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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
462 | B | Appleman and Card Game | PROGRAMMING | 1,300 | [
"greedy"
] | null | null | Appleman has *n* cards. Each card has an uppercase letter written on it. Toastman must choose *k* cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card *i* you should calculate how much Toastman's cards have the letter equal to letter on *i*th, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get? | The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The next line contains *n* uppercase letters without spaces — the *i*-th letter describes the *i*-th card of the Appleman. | Print a single integer – the answer to the problem. | [
"15 10\nDZFDFZDFDDDDDDF\n",
"6 4\nYJSNPI\n"
] | [
"82\n",
"4\n"
] | In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin. | 1,000 | [
{
"input": "15 10\nDZFDFZDFDDDDDDF",
"output": "82"
},
{
"input": "6 4\nYJSNPI",
"output": "4"
},
{
"input": "5 3\nAOWBY",
"output": "3"
},
{
"input": "1 1\nV",
"output": "1"
},
{
"input": "2 1\nWT",
"output": "1"
},
{
"input": "2 2\nBL",
"output": "2"
},
{
"input": "5 1\nFACJT",
"output": "1"
},
{
"input": "5 5\nMJDIJ",
"output": "7"
},
{
"input": "15 5\nAZBIPTOFTJCJJIK",
"output": "13"
},
{
"input": "100 1\nEVEEVEEEGGECFEHEFVFVFHVHEEEEEFCVEEEEEEVFVEEVEEHEEVEFEVVEFEEEFEVECEHGHEEFGEEVCEECCECEFHEVEEEEEEGEEHVH",
"output": "1"
},
{
"input": "100 15\nKKTFFUTFCKUIKKKKFIFFKTUKUUKUKKIKKKTIFKTKUCFFKKKIIKKKKKKTFKFKKIRKKKFKUUKIKUUUFFKKKKTUZKITUIKKIKUKKTIK",
"output": "225"
},
{
"input": "100 50\nYYIYYAAAIEAAYAYAEAIIIAAEAAYEAEYYYIAEYAYAYYAAAIAYAEAAYAYYIYAAYYAAAAAAIYYYAAYAAEAAYAIEIYIYAYAYAYIIAAEY",
"output": "1972"
},
{
"input": "100 90\nFAFAOOAOOAFAOTFAFAFFATAAAOFAAOAFBAAAFBOAOFFFOAOAFAPFOFAOFAAFOAAAAFAAFOFAAOFPPAAOOAAOOFFOFFFOFAOTOFAF",
"output": "2828"
},
{
"input": "100 99\nBFFBBFBFBQFFFFFQBFFBFFBQFBFQFBBFQFFFBFFFBFQFQFBFFBBFYQFBFFFFFFFBQQFQBFBQBQFFFBQQFFFBQFYFBFBFFFBBBQQY",
"output": "3713"
},
{
"input": "100 100\nMQSBDAJABILIBCUEOWGWCEXMUTEYQKAIWGINXVQEOFDUBSVULROQHQRZZAALVQFEFRAAAYUIMGCAFQGIAEFBETRECGSFQJNXHHDN",
"output": "514"
},
{
"input": "100 50\nBMYIXQSJNHGFVFPJBIOBXIKSFNUFPVODCUBQYSIIQNVNXXCWXWRHKFEUPPIIDDGRDBJLZDCBMNJMYRMWFIHOSTDJJHXHPNRKWNFD",
"output": "328"
},
{
"input": "100 50\nENFNEMLJEMDMFMNNGNIMNINALGLLLAEMENEMNLMMIEIJNAINBJEJMFJLLIMINELGFLAIAMJMHMGNLIEFJIEEFEFGLLLDLMEAEIMM",
"output": "748"
}
] | 1,647,755,895 | 2,147,483,647 | Python 3 | OK | TESTS | 41 | 61 | 512,000 | n, k = map(int, input().split())
s = list(str(input()))
dict = {}
for i in s:
if i in dict:
dict[i] += 1
else :
dict[i] = 1
p = list(dict.values())
p.sort(reverse = True)
ans = 0
for i in p:
if i <= k:
k -= i
ans += i*i
else :
ans += k*k
break
print(ans)
| Title: Appleman and Card Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Appleman has *n* cards. Each card has an uppercase letter written on it. Toastman must choose *k* cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card *i* you should calculate how much Toastman's cards have the letter equal to letter on *i*th, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=105). The next line contains *n* uppercase letters without spaces — the *i*-th letter describes the *i*-th card of the Appleman.
Output Specification:
Print a single integer – the answer to the problem.
Demo Input:
['15 10\nDZFDFZDFDDDDDDF\n', '6 4\nYJSNPI\n']
Demo Output:
['82\n', '4\n']
Note:
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin. | ```python
n, k = map(int, input().split())
s = list(str(input()))
dict = {}
for i in s:
if i in dict:
dict[i] += 1
else :
dict[i] = 1
p = list(dict.values())
p.sort(reverse = True)
ans = 0
for i in p:
if i <= k:
k -= i
ans += i*i
else :
ans += k*k
break
print(ans)
``` | 3 | |
486 | A | Calculating Function | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | For a positive integer *n* let's define a function *f*:
*f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n*
Your task is to calculate *f*(*n*) for a given integer *n*. | The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015). | Print *f*(*n*) in a single line. | [
"4\n",
"5\n"
] | [
"2\n",
"-3\n"
] | *f*(4) = - 1 + 2 - 3 + 4 = 2
*f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3 | 500 | [
{
"input": "4",
"output": "2"
},
{
"input": "5",
"output": "-3"
},
{
"input": "1000000000",
"output": "500000000"
},
{
"input": "1000000001",
"output": "-500000001"
},
{
"input": "1000000000000000",
"output": "500000000000000"
},
{
"input": "100",
"output": "50"
},
{
"input": "101",
"output": "-51"
},
{
"input": "102",
"output": "51"
},
{
"input": "103",
"output": "-52"
},
{
"input": "104",
"output": "52"
},
{
"input": "105",
"output": "-53"
},
{
"input": "106",
"output": "53"
},
{
"input": "107",
"output": "-54"
},
{
"input": "108",
"output": "54"
},
{
"input": "109",
"output": "-55"
},
{
"input": "208170109961052",
"output": "104085054980526"
},
{
"input": "46017661651072",
"output": "23008830825536"
},
{
"input": "4018154546667",
"output": "-2009077273334"
},
{
"input": "288565475053",
"output": "-144282737527"
},
{
"input": "3052460231",
"output": "-1526230116"
},
{
"input": "29906716",
"output": "14953358"
},
{
"input": "87897701693326",
"output": "43948850846663"
},
{
"input": "8240",
"output": "4120"
},
{
"input": "577935",
"output": "-288968"
},
{
"input": "62",
"output": "31"
},
{
"input": "1",
"output": "-1"
},
{
"input": "2",
"output": "1"
},
{
"input": "9999999999999",
"output": "-5000000000000"
},
{
"input": "1000000000000",
"output": "500000000000"
},
{
"input": "99999999999999",
"output": "-50000000000000"
},
{
"input": "999999999999999",
"output": "-500000000000000"
},
{
"input": "42191359342",
"output": "21095679671"
},
{
"input": "100000000000000",
"output": "50000000000000"
},
{
"input": "145645214654154",
"output": "72822607327077"
},
{
"input": "4294967296",
"output": "2147483648"
},
{
"input": "3037000499",
"output": "-1518500250"
},
{
"input": "10000000000001",
"output": "-5000000000001"
},
{
"input": "100000017040846",
"output": "50000008520423"
},
{
"input": "98979894985999",
"output": "-49489947493000"
}
] | 1,695,374,083 | 2,147,483,647 | Python 3 | OK | TESTS | 39 | 46 | 0 | # -*- coding: utf-8 -*-
"""
Created on Fri Sep 22 16:46:00 2023
@author: 2300011413
"""
n=int(input())
if n%2==0:
print(n//2)
else:
print((n-1)//2-n) | Title: Calculating Function
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For a positive integer *n* let's define a function *f*:
*f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n*
Your task is to calculate *f*(*n*) for a given integer *n*.
Input Specification:
The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015).
Output Specification:
Print *f*(*n*) in a single line.
Demo Input:
['4\n', '5\n']
Demo Output:
['2\n', '-3\n']
Note:
*f*(4) = - 1 + 2 - 3 + 4 = 2
*f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3 | ```python
# -*- coding: utf-8 -*-
"""
Created on Fri Sep 22 16:46:00 2023
@author: 2300011413
"""
n=int(input())
if n%2==0:
print(n//2)
else:
print((n-1)//2-n)
``` | 3 | |
455 | A | Boredom | PROGRAMMING | 1,500 | [
"dp"
] | null | null | Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.
Given a sequence *a* consisting of *n* integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it *a**k*) and delete it, at that all elements equal to *a**k*<=+<=1 and *a**k*<=-<=1 also must be deleted from the sequence. That step brings *a**k* points to the player.
Alex is a perfectionist, so he decided to get as many points as possible. Help him. | The first line contains integer *n* (1<=≤<=*n*<=≤<=105) that shows how many numbers are in Alex's sequence.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105). | Print a single integer — the maximum number of points that Alex can earn. | [
"2\n1 2\n",
"3\n1 2 3\n",
"9\n1 2 1 3 2 2 2 2 3\n"
] | [
"2\n",
"4\n",
"10\n"
] | Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points. | 500 | [
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n1 2 3",
"output": "4"
},
{
"input": "9\n1 2 1 3 2 2 2 2 3",
"output": "10"
},
{
"input": "5\n3 3 4 5 4",
"output": "11"
},
{
"input": "5\n5 3 5 3 4",
"output": "16"
},
{
"input": "5\n4 2 3 2 5",
"output": "9"
},
{
"input": "10\n10 5 8 9 5 6 8 7 2 8",
"output": "46"
},
{
"input": "10\n1 1 1 1 1 1 2 3 4 4",
"output": "14"
},
{
"input": "100\n6 6 8 9 7 9 6 9 5 7 7 4 5 3 9 1 10 3 4 5 8 9 6 5 6 4 10 9 1 4 1 7 1 4 9 10 8 2 9 9 10 5 8 9 5 6 8 7 2 8 7 6 2 6 10 8 6 2 5 5 3 2 8 8 5 3 6 2 1 4 7 2 7 3 7 4 10 10 7 5 4 7 5 10 7 1 1 10 7 7 7 2 3 4 2 8 4 7 4 4",
"output": "296"
},
{
"input": "100\n6 1 5 7 10 10 2 7 3 7 2 10 7 6 3 5 5 5 3 7 2 4 2 7 7 4 2 8 2 10 4 7 9 1 1 7 9 7 1 10 10 9 5 6 10 1 7 5 8 1 1 5 3 10 2 4 3 5 2 7 4 9 5 10 1 3 7 6 6 9 3 6 6 10 1 10 6 1 10 3 4 1 7 9 2 7 8 9 3 3 2 4 6 6 1 2 9 4 1 2",
"output": "313"
},
{
"input": "100\n7 6 3 8 8 3 10 5 3 8 6 4 6 9 6 7 3 9 10 7 5 5 9 10 7 2 3 8 9 5 4 7 9 3 6 4 9 10 7 6 8 7 6 6 10 3 7 4 5 7 7 5 1 5 4 8 7 3 3 4 7 8 5 9 2 2 3 1 6 4 6 6 6 1 7 10 7 4 5 3 9 2 4 1 5 10 9 3 9 6 8 5 2 1 10 4 8 5 10 9",
"output": "298"
},
{
"input": "100\n2 10 9 1 2 6 7 2 2 8 9 9 9 5 6 2 5 1 1 10 7 4 5 5 8 1 9 4 10 1 9 3 1 8 4 10 8 8 2 4 6 5 1 4 2 2 1 2 8 5 3 9 4 10 10 7 8 6 1 8 2 6 7 1 6 7 3 10 10 3 7 7 6 9 6 8 8 10 4 6 4 3 3 3 2 3 10 6 8 5 5 10 3 7 3 1 1 1 5 5",
"output": "312"
},
{
"input": "100\n4 9 7 10 4 7 2 6 1 9 1 8 7 5 5 7 6 7 9 8 10 5 3 5 7 10 3 2 1 3 8 9 4 10 4 7 6 4 9 6 7 1 9 4 3 5 8 9 2 7 10 5 7 5 3 8 10 3 8 9 3 4 3 10 6 5 1 8 3 2 5 8 4 7 5 3 3 2 6 9 9 8 2 7 6 3 2 2 8 8 4 5 6 9 2 3 2 2 5 2",
"output": "287"
},
{
"input": "100\n4 8 10 1 8 8 8 1 10 3 1 8 6 8 6 1 10 3 3 3 3 7 2 1 1 6 10 1 7 9 8 10 3 8 6 2 1 6 5 6 10 8 9 7 4 3 10 5 3 9 10 5 10 8 8 5 7 8 9 5 3 9 9 2 7 8 1 10 4 9 2 8 10 10 5 8 5 1 7 3 4 5 2 5 9 3 2 5 6 2 3 10 1 5 9 6 10 4 10 8",
"output": "380"
},
{
"input": "100\n4 8 10 1 8 8 8 1 10 3 1 8 6 8 6 1 10 3 3 3 3 7 2 1 1 6 10 1 7 9 8 10 3 8 6 2 1 6 5 6 10 8 9 7 4 3 10 5 3 9 10 5 10 8 8 5 7 8 9 5 3 9 9 2 7 8 1 10 4 9 2 8 10 10 5 8 5 1 7 3 4 5 2 5 9 3 2 5 6 2 3 10 1 5 9 6 10 4 10 8",
"output": "380"
},
{
"input": "100\n10 5 8 4 4 4 1 4 5 8 3 10 2 4 1 10 8 1 1 6 8 4 2 9 1 3 1 7 7 9 3 5 5 8 6 9 9 4 8 1 3 3 2 6 1 5 4 5 3 5 5 6 7 5 7 9 3 5 4 9 2 6 8 1 1 7 7 3 8 9 8 7 3 2 4 1 6 1 3 9 4 2 2 8 5 10 1 8 8 5 1 5 6 9 4 5 6 5 10 2",
"output": "265"
},
{
"input": "100\n7 5 1 8 5 6 6 2 6 2 7 7 3 6 2 4 4 2 10 2 2 2 10 6 6 1 5 10 9 1 5 9 8 9 4 1 10 5 7 5 7 6 4 8 8 1 7 8 3 8 2 1 8 4 10 3 5 6 6 10 9 6 5 1 10 7 6 9 9 2 10 10 9 1 2 1 7 7 4 10 1 10 5 5 3 8 9 8 1 4 10 2 4 5 4 4 1 6 2 9",
"output": "328"
},
{
"input": "100\n5 6 10 7 1 7 10 1 9 1 5 1 4 1 3 3 7 9 1 6 1 6 5 7 1 6 3 1 3 6 3 8 2 4 1 5 2 10 7 3 10 4 10 1 5 4 2 9 7 9 5 7 10 4 1 4 8 9 3 1 3 7 7 4 3 7 7 10 6 9 5 5 6 5 3 9 8 8 5 5 4 10 9 4 10 4 1 8 3 5 4 10 9 3 10 4 10 7 10 9",
"output": "324"
},
{
"input": "10\n7 4 5 3 9 1 10 3 4 5",
"output": "34"
},
{
"input": "10\n8 9 6 5 6 4 10 9 1 4",
"output": "39"
},
{
"input": "10\n1 7 1 4 9 10 8 2 9 9",
"output": "40"
},
{
"input": "1\n100000",
"output": "100000"
}
] | 1,700,050,090 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 10 | 108 | 12,800,000 | n = int(input())
array = sorted(list(map(int,input().split())))
lst = ['s']+list(set(array))
dic = {0:0}
for i in array:
if i in dic:
dic[i] += 1
else:
dic[i] = 1
m = len(dic)
dp = [0]*m
for i in range(1,m):
if lst[i-1] != lst[i] - 1:
dp[i] = dp[i-1] + lst[i]*dic[lst[i]]
else:
dp[i] = max(dp[i-2] + lst[i]*dic[lst[i]],dp[i-3] + lst[i]*dic[lst[i]])
print(max(dp[-1],dp[-2])) | Title: Boredom
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.
Given a sequence *a* consisting of *n* integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it *a**k*) and delete it, at that all elements equal to *a**k*<=+<=1 and *a**k*<=-<=1 also must be deleted from the sequence. That step brings *a**k* points to the player.
Alex is a perfectionist, so he decided to get as many points as possible. Help him.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) that shows how many numbers are in Alex's sequence.
The second line contains *n* integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105).
Output Specification:
Print a single integer — the maximum number of points that Alex can earn.
Demo Input:
['2\n1 2\n', '3\n1 2 3\n', '9\n1 2 1 3 2 2 2 2 3\n']
Demo Output:
['2\n', '4\n', '10\n']
Note:
Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points. | ```python
n = int(input())
array = sorted(list(map(int,input().split())))
lst = ['s']+list(set(array))
dic = {0:0}
for i in array:
if i in dic:
dic[i] += 1
else:
dic[i] = 1
m = len(dic)
dp = [0]*m
for i in range(1,m):
if lst[i-1] != lst[i] - 1:
dp[i] = dp[i-1] + lst[i]*dic[lst[i]]
else:
dp[i] = max(dp[i-2] + lst[i]*dic[lst[i]],dp[i-3] + lst[i]*dic[lst[i]])
print(max(dp[-1],dp[-2]))
``` | 0 | |
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\nRHVRWGODYWWTYZFLFYKCVUFFRTQDINKNWPKFHZBFWBHWINWJW",
"output": "1"
},
{
"input": "4\nSAKE\nSAKE\n13\n2",
"output": "4"
},
{
"input": "2\n0\n17",
"output": "2"
},
{
"input": "1\n0",
"output": "1"
}
] | 1,540,759,962 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 216 | 0 | n=int(input())
l=[]
p=['0','1','2','3','4','5','6','7','8','9']
s=0
for i in range (n):
x=input()
l.append(x)
if x[0] not in (p) :
s=s+1
print(s) | 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
n=int(input())
l=[]
p=['0','1','2','3','4','5','6','7','8','9']
s=0
for i in range (n):
x=input()
l.append(x)
if x[0] not in (p) :
s=s+1
print(s)
``` | 0 |
630 | K | Indivisibility | PROGRAMMING | 1,500 | [
"math",
"number theory"
] | null | null | IT City company developing computer games decided to upgrade its way to reward its employees. Now it looks the following way. After a new game release users start buying it actively, and the company tracks the number of sales with precision to each transaction. Every time when the next number of sales is not divisible by any number from 2 to 10 every developer of this game gets a small bonus.
A game designer Petya knows that the company is just about to release a new game that was partly developed by him. On the basis of his experience he predicts that *n* people will buy the game during the first month. Now Petya wants to determine how many times he will get the bonus. Help him to know it. | The only line of the input contains one integer *n* (1<=≤<=*n*<=≤<=1018) — the prediction on the number of people who will buy the game. | Output one integer showing how many numbers from 1 to *n* are not divisible by any number from 2 to 10. | [
"12\n"
] | [
"2"
] | none | 0 | [
{
"input": "12",
"output": "2"
},
{
"input": "2519",
"output": "576"
},
{
"input": "2521",
"output": "577"
},
{
"input": "1",
"output": "1"
},
{
"input": "314159265",
"output": "71807832"
},
{
"input": "718281828459045235",
"output": "164178703647781768"
},
{
"input": "1000000000000000000",
"output": "228571428571428571"
},
{
"input": "987654321234567890",
"output": "225749559139329804"
},
{
"input": "3628800",
"output": "829440"
},
{
"input": "504000000000000000",
"output": "115200000000000000"
}
] | 1,690,717,444 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 10 | 46 | 0 | n = int(input())
print(n-n//2-n//3-n//5-n//7+n//6+n//10+n//14+n//15+n//21+n//35-n//30-n//42-n//70-n//105+n//210) | Title: Indivisibility
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
IT City company developing computer games decided to upgrade its way to reward its employees. Now it looks the following way. After a new game release users start buying it actively, and the company tracks the number of sales with precision to each transaction. Every time when the next number of sales is not divisible by any number from 2 to 10 every developer of this game gets a small bonus.
A game designer Petya knows that the company is just about to release a new game that was partly developed by him. On the basis of his experience he predicts that *n* people will buy the game during the first month. Now Petya wants to determine how many times he will get the bonus. Help him to know it.
Input Specification:
The only line of the input contains one integer *n* (1<=≤<=*n*<=≤<=1018) — the prediction on the number of people who will buy the game.
Output Specification:
Output one integer showing how many numbers from 1 to *n* are not divisible by any number from 2 to 10.
Demo Input:
['12\n']
Demo Output:
['2']
Note:
none | ```python
n = int(input())
print(n-n//2-n//3-n//5-n//7+n//6+n//10+n//14+n//15+n//21+n//35-n//30-n//42-n//70-n//105+n//210)
``` | 3 | |
141 | A | Amusing Joke | PROGRAMMING | 800 | [
"implementation",
"sortings",
"strings"
] | null | null | So, the New Year holidays are over. Santa Claus and his colleagues can take a rest and have guests at last. When two "New Year and Christmas Men" meet, thear assistants cut out of cardboard the letters from the guest's name and the host's name in honor of this event. Then the hung the letters above the main entrance. One night, when everyone went to bed, someone took all the letters of our characters' names. Then he may have shuffled the letters and put them in one pile in front of the door.
The next morning it was impossible to find the culprit who had made the disorder. But everybody wondered whether it is possible to restore the names of the host and his guests from the letters lying at the door? That is, we need to verify that there are no extra letters, and that nobody will need to cut more letters.
Help the "New Year and Christmas Men" and their friends to cope with this problem. You are given both inscriptions that hung over the front door the previous night, and a pile of letters that were found at the front door next morning. | The input file consists of three lines: the first line contains the guest's name, the second line contains the name of the residence host and the third line contains letters in a pile that were found at the door in the morning. All lines are not empty and contain only uppercase Latin letters. The length of each line does not exceed 100. | Print "YES" without the quotes, if the letters in the pile could be permuted to make the names of the "New Year and Christmas Men". Otherwise, print "NO" without the quotes. | [
"SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS\n",
"PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI\n",
"BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | In the first sample the letters written in the last line can be used to write the names and there won't be any extra letters left.
In the second sample letter "P" is missing from the pile and there's an extra letter "L".
In the third sample there's an extra letter "L". | 500 | [
{
"input": "SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS",
"output": "YES"
},
{
"input": "PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI",
"output": "NO"
},
{
"input": "BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER",
"output": "NO"
},
{
"input": "B\nA\nAB",
"output": "YES"
},
{
"input": "ONDOL\nJNPB\nONLNJBODP",
"output": "YES"
},
{
"input": "Y\nW\nYW",
"output": "YES"
},
{
"input": "OI\nM\nIMO",
"output": "YES"
},
{
"input": "VFQRWWWACX\nGHZJPOQUSXRAQDGOGMR\nOPAWDOUSGWWCGQXXQAZJRQRGHRMVF",
"output": "YES"
},
{
"input": "JUTCN\nPIGMZOPMEUFADQBW\nNWQGZMAIPUPOMCDUB",
"output": "NO"
},
{
"input": "Z\nO\nZOCNDOLTBZKQLTBOLDEGXRHZGTTPBJBLSJCVSVXISQZCSFDEBXRCSGBGTHWOVIXYHACAGBRYBKBJAEPIQZHVEGLYH",
"output": "NO"
},
{
"input": "IQ\nOQ\nQOQIGGKFNHJSGCGM",
"output": "NO"
},
{
"input": "ROUWANOPNIGTVMIITVMZ\nOQTUPZMTKUGY\nVTVNGZITGPUNPMQOOATUUIYIWMMKZOTR",
"output": "YES"
},
{
"input": "OVQELLOGFIOLEHXMEMBJDIGBPGEYFG\nJNKFPFFIJOFHRIFHXEWYZOPDJBZTJZKBWQTECNHRFSJPJOAPQT\nYAIPFFFEXJJNEJPLREIGODEGQZVMCOBDFKWTMWJSBEBTOFFQOHIQJLHFNXIGOHEZRZLFOKJBJPTPHPGY",
"output": "YES"
},
{
"input": "NBJGVNGUISUXQTBOBKYHQCOOVQWUXWPXBUDLXPKX\nNSFQDFUMQDQWQ\nWXKKVNTDQQFXCUQBIMQGQHSLVGWSBFYBUPOWPBDUUJUXQNOQDNXOX",
"output": "YES"
},
{
"input": "IJHHGKCXWDBRWJUPRDBZJLNTTNWKXLUGJSBWBOAUKWRAQWGFNL\nNJMWRMBCNPHXTDQQNZ\nWDNJRCLILNQRHWBANLTXWMJBPKUPGKJDJZAQWKTZFBRCTXHHBNXRGUQUNBNMWODGSJWW",
"output": "YES"
},
{
"input": "SRROWANGUGZHCIEFYMQVTWVOMDWPUZJFRDUMVFHYNHNTTGNXCJ\nDJYWGLBFCCECXFHOLORDGDCNRHPWXNHXFCXQCEZUHRRNAEKUIX\nWCUJDNYHNHYOPWMHLDCDYRWBVOGHFFUKOZTXJRXJHRGWICCMRNEVNEGQWTZPNFCSHDRFCFQDCXMHTLUGZAXOFNXNVGUEXIACRERU",
"output": "YES"
},
{
"input": "H\nJKFGHMIAHNDBMFXWYQLZRSVNOTEGCQSVUBYUOZBTNKTXPFQDCMKAGFITEUGOYDFIYQIORMFJEOJDNTFVIQEBICSNGKOSNLNXJWC\nBQSVDOGIHCHXSYNYTQFCHNJGYFIXTSOQINZOKSVQJMTKNTGFNXAVTUYEONMBQMGJLEWJOFGEARIOPKFUFCEMUBRBDNIIDFZDCLWK",
"output": "YES"
},
{
"input": "DSWNZRFVXQ\nPVULCZGOOU\nUOLVZXNUPOQRZGWFVDSCANQTCLEIE",
"output": "NO"
},
{
"input": "EUHTSCENIPXLTSBMLFHD\nIZAVSZPDLXOAGESUSE\nLXAELAZ",
"output": "NO"
},
{
"input": "WYSJFEREGELSKRQRXDXCGBODEFZVSI\nPEJKMGFLBFFDWRCRFSHVEFLEBTJCVCHRJTLDTISHPOGFWPLEWNYJLMXWIAOTYOXMV\nHXERTZWLEXTPIOTFRVMEJVYFFJLRPFMXDEBNSGCEOFFCWTKIDDGCFYSJKGLHBORWEPLDRXRSJYBGASSVCMHEEJFLVI",
"output": "NO"
},
{
"input": "EPBMDIUQAAUGLBIETKOKFLMTCVEPETWJRHHYKCKU\nHGMAETVPCFZYNNKDQXVXUALHYLOTCHM\nECGXACVKEYMCEDOTMKAUFHLHOMT",
"output": "NO"
},
{
"input": "NUBKQEJHALANSHEIFUZHYEZKKDRFHQKAJHLAOWTZIMOCWOVVDW\nEFVOBIGAUAUSQGVSNBKNOBDMINODMFSHDL\nKLAMKNTHBFFOHVKWICHBKNDDQNEISODUSDNLUSIOAVWY",
"output": "NO"
},
{
"input": "VXINHOMEQCATZUGAJEIUIZZLPYFGUTVLNBNWCUVMEENUXKBWBGZTMRJJVJDLVSLBABVCEUDDSQFHOYPYQTWVAGTWOLKYISAGHBMC\nZMRGXPZSHOGCSAECAPGVOIGCWEOWWOJXLGYRDMPXBLOKZVRACPYQLEQGFQCVYXAGBEBELUTDAYEAGPFKXRULZCKFHZCHVCWIRGPK\nRCVUXGQVNWFGRUDLLENNDQEJHYYVWMKTLOVIPELKPWCLSQPTAXAYEMGWCBXEVAIZGGDDRBRT",
"output": "NO"
},
{
"input": "PHBDHHWUUTZAHELGSGGOPOQXSXEZIXHZTOKYFBQLBDYWPVCNQSXHEAXRRPVHFJBVBYCJIFOTQTWSUOWXLKMVJJBNLGTVITWTCZZ\nFUPDLNVIHRWTEEEHOOEC\nLOUSUUSZCHJBPEWIILUOXEXRQNCJEGTOBRVZLTTZAHTKVEJSNGHFTAYGY",
"output": "NO"
},
{
"input": "GDSLNIIKTO\nJF\nPDQYFKDTNOLI",
"output": "NO"
},
{
"input": "AHOKHEKKPJLJIIWJRCGY\nORELJCSIX\nZVWPXVFWFSWOXXLIHJKPXIOKRELYE",
"output": "NO"
},
{
"input": "ZWCOJFORBPHXCOVJIDPKVECMHVHCOC\nTEV\nJVGTBFTLFVIEPCCHODOFOMCVZHWXVCPEH",
"output": "NO"
},
{
"input": "AGFIGYWJLVMYZGNQHEHWKJIAWBPUAQFERMCDROFN\nPMJNHMVNRGCYZAVRWNDSMLSZHFNYIUWFPUSKKIGU\nMCDVPPRXGUAYLSDRHRURZASXUWZSIIEZCPXUVEONKNGNWRYGOSFMCKESMVJZHWWUCHWDQMLASLNNMHAU",
"output": "NO"
},
{
"input": "XLOWVFCZSSXCSYQTIIDKHNTKNKEEDFMDZKXSPVLBIDIREDUAIN\nZKIWNDGBISDB\nSLPKLYFYSRNRMOSWYLJJDGFFENPOXYLPZFTQDANKBDNZDIIEWSUTTKYBKVICLG",
"output": "NO"
},
{
"input": "PMUKBTRKFIAYVGBKHZHUSJYSSEPEOEWPOSPJLWLOCTUYZODLTUAFCMVKGQKRRUSOMPAYOTBTFPXYAZXLOADDEJBDLYOTXJCJYTHA\nTWRRAJLCQJTKOKWCGUH\nEWDPNXVCXWCDQCOYKKSOYTFSZTOOPKPRDKFJDETKSRAJRVCPDOBWUGPYRJPUWJYWCBLKOOTUPBESTOFXZHTYLLMCAXDYAEBUTAHM",
"output": "NO"
},
{
"input": "QMIMGQRQDMJDPNFEFXSXQMCHEJKTWCTCVZPUAYICOIRYOWKUSIWXJLHDYWSBOITHTMINXFKBKAWZTXXBJIVYCRWKXNKIYKLDDXL\nV\nFWACCXBVDOJFIUAVYRALBYJKXXWIIFORRUHKHCXLDBZMXIYJWISFEAWTIQFIZSBXMKNOCQKVKRWDNDAMQSTKYLDNYVTUCGOJXJTW",
"output": "NO"
},
{
"input": "XJXPVOOQODELPPWUISSYVVXRJTYBPDHJNENQEVQNVFIXSESKXVYPVVHPMOSX\nLEXOPFPVPSZK\nZVXVPYEYOYXVOISVLXPOVHEQVXPNQJIOPFDTXEUNMPEPPHELNXKKWSVSOXSBPSJDPVJVSRFQ",
"output": "YES"
},
{
"input": "OSKFHGYNQLSRFSAHPXKGPXUHXTRBJNAQRBSSWJVEENLJCDDHFXVCUNPZAIVVO\nFNUOCXAGRRHNDJAHVVLGGEZQHWARYHENBKHP\nUOEFNWVXCUNERLKVTHAGPSHKHDYFPYWZHJKHQLSNFBJHVJANRXCNSDUGVDABGHVAOVHBJZXGRACHRXEGNRPQEAPORQSILNXFS",
"output": "YES"
},
{
"input": "VYXYVVACMLPDHONBUTQFZTRREERBLKUJYKAHZRCTRLRCLOZYWVPBRGDQPFPQIF\nFE\nRNRPEVDRLYUQFYRZBCQLCYZEABKLRXCJLKVZBVFUEYRATOMDRTHFPGOWQVTIFPPH",
"output": "YES"
},
{
"input": "WYXUZQJQNLASEGLHPMSARWMTTQMQLVAZLGHPIZTRVTCXDXBOLNXZPOFCTEHCXBZ\nBLQZRRWP\nGIQZXPLTTMNHQVWPPEAPLOCDMBSTHRCFLCQRRZXLVAOQEGZBRUZJXXZTMAWLZHSLWNQTYXB",
"output": "YES"
},
{
"input": "MKVJTSSTDGKPVVDPYSRJJYEVGKBMSIOKHLZQAEWLRIBINVRDAJIBCEITKDHUCCVY\nPUJJQFHOGZKTAVNUGKQUHMKTNHCCTI\nQVJKUSIGTSVYUMOMLEGHWYKSKQTGATTKBNTKCJKJPCAIRJIRMHKBIZISEGFHVUVQZBDERJCVAKDLNTHUDCHONDCVVJIYPP",
"output": "YES"
},
{
"input": "OKNJOEYVMZXJMLVJHCSPLUCNYGTDASKSGKKCRVIDGEIBEWRVBVRVZZTLMCJLXHJIA\nDJBFVRTARTFZOWN\nAGHNVUNJVCPLWSVYBJKZSVTFGLELZASLWTIXDDJXCZDICTVIJOTMVEYOVRNMJGRKKHRMEBORAKFCZJBR",
"output": "YES"
},
{
"input": "OQZACLPSAGYDWHFXDFYFRRXWGIEJGSXWUONAFWNFXDTGVNDEWNQPHUXUJNZWWLBPYL\nOHBKWRFDRQUAFRCMT\nWIQRYXRJQWWRUWCYXNXALKFZGXFTLOODWRDPGURFUFUQOHPWBASZNVWXNCAGHWEHFYESJNFBMNFDDAPLDGT",
"output": "YES"
},
{
"input": "OVIRQRFQOOWVDEPLCJETWQSINIOPLTLXHSQWUYUJNFBMKDNOSHNJQQCDHZOJVPRYVSV\nMYYDQKOOYPOOUELCRIT\nNZSOTVLJTTVQLFHDQEJONEOUOFOLYVSOIYUDNOSIQVIRMVOERCLMYSHPCQKIDRDOQPCUPQBWWRYYOXJWJQPNKH",
"output": "YES"
},
{
"input": "WGMBZWNMSJXNGDUQUJTCNXDSJJLYRDOPEGPQXYUGBESDLFTJRZDDCAAFGCOCYCQMDBWK\nYOBMOVYTUATTFGJLYUQD\nDYXVTLQCYFJUNJTUXPUYOPCBCLBWNSDUJRJGWDOJDSQAAMUOJWSYERDYDXYTMTOTMQCGQZDCGNFBALGGDFKZMEBG",
"output": "YES"
},
{
"input": "CWLRBPMEZCXAPUUQFXCUHAQTLPBTXUUKWVXKBHKNSSJFEXLZMXGVFHHVTPYAQYTIKXJJE\nMUFOSEUEXEQTOVLGDSCWM\nJUKEQCXOXWEHCGKFPBIGMWVJLXUONFXBYTUAXERYTXKCESKLXAEHVPZMMUFTHLXTTZSDMBJLQPEUWCVUHSQQVUASPF",
"output": "YES"
},
{
"input": "IDQRX\nWETHO\nODPDGBHVUVSSISROHQJTUKPUCLXABIZQQPPBPKOSEWGEHRSRRNBAVLYEMZISMWWGKHVTXKUGUXEFBSWOIWUHRJGMWBMHQLDZHBWA",
"output": "NO"
},
{
"input": "IXFDY\nJRMOU\nDF",
"output": "NO"
},
{
"input": "JPSPZ\nUGCUB\nJMZZZZZZZZ",
"output": "NO"
},
{
"input": "AC\nA\nBBA",
"output": "NO"
},
{
"input": "UIKWWKXLSHTOOZOVGXKYSOJEHAUEEG\nKZXQDWJJWRXFHKJDQHJK\nXMZHTFOGEXAUJXXJUYVJIFOTKLZHDKELJWERHMGAWGKWAQKEKHIDWGGZVYOHKXRPWSJDPESFJUMKQYWBYUTHQYEFZUGKQOBHYDWB",
"output": "NO"
},
{
"input": "PXWRXRPFLR\nPJRWWXIVHODV\nXW",
"output": "NO"
},
{
"input": "CHTAZVHGSHCVIBK\nEQINEBKXEPYJSAZIMLDF\nZCZZZZDZMCZZEZDZZEZZZZQZZBZZZOZZCZE",
"output": "NO"
},
{
"input": "GXPZFSELJJNDAXYRV\nUYBKPMVBSOVOJWMONLTJOJCNQKMTAHEWLHOWIIBH\nHCWNFWJPEJIWOVPTBMVCRJLSISSVNOHCKLBFMIUAIMASQWPXEYXBOXQGFEMYJLBKDCZIMJNHOJEDGGANIVYKQTUOSOVOPWHVJGXH",
"output": "NO"
},
{
"input": "LFGJCJJDUTUP\nOVSBILTIYCJCRHKCIXCETJQJJ\nGIJJTJCLTJJJ",
"output": "NO"
},
{
"input": "GIO\nPRL\nPRL",
"output": "NO"
},
{
"input": "A\nB\nABC",
"output": "NO"
},
{
"input": "KKK\nKKK\nZZZZZ",
"output": "NO"
},
{
"input": "ZMYGQLDBLAPN\nZFJBKWHROVNPSJQUDFTHOCGREUFLYIWYICD\nZMJZZEDAZANKZZZZZZEZZBZDZZZZZZKHZZFZZZDZNZMDZZA",
"output": "NO"
}
] | 1,680,626,002 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 92 | 0 | x1=input()
x2=input()
x3=input()
c=0
guest=x1+x2
guest_d={}
letter={}
for i in guest:
guest_d[i]=guest.count(i)
for i in letter:
letter[i]=x3.count(i)
if guest_d==letter:
print("YES")
else:
print("NO") | Title: Amusing Joke
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
So, the New Year holidays are over. Santa Claus and his colleagues can take a rest and have guests at last. When two "New Year and Christmas Men" meet, thear assistants cut out of cardboard the letters from the guest's name and the host's name in honor of this event. Then the hung the letters above the main entrance. One night, when everyone went to bed, someone took all the letters of our characters' names. Then he may have shuffled the letters and put them in one pile in front of the door.
