min_count == float('inf') else min_count)
``` | 3 | |
999 | B | Reversing Encryption | PROGRAMMING | 900 | [
"implementation"
] | null | null | A string $s$ of length $n$ can be encrypted by the following algorithm:
- iterate over all divisors of $n$ in decreasing order (i.e. from $n$ to $1$), - for each divisor $d$, reverse the substring $s[1 \dots d]$ (i.e. the substring which starts at position $1$ and ends at position $d$).
For example, the above algorithm applied to the string $s$="codeforces" leads to the following changes: "codeforces" $\to$ "secrofedoc" $\to$ "orcesfedoc" $\to$ "rocesfedoc" $\to$ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because $d=1$).
You are given the encrypted string $t$. Your task is to decrypt this string, i.e., to find a string $s$ such that the above algorithm results in string $t$. It can be proven that this string $s$ always exists and is unique. | The first line of input consists of a single integer $n$ ($1 \le n \le 100$) — the length of the string $t$. The second line of input consists of the string $t$. The length of $t$ is $n$, and it consists only of lowercase Latin letters. | Print a string $s$ such that the above algorithm results in $t$. | [
"10\nrocesfedoc\n",
"16\nplmaetwoxesisiht\n",
"1\nz\n"
] | [
"codeforces\n",
"thisisexampletwo\n",
"z\n"
] | The first example is described in the problem statement. | 0 | [
{
"input": "10\nrocesfedoc",
"output": "codeforces"
},
{
"input": "16\nplmaetwoxesisiht",
"output": "thisisexampletwo"
},
{
"input": "1\nz",
"output": "z"
},
{
"input": "2\nir",
"output": "ri"
},
{
"input": "3\nilj",
"output": "jli"
},
{
"input": "4\nj... | 1,648,591,851 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 20 | 46 | 0 | # c4ts0up
n = int(input())
cad = input()
divs = []
for i in range(1, n+1, 1):
if (n%i == 0):
divs.append(i)
for d in divs:
rev = cad[:d]
rev = rev[::-1]
normal = cad[d:]
cad = rev + normal
print(cad)
| Title: Reversing Encryption
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string $s$ of length $n$ can be encrypted by the following algorithm:
- iterate over all divisors of $n$ in decreasing order (i.e. from $n$ to $1$), - for each divisor $d$, reverse the substring $s[1 \dots d]$ (i.e. the substring which starts at position $1$ and ends at position $d$).
For example, the above algorithm applied to the string $s$="codeforces" leads to the following changes: "codeforces" $\to$ "secrofedoc" $\to$ "orcesfedoc" $\to$ "rocesfedoc" $\to$ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because $d=1$).
You are given the encrypted string $t$. Your task is to decrypt this string, i.e., to find a string $s$ such that the above algorithm results in string $t$. It can be proven that this string $s$ always exists and is unique.
Input Specification:
The first line of input consists of a single integer $n$ ($1 \le n \le 100$) — the length of the string $t$. The second line of input consists of the string $t$. The length of $t$ is $n$, and it consists only of lowercase Latin letters.
Output Specification:
Print a string $s$ such that the above algorithm results in $t$.
Demo Input:
['10\nrocesfedoc\n', '16\nplmaetwoxesisiht\n', '1\nz\n']
Demo Output:
['codeforces\n', 'thisisexampletwo\n', 'z\n']
Note:
The first example is described in the problem statement. | ```python
# c4ts0up
n = int(input())
cad = input()
divs = []
for i in range(1, n+1, 1):
if (n%i == 0):
divs.append(i)
for d in divs:
rev = cad[:d]
rev = rev[::-1]
normal = cad[d:]
cad = rev + normal
print(cad)
``` | 3 | |
624 | A | Save Luke | PROGRAMMING | 800 | [
"math"
] | null | null | Luke Skywalker got locked up in a rubbish shredder between two presses. R2D2 is already working on his rescue, but Luke needs to stay alive as long as possible. For simplicity we will assume that everything happens on a straight line, the presses are initially at coordinates 0 and *L*, and they move towards each other with speed *v*1 and *v*2, respectively. Luke has width *d* and is able to choose any position between the presses. Luke dies as soon as the distance between the presses is less than his width. Your task is to determine for how long Luke can stay alive. | The first line of the input contains four integers *d*, *L*, *v*1, *v*2 (1<=≤<=*d*,<=*L*,<=*v*1,<=*v*2<=≤<=10<=000,<=*d*<=<<=*L*) — Luke's width, the initial position of the second press and the speed of the first and second presses, respectively. | Print a single real value — the maximum period of time Luke can stay alive for. Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if . | [
"2 6 2 2\n",
"1 9 1 2\n"
] | [
"1.00000000000000000000\n",
"2.66666666666666650000\n"
] | In the first sample Luke should stay exactly in the middle of the segment, that is at coordinates [2;4], as the presses move with the same speed.
In the second sample he needs to occupy the position <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/71395c777960eaded59a9fdc428a9625f152605b.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In this case both presses move to his edges at the same time. | 500 | [
{
"input": "2 6 2 2",
"output": "1.00000000000000000000"
},
{
"input": "1 9 1 2",
"output": "2.66666666666666650000"
},
{
"input": "1 10000 1 1",
"output": "4999.50000000000000000000"
},
{
"input": "9999 10000 10000 10000",
"output": "0.00005000000000000000"
},
{
... | 1,630,511,562 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 109 | 22,220,800 | from sys import stdin, stdout
from collections import defaultdict
def read():
return stdin.readline().rstrip()
def read_int():
return int(read())
def read_ints():
return list(map(int, read().split()))
def solve():
d,L,v1,v2=read_ints()
print("%.15f" % (L-d)/(v1+v2))
solve()
| Title: Save Luke
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luke Skywalker got locked up in a rubbish shredder between two presses. R2D2 is already working on his rescue, but Luke needs to stay alive as long as possible. For simplicity we will assume that everything happens on a straight line, the presses are initially at coordinates 0 and *L*, and they move towards each other with speed *v*1 and *v*2, respectively. Luke has width *d* and is able to choose any position between the presses. Luke dies as soon as the distance between the presses is less than his width. Your task is to determine for how long Luke can stay alive.
Input Specification:
The first line of the input contains four integers *d*, *L*, *v*1, *v*2 (1<=≤<=*d*,<=*L*,<=*v*1,<=*v*2<=≤<=10<=000,<=*d*<=<<=*L*) — Luke's width, the initial position of the second press and the speed of the first and second presses, respectively.
Output Specification:
Print a single real value — the maximum period of time Luke can stay alive for. Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=6.
Namely: let's assume that your answer is *a*, and the answer of the jury is *b*. The checker program will consider your answer correct, if .
Demo Input:
['2 6 2 2\n', '1 9 1 2\n']
Demo Output:
['1.00000000000000000000\n', '2.66666666666666650000\n']
Note:
In the first sample Luke should stay exactly in the middle of the segment, that is at coordinates [2;4], as the presses move with the same speed.
In the second sample he needs to occupy the position <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/71395c777960eaded59a9fdc428a9625f152605b.png" style="max-width: 100.0%;max-height: 100.0%;"/>. In this case both presses move to his edges at the same time. | ```python
from sys import stdin, stdout
from collections import defaultdict
def read():
return stdin.readline().rstrip()
def read_int():
return int(read())
def read_ints():
return list(map(int, read().split()))
def solve():
d,L,v1,v2=read_ints()
print("%.15f" % (L-d)/(v1+v2))
solve()
``` | -1 | |
239 | A | Two Bags of Potatoes | PROGRAMMING | 1,200 | [
"greedy",
"implementation",
"math"
] | null | null | Valera had two bags of potatoes, the first of these bags contains *x* (*x*<=≥<=1) potatoes, and the second — *y* (*y*<=≥<=1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains *x* potatoes) Valera lost. Valera remembers that the total amount of potatoes (*x*<=+<=*y*) in the two bags, firstly, was not gerater than *n*, and, secondly, was divisible by *k*.
Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order. | The first line of input contains three integers *y*, *k*, *n* (1<=≤<=*y*,<=*k*,<=*n*<=≤<=109; <=≤<=105). | Print the list of whitespace-separated integers — all possible values of *x* in ascending order. You should print each possible value of *x* exactly once.