The next morning it was impossible to find the culprit who had made the disorder. But everybody wondered whether it is possible to restore the names of the host and his guests from the letters lying at the door? That is, we need to verify that there are no extra letters, and that nobody will need to cut more letters.
Help the "New Year and Christmas Men" and their friends to cope with this problem. You are given both inscriptions that hung over the front door the previous night, and a pile of letters that were found at the front door next morning.
Input Specification:
The input file consists of three lines: the first line contains the guest's name, the second line contains the name of the residence host and the third line contains letters in a pile that were found at the door in the morning. All lines are not empty and contain only uppercase Latin letters. The length of each line does not exceed 100.
Output Specification:
Print "YES" without the quotes, if the letters in the pile could be permuted to make the names of the "New Year and Christmas Men". Otherwise, print "NO" without the quotes.
Demo Input:
['SANTACLAUS\nDEDMOROZ\nSANTAMOROZDEDCLAUS\n', 'PAPAINOEL\nJOULUPUKKI\nJOULNAPAOILELUPUKKI\n', 'BABBONATALE\nFATHERCHRISTMAS\nBABCHRISTMASBONATALLEFATHER\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
In the first sample the letters written in the last line can be used to write the names and there won't be any extra letters left.
In the second sample letter "P" is missing from the pile and there's an extra letter "L".
In the third sample there's an extra letter "L". | ```python
x1=input()
x2=input()
x3=input()
c=0
guest=x1+x2
guest_d={}
letter={}
for i in guest:
guest_d[i]=guest.count(i)
for i in letter:
letter[i]=x3.count(i)
if guest_d==letter:
print("YES")
else:
print("NO")
``` | 0 | |
868 | A | Bark to Unlock | PROGRAMMING | 900 | [
"brute force",
"implementation",
"strings"
] | null | null | As technologies develop, manufacturers are making the process of unlocking a phone as user-friendly as possible. To unlock its new phone, Arkady's pet dog Mu-mu has to bark the password once. The phone represents a password as a string of two lowercase English letters.
Mu-mu's enemy Kashtanka wants to unlock Mu-mu's phone to steal some sensible information, but it can only bark *n* distinct words, each of which can be represented as a string of two lowercase English letters. Kashtanka wants to bark several words (not necessarily distinct) one after another to pronounce a string containing the password as a substring. Tell if it's possible to unlock the phone in this way, or not. | The first line contains two lowercase English letters — the password on the phone.
The second line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of words Kashtanka knows.
The next *n* lines contain two lowercase English letters each, representing the words Kashtanka knows. The words are guaranteed to be distinct. | Print "YES" if Kashtanka can bark several words in a line forming a string containing the password, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower). | [
"ya\n4\nah\noy\nto\nha\n",
"hp\n2\nht\ntp\n",
"ah\n1\nha\n"
] | [
"YES\n",
"NO\n",
"YES\n"
] | In the first example the password is "ya", and Kashtanka can bark "oy" and then "ah", and then "ha" to form the string "oyahha" which contains the password. So, the answer is "YES".
In the second example Kashtanka can't produce a string containing password as a substring. Note that it can bark "ht" and then "tp" producing "http", but it doesn't contain the password "hp" as a substring.
In the third example the string "hahahaha" contains "ah" as a substring. | 250 | [
{
"input": "ya\n4\nah\noy\nto\nha",
"output": "YES"
},
{
"input": "hp\n2\nht\ntp",
"output": "NO"
},
{
"input": "ah\n1\nha",
"output": "YES"
},
{
"input": "bb\n4\nba\nab\naa\nbb",
"output": "YES"
},
{
"input": "bc\n4\nca\nba\nbb\ncc",
"output": "YES"
},
{
"input": "ba\n4\ncd\nad\ncc\ncb",
"output": "YES"
},
{
"input": "pg\n4\nzl\nxs\ndi\nxn",
"output": "NO"
},
{
"input": "bn\n100\ndf\nyb\nze\nml\nyr\nof\nnw\nfm\ndw\nlv\nzr\nhu\nzt\nlw\nld\nmo\nxz\ntp\nmr\nou\nme\npx\nvp\nes\nxi\nnr\nbx\nqc\ngm\njs\nkn\ntw\nrq\nkz\nuc\nvc\nqr\nab\nna\nro\nya\nqy\ngu\nvk\nqk\ngs\nyq\nop\nhw\nrj\neo\nlz\nbh\nkr\nkb\nma\nrd\nza\nuf\nhq\nmc\nmn\nti\nwn\nsh\nax\nsi\nnd\ntz\ndu\nfj\nkl\nws\now\nnf\nvr\nye\nzc\niw\nfv\nkv\noo\nsm\nbc\nrs\nau\nuz\nuv\ngh\nsu\njn\ndz\nrl\nwj\nbk\nzl\nas\nms\nit\nwu",
"output": "YES"
},
{
"input": "bb\n1\naa",
"output": "NO"
},
{
"input": "qm\n25\nqw\nwe\ner\nrt\nty\nyu\nui\nio\nop\npa\nas\nsd\ndf\nfg\ngh\nhj\njk\nkl\nlz\nzx\nxc\ncv\nvb\nbn\nnm",
"output": "NO"
},
{
"input": "mq\n25\nqw\nwe\ner\nrt\nty\nyu\nui\nio\nop\npa\nas\nsd\ndf\nfg\ngh\nhj\njk\nkl\nlz\nzx\nxc\ncv\nvb\nbn\nnm",
"output": "YES"
},
{
"input": "aa\n1\naa",
"output": "YES"
},
{
"input": "bb\n1\nbb",
"output": "YES"
},
{
"input": "ba\n1\ncc",
"output": "NO"
},
{
"input": "ha\n1\nha",
"output": "YES"
},
{
"input": "aa\n1\naa",
"output": "YES"
},
{
"input": "ez\n1\njl",
"output": "NO"
},
{
"input": "aa\n2\nab\nba",
"output": "YES"
},
{
"input": "aa\n2\nca\ncc",
"output": "NO"
},
{
"input": "dd\n2\nac\ndc",
"output": "NO"
},
{
"input": "qc\n2\nyc\nkr",
"output": "NO"
},
{
"input": "aa\n3\nba\nbb\nab",
"output": "YES"
},
{
"input": "ca\n3\naa\nbb\nab",
"output": "NO"
},
{
"input": "ca\n3\nbc\nbd\nca",
"output": "YES"
},
{
"input": "dd\n3\nmt\nrg\nxl",
"output": "NO"
},
{
"input": "be\n20\nad\ncd\ncb\ndb\ndd\naa\nab\nca\nae\ned\ndc\nbb\nba\nda\nee\nea\ncc\nac\nec\neb",
"output": "YES"
},
{
"input": "fc\n20\nca\nbb\nce\nfd\nde\nfa\ncc\nec\nfb\nfc\nff\nbe\ncf\nba\ndb\ned\naf\nae\nda\nef",
"output": "YES"
},
{
"input": "ca\n20\ndc\naf\ndf\neg\naa\nbc\nea\nbd\nab\ndb\ngc\nfb\nba\nbe\nee\ngf\ncf\nag\nga\nca",
"output": "YES"
},
{
"input": "ke\n20\nzk\nra\nbq\nqz\nwt\nzg\nmz\nuk\nge\nuv\nud\nfd\neh\ndm\nsk\nki\nfv\ntp\nat\nfb",
"output": "YES"
},
{
"input": "hh\n50\nag\nhg\ndg\nfh\neg\ngh\ngd\nda\nbh\nab\nhf\ndc\nhb\nfe\nad\nec\nac\nfd\nca\naf\ncg\nhd\neb\nce\nhe\nha\ngb\nea\nae\nfb\nff\nbe\nch\nhh\nee\nde\nge\ngf\naa\ngg\neh\ned\nbf\nfc\nah\nga\nbd\ncb\nbg\nbc",
"output": "YES"
},
{
"input": "id\n50\nhi\ndc\nfg\nee\ngi\nhc\nac\nih\ndg\nfc\nde\ned\nie\neb\nic\ncf\nib\nfa\ngc\nba\nbe\nga\nha\nhg\nia\ndf\nab\nei\neh\nad\nii\nci\ndh\nec\nif\ndi\nbg\nag\nhe\neg\nca\nae\ndb\naa\nid\nfh\nhh\ncc\nfb\ngb",
"output": "YES"
},
{
"input": "fe\n50\nje\nbi\nbg\ngc\nfb\nig\ndf\nji\ndg\nfe\nfc\ncf\ngf\nai\nhe\nac\nch\nja\ngh\njf\nge\ncb\nij\ngb\ncg\naf\neh\nee\nhd\njd\njb\nii\nca\nci\nga\nab\nhi\nag\nfj\nej\nfi\nie\ndj\nfg\nef\njc\njg\njh\nhf\nha",
"output": "YES"
},
{
"input": "rn\n50\nba\nec\nwg\nao\nlk\nmz\njj\ncf\nfa\njk\ndy\nsz\njs\nzr\nqv\ntx\nwv\nrd\nqw\nls\nrr\nvt\nrx\nkc\neh\nnj\niq\nyi\nkh\nue\nnv\nkz\nrn\nes\nua\nzf\nvu\nll\neg\nmj\ncz\nzj\nxz\net\neb\nci\nih\nig\nam\nvd",
"output": "YES"
},
{
"input": "ee\n100\nah\nfb\ncd\nbi\nii\nai\nid\nag\nie\nha\ndi\nec\nae\nce\njb\ndg\njg\ngd\ngf\nda\nih\nbd\nhj\ngg\nhb\ndf\ned\nfh\naf\nja\nci\nfc\nic\nji\nac\nhi\nfj\nch\nbc\njd\naa\nff\nad\ngj\nej\nde\nee\nhe\ncf\nga\nia\ncg\nbb\nhc\nbe\ngi\njf\nbg\naj\njj\nbh\nfe\ndj\nef\ngb\nge\ndb\nig\ncj\ndc\nij\njh\nei\ndd\nib\nhf\neg\nbf\nfg\nab\ngc\nfd\nhd\ngh\neh\njc\neb\nhh\nca\nje\nbj\nif\nea\nhg\nfa\ncc\nba\ndh\ncb\nfi",
"output": "YES"
},
{
"input": "if\n100\njd\nbc\nje\nhi\nga\nde\nkb\nfc\ncd\ngd\naj\ncb\nei\nbf\ncf\ndk\ndb\ncg\nki\ngg\nkg\nfa\nkj\nii\njf\njg\ngb\nbh\nbg\neh\nhj\nhb\ndg\ndj\njc\njb\nce\ndi\nig\nci\ndf\nji\nhc\nfk\naf\nac\ngk\nhd\nae\nkd\nec\nkc\neb\nfh\nij\nie\nca\nhh\nkf\nha\ndd\nif\nef\nih\nhg\nej\nfe\njk\nea\nib\nck\nhf\nak\ngi\nch\ndc\nba\nke\nad\nka\neg\njh\nja\ngc\nfd\ncc\nab\ngj\nik\nfg\nbj\nhe\nfj\nge\ngh\nhk\nbk\ned\nid\nfi",
"output": "YES"
},
{
"input": "kd\n100\nek\nea\nha\nkf\nkj\ngh\ndl\nfj\nal\nga\nlj\nik\ngd\nid\ncb\nfh\ndk\nif\nbh\nkb\nhc\nej\nhk\ngc\ngb\nef\nkk\nll\nlf\nkh\ncl\nlh\njj\nil\nhh\nci\ndb\ndf\ngk\njg\nch\nbd\ncg\nfg\nda\neb\nlg\ndg\nbk\nje\nbg\nbl\njl\ncj\nhb\nei\naa\ngl\nka\nfa\nfi\naf\nkc\nla\ngi\nij\nib\nle\ndi\nck\nag\nlc\nca\nge\nie\nlb\nke\nii\nae\nig\nic\nhe\ncf\nhd\nak\nfb\nhi\ngf\nad\nba\nhg\nbi\nkl\nac\ngg\ngj\nbe\nlk\nld\naj",
"output": "YES"
},
{
"input": "ab\n1\nab",
"output": "YES"
},
{
"input": "ya\n1\nya",
"output": "YES"
},
{
"input": "ay\n1\nyb",
"output": "NO"
},
{
"input": "ax\n2\nii\nxa",
"output": "YES"
},
{
"input": "hi\n1\nhi",
"output": "YES"
},
{
"input": "ag\n1\nag",
"output": "YES"
},
{
"input": "th\n1\nth",
"output": "YES"
},
{
"input": "sb\n1\nsb",
"output": "YES"
},
{
"input": "hp\n1\nhp",
"output": "YES"
},
{
"input": "ah\n1\nah",
"output": "YES"
},
{
"input": "ta\n1\nta",
"output": "YES"
},
{
"input": "tb\n1\ntb",
"output": "YES"
},
{
"input": "ab\n5\nca\nda\nea\nfa\nka",
"output": "NO"
},
{
"input": "ac\n1\nac",
"output": "YES"
},
{
"input": "ha\n2\nha\nzz",
"output": "YES"
},
{
"input": "ok\n1\nok",
"output": "YES"
},
{
"input": "bc\n1\nbc",
"output": "YES"
},
{
"input": "az\n1\nzz",
"output": "NO"
},
{
"input": "ab\n2\nba\ntt",
"output": "YES"
},
{
"input": "ah\n2\nap\nhp",
"output": "NO"
},
{
"input": "sh\n1\nsh",
"output": "YES"
},
{
"input": "az\n1\nby",
"output": "NO"
},
{
"input": "as\n1\nas",
"output": "YES"
},
{
"input": "ab\n2\nab\ncd",
"output": "YES"
},
{
"input": "ab\n2\nxa\nza",
"output": "NO"
},
{
"input": "ab\n2\net\nab",
"output": "YES"
},
{
"input": "ab\n1\naa",
"output": "NO"
},
{
"input": "ab\n2\nab\nde",
"output": "YES"
},
{
"input": "ah\n2\nba\nha",
"output": "YES"
},
{
"input": "ha\n3\ndd\ncc\nha",
"output": "YES"
},
{
"input": "oo\n1\nox",
"output": "NO"
},
{
"input": "ab\n2\nax\nbx",
"output": "NO"
},
{
"input": "ww\n4\nuw\now\npo\nko",
"output": "NO"
},
{
"input": "ay\n1\nay",
"output": "YES"
},
{
"input": "yo\n1\nyo",
"output": "YES"
},
{
"input": "ba\n1\nba",
"output": "YES"
},
{
"input": "qw\n1\nqw",
"output": "YES"
},
{
"input": "la\n1\nla",
"output": "YES"
},
{
"input": "ab\n2\nbb\nbc",
"output": "NO"
},
{
"input": "aa\n2\nab\nac",
"output": "NO"
},
{
"input": "ah\n2\nbb\nha",
"output": "YES"
},
{
"input": "ya\n42\nab\nac\nad\nae\naf\nag\nah\nai\nak\naj\nba\nbc\nbd\nbe\nbf\nbg\nbh\nbi\nbk\nbj\ncb\nca\ncd\nce\ncf\ncg\nch\nci\nck\ncj\ndb\ndc\nda\nde\ndf\ndg\ndh\ndi\ndk\ndj\nef\nek",
"output": "NO"
},
{
"input": "ab\n3\nab\nxx\nyy",
"output": "YES"
},
{
"input": "ab\n2\nab\ncc",
"output": "YES"
},
{
"input": "sa\n2\nxx\nas",
"output": "YES"
},
{
"input": "ma\n1\nma",
"output": "YES"
},
{
"input": "ba\n1\nbb",
"output": "NO"
},
{
"input": "bc\n1\nab",
"output": "NO"
},
{
"input": "fa\n1\nfa",
"output": "YES"
},
{
"input": "ap\n1\nap",
"output": "YES"
},
{
"input": "ab\n1\nbb",
"output": "NO"
},
{
"input": "bk\n1\nbk",
"output": "YES"
},
{
"input": "xy\n2\nxy\naa",
"output": "YES"
},
{
"input": "ab\n2\nza\nbz",
"output": "YES"
}
] | 1,602,282,855 | 2,147,483,647 | PyPy 3 | OK | TESTS | 90 | 140 | 0 | pw = input()
n = int(input())
words = [input() for i in range(n)]
match1 = False
match2 = False
if pw in words:
print("YES")
else:
for word in words:
if pw[0] == word[1]:
match1 = True
if pw[1] == word[0]:
match2 = True
if match1 and match2:
print("YES")
else:
print("NO") | Title: Bark to Unlock
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
As technologies develop, manufacturers are making the process of unlocking a phone as user-friendly as possible. To unlock its new phone, Arkady's pet dog Mu-mu has to bark the password once. The phone represents a password as a string of two lowercase English letters.
Mu-mu's enemy Kashtanka wants to unlock Mu-mu's phone to steal some sensible information, but it can only bark *n* distinct words, each of which can be represented as a string of two lowercase English letters. Kashtanka wants to bark several words (not necessarily distinct) one after another to pronounce a string containing the password as a substring. Tell if it's possible to unlock the phone in this way, or not.
Input Specification:
The first line contains two lowercase English letters — the password on the phone.
The second line contains single integer *n* (1<=≤<=*n*<=≤<=100) — the number of words Kashtanka knows.
The next *n* lines contain two lowercase English letters each, representing the words Kashtanka knows. The words are guaranteed to be distinct.
Output Specification:
Print "YES" if Kashtanka can bark several words in a line forming a string containing the password, and "NO" otherwise.
You can print each letter in arbitrary case (upper or lower).
Demo Input:
['ya\n4\nah\noy\nto\nha\n', 'hp\n2\nht\ntp\n', 'ah\n1\nha\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n']
Note:
In the first example the password is "ya", and Kashtanka can bark "oy" and then "ah", and then "ha" to form the string "oyahha" which contains the password. So, the answer is "YES".
In the second example Kashtanka can't produce a string containing password as a substring. Note that it can bark "ht" and then "tp" producing "http", but it doesn't contain the password "hp" as a substring.
In the third example the string "hahahaha" contains "ah" as a substring. | ```python
pw = input()
n = int(input())
words = [input() for i in range(n)]
match1 = False
match2 = False
if pw in words:
print("YES")
else:
for word in words:
if pw[0] == word[1]:
match1 = True
if pw[1] == word[0]:
match2 = True
if match1 and match2:
print("YES")
else:
print("NO")
``` | 3 | |
558 | A | Lala Land and Apple Trees | PROGRAMMING | 1,100 | [
"brute force",
"implementation",
"sortings"
] | null | null | Amr lives in Lala Land. Lala Land is a very beautiful country that is located on a coordinate line. Lala Land is famous with its apple trees growing everywhere.
Lala Land has exactly *n* apple trees. Tree number *i* is located in a position *x**i* and has *a**i* apples growing on it. Amr wants to collect apples from the apple trees. Amr currently stands in *x*<==<=0 position. At the beginning, he can choose whether to go right or left. He'll continue in his direction until he meets an apple tree he didn't visit before. He'll take all of its apples and then reverse his direction, continue walking in this direction until he meets another apple tree he didn't visit before and so on. In the other words, Amr reverses his direction when visiting each new apple tree. Amr will stop collecting apples when there are no more trees he didn't visit in the direction he is facing.
What is the maximum number of apples he can collect? | The first line contains one number *n* (1<=≤<=*n*<=≤<=100), the number of apple trees in Lala Land.
The following *n* lines contains two integers each *x**i*, *a**i* (<=-<=105<=≤<=*x**i*<=≤<=105, *x**i*<=≠<=0, 1<=≤<=*a**i*<=≤<=105), representing the position of the *i*-th tree and number of apples on it.
It's guaranteed that there is at most one apple tree at each coordinate. It's guaranteed that no tree grows in point 0. | Output the maximum number of apples Amr can collect. | [
"2\n-1 5\n1 5\n",
"3\n-2 2\n1 4\n-1 3\n",
"3\n1 9\n3 5\n7 10\n"
] | [
"10",
"9",
"9"
] | In the first sample test it doesn't matter if Amr chose at first to go left or right. In both cases he'll get all the apples.
In the second sample test the optimal solution is to go left to *x* = - 1, collect apples from there, then the direction will be reversed, Amr has to go to *x* = 1, collect apples from there, then the direction will be reversed and Amr goes to the final tree *x* = - 2.
In the third sample test the optimal solution is to go right to *x* = 1, collect apples from there, then the direction will be reversed and Amr will not be able to collect anymore apples because there are no apple trees to his left. | 500 | [
{
"input": "2\n-1 5\n1 5",
"output": "10"
},
{
"input": "3\n-2 2\n1 4\n-1 3",
"output": "9"
},
{
"input": "3\n1 9\n3 5\n7 10",
"output": "9"
},
{
"input": "1\n1 1",
"output": "1"
},
{
"input": "4\n10000 100000\n-1000 100000\n-2 100000\n-1 100000",
"output": "300000"
},
{
"input": "1\n-1 1",
"output": "1"
},
{
"input": "27\n-30721 24576\n-6620 92252\n88986 24715\n-94356 10509\n-6543 29234\n-68554 69530\n39176 96911\n67266 99669\n95905 51002\n-94093 92134\n65382 23947\n-6525 79426\n-448 67531\n-70083 26921\n-86333 50029\n48924 8036\n-27228 5349\n6022 10691\n-13840 56735\n50398 58794\n-63258 45557\n-27792 77057\n98295 1203\n-51294 18757\n35037 61941\n-30112 13076\n82334 20463",
"output": "1036452"
},
{
"input": "18\n-18697 44186\n56333 51938\n-75688 49735\n77762 14039\n-43996 81060\n69700 49107\n74532 45568\n-94476 203\n-92347 90745\n58921 44650\n57563 63561\n44630 8486\n35750 5999\n3249 34202\n75358 68110\n-33245 60458\n-88148 2342\n87856 85532",
"output": "632240"
},
{
"input": "28\n49728 91049\n-42863 4175\n-89214 22191\n77977 16965\n-42960 87627\n-84329 97494\n89270 75906\n-13695 28908\n-72279 13607\n-97327 87062\n-58682 32094\n39108 99936\n29304 93784\n-63886 48237\n-77359 57648\n-87013 79017\n-41086 35033\n-60613 83555\n-48955 56816\n-20568 26802\n52113 25160\n-88885 45294\n22601 42971\n62693 65662\n-15985 5357\n86671 8522\n-59921 11271\n-79304 25044",
"output": "891593"
},
{
"input": "25\n5704 67795\n6766 31836\n-41715 89987\n76854 9848\n11648 90020\n-79763 10107\n96971 92636\n-64205 71937\n87997 38273\n-9782 57187\n22186 6905\n-41130 40258\n-28403 66579\n19578 43375\n35735 52929\n-52417 89388\n-89430 1939\n9401 43491\n-11228 10112\n-86859 16024\n-51486 33467\n-80578 65080\n-52820 98445\n-89165 7657\n-97106 79422",
"output": "1109655"
},
{
"input": "16\n-41732 47681\n44295 28942\n-75194 99827\n69982 18020\n-75378 22026\n80032 22908\n-34879 41113\n36257 48574\n-35882 84333\n29646 71151\n-86214 80886\n72724 39364\n-42529 60880\n29150 29921\n-8471 80781\n79387 70834",
"output": "847241"
},
{
"input": "3\n-94146 4473\n28707 99079\n-4153 8857",
"output": "112409"
},
{
"input": "3\n-3 3\n-2 2\n-1 1",
"output": "1"
},
{
"input": "2\n100000 3\n-100000 9",
"output": "12"
},
{
"input": "2\n-100000 100000\n100000 99999",
"output": "199999"
}
] | 1,691,095,514 | 2,147,483,647 | Python 3 | OK | TESTS | 46 | 46 | 0 | n = int(input())
l = []
l2 = []
l3 = []
for i in range(0,n) :
l.append(list(map(int,input().split(" "))))
for k in l :
if k[0] > 0 :
l2.append(k)
else :
l3.append(k)
l2.sort()
s = 0
lcv = min(len(l2),len(l3))
l3.sort(reverse=True)
if len(l2) == len(l3) :
for i in range(0,len(l2)) :
s += (l2[i][1] + l3[i][1] )
elif len(l3) > len(l2) :
while lcv >= 0 :
lcv -=1
s += l3[0][1]
l3.pop(0)
if len(l2) == 0 :
break
else :
s+= l2[0][1]
l2.pop(0)
else :
while lcv >= 0 :
lcv -=1
s += l2[0][1]
l2.pop(0)
if len(l3) == 0 :
break
else :
s+= l3[0][1]
l3.pop(0)
print(s)
| Title: Lala Land and Apple Trees
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Amr lives in Lala Land. Lala Land is a very beautiful country that is located on a coordinate line. Lala Land is famous with its apple trees growing everywhere.
Lala Land has exactly *n* apple trees. Tree number *i* is located in a position *x**i* and has *a**i* apples growing on it. Amr wants to collect apples from the apple trees. Amr currently stands in *x*<==<=0 position. At the beginning, he can choose whether to go right or left. He'll continue in his direction until he meets an apple tree he didn't visit before. He'll take all of its apples and then reverse his direction, continue walking in this direction until he meets another apple tree he didn't visit before and so on. In the other words, Amr reverses his direction when visiting each new apple tree. Amr will stop collecting apples when there are no more trees he didn't visit in the direction he is facing.
What is the maximum number of apples he can collect?