If there are no such values of *x* print a single integer -1. | [
"10 1 10\n",
"10 6 40\n"
] | [
"-1\n",
"2 8 14 20 26 \n"
] | none | 500 | [
{
"input": "10 1 10",
"output": "-1"
},
{
"input": "10 6 40",
"output": "2 8 14 20 26 "
},
{
"input": "10 1 20",
"output": "1 2 3 4 5 6 7 8 9 10 "
},
{
"input": "1 10000 1000000000",
"output": "9999 19999 29999 39999 49999 59999 69999 79999 89999 99999 109999 119999 12999... | 1,590,593,522 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 186 | 307,200 | a,b,c = map(int, input().split())
if(c-a<=0):
print(-1)
else:
rem = a%b
x= []
p = c-a-rem
k=0
while(p>=0):
x.append(p)
k+=1
p-=b
x.reverse()
for i in range(len(x)):
print(x[i], end=" ")
if(k==0):
print(-1) | Title: Two Bags of Potatoes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera had two bags of potatoes, the first of these bags contains *x* (*x*<=≥<=1) potatoes, and the second — *y* (*y*<=≥<=1) potatoes. Valera — very scattered boy, so the first bag of potatoes (it contains *x* potatoes) Valera lost. Valera remembers that the total amount of potatoes (*x*<=+<=*y*) in the two bags, firstly, was not gerater than *n*, and, secondly, was divisible by *k*.
Help Valera to determine how many potatoes could be in the first bag. Print all such possible numbers in ascending order.
Input Specification:
The first line of input contains three integers *y*, *k*, *n* (1<=≤<=*y*,<=*k*,<=*n*<=≤<=109; <=≤<=105).
Output Specification:
Print the list of whitespace-separated integers — all possible values of *x* in ascending order. You should print each possible value of *x* exactly once.
If there are no such values of *x* print a single integer -1.
Demo Input:
['10 1 10\n', '10 6 40\n']
Demo Output:
['-1\n', '2 8 14 20 26 \n']
Note:
none | ```python
a,b,c = map(int, input().split())
if(c-a<=0):
print(-1)
else:
rem = a%b
x= []
p = c-a-rem
k=0
while(p>=0):
x.append(p)
k+=1
p-=b
x.reverse()
for i in range(len(x)):
print(x[i], end=" ")
if(k==0):
print(-1)
``` | 0 | |
755 | B | PolandBall and Game | PROGRAMMING | 1,100 | [
"binary search",
"data structures",
"games",
"greedy",
"sortings",
"strings"
] | null | null | PolandBall is playing a game with EnemyBall. The rules are simple. Players have to say words in turns. You cannot say a word which was already said. PolandBall starts. The Ball which can't say a new word loses.
You're given two lists of words familiar to PolandBall and EnemyBall. Can you determine who wins the game, if both play optimally? | The first input line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=103) — number of words PolandBall and EnemyBall know, respectively.
Then *n* strings follow, one per line — words familiar to PolandBall.
Then *m* strings follow, one per line — words familiar to EnemyBall.
Note that one Ball cannot know a word more than once (strings are unique), but some words can be known by both players.
Each word is non-empty and consists of no more than 500 lowercase English alphabet letters. | In a single line of print the answer — "YES" if PolandBall wins and "NO" otherwise. Both Balls play optimally. | [
"5 1\npolandball\nis\na\ncool\ncharacter\nnope\n",
"2 2\nkremowka\nwadowicka\nkremowka\nwiedenska\n",
"1 2\na\na\nb\n"
] | [
"YES",
"YES",
"NO"
] | In the first example PolandBall knows much more words and wins effortlessly.
In the second example if PolandBall says kremowka first, then EnemyBall cannot use that word anymore. EnemyBall can only say wiedenska. PolandBall says wadowicka and wins. | 1,000 | [
{
"input": "5 1\npolandball\nis\na\ncool\ncharacter\nnope",
"output": "YES"
},
{
"input": "2 2\nkremowka\nwadowicka\nkremowka\nwiedenska",
"output": "YES"
},
{
"input": "1 2\na\na\nb",
"output": "NO"
},
{
"input": "2 2\na\nb\nb\nc",
"output": "YES"
},
{
"input": "... | 1,674,233,677 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 109 | 1,638,400 | x, y = [int(x) for x in input().split()]
lisx = []
for i in range(x):
pala = str(input())
lisx.append(pala)
lisy = []
for i in range(y):
pala = str(input())
lisy.append(pala)
iguais = 0
for i in lisx:
if i in lisy:
iguais += 1
if iguais % 2 == 0:
if x > y:
print('YES')
else:
print('NO')
else:
if x+1 > y:
print('YES')
else:
print('NO') | Title: PolandBall and Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
PolandBall is playing a game with EnemyBall. The rules are simple. Players have to say words in turns. You cannot say a word which was already said. PolandBall starts. The Ball which can't say a new word loses.
You're given two lists of words familiar to PolandBall and EnemyBall. Can you determine who wins the game, if both play optimally?
Input Specification:
The first input line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=103) — number of words PolandBall and EnemyBall know, respectively.
Then *n* strings follow, one per line — words familiar to PolandBall.
Then *m* strings follow, one per line — words familiar to EnemyBall.
Note that one Ball cannot know a word more than once (strings are unique), but some words can be known by both players.
Each word is non-empty and consists of no more than 500 lowercase English alphabet letters.
Output Specification:
In a single line of print the answer — "YES" if PolandBall wins and "NO" otherwise. Both Balls play optimally.
Demo Input:
['5 1\npolandball\nis\na\ncool\ncharacter\nnope\n', '2 2\nkremowka\nwadowicka\nkremowka\nwiedenska\n', '1 2\na\na\nb\n']
Demo Output:
['YES', 'YES', 'NO']
Note:
In the first example PolandBall knows much more words and wins effortlessly.
In the second example if PolandBall says kremowka first, then EnemyBall cannot use that word anymore. EnemyBall can only say wiedenska. PolandBall says wadowicka and wins. | ```python
x, y = [int(x) for x in input().split()]
lisx = []
for i in range(x):
pala = str(input())
lisx.append(pala)
lisy = []
for i in range(y):
pala = str(input())
lisy.append(pala)
iguais = 0
for i in lisx:
if i in lisy:
iguais += 1
if iguais % 2 == 0:
if x > y:
print('YES')
else:
print('NO')
else:
if x+1 > y:
print('YES')
else:
print('NO')
``` | 3 | |
32 | A | Reconnaissance | PROGRAMMING | 800 | [
"brute force"
] | A. Reconnaissance | 2 | 256 | According to the regulations of Berland's army, a reconnaissance unit should consist of exactly two soldiers. Since these two soldiers shouldn't differ much, their heights can differ by at most *d* centimeters. Captain Bob has *n* soldiers in his detachment. Their heights are *a*1,<=*a*2,<=...,<=*a**n* centimeters. Some soldiers are of the same height. Bob wants to know, how many ways exist to form a reconnaissance unit of two soldiers from his detachment.
Ways (1,<=2) and (2,<=1) should be regarded as different. | The first line contains two integers *n* and *d* (1<=≤<=*n*<=≤<=1000,<=1<=≤<=*d*<=≤<=109) — amount of soldiers in Bob's detachment and the maximum allowed height difference respectively. The second line contains *n* space-separated integers — heights of all the soldiers in Bob's detachment. These numbers don't exceed 109. | Output one number — amount of ways to form a reconnaissance unit of two soldiers, whose height difference doesn't exceed *d*. | [
"5 10\n10 20 50 60 65\n",
"5 1\n55 30 29 31 55\n"
] | [
"6\n",
"6\n"
] | none | 500 | [
{
"input": "5 10\n10 20 50 60 65",
"output": "6"
},
{
"input": "5 1\n55 30 29 31 55",
"output": "6"
},
{
"input": "6 10\n4 6 4 1 9 3",
"output": "30"
},
{
"input": "7 100\n19 1694 261 162 1 234 513",
"output": "8"
},
{
"input": "8 42\n37 53 74 187 568 22 5 65",
... | 1,627,481,411 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 92 | 6,656,000 | [n , d] = map(int, input().split())
nums = list(map(int, input().split()))
x = 0
for i in range(n):
for j in range(n):
if abs(i - j) <= 10:
x += 1
print(x)
| Title: Reconnaissance
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
According to the regulations of Berland's army, a reconnaissance unit should consist of exactly two soldiers. Since these two soldiers shouldn't differ much, their heights can differ by at most *d* centimeters. Captain Bob has *n* soldiers in his detachment. Their heights are *a*1,<=*a*2,<=...,<=*a**n* centimeters. Some soldiers are of the same height. Bob wants to know, how many ways exist to form a reconnaissance unit of two soldiers from his detachment.
Ways (1,<=2) and (2,<=1) should be regarded as different.
Input Specification:
The first line contains two integers *n* and *d* (1<=≤<=*n*<=≤<=1000,<=1<=≤<=*d*<=≤<=109) — amount of soldiers in Bob's detachment and the maximum allowed height difference respectively. The second line contains *n* space-separated integers — heights of all the soldiers in Bob's detachment. These numbers don't exceed 109.
Output Specification:
Output one number — amount of ways to form a reconnaissance unit of two soldiers, whose height difference doesn't exceed *d*.
Demo Input:
['5 10\n10 20 50 60 65\n', '5 1\n55 30 29 31 55\n']
Demo Output:
['6\n', '6\n']
Note:
none | ```python
[n , d] = map(int, input().split())
nums = list(map(int, input().split()))
x = 0
for i in range(n):
for j in range(n):
if abs(i - j) <= 10:
x += 1
print(x)
``` | 0 |
34 | B | Sale | PROGRAMMING | 900 | [
"greedy",
"sortings"
] | B. Sale | 2 | 256 | Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn. | The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets. | Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets. | [
"5 3\n-6 0 35 -2 4\n",
"4 2\n7 0 0 -7\n"
] | [
"8\n",
"7\n"
] | none | 1,000 | [
{
"input": "5 3\n-6 0 35 -2 4",
"output": "8"
},
{
"input": "4 2\n7 0 0 -7",
"output": "7"
},
{
"input": "6 6\n756 -611 251 -66 572 -818",
"output": "1495"
},
{
"input": "5 5\n976 437 937 788 518",
"output": "0"
},
{
"input": "5 3\n-2 -2 -2 -2 -2",
"output": "... | 1,559,985,896 | 2,147,483,647 | Python 3 | OK | TESTS | 25 | 218 | 0 | # hamed damirchi 95222031
n, m = [int(x) for x in input().split()]
TV_prices = [int(x) for x in input().split()]
max_benefit = 0
TV_prices_minuses = []
for i in TV_prices:
if i<0:
TV_prices_minuses.append(abs(i))
TV_prices_minuses.sort()
TV_prices_minuses.reverse()
for i in range(m):
if len(TV_prices_minuses)>i:
max_benefit+=TV_prices_minuses[i]
print(max_benefit)
| Title: Sale
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob got to a sale of old TV sets. There were *n* TV sets at that sale. TV set with index *i* costs *a**i* bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most *m* TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*m*<=≤<=*n*<=≤<=100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains *n* space-separated integers *a**i* (<=-<=1000<=≤<=*a**i*<=≤<=1000) — prices of the TV sets.
Output Specification:
Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most *m* TV sets.
Demo Input:
['5 3\n-6 0 35 -2 4\n', '4 2\n7 0 0 -7\n']
Demo Output:
['8\n', '7\n']
Note:
none | ```python
# hamed damirchi 95222031
n, m = [int(x) for x in input().split()]
TV_prices = [int(x) for x in input().split()]
max_benefit = 0
TV_prices_minuses = []
for i in TV_prices:
if i<0:
TV_prices_minuses.append(abs(i))
TV_prices_minuses.sort()
TV_prices_minuses.reverse()
for i in range(m):
if len(TV_prices_minuses)>i:
max_benefit+=TV_prices_minuses[i]
print(max_benefit)
``` | 3.9455 |
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"
},
{
... | 1,683,206,263 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 40 | 62 | 0 | def pts(p, t):
return max(3 * p // 10, p - p * t // 250)
a, b, c, d = map(int, input().split())
m, v = pts(a, c), pts(b, d)
if m == v:
print("Tie")
else:
print(["Vasya","Misha"][m > v]) | 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
def pts(p, t):
return max(3 * p // 10, p - p * t // 250)
a, b, c, d = map(int, input().split())
m, v = pts(a, c), pts(b, d)
if m == v:
print("Tie")
else:
print(["Vasya","Misha"][m > v])
``` | 3 | |
58 | A | Chat room | PROGRAMMING | 1,000 | [
"greedy",
"strings"
] | A. Chat room | 1 | 256 | Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*. | The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters. | If Vasya managed to say hello, print "YES", otherwise print "NO". | [
"ahhellllloou\n",
"hlelo\n"
] | [
"YES\n",
"NO\n"
] | none | 500 | [
{
"input": "ahhellllloou",
"output": "YES"
},
{
"input": "hlelo",
"output": "NO"
},
{
"input": "helhcludoo",
"output": "YES"
},
{
"input": "hehwelloho",
"output": "YES"
},
{
"input": "pnnepelqomhhheollvlo",
"output": "YES"
},
{
"input": "tymbzjyqhymeda... | 1,678,385,228 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 40 | 62 | 0 | s = input()
f = []
for i in range(len(s)):
if s[i] == "h" and "h" not in f: f.append(s[i])
if s[i] == "e" and "e" not in f and "h" in f: f.append(s[i])
if s[i] == "l" and f.count("l") < 2 and "h" in f and "e" in f: f.append(s[i])
if s[i] == "o" and "o" not in f and "h" in f and "e" in f and f.count("l") == 2: f.append(s[i])
res = "".join(f)
if res == "hello":
print("YES")
else:
print("NO") | Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
s = input()
f = []
for i in range(len(s)):
if s[i] == "h" and "h" not in f: f.append(s[i])
if s[i] == "e" and "e" not in f and "h" in f: f.append(s[i])
if s[i] == "l" and f.count("l") < 2 and "h" in f and "e" in f: f.append(s[i])
if s[i] == "o" and "o" not in f and "h" in f and "e" in f and f.count("l") == 2: f.append(s[i])
res = "".join(f)
if res == "hello":
print("YES")
else:
print("NO")
``` | 3.969 |
437 | A | The Child and Homework | PROGRAMMING | 1,300 | [
"implementation"
] | null | null | Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices: A, B, C and D. Each choice has a description, and the child should find out the only one that is correct.
Fortunately the child knows how to solve such complicated test. The child will follow the algorithm:
- If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great. - If there is exactly one great choice then the child chooses it. Otherwise the child chooses C (the child think it is the luckiest choice).
You are given a multiple-choice questions, can you predict child's choose? | The first line starts with "A." (without quotes), then followed the description of choice A. The next three lines contains the descriptions of the other choices in the same format. They are given in order: B, C, D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length.
Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_". | Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes). | [
"A.VFleaKing_is_the_author_of_this_problem\nB.Picks_is_the_author_of_this_problem\nC.Picking_is_the_author_of_this_problem\nD.Ftiasch_is_cute\n",
"A.ab\nB.abcde\nC.ab\nD.abc\n",
"A.c\nB.cc\nC.c\nD.c\n"
] | [
"D\n",
"C\n",
"B\n"
] | In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choice D (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will choose D.
In the second sample, no choice is great, so the child will choose the luckiest choice C.
In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will choose B. | 500 | [
{
"input": "A.VFleaKing_is_the_author_of_this_problem\nB.Picks_is_the_author_of_this_problem\nC.Picking_is_the_author_of_this_problem\nD.Ftiasch_is_cute",
"output": "D"
},
{
"input": "A.ab\nB.abcde\nC.ab\nD.abc",
"output": "C"
},
{
"input": "A.c\nB.cc\nC.c\nD.c",
"output": "B"
},
... | 1,612,492,950 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 2 | 109 | 2,355,200 | from math import *
from collections import *
from sys import *
import string
import re
t=stdin.readline
p=stdout.write
def GI(): return map(int, input().strip().split())
def GS(): return map(str, t().strip().split())
def IL(): return list(map(int, t().strip().split()))
def SL(): return list(map(str, t().strip().split()))
a=sorted([(len(t())-2,i) for i in 'ABCD'])
c=0
if a[0][0]*2<=a[1][0]: c+=1
if a[-2][0]*2<=a[-1][0]: c+=2
print(['C',a[0][1],a[-1][1],'C'][c]) | Title: The Child and Homework
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once upon a time a child got a test consisting of multiple-choice questions as homework. A multiple-choice question consists of four choices: A, B, C and D. Each choice has a description, and the child should find out the only one that is correct.