Input Specification:
The first line contains one number *n* (1<=≤<=*n*<=≤<=100), the number of apple trees in Lala Land.
The following *n* lines contains two integers each *x**i*, *a**i* (<=-<=105<=≤<=*x**i*<=≤<=105, *x**i*<=≠<=0, 1<=≤<=*a**i*<=≤<=105), representing the position of the *i*-th tree and number of apples on it.
It's guaranteed that there is at most one apple tree at each coordinate. It's guaranteed that no tree grows in point 0.
Output Specification:
Output the maximum number of apples Amr can collect.
Demo Input:
['2\n-1 5\n1 5\n', '3\n-2 2\n1 4\n-1 3\n', '3\n1 9\n3 5\n7 10\n']
Demo Output:
['10', '9', '9']
Note:
In the first sample test it doesn't matter if Amr chose at first to go left or right. In both cases he'll get all the apples.
In the second sample test the optimal solution is to go left to *x* = - 1, collect apples from there, then the direction will be reversed, Amr has to go to *x* = 1, collect apples from there, then the direction will be reversed and Amr goes to the final tree *x* = - 2.
In the third sample test the optimal solution is to go right to *x* = 1, collect apples from there, then the direction will be reversed and Amr will not be able to collect anymore apples because there are no apple trees to his left. | ```python
n = int(input())
l = []
l2 = []
l3 = []
for i in range(0,n) :
l.append(list(map(int,input().split(" "))))
for k in l :
if k[0] > 0 :
l2.append(k)
else :
l3.append(k)
l2.sort()
s = 0
lcv = min(len(l2),len(l3))
l3.sort(reverse=True)
if len(l2) == len(l3) :
for i in range(0,len(l2)) :
s += (l2[i][1] + l3[i][1] )
elif len(l3) > len(l2) :
while lcv >= 0 :
lcv -=1
s += l3[0][1]
l3.pop(0)
if len(l2) == 0 :
break
else :
s+= l2[0][1]
l2.pop(0)
else :
while lcv >= 0 :
lcv -=1
s += l2[0][1]
l2.pop(0)
if len(l3) == 0 :
break
else :
s+= l3[0][1]
l3.pop(0)
print(s)
``` | 3 | |
909 | B | Segments | PROGRAMMING | 1,300 | [
"constructive algorithms",
"math"
] | null | null | You are given an integer *N*. Consider all possible segments on the coordinate axis with endpoints at integer points with coordinates between 0 and *N*, inclusive; there will be of them.
You want to draw these segments in several layers so that in each layer the segments don't overlap (they might touch at the endpoints though). You can not move the segments to a different location on the coordinate axis.
Find the minimal number of layers you have to use for the given *N*. | The only input line contains a single integer *N* (1<=≤<=*N*<=≤<=100). | Output a single integer - the minimal number of layers required to draw the segments for the given *N*. | [
"2\n",
"3\n",
"4\n"
] | [
"2\n",
"4\n",
"6\n"
] | As an example, here are the segments and their optimal arrangement into layers for *N* = 4. | 1,000 | [
{
"input": "2",
"output": "2"
},
{
"input": "3",
"output": "4"
},
{
"input": "4",
"output": "6"
},
{
"input": "21",
"output": "121"
},
{
"input": "100",
"output": "2550"
},
{
"input": "1",
"output": "1"
},
{
"input": "5",
"output": "9"
},
{
"input": "6",
"output": "12"
},
{
"input": "7",
"output": "16"
},
{
"input": "8",
"output": "20"
},
{
"input": "9",
"output": "25"
},
{
"input": "10",
"output": "30"
},
{
"input": "11",
"output": "36"
},
{
"input": "12",
"output": "42"
},
{
"input": "13",
"output": "49"
},
{
"input": "14",
"output": "56"
},
{
"input": "15",
"output": "64"
},
{
"input": "16",
"output": "72"
},
{
"input": "17",
"output": "81"
},
{
"input": "18",
"output": "90"
},
{
"input": "19",
"output": "100"
},
{
"input": "20",
"output": "110"
},
{
"input": "22",
"output": "132"
},
{
"input": "23",
"output": "144"
},
{
"input": "24",
"output": "156"
},
{
"input": "25",
"output": "169"
},
{
"input": "26",
"output": "182"
},
{
"input": "27",
"output": "196"
},
{
"input": "28",
"output": "210"
},
{
"input": "29",
"output": "225"
},
{
"input": "30",
"output": "240"
},
{
"input": "31",
"output": "256"
},
{
"input": "32",
"output": "272"
},
{
"input": "33",
"output": "289"
},
{
"input": "34",
"output": "306"
},
{
"input": "35",
"output": "324"
},
{
"input": "36",
"output": "342"
},
{
"input": "37",
"output": "361"
},
{
"input": "38",
"output": "380"
},
{
"input": "39",
"output": "400"
},
{
"input": "40",
"output": "420"
},
{
"input": "41",
"output": "441"
},
{
"input": "42",
"output": "462"
},
{
"input": "43",
"output": "484"
},
{
"input": "44",
"output": "506"
},
{
"input": "45",
"output": "529"
},
{
"input": "46",
"output": "552"
},
{
"input": "47",
"output": "576"
},
{
"input": "48",
"output": "600"
},
{
"input": "49",
"output": "625"
},
{
"input": "50",
"output": "650"
},
{
"input": "51",
"output": "676"
},
{
"input": "52",
"output": "702"
},
{
"input": "53",
"output": "729"
},
{
"input": "54",
"output": "756"
},
{
"input": "55",
"output": "784"
},
{
"input": "56",
"output": "812"
},
{
"input": "57",
"output": "841"
},
{
"input": "58",
"output": "870"
},
{
"input": "59",
"output": "900"
},
{
"input": "60",
"output": "930"
},
{
"input": "61",
"output": "961"
},
{
"input": "62",
"output": "992"
},
{
"input": "63",
"output": "1024"
},
{
"input": "64",
"output": "1056"
},
{
"input": "65",
"output": "1089"
},
{
"input": "66",
"output": "1122"
},
{
"input": "67",
"output": "1156"
},
{
"input": "68",
"output": "1190"
},
{
"input": "69",
"output": "1225"
},
{
"input": "70",
"output": "1260"
},
{
"input": "71",
"output": "1296"
},
{
"input": "72",
"output": "1332"
},
{
"input": "73",
"output": "1369"
},
{
"input": "74",
"output": "1406"
},
{
"input": "75",
"output": "1444"
},
{
"input": "76",
"output": "1482"
},
{
"input": "77",
"output": "1521"
},
{
"input": "78",
"output": "1560"
},
{
"input": "79",
"output": "1600"
},
{
"input": "80",
"output": "1640"
},
{
"input": "81",
"output": "1681"
},
{
"input": "82",
"output": "1722"
},
{
"input": "83",
"output": "1764"
},
{
"input": "84",
"output": "1806"
},
{
"input": "85",
"output": "1849"
},
{
"input": "86",
"output": "1892"
},
{
"input": "87",
"output": "1936"
},
{
"input": "88",
"output": "1980"
},
{
"input": "89",
"output": "2025"
},
{
"input": "90",
"output": "2070"
},
{
"input": "91",
"output": "2116"
},
{
"input": "92",
"output": "2162"
},
{
"input": "93",
"output": "2209"
},
{
"input": "94",
"output": "2256"
},
{
"input": "95",
"output": "2304"
},
{
"input": "96",
"output": "2352"
},
{
"input": "97",
"output": "2401"
},
{
"input": "98",
"output": "2450"
},
{
"input": "99",
"output": "2500"
},
{
"input": "1",
"output": "1"
},
{
"input": "5",
"output": "9"
}
] | 1,515,680,938 | 1,738 | Python 3 | OK | TESTS | 102 | 62 | 5,632,000 | n = int(input())
n += 1
n = n*n
n = n//4
print(n) | Title: Segments
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an integer *N*. Consider all possible segments on the coordinate axis with endpoints at integer points with coordinates between 0 and *N*, inclusive; there will be of them.
You want to draw these segments in several layers so that in each layer the segments don't overlap (they might touch at the endpoints though). You can not move the segments to a different location on the coordinate axis.
Find the minimal number of layers you have to use for the given *N*.
Input Specification:
The only input line contains a single integer *N* (1<=≤<=*N*<=≤<=100).
Output Specification:
Output a single integer - the minimal number of layers required to draw the segments for the given *N*.
Demo Input:
['2\n', '3\n', '4\n']
Demo Output:
['2\n', '4\n', '6\n']
Note:
As an example, here are the segments and their optimal arrangement into layers for *N* = 4. | ```python
n = int(input())
n += 1
n = n*n
n = n//4
print(n)
``` | 3 | |
330 | A | Cakeminator | PROGRAMMING | 800 | [
"brute force",
"implementation"
] | null | null | You are given a rectangular cake, represented as an *r*<=×<=*c* grid. Each cell either has an evil strawberry, or is empty. For example, a 3<=×<=4 cake may look as follows:
The cakeminator is going to eat the cake! Each time he eats, he chooses a row or a column that does not contain any evil strawberries and contains at least one cake cell that has not been eaten before, and eats all the cake cells there. He may decide to eat any number of times.
Please output the maximum number of cake cells that the cakeminator can eat. | The first line contains two integers *r* and *c* (2<=≤<=*r*,<=*c*<=≤<=10), denoting the number of rows and the number of columns of the cake. The next *r* lines each contains *c* characters — the *j*-th character of the *i*-th line denotes the content of the cell at row *i* and column *j*, and is either one of these:
- '.' character denotes a cake cell with no evil strawberry; - 'S' character denotes a cake cell with an evil strawberry. | Output the maximum number of cake cells that the cakeminator can eat. | [
"3 4\nS...\n....\n..S.\n"
] | [
"8\n"
] | For the first example, one possible way to eat the maximum number of cake cells is as follows (perform 3 eats). | 500 | [
{
"input": "3 4\nS...\n....\n..S.",
"output": "8"
},
{
"input": "2 2\n..\n..",
"output": "4"
},
{
"input": "2 2\nSS\nSS",
"output": "0"
},
{
"input": "7 3\nS..\nS..\nS..\nS..\nS..\nS..\nS..",
"output": "14"
},
{
"input": "3 5\n..S..\nSSSSS\n..S..",
"output": "0"
},
{
"input": "10 10\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS\nSSSSSSSSSS",
"output": "0"
},
{
"input": "10 10\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS\nS...SSSSSS",
"output": "30"
},
{
"input": "10 10\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..\n....S..S..",
"output": "80"
},
{
"input": "9 5\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS\nSSSSS",
"output": "0"
},
{
"input": "9 9\n...S.....\nS.S.....S\n.S....S..\n.S.....SS\n.........\n..S.S..S.\n.SS......\n....S....\n..S...S..",
"output": "17"
},
{
"input": "5 6\nSSSSSS\nSSSSSS\nSSSSSS\nSS.S..\nS.S.SS",
"output": "0"
},
{
"input": "9 8\n........\n.......S\n........\nS.......\n........\n........\nS.......\n........\n.......S",
"output": "64"
},
{
"input": "9 7\n......S\n......S\nS.S.S..\n.......\n.......\n.S.....\n.S....S\n..S....\n.S....S",
"output": "28"
},
{
"input": "10 10\n.....S....\n....SS..S.\n.S...S....\n........SS\n.S.......S\nSS..S.....\n.SS.....SS\nS..S......\n.......SSS\nSSSSS....S",
"output": "10"
},
{
"input": "6 7\n..S.SS.\n......S\n....S.S\nSS..S..\nS..SS.S\n.....S.",
"output": "0"
},
{
"input": "10 6\n.SSSSS\nSSS.SS\nSSSSSS\nS.SSSS\nSSSSS.\nS.SSSS\nSS.SSS\n.SSS.S\n.SSS..\nSS..SS",
"output": "0"
},
{
"input": "2 2\n..\n..",
"output": "4"
},
{
"input": "3 2\nS.\n.S\nS.",
"output": "0"
},
{
"input": "3 2\nS.\n.S\nS.",
"output": "0"
},
{
"input": "4 3\n.S.\nS.S\n.S.\nS.S",
"output": "0"
},
{
"input": "2 3\n...\nSSS",
"output": "3"
},
{
"input": "2 4\nS.SS\nS.SS",
"output": "2"
},
{
"input": "2 2\n..\n.S",
"output": "3"
},
{
"input": "3 2\n.S\n.S\nSS",
"output": "0"
},
{
"input": "2 4\nSS.S\n..S.",
"output": "0"
},
{
"input": "2 3\n...\nS..",
"output": "5"
}
] | 1,672,846,272 | 2,147,483,647 | PyPy 3 | OK | TESTS | 42 | 186 | 0 | a , b = [int(x) for x in input().split()]
ls =[]
for i in range(0,a):
ls.append(input())
lt = [""] * b
p = 0
for i in range(0,len(ls)):
if "S" not in ls[i]:
p+= len(ls[i])
ls [i] = b*"a"
for i in range(0,b):
for j in range(0,a):
lt[i] += str(ls[j])[i]
for x in lt:
if "S" not in x: p+= x.count(".")
print(p)
| Title: Cakeminator
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a rectangular cake, represented as an *r*<=×<=*c* grid. Each cell either has an evil strawberry, or is empty. For example, a 3<=×<=4 cake may look as follows:
The cakeminator is going to eat the cake! Each time he eats, he chooses a row or a column that does not contain any evil strawberries and contains at least one cake cell that has not been eaten before, and eats all the cake cells there. He may decide to eat any number of times.
Please output the maximum number of cake cells that the cakeminator can eat.
Input Specification:
The first line contains two integers *r* and *c* (2<=≤<=*r*,<=*c*<=≤<=10), denoting the number of rows and the number of columns of the cake. The next *r* lines each contains *c* characters — the *j*-th character of the *i*-th line denotes the content of the cell at row *i* and column *j*, and is either one of these:
- '.' character denotes a cake cell with no evil strawberry; - 'S' character denotes a cake cell with an evil strawberry.
Output Specification:
Output the maximum number of cake cells that the cakeminator can eat.
Demo Input:
['3 4\nS...\n....\n..S.\n']
Demo Output:
['8\n']
Note:
For the first example, one possible way to eat the maximum number of cake cells is as follows (perform 3 eats). | ```python
a , b = [int(x) for x in input().split()]
ls =[]
for i in range(0,a):
ls.append(input())
lt = [""] * b
p = 0
for i in range(0,len(ls)):
if "S" not in ls[i]:
p+= len(ls[i])
ls [i] = b*"a"
for i in range(0,b):
for j in range(0,a):
lt[i] += str(ls[j])[i]
for x in lt:
if "S" not in x: p+= x.count(".")
print(p)
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Bear Limak has *n* colored balls, arranged in one long row. Balls are numbered 1 through *n*, from left to right. There are *n* possible colors, also numbered 1 through *n*. The *i*-th ball has color *t**i*.
For a fixed interval (set of consecutive elements) of balls we can define a dominant color. It's a color occurring the biggest number of times in the interval. In case of a tie between some colors, the one with the smallest number (index) is chosen as dominant.
There are non-empty intervals in total. For each color, your task is to count the number of intervals in which this color is dominant. | The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=5000) — the number of balls.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=*n*) where *t**i* is the color of the *i*-th ball. | Print *n* integers. The *i*-th of them should be equal to the number of intervals where *i* is a dominant color. | [
"4\n1 2 1 2\n",
"3\n1 1 1\n"
] | [
"7 3 0 0 \n",
"6 0 0 \n"
] | In the first sample, color 2 is dominant in three intervals:
- An interval [2, 2] contains one ball. This ball's color is 2 so it's clearly a dominant color. - An interval [4, 4] contains one ball, with color 2 again. - An interval [2, 4] contains two balls of color 2 and one ball of color 1.
There are 7 more intervals and color 1 is dominant in all of them. | 0 | [
{
"input": "4\n1 2 1 2",
"output": "7 3 0 0 "
},
{
"input": "3\n1 1 1",
"output": "6 0 0 "
},
{
"input": "10\n9 1 5 2 9 2 9 2 1 1",
"output": "18 30 0 0 1 0 0 0 6 0 "
},
{
"input": "50\n17 13 19 19 19 34 32 24 24 13 34 17 19 19 7 32 19 13 13 30 19 34 34 28 41 24 24 47 22 34 21 21 30 7 22 21 32 19 34 19 34 22 7 28 6 13 19 30 13 30",
"output": "0 0 0 0 0 22 40 0 0 0 0 0 98 0 0 0 5 0 675 0 165 9 0 61 0 0 0 5 0 6 0 4 0 183 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 "
},
{
"input": "150\n28 124 138 71 71 18 78 136 138 93 145 93 18 15 71 47 47 64 18 72 138 72 18 150 7 71 109 149 18 115 149 149 15 78 124 27 72 124 28 108 138 109 108 111 148 138 78 27 28 150 138 65 15 145 109 47 102 62 28 7 115 108 102 149 150 27 111 64 149 124 13 21 108 64 7 15 72 72 124 47 102 28 109 18 124 28 111 138 7 13 21 62 136 62 13 64 71 7 130 47 77 65 71 148 15 93 64 65 28 65 13 78 78 47 115 138 28 115 72 136 124 145 150 62 105 78 71 102 109 150 27 130 62 7 93 72 93 62 7 124 72 21 62 18 62 7 108 78 148 149",
"output": "0 0 0 0 0 0 1863 0 0 0 0 0 604 0 97 0 0 1026 0 0 12 0 0 0 0 0 208 2982 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 67 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1400 0 158 371 0 0 0 0 0 92 296 0 0 0 0 1 199 0 0 0 0 0 0 0 0 0 0 0 0 0 0 48 0 0 0 0 0 0 0 0 151 0 0 1 0 0 230 37 0 6 0 0 0 44 0 0 0 0 0 0 0 0 174 0 0 0 0 0 2 0 0 0 0 0 4 0 729 0 0 0 0 0 0 4 0 0 4 486 29 "
},
{
"input": "1\n1",
"output": "1 "
},
{
"input": "2\n2 1",
"output": "2 1 "
}
] | 1,689,441,131 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 62 | 0 | print("_RANDOM_GUESS_1689441131.7455006")# 1689441131.7455204 | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bear Limak has *n* colored balls, arranged in one long row. Balls are numbered 1 through *n*, from left to right. There are *n* possible colors, also numbered 1 through *n*. The *i*-th ball has color *t**i*.
For a fixed interval (set of consecutive elements) of balls we can define a dominant color. It's a color occurring the biggest number of times in the interval. In case of a tie between some colors, the one with the smallest number (index) is chosen as dominant.
There are non-empty intervals in total. For each color, your task is to count the number of intervals in which this color is dominant.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=5000) — the number of balls.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=*n*) where *t**i* is the color of the *i*-th ball.
Output Specification:
Print *n* integers. The *i*-th of them should be equal to the number of intervals where *i* is a dominant color.
Demo Input:
['4\n1 2 1 2\n', '3\n1 1 1\n']
Demo Output:
['7 3 0 0 \n', '6 0 0 \n']
Note:
In the first sample, color 2 is dominant in three intervals:
- An interval [2, 2] contains one ball. This ball's color is 2 so it's clearly a dominant color. - An interval [4, 4] contains one ball, with color 2 again. - An interval [2, 4] contains two balls of color 2 and one ball of color 1.
There are 7 more intervals and color 1 is dominant in all of them. | ```python
print("_RANDOM_GUESS_1689441131.7455006")# 1689441131.7455204
``` | 0 | |
373 | B | Making Sequences is Fun | PROGRAMMING | 1,600 | [
"binary search",
"implementation",
"math"
] | null | null | We'll define *S*(*n*) for positive integer *n* as follows: the number of the *n*'s digits in the decimal base. For example, *S*(893)<==<=3, *S*(114514)<==<=6.
You want to make a consecutive integer sequence starting from number *m* (*m*,<=*m*<=+<=1,<=...). But you need to pay *S*(*n*)·*k* to add the number *n* to the sequence.
You can spend a cost up to *w*, and you want to make the sequence as long as possible. Write a program that tells sequence's maximum length. | The first line contains three integers *w* (1<=≤<=*w*<=≤<=1016), *m* (1<=≤<=*m*<=≤<=1016), *k* (1<=≤<=*k*<=≤<=109).
Please, do not write 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. | The first line should contain a single integer — the answer to the problem. | [
"9 1 1\n",
"77 7 7\n",
"114 5 14\n",
"1 1 2\n"
] | [
"9\n",
"7\n",
"6\n",
"0\n"
] | none | 1,000 | [
{
"input": "9 1 1",
"output": "9"
},
{
"input": "77 7 7",
"output": "7"
},
{
"input": "114 5 14",
"output": "6"
},
{
"input": "1 1 2",
"output": "0"
},
{
"input": "462 183 8",
"output": "19"
},
{
"input": "462 183 8",
"output": "19"
},
{
"input": "462 183 8",
"output": "19"
},
{
"input": "462 183 8",
"output": "19"
},
{
"input": "462 183 8",
"output": "19"
},
{
"input": "462 183 8",
"output": "19"
},
{
"input": "462 183 8",
"output": "19"
},
{
"input": "10000000000000000 1 1",
"output": "674074074074073"
},
{
"input": "2155990066796462 2710473050636183 563261158",
"output": "239230"
},
{
"input": "552719169048748 7822730034794389 374302919",
"output": "92291"
},
{
"input": "6895788044242644 3474119895793364 732614708",
"output": "588285"
},
{
"input": "9230573804495452 8577408935470783 90893866",
"output": "6347082"
},
{
"input": "5363062262667637 2932858345469643 449197576",
"output": "746200"
},
{
"input": "1951414609070432 8979282171178729 260239468",
"output": "468658"
},
{
"input": "6289344232651169 4170875805886064 618522710",
"output": "635520"
},
{
"input": "111984789523940 8017182602114316 976839488",
"output": "7164"
},
{
"input": "3663662721733869 4845943245608254 787880219",
"output": "290626"
},
{
"input": "10000000000000000 10000000000000000 1",
"output": "588235294117647"
},
{
"input": "10000000000000000 9999999999999999 1",
"output": "588235294117647"
},
{
"input": "9999999999999999 1 1",
"output": "674074074074073"
},
{
"input": "3 1 4",
"output": "0"
},
{
"input": "9999999999999998 1 1",
"output": "674074074074072"
},
{
"input": "1 1000000000000000 1",
"output": "0"
},
{
"input": "1 1000000000000000 1000000000",
"output": "0"
},
{
"input": "10000000000000000 10000000000000000 1000000000",
"output": "588235"
},
{
"input": "10000000000000000 10000000000000000 1",
"output": "588235294117647"
},
{
"input": "121212121 3434343434 56",
"output": "216450"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "98 55 1",
"output": "47"
},
{
"input": "1 10000000000000000 123456",
"output": "0"
},
{
"input": "10000000000000000 1000000000000001 1000000000",
"output": "625000"
}
] | 1,609,256,093 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 108 | 307,200 | def cnt(x, y):
return y-x
w, m, k = map(int, input().split())
p, d, res = 1, 0, 0
while p <= m:
p *= 10
d += 1
while cnt(m, p)*d*k <= w:
w -= cnt(m, p)*d*k
res += cnt(m, p)
m = p
p *= 10
d += 1
lo, hi = m, p
while hi-lo > 1:
mid = (lo+hi)//2
if (cnt(m, mid)*d*k <= w): lo = mid
else: hi = mid
res += cnt(m, lo)
print(res)
| Title: Making Sequences is Fun
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We'll define *S*(*n*) for positive integer *n* as follows: the number of the *n*'s digits in the decimal base. For example, *S*(893)<==<=3, *S*(114514)<==<=6.
You want to make a consecutive integer sequence starting from number *m* (*m*,<=*m*<=+<=1,<=...). But you need to pay *S*(*n*)·*k* to add the number *n* to the sequence.
You can spend a cost up to *w*, and you want to make the sequence as long as possible. Write a program that tells sequence's maximum length.
Input Specification:
The first line contains three integers *w* (1<=≤<=*w*<=≤<=1016), *m* (1<=≤<=*m*<=≤<=1016), *k* (1<=≤<=*k*<=≤<=109).
Please, do not write the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output Specification:
The first line should contain a single integer — the answer to the problem.