Fortunately the child knows how to solve such complicated test. The child will follow the algorithm:
- If there is some choice whose description at least twice shorter than all other descriptions, or at least twice longer than all other descriptions, then the child thinks the choice is great. - If there is exactly one great choice then the child chooses it. Otherwise the child chooses C (the child think it is the luckiest choice).
You are given a multiple-choice questions, can you predict child's choose?
Input Specification:
The first line starts with "A." (without quotes), then followed the description of choice A. The next three lines contains the descriptions of the other choices in the same format. They are given in order: B, C, D. Please note, that the description goes after prefix "X.", so the prefix mustn't be counted in description's length.
Each description is non-empty and consists of at most 100 characters. Each character can be either uppercase English letter or lowercase English letter, or "_".
Output Specification:
Print a single line with the child's choice: "A", "B", "C" or "D" (without quotes).
Demo Input:
['A.VFleaKing_is_the_author_of_this_problem\nB.Picks_is_the_author_of_this_problem\nC.Picking_is_the_author_of_this_problem\nD.Ftiasch_is_cute\n', 'A.ab\nB.abcde\nC.ab\nD.abc\n', 'A.c\nB.cc\nC.c\nD.c\n']
Demo Output:
['D\n', 'C\n', 'B\n']
Note:
In the first sample, the first choice has length 39, the second one has length 35, the third one has length 37, and the last one has length 15. The choice D (length 15) is twice shorter than all other choices', so it is great choice. There is no other great choices so the child will choose D.
In the second sample, no choice is great, so the child will choose the luckiest choice C.
In the third sample, the choice B (length 2) is twice longer than all other choices', so it is great choice. There is no other great choices so the child will choose B. | ```python
from math import *
from collections import *
from sys import *
import string
import re
t=stdin.readline
p=stdout.write
def GI(): return map(int, input().strip().split())
def GS(): return map(str, t().strip().split())
def IL(): return list(map(int, t().strip().split()))
def SL(): return list(map(str, t().strip().split()))
a=sorted([(len(t())-2,i) for i in 'ABCD'])
c=0
if a[0][0]*2<=a[1][0]: c+=1
if a[-2][0]*2<=a[-1][0]: c+=2
print(['C',a[0][1],a[-1][1],'C'][c])
``` | 0 | |
994 | A | Fingerprints | PROGRAMMING | 800 | [
"implementation"
] | null | null | You are locked in a room with a door that has a keypad with 10 keys corresponding to digits from 0 to 9. To escape from the room, you need to enter a correct code. You also have a sequence of digits.
Some keys on the keypad have fingerprints. You believe the correct code is the longest not necessarily contiguous subsequence of the sequence you have that only contains digits with fingerprints on the corresponding keys. Find such code. | The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10$) representing the number of digits in the sequence you have and the number of keys on the keypad that have fingerprints.
The next line contains $n$ distinct space-separated integers $x_1, x_2, \ldots, x_n$ ($0 \le x_i \le 9$) representing the sequence.
The next line contains $m$ distinct space-separated integers $y_1, y_2, \ldots, y_m$ ($0 \le y_i \le 9$) — the keys with fingerprints. | In a single line print a space-separated sequence of integers representing the code. If the resulting sequence is empty, both printing nothing and printing a single line break is acceptable. | [
"7 3\n3 5 7 1 6 2 8\n1 2 7\n",
"4 4\n3 4 1 0\n0 1 7 9\n"
] | [
"7 1 2\n",
"1 0\n"
] | In the first example, the only digits with fingerprints are $1$, $2$ and $7$. All three of them appear in the sequence you know, $7$ first, then $1$ and then $2$. Therefore the output is 7 1 2. Note that the order is important, and shall be the same as the order in the original sequence.
In the second example digits $0$, $1$, $7$ and $9$ have fingerprints, however only $0$ and $1$ appear in the original sequence. $1$ appears earlier, so the output is 1 0. Again, the order is important. | 500 | [
{
"input": "7 3\n3 5 7 1 6 2 8\n1 2 7",
"output": "7 1 2"
},
{
"input": "4 4\n3 4 1 0\n0 1 7 9",
"output": "1 0"
},
{
"input": "9 4\n9 8 7 6 5 4 3 2 1\n2 4 6 8",
"output": "8 6 4 2"
},
{
"input": "10 5\n3 7 1 2 4 6 9 0 5 8\n4 3 0 7 9",
"output": "3 7 4 9 0"
},
{
"... | 1,593,437,267 | 2,147,483,647 | PyPy 3 | OK | TESTS | 31 | 155 | 20,172,800 | n,m=input().split()
x=list(input().split())
y=list(input().split())
print(*filter(lambda x:x in y,x)) | Title: Fingerprints
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are locked in a room with a door that has a keypad with 10 keys corresponding to digits from 0 to 9. To escape from the room, you need to enter a correct code. You also have a sequence of digits.
Some keys on the keypad have fingerprints. You believe the correct code is the longest not necessarily contiguous subsequence of the sequence you have that only contains digits with fingerprints on the corresponding keys. Find such code.
Input Specification:
The first line contains two integers $n$ and $m$ ($1 \le n, m \le 10$) representing the number of digits in the sequence you have and the number of keys on the keypad that have fingerprints.
The next line contains $n$ distinct space-separated integers $x_1, x_2, \ldots, x_n$ ($0 \le x_i \le 9$) representing the sequence.
The next line contains $m$ distinct space-separated integers $y_1, y_2, \ldots, y_m$ ($0 \le y_i \le 9$) — the keys with fingerprints.
Output Specification:
In a single line print a space-separated sequence of integers representing the code. If the resulting sequence is empty, both printing nothing and printing a single line break is acceptable.
Demo Input:
['7 3\n3 5 7 1 6 2 8\n1 2 7\n', '4 4\n3 4 1 0\n0 1 7 9\n']
Demo Output:
['7 1 2\n', '1 0\n']
Note:
In the first example, the only digits with fingerprints are $1$, $2$ and $7$. All three of them appear in the sequence you know, $7$ first, then $1$ and then $2$. Therefore the output is 7 1 2. Note that the order is important, and shall be the same as the order in the original sequence.
In the second example digits $0$, $1$, $7$ and $9$ have fingerprints, however only $0$ and $1$ appear in the original sequence. $1$ appears earlier, so the output is 1 0. Again, the order is important. | ```python
n,m=input().split()
x=list(input().split())
y=list(input().split())
print(*filter(lambda x:x in y,x))
``` | 3 | |
797 | B | Odd sum | PROGRAMMING | 1,400 | [
"dp",
"greedy",
"implementation"
] | null | null | You are given sequence *a*1,<=*a*2,<=...,<=*a**n* of integer numbers of length *n*. Your task is to find such subsequence that its sum is odd and maximum among all such subsequences. It's guaranteed that given sequence contains subsequence with odd sum.
Subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
You should write a program which finds sum of the best subsequence. | The first line contains integer number *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=104<=≤<=*a**i*<=≤<=104). The sequence contains at least one subsequence with odd sum. | Print sum of resulting subseqeuence. | [
"4\n-2 2 -3 1\n",
"3\n2 -5 -3\n"
] | [
"3\n",
"-1\n"
] | In the first example sum of the second and the fourth elements is 3. | 0 | [
{
"input": "4\n-2 2 -3 1",
"output": "3"
},
{
"input": "3\n2 -5 -3",
"output": "-1"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "1\n-1",
"output": "-1"
},
{
"input": "15\n-6004 4882 9052 413 6056 4306 9946 -4616 -6135 906 -1718 5252 -2866 9061 4046",
"ou... | 1,612,124,936 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 0 | 61 | 0 | import sys
length = input('size:')
inputdata = input('input:').split(' ')
LOD = -10000
sum = 0
for i in inputdata:
if(int(i) > 0):
sum += int(i)
elif(int(i) < LOD):
LOD = int(i)
print(sum)
| Title: Odd sum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given sequence *a*1,<=*a*2,<=...,<=*a**n* of integer numbers of length *n*. Your task is to find such subsequence that its sum is odd and maximum among all such subsequences. It's guaranteed that given sequence contains subsequence with odd sum.
Subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
You should write a program which finds sum of the best subsequence.
Input Specification:
The first line contains integer number *n* (1<=≤<=*n*<=≤<=105).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=104<=≤<=*a**i*<=≤<=104). The sequence contains at least one subsequence with odd sum.
Output Specification:
Print sum of resulting subseqeuence.
Demo Input:
['4\n-2 2 -3 1\n', '3\n2 -5 -3\n']
Demo Output:
['3\n', '-1\n']
Note:
In the first example sum of the second and the fourth elements is 3. | ```python
import sys
length = input('size:')
inputdata = input('input:').split(' ')
LOD = -10000
sum = 0
for i in inputdata:
if(int(i) > 0):
sum += int(i)
elif(int(i) < LOD):
LOD = int(i)
print(sum)
``` | 0 | |
484 | A | Bits | PROGRAMMING | 1,700 | [
"bitmasks",
"constructive algorithms"
] | null | null | Let's denote as the number of bits set ('1' bits) in the binary representation of the non-negative integer *x*.
You are given multiple queries consisting of pairs of integers *l* and *r*. For each query, find the *x*, such that *l*<=≤<=*x*<=≤<=*r*, and is maximum possible. If there are multiple such numbers find the smallest of them. | The first line contains integer *n* — the number of queries (1<=≤<=*n*<=≤<=10000).
Each of the following *n* lines contain two integers *l**i*,<=*r**i* — the arguments for the corresponding query (0<=≤<=*l**i*<=≤<=*r**i*<=≤<=1018). | For each query print the answer in a separate line. | [
"3\n1 2\n2 4\n1 10\n"
] | [
"1\n3\n7\n"
] | The binary representations of numbers from 1 to 10 are listed below:
1<sub class="lower-index">10</sub> = 1<sub class="lower-index">2</sub>
2<sub class="lower-index">10</sub> = 10<sub class="lower-index">2</sub>
3<sub class="lower-index">10</sub> = 11<sub class="lower-index">2</sub>
4<sub class="lower-index">10</sub> = 100<sub class="lower-index">2</sub>
5<sub class="lower-index">10</sub> = 101<sub class="lower-index">2</sub>
6<sub class="lower-index">10</sub> = 110<sub class="lower-index">2</sub>
7<sub class="lower-index">10</sub> = 111<sub class="lower-index">2</sub>
8<sub class="lower-index">10</sub> = 1000<sub class="lower-index">2</sub>
9<sub class="lower-index">10</sub> = 1001<sub class="lower-index">2</sub>
10<sub class="lower-index">10</sub> = 1010<sub class="lower-index">2</sub> | 500 | [
{
"input": "3\n1 2\n2 4\n1 10",
"output": "1\n3\n7"
},
{
"input": "55\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n2 2\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n3 3\n3 4\n3 5\n3 6\n3 7\n3 8\n3 9\n3 10\n4 4\n4 5\n4 6\n4 7\n4 8\n4 9\n4 10\n5 5\n5 6\n5 7\n5 8\n5 9\n5 10\n6 6\n6 7\n6 8\n6 9\n6 10... | 1,564,474,188 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 3 | 358 | 9,113,600 | from sys import stdin
#stdin = open("sum.in", 'r')
hat = stdin.readlines()
# print(hat)
nn = int(hat[0])
def x(n, ind):
return n & ~(1 << ind-1)
def z(n):
cheaker = 0
l = len(str(bin(n)))
for i in range(l-2, 0, -1):
if (n >> i-1) & 1 == 1:
cheaker += 1
if cheaker == l-2:
return [True, cheaker]
else:
return [False, cheaker]
for j in range(nn):
#print("hell")
flag = True
a, b = map(int, hat[1+j].split())
#print(j, a, b)
mine = 0
if a == b:
print(a)
continue
elif z(b)[0] and z(a)[0]:
#print(z(b)[0] , z(a)[0])
print(b)
continue
for i in range(60, 0, -1):
#print("aaaa")
if 2**i-1 == b:
print(2**i-1)
flag = False
break
elif a < 2**i <= b:
print(2**i-1)
flag = False
break
elif 2**i <= b:
# print(i)
mine = 2**(i+1)-1
y = len(str(bin(mine)))-3
while y:
#print("hea")
mine = mine & ~(1 << y-1)
y -= 1
if a < mine <= b:
print(mine)
flag = False
break
if flag:
#print("yess")
if z(b)[1] == z(a)[1]:
#print("Asdasdasdasd")
print(a)
elif a % 2 == 0 and b % 2 != 0:
print(b)
elif a % 2 != 0 and b % 2 == 0:
print(a)
elif b % 2 == 0:
print(b-1)
else:
print(b)
| Title: Bits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's denote as the number of bits set ('1' bits) in the binary representation of the non-negative integer *x*.
You are given multiple queries consisting of pairs of integers *l* and *r*. For each query, find the *x*, such that *l*<=≤<=*x*<=≤<=*r*, and is maximum possible. If there are multiple such numbers find the smallest of them.
Input Specification:
The first line contains integer *n* — the number of queries (1<=≤<=*n*<=≤<=10000).
Each of the following *n* lines contain two integers *l**i*,<=*r**i* — the arguments for the corresponding query (0<=≤<=*l**i*<=≤<=*r**i*<=≤<=1018).
Output Specification:
For each query print the answer in a separate line.
Demo Input:
['3\n1 2\n2 4\n1 10\n']
Demo Output:
['1\n3\n7\n']
Note:
The binary representations of numbers from 1 to 10 are listed below:
1<sub class="lower-index">10</sub> = 1<sub class="lower-index">2</sub>
2<sub class="lower-index">10</sub> = 10<sub class="lower-index">2</sub>
3<sub class="lower-index">10</sub> = 11<sub class="lower-index">2</sub>
4<sub class="lower-index">10</sub> = 100<sub class="lower-index">2</sub>
5<sub class="lower-index">10</sub> = 101<sub class="lower-index">2</sub>
6<sub class="lower-index">10</sub> = 110<sub class="lower-index">2</sub>
7<sub class="lower-index">10</sub> = 111<sub class="lower-index">2</sub>
8<sub class="lower-index">10</sub> = 1000<sub class="lower-index">2</sub>
9<sub class="lower-index">10</sub> = 1001<sub class="lower-index">2</sub>
10<sub class="lower-index">10</sub> = 1010<sub class="lower-index">2</sub> | ```python
from sys import stdin
#stdin = open("sum.in", 'r')
hat = stdin.readlines()
# print(hat)
nn = int(hat[0])
def x(n, ind):
return n & ~(1 << ind-1)
def z(n):
cheaker = 0
l = len(str(bin(n)))
for i in range(l-2, 0, -1):
if (n >> i-1) & 1 == 1:
cheaker += 1
if cheaker == l-2:
return [True, cheaker]
else:
return [False, cheaker]
for j in range(nn):
#print("hell")
flag = True
a, b = map(int, hat[1+j].split())
#print(j, a, b)
mine = 0
if a == b:
print(a)
continue
elif z(b)[0] and z(a)[0]:
#print(z(b)[0] , z(a)[0])
print(b)
continue
for i in range(60, 0, -1):
#print("aaaa")
if 2**i-1 == b:
print(2**i-1)
flag = False
break
elif a < 2**i <= b:
print(2**i-1)
flag = False
break
elif 2**i <= b:
# print(i)
mine = 2**(i+1)-1
y = len(str(bin(mine)))-3
while y:
#print("hea")
mine = mine & ~(1 << y-1)
y -= 1
if a < mine <= b:
print(mine)
flag = False
break
if flag:
#print("yess")
if z(b)[1] == z(a)[1]:
#print("Asdasdasdasd")
print(a)
elif a % 2 == 0 and b % 2 != 0:
print(b)
elif a % 2 != 0 and b % 2 == 0:
print(a)
elif b % 2 == 0:
print(b-1)
else:
print(b)
``` | 0 | |
325 | B | Stadium and Games | PROGRAMMING | 1,800 | [
"binary search",
"math"
] | null | null | Daniel is organizing a football tournament. He has come up with the following tournament format:
1. In the first several (possibly zero) stages, while the number of teams is even, they split in pairs and play one game for each pair. At each stage the loser of each pair is eliminated (there are no draws). Such stages are held while the number of teams is even. 1. Eventually there will be an odd number of teams remaining. If there is one team remaining, it will be declared the winner, and the tournament ends. Otherwise each of the remaining teams will play with each other remaining team once in round robin tournament (if there are *x* teams, there will be games), and the tournament ends.