Demo Input:
['9 1 1\n', '77 7 7\n', '114 5 14\n', '1 1 2\n']
Demo Output:
['9\n', '7\n', '6\n', '0\n']
Note:
none | ```python
def cnt(x, y):
return y-x
w, m, k = map(int, input().split())
p, d, res = 1, 0, 0
while p <= m:
p *= 10
d += 1
while cnt(m, p)*d*k <= w:
w -= cnt(m, p)*d*k
res += cnt(m, p)
m = p
p *= 10
d += 1
lo, hi = m, p
while hi-lo > 1:
mid = (lo+hi)//2
if (cnt(m, mid)*d*k <= w): lo = mid
else: hi = mid
res += cnt(m, lo)
print(res)
``` | 3 | |
112 | A | Petya and Strings | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Petya and Strings | 2 | 256 | Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison. | Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters. | If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared. | [
"aaaa\naaaA\n",
"abs\nAbz\n",
"abcdefg\nAbCdEfF\n"
] | [
"0\n",
"-1\n",
"1\n"
] | If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
- http://en.wikipedia.org/wiki/Lexicographical_order | 500 | [
{
"input": "aaaa\naaaA",
"output": "0"
},
{
"input": "abs\nAbz",
"output": "-1"
},
{
"input": "abcdefg\nAbCdEfF",
"output": "1"
},
{
"input": "asadasdasd\nasdwasdawd",
"output": "-1"
},
{
"input": "aslkjlkasdd\nasdlkjdajwi",
"output": "1"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "0"
},
{
"input": "aAaaaAAaAaaAzZsssSsdDfeEaeqZlpP\nAaaaAaaAaaAaZzSSSSsDdFeeAeQZLpp",
"output": "0"
},
{
"input": "bwuEhEveouaTECagLZiqmUdxEmhRSOzMauJRWLQMppZOumxhAmwuGeDIkvkBLvMXwUoFmpAfDprBcFtEwOULcZWRQhcTbTbX\nHhoDWbcxwiMnCNexOsKsujLiSGcLllXOkRSbnOzThAjnnliLYFFmsYkOfpTxRNEfBsoUHfoLTiqAINRPxWRqrTJhgfkKcDOH",
"output": "-1"
},
{
"input": "kGWUuguKzcvxqKTNpxeDWXpXkrXDvGMFGoXKDfPBZvWSDUyIYBynbKOUonHvmZaKeirUhfmVRKtGhAdBfKMWXDUoqvbfpfHYcg\ncvOULleuIIiYVVxcLZmHVpNGXuEpzcWZZWyMOwIwbpkKPwCfkVbKkUuosvxYCKjqfVmHfJKbdrsAcatPYgrCABaFcoBuOmMfFt",
"output": "1"
},
{
"input": "nCeNVIzHqPceNhjHeHvJvgBsNFiXBATRrjSTXJzhLMDMxiJztphxBRlDlqwDFImWeEPkggZCXSRwelOdpNrYnTepiOqpvkr\nHJbjJFtlvNxIbkKlxQUwmZHJFVNMwPAPDRslIoXISBYHHfymyIaQHLgECPxAmqnOCizwXnIUBRmpYUBVPenoUKhCobKdOjL",
"output": "1"
},
{
"input": "ttXjenUAlfixytHEOrPkgXmkKTSGYuyVXGIHYmWWYGlBYpHkujueqBSgjLguSgiMGJWATIGEUjjAjKXdMiVbHozZUmqQtFrT\nJziDBFBDmDJCcGqFsQwDFBYdOidLxxhBCtScznnDgnsiStlWFnEXQrJxqTXKPxZyIGfLIToETKWZBPUIBmLeImrlSBWCkTNo",
"output": "1"
},
{
"input": "AjQhPqSVhwQQjcgCycjKorWBgFCRuQBwgdVuAPSMJAvTyxGVuFHjfJzkKfsmfhFbKqFrFIohSZBbpjgEHebezmVlGLTPSCTMf\nXhxWuSnMmKFrCUOwkTUmvKAfbTbHWzzOTzxJatLLCdlGnHVaBUnxDlsqpvjLHMThOPAFBggVKDyKBrZAmjnjrhHlrnSkyzBja",
"output": "-1"
},
{
"input": "HCIgYtnqcMyjVngziNflxKHtdTmcRJhzMAjFAsNdWXFJYEhiTzsQUtFNkAbdrFBRmvLirkuirqTDvIpEfyiIqkrwsjvpPWTEdI\nErqiiWKsmIjyZuzgTlTqxYZwlrpvRyaVhRTOYUqtPMVGGtWOkDCOOQRKrkkRzPftyQCkYkzKkzTPqqXmeZhvvEEiEhkdOmoMvy",
"output": "1"
},
{
"input": "mtBeJYILXcECGyEVSyzLFdQJbiVnnfkbsYYsdUJSIRmyzLfTTtFwIBmRLVnwcewIqcuydkcLpflHAFyDaToLiFMgeHvQorTVbI\nClLvyejznjbRfCDcrCzkLvqQaGzTjwmWONBdCctJAPJBcQrcYvHaSLQgPIJbmkFBhFzuQLBiRzAdNHulCjIAkBvZxxlkdzUWLR",
"output": "1"
},
{
"input": "tjucSbGESVmVridTBjTmpVBCwwdWKBPeBvmgdxgIVLwQxveETnSdxkTVJpXoperWSgdpPMKNmwDiGeHfxnuqaDissgXPlMuNZIr\nHfjOOJhomqNIKHvqSgfySjlsWJQBuWYwhLQhlZYlpZwboMpoLoluGsBmhhlYgeIouwdkPfiaAIrkYRlxtiFazOPOllPsNZHcIZd",
"output": "1"
},
{
"input": "AanbDfbZNlUodtBQlvPMyomStKNhgvSGhSbTdabxGFGGXCdpsJDimsAykKjfBDPMulkhBMsqLmVKLDoesHZsRAEEdEzqigueXInY\ncwfyjoppiJNrjrOLNZkqcGimrpTsiyFBVgMWEPXsMrxLJDDbtYzerXiFGuLBcQYitLdqhGHBpdjRnkUegmnwhGHAKXGyFtscWDSI",
"output": "-1"
},
{
"input": "HRfxniwuJCaHOcaOVgjOGHXKrwxrDQxJpppeGDXnTAowyKbCsCQPbchCKeTWOcKbySSYnoaTJDnmRcyGPbfXJyZoPcARHBu\nxkLXvwkvGIWSQaFTznLOctUXNuzzBBOlqvzmVfTSejekTAlwidRrsxkbZTsGGeEWxCXHzqWVuLGoCyrGjKkQoHqduXwYQKC",
"output": "-1"
},
{
"input": "OjYwwNuPESIazoyLFREpObIaMKhCaKAMWMfRGgucEuyNYRantwdwQkmflzfqbcFRaXBnZoIUGsFqXZHGKwlaBUXABBcQEWWPvkjW\nRxLqGcTTpBwHrHltCOllnTpRKLDofBUqqHxnOtVWPgvGaeHIevgUSOeeDOJubfqonFpVNGVbHFcAhjnyFvrrqnRgKhkYqQZmRfUl",
"output": "-1"
},
{
"input": "tatuhQPIzjptlzzJpCAPXSRTKZRlwgfoCIsFjJquRoIDyZZYRSPdFUTjjUPhLBBfeEIfLQpygKXRcyQFiQsEtRtLnZErBqW\ntkHUjllbafLUWhVCnvblKjgYIEoHhsjVmrDBmAWbvtkHxDbRFvsXAjHIrujaDbYwOZmacknhZPeCcorbRgHjjgAgoJdjvLo",
"output": "-1"
},
{
"input": "cymCPGqdXKUdADEWDdUaLEEMHiXHsdAZuDnJDMUvxvrLRBrPSDpXPAgMRoGplLtniFRTomDTAHXWAdgUveTxaqKVSvnOyhOwiRN\nuhmyEWzapiRNPFDisvHTbenXMfeZaHqOFlKjrfQjUBwdFktNpeiRoDWuBftZLcCZZAVfioOihZVNqiNCNDIsUdIhvbcaxpTRWoV",
"output": "-1"
},
{
"input": "sSvpcITJAwghVfJaLKBmyjOkhltTGjYJVLWCYMFUomiJaKQYhXTajvZVHIMHbyckYROGQZzjWyWCcnmDmrkvTKfHSSzCIhsXgEZa\nvhCXkCwAmErGVBPBAnkSYEYvseFKbWSktoqaHYXUmYkHfOkRwuEyBRoGoBrOXBKVxXycjZGStuvDarnXMbZLWrbjrisDoJBdSvWJ",
"output": "-1"
},
{
"input": "hJDANKUNBisOOINDsTixJmYgHNogtpwswwcvVMptfGwIjvqgwTYFcqTdyAqaqlnhOCMtsnWXQqtjFwQlEcBtMFAtSqnqthVb\nrNquIcjNWESjpPVWmzUJFrelpUZeGDmSvCurCqVmKHKVAAPkaHksniOlzjiKYIJtvbuQWZRufMebpTFPqyxIWWjfPaWYiNlK",
"output": "-1"
},
{
"input": "ycLoapxsfsDTHMSfAAPIUpiEhQKUIXUcXEiopMBuuZLHtfPpLmCHwNMNQUwsEXxCEmKHTBSnKhtQhGWUvppUFZUgSpbeChX\ndCZhgVXofkGousCzObxZSJwXcHIaqUDSCPKzXntcVmPxtNcXmVcjsetZYxedmgQzXTZHMvzjoaXCMKsncGciSDqQWIIRlys",
"output": "1"
},
{
"input": "nvUbnrywIePXcoukIhwTfUVcHUEgXcsMyNQhmMlTltZiCooyZiIKRIGVHMCnTKgzXXIuvoNDEZswKoACOBGSyVNqTNQqMhAG\nplxuGSsyyJjdvpddrSebOARSAYcZKEaKjqbCwvjhNykuaECoQVHTVFMKXwvrQXRaqXsHsBaGVhCxGRxNyGUbMlxOarMZNXxy",
"output": "-1"
},
{
"input": "EncmXtAblQzcVRzMQqdDqXfAhXbtJKQwZVWyHoWUckohnZqfoCmNJDzexFgFJYrwNHGgzCJTzQQFnxGlhmvQTpicTkEeVICKac\nNIUNZoMLFMyAjVgQLITELJSodIXcGSDWfhFypRoGYuogJpnqGTotWxVqpvBHjFOWcDRDtARsaHarHaOkeNWEHGTaGOFCOFEwvK",
"output": "-1"
},
{
"input": "UG\nak",
"output": "1"
},
{
"input": "JZR\nVae",
"output": "-1"
},
{
"input": "a\nZ",
"output": "-1"
},
{
"input": "rk\nkv",
"output": "1"
},
{
"input": "RvuT\nbJzE",
"output": "1"
},
{
"input": "PPS\nydq",
"output": "-1"
},
{
"input": "q\nq",
"output": "0"
},
{
"input": "peOw\nIgSJ",
"output": "1"
},
{
"input": "PyK\noKN",
"output": "1"
},
{
"input": "O\ni",
"output": "1"
},
{
"input": "NmGY\npDlP",
"output": "-1"
},
{
"input": "nG\nZf",
"output": "-1"
},
{
"input": "m\na",
"output": "1"
},
{
"input": "MWyB\nWZEV",
"output": "-1"
},
{
"input": "Gre\nfxc",
"output": "1"
},
{
"input": "Ooq\nwap",
"output": "-1"
},
{
"input": "XId\nlbB",
"output": "1"
},
{
"input": "lfFpECEqUMEOJhipvkZjDPcpDNJedOVXiSMgBvBZbtfzIKekcvpWPCazKAhJyHircRtgcBIJwwstpHaLAgxFOngAWUZRgCef\nLfFPEcequmeojHIpVkzjDPcpdNJEDOVXiSmGBVBZBtfZikEKcvPwpCAzKAHJyHIrCRTgCbIJWwSTphALagXfOnGAwUzRGcEF",
"output": "0"
},
{
"input": "DQBdtSEDtFGiNRUeJNbOIfDZnsryUlzJHGTXGFXnwsVyxNtLgmklmFvRCzYETBVdmkpJJIvIOkMDgCFHZOTODiYrkwXd\nDQbDtsEdTFginRUEJNBOIfdZnsryulZJHGtxGFxnwSvYxnTLgmKlmFVRCzyEtBVdmKpJjiVioKMDgCFhzoTODiYrKwXD",
"output": "0"
},
{
"input": "tYWRijFQSzHBpCjUzqBtNvBKyzZRnIdWEuyqnORBQTLyOQglIGfYJIRjuxnbLvkqZakNqPiGDvgpWYkfxYNXsdoKXZtRkSasfa\nTYwRiJfqsZHBPcJuZQBTnVbkyZZRnidwEuYQnorbQTLYOqGligFyjirJUxnblVKqZaknQpigDVGPwyKfxyNXSDoKxztRKSaSFA",
"output": "0"
},
{
"input": "KhScXYiErQIUtmVhNTCXSLAviefIeHIIdiGhsYnPkSBaDTvMkyanfMLBOvDWgRybLtDqvXVdVjccNunDyijhhZEAKBrdz\nkHsCXyiErqIuTMVHNTCxSLaViEFIEhIIDiGHsYNpKsBAdTvMKyANFMLBovdwGRYbLtdQVxvDVJCcNUndYiJHhzeakBrdZ",
"output": "0"
},
{
"input": "cpPQMpjRQJKQVXjWDYECXbagSmNcVfOuBWNZxihdERraVuiOpSVDCPgTGuSQALNoVjySceHcKXwOEpSzXrEqWwwrYeppNiWhDVg\nCPPqmPjRqJkQvxJwdyECXBAGsMNcVfOuBWNzxIhderRavUiOpSvDCpGTgusqAlNovjyScEhCKXwoePSZxrEQwWwryEPPniWHDvG",
"output": "0"
},
{
"input": "SajcCGMepaLjZIWLRBGFcrZRCRvvoCsIyKsQerbrwsIamxxpRmQSZSalasJLVFbCHCuXJlubciQAvLxXYBazLsMKLHLdDQ\nsaJcCgmEpaLJziWlrBgFcRzrCrVVOcSIykSQerBrwSIamxxPrMqSzSalASjLVFbChCUxjLUbCIQAVlxxybAZLsmkLhLDdQ",
"output": "0"
},
{
"input": "kigPrWNTOUNDBskAfefjhHYZNYdnfZWuXWzHiBxFQryBbAkPtenFwWvCSTYGpzOntUNzNUhxRWjKmicTwLwJAnbAxj\nkigpRWntOUNdBsKaFEFjhhYZnYDNfzWuXwZhibxFQRybbakPteNfwwvcStyGPzoNTunznuHXrWjKMIctWLWJANBAxJ",
"output": "0"
},
{
"input": "nTomZZuTTRTAAPoUsySVFGElrpQRNLjqvFmcYytiheQnjUhPLnqNBiYtQkljbcvmjuNAVKbvQOWpqqFlQhAhULIhquoCnjUI\nntOmzZuttrtAAPOUSySVFgeLRPQrNLjQvfmCyYTiHEQnjuHPlNQNbIYtqKLJBCVmjunavkbvQOWPQQFlqHaHULIHQuOcnJUi",
"output": "0"
},
{
"input": "abac\nadaa",
"output": "-1"
},
{
"input": "Bbc\nabc",
"output": "1"
},
{
"input": "aaaba\naaaab",
"output": "1"
}
] | 1,699,384,280 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 60 | 0 | a=str(input())
b=str(input())
if len(a)==len(b) :
print(0)
else:
if len(a)>len(b) :
print(1)
else:
print(-1) | Title: Petya and Strings
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Petya loves presents. His mum bought him two strings of the same size for his birthday. The strings consist of uppercase and lowercase Latin letters. Now Petya wants to compare those two strings lexicographically. The letters' case does not matter, that is an uppercase letter is considered equivalent to the corresponding lowercase letter. Help Petya perform the comparison.
Input Specification:
Each of the first two lines contains a bought string. The strings' lengths range from 1 to 100 inclusive. It is guaranteed that the strings are of the same length and also consist of uppercase and lowercase Latin letters.
Output Specification:
If the first string is less than the second one, print "-1". If the second string is less than the first one, print "1". If the strings are equal, print "0". Note that the letters' case is not taken into consideration when the strings are compared.
Demo Input:
['aaaa\naaaA\n', 'abs\nAbz\n', 'abcdefg\nAbCdEfF\n']
Demo Output:
['0\n', '-1\n', '1\n']
Note:
If you want more formal information about the lexicographical order (also known as the "dictionary order" or "alphabetical order"), you can visit the following site:
- http://en.wikipedia.org/wiki/Lexicographical_order | ```python
a=str(input())
b=str(input())
if len(a)==len(b) :
print(0)
else:
if len(a)>len(b) :
print(1)
else:
print(-1)
``` | 0 |
689 | D | Friends and Subsequences | PROGRAMMING | 2,100 | [
"binary search",
"data structures"
] | null | null | Mike and !Mike are old childhood rivals, they are opposite in everything they do, except programming. Today they have a problem they cannot solve on their own, but together (with you) — who knows?
Every one of them has an integer sequences *a* and *b* of length *n*. Being given a query of the form of pair of integers (*l*,<=*r*), Mike can instantly tell the value of while !Mike can instantly tell the value of .
Now suppose a robot (you!) asks them all possible different queries of pairs of integers (*l*,<=*r*) (1<=≤<=*l*<=≤<=*r*<=≤<=*n*) (so he will make exactly *n*(*n*<=+<=1)<=/<=2 queries) and counts how many times their answers coincide, thus for how many pairs is satisfied.
How many occasions will the robot count? | The first line contains only integer *n* (1<=≤<=*n*<=≤<=200<=000).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109) — the sequence *a*.
The third line contains *n* integer numbers *b*1,<=*b*2,<=...,<=*b**n* (<=-<=109<=≤<=*b**i*<=≤<=109) — the sequence *b*. | Print the only integer number — the number of occasions the robot will count, thus for how many pairs is satisfied. | [
"6\n1 2 3 2 1 4\n6 7 1 2 3 2\n",
"3\n3 3 3\n1 1 1\n"
] | [
"2\n",
"0\n"
] | The occasions in the first sample case are:
1.*l* = 4,*r* = 4 since *max*{2} = *min*{2}.
2.*l* = 4,*r* = 5 since *max*{2, 1} = *min*{2, 3}.
There are no occasions in the second sample case since Mike will answer 3 to any query pair, but !Mike will always answer 1. | 2,000 | [
{
"input": "6\n1 2 3 2 1 4\n6 7 1 2 3 2",
"output": "2"
},
{
"input": "3\n3 3 3\n1 1 1",
"output": "0"
},
{
"input": "17\n714413739 -959271262 714413739 -745891378 926207665 -404845105 -404845105 -959271262 -189641729 -670860364 714413739 -189641729 192457837 -745891378 -670860364 536388097 -959271262\n-417715348 -959271262 -959271262 714413739 -189641729 571055593 571055593 571055593 -417715348 -417715348 192457837 -745891378 536388097 571055593 -189641729 571055593 -670860364",
"output": "1"
},
{
"input": "1\n509658558\n509658558",
"output": "1"
},
{
"input": "1\n509658558\n-544591380",
"output": "0"
},
{
"input": "3\n1 1 1\n2 2 2",
"output": "0"
}
] | 1,654,092,441 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 94 | 982 | 456,908,800 |
dx=[[0]*(1<<20) for _ in range(20)]
dy=[[0]*(1<<20) for _ in range(20)]
lg=[0]*(1<<20)
def mx(l,r):
p=lg[r-l+1]
return max(dx[p][l],dx[p][r-(1<<p)+1])
def mn(l,r):
p=lg[r-l+1]
return min(dy[p][l],dy[p][r-(1<<p)+1])
if __name__ == '__main__':
n=int(input())
dx[0][1:n]=[int(x) for x in input().split()]
dy[0][1:n]=[int(x) for x in input().split()]
for p in range(1,20):
if (1<<p)>n:
break
for i in range(1,n+2-(1<<p)):
dx[p][i]=max(dx[p-1][i],dx[p-1][i+(1<<(p-1))])
dy[p][i]=min(dy[p-1][i],dy[p-1][i+(1<<(p-1))])
for i in range(2,n+1):
lg[i]=lg[i>>1]+1
ans=0
for i in range(1,n+1):
if mx(i,i)<=mn(i,i):
l=i-1
r=i
p=1<<lg[n-i+1]
while p:
if p+l<=n and mx(i,l+p)<mn(i,l+p):l+=p
if p+r<=n and mx(i,r+p)<=mn(i,r+p):r+=p
p>>=1
ans+=r-l
print(ans)
| Title: Friends and Subsequences
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mike and !Mike are old childhood rivals, they are opposite in everything they do, except programming. Today they have a problem they cannot solve on their own, but together (with you) — who knows?
Every one of them has an integer sequences *a* and *b* of length *n*. Being given a query of the form of pair of integers (*l*,<=*r*), Mike can instantly tell the value of while !Mike can instantly tell the value of .
Now suppose a robot (you!) asks them all possible different queries of pairs of integers (*l*,<=*r*) (1<=≤<=*l*<=≤<=*r*<=≤<=*n*) (so he will make exactly *n*(*n*<=+<=1)<=/<=2 queries) and counts how many times their answers coincide, thus for how many pairs is satisfied.
How many occasions will the robot count?
Input Specification:
The first line contains only integer *n* (1<=≤<=*n*<=≤<=200<=000).
The second line contains *n* integer numbers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109) — the sequence *a*.
The third line contains *n* integer numbers *b*1,<=*b*2,<=...,<=*b**n* (<=-<=109<=≤<=*b**i*<=≤<=109) — the sequence *b*.
Output Specification:
Print the only integer number — the number of occasions the robot will count, thus for how many pairs is satisfied.
Demo Input:
['6\n1 2 3 2 1 4\n6 7 1 2 3 2\n', '3\n3 3 3\n1 1 1\n']
Demo Output:
['2\n', '0\n']
Note:
The occasions in the first sample case are:
1.*l* = 4,*r* = 4 since *max*{2} = *min*{2}.
2.*l* = 4,*r* = 5 since *max*{2, 1} = *min*{2, 3}.
There are no occasions in the second sample case since Mike will answer 3 to any query pair, but !Mike will always answer 1. | ```python
dx=[[0]*(1<<20) for _ in range(20)]
dy=[[0]*(1<<20) for _ in range(20)]
lg=[0]*(1<<20)
def mx(l,r):
p=lg[r-l+1]
return max(dx[p][l],dx[p][r-(1<<p)+1])
def mn(l,r):
p=lg[r-l+1]
return min(dy[p][l],dy[p][r-(1<<p)+1])
if __name__ == '__main__':
n=int(input())
dx[0][1:n]=[int(x) for x in input().split()]
dy[0][1:n]=[int(x) for x in input().split()]
for p in range(1,20):
if (1<<p)>n:
break
for i in range(1,n+2-(1<<p)):
dx[p][i]=max(dx[p-1][i],dx[p-1][i+(1<<(p-1))])
dy[p][i]=min(dy[p-1][i],dy[p-1][i+(1<<(p-1))])
for i in range(2,n+1):
lg[i]=lg[i>>1]+1
ans=0
for i in range(1,n+1):
if mx(i,i)<=mn(i,i):
l=i-1
r=i
p=1<<lg[n-i+1]
while p:
if p+l<=n and mx(i,l+p)<mn(i,l+p):l+=p
if p+r<=n and mx(i,r+p)<=mn(i,r+p):r+=p
p>>=1
ans+=r-l
print(ans)
``` | 3 | |
938 | A | Word Correction | PROGRAMMING | 800 | [
"implementation"
] | null | null | Victor tries to write his own text editor, with word correction included. However, the rules of word correction are really strange.
Victor thinks that if a word contains two consecutive vowels, then it's kinda weird and it needs to be replaced. So the word corrector works in such a way: as long as there are two consecutive vowels in the word, it deletes the first vowel in a word such that there is another vowel right before it. If there are no two consecutive vowels in the word, it is considered to be correct.
You are given a word *s*. Can you predict what will it become after correction?
In this problem letters a, e, i, o, u and y are considered to be vowels. | The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of letters in word *s* before the correction.
The second line contains a string *s* consisting of exactly *n* lowercase Latin letters — the word before the correction. | Output the word *s* after the correction. | [
"5\nweird\n",
"4\nword\n",
"5\naaeaa\n"
] | [
"werd\n",
"word\n",
"a\n"
] | Explanations of the examples:
1. There is only one replace: weird <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> werd;1. No replace needed since there are no two consecutive vowels;1. aaeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> a. | 0 | [
{
"input": "5\nweird",
"output": "werd"
},
{
"input": "4\nword",
"output": "word"
},
{
"input": "5\naaeaa",
"output": "a"
},
{
"input": "100\naaaaabbbbboyoyoyoyoyacadabbbbbiuiufgiuiuaahjabbbklboyoyoyoyoyaaaaabbbbbiuiuiuiuiuaaaaabbbbbeyiyuyzyw",
"output": "abbbbbocadabbbbbifgihjabbbklbobbbbbibbbbbezyw"
},
{
"input": "69\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "12\nmmmmmmmmmmmm",
"output": "mmmmmmmmmmmm"
},
{
"input": "18\nyaywptqwuyiqypwoyw",
"output": "ywptqwuqypwow"
},
{
"input": "85\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "13\nmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmm"
},
{
"input": "10\nmmmmmmmmmm",
"output": "mmmmmmmmmm"
},
{
"input": "11\nmmmmmmmmmmm",
"output": "mmmmmmmmmmm"
},
{
"input": "15\nmmmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmmmm"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "14\nmmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmmm"
},
{
"input": "33\nmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm",
"output": "mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm"
},
{
"input": "79\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "90\nbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "2\naa",
"output": "a"
},
{
"input": "18\niuiuqpyyaoaetiwliu",
"output": "iqpytiwli"
},
{
"input": "5\nxxxxx",
"output": "xxxxx"
},
{
"input": "6\nxxxahg",
"output": "xxxahg"
},
{
"input": "3\nzcv",
"output": "zcv"
},
{
"input": "4\naepo",
"output": "apo"
},
{
"input": "5\nqqqqq",
"output": "qqqqq"
},
{
"input": "6\naaaaaa",
"output": "a"
},
{
"input": "4\naeta",
"output": "ata"
},
{
"input": "20\nttyttlwaoieulyiluuri",
"output": "ttyttlwalyluri"
},
{
"input": "1\nb",
"output": "b"
},
{
"input": "3\nanc",
"output": "anc"
},
{
"input": "1\ne",
"output": "e"
},
{
"input": "3\naie",
"output": "a"
},
{
"input": "3\nvio",
"output": "vi"
},
{
"input": "2\nea",
"output": "e"
},
{
"input": "3\nuas",
"output": "us"
},
{
"input": "2\nba",
"output": "ba"
},
{
"input": "2\naq",
"output": "aq"
},
{
"input": "2\nya",
"output": "y"
},
{
"input": "2\nou",
"output": "o"
},
{
"input": "2\nbb",
"output": "bb"
},
{
"input": "7\nayylmao",
"output": "alma"
},
{
"input": "2\nab",
"output": "ab"
},
{
"input": "19\nyuouiyaoiiweqrryqqp",
"output": "yweqrryqqp"
},
{
"input": "25\niqypwqpriiioetiuqqqttouei",
"output": "iqypwqpritiqqqtto"
},
{
"input": "100\naaaaabbbbboyoyoyoyoyacadabbbbbiuiufgiuiuaahjabbbklboyoyoyoyoyaaaaabbbbbiuiuiuiuiuaaaaabbbbbeyiyuyzyz",
"output": "abbbbbocadabbbbbifgihjabbbklbobbbbbibbbbbezyz"
},
{
"input": "17\naccccccccccccccca",
"output": "accccccccccccccca"
},
{
"input": "5\nababa",
"output": "ababa"
},
{
"input": "10\naaaaaaaaaa",
"output": "a"
},
{
"input": "22\naaaaabbbbboyoyoyoyoyac",
"output": "abbbbboc"
},
{
"input": "7\nmahmoud",
"output": "mahmod"
}
] | 1,518,879,699 | 2,147,483,647 | Python 3 | OK | TESTS | 49 | 77 | 5,632,000 | vowel="aeiouy"
n=int(input().strip())
s=input().strip()
s="3"+s
count=0
out=""
turn=True
for c in s:
if c not in vowel:
out+=c
turn=True
else:
if turn:
out+=c
turn=False
print(out[1:])
| Title: Word Correction
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Victor tries to write his own text editor, with word correction included. However, the rules of word correction are really strange.
Victor thinks that if a word contains two consecutive vowels, then it's kinda weird and it needs to be replaced. So the word corrector works in such a way: as long as there are two consecutive vowels in the word, it deletes the first vowel in a word such that there is another vowel right before it. If there are no two consecutive vowels in the word, it is considered to be correct.
You are given a word *s*. Can you predict what will it become after correction?
In this problem letters a, e, i, o, u and y are considered to be vowels.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=100) — the number of letters in word *s* before the correction.
The second line contains a string *s* consisting of exactly *n* lowercase Latin letters — the word before the correction.
Output Specification:
Output the word *s* after the correction.
Demo Input:
['5\nweird\n', '4\nword\n', '5\naaeaa\n']
Demo Output:
['werd\n', 'word\n', 'a\n']
Note:
Explanations of the examples:
1. There is only one replace: weird <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> werd;1. No replace needed since there are no two consecutive vowels;1. aaeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aeaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aaa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> aa <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> a. | ```python
vowel="aeiouy"
n=int(input().strip())
s=input().strip()
s="3"+s
count=0
out=""
turn=True
for c in s:
if c not in vowel:
out+=c
turn=True
else:
if turn:
out+=c
turn=False
print(out[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\n1 2 2 2 4 5 6 7 8 10",
"output": "376"
},
{
"input": "10\n161821 171308 228496 397870 431255 542924 718337 724672 888642 892002",
"output": "40204082"
},
{
"input": "1\n397870",
"output": "397870"
},
{
"input": "1\n1000000",
"output": "1000000"
},
{
"input": "10\n10 8 7 6 5 4 2 2 2 1",
"output": "376"
},
{
"input": "10\n892002 888642 724672 718337 542924 431255 397870 228496 171308 161821",
"output": "40204082"
},
{
"input": "10\n5 2 6 10 10 10 10 2 2 5",
"output": "485"
},
{
"input": "10\n431255 724672 228496 397870 397870 397870 397870 724672 888642 431255",
"output": "36742665"
},
{
"input": "10\n2 2 2 5 5 6 10 10 10 10",
"output": "485"
},
{
"input": "10\n228496 397870 397870 397870 397870 431255 431255 724672 724672 888642",
"output": "36742665"
},
{
"input": "10\n10 10 10 10 6 5 5 2 2 2",
"output": "485"
},
{
"input": "10\n888642 724672 724672 431255 431255 397870 397870 397870 397870 228496",
"output": "36742665"
},
{
"input": "10\n10 10 10 10 10 10 10 10 10 10",
"output": "640"
},
{
"input": "10\n1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000 1000000",
"output": "64000000"
},
{
"input": "1\n397870",
"output": "397870"
},
{
"input": "2\n1 2",
"output": "6"
},
{
"input": "2\n2 3",
"output": "10"
},
{
"input": "2\n1 1",
"output": "4"
}
] | 1,512,551,136 | 2,147,483,647 | Python 3 | OK | TESTS | 36 | 421 | 26,214,400 | n=int(input())
N=[int(i) for i in input().split()]
number=0
if n==1:
print(N[0])
elif n==2:
print(2*sum(N))
else:
N.sort(reverse=True)
for i in range(n):
number+=N[i]*(n-i+1)
print(number-N[0])
| 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())
N=[int(i) for i in input().split()]
number=0
if n==1:
print(N[0])
elif n==2:
print(2*sum(N))
else:
N.sort(reverse=True)
for i in range(n):
number+=N[i]*(n-i+1)
print(number-N[0])
``` | 3 | |
131 | A | cAPS lOCK | PROGRAMMING | 1,000 | [
"implementation",
"strings"
] | null | null | wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
- either it only contains uppercase letters; - or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged. | The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive. | Print the result of the given word's processing. | [
"cAPS\n",
"Lock\n"
] | [
"Caps",
"Lock\n"
] | none | 500 | [
{
"input": "cAPS",
"output": "Caps"
},
{
"input": "Lock",
"output": "Lock"
},
{
"input": "cAPSlOCK",
"output": "cAPSlOCK"
},
{
"input": "CAPs",
"output": "CAPs"
},
{
"input": "LoCK",
"output": "LoCK"
},
{
"input": "OOPS",
"output": "oops"
},
{
"input": "oops",
"output": "oops"
},
{
"input": "a",
"output": "A"
},
{
"input": "A",
"output": "a"
},
{
"input": "aA",
"output": "Aa"
},
{
"input": "Zz",
"output": "Zz"
},
{
"input": "Az",
"output": "Az"
},
{
"input": "zA",
"output": "Za"
},
{
"input": "AAA",
"output": "aaa"
},
{
"input": "AAa",
"output": "AAa"
},
{
"input": "AaR",
"output": "AaR"
},
{
"input": "Tdr",
"output": "Tdr"
},
{
"input": "aTF",
"output": "Atf"
},
{
"input": "fYd",
"output": "fYd"
},
{
"input": "dsA",
"output": "dsA"
},
{
"input": "fru",
"output": "fru"
},
{
"input": "hYBKF",
"output": "Hybkf"
},
{
"input": "XweAR",
"output": "XweAR"
},
{
"input": "mogqx",
"output": "mogqx"
},
{
"input": "eOhEi",
"output": "eOhEi"
},
{
"input": "nkdku",
"output": "nkdku"
},
{
"input": "zcnko",
"output": "zcnko"
},
{
"input": "lcccd",
"output": "lcccd"
},
{
"input": "vwmvg",
"output": "vwmvg"
},
{
"input": "lvchf",
"output": "lvchf"
},
{
"input": "IUNVZCCHEWENCHQQXQYPUJCRDZLUXCLJHXPHBXEUUGNXOOOPBMOBRIBHHMIRILYJGYYGFMTMFSVURGYHUWDRLQVIBRLPEVAMJQYO",
"output": "iunvzcchewenchqqxqypujcrdzluxcljhxphbxeuugnxooopbmobribhhmirilyjgyygfmtmfsvurgyhuwdrlqvibrlpevamjqyo"
},
{
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"output": "obhszcamdxejwozlkxqkivxuuqjkjlmmfnbpxaefxgvnskqljgxhuxhgcotesivksfmvvxfvmtekacriwalaggmcgfexqknymrtg"
},
{
"input": "IKJYZIKROIYUUCTHSVSKZTETNNOCMAUBLFJCEVANCADASMZRCNLBZPQRXESHEEMOMEPCHROSRTNBIDXYMEPJSIXSZQEBTEKKUHFS",
"output": "ikjyzikroiyuucthsvskztetnnocmaublfjcevancadasmzrcnlbzpqrxesheemomepchrosrtnbidxymepjsixszqebtekkuhfs"
},
{
"input": "cTKDZNWVYRTFPQLDAUUNSPKTDJTUPPFPRXRSINTVFVNNQNKXWUZUDHZBUSOKTABUEDQKUIVRTTVUREEOBJTSDKJKVEGFXVHXEYPE",
"output": "Ctkdznwvyrtfpqldauunspktdjtuppfprxrsintvfvnnqnkxwuzudhzbusoktabuedqkuivrttvureeobjtsdkjkvegfxvhxeype"
},
{
"input": "uCKJZRGZJCPPLEEYJTUNKOQSWGBMTBQEVPYFPIPEKRVYQNTDPANOIXKMPINNFUSZWCURGBDPYTEKBEKCPMVZPMWAOSHJYMGKOMBQ",
"output": "Uckjzrgzjcppleeyjtunkoqswgbmtbqevpyfpipekrvyqntdpanoixkmpinnfuszwcurgbdpytekbekcpmvzpmwaoshjymgkombq"
},
{
"input": "KETAXTSWAAOBKUOKUQREHIOMVMMRSAEWKGXZKRASwTVNSSFSNIWYNPSTMRADOADEEBURRHPOOBIEUIBGYDJCEKPNLEUCANZYJKMR",
"output": "KETAXTSWAAOBKUOKUQREHIOMVMMRSAEWKGXZKRASwTVNSSFSNIWYNPSTMRADOADEEBURRHPOOBIEUIBGYDJCEKPNLEUCANZYJKMR"
},
{
"input": "ZEKGDMWJPVUWFlNXRLUmWKLMMYSLRQQIBRWDPKWITUIMZYYKOEYGREKHHZRZZUFPVTNIHKGTCCTLOKSZITXXZDMPITHNZUIGDZLE",
"output": "ZEKGDMWJPVUWFlNXRLUmWKLMMYSLRQQIBRWDPKWITUIMZYYKOEYGREKHHZRZZUFPVTNIHKGTCCTLOKSZITXXZDMPITHNZUIGDZLE"
},
{
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"output": "TcMbVPCFvnNkCEUUCIFLgBJeCOKuJhIGwXFrhAZjuAhBraMSchBfWwIuHAEbgJOFzGtxDLDXzDSaPCFujGGxgxdlHUIQYRrMFCgJ"
},
{
"input": "xFGqoLILNvxARKuIntPfeukFtMbvzDezKpPRAKkIoIvwqNXnehRVwkkXYvuRCeoieBaBfTjwsYhDeCLvBwktntyluoxCYVioXGdm",
"output": "xFGqoLILNvxARKuIntPfeukFtMbvzDezKpPRAKkIoIvwqNXnehRVwkkXYvuRCeoieBaBfTjwsYhDeCLvBwktntyluoxCYVioXGdm"
},
{
"input": "udvqolbxdwbkijwvhlyaelhynmnfgszbhgshlcwdkaibceqomzujndixuzivlsjyjqxzxodzbukxxhwwultvekdfntwpzlhhrIjm",
"output": "udvqolbxdwbkijwvhlyaelhynmnfgszbhgshlcwdkaibceqomzujndixuzivlsjyjqxzxodzbukxxhwwultvekdfntwpzlhhrIjm"
},
{
"input": "jgpwhetqqoncighgzbbaLwwwxkxivuwtokehrgprfgewzcwxkavwoflcgsgbhoeamzbefzoonwsyzisetoydrpufktzgbaycgaeg",
"output": "jgpwhetqqoncighgzbbaLwwwxkxivuwtokehrgprfgewzcwxkavwoflcgsgbhoeamzbefzoonwsyzisetoydrpufktzgbaycgaeg"
},
{
"input": "vyujsazdstbnkxeunedfbolicojzjpufgfemhtmdrswvmuhoivjvonacefqenbqudelmdegxqtbwezsbydmanzutvdgkgrjxzlnc",
"output": "vyujsazdstbnkxeunedfbolicojzjpufgfemhtmdrswvmuhoivjvonacefqenbqudelmdegxqtbwezsbydmanzutvdgkgrjxzlnc"
},
{
"input": "pivqnuqkaofcduvbttztjbuavrqwiqrwkfncmvatoxruelyoecnkpqraiahumiaiqeyjapbqyrsxcdgjbihivtqezvasfmzntdfv",
"output": "pivqnuqkaofcduvbttztjbuavrqwiqrwkfncmvatoxruelyoecnkpqraiahumiaiqeyjapbqyrsxcdgjbihivtqezvasfmzntdfv"
},
{
"input": "upvtbsxswbohxshdrbjxcungzquhuomgxwlryvshshsfvqbrxvcikbglietlpqkiwbhiqpmdwuqosbtdvyxekkaqiwrbsibpoect",
"output": "upvtbsxswbohxshdrbjxcungzquhuomgxwlryvshshsfvqbrxvcikbglietlpqkiwbhiqpmdwuqosbtdvyxekkaqiwrbsibpoect"
},
{
"input": "cTKDZNWVYRTFPQLDAUUNSPKTDJTUPPFPRXRSINTVFVNNQNKXWUZUDHZBUSOKTABUEDQKUIVRTTVUREEOBJTSDKJKVEGFXVHXEYPE",
"output": "Ctkdznwvyrtfpqldauunspktdjtuppfprxrsintvfvnnqnkxwuzudhzbusoktabuedqkuivrttvureeobjtsdkjkvegfxvhxeype"
},
{
"input": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu",
"output": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu"
},
{
"input": "aBACABa",
"output": "aBACABa"
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAaa",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAaa"
},
{
"input": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA"
},
{
"input": "dDDDDDDDDDDDDD",
"output": "Dddddddddddddd"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "z",
"output": "Z"
},
{
"input": "AZ",
"output": "az"
},
{
"input": "Z",
"output": "z"
},
{
"input": "aAAAA",
"output": "Aaaaa"
},
{
"input": "F",
"output": "f"
}
] | 1,693,164,909 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 56 | 62 | 0 | a = input()
def caps(a):
f = True
for i in range(1, len(a)):
if a[i].islower() == 1:
f = False
if f == 1:
a = a.swapcase()
return a
else:
return a
print(caps(a))
| Title: cAPS lOCK
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
- either it only contains uppercase letters; - or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input Specification:
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output Specification:
Print the result of the given word's processing.