For example, if there were 20 teams initially, they would begin by playing 10 games. So, 10 teams would be eliminated, and the remaining 10 would play 5 games. Then the remaining 5 teams would play 10 games in a round robin tournament. In total there would be 10+5+10=25 games.
Daniel has already booked the stadium for *n* games. Help him to determine how many teams he should invite so that the tournament needs exactly *n* games. You should print all possible numbers of teams that will yield exactly *n* games in ascending order, or -1 if there are no such numbers. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1018), the number of games that should be played.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier. | Print all possible numbers of invited teams in ascending order, one per line. If exactly *n* games cannot be played, output one number: -1. | [
"3\n",
"25\n",
"2\n"
] | [
"3\n4\n",
"20\n",
"-1\n"
] | none | 1,000 | [
{
"input": "3",
"output": "3\n4"
},
{
"input": "25",
"output": "20"
},
{
"input": "2",
"output": "-1"
},
{
"input": "1",
"output": "2"
},
{
"input": "15",
"output": "10\n16"
},
{
"input": "314",
"output": "-1"
},
{
"input": "524800",
"o... | 1,444,813,873 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 60 | 0 | n = int(input())
hor1 = [int(read) for read in input().split()]
hor2 = [int(read) for read in input().split()]
vert = [int(read) for read in input().split()]
pref1 = [0] * (n)
pref2 = [0] * (n)
for i in range(1, n):
pref1[i] = pref1[i-1] + hor1[i-1]
pref2[i] = pref2[i-1] + hor2[i-1]
#print(pref1)
#print(pref2)
mini = 10**9
prev = 10**9
ans = set()
for i in range(n):
x = pref1[i] + (pref2[n-1] - pref2[i]) + vert[i]
#print(i, x, pref1[i], pref2[n-1] - pref2[i], vert[i])
if x <= mini:
prev = mini
mini = x
elif x < prev:
prev = x
print(mini + prev) | Title: Stadium and Games
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Daniel is organizing a football tournament. He has come up with the following tournament format:
1. In the first several (possibly zero) stages, while the number of teams is even, they split in pairs and play one game for each pair. At each stage the loser of each pair is eliminated (there are no draws). Such stages are held while the number of teams is even. 1. Eventually there will be an odd number of teams remaining. If there is one team remaining, it will be declared the winner, and the tournament ends. Otherwise each of the remaining teams will play with each other remaining team once in round robin tournament (if there are *x* teams, there will be games), and the tournament ends.
For example, if there were 20 teams initially, they would begin by playing 10 games. So, 10 teams would be eliminated, and the remaining 10 would play 5 games. Then the remaining 5 teams would play 10 games in a round robin tournament. In total there would be 10+5+10=25 games.
Daniel has already booked the stadium for *n* games. Help him to determine how many teams he should invite so that the tournament needs exactly *n* games. You should print all possible numbers of teams that will yield exactly *n* games in ascending order, or -1 if there are no such numbers.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1018), the number of games that should be played.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print all possible numbers of invited teams in ascending order, one per line. If exactly *n* games cannot be played, output one number: -1.
Demo Input:
['3\n', '25\n', '2\n']
Demo Output:
['3\n4\n', '20\n', '-1\n']
Note:
none | ```python
n = int(input())
hor1 = [int(read) for read in input().split()]
hor2 = [int(read) for read in input().split()]
vert = [int(read) for read in input().split()]
pref1 = [0] * (n)
pref2 = [0] * (n)
for i in range(1, n):
pref1[i] = pref1[i-1] + hor1[i-1]
pref2[i] = pref2[i-1] + hor2[i-1]
#print(pref1)
#print(pref2)
mini = 10**9
prev = 10**9
ans = set()
for i in range(n):
x = pref1[i] + (pref2[n-1] - pref2[i]) + vert[i]
#print(i, x, pref1[i], pref2[n-1] - pref2[i], vert[i])
if x <= mini:
prev = mini
mini = x
elif x < prev:
prev = x
print(mini + prev)
``` | -1 | |
509 | A | Maximum in Table | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | An *n*<=×<=*n* table *a* is defined as follows:
- The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1.
These conditions define all the values in the table.
You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above. | The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table. | Print a single line containing a positive integer *m* — the maximum value in the table. | [
"1\n",
"5\n"
] | [
"1",
"70"
] | In the second test the rows of the table look as follows: | 0 | [
{
"input": "1",
"output": "1"
},
{
"input": "5",
"output": "70"
},
{
"input": "2",
"output": "2"
},
{
"input": "3",
"output": "6"
},
{
"input": "4",
"output": "20"
},
{
"input": "6",
"output": "252"
},
{
"input": "7",
"output": "924"
... | 1,624,020,562 | 2,147,483,647 | PyPy 3 | OK | TESTS | 10 | 77 | 0 | n = int(input())
lst = [[1] * n for _ in range(n)]
for i in range (1, n):
for j in range(1, n):
lst[i][j] = lst[i-1][j] + lst[i][j-1]
print(lst[n-1][n-1]) | Title: Maximum in Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
An *n*<=×<=*n* table *a* is defined as follows:
- The first row and the first column contain ones, that is: *a**i*,<=1<==<=*a*1,<=*i*<==<=1 for all *i*<==<=1,<=2,<=...,<=*n*. - Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula *a**i*,<=*j*<==<=*a**i*<=-<=1,<=*j*<=+<=*a**i*,<=*j*<=-<=1.
These conditions define all the values in the table.
You are given a number *n*. You need to determine the maximum value in the *n*<=×<=*n* table defined by the rules above.
Input Specification:
The only line of input contains a positive integer *n* (1<=≤<=*n*<=≤<=10) — the number of rows and columns of the table.
Output Specification:
Print a single line containing a positive integer *m* — the maximum value in the table.
Demo Input:
['1\n', '5\n']
Demo Output:
['1', '70']
Note:
In the second test the rows of the table look as follows: | ```python
n = int(input())
lst = [[1] * n for _ in range(n)]
for i in range (1, n):
for j in range(1, n):
lst[i][j] = lst[i-1][j] + lst[i][j-1]
print(lst[n-1][n-1])
``` | 3 | |
907 | A | Masha and Bears | PROGRAMMING | 1,300 | [
"brute force",
"implementation"
] | null | null | A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size *a* can climb into some car with size *b* if and only if *a*<=≤<=*b*, he or she likes it if and only if he can climb into this car and 2*a*<=≥<=*b*.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars. | You are given four integers *V*1, *V*2, *V*3, *V**m*(1<=≤<=*V**i*<=≤<=100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that *V*1<=><=*V*2<=><=*V*3. | Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes). | [
"50 30 10 10\n",
"100 50 10 21\n"
] | [
"50\n30\n10\n",
"-1\n"
] | In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20. | 500 | [
{
"input": "50 30 10 10",
"output": "50\n30\n10"
},
{
"input": "100 50 10 21",
"output": "-1"
},
{
"input": "100 50 19 10",
"output": "100\n50\n19"
},
{
"input": "99 50 25 49",
"output": "100\n99\n49"
},
{
"input": "3 2 1 1",
"output": "4\n3\n1"
},
{
"... | 1,514,044,040 | 6,140 | Python 3 | OK | TESTS | 32 | 61 | 5,632,000 | a=input().split()
for i in range(4):
a[i]=int(a[i])
if a[3]>=a[1] or a[2]*2<a[3] or a[3]*2<a[2]:
print(-1)
else:
print(a[0]*2)
print(a[1]*2)
if a[3]>=a[2]:
print(a[3])
else:
print(a[2])
| Title: Masha and Bears
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size *a* can climb into some car with size *b* if and only if *a*<=≤<=*b*, he or she likes it if and only if he can climb into this car and 2*a*<=≥<=*b*.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input Specification:
You are given four integers *V*1, *V*2, *V*3, *V**m*(1<=≤<=*V**i*<=≤<=100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that *V*1<=><=*V*2<=><=*V*3.