Demo Input:
['cAPS\n', 'Lock\n']
Demo Output:
['Caps', 'Lock\n']
Note:
none | ```python
a = input()
def caps(a):
f = True
for i in range(1, len(a)):
if a[i].islower() == 1:
f = False
if f == 1:
a = a.swapcase()
return a
else:
return a
print(caps(a))
``` | 3 | |
82 | A | Double Cola | PROGRAMMING | 1,100 | [
"implementation",
"math"
] | A. Double Cola | 1 | 256 | Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon. | The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers. | Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially. | [
"1\n",
"6\n",
"1802\n"
] | [
"Sheldon\n",
"Sheldon\n",
"Penny\n"
] | none | 500 | [
{
"input": "1",
"output": "Sheldon"
},
{
"input": "6",
"output": "Sheldon"
},
{
"input": "1802",
"output": "Penny"
},
{
"input": "1",
"output": "Sheldon"
},
{
"input": "2",
"output": "Leonard"
},
{
"input": "3",
"output": "Penny"
},
{
"input": "4",
"output": "Rajesh"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "10",
"output": "Penny"
},
{
"input": "534",
"output": "Rajesh"
},
{
"input": "5033",
"output": "Howard"
},
{
"input": "10010",
"output": "Howard"
},
{
"input": "500000000",
"output": "Penny"
},
{
"input": "63",
"output": "Rajesh"
},
{
"input": "841",
"output": "Leonard"
},
{
"input": "3667",
"output": "Penny"
},
{
"input": "38614",
"output": "Howard"
},
{
"input": "282798",
"output": "Rajesh"
},
{
"input": "9266286",
"output": "Rajesh"
},
{
"input": "27385966",
"output": "Leonard"
},
{
"input": "121580142",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "300",
"output": "Howard"
},
{
"input": "1745",
"output": "Leonard"
},
{
"input": "8302",
"output": "Rajesh"
},
{
"input": "184518",
"output": "Sheldon"
},
{
"input": "1154414",
"output": "Rajesh"
},
{
"input": "28643950",
"output": "Leonard"
},
{
"input": "159222638",
"output": "Howard"
},
{
"input": "24",
"output": "Penny"
},
{
"input": "505",
"output": "Penny"
},
{
"input": "4425",
"output": "Rajesh"
},
{
"input": "12079",
"output": "Sheldon"
},
{
"input": "469726",
"output": "Penny"
},
{
"input": "3961838",
"output": "Penny"
},
{
"input": "57710446",
"output": "Leonard"
},
{
"input": "80719470",
"output": "Howard"
},
{
"input": "1000000000",
"output": "Penny"
},
{
"input": "999999999",
"output": "Penny"
},
{
"input": "999999998",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
}
] | 1,608,708,040 | 2,147,483,647 | Python 3 | OK | TESTS | 41 | 109 | 307,200 | n = int(input())
a1 = 'Sheldon'
a2 = 'Leonard'
a3 = 'Penny'
a4 = 'Rajesh'
a5 = 'Howard'
t = 0
k = 0
p = 1
f = 0
while t - 4 <= n:
p = 5 * 2 ** k
k = k + 1
t = p
if t - 4 == n:
p = t - 4
break
elif t - 4 > n:
p = p / 2 - 4
k = k - 1
break
cnt_a = 2 ** (k - 1)
while n - p >= 0:
f += 1
p = p + cnt_a
if f == 1:
print(a1)
elif f == 2:
print(a2)
elif f == 3:
print(a3)
elif f == 4:
print(a4)
elif f == 5:
print(a5) | Title: Double Cola
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input Specification:
The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output Specification:
Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Demo Input:
['1\n', '6\n', '1802\n']
Demo Output:
['Sheldon\n', 'Sheldon\n', 'Penny\n']
Note:
none | ```python
n = int(input())
a1 = 'Sheldon'
a2 = 'Leonard'
a3 = 'Penny'
a4 = 'Rajesh'
a5 = 'Howard'
t = 0
k = 0
p = 1
f = 0
while t - 4 <= n:
p = 5 * 2 ** k
k = k + 1
t = p
if t - 4 == n:
p = t - 4
break
elif t - 4 > n:
p = p / 2 - 4
k = k - 1
break
cnt_a = 2 ** (k - 1)
while n - p >= 0:
f += 1
p = p + cnt_a
if f == 1:
print(a1)
elif f == 2:
print(a2)
elif f == 3:
print(a3)
elif f == 4:
print(a4)
elif f == 5:
print(a5)
``` | 3.944928 |
41 | A | Translation | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Translation | 2 | 256 | The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly. | The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols. | If the word *t* is a word *s*, written reversely, print YES, otherwise print NO. | [
"code\nedoc\n",
"abb\naba\n",
"code\ncode\n"
] | [
"YES\n",
"NO\n",
"NO\n"
] | none | 500 | [
{
"input": "code\nedoc",
"output": "YES"
},
{
"input": "abb\naba",
"output": "NO"
},
{
"input": "code\ncode",
"output": "NO"
},
{
"input": "abacaba\nabacaba",
"output": "YES"
},
{
"input": "q\nq",
"output": "YES"
},
{
"input": "asrgdfngfnmfgnhweratgjkk\nasrgdfngfnmfgnhweratgjkk",
"output": "NO"
},
{
"input": "z\na",
"output": "NO"
},
{
"input": "asd\ndsa",
"output": "YES"
},
{
"input": "abcdef\nfecdba",
"output": "NO"
},
{
"input": "ywjjbirapvskozubvxoemscfwl\ngnduubaogtfaiowjizlvjcu",
"output": "NO"
},
{
"input": "mfrmqxtzvgaeuleubcmcxcfqyruwzenguhgrmkuhdgnhgtgkdszwqyd\nmfxufheiperjnhyczclkmzyhcxntdfskzkzdwzzujdinf",
"output": "NO"
},
{
"input": "bnbnemvybqizywlnghlykniaxxxlkhftppbdeqpesrtgkcpoeqowjwhrylpsziiwcldodcoonpimudvrxejjo\ntiynnekmlalogyvrgptbinkoqdwzuiyjlrldxhzjmmp",
"output": "NO"
},
{
"input": "pwlpubwyhzqvcitemnhvvwkmwcaawjvdiwtoxyhbhbxerlypelevasmelpfqwjk\nstruuzebbcenziscuoecywugxncdwzyfozhljjyizpqcgkyonyetarcpwkqhuugsqjuixsxptmbnlfupdcfigacdhhrzb",
"output": "NO"
},
{
"input": "gdvqjoyxnkypfvdxssgrihnwxkeojmnpdeobpecytkbdwujqfjtxsqspxvxpqioyfagzjxupqqzpgnpnpxcuipweunqch\nkkqkiwwasbhezqcfeceyngcyuogrkhqecwsyerdniqiocjehrpkljiljophqhyaiefjpavoom",
"output": "NO"
},
{
"input": "umeszdawsvgkjhlqwzents\nhxqhdungbylhnikwviuh",
"output": "NO"
},
{
"input": "juotpscvyfmgntshcealgbsrwwksgrwnrrbyaqqsxdlzhkbugdyx\nibqvffmfktyipgiopznsqtrtxiijntdbgyy",
"output": "NO"
},
{
"input": "zbwueheveouatecaglziqmudxemhrsozmaujrwlqmppzoumxhamwugedikvkblvmxwuofmpafdprbcftew\nulczwrqhctbtbxrhhodwbcxwimncnexosksujlisgclllxokrsbnozthajnnlilyffmsyko",
"output": "NO"
},
{
"input": "nkgwuugukzcv\nqktnpxedwxpxkrxdvgmfgoxkdfpbzvwsduyiybynbkouonhvmzakeiruhfmvrktghadbfkmwxduoqv",
"output": "NO"
},
{
"input": "incenvizhqpcenhjhehvjvgbsnfixbatrrjstxjzhlmdmxijztphxbrldlqwdfimweepkggzcxsrwelodpnryntepioqpvk\ndhjbjjftlvnxibkklxquwmzhjfvnmwpapdrslioxisbyhhfymyiaqhlgecpxamqnocizwxniubrmpyubvpenoukhcobkdojlybxd",
"output": "NO"
},
{
"input": "w\nw",
"output": "YES"
},
{
"input": "vz\nzv",
"output": "YES"
},
{
"input": "ry\nyr",
"output": "YES"
},
{
"input": "xou\nuox",
"output": "YES"
},
{
"input": "axg\ngax",
"output": "NO"
},
{
"input": "zdsl\nlsdz",
"output": "YES"
},
{
"input": "kudl\nldku",
"output": "NO"
},
{
"input": "zzlzwnqlcl\nlclqnwzlzz",
"output": "YES"
},
{
"input": "vzzgicnzqooejpjzads\nsdazjpjeooqzncigzzv",
"output": "YES"
},
{
"input": "raqhmvmzuwaykjpyxsykr\nxkysrypjkyawuzmvmhqar",
"output": "NO"
},
{
"input": "ngedczubzdcqbxksnxuavdjaqtmdwncjnoaicvmodcqvhfezew\nwezefhvqcdomvciaonjcnwdmtqajdvauxnskxbqcdzbuzcdegn",
"output": "YES"
},
{
"input": "muooqttvrrljcxbroizkymuidvfmhhsjtumksdkcbwwpfqdyvxtrlymofendqvznzlmim\nmimlznzvqdnefomylrtxvydqfpwwbckdskmutjshhmfvdiumykziorbxcjlrrvttqooum",
"output": "YES"
},
{
"input": "vxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaivg\ngviayyikkitmuomcpiakhbxszgbnhvwyzkftwoagzixaearxpjacrnvpvbuzenvovehkmmxvblqyxvctroddksdsgebcmlluqpxv",
"output": "YES"
},
{
"input": "mnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfdc\ncdfmkdgrdptkpewbsqvszipgxvgvuiuzbkkwuowbafkikgvnqdkxnayzdjygvezmtsgywnupocdntipiyiorblqkrzjpzatxahnm",
"output": "NO"
},
{
"input": "dgxmzbqofstzcdgthbaewbwocowvhqpinehpjatnnbrijcolvsatbblsrxabzrpszoiecpwhfjmwuhqrapvtcgvikuxtzbftydkw\nwkdytfbztxukivgctvparqhuwmjfhwpceiozsprzbaxrslbbqasvlocjirbnntajphenipthvwocowbweabhtgdcztsfoqbzmxgd",
"output": "NO"
},
{
"input": "gxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwgeh\nhegwxvocotmzstqfbmpjvijgkcyodlxyjawrpkczpmdspsuhoiruavnnnuwvtwohglkdxjetshkboalvzqbgjgthoteceixioxg",
"output": "YES"
},
{
"input": "sihxuwvmaambplxvjfoskinghzicyfqebjtkysotattkahssumfcgrkheotdxwjckpvapbkaepqrxseyfrwtyaycmrzsrsngkh\nhkgnsrszrmcyaytwrfyesxrqpeakbpavpkcjwxdtoehkrgcfmusshakttatosyktjbeqfycizhgniksofjvxlpbmaamvwuxhis",
"output": "YES"
},
{
"input": "ycnahksbughnonldzrhkysujmylcgcfuludjvjiahtkyzqvkopzqcnwhltbzfugzojqkjjlggmvnultascmygelkiktmfieok\nkoeifmtkiklegkmcsatlunvmggkjjlqjozgufzbtlhwncqzpokvqzykthaijvjdulufcgclymjusyyhrzdlnonhgubskhancy",
"output": "NO"
},
{
"input": "wbqasaehtkfojruzyhrlgwmtyiovmzyfifslvlemhqheyaelzwnthrenjsbmntwaoryzwfbxmscmypvxlfmzpnkkjlvwvmtz\nztmvwvljkknpzmflxvpymcsmxbfwzyroawtnmbsjnerhtnwzleayehqhmelvlsfifyzmvoiytmwglrhyzurjofktheasaqbw",
"output": "YES"
},
{
"input": "imippqurprbhfugngtgifelytadegwrgaefnfhbjjnmzikvjaccotqzemufqieqldgnbmviisgkynzeldlhqxuqphjfmyij\njiymfjhpquxqhldleznykgsiivmbngdlqeiqfumezqtoccajvkizmnjjbhfnfeagrwgedatylefigtgngufhbrpruqppimi",
"output": "YES"
},
{
"input": "bikydffiuisckpvzqlteqfhegsagimodb\nbdomigasgehfqetlqzvpkcsiuiffdykib",
"output": "YES"
}
] | 1,607,141,929 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 156 | 0 | s=input()
if reversed(s)==(input()):
print("YES")
else:
print("NO") | Title: Translation
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
The translation from the Berland language into the Birland language is not an easy task. Those languages are very similar: a berlandish word differs from a birlandish word with the same meaning a little: it is spelled (and pronounced) reversely. For example, a Berlandish word code corresponds to a Birlandish word edoc. However, it's easy to make a mistake during the «translation». Vasya translated word *s* from Berlandish into Birlandish as *t*. Help him: find out if he translated the word correctly.
Input Specification:
The first line contains word *s*, the second line contains word *t*. The words consist of lowercase Latin letters. The input data do not consist unnecessary spaces. The words are not empty and their lengths do not exceed 100 symbols.
Output Specification:
If the word *t* is a word *s*, written reversely, print YES, otherwise print NO.
Demo Input:
['code\nedoc\n', 'abb\naba\n', 'code\ncode\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none | ```python
s=input()
if reversed(s)==(input()):
print("YES")
else:
print("NO")
``` | 0 |
150 | A | Win or Freeze | PROGRAMMING | 1,400 | [
"games",
"math",
"number theory"
] | null | null | You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer *q*. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move. | The first line contains the only integer *q* (1<=≤<=*q*<=≤<=1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator. | In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them. | [
"6\n",
"30\n",
"1\n"
] | [
"2\n",
"1\n6\n",
"1\n0\n"
] | Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory. | 500 | [
{
"input": "6",
"output": "2"
},
{
"input": "30",
"output": "1\n6"
},
{
"input": "1",
"output": "1\n0"
},
{
"input": "2",
"output": "1\n0"
},
{
"input": "3",
"output": "1\n0"
},
{
"input": "5",
"output": "1\n0"
},
{
"input": "445538663413",
"output": "1\n0"
},
{
"input": "5138168457911",
"output": "2"
},
{
"input": "472670214391",
"output": "1\n23020027"
},
{
"input": "1468526771489",
"output": "1\n613783"
},
{
"input": "1307514188557",
"output": "1\n39283"
},
{
"input": "8110708459517",
"output": "2"
},
{
"input": "7938986881993",
"output": "1\n378028993"
},
{
"input": "34280152201",
"output": "2"
},
{
"input": "8587340257",
"output": "1\n9409"
},
{
"input": "1245373417369",
"output": "1\n908209"
},
{
"input": "48855707",
"output": "1\n18769"
},
{
"input": "614125",
"output": "1\n25"
},
{
"input": "1716443237161",
"output": "1\n5329"
},
{
"input": "49380563",
"output": "1\n289"
},
{
"input": "2975",
"output": "1\n25"
},
{
"input": "5839252225",
"output": "1\n25"
},
{
"input": "266418",
"output": "1\n9"
},
{
"input": "319757451841",
"output": "1\n289"
},
{
"input": "3047527844089",
"output": "2"
},
{
"input": "1408514752349",
"output": "1\n72361"
},
{
"input": "9999925100701",
"output": "1\n0"
},
{
"input": "9999926826034",
"output": "2"
},
{
"input": "6599669076000",
"output": "1\n4"
},
{
"input": "324",
"output": "1\n4"
},
{
"input": "99",
"output": "1\n9"
},
{
"input": "4",
"output": "2"
},
{
"input": "8",
"output": "1\n4"
},
{
"input": "1000000000000",
"output": "1\n4"
},
{
"input": "9",
"output": "2"
},
{
"input": "50",
"output": "1\n25"
},
{
"input": "16",
"output": "1\n4"
},
{
"input": "25",
"output": "2"
},
{
"input": "44",
"output": "1\n4"
},
{
"input": "388",
"output": "1\n4"
},
{
"input": "2000000014",
"output": "2"
},
{
"input": "128",
"output": "1\n4"
},
{
"input": "27",
"output": "1\n9"
},
{
"input": "274875809788",
"output": "1\n4"
},
{
"input": "64",
"output": "1\n4"
},
{
"input": "802241960524",
"output": "1\n4"
},
{
"input": "401120980262",
"output": "2"
},
{
"input": "236",
"output": "1\n4"
},
{
"input": "7420738134810",
"output": "1\n6"
},
{
"input": "12",
"output": "1\n4"
},
{
"input": "57461344602",
"output": "1\n6"
},
{
"input": "81",
"output": "1\n9"
},
{
"input": "30971726",
"output": "2"
}
] | 1,524,398,129 | 2,147,483,647 | Python 3 | OK | TESTS | 53 | 498 | 7,372,800 | import math
q = int(input())
c = 0
ans = 1
while q % 2 == 0:
q //= 2
if c == 2:
print("1\n"+str(ans), '')
exit()
c +=1
ans *= 2
for i in range(3, int(math.sqrt(q)+1), 2):
while q%i == 0:
q //= i
if c == 2:
print("1\n"+str(ans), '')
exit()
c+=1
ans *=i
if q > 2:
if c == 2:
print("1\n"+str(ans), '')
exit()
c += 1
if c <= 1:
print('1\n0', '')
else:
print('2')
| Title: Win or Freeze
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You can't possibly imagine how cold our friends are this winter in Nvodsk! Two of them play the following game to warm up: initially a piece of paper has an integer *q*. During a move a player should write any integer number that is a non-trivial divisor of the last written number. Then he should run this number of circles around the hotel. Let us remind you that a number's divisor is called non-trivial if it is different from one and from the divided number itself.
The first person who can't make a move wins as he continues to lie in his warm bed under three blankets while the other one keeps running. Determine which player wins considering that both players play optimally. If the first player wins, print any winning first move.
Input Specification:
The first line contains the only integer *q* (1<=≤<=*q*<=≤<=1013).
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Output Specification:
In the first line print the number of the winning player (1 or 2). If the first player wins then the second line should contain another integer — his first move (if the first player can't even make the first move, print 0). If there are multiple solutions, print any of them.
Demo Input:
['6\n', '30\n', '1\n']
Demo Output:
['2\n', '1\n6\n', '1\n0\n']
Note:
Number 6 has only two non-trivial divisors: 2 and 3. It is impossible to make a move after the numbers 2 and 3 are written, so both of them are winning, thus, number 6 is the losing number. A player can make a move and write number 6 after number 30; 6, as we know, is a losing number. Thus, this move will bring us the victory. | ```python
import math
q = int(input())
c = 0
ans = 1
while q % 2 == 0:
q //= 2
if c == 2:
print("1\n"+str(ans), '')
exit()
c +=1
ans *= 2
for i in range(3, int(math.sqrt(q)+1), 2):
while q%i == 0:
q //= i
if c == 2:
print("1\n"+str(ans), '')
exit()
c+=1
ans *=i
if q > 2:
if c == 2:
print("1\n"+str(ans), '')
exit()
c += 1
if c <= 1:
print('1\n0', '')
else:
print('2')
``` | 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",
"output": "1"
},
{
"input": "12345 67890123456789123457",
"output": "1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158 8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158",
"output": "8392739158839273915883927391588392739158839273915883927391588392739158839273915883927391588392739158"
},
{
"input": "1 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1"
},
{
"input": "8328748239473982794239847237438782379810988324751 9328748239473982794239847237438782379810988324751",
"output": "1"
},
{
"input": "1029398958432734901284327523909481928483573793 1029398958432734901284327523909481928483573794",
"output": "1"
},
{
"input": "10000 1000000000",
"output": "1"
},
{
"input": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "11210171722243 65715435710585778347",
"output": "1"
},
{
"input": "2921881079263974825226940825843 767693191032295360887755303860323261471",
"output": "1"
},
{
"input": "8025352957265704896940312528736939363590612908210603 96027920417708260814607687034511406492969694925539085",
"output": "1"
},
{
"input": "23510978780782786207241069904470895053213996267165977112058175452757132930 210352653280909370107314249722987050753257161175393375412301228883856435481424",
"output": "1"
},
{
"input": "8150070767079366215626260746398623663859344142817267779361251788637547414925170226504788118262 49924902262298336032630839998470954964895251605110946547855439236151401194070172107435992986913614",
"output": "1"
},
{
"input": "15943150466658398903 15943150466658398903",
"output": "15943150466658398903"
},
{
"input": "410470228200245407491525399055972 410470228200245407491525399055972",
"output": "410470228200245407491525399055972"
},
{
"input": "51894705655711504622197349350106792045098781545973899451307 51894705655711504622197349350106792045098781545973899451307",
"output": "51894705655711504622197349350106792045098781545973899451307"
},
{
"input": "60353594589897438036015726222485085035927634677598681595162804007836722215668410 60353594589897438036015726222485085035927634677598681595162804007836722215668410",
"output": "60353594589897438036015726222485085035927634677598681595162804007836722215668410"
},
{
"input": "761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535 761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535",
"output": "761733780145118977868180796896376577405349682060892737466239031663476251177476275459280340045369535"
},
{
"input": "1 2000000000",
"output": "1"
},
{
"input": "13 1928834874",
"output": "1"
},
{
"input": "87 2938984237482934238",
"output": "1"
},
{
"input": "213 413",
"output": "1"
},
{
"input": "3 4",
"output": "1"
}
] | 1,631,521,017 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 124 | 22,220,800 | # your code goes here
def hcfnaive(a,b):
if(b==0):
return a
else:
return hcfnaive(b,a%b)
a = input()
b= input()
# prints 12
print ("The gcd of 60 and 48 is : ",end="")
print (hcfnaive(a,b)) | 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
# your code goes here
def hcfnaive(a,b):
if(b==0):
return a
else:
return hcfnaive(b,a%b)
a = input()
b= input()
# prints 12
print ("The gcd of 60 and 48 is : ",end="")
print (hcfnaive(a,b))
``` | -1 | |
200 | B | Drinks | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink. | The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space. | Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4. | [
"3\n50 50 100\n",
"4\n0 25 50 75\n"
] | [
"66.666666666667\n",
"37.500000000000\n"
] | Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent. | 500 | [
{
"input": "3\n50 50 100",
"output": "66.666666666667"
},
{
"input": "4\n0 25 50 75",
"output": "37.500000000000"
},
{
"input": "3\n0 1 8",
"output": "3.000000000000"
},
{
"input": "5\n96 89 93 95 70",
"output": "88.600000000000"
},
{
"input": "7\n62 41 78 4 38 39 75",
"output": "48.142857142857"
},
{
"input": "13\n2 22 7 0 1 17 3 17 11 2 21 26 22",
"output": "11.615384615385"
},
{
"input": "21\n5 4 11 7 0 5 45 21 0 14 51 6 0 16 10 19 8 9 7 12 18",
"output": "12.761904761905"
},
{
"input": "26\n95 70 93 74 94 70 91 70 39 79 80 57 87 75 37 93 48 67 51 90 85 26 23 64 66 84",
"output": "69.538461538462"
},
{
"input": "29\n84 99 72 96 83 92 95 98 97 93 76 84 99 93 81 76 93 99 99 100 95 100 96 95 97 100 71 98 94",
"output": "91.551724137931"
},
{
"input": "33\n100 99 100 100 99 99 99 100 100 100 99 99 99 100 100 100 100 99 100 99 100 100 97 100 100 100 100 100 100 100 98 98 100",
"output": "99.515151515152"
},
{
"input": "34\n14 9 10 5 4 26 18 23 0 1 0 20 18 15 2 2 3 5 14 1 9 4 2 15 7 1 7 19 10 0 0 11 0 2",
"output": "8.147058823529"
},
{
"input": "38\n99 98 100 100 99 92 99 99 98 84 88 94 86 99 93 100 98 99 65 98 85 84 64 97 96 89 79 96 91 84 99 93 72 96 94 97 96 93",
"output": "91.921052631579"
},
{
"input": "52\n100 94 99 98 99 99 99 95 97 97 98 100 100 98 97 100 98 90 100 99 97 94 90 98 100 100 90 99 100 95 98 95 94 85 97 94 96 94 99 99 99 98 100 100 94 99 99 100 98 87 100 100",
"output": "97.019230769231"
},
{
"input": "58\n10 70 12 89 1 82 100 53 40 100 21 69 92 91 67 66 99 77 25 48 8 63 93 39 46 79 82 14 44 42 1 79 0 69 56 73 67 17 59 4 65 80 20 60 77 52 3 61 16 76 33 18 46 100 28 59 9 6",
"output": "50.965517241379"
},
{
"input": "85\n7 8 1 16 0 15 1 7 0 11 15 6 2 12 2 8 9 8 2 0 3 7 15 7 1 8 5 7 2 26 0 3 11 1 8 10 31 0 7 6 1 8 1 0 9 14 4 8 7 16 9 1 0 16 10 9 6 1 1 4 2 7 4 5 4 1 20 6 16 16 1 1 10 17 8 12 14 19 3 8 1 7 10 23 10",
"output": "7.505882352941"
},
{
"input": "74\n5 3 0 7 13 10 12 10 18 5 0 18 2 13 7 17 2 7 5 2 40 19 0 2 2 3 0 45 4 20 0 4 2 8 1 19 3 9 17 1 15 0 16 1 9 4 0 9 32 2 6 18 11 18 1 15 16 12 7 19 5 3 9 28 26 8 3 10 33 29 4 13 28 6",
"output": "10.418918918919"
},
{
"input": "98\n42 9 21 11 9 11 22 12 52 20 10 6 56 9 26 27 1 29 29 14 38 17 41 21 7 45 15 5 29 4 51 20 6 8 34 17 13 53 30 45 0 10 16 41 4 5 6 4 14 2 31 6 0 11 13 3 3 43 13 36 51 0 7 16 28 23 8 36 30 22 8 54 21 45 39 4 50 15 1 30 17 8 18 10 2 20 16 50 6 68 15 6 38 7 28 8 29 41",
"output": "20.928571428571"
},
{
"input": "99\n60 65 40 63 57 44 30 84 3 10 39 53 40 45 72 20 76 11 61 32 4 26 97 55 14 57 86 96 34 69 52 22 26 79 31 4 21 35 82 47 81 28 72 70 93 84 40 4 69 39 83 58 30 7 32 73 74 12 92 23 61 88 9 58 70 32 75 40 63 71 46 55 39 36 14 97 32 16 95 41 28 20 85 40 5 50 50 50 75 6 10 64 38 19 77 91 50 72 96",
"output": "49.191919191919"
},
{
"input": "99\n100 88 40 30 81 80 91 98 69 73 88 96 79 58 14 100 87 84 52 91 83 88 72 83 99 35 54 80 46 79 52 72 85 32 99 39 79 79 45 83 88 50 75 75 50 59 65 75 97 63 92 58 89 46 93 80 89 33 69 86 99 99 66 85 72 74 79 98 85 95 46 63 77 97 49 81 89 39 70 76 68 91 90 56 31 93 51 87 73 95 74 69 87 95 57 68 49 95 92",
"output": "73.484848484848"
},
{
"input": "100\n18 15 17 0 3 3 0 4 1 8 2 22 7 21 5 0 0 8 3 16 1 0 2 9 9 3 10 8 17 20 5 4 8 12 2 3 1 1 3 2 23 0 1 0 5 7 4 0 1 3 3 4 25 2 2 14 8 4 9 3 0 11 0 3 12 3 14 16 7 7 14 1 17 9 0 35 42 12 3 1 25 9 3 8 5 3 2 8 22 14 11 6 3 9 6 8 7 7 4 6",
"output": "7.640000000000"
},
{
"input": "100\n88 77 65 87 100 63 91 96 92 89 77 95 76 80 84 83 100 71 85 98 26 54 74 78 69 59 96 86 88 91 95 26 52 88 64 70 84 81 76 84 94 82 100 66 97 98 43 94 59 94 100 80 98 73 69 83 94 70 74 79 91 31 62 88 69 55 62 97 40 64 62 83 87 85 50 90 69 72 67 49 100 51 69 96 81 90 83 91 86 34 79 69 100 66 97 98 47 97 74 100",
"output": "77.660000000000"
},
{
"input": "100\n91 92 90 91 98 84 85 96 83 98 99 87 94 70 87 75 86 90 89 88 82 83 91 94 88 86 90 99 100 98 97 75 95 99 95 100 91 92 76 93 95 97 88 93 95 81 96 89 88 100 98 87 90 96 100 99 58 90 96 77 92 82 100 100 93 93 98 99 79 88 97 95 98 66 96 83 96 100 99 92 98 98 92 93 100 97 98 100 98 97 100 100 94 90 99 100 98 79 80 81",
"output": "91.480000000000"
},
{
"input": "1\n0",
"output": "0.000000000000"
},
{
"input": "1\n100",
"output": "100.000000000000"
},
{
"input": "1\n78",
"output": "78.000000000000"
},
{
"input": "2\n0 100",
"output": "50.000000000000"
},
{
"input": "2\n100 100",
"output": "100.000000000000"
},
{
"input": "5\n0 0 0 0 1",
"output": "0.200000000000"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 99",
"output": "99.990000000000"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1",
"output": "0.010000000000"
},
{
"input": "5\n100 100 100 100 100",
"output": "100.000000000000"
}
] | 1,696,485,954 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 60 | 0 | x = int(input())
for _ in range(x):
i = int(input())
z = i / x
print(z)
| Title: Drinks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Vasya loves orange juice very much. That's why any food and drink in his kitchen necessarily contains orange juice. There are *n* drinks in his fridge, the volume fraction of orange juice in the *i*-th drink equals *p**i* percent.
One day Vasya decided to make himself an orange cocktail. He took equal proportions of each of the *n* drinks and mixed them. Then he wondered, how much orange juice the cocktail has.
Find the volume fraction of orange juice in the final drink.
Input Specification:
The first input line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of orange-containing drinks in Vasya's fridge. The second line contains *n* integers *p**i* (0<=≤<=*p**i*<=≤<=100) — the volume fraction of orange juice in the *i*-th drink, in percent. The numbers are separated by a space.
Output Specification:
Print the volume fraction in percent of orange juice in Vasya's cocktail. The answer will be considered correct if the absolute or relative error does not exceed 10<=<=-<=4.
Demo Input:
['3\n50 50 100\n', '4\n0 25 50 75\n']
Demo Output:
['66.666666666667\n', '37.500000000000\n']
Note:
Note to the first sample: let's assume that Vasya takes *x* milliliters of each drink from the fridge. Then the volume of pure juice in the cocktail will equal <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/c1fac6e64d3a8ee6a5ac138cbe51e60039b22473.png" style="max-width: 100.0%;max-height: 100.0%;"/> milliliters. The total cocktail's volume equals 3·*x* milliliters, so the volume fraction of the juice in the cocktail equals <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ceb0664e55a1f9f5fa1243ec74680a4665a4d58d.png" style="max-width: 100.0%;max-height: 100.0%;"/>, that is, 66.(6) percent. | ```python
x = int(input())
for _ in range(x):
i = int(input())
z = i / x
print(z)
``` | -1 | |
1,008 | A | Romaji | PROGRAMMING | 900 | [
"implementation",
"strings"
] | null | null | Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.
In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.
Help Vitya find out if a word $s$ is Berlanese. | The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters. | Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".
You can print each letter in any case (upper or lower). | [
"sumimasen\n",
"ninja\n",
"codeforces\n"
] | [
"YES\n",
"YES\n",
"NO\n"
] | In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.
In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese. | 500 | [
{
"input": "sumimasen",
"output": "YES"
},
{
"input": "ninja",
"output": "YES"
},
{
"input": "codeforces",
"output": "NO"
},
{
"input": "auuaoonntanonnuewannnnpuuinniwoonennyolonnnvienonpoujinndinunnenannmuveoiuuhikucuziuhunnnmunzancenen",
"output": "YES"
},
{
"input": "n",
"output": "YES"
},
{
"input": "necnei",
"output": "NO"
},
{
"input": "nternn",
"output": "NO"
},
{
"input": "aucunuohja",
"output": "NO"
},
{
"input": "a",
"output": "YES"
},
{
"input": "b",
"output": "NO"
},
{
"input": "nn",
"output": "YES"
},
{
"input": "nnnzaaa",
"output": "YES"
},
{
"input": "zn",
"output": "NO"
},
{
"input": "ab",
"output": "NO"
},
{
"input": "aaaaaaaaaa",
"output": "YES"
},
{
"input": "aaaaaaaaab",
"output": "NO"
},
{
"input": "aaaaaaaaan",
"output": "YES"
},
{
"input": "baaaaaaaaa",
"output": "YES"
},
{
"input": "naaaaaaaaa",
"output": "YES"
},
{
"input": "nbaaaaaaaa",
"output": "YES"
},
{
"input": "bbaaaaaaaa",
"output": "NO"
},
{
"input": "bnaaaaaaaa",
"output": "NO"
},
{
"input": "eonwonojannonnufimiiniewuqaienokacevecinfuqihatenhunliquuyebayiaenifuexuanenuaounnboancaeowonu",
"output": "YES"
},
{
"input": "uixinnepnlinqaingieianndeakuniooudidonnnqeaituioeneiroionxuowudiooonayenfeonuino",
"output": "NO"
},
{
"input": "nnnnnyigaveteononnnnxaalenxuiiwannntoxonyoqonlejuoxuoconnnentoinnul",
"output": "NO"
},
{
"input": "ndonneasoiunhomuunnhuitonnntunntoanerekonoupunanuauenu",
"output": "YES"
},
{
"input": "anujemogawautiedoneobninnibonuunaoennnyoorufonxionntinimiboonununnnnnleenqunminzayoutanlalo",
"output": "NO"
},
{
"input": "y",
"output": "NO"
},
{
"input": "by",
"output": "NO"
},
{
"input": "yy",
"output": "NO"
},
{
"input": "nbn",
"output": "NO"
},
{
"input": "nz",
"output": "NO"
},
{
"input": "king",
"output": "NO"
},
{
"input": "g",
"output": "NO"
},
{
"input": "az",
"output": "NO"
},
{
"input": "x",
"output": "NO"
},
{
"input": "z",
"output": "NO"
},
{
"input": "yn",
"output": "NO"
},
{
"input": "aeo",
"output": "YES"
},
{
"input": "nb",
"output": "NO"
},
{
"input": "npn",
"output": "NO"
},
{
"input": "kini",
"output": "YES"
},
{
"input": "pya",
"output": "NO"
},
{
"input": "m",
"output": "NO"
},
{
"input": "p",
"output": "NO"
},
{
"input": "aaaaaaaak",
"output": "NO"
},
{
"input": "aab",
"output": "NO"
},
{
"input": "d",
"output": "NO"
},
{
"input": "at",
"output": "NO"
},
{
"input": "aaaaaak",
"output": "NO"
},
{
"input": "aaz",
"output": "NO"
},
{
"input": "aaab",
"output": "NO"
},
{
"input": "s",
"output": "NO"
},
{
"input": "nzzen",
"output": "NO"
},
{
"input": "aeionnhhhn",
"output": "NO"
},
{
"input": "h",
"output": "NO"
},
{
"input": "ny",
"output": "NO"
}
] | 1,645,366,047 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 62 | 0 | s=input()+'*'
ans= 'YES'
for i in range (len(s)-1):
if s[i] not in 'aeioun' and s[i+1] not in 'aeioun':
ans= 'NO'
break
print (ans)
| Title: Romaji
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.
In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.
Help Vitya find out if a word $s$ is Berlanese.
Input Specification:
The first line of the input contains the string $s$ consisting of $|s|$ ($1\leq |s|\leq 100$) lowercase Latin letters.
Output Specification:
Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".
You can print each letter in any case (upper or lower).
Demo Input:
['sumimasen\n', 'ninja\n', 'codeforces\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.
In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese. | ```python
s=input()+'*'
ans= 'YES'
for i in range (len(s)-1):
if s[i] not in 'aeioun' and s[i+1] not in 'aeioun':
ans= 'NO'
break
print (ans)
``` | 0 | |
897 | A | Scarborough Fair | PROGRAMMING | 800 | [
"implementation"
] | null | null | Parsley, sage, rosemary and thyme.
Remember me to one who lives there.
He once was the true love of mine.
Willem is taking the girl to the highest building in island No.28, however, neither of them knows how to get there.
Willem asks his friend, Grick for directions, Grick helped them, and gave them a task.
Although the girl wants to help, Willem insists on doing it by himself.
Grick gave Willem a string of length *n*.
Willem needs to do *m* operations, each operation has four parameters *l*,<=*r*,<=*c*1,<=*c*2, which means that all symbols *c*1 in range [*l*,<=*r*] (from *l*-th to *r*-th, including *l* and *r*) are changed into *c*2. String is 1-indexed.
Grick wants to know the final string after all the *m* operations. | The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
The second line contains a string *s* of length *n*, consisting of lowercase English letters.
Each of the next *m* lines contains four parameters *l*,<=*r*,<=*c*1,<=*c*2 (1<=≤<=*l*<=≤<=*r*<=≤<=*n*, *c*1,<=*c*2 are lowercase English letters), separated by space. | Output string *s* after performing *m* operations described above. | [
"3 1\nioi\n1 1 i n\n",
"5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g\n"
] | [
"noi",
"gaaak"
] | For the second example:
After the first operation, the string is wxxak.
After the second operation, the string is waaak.
After the third operation, the string is gaaak. | 500 | [
{
"input": "3 1\nioi\n1 1 i n",
"output": "noi"
},
{
"input": "5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g",
"output": "gaaak"
},
{
"input": "9 51\nbhfbdcgff\n2 3 b b\n2 8 e f\n3 8 g f\n5 7 d a\n1 5 e b\n3 4 g b\n6 7 c d\n3 6 e g\n3 6 e h\n5 6 a e\n7 9 a c\n4 9 a h\n3 7 c b\n6 9 b g\n1 7 h b\n4 5 a e\n3 9 f a\n1 2 c h\n4 8 a c\n3 5 e d\n3 4 g f\n2 3 d h\n2 3 d e\n1 7 d g\n2 6 e g\n2 3 d g\n5 5 h h\n2 8 g d\n8 9 a f\n5 9 c e\n1 7 f d\n1 6 e e\n5 7 c a\n8 9 b b\n2 6 e b\n6 6 g h\n1 2 b b\n1 5 a f\n5 8 f h\n1 5 e g\n3 9 f h\n6 8 g a\n4 6 h g\n1 5 f a\n5 6 a c\n4 8 e d\n1 4 d g\n7 8 b f\n5 6 h b\n3 9 c e\n1 9 b a",
"output": "aahaddddh"
},
{
"input": "28 45\ndcbbaddjhbeefjadjchgkhgggfha\n10 25 c a\n13 19 a f\n12 28 e d\n12 27 e a\n9 20 b e\n7 17 g d\n22 26 j j\n8 16 c g\n14 16 a d\n3 10 f c\n10 26 d b\n8 17 i e\n10 19 d i\n6 21 c j\n7 22 b k\n17 19 a i\n4 18 j k\n8 25 a g\n10 27 j e\n9 18 g d\n16 23 h a\n17 26 k e\n8 16 h f\n1 15 d f\n22 28 k k\n11 20 c k\n6 11 b h\n17 17 e i\n15 22 g h\n8 18 c f\n4 16 e a\n8 25 b c\n6 24 d g\n5 9 f j\n12 19 i h\n4 25 e f\n15 25 c j\n15 27 e e\n11 20 b f\n19 27 e k\n2 21 d a\n9 27 k e\n14 24 b a\n3 6 i g\n2 26 k f",
"output": "fcbbajjfjaaefefehfahfagggfha"
},
{
"input": "87 5\nnfinedeojadjmgafnaogekfjkjfncnliagfchjfcmellgigjjcaaoeakdolchjcecljdeblmheimkibkgdkcdml\n47 56 a k\n51 81 o d\n5 11 j h\n48 62 j d\n16 30 k m",
"output": "nfinedeohadjmgafnaogemfjmjfncnliagfchjfcmellgigddckkdekkddlchdcecljdeblmheimkibkgdkcdml"
},
{
"input": "5 16\nacfbb\n1 2 e f\n2 5 a f\n2 3 b e\n4 4 f a\n2 3 f a\n1 2 b e\n4 5 c d\n2 4 e c\n1 4 e a\n1 3 d c\n3 5 e b\n3 5 e b\n2 2 e d\n1 3 e c\n3 3 a e\n1 5 a a",
"output": "acebb"
},
{
"input": "94 13\nbcaaaaaaccacddcdaacbdaabbcbaddbccbccbbbddbadddcccbddadddaadbdababadaacdcdbcdadabdcdcbcbcbcbbcd\n52 77 d d\n21 92 d b\n45 48 c b\n20 25 d a\n57 88 d b\n3 91 b d\n64 73 a a\n5 83 b d\n2 69 c c\n28 89 a b\n49 67 c b\n41 62 a c\n49 87 b c",
"output": "bcaaaaaaccacddcdaacddaaddcdbdddccdccddddddbdddddcdddcdddccdddcdcdcdcccdcddcdcdcddcdcdcdcdcdbcd"
},
{
"input": "67 39\nacbcbccccbabaabcabcaaaaaaccbcbbcbaaaacbbcccbcbabbcacccbbabbabbabaac\n4 36 a b\n25 38 a a\n3 44 b c\n35 57 b a\n4 8 a c\n20 67 c a\n30 66 b b\n27 40 a a\n2 56 a b\n10 47 c a\n22 65 c b\n29 42 a b\n1 46 c b\n57 64 b c\n20 29 b a\n14 51 c a\n12 55 b b\n20 20 a c\n2 57 c a\n22 60 c b\n16 51 c c\n31 64 a c\n17 30 c a\n23 36 c c\n28 67 a c\n37 40 a c\n37 50 b c\n29 48 c b\n2 34 b c\n21 53 b a\n26 63 a c\n23 28 c a\n51 56 c b\n32 61 b b\n64 67 b b\n21 67 b c\n8 53 c c\n40 62 b b\n32 38 c c",
"output": "accccccccaaaaaaaaaaaaaaaaaaaccccccccccccccccccccccccccccccccccccccc"
},
{
"input": "53 33\nhhcbhfafeececbhadfbdbehdfacfchbhdbfebdfeghebfcgdhehfh\n27 41 h g\n18 35 c b\n15 46 h f\n48 53 e g\n30 41 b c\n12 30 b f\n10 37 e f\n18 43 a h\n10 52 d a\n22 48 c e\n40 53 f d\n7 12 b h\n12 51 f a\n3 53 g a\n19 41 d h\n22 29 b h\n2 30 a b\n26 28 e h\n25 35 f a\n19 31 h h\n44 44 d e\n19 22 e c\n29 44 d h\n25 33 d h\n3 53 g c\n18 44 h b\n19 28 f e\n3 22 g h\n8 17 c a\n37 51 d d\n3 28 e h\n27 50 h h\n27 46 f b",
"output": "hhcbhfbfhfababbbbbbbbbbbbbbbbbeaaeaaeaaeabebdeaahahdh"
},
{
"input": "83 10\nfhbecdgadecabbbecedcgfdcefcbgechbedagecgdgfgdaahchdgchbeaedgafdefecdchceececfcdhcdh\n9 77 e e\n26 34 b g\n34 70 b a\n40 64 e g\n33 78 h f\n14 26 a a\n17 70 d g\n56 65 a c\n8 41 d c\n11 82 c b",
"output": "fhbecdgacebabbbebegbgfgbefbggebhgegagebgggfggaafbfggbfagbgggbfggfebgbfbeebebfbdhbdh"
},
{
"input": "1 4\ne\n1 1 c e\n1 1 e a\n1 1 e c\n1 1 d a",
"output": "a"
},
{
"input": "71 21\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\n61 61 a a\n32 56 a a\n10 67 a a\n7 32 a a\n26 66 a a\n41 55 a a\n49 55 a a\n4 61 a a\n53 59 a a\n37 58 a a\n7 63 a a\n39 40 a a\n51 64 a a\n27 37 a a\n22 71 a a\n4 45 a a\n7 8 a a\n43 46 a a\n19 28 a a\n51 54 a a\n14 67 a a",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "30 4\neaaddabedcbbcccddbabdecadcecce\n2 17 c a\n16 29 e e\n16 21 c b\n7 11 b c",
"output": "eaaddacedacbaaaddbabdecadcecce"
},
{
"input": "48 30\naaaabaabbaababbbaabaabaababbabbbaabbbaabaaaaaaba\n3 45 a b\n1 14 a a\n15 32 a b\n37 47 a b\n9 35 a b\n36 39 b b\n6 26 a b\n36 44 a a\n28 44 b a\n29 31 b a\n20 39 a a\n45 45 a b\n21 32 b b\n7 43 a b\n14 48 a b\n14 33 a b\n39 44 a a\n9 36 b b\n4 23 b b\n9 42 b b\n41 41 b a\n30 47 a b\n8 42 b a\n14 38 b b\n3 15 a a\n35 47 b b\n14 34 a b\n38 43 a b\n1 35 b a\n16 28 b a",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbb"
},
{
"input": "89 29\nbabaabaaabaaaababbbbbbbabbbaaaaababbaababababbababaaabbababaaabbbbaaabaaaaaabaaabaabbabab\n39 70 b b\n3 56 b b\n5 22 b a\n4 39 a b\n41 87 b b\n34 41 a a\n10 86 a b\n29 75 a b\n2 68 a a\n27 28 b b\n42 51 b a\n18 61 a a\n6 67 b a\n47 63 a a\n8 68 a b\n4 74 b a\n19 65 a b\n8 55 a b\n5 30 a a\n3 65 a b\n16 57 a b\n34 56 b a\n1 70 a b\n59 68 b b\n29 57 b a\n47 49 b b\n49 73 a a\n32 61 b b\n29 42 a a",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbaaaabbbbbbbbbbbbbab"
},
{
"input": "59 14\nfbebcfabdefbaaedcefdeecababcabebadfbccaaedaebfdaefdbbcbebbe\n5 32 e f\n8 46 e e\n31 43 e f\n3 10 e a\n53 54 f d\n55 59 d a\n39 58 e b\n54 56 f a\n9 40 b e\n28 37 d a\n7 35 e b\n7 56 c f\n23 26 e a\n15 44 e d",
"output": "fbabcfabdffbaafdfffdfffababfabfbaafdffaafdabbfdabfdbbfbbbbe"
},
{
"input": "7 17\nbbaabab\n3 5 a b\n5 7 a a\n5 5 a a\n4 4 b a\n7 7 a a\n5 6 b b\n1 3 b a\n6 7 a b\n4 6 a b\n6 6 a a\n2 4 b a\n1 7 b a\n4 6 b b\n2 5 b b\n2 5 a b\n1 4 a a\n4 4 b a",
"output": "abbabaa"
},
{
"input": "100 1\ndebaaagbfdgehagadabfgheegggfghghgeeeabgceffeffggcbcegfgebbdhebhfagcgadcbdbabddbcadgbgdebdfehceehcaef\n13 99 f c",
"output": "debaaagbfdgehagadabcgheegggcghghgeeeabgcecceccggcbcegcgebbdhebhcagcgadcbdbabddbcadgbgdebdcehceehcaef"
},
{
"input": "1 1\na\n1 1 a b",
"output": "b"
},
{
"input": "100 1\naaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa\n1 100 a b",
"output": "bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "2 2\naa\n1 2 a b\n1 2 b c",
"output": "cc"
},
{
"input": "3 3\naaa\n1 3 a b\n1 3 b c\n1 3 c d",
"output": "ddd"
},
{
"input": "2 2\naa\n2 2 a b\n1 1 a b",
"output": "bb"
}
] | 1,618,161,839 | 2,147,483,647 | PyPy 3 | OK | TESTS | 47 | 108 | 0 | n, m = list(map(int,input().split()))
a = input()
for i in range(m):
l,r,c1,c2 = input().split()
l = int(l)
r = int(r)
a = a[:l-1] + a[l-1:r].replace(c1,c2) + a[r:]
print(a) | Title: Scarborough Fair
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Parsley, sage, rosemary and thyme.
Remember me to one who lives there.
He once was the true love of mine.
Willem is taking the girl to the highest building in island No.28, however, neither of them knows how to get there.
Willem asks his friend, Grick for directions, Grick helped them, and gave them a task.
Although the girl wants to help, Willem insists on doing it by himself.
Grick gave Willem a string of length *n*.
Willem needs to do *m* operations, each operation has four parameters *l*,<=*r*,<=*c*1,<=*c*2, which means that all symbols *c*1 in range [*l*,<=*r*] (from *l*-th to *r*-th, including *l* and *r*) are changed into *c*2. String is 1-indexed.
Grick wants to know the final string after all the *m* operations.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
The second line contains a string *s* of length *n*, consisting of lowercase English letters.
Each of the next *m* lines contains four parameters *l*,<=*r*,<=*c*1,<=*c*2 (1<=≤<=*l*<=≤<=*r*<=≤<=*n*, *c*1,<=*c*2 are lowercase English letters), separated by space.
Output Specification:
Output string *s* after performing *m* operations described above.
Demo Input:
['3 1\nioi\n1 1 i n\n', '5 3\nwxhak\n3 3 h x\n1 5 x a\n1 3 w g\n']
Demo Output:
['noi', 'gaaak']
Note:
For the second example:
After the first operation, the string is wxxak.
After the second operation, the string is waaak.
After the third operation, the string is gaaak. | ```python
n, m = list(map(int,input().split()))
a = input()
for i in range(m):
l,r,c1,c2 = input().split()
l = int(l)
r = int(r)
a = a[:l-1] + a[l-1:r].replace(c1,c2) + a[r:]
print(a)
``` | 3 | |
817 | C | Really Big Numbers | PROGRAMMING | 1,600 | [
"binary search",
"brute force",
"dp",
"math"
] | null | null | Ivan likes to learn different things about numbers, but he is especially interested in really big numbers. Ivan thinks that a positive integer number *x* is really big if the difference between *x* and the sum of its digits (in decimal representation) is not less than *s*. To prove that these numbers may have different special properties, he wants to know how rare (or not rare) they are — in fact, he needs to calculate the quantity of really big numbers that are not greater than *n*.
Ivan tried to do the calculations himself, but soon realized that it's too difficult for him. So he asked you to help him in calculations. | The first (and the only) line contains two integers *n* and *s* (1<=≤<=*n*,<=*s*<=≤<=1018). | Print one integer — the quantity of really big numbers that are not greater than *n*. | [
"12 1\n",
"25 20\n",
"10 9\n"
] | [
"3\n",
"0\n",
"1\n"
] | In the first example numbers 10, 11 and 12 are really big.
In the second example there are no really big numbers that are not greater than 25 (in fact, the first really big number is 30: 30 - 3 ≥ 20).
In the third example 10 is the only really big number (10 - 1 ≥ 9). | 0 | [
{
"input": "12 1",
"output": "3"
},
{
"input": "25 20",
"output": "0"
},
{
"input": "10 9",
"output": "1"
},
{
"input": "300 1000",
"output": "0"
},
{
"input": "500 1000",
"output": "0"
},
{
"input": "1000 2000",
"output": "0"
},
{
"input": "10000 1000",
"output": "8991"
},
{
"input": "1000000000000000000 1000000000000000000",
"output": "0"
},
{
"input": "1000000000000000000 100000000000000000",
"output": "899999999999999991"
},
{
"input": "1000000000000000000 10000000000000000",
"output": "989999999999999991"
},
{
"input": "1000000000000000000 1000000000000000",
"output": "998999999999999991"
},
{
"input": "1000000000000000000 100000000000000",
"output": "999899999999999991"
},
{
"input": "1000000000000000000 200000000000000000",
"output": "799999999999999991"
},
{
"input": "10 5",
"output": "1"
},
{
"input": "20 5",
"output": "11"
},
{
"input": "20 9",
"output": "11"
},
{
"input": "100 9",
"output": "91"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "130 118",
"output": "1"
},
{
"input": "190 181",
"output": "0"
},
{
"input": "1999 1971",
"output": "10"
},
{
"input": "100 99",
"output": "1"
},
{
"input": "6909094398 719694282",
"output": "6189400069"
},
{
"input": "260 258",
"output": "0"
},
{
"input": "35 19",
"output": "6"
},
{
"input": "100 87",
"output": "1"
},
{
"input": "91 89",
"output": "0"
},
{
"input": "109 89",
"output": "10"
},
{
"input": "109 91",
"output": "10"
},
{
"input": "20331 11580",
"output": "8732"
},
{
"input": "405487470 255750281",
"output": "149737161"
},
{
"input": "17382 12863",
"output": "4493"
},
{
"input": "19725 14457",
"output": "5246"
},
{
"input": "24848 15384",
"output": "9449"
},
{
"input": "25727 15982",
"output": "9728"
},
{
"input": "109 90",
"output": "10"
},
{
"input": "1000000000000000000 999999999999999999",
"output": "1"
},
{
"input": "1000000000000000000 999999999999999998",
"output": "1"
},
{
"input": "1009 980",
"output": "10"
},
{
"input": "999999999999999999 999999999999999838",
"output": "0"
},
{
"input": "1000000000000000000 99999999999999800",
"output": "900000000000000061"
},
{
"input": "8785369357 3377262261",
"output": "5408107058"
},
{
"input": "110 109",
"output": "0"
},
{
"input": "999 777",
"output": "200"
},
{
"input": "327170000015578 77230000029054",
"output": "249939999986479"
},
{
"input": "12515000022229 1791000022317",
"output": "10723999999880"
},
{
"input": "9999999999999 9999999999882",
"output": "10"
},
{
"input": "213 196",
"output": "14"
},
{
"input": "92 82",
"output": "0"
},
{
"input": "148 136",
"output": "0"
},
{
"input": "8 9",
"output": "0"
},
{
"input": "309 299",
"output": "0"
},
{
"input": "9999 9963",
"output": "10"
},
{
"input": "82 81",
"output": "0"
},
{
"input": "9999999 9999936",
"output": "10"
},
{
"input": "171 155",
"output": "2"
},
{
"input": "999 972",
"output": "10"
},
{
"input": "999999999999 999999999891",
"output": "10"
},
{
"input": "9 9",
"output": "0"
},
{
"input": "6900 6885",
"output": "1"
},
{
"input": "96 57",
"output": "27"
},
{
"input": "5 4",
"output": "0"
},
{
"input": "17386 5814",
"output": "11557"
},
{
"input": "493679757404593 316259583979965",
"output": "177420173424564"
},
{
"input": "18474 9478",
"output": "8975"
},
{
"input": "270091571496186 250931112649966",
"output": "19160458846177"
},
{
"input": "565751690089037 381448507916936",
"output": "184303182172038"
},
{
"input": "19 10",
"output": "0"
}
] | 1,694,528,809 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 30 | 0 | n, s = tuple(map(int,input().split()))
def get_first_after(n):
for i in range(1, 9):
if (n + i) % 9 == 0:
return n + i
if s % 9 == 0:
ans = n - (s / 9) * 10
print(max(0, int(ans + 1)))
else:
s = get_first_after(s)
ans = n - (s / 9) * 10
print(max(0, int(ans + 1)))
| Title: Really Big Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ivan likes to learn different things about numbers, but he is especially interested in really big numbers. Ivan thinks that a positive integer number *x* is really big if the difference between *x* and the sum of its digits (in decimal representation) is not less than *s*. To prove that these numbers may have different special properties, he wants to know how rare (or not rare) they are — in fact, he needs to calculate the quantity of really big numbers that are not greater than *n*.
Ivan tried to do the calculations himself, but soon realized that it's too difficult for him. So he asked you to help him in calculations.
Input Specification:
The first (and the only) line contains two integers *n* and *s* (1<=≤<=*n*,<=*s*<=≤<=1018).
Output Specification:
Print one integer — the quantity of really big numbers that are not greater than *n*.
Demo Input:
['12 1\n', '25 20\n', '10 9\n']
Demo Output:
['3\n', '0\n', '1\n']
Note:
In the first example numbers 10, 11 and 12 are really big.
In the second example there are no really big numbers that are not greater than 25 (in fact, the first really big number is 30: 30 - 3 ≥ 20).
In the third example 10 is the only really big number (10 - 1 ≥ 9). | ```python
n, s = tuple(map(int,input().split()))
def get_first_after(n):
for i in range(1, 9):
if (n + i) % 9 == 0:
return n + i
if s % 9 == 0:
ans = n - (s / 9) * 10
print(max(0, int(ans + 1)))
else:
s = get_first_after(s)
ans = n - (s / 9) * 10
print(max(0, int(ans + 1)))
``` | 0 | |
333 | B | Chips | PROGRAMMING | 1,800 | [
"greedy"
] | null | null | Gerald plays the following game. He has a checkered field of size *n*<=×<=*n* cells, where *m* various cells are banned. Before the game, he has to put a few chips on some border (but not corner) board cells. Then for *n*<=-<=1 minutes, Gerald every minute moves each chip into an adjacent cell. He moves each chip from its original edge to the opposite edge. Gerald loses in this game in each of the three cases:
- At least one of the chips at least once fell to the banned cell. - At least once two chips were on the same cell. - At least once two chips swapped in a minute (for example, if you stand two chips on two opposite border cells of a row with even length, this situation happens in the middle of the row).
In that case he loses and earns 0 points. When nothing like that happened, he wins and earns the number of points equal to the number of chips he managed to put on the board. Help Gerald earn the most points. | The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=1000, 0<=≤<=*m*<=≤<=105) — the size of the field and the number of banned cells. Next *m* lines each contain two space-separated integers. Specifically, the *i*-th of these lines contains numbers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*) — the coordinates of the *i*-th banned cell. All given cells are distinct.
Consider the field rows numbered from top to bottom from 1 to *n*, and the columns — from left to right from 1 to *n*. | Print a single integer — the maximum points Gerald can earn in this game. | [
"3 1\n2 2\n",
"3 0\n",
"4 3\n3 1\n3 2\n3 3\n"
] | [
"0\n",
"1\n",
"1\n"
] | In the first test the answer equals zero as we can't put chips into the corner cells.
In the second sample we can place one chip into either cell (1, 2), or cell (3, 2), or cell (2, 1), or cell (2, 3). We cannot place two chips.
In the third sample we can only place one chip into either cell (2, 1), or cell (2, 4). | 1,000 | [
{
"input": "3 1\n2 2",
"output": "0"
},
{
"input": "3 0",
"output": "1"
},
{
"input": "4 3\n3 1\n3 2\n3 3",
"output": "1"
},
{
"input": "2 1\n1 1",
"output": "0"
},
{
"input": "2 3\n1 2\n2 1\n2 2",
"output": "0"
},
{
"input": "5 1\n3 2",
"output": "4"
},
{
"input": "5 1\n2 3",
"output": "4"
},
{
"input": "1000 0",
"output": "1996"
},
{
"input": "999 0",
"output": "1993"
},
{
"input": "5 5\n3 2\n5 4\n3 3\n2 3\n1 2",
"output": "1"
},
{
"input": "5 5\n3 2\n1 4\n5 1\n4 5\n3 1",
"output": "2"
},
{
"input": "5 5\n2 2\n5 3\n2 3\n5 1\n4 4",
"output": "1"
},
{
"input": "6 5\n2 6\n6 5\n3 1\n2 2\n1 2",
"output": "4"
},
{
"input": "6 5\n2 6\n5 2\n4 3\n6 6\n2 5",
"output": "2"
},
{
"input": "6 5\n2 1\n6 4\n2 2\n4 3\n4 1",
"output": "3"
}
] | 1,488,614,318 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 9 | 1,000 | 28,672,000 | n, m = map(int, input().split())
ans = 0
row = [1] * n * 2
for x in range(m):
a, b = [int(x) for x in input().split()]
row[a - 1] = 0
row[b - 1 + n] = 0
for i in range(1, (n + 1) // 2):
j = n - 1 - i
if i == j:
ans += min(1, row[i] + row[i + n])
else:
ans += row[i] + row[j] + row[i + n] + row[j + n]
print (ans) | Title: Chips
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Gerald plays the following game. He has a checkered field of size *n*<=×<=*n* cells, where *m* various cells are banned. Before the game, he has to put a few chips on some border (but not corner) board cells. Then for *n*<=-<=1 minutes, Gerald every minute moves each chip into an adjacent cell. He moves each chip from its original edge to the opposite edge. Gerald loses in this game in each of the three cases:
- At least one of the chips at least once fell to the banned cell. - At least once two chips were on the same cell. - At least once two chips swapped in a minute (for example, if you stand two chips on two opposite border cells of a row with even length, this situation happens in the middle of the row).
In that case he loses and earns 0 points. When nothing like that happened, he wins and earns the number of points equal to the number of chips he managed to put on the board. Help Gerald earn the most points.
Input Specification:
The first line contains two space-separated integers *n* and *m* (2<=≤<=*n*<=≤<=1000, 0<=≤<=*m*<=≤<=105) — the size of the field and the number of banned cells. Next *m* lines each contain two space-separated integers. Specifically, the *i*-th of these lines contains numbers *x**i* and *y**i* (1<=≤<=*x**i*,<=*y**i*<=≤<=*n*) — the coordinates of the *i*-th banned cell. All given cells are distinct.
Consider the field rows numbered from top to bottom from 1 to *n*, and the columns — from left to right from 1 to *n*.
Output Specification:
Print a single integer — the maximum points Gerald can earn in this game.
Demo Input:
['3 1\n2 2\n', '3 0\n', '4 3\n3 1\n3 2\n3 3\n']
Demo Output:
['0\n', '1\n', '1\n']
Note:
In the first test the answer equals zero as we can't put chips into the corner cells.
In the second sample we can place one chip into either cell (1, 2), or cell (3, 2), or cell (2, 1), or cell (2, 3). We cannot place two chips.