Output Specification:
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Demo Input:
['50 30 10 10\n', '100 50 10 21\n']
Demo Output:
['50\n30\n10\n', '-1\n']
Note:
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20. | ```python
a=input().split()
for i in range(4):
a[i]=int(a[i])
if a[3]>=a[1] or a[2]*2<a[3] or a[3]*2<a[2]:
print(-1)
else:
print(a[0]*2)
print(a[1]*2)
if a[3]>=a[2]:
print(a[3])
else:
print(a[2])
``` | 3 | |
831 | A | Unimodal Array | PROGRAMMING | 1,000 | [
"implementation"
] | null | null | Array of integers is unimodal, if:
- it is strictly increasing in the beginning; - after that it is constant; - after that it is strictly decreasing.
The first block (increasing) and the last block (decreasing) may be absent. It is allowed that both of this blocks are absent.
For example, the following three arrays are unimodal: [5,<=7,<=11,<=11,<=2,<=1], [4,<=4,<=2], [7], but the following three are not unimodal: [5,<=5,<=6,<=6,<=1], [1,<=2,<=1,<=2], [4,<=5,<=5,<=6].
Write a program that checks if an array is unimodal. | The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1<=000) — the elements of the array. | Print "YES" if the given array is unimodal. Otherwise, print "NO".
You can output each letter in any case (upper or lower). | [
"6\n1 5 5 5 4 2\n",
"5\n10 20 30 20 10\n",
"4\n1 2 1 2\n",
"7\n3 3 3 3 3 3 3\n"
] | [
"YES\n",
"YES\n",
"NO\n",
"YES\n"
] | In the first example the array is unimodal, because it is strictly increasing in the beginning (from position 1 to position 2, inclusively), that it is constant (from position 2 to position 4, inclusively) and then it is strictly decreasing (from position 4 to position 6, inclusively). | 500 | [
{
"input": "6\n1 5 5 5 4 2",
"output": "YES"
},
{
"input": "5\n10 20 30 20 10",
"output": "YES"
},
{
"input": "4\n1 2 1 2",
"output": "NO"
},
{
"input": "7\n3 3 3 3 3 3 3",
"output": "YES"
},
{
"input": "6\n5 7 11 11 2 1",
"output": "YES"
},
{
"input":... | 1,642,658,269 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 31 | 0 | s=str(input())
i=0
count=0
for i in range(len(s-1)):
if s[i]<s[i+1]:
count+=1
else:
break
for i in range(i,len(s-1)):
if s[i]==s[i+1]:
count+=1
else:
break
for i in range(i,len(s-1)):
if s[i]>s[i+1]:
count+=1
else:
break
if count==len(s):
print('YES')
else:
print('NO')
| Title: Unimodal Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Array of integers is unimodal, if:
- it is strictly increasing in the beginning; - after that it is constant; - after that it is strictly decreasing.
The first block (increasing) and the last block (decreasing) may be absent. It is allowed that both of this blocks are absent.
For example, the following three arrays are unimodal: [5,<=7,<=11,<=11,<=2,<=1], [4,<=4,<=2], [7], but the following three are not unimodal: [5,<=5,<=6,<=6,<=1], [1,<=2,<=1,<=2], [4,<=5,<=5,<=6].
Write a program that checks if an array is unimodal.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1<=000) — the elements of the array.
Output Specification:
Print "YES" if the given array is unimodal. Otherwise, print "NO".
You can output each letter in any case (upper or lower).
Demo Input:
['6\n1 5 5 5 4 2\n', '5\n10 20 30 20 10\n', '4\n1 2 1 2\n', '7\n3 3 3 3 3 3 3\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n', 'YES\n']
Note:
In the first example the array is unimodal, because it is strictly increasing in the beginning (from position 1 to position 2, inclusively), that it is constant (from position 2 to position 4, inclusively) and then it is strictly decreasing (from position 4 to position 6, inclusively). | ```python
s=str(input())
i=0
count=0
for i in range(len(s-1)):
if s[i]<s[i+1]:
count+=1
else:
break
for i in range(i,len(s-1)):
if s[i]==s[i+1]:
count+=1
else:
break
for i in range(i,len(s-1)):
if s[i]>s[i+1]:
count+=1
else:
break
if count==len(s):
print('YES')
else:
print('NO')
``` | -1 | |
472 | A | Design Tutorial: Learn from Math | PROGRAMMING | 800 | [
"math",
"number theory"
] | null | null | One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.
For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem.
You are given an integer *n* no less than 12, express it as a sum of two composite numbers. | The only line contains an integer *n* (12<=≤<=*n*<=≤<=106). | Output two composite integers *x* and *y* (1<=<<=*x*,<=*y*<=<<=*n*) such that *x*<=+<=*y*<==<=*n*. If there are multiple solutions, you can output any of them. | [
"12\n",
"15\n",
"23\n",
"1000000\n"
] | [
"4 8\n",
"6 9\n",
"8 15\n",
"500000 500000\n"
] | In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well.
In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number. | 500 | [
{
"input": "12",
"output": "4 8"
},
{
"input": "15",
"output": "6 9"
},
{
"input": "23",
"output": "8 15"
},
{
"input": "1000000",
"output": "500000 500000"
},
{
"input": "63874",
"output": "4 63870"
},
{
"input": "14568",
"output": "4 14564"
},
... | 1,682,350,176 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 46 | 0 | n=int(input())
if n%2==0:
if n-8>=4:
print(8,n-8,sep=" ")
else:
if n-9>=4:
print(9,n-9,sep=" ") | Title: Design Tutorial: Learn from Math
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.
For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem.
You are given an integer *n* no less than 12, express it as a sum of two composite numbers.
Input Specification:
The only line contains an integer *n* (12<=≤<=*n*<=≤<=106).
Output Specification:
Output two composite integers *x* and *y* (1<=<<=*x*,<=*y*<=<<=*n*) such that *x*<=+<=*y*<==<=*n*. If there are multiple solutions, you can output any of them.
Demo Input:
['12\n', '15\n', '23\n', '1000000\n']
Demo Output:
['4 8\n', '6 9\n', '8 15\n', '500000 500000\n']
Note:
In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well.
In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number. | ```python
n=int(input())
if n%2==0:
if n-8>=4:
print(8,n-8,sep=" ")
else:
if n-9>=4:
print(9,n-9,sep=" ")
``` | 3 | |
13 | A | Numbers | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Numbers | 1 | 64 | Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number *A* written in all bases from 2 to *A*<=-<=1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10. | Input contains one integer number *A* (3<=≤<=*A*<=≤<=1000). | Output should contain required average value in format «X/Y», where X is the numerator and Y is the denominator. | [
"5\n",
"3\n"
] | [
"7/3\n",
"2/1\n"
] | In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively. | 0 | [
{
"input": "5",
"output": "7/3"
},
{
"input": "3",
"output": "2/1"
},
{
"input": "1000",
"output": "90132/499"
},
{
"input": "927",
"output": "155449/925"
},
{
"input": "260",
"output": "6265/129"
},
{
"input": "131",
"output": "3370/129"
},
{
... | 1,693,850,860 | 2,147,483,647 | Python 3 | OK | TESTS | 99 | 92 | 0 | import math
from math import log
from math import floor
n = int(input())
ans = 0
for i in range(2,n):
cnt = n
while cnt>0:
ans+=(cnt%i)
cnt//=i
# res = floor(log(n)/log(i))+1
# ans.append(res)
# print(ans)
res = ans
res1 = n-2
while res1>0:
p = res1
res1=res%res1
res=p
print(str(ans//res)+'/'+str((n-2)//res))
# if len(ans)==1:
# print(f'{sum(ans)}/1')
# else:
# print(f'{sum(ans)}/{max(ans)}') | Title: Numbers
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Little Petya likes numbers a lot. He found that number 123 in base 16 consists of two digits: the first is 7 and the second is 11. So the sum of digits of 123 in base 16 is equal to 18.