In the third sample we can only place one chip into either cell (2, 1), or cell (2, 4). | ```python
n, m = map(int, input().split())
ans = 0
row = [1] * n * 2
for x in range(m):
a, b = [int(x) for x in input().split()]
row[a - 1] = 0
row[b - 1 + n] = 0
for i in range(1, (n + 1) // 2):
j = n - 1 - i
if i == j:
ans += min(1, row[i] + row[i + n])
else:
ans += row[i] + row[j] + row[i + n] + row[j + n]
print (ans)
``` | 0 | |
152 | B | Steps | PROGRAMMING | 1,300 | [
"binary search",
"implementation"
] | null | null | One day Vasya went out for a walk in the yard but there weren't any of his friends outside and he had no one to play touch and run. But the boy didn't lose the high spirits and decided to play touch and run with himself. You may ask: "How did he do that?" The answer is simple.
Vasya noticed that the yard is a rectangular *n*<=×<=*m* field. The squares have coordinates (*x*,<=*y*) (1<=≤<=*x*<=≤<=*n*,<=1<=≤<=*y*<=≤<=*m*), where *x* is the index of the row and *y* is the index of the column.
Initially Vasya stands in the square with coordinates (*x**c*,<=*y**c*). To play, he has got a list of *k* vectors (*dx**i*,<=*dy**i*) of non-zero length. The game goes like this. The boy considers all vectors in the order from 1 to *k*, and consecutively chooses each vector as the current one. After the boy has chosen a current vector, he makes the maximally possible number of valid steps in the vector's direction (it is possible that he makes zero steps).
A step is defined as one movement from the square where the boy is standing now, in the direction of the current vector. That is, if Vasya is positioned in square (*x*,<=*y*), and the current vector is (*dx*,<=*dy*), one step moves Vasya to square (*x*<=+<=*dx*,<=*y*<=+<=*dy*). A step is considered valid, if the boy does not go out of the yard if he performs the step.
Vasya stepped on and on, on and on until he ran out of vectors in his list. Ha had been stepping for so long that he completely forgot how many steps he had made. Help the boy and count how many steps he had made. | The first input line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=109) — the yard's sizes. The second line contains integers *x**c* and *y**c* — the initial square's coordinates (1<=≤<=*x**c*<=≤<=*n*,<=1<=≤<=*y**c*<=≤<=*m*).
The third line contains an integer *k* (1<=≤<=*k*<=≤<=104) — the number of vectors. Then follow *k* lines, each of them contains two integers *dx**i* and *dy**i* (|*dx**i*|,<=|*dy**i*|<=≤<=109,<=|*dx*|<=+<=|*dy*|<=≥<=1). | Print the single number — the number of steps Vasya had made.
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator. | [
"4 5\n1 1\n3\n1 1\n1 1\n0 -2\n",
"10 10\n1 2\n1\n-1 0\n"
] | [
"4\n",
"0\n"
] | In the first sample Vasya is initially positioned at square (1, 1) and makes 3 steps by the first vector (1, 1). So, he consecutively visits the squares (2, 2), (3, 3), (4, 4). Then he makes 0 steps by the second vector (1, 1). He makes 1 more step by the third vector (0, - 2) and he ends up in square (4, 2). Overall, Vasya makes 4 steps.
In the second sample Vasya is initially positioned in square (1, 2) and makes 0 steps by vector ( - 1, 0), as the square with coordinates (0, 2) is located outside the yard. | 1,000 | [
{
"input": "4 5\n1 1\n3\n1 1\n1 1\n0 -2",
"output": "4"
},
{
"input": "10 10\n1 2\n1\n-1 0",
"output": "0"
},
{
"input": "10 20\n10 3\n10\n-2 -6\n-1 0\n-8 0\n0 5\n-1 3\n16 -16\n-1 9\n0 -18\n9 -1\n-9 5",
"output": "13"
},
{
"input": "20 10\n14 4\n10\n6 0\n-7 -7\n12 -2\n-4 9\n20 3\n-1 -16\n0 2\n-1 1\n20 0\n-1 1",
"output": "4"
},
{
"input": "1000 2000\n226 566\n20\n0 -100\n-100 100\n100 0\n42 0\n-100 -79\n-66 -16\n0 -7\n-1 0\n0 100\n100 91\n99 0\n1 0\n-100 0\n70 -100\n-100 100\n100 1\n66 0\n-100 1\n-47 -100\n-42 0",
"output": "708"
},
{
"input": "1 1\n1 1\n1\n1 1",
"output": "0"
},
{
"input": "3 3\n2 2\n1\n2 0",
"output": "0"
}
] | 1,604,981,105 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 2 | 186 | 0 | n, m = map(int, input().split())
x, y = map(int, input().split())
length = int(input())
steps = 0
for i in range(length):
dx, dy = map(int, input().split())
while y + dy>=1 and y + dy<=n and dx + x>=1 and dx + x<=m:
x+= dx
y+= dy
steps+=1
print(steps)
| Title: Steps
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya went out for a walk in the yard but there weren't any of his friends outside and he had no one to play touch and run. But the boy didn't lose the high spirits and decided to play touch and run with himself. You may ask: "How did he do that?" The answer is simple.
Vasya noticed that the yard is a rectangular *n*<=×<=*m* field. The squares have coordinates (*x*,<=*y*) (1<=≤<=*x*<=≤<=*n*,<=1<=≤<=*y*<=≤<=*m*), where *x* is the index of the row and *y* is the index of the column.
Initially Vasya stands in the square with coordinates (*x**c*,<=*y**c*). To play, he has got a list of *k* vectors (*dx**i*,<=*dy**i*) of non-zero length. The game goes like this. The boy considers all vectors in the order from 1 to *k*, and consecutively chooses each vector as the current one. After the boy has chosen a current vector, he makes the maximally possible number of valid steps in the vector's direction (it is possible that he makes zero steps).
A step is defined as one movement from the square where the boy is standing now, in the direction of the current vector. That is, if Vasya is positioned in square (*x*,<=*y*), and the current vector is (*dx*,<=*dy*), one step moves Vasya to square (*x*<=+<=*dx*,<=*y*<=+<=*dy*). A step is considered valid, if the boy does not go out of the yard if he performs the step.
Vasya stepped on and on, on and on until he ran out of vectors in his list. Ha had been stepping for so long that he completely forgot how many steps he had made. Help the boy and count how many steps he had made.
Input Specification:
The first input line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=109) — the yard's sizes. The second line contains integers *x**c* and *y**c* — the initial square's coordinates (1<=≤<=*x**c*<=≤<=*n*,<=1<=≤<=*y**c*<=≤<=*m*).
The third line contains an integer *k* (1<=≤<=*k*<=≤<=104) — the number of vectors. Then follow *k* lines, each of them contains two integers *dx**i* and *dy**i* (|*dx**i*|,<=|*dy**i*|<=≤<=109,<=|*dx*|<=+<=|*dy*|<=≥<=1).
Output Specification:
Print the single number — the number of steps Vasya had made.
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specificator.
Demo Input:
['4 5\n1 1\n3\n1 1\n1 1\n0 -2\n', '10 10\n1 2\n1\n-1 0\n']
Demo Output:
['4\n', '0\n']
Note:
In the first sample Vasya is initially positioned at square (1, 1) and makes 3 steps by the first vector (1, 1). So, he consecutively visits the squares (2, 2), (3, 3), (4, 4). Then he makes 0 steps by the second vector (1, 1). He makes 1 more step by the third vector (0, - 2) and he ends up in square (4, 2). Overall, Vasya makes 4 steps.
In the second sample Vasya is initially positioned in square (1, 2) and makes 0 steps by vector ( - 1, 0), as the square with coordinates (0, 2) is located outside the yard. | ```python
n, m = map(int, input().split())
x, y = map(int, input().split())
length = int(input())
steps = 0
for i in range(length):
dx, dy = map(int, input().split())
while y + dy>=1 and y + dy<=n and dx + x>=1 and dx + x<=m:
x+= dx
y+= dy
steps+=1
print(steps)
``` | 0 | |
295 | A | Greg and Array | PROGRAMMING | 1,400 | [
"data structures",
"implementation"
] | null | null | Greg has an array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n* and *m* operations. Each operation looks as: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). To apply operation *i* to the array means to increase all array elements with numbers *l**i*,<=*l**i*<=+<=1,<=...,<=*r**i* by value *d**i*.
Greg wrote down *k* queries on a piece of paper. Each query has the following form: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*). That means that one should apply operations with numbers *x**i*,<=*x**i*<=+<=1,<=...,<=*y**i* to the array.
Now Greg is wondering, what the array *a* will be after all the queries are executed. Help Greg. | The first line contains integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=105). The second line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=105) — the initial array.
Next *m* lines contain operations, the operation number *i* is written as three integers: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*), (0<=≤<=*d**i*<=≤<=105).
Next *k* lines contain the queries, the query number *i* is written as two integers: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*).
The numbers in the lines are separated by single spaces. | On a single line print *n* integers *a*1,<=*a*2,<=...,<=*a**n* — the array after executing all the queries. Separate the printed numbers by spaces.
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 of the %I64d specifier. | [
"3 3 3\n1 2 3\n1 2 1\n1 3 2\n2 3 4\n1 2\n1 3\n2 3\n",
"1 1 1\n1\n1 1 1\n1 1\n",
"4 3 6\n1 2 3 4\n1 2 1\n2 3 2\n3 4 4\n1 2\n1 3\n2 3\n1 2\n1 3\n2 3\n"
] | [
"9 18 17\n",
"2\n",
"5 18 31 20\n"
] | none | 500 | [
{
"input": "3 3 3\n1 2 3\n1 2 1\n1 3 2\n2 3 4\n1 2\n1 3\n2 3",
"output": "9 18 17"
},
{
"input": "1 1 1\n1\n1 1 1\n1 1",
"output": "2"
},
{
"input": "4 3 6\n1 2 3 4\n1 2 1\n2 3 2\n3 4 4\n1 2\n1 3\n2 3\n1 2\n1 3\n2 3",
"output": "5 18 31 20"
},
{
"input": "1 1 1\n0\n1 1 0\n1 1",
"output": "0"
}
] | 1,649,032,935 | 2,147,483,647 | Python 3 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
int m = scan.nextInt();
int k = scan.nextInt();
long[] arr = new long[n + 1];
for (int i = 0; i < n; i++) {
arr[i] = scan.nextInt();
}
Node[] nodes = new Node[m + 1];
for (int i = 0; i < m; i++) {
int l = scan.nextInt();
int r = scan.nextInt();
int d = scan.nextInt();
Node cur = new Node(l,r,d);
nodes[i] = cur;
}
long[] temp = new long[m+1];
for (int i = 0; i < k; i++) {
int x = scan.nextInt();
int y = scan.nextInt();
temp[x]++;
if (y + 1 <= m) {
temp[y+1]--;
}
}
for(int i = 1; i <= m; i++){
temp[i] = temp[i] + temp[i-1];
}
long[] sum = new long[n+1];
for(int i = 1; i <= m; i++){
int x = nodes[i].l;
int y = nodes[i].r;
int d=nodes[i].d;
sum[x] = sum[x] + d*temp[i];
if(y+1<=n){
sum[y+1] = sum[y+1] - d*temp[i];
}
}
for(int i=1;i<=n;i++){
sum[i] += sum[i-1];
}
for(int i=1;i<=n;i++) {
System.out.print((arr[i]+sum[i])+" ");
}
}
public static class Node {
int l;
int r;
int d;
Node(int l, int r, int d) {
this.l = l;
this.r = r;
this.d = d;
}
}
}
| Title: Greg and Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Greg has an array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n* and *m* operations. Each operation looks as: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). To apply operation *i* to the array means to increase all array elements with numbers *l**i*,<=*l**i*<=+<=1,<=...,<=*r**i* by value *d**i*.
Greg wrote down *k* queries on a piece of paper. Each query has the following form: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*). That means that one should apply operations with numbers *x**i*,<=*x**i*<=+<=1,<=...,<=*y**i* to the array.
Now Greg is wondering, what the array *a* will be after all the queries are executed. Help Greg.
Input Specification:
The first line contains integers *n*, *m*, *k* (1<=≤<=*n*,<=*m*,<=*k*<=≤<=105). The second line contains *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=105) — the initial array.
Next *m* lines contain operations, the operation number *i* is written as three integers: *l**i*, *r**i*, *d**i*, (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*), (0<=≤<=*d**i*<=≤<=105).
Next *k* lines contain the queries, the query number *i* is written as two integers: *x**i*, *y**i*, (1<=≤<=*x**i*<=≤<=*y**i*<=≤<=*m*).
The numbers in the lines are separated by single spaces.
Output Specification:
On a single line print *n* integers *a*1,<=*a*2,<=...,<=*a**n* — the array after executing all the queries. Separate the printed numbers by spaces.
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 of the %I64d specifier.
Demo Input:
['3 3 3\n1 2 3\n1 2 1\n1 3 2\n2 3 4\n1 2\n1 3\n2 3\n', '1 1 1\n1\n1 1 1\n1 1\n', '4 3 6\n1 2 3 4\n1 2 1\n2 3 2\n3 4 4\n1 2\n1 3\n2 3\n1 2\n1 3\n2 3\n']
Demo Output:
['9 18 17\n', '2\n', '5 18 31 20\n']
Note:
none | ```python
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
int m = scan.nextInt();
int k = scan.nextInt();
long[] arr = new long[n + 1];
for (int i = 0; i < n; i++) {
arr[i] = scan.nextInt();
}
Node[] nodes = new Node[m + 1];
for (int i = 0; i < m; i++) {
int l = scan.nextInt();
int r = scan.nextInt();
int d = scan.nextInt();
Node cur = new Node(l,r,d);
nodes[i] = cur;
}
long[] temp = new long[m+1];
for (int i = 0; i < k; i++) {
int x = scan.nextInt();
int y = scan.nextInt();
temp[x]++;
if (y + 1 <= m) {
temp[y+1]--;
}
}
for(int i = 1; i <= m; i++){
temp[i] = temp[i] + temp[i-1];
}
long[] sum = new long[n+1];
for(int i = 1; i <= m; i++){
int x = nodes[i].l;
int y = nodes[i].r;
int d=nodes[i].d;
sum[x] = sum[x] + d*temp[i];
if(y+1<=n){
sum[y+1] = sum[y+1] - d*temp[i];
}
}
for(int i=1;i<=n;i++){
sum[i] += sum[i-1];
}
for(int i=1;i<=n;i++) {
System.out.print((arr[i]+sum[i])+" ");
}
}
public static class Node {
int l;
int r;
int d;
Node(int l, int r, int d) {
this.l = l;
this.r = r;
this.d = d;
}
}
}
``` | -1 | |
320 | B | Ping-Pong (Easy Version) | PROGRAMMING | 1,500 | [
"dfs and similar",
"graphs"
] | null | null | In this problem at each moment you have a set of intervals. You can move from interval (*a*,<=*b*) from our set to interval (*c*,<=*d*) from our set if and only if *c*<=<<=*a*<=<<=*d* or *c*<=<<=*b*<=<<=*d*. Also there is a path from interval *I*1 from our set to interval *I*2 from our set if there is a sequence of successive moves starting from *I*1 so that we can reach *I*2.
Your program should handle the queries of the following two types:
1. "1 x y" (*x*<=<<=*y*) — add the new interval (*x*,<=*y*) to the set of intervals. The length of the new interval is guaranteed to be strictly greater than all the previous intervals.1. "2 a b" (*a*<=≠<=*b*) — answer the question: is there a path from *a*-th (one-based) added interval to *b*-th (one-based) added interval?
Answer all the queries. Note, that initially you have an empty set of intervals. | The first line of the input contains integer *n* denoting the number of queries, (1<=≤<=*n*<=≤<=100). Each of the following lines contains a query as described above. All numbers in the input are integers and don't exceed 109 by their absolute value.
It's guaranteed that all queries are correct. | For each query of the second type print "YES" or "NO" on a separate line depending on the answer. | [
"5\n1 1 5\n1 5 11\n2 1 2\n1 2 9\n2 1 2\n"
] | [
"NO\nYES\n"
] | none | 1,000 | [
{
"input": "5\n1 1 5\n1 5 11\n2 1 2\n1 2 9\n2 1 2",
"output": "NO\nYES"
},
{
"input": "10\n1 -311 -186\n1 -1070 -341\n1 -1506 -634\n1 688 1698\n2 2 4\n1 70 1908\n2 1 2\n2 2 4\n1 -1053 1327\n2 5 4",
"output": "NO\nNO\nNO\nYES"
},
{
"input": "10\n1 -1365 -865\n1 1244 1834\n2 1 2\n1 -1508 -752\n2 3 2\n2 2 1\n1 -779 595\n1 -1316 877\n2 2 1\n1 -698 1700",
"output": "NO\nNO\nNO\nNO"
},
{
"input": "20\n1 1208 1583\n1 -258 729\n1 -409 1201\n1 194 1938\n1 -958 1575\n1 -1466 1752\n2 1 2\n2 1 2\n2 6 5\n1 -1870 1881\n1 -2002 2749\n1 -2002 2984\n1 -2002 3293\n2 2 4\n2 8 10\n2 9 6\n1 -2002 3572\n1 -2002 4175\n1 -2002 4452\n1 -2002 4605",
"output": "YES\nYES\nYES\nYES\nYES\nNO"
},
{
"input": "9\n1 1 4\n1 5 20\n1 11 30\n1 29 60\n1 59 100\n1 100 200\n2 1 5\n2 1 6\n2 2 5",
"output": "NO\nNO\nYES"
}
] | 1,621,488,090 | 2,147,483,647 | Python 3 | OK | TESTS | 24 | 154 | 0 |
n = 0;
def dfs(i):
vis[i] = 1;
for k in range(n+1):
if vis[k] == 1:
continue;
if (a[i] > a[k] and a[i] < b[k]):
dfs(k);
elif(b[i] > a[k] and b[i] < b[k]):
dfs(k);
k = int(input());
a = [0 for i in range(k+1)]
b = [0 for i in range(k+1)]
vis = [0 for i in range(k+1)]
for i in range(1, k+1):
m, x, y = map(int, input().split());
if m == 1:
n += 1
a[n] = x
b[n] = y
else:
for j in range(k+1):
vis[j] = 0
dfs(x)
if(vis[y] == 1):
print("YES")
else:
print("NO")
| Title: Ping-Pong (Easy Version)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem at each moment you have a set of intervals. You can move from interval (*a*,<=*b*) from our set to interval (*c*,<=*d*) from our set if and only if *c*<=<<=*a*<=<<=*d* or *c*<=<<=*b*<=<<=*d*. Also there is a path from interval *I*1 from our set to interval *I*2 from our set if there is a sequence of successive moves starting from *I*1 so that we can reach *I*2.
Your program should handle the queries of the following two types:
1. "1 x y" (*x*<=<<=*y*) — add the new interval (*x*,<=*y*) to the set of intervals. The length of the new interval is guaranteed to be strictly greater than all the previous intervals.1. "2 a b" (*a*<=≠<=*b*) — answer the question: is there a path from *a*-th (one-based) added interval to *b*-th (one-based) added interval?
Answer all the queries. Note, that initially you have an empty set of intervals.
Input Specification:
The first line of the input contains integer *n* denoting the number of queries, (1<=≤<=*n*<=≤<=100). Each of the following lines contains a query as described above. All numbers in the input are integers and don't exceed 109 by their absolute value.
It's guaranteed that all queries are correct.
Output Specification:
For each query of the second type print "YES" or "NO" on a separate line depending on the answer.
Demo Input:
['5\n1 1 5\n1 5 11\n2 1 2\n1 2 9\n2 1 2\n']
Demo Output:
['NO\nYES\n']
Note:
none | ```python
n = 0;
def dfs(i):
vis[i] = 1;
for k in range(n+1):
if vis[k] == 1:
continue;
if (a[i] > a[k] and a[i] < b[k]):
dfs(k);
elif(b[i] > a[k] and b[i] < b[k]):
dfs(k);
k = int(input());
a = [0 for i in range(k+1)]
b = [0 for i in range(k+1)]
vis = [0 for i in range(k+1)]
for i in range(1, k+1):
m, x, y = map(int, input().split());
if m == 1:
n += 1
a[n] = x
b[n] = y
else:
for j in range(k+1):
vis[j] = 0
dfs(x)
if(vis[y] == 1):
print("YES")
else:
print("NO")
``` | 3 | |
25 | A | IQ test | PROGRAMMING | 1,300 | [
"brute force"
] | A. IQ test | 2 | 256 | Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness. | The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness. | Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order. | [
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] | [
"3\n",
"2\n"
] | none | 0 | [
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "2"
},
{
"input": "4\n75 13 94 77",
"output": "3"
},
{
"input": "4\n97 8 27 3",
"output": "2"
},
{
"input": "10\n95 51 12 91 85 3 1 31 25 7",
"output": "3"
},
{
"input": "20\n88 96 66 51 14 88 2 92 18 72 18 88 20 30 4 82 90 100 24 46",
"output": "4"
},
{
"input": "30\n20 94 56 50 10 98 52 32 14 22 24 60 4 8 98 46 34 68 82 82 98 90 50 20 78 49 52 94 64 36",
"output": "26"
},
{
"input": "50\n79 27 77 57 37 45 27 49 65 33 57 21 71 19 75 85 65 61 23 97 85 9 23 1 9 3 99 77 77 21 79 69 15 37 15 7 93 81 13 89 91 31 45 93 15 97 55 80 85 83",
"output": "48"
},
{
"input": "60\n46 11 73 65 3 69 3 53 43 53 97 47 55 93 31 75 35 3 9 73 23 31 3 81 91 79 61 21 15 11 11 11 81 7 83 75 39 87 83 59 89 55 93 27 49 67 67 29 1 93 11 17 9 19 35 21 63 31 31 25",
"output": "1"
},
{
"input": "70\n28 42 42 92 64 54 22 38 38 78 62 38 4 38 14 66 4 92 66 58 94 26 4 44 41 88 48 82 44 26 74 44 48 4 16 92 34 38 26 64 94 4 30 78 50 54 12 90 8 16 80 98 28 100 74 50 36 42 92 18 76 98 8 22 2 50 58 50 64 46",
"output": "25"
},
{
"input": "100\n43 35 79 53 13 91 91 45 65 83 57 9 42 39 85 45 71 51 61 59 31 13 63 39 25 21 79 39 91 67 21 61 97 75 93 83 29 79 59 97 11 37 63 51 39 55 91 23 21 17 47 23 35 75 49 5 69 99 5 7 41 17 25 89 15 79 21 63 53 81 43 91 59 91 69 99 85 15 91 51 49 37 65 7 89 81 21 93 61 63 97 93 45 17 13 69 57 25 75 73",
"output": "13"
},
{
"input": "100\n50 24 68 60 70 30 52 22 18 74 68 98 20 82 4 46 26 68 100 78 84 58 74 98 38 88 68 86 64 80 82 100 20 22 98 98 52 6 94 10 48 68 2 18 38 22 22 82 44 20 66 72 36 58 64 6 36 60 4 96 76 64 12 90 10 58 64 60 74 28 90 26 24 60 40 58 2 16 76 48 58 36 82 60 24 44 4 78 28 38 8 12 40 16 38 6 66 24 31 76",
"output": "99"
},
{
"input": "100\n47 48 94 48 14 18 94 36 96 22 12 30 94 20 48 98 40 58 2 94 8 36 98 18 98 68 2 60 76 38 18 100 8 72 100 68 2 86 92 72 58 16 48 14 6 58 72 76 6 88 80 66 20 28 74 62 86 68 90 86 2 56 34 38 56 90 4 8 76 44 32 86 12 98 38 34 54 92 70 94 10 24 82 66 90 58 62 2 32 58 100 22 58 72 2 22 68 72 42 14",
"output": "1"
},
{
"input": "99\n38 20 68 60 84 16 28 88 60 48 80 28 4 92 70 60 46 46 20 34 12 100 76 2 40 10 8 86 6 80 50 66 12 34 14 28 26 70 46 64 34 96 10 90 98 96 56 88 50 74 70 94 2 94 24 66 68 46 22 30 6 10 64 32 88 14 98 100 64 58 50 18 50 50 8 38 8 16 54 2 60 54 62 84 92 98 4 72 66 26 14 88 99 16 10 6 88 56 22",
"output": "93"
},
{
"input": "99\n50 83 43 89 53 47 69 1 5 37 63 87 95 15 55 95 75 89 33 53 89 75 93 75 11 85 49 29 11 97 49 67 87 11 25 37 97 73 67 49 87 43 53 97 43 29 53 33 45 91 37 73 39 49 59 5 21 43 87 35 5 63 89 57 63 47 29 99 19 85 13 13 3 13 43 19 5 9 61 51 51 57 15 89 13 97 41 13 99 79 13 27 97 95 73 33 99 27 23",
"output": "1"
},
{
"input": "98\n61 56 44 30 58 14 20 24 88 28 46 56 96 52 58 42 94 50 46 30 46 80 72 88 68 16 6 60 26 90 10 98 76 20 56 40 30 16 96 20 88 32 62 30 74 58 36 76 60 4 24 36 42 54 24 92 28 14 2 74 86 90 14 52 34 82 40 76 8 64 2 56 10 8 78 16 70 86 70 42 70 74 22 18 76 98 88 28 62 70 36 72 20 68 34 48 80 98",
"output": "1"
},
{
"input": "98\n66 26 46 42 78 32 76 42 26 82 8 12 4 10 24 26 64 44 100 46 94 64 30 18 88 28 8 66 30 82 82 28 74 52 62 80 80 60 94 86 64 32 44 88 92 20 12 74 94 28 34 58 4 22 16 10 94 76 82 58 40 66 22 6 30 32 92 54 16 76 74 98 18 48 48 30 92 2 16 42 84 74 30 60 64 52 50 26 16 86 58 96 79 60 20 62 82 94",
"output": "93"
},
{
"input": "95\n9 31 27 93 17 77 75 9 9 53 89 39 51 99 5 1 11 39 27 49 91 17 27 79 81 71 37 75 35 13 93 4 99 55 85 11 23 57 5 43 5 61 15 35 23 91 3 81 99 85 43 37 39 27 5 67 7 33 75 59 13 71 51 27 15 93 51 63 91 53 43 99 25 47 17 71 81 15 53 31 59 83 41 23 73 25 91 91 13 17 25 13 55 57 29",
"output": "32"
},
{
"input": "100\n91 89 81 45 53 1 41 3 77 93 55 97 55 97 87 27 69 95 73 41 93 21 75 35 53 56 5 51 87 59 91 67 33 3 99 45 83 17 97 47 75 97 7 89 17 99 23 23 81 25 55 97 27 35 69 5 77 35 93 19 55 59 37 21 31 37 49 41 91 53 73 69 7 37 37 39 17 71 7 97 55 17 47 23 15 73 31 39 57 37 9 5 61 41 65 57 77 79 35 47",
"output": "26"
},
{
"input": "99\n38 56 58 98 80 54 26 90 14 16 78 92 52 74 40 30 84 14 44 80 16 90 98 68 26 24 78 72 42 16 84 40 14 44 2 52 50 2 12 96 58 66 8 80 44 52 34 34 72 98 74 4 66 74 56 21 8 38 76 40 10 22 48 32 98 34 12 62 80 68 64 82 22 78 58 74 20 22 48 56 12 38 32 72 6 16 74 24 94 84 26 38 18 24 76 78 98 94 72",
"output": "56"
},
{
"input": "100\n44 40 6 40 56 90 98 8 36 64 76 86 98 76 36 92 6 30 98 70 24 98 96 60 24 82 88 68 86 96 34 42 58 10 40 26 56 10 88 58 70 32 24 28 14 82 52 12 62 36 70 60 52 34 74 30 78 76 10 16 42 94 66 90 70 38 52 12 58 22 98 96 14 68 24 70 4 30 84 98 8 50 14 52 66 34 100 10 28 100 56 48 38 12 38 14 91 80 70 86",
"output": "97"
},
{
"input": "100\n96 62 64 20 90 46 56 90 68 36 30 56 70 28 16 64 94 34 6 32 34 50 94 22 90 32 40 2 72 10 88 38 28 92 20 26 56 80 4 100 100 90 16 74 74 84 8 2 30 20 80 32 16 46 92 56 42 12 96 64 64 42 64 58 50 42 74 28 2 4 36 32 70 50 54 92 70 16 45 76 28 16 18 50 48 2 62 94 4 12 52 52 4 100 70 60 82 62 98 42",
"output": "79"
},
{
"input": "99\n14 26 34 68 90 58 50 36 8 16 18 6 2 74 54 20 36 84 32 50 52 2 26 24 3 64 20 10 54 26 66 44 28 72 4 96 78 90 96 86 68 28 94 4 12 46 100 32 22 36 84 32 44 94 76 94 4 52 12 30 74 4 34 64 58 72 44 16 70 56 54 8 14 74 8 6 58 62 98 54 14 40 80 20 36 72 28 98 20 58 40 52 90 64 22 48 54 70 52",
"output": "25"
},
{
"input": "95\n82 86 30 78 6 46 80 66 74 72 16 24 18 52 52 38 60 36 86 26 62 28 22 46 96 26 94 84 20 46 66 88 76 32 12 86 74 18 34 88 4 48 94 6 58 6 100 82 4 24 88 32 54 98 34 48 6 76 42 88 42 28 100 4 22 2 10 66 82 54 98 20 60 66 38 98 32 47 86 58 6 100 12 46 2 42 8 84 78 28 24 70 34 28 86",
"output": "78"
},
{
"input": "90\n40 50 8 42 76 24 58 42 26 68 20 48 54 12 34 84 14 36 32 88 6 50 96 56 20 92 48 16 40 34 96 46 20 84 30 50 20 98 8 44 96 42 8 76 70 38 84 30 40 88 84 72 2 22 52 58 16 62 100 66 80 40 50 32 14 62 88 72 22 99 76 50 84 82 8 82 98 46 26 40 2 98 18 78 30 72 70 18 34 68",
"output": "70"
},
{
"input": "80\n81 43 87 1 55 43 53 61 27 19 43 13 89 9 33 83 75 55 97 71 91 37 95 5 21 69 81 93 95 69 31 83 55 7 97 7 79 57 8 61 27 85 49 1 15 97 63 79 29 73 41 85 5 41 31 93 67 11 63 59 15 99 91 77 43 69 23 23 81 73 19 1 67 51 1 75 99 67 3 81",
"output": "39"
},
{
"input": "98\n13 83 61 27 35 1 85 95 97 73 95 65 73 45 5 43 27 83 91 19 11 3 85 59 9 39 69 23 45 7 51 85 5 71 5 95 1 51 75 3 43 57 3 11 33 71 21 99 47 41 87 39 71 87 31 85 91 49 83 5 49 85 47 91 55 99 33 23 31 23 23 73 29 77 55 31 25 5 81 49 91 15 15 39 87 5 9 40 69 47 29 33 11 21 49 79 51 83",
"output": "88"
},
{
"input": "3\n100 100 1",
"output": "3"
}
] | 1,625,172,778 | 2,147,483,647 | Python 3 | OK | TESTS | 32 | 154 | 0 | n = int(input())
a = list(map(int, input().split()))
b = []
for i in range(n):
b.append(a[i] % 2)
for i in range(len(b)-1):
if b[i] != b[i+1] and b.count(b[i]) == 1:
print(i + 1)
exit()
print(len(b))
| Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none | ```python
n = int(input())
a = list(map(int, input().split()))
b = []
for i in range(n):
b.append(a[i] % 2)
for i in range(len(b)-1):
if b[i] != b[i+1] and b.count(b[i]) == 1:
print(i + 1)
exit()
print(len(b))
``` | 3.9615 |
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\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,661,807,493 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 30 | 0 | import numpy as np
n = int(input())
l = [0]*3
x = [0]*3
for i in range(n):
l=list(map(lambda x,y:x+y,l,list(map(int,input().split()))))
if l==x:
print("YES")
else:
print("NO")
| Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none | ```python
import numpy as np
n = int(input())
l = [0]*3
x = [0]*3
for i in range(n):
l=list(map(lambda x,y:x+y,l,list(map(int,input().split()))))
if l==x:
print("YES")
else:
print("NO")
``` | -1 |
350 | A | TL | PROGRAMMING | 1,200 | [
"brute force",
"greedy",
"implementation"
] | null | null | Valera wanted to prepare a Codesecrof round. He's already got one problem and he wants to set a time limit (TL) on it.