Now he wonders what is an average value of sum of digits of the number *A* written in all bases from 2 to *A*<=-<=1.
Note that all computations should be done in base 10. You should find the result as an irreducible fraction, written in base 10.
Input Specification:
Input contains one integer number *A* (3<=≤<=*A*<=≤<=1000).
Output Specification:
Output should contain required average value in format «X/Y», where X is the numerator and Y is the denominator.
Demo Input:
['5\n', '3\n']
Demo Output:
['7/3\n', '2/1\n']
Note:
In the first sample number 5 written in all bases from 2 to 4 looks so: 101, 12, 11. Sums of digits are 2, 3 and 2, respectively. | ```python
import math
from math import log
from math import floor
n = int(input())
ans = 0
for i in range(2,n):
cnt = n
while cnt>0:
ans+=(cnt%i)
cnt//=i
# res = floor(log(n)/log(i))+1
# ans.append(res)
# print(ans)
res = ans
res1 = n-2
while res1>0:
p = res1
res1=res%res1
res=p
print(str(ans//res)+'/'+str((n-2)//res))
# if len(ans)==1:
# print(f'{sum(ans)}/1')
# else:
# print(f'{sum(ans)}/{max(ans)}')
``` | 3.954 |
1,006 | B | Polycarp's Practice | PROGRAMMING | 1,200 | [
"greedy",
"implementation",
"sortings"
] | null | null | Polycarp is practicing his problem solving skill. He has a list of $n$ problems with difficulties $a_1, a_2, \dots, a_n$, respectively. His plan is to practice for exactly $k$ days. Each day he has to solve at least one problem from his list. Polycarp solves the problems in the order they are given in his list, he cannot skip any problem from his list. He has to solve all $n$ problems in exactly $k$ days.
Thus, each day Polycarp solves a contiguous sequence of (consecutive) problems from the start of the list. He can't skip problems or solve them multiple times. As a result, in $k$ days he will solve all the $n$ problems.
The profit of the $j$-th day of Polycarp's practice is the maximum among all the difficulties of problems Polycarp solves during the $j$-th day (i.e. if he solves problems with indices from $l$ to $r$ during a day, then the profit of the day is $\max\limits_{l \le i \le r}a_i$). The total profit of his practice is the sum of the profits over all $k$ days of his practice.
You want to help Polycarp to get the maximum possible total profit over all valid ways to solve problems. Your task is to distribute all $n$ problems between $k$ days satisfying the conditions above in such a way, that the total profit is maximum.
For example, if $n = 8, k = 3$ and $a = [5, 4, 2, 6, 5, 1, 9, 2]$, one of the possible distributions with maximum total profit is: $[5, 4, 2], [6, 5], [1, 9, 2]$. Here the total profit equals $5 + 6 + 9 = 20$. | The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2000$) — the number of problems and the number of days, respectively.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2000$) — difficulties of problems in Polycarp's list, in the order they are placed in the list (i.e. in the order Polycarp will solve them). | In the first line of the output print the maximum possible total profit.
In the second line print exactly $k$ positive integers $t_1, t_2, \dots, t_k$ ($t_1 + t_2 + \dots + t_k$ must equal $n$), where $t_j$ means the number of problems Polycarp will solve during the $j$-th day in order to achieve the maximum possible total profit of his practice.
If there are many possible answers, you may print any of them. | [
"8 3\n5 4 2 6 5 1 9 2\n",
"5 1\n1 1 1 1 1\n",
"4 2\n1 2000 2000 2\n"
] | [
"20\n3 2 3",
"1\n5\n",
"4000\n2 2\n"
] | The first example is described in the problem statement.
In the second example there is only one possible distribution.
In the third example the best answer is to distribute problems in the following way: $[1, 2000], [2000, 2]$. The total profit of this distribution is $2000 + 2000 = 4000$. | 0 | [
{
"input": "8 3\n5 4 2 6 5 1 9 2",
"output": "20\n4 1 3"
},
{
"input": "5 1\n1 1 1 1 1",
"output": "1\n5"
},
{
"input": "4 2\n1 2000 2000 2",
"output": "4000\n2 2"
},
{
"input": "1 1\n2000",
"output": "2000\n1"
},
{
"input": "1 1\n1234",
"output": "1234\n1"
... | 1,667,536,842 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 3 | 62 | 1,433,600 | import math
n, k = map(int, input().split())
li = list(map(int, input().split()))
ans = 0
for i in range(0,n,math.ceil(n/k)):
ans += max(li[i:i+k])
print(ans)
c = math.ceil(n/k)
x = n
while x > 0:
if x > c:
x -= c
print(c, end=" ")
else:
print(x)
x = 0
| Title: Polycarp's Practice
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp is practicing his problem solving skill. He has a list of $n$ problems with difficulties $a_1, a_2, \dots, a_n$, respectively. His plan is to practice for exactly $k$ days. Each day he has to solve at least one problem from his list. Polycarp solves the problems in the order they are given in his list, he cannot skip any problem from his list. He has to solve all $n$ problems in exactly $k$ days.
Thus, each day Polycarp solves a contiguous sequence of (consecutive) problems from the start of the list. He can't skip problems or solve them multiple times. As a result, in $k$ days he will solve all the $n$ problems.
The profit of the $j$-th day of Polycarp's practice is the maximum among all the difficulties of problems Polycarp solves during the $j$-th day (i.e. if he solves problems with indices from $l$ to $r$ during a day, then the profit of the day is $\max\limits_{l \le i \le r}a_i$). The total profit of his practice is the sum of the profits over all $k$ days of his practice.
You want to help Polycarp to get the maximum possible total profit over all valid ways to solve problems. Your task is to distribute all $n$ problems between $k$ days satisfying the conditions above in such a way, that the total profit is maximum.
For example, if $n = 8, k = 3$ and $a = [5, 4, 2, 6, 5, 1, 9, 2]$, one of the possible distributions with maximum total profit is: $[5, 4, 2], [6, 5], [1, 9, 2]$. Here the total profit equals $5 + 6 + 9 = 20$.
Input Specification:
The first line of the input contains two integers $n$ and $k$ ($1 \le k \le n \le 2000$) — the number of problems and the number of days, respectively.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2000$) — difficulties of problems in Polycarp's list, in the order they are placed in the list (i.e. in the order Polycarp will solve them).
Output Specification:
In the first line of the output print the maximum possible total profit.
In the second line print exactly $k$ positive integers $t_1, t_2, \dots, t_k$ ($t_1 + t_2 + \dots + t_k$ must equal $n$), where $t_j$ means the number of problems Polycarp will solve during the $j$-th day in order to achieve the maximum possible total profit of his practice.
If there are many possible answers, you may print any of them.
Demo Input:
['8 3\n5 4 2 6 5 1 9 2\n', '5 1\n1 1 1 1 1\n', '4 2\n1 2000 2000 2\n']
Demo Output:
['20\n3 2 3', '1\n5\n', '4000\n2 2\n']
Note:
The first example is described in the problem statement.
In the second example there is only one possible distribution.
In the third example the best answer is to distribute problems in the following way: $[1, 2000], [2000, 2]$. The total profit of this distribution is $2000 + 2000 = 4000$. | ```python
import math
n, k = map(int, input().split())
li = list(map(int, input().split()))
ans = 0
for i in range(0,n,math.ceil(n/k)):
ans += max(li[i:i+k])
print(ans)
c = math.ceil(n/k)
x = n
while x > 0:
if x > c:
x -= c
print(c, end=" ")
else:
print(x)
x = 0
``` | 0 | |
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... | 1,652,615,819 | 2,147,483,647 | PyPy 3 | OK | TESTS | 52 | 280 | 32,665,600 | z=lambda :tuple(map(int,input().split()));s,s1=z();a,b=z();print("YES" if z()[a-1]<z()[-b] else "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
z=lambda :tuple(map(int,input().split()));s,s1=z();a,b=z();print("YES" if z()[a-1]<z()[-b] else "NO")
``` | 3 |
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