Valera has written *n* correct solutions. For each correct solution, he knows its running time (in seconds). Valera has also wrote *m* wrong solutions and for each wrong solution he knows its running time (in seconds).
Let's suppose that Valera will set *v* seconds TL in the problem. Then we can say that a solution passes the system testing if its running time is at most *v* seconds. We can also say that a solution passes the system testing with some "extra" time if for its running time, *a* seconds, an inequality 2*a*<=≤<=*v* holds.
As a result, Valera decided to set *v* seconds TL, that the following conditions are met:
1. *v* is a positive integer; 1. all correct solutions pass the system testing; 1. at least one correct solution passes the system testing with some "extra" time; 1. all wrong solutions do not pass the system testing; 1. value *v* is minimum among all TLs, for which points 1, 2, 3, 4 hold.
Help Valera and find the most suitable TL or else state that such TL doesn't exist. | The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains *n* space-separated positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the running time of each of the *n* correct solutions in seconds. The third line contains *m* space-separated positive integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=100) — the running time of each of *m* wrong solutions in seconds. | If there is a valid TL value, print it. Otherwise, print -1. | [
"3 6\n4 5 2\n8 9 6 10 7 11\n",
"3 1\n3 4 5\n6\n"
] | [
"5",
"-1\n"
] | none | 500 | [
{
"input": "3 6\n4 5 2\n8 9 6 10 7 11",
"output": "5"
},
{
"input": "3 1\n3 4 5\n6",
"output": "-1"
},
{
"input": "2 5\n45 99\n49 41 77 83 45",
"output": "-1"
},
{
"input": "50 50\n18 13 5 34 10 36 36 12 15 11 16 17 14 36 23 45 32 24 31 18 24 32 7 1 31 3 49 8 16 23 3 39 47 43 42 38 40 22 41 1 49 47 9 8 19 15 29 30 16 18\n91 58 86 51 94 94 73 84 98 69 74 56 52 80 88 61 53 99 88 50 55 95 65 84 87 79 51 52 69 60 74 73 93 61 73 59 64 56 95 78 86 72 79 70 93 78 54 61 71 50",
"output": "49"
},
{
"input": "55 44\n93 17 74 15 34 16 41 80 26 54 94 94 86 93 20 44 63 72 39 43 67 4 37 49 76 94 5 51 64 74 11 47 77 97 57 30 42 72 71 26 8 14 67 64 49 57 30 23 40 4 76 78 87 78 79\n38 55 17 65 26 7 36 65 48 28 49 93 18 98 31 90 26 57 1 26 88 56 48 56 23 13 8 67 80 2 51 3 21 33 20 54 2 45 21 36 3 98 62 2",
"output": "-1"
},
{
"input": "32 100\n30 8 4 35 18 41 18 12 33 39 39 18 39 19 33 46 45 33 34 27 14 39 40 21 38 9 42 35 27 10 14 14\n65 49 89 64 47 78 59 52 73 51 84 82 88 63 91 99 67 87 53 99 75 47 85 82 58 47 80 50 65 91 83 90 77 52 100 88 97 74 98 99 50 93 65 61 65 65 65 96 61 51 84 67 79 90 92 83 100 100 100 95 80 54 77 51 98 64 74 62 60 96 73 74 94 55 89 60 92 65 74 79 66 81 53 47 71 51 54 85 74 97 68 72 88 94 100 85 65 63 65 90",
"output": "46"
},
{
"input": "1 50\n7\n65 52 99 78 71 19 96 72 80 15 50 94 20 35 79 95 44 41 45 53 77 50 74 66 59 96 26 84 27 48 56 84 36 78 89 81 67 34 79 74 99 47 93 92 90 96 72 28 78 66",
"output": "14"
},
{
"input": "1 1\n4\n9",
"output": "8"
},
{
"input": "1 1\n2\n4",
"output": "-1"
},
{
"input": "22 56\n49 20 42 68 15 46 98 78 82 8 7 33 50 30 75 96 36 88 35 99 19 87\n15 18 81 24 35 89 25 32 23 3 48 24 52 69 18 32 23 61 48 98 50 38 5 17 70 20 38 32 49 54 68 11 51 81 46 22 19 59 29 38 45 83 18 13 91 17 84 62 25 60 97 32 23 13 83 58",
"output": "-1"
},
{
"input": "1 1\n50\n100",
"output": "-1"
},
{
"input": "1 1\n49\n100",
"output": "98"
},
{
"input": "1 1\n100\n100",
"output": "-1"
},
{
"input": "1 1\n99\n100",
"output": "-1"
},
{
"input": "8 4\n1 2 49 99 99 95 78 98\n100 100 100 100",
"output": "99"
},
{
"input": "68 85\n43 55 2 4 72 45 19 56 53 81 18 90 11 87 47 8 94 88 24 4 67 9 21 70 25 66 65 27 46 13 8 51 65 99 37 43 71 59 71 79 32 56 49 43 57 85 95 81 40 28 60 36 72 81 60 40 16 78 61 37 29 26 15 95 70 27 50 97\n6 6 48 72 54 31 1 50 29 64 93 9 29 93 66 63 25 90 52 1 66 13 70 30 24 87 32 90 84 72 44 13 25 45 31 16 92 60 87 40 62 7 20 63 86 78 73 88 5 36 74 100 64 34 9 5 62 29 58 48 81 46 84 56 27 1 60 14 54 88 31 93 62 7 9 69 27 48 10 5 33 10 53 66 2",
"output": "-1"
},
{
"input": "5 100\n1 1 1 1 1\n77 53 38 29 97 33 64 17 78 100 27 12 42 44 20 24 44 68 58 57 65 90 8 24 4 6 74 68 61 43 25 69 8 62 36 85 67 48 69 30 35 41 42 12 87 66 50 92 53 76 38 67 85 7 80 78 53 76 94 8 37 50 4 100 4 71 10 48 34 47 83 42 25 81 64 72 25 51 53 75 43 98 53 77 94 38 81 15 89 91 72 76 7 36 27 41 88 18 19 75",
"output": "2"
},
{
"input": "3 3\n2 3 4\n8 9 10",
"output": "4"
},
{
"input": "2 1\n2 3\n15",
"output": "4"
},
{
"input": "2 1\n2 4\n4",
"output": "-1"
},
{
"input": "2 3\n4 5\n10 11 12",
"output": "8"
},
{
"input": "3 1\n2 3 3\n5",
"output": "4"
},
{
"input": "2 1\n9 10\n100",
"output": "18"
},
{
"input": "3 3\n3 12 15\n7 8 9",
"output": "-1"
},
{
"input": "2 2\n3 5\n7 8",
"output": "6"
},
{
"input": "3 3\n4 5 6\n10 11 12",
"output": "8"
},
{
"input": "3 5\n2 3 3\n6 6 6 6 2",
"output": "-1"
},
{
"input": "3 6\n4 5 3\n8 9 7 10 7 11",
"output": "6"
},
{
"input": "3 6\n4 5 2\n8 9 6 10 7 4",
"output": "-1"
},
{
"input": "2 1\n4 6\n10",
"output": "8"
},
{
"input": "1 2\n1\n3 1",
"output": "-1"
},
{
"input": "2 1\n5 6\n20",
"output": "10"
},
{
"input": "2 1\n1 5\n5",
"output": "-1"
},
{
"input": "3 2\n10 20 30\n30 40",
"output": "-1"
},
{
"input": "2 2\n5 6\n7 100",
"output": "-1"
},
{
"input": "2 1\n2 5\n7",
"output": "5"
},
{
"input": "1 1\n5\n20",
"output": "10"
},
{
"input": "2 1\n10 11\n100",
"output": "20"
},
{
"input": "1 1\n1\n10",
"output": "2"
},
{
"input": "1 1\n10\n100",
"output": "20"
}
] | 1,435,334,578 | 2,147,483,647 | Python 3 | OK | TESTS | 45 | 122 | 0 | input()
a = [int(i) for i in input().split()]
b = [int(i) for i in input().split()]
v, p, q = min(a), max(a), min(b)
ans = max(2 * v, p)
print(ans if q > ans else -1)
| Title: TL
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera wanted to prepare a Codesecrof round. He's already got one problem and he wants to set a time limit (TL) on it.
Valera has written *n* correct solutions. For each correct solution, he knows its running time (in seconds). Valera has also wrote *m* wrong solutions and for each wrong solution he knows its running time (in seconds).
Let's suppose that Valera will set *v* seconds TL in the problem. Then we can say that a solution passes the system testing if its running time is at most *v* seconds. We can also say that a solution passes the system testing with some "extra" time if for its running time, *a* seconds, an inequality 2*a*<=≤<=*v* holds.
As a result, Valera decided to set *v* seconds TL, that the following conditions are met:
1. *v* is a positive integer; 1. all correct solutions pass the system testing; 1. at least one correct solution passes the system testing with some "extra" time; 1. all wrong solutions do not pass the system testing; 1. value *v* is minimum among all TLs, for which points 1, 2, 3, 4 hold.
Help Valera and find the most suitable TL or else state that such TL doesn't exist.
Input Specification:
The first line contains two integers *n*, *m* (1<=≤<=*n*,<=*m*<=≤<=100). The second line contains *n* space-separated positive integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the running time of each of the *n* correct solutions in seconds. The third line contains *m* space-separated positive integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**i*<=≤<=100) — the running time of each of *m* wrong solutions in seconds.
Output Specification:
If there is a valid TL value, print it. Otherwise, print -1.
Demo Input:
['3 6\n4 5 2\n8 9 6 10 7 11\n', '3 1\n3 4 5\n6\n']
Demo Output:
['5', '-1\n']
Note:
none | ```python
input()
a = [int(i) for i in input().split()]
b = [int(i) for i in input().split()]
v, p, q = min(a), max(a), min(b)
ans = max(2 * v, p)
print(ans if q > ans else -1)
``` | 3 | |
52 | A | 123-sequence | PROGRAMMING | 900 | [
"implementation"
] | A. 123-sequence | 2 | 256 | There is a given sequence of integers *a*1,<=*a*2,<=...,<=*a**n*, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=106). The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3). | Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal. | [
"9\n1 3 2 2 2 1 1 2 3\n"
] | [
"5\n"
] | In the example all the numbers equal to 1 and 3 should be replaced by 2. | 500 | [
{
"input": "9\n1 3 2 2 2 1 1 2 3",
"output": "5"
},
{
"input": "6\n3 3 2 2 1 3",
"output": "3"
},
{
"input": "12\n3 1 3 1 2 1 3 2 2 1 2 1",
"output": "7"
},
{
"input": "15\n3 2 1 1 1 1 3 2 2 3 3 1 2 3 2",
"output": "10"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "2\n3 2",
"output": "1"
},
{
"input": "2\n3 1",
"output": "1"
},
{
"input": "18\n2 3 2 1 2 3 2 1 2 3 3 3 1 2 3 3 3 2",
"output": "10"
},
{
"input": "30\n2 1 3 2 3 2 2 2 2 3 2 2 3 2 1 1 3 1 3 2 1 2 3 1 1 3 3 1 3 1",
"output": "19"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n3",
"output": "0"
}
] | 1,695,534,001 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 23 | 1,122 | 98,611,200 | n=int(input())
l=list(map(int,input().split()))
print(min((l.count(1)+l.count(3)),(l.count(1)+l.count(2)),(l.count(2)+l.count(3))))
| Title: 123-sequence
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
There is a given sequence of integers *a*1,<=*a*2,<=...,<=*a**n*, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=106). The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3).
Output Specification:
Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal.
Demo Input:
['9\n1 3 2 2 2 1 1 2 3\n']
Demo Output:
['5\n']
Note:
In the example all the numbers equal to 1 and 3 should be replaced by 2. | ```python
n=int(input())
l=list(map(int,input().split()))
print(min((l.count(1)+l.count(3)),(l.count(1)+l.count(2)),(l.count(2)+l.count(3))))
``` | 3.535822 |
0 | none | none | none | 0 | [
"none"
] | null | null | You are given an *n*<=×<=*m* rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
we obtain the table:
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good. | The first line contains two integers — *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Next *n* lines contain *m* small English letters each — the characters of the table. | Print a single number — the minimum number of columns that you need to remove in order to make the table good. | [
"1 10\ncodeforces\n",
"4 4\ncase\ncare\ntest\ncode\n",
"5 4\ncode\nforc\nesco\ndefo\nrces\n"
] | [
"0\n",
"2\n",
"4\n"
] | In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings *s* and *t* have equal length. Then, *s* is lexicographically larger than *t* if they are not equal and the character following the largest common prefix of *s* and *t* (the prefix may be empty) in *s* is alphabetically larger than the corresponding character of *t*. | 0 | [
{
"input": "1 10\ncodeforces",
"output": "0"
},
{
"input": "4 4\ncase\ncare\ntest\ncode",
"output": "2"
},
{
"input": "5 4\ncode\nforc\nesco\ndefo\nrces",
"output": "4"
},
{
"input": "2 2\nfb\nye",
"output": "0"
},
{
"input": "5 5\nrzrzh\nrzrzh\nrzrzh\nrzrzh\nrzrzh",
"output": "0"
},
{
"input": "10 10\nddorannorz\nmdrnzqvqgo\ngdtdjmlsuf\neoxbrntqdp\nhribwlslgo\newlqrontvk\nnxibmnawnh\nvxiwdjvdom\nhyhhewmzmp\niysgvzayst",
"output": "1"
},
{
"input": "9 7\nygqartj\nlgwxlqv\nancjjpr\nwnnhkpx\ncnnhvty\nxsfrbqp\nxsolyne\nbsoojiq\nxstetjb",
"output": "1"
},
{
"input": "4 50\nulkteempxafxafcvfwmwhsixwzgbmubcqqceevbbwijeerqbsj\neyqxsievaratndjoekltlqwppfgcukjwxdxexhejbfhzklppkk\npskatxpbjdbmjpwhussetytneohgzxgirluwnbraxtxmaupuid\neappatavdzktqlrjqttmwwroathnulubpjgsjazcycecwmxwvn",
"output": "20"
},
{
"input": "5 50\nvlrkwhvbigkhihwqjpvmohdsszvndheqlmdsspkkxxiedobizr\nmhnzwdefqmttclfxocdmvvtdjtvqhmdllrtrrlnewuqowmtrmp\nrihlhxrqfhpcddslxepesvjqmlqgwyehvxjcsytevujfegeewh\nqrdyiymanvbdjomyruspreihahjhgkcixwowfzczundxqydldq\nkgnrbjlrmkuoiuzeiqwhnyjpuzfnsinqiamlnuzksrdnlvaxjd",
"output": "50"
},
{
"input": "100 1\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\ni\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nv\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx\nx",
"output": "0"
},
{
"input": "1 100\nteloaetuldspjqdlcktjlishwynmjjhlomvemhoyyplbltfwmrlnazbbjvyvwvoxjvvoadkznvxqubgwesoxrznvbdizjdzixecb",
"output": "0"
},
{
"input": "4 100\ngdgmmejiigzsmlarrnfsypvlbutvoxazcigpcospgztqkowfhhbnnbxxrbmwbxwkvxlxzabjjjdtbebedukdelooqlxnadjwjpnp\ndmtsnsbsfdzqvydxcdcnkkfcbseicxhvclsligbhqlkccwujpirymoshkqcosbtlxdypsdqkqaolrqtiibymavcwmbfryttdckhw\njtdmpdljcpciuuoznvqqmafvoqychzfslmwqouuarxctunlzegxsucrwontjplkkxhgixgdbvnewphnatxnwqxqshcexpxlyjuwq\nvhdtvexkyhjmahejbhyfeaompjkdvxmbtdbmvrxrixhnkkjgyvfbxlviatbikfejnqhkrtafftmsgyznpsfxsnzaqdzhxridzilo",
"output": "4"
},
{
"input": "5 2\nab\ndc\ngd\ngc\nhx",
"output": "1"
},
{
"input": "2 1\nb\na",
"output": "1"
},
{
"input": "4 3\nabd\nacc\nbdx\nbcy",
"output": "2"
},
{
"input": "4 2\nad\nac\nbd\nbc",
"output": "1"
},
{
"input": "3 3\naac\nabb\nbba",
"output": "0"
},
{
"input": "3 3\nadc\nbdb\ncda",
"output": "0"
},
{
"input": "4 3\nabz\ndby\ngbx\nhbw",
"output": "0"
},
{
"input": "3 3\naac\nbab\ncaa",
"output": "0"
},
{
"input": "5 6\naaaeaa\nababab\nacacac\nadadad\naeaeae",
"output": "0"
},
{
"input": "3 2\naa\nbc\nba",
"output": "1"
},
{
"input": "3 3\naaz\ndaa\nbaa",
"output": "2"
},
{
"input": "3 3\nabc\nbbb\ncba",
"output": "0"
},
{
"input": "2 3\naxc\nbxa",
"output": "0"
}
] | 1,689,336,186 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | print("_RANDOM_GUESS_1689336185.798442")# 1689336185.798464 | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an *n*<=×<=*m* rectangular table consisting of lower case English letters. In one operation you can completely remove one column from the table. The remaining parts are combined forming a new table. For example, after removing the second column from the table
we obtain the table:
A table is called good if its rows are ordered from top to bottom lexicographically, i.e. each row is lexicographically no larger than the following one. Determine the minimum number of operations of removing a column needed to make a given table good.
Input Specification:
The first line contains two integers — *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Next *n* lines contain *m* small English letters each — the characters of the table.
Output Specification:
Print a single number — the minimum number of columns that you need to remove in order to make the table good.
Demo Input:
['1 10\ncodeforces\n', '4 4\ncase\ncare\ntest\ncode\n', '5 4\ncode\nforc\nesco\ndefo\nrces\n']
Demo Output:
['0\n', '2\n', '4\n']
Note:
In the first sample the table is already good.
In the second sample you may remove the first and third column.
In the third sample you have to remove all the columns (note that the table where all rows are empty is considered good by definition).
Let strings *s* and *t* have equal length. Then, *s* is lexicographically larger than *t* if they are not equal and the character following the largest common prefix of *s* and *t* (the prefix may be empty) in *s* is alphabetically larger than the corresponding character of *t*. | ```python
print("_RANDOM_GUESS_1689336185.798442")# 1689336185.798464
``` | 0 | |
772 | A | Voltage Keepsake | PROGRAMMING | 1,800 | [
"binary search",
"math"
] | null | null | You have *n* devices that you want to use simultaneously.
The *i*-th device uses *a**i* units of power per second. This usage is continuous. That is, in λ seconds, the device will use λ·*a**i* units of power. The *i*-th device currently has *b**i* units of power stored. All devices can store an arbitrary amount of power.
You have a single charger that can plug to any single device. The charger will add *p* units of power per second to a device. This charging is continuous. That is, if you plug in a device for λ seconds, it will gain λ·*p* units of power. You can switch which device is charging at any arbitrary unit of time (including real numbers), and the time it takes to switch is negligible.
You are wondering, what is the maximum amount of time you can use the devices until one of them hits 0 units of power.
If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power. | The first line contains two integers, *n* and *p* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*p*<=≤<=109) — the number of devices and the power of the charger.
This is followed by *n* lines which contain two integers each. Line *i* contains the integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=100<=000) — the power of the device and the amount of power stored in the device in the beginning. | If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.
Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=4.
Namely, let's assume that your answer is *a* and the answer of the jury is *b*. The checker program will consider your answer correct if . | [
"2 1\n2 2\n2 1000\n",
"1 100\n1 1\n",
"3 5\n4 3\n5 2\n6 1\n"
] | [
"2.0000000000",
"-1\n",
"0.5000000000"
] | In sample test 1, you can charge the first device for the entire time until it hits zero power. The second device has enough power to last this time without being charged.
In sample test 2, you can use the device indefinitely.
In sample test 3, we can charge the third device for 2 / 5 of a second, then switch to charge the second device for a 1 / 10 of a second. | 500 | [
{
"input": "2 1\n2 2\n2 1000",
"output": "2.0000000000"
},
{
"input": "1 100\n1 1",
"output": "-1"
},
{
"input": "3 5\n4 3\n5 2\n6 1",
"output": "0.5000000000"
},
{
"input": "1 1\n1 87",
"output": "-1"
},
{
"input": "1 1\n100 77",
"output": "0.7777777778"
},
{
"input": "5 10\n3 81\n3 49\n1 20\n1 12\n1 30",
"output": "-1"
},
{
"input": "5 10\n4 3\n1 54\n2 57\n2 31\n1 99",
"output": "-1"
},
{
"input": "5 10\n2 81\n3 31\n4 49\n1 35\n1 67",
"output": "263.0000000000"
},
{
"input": "10 1\n1 92\n1 92\n1 92\n1 92\n1 92\n1 92\n1 92\n1 92\n1 92\n1 92",
"output": "102.2222222222"
},
{
"input": "10 1\n1 16\n1 16\n1 16\n1 16\n1 16\n1 16\n1 16\n1 16\n1 16\n1 16",
"output": "17.7777777778"
},
{
"input": "10 1\n1 40\n1 40\n1 40\n1 40\n1 40\n1 40\n1 40\n1 40\n1 40\n1 40",
"output": "44.4444444444"
},
{
"input": "2 1\n1 10\n1 10",
"output": "20.0000000000"
},
{
"input": "20 16807\n75250 50074\n43659 8931\n11273 27545\n50879 77924\n37710 64441\n38166 84493\n43043 7988\n22504 82328\n31730 78841\n42613 44304\n33170 17710\n97158 29561\n70934 93100\n80279 51817\n95336 99098\n7827 13513\n29268 23811\n77634 80980\n79150 36580\n58822 11968",
"output": "0.2244225704"
}
] | 1,696,479,639 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 31 | 0 | n, x = map(int, input().split())
sas1 = -x
bob1 = 0
de1 = [list(map(int, input().split())) for _ in range(n)]
for i in de:
sas1 += i[0]
bob1 += i[1]
de.sort(key=lambda k: k[0] / k[1])
for d, e in de:
if bob1 * d < sas1 * e:
sas1 -= d
bob1 -= e
print(bob / sas if sas > 0 else -1)
| Title: Voltage Keepsake
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have *n* devices that you want to use simultaneously.
The *i*-th device uses *a**i* units of power per second. This usage is continuous. That is, in λ seconds, the device will use λ·*a**i* units of power. The *i*-th device currently has *b**i* units of power stored. All devices can store an arbitrary amount of power.
You have a single charger that can plug to any single device. The charger will add *p* units of power per second to a device. This charging is continuous. That is, if you plug in a device for λ seconds, it will gain λ·*p* units of power. You can switch which device is charging at any arbitrary unit of time (including real numbers), and the time it takes to switch is negligible.
You are wondering, what is the maximum amount of time you can use the devices until one of them hits 0 units of power.
If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.
Input Specification:
The first line contains two integers, *n* and *p* (1<=≤<=*n*<=≤<=100<=000, 1<=≤<=*p*<=≤<=109) — the number of devices and the power of the charger.
This is followed by *n* lines which contain two integers each. Line *i* contains the integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=100<=000) — the power of the device and the amount of power stored in the device in the beginning.
Output Specification:
If you can use the devices indefinitely, print -1. Otherwise, print the maximum amount of time before any one device hits 0 power.
Your answer will be considered correct if its absolute or relative error does not exceed 10<=-<=4.
Namely, let's assume that your answer is *a* and the answer of the jury is *b*. The checker program will consider your answer correct if .
Demo Input:
['2 1\n2 2\n2 1000\n', '1 100\n1 1\n', '3 5\n4 3\n5 2\n6 1\n']
Demo Output:
['2.0000000000', '-1\n', '0.5000000000']
Note:
In sample test 1, you can charge the first device for the entire time until it hits zero power. The second device has enough power to last this time without being charged.
In sample test 2, you can use the device indefinitely.
In sample test 3, we can charge the third device for 2 / 5 of a second, then switch to charge the second device for a 1 / 10 of a second. | ```python
n, x = map(int, input().split())
sas1 = -x
bob1 = 0
de1 = [list(map(int, input().split())) for _ in range(n)]
for i in de:
sas1 += i[0]
bob1 += i[1]
de.sort(key=lambda k: k[0] / k[1])
for d, e in de:
if bob1 * d < sas1 * e:
sas1 -= d
bob1 -= e
print(bob / sas if sas > 0 else -1)
``` | -1 | |
129 | A | Cookies | PROGRAMMING | 900 | [
"implementation"
] | null | null | Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even? | The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag. | Print in the only line the only number — the sought number of ways. If there are no such ways print 0. | [
"1\n1\n",
"10\n1 2 2 3 4 4 4 2 2 2\n",
"11\n2 2 2 2 2 2 2 2 2 2 99\n"
] | [
"1\n",
"8\n",
"1\n"
] | In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies.
In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total.
In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies. | 500 | [
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 2 2 3 4 4 4 2 2 2",
"output": "8"
},
{
"input": "11\n2 2 2 2 2 2 2 2 2 2 99",
"output": "1"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n2 2",
"output": "2"
},
{
"input": "2\n1 2",
"output": "1"
},
{
"input": "7\n7 7 7 7 7 7 7",
"output": "7"
},
{
"input": "8\n1 2 3 4 5 6 7 8",
"output": "4"
},
{
"input": "100\n1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2 1 1 1 1 1 2 2 2 2 2",
"output": "50"
},
{
"input": "99\n99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99 100 99",
"output": "49"
},
{
"input": "82\n43 44 96 33 23 42 33 66 53 87 8 90 43 91 40 88 51 18 48 62 59 10 22 20 54 6 13 63 2 56 31 52 98 42 54 32 26 77 9 24 33 91 16 30 39 34 78 82 73 90 12 15 67 76 30 18 44 86 84 98 65 54 100 79 28 34 40 56 11 43 72 35 86 59 89 40 30 33 7 19 44 15",
"output": "50"
},
{
"input": "17\n50 14 17 77 74 74 38 76 41 27 45 29 66 98 38 73 38",
"output": "7"
},
{
"input": "94\n81 19 90 99 26 11 86 44 78 36 80 59 99 90 78 72 71 20 94 56 42 40 71 84 10 85 10 70 52 27 39 55 90 16 48 25 7 79 99 100 38 10 99 56 3 4 78 9 16 57 14 40 52 54 57 70 30 86 56 84 97 60 59 69 49 66 23 92 90 46 86 73 53 47 1 83 14 20 24 66 13 45 41 14 86 75 55 88 48 95 82 24 47 87",
"output": "39"
},
{
"input": "88\n64 95 12 90 40 65 98 45 52 54 79 7 81 25 98 19 68 82 41 53 35 50 5 22 32 21 8 39 8 6 72 27 81 30 12 79 21 42 60 2 66 87 46 93 62 78 52 71 76 32 78 94 86 85 55 15 34 76 41 20 32 26 94 81 89 45 74 49 11 40 40 39 49 46 80 85 90 23 80 40 86 58 70 26 48 93 23 53",
"output": "37"
},
{
"input": "84\n95 9 43 43 13 84 60 90 1 8 97 99 54 34 59 83 33 15 51 26 40 12 66 65 19 30 29 78 92 60 25 13 19 84 71 73 12 24 54 49 16 41 11 40 57 59 34 40 39 9 71 83 1 77 79 53 94 47 78 55 77 85 29 52 80 90 53 77 97 97 27 79 28 23 83 25 26 22 49 86 63 56 3 32",
"output": "51"
},
{
"input": "47\n61 97 76 94 91 22 2 68 62 73 90 47 16 79 44 71 98 68 43 6 53 52 40 27 68 67 43 96 14 91 60 61 96 24 97 13 32 65 85 96 81 77 34 18 23 14 80",
"output": "21"
},
{
"input": "69\n71 1 78 74 58 89 30 6 100 90 22 61 11 59 14 74 27 25 78 61 45 19 25 33 37 4 52 43 53 38 9 100 56 67 69 38 76 91 63 60 93 52 28 61 9 98 8 14 57 63 89 64 98 51 36 66 36 86 13 82 50 91 52 64 86 78 78 83 81",
"output": "37"
},
{
"input": "52\n38 78 36 75 19 3 56 1 39 97 24 79 84 16 93 55 96 64 12 24 1 86 80 29 12 32 36 36 73 39 76 65 53 98 30 20 28 8 86 43 70 22 75 69 62 65 81 25 53 40 71 59",
"output": "28"
},
{
"input": "74\n81 31 67 97 26 75 69 81 11 13 13 74 77 88 52 20 52 64 66 75 72 28 41 54 26 75 41 91 75 15 18 36 13 83 63 61 14 48 53 63 19 67 35 48 23 65 73 100 44 55 92 88 99 17 73 25 83 7 31 89 12 80 98 39 42 75 14 29 81 35 77 87 33 94",
"output": "47"
},
{
"input": "44\n46 56 31 31 37 71 94 2 14 100 45 72 36 72 80 3 38 54 42 98 50 32 31 42 62 31 45 50 95 100 18 17 64 22 18 25 52 56 70 57 43 40 81 28",
"output": "15"
},
{
"input": "22\n28 57 40 74 51 4 45 84 99 12 95 14 92 60 47 81 84 51 31 91 59 42",
"output": "11"
},
{
"input": "59\n73 45 94 76 41 49 65 13 74 66 36 25 47 75 40 23 92 72 11 32 32 8 81 26 68 56 41 8 76 47 96 55 70 11 84 14 83 18 70 22 30 39 28 100 48 11 92 45 78 69 86 1 54 90 98 91 13 17 35",
"output": "33"
},
{
"input": "63\n20 18 44 94 68 57 16 43 74 55 68 24 21 95 76 84 50 50 47 86 86 12 58 55 28 72 86 18 34 45 81 88 3 72 41 9 60 90 81 93 12 6 9 6 2 41 1 7 9 29 81 14 64 80 20 36 67 54 7 5 35 81 22",
"output": "37"
},
{
"input": "28\n49 84 48 19 44 91 11 82 96 95 88 90 71 82 87 25 31 23 18 13 98 45 26 65 35 12 31 14",
"output": "15"
},
{
"input": "61\n34 18 28 64 28 45 9 77 77 20 63 92 79 16 16 100 86 2 91 91 57 15 31 95 10 88 84 5 82 83 53 98 59 17 97 80 76 80 81 3 91 81 87 93 61 46 10 49 6 22 21 75 63 89 21 81 30 19 67 38 77",
"output": "35"
},
{
"input": "90\n41 90 43 1 28 75 90 50 3 70 76 64 81 63 25 69 83 82 29 91 59 66 21 61 7 55 72 49 38 69 72 20 64 58 30 81 61 29 96 14 39 5 100 20 29 98 75 29 44 78 97 45 26 77 73 59 22 99 41 6 3 96 71 20 9 18 96 18 90 62 34 78 54 5 41 6 73 33 2 54 26 21 18 6 45 57 43 73 95 75",
"output": "42"
},
{
"input": "45\n93 69 4 27 20 14 71 48 79 3 32 26 49 30 57 88 13 56 49 61 37 32 47 41 41 70 45 68 82 18 8 6 25 20 15 13 71 99 28 6 52 34 19 59 26",
"output": "23"
},
{
"input": "33\n29 95 48 49 91 10 83 71 47 25 66 36 51 12 34 10 54 74 41 96 89 26 89 1 42 33 1 62 9 32 49 65 78",
"output": "15"
},
{
"input": "34\n98 24 42 36 41 82 28 58 89 34 77 70 76 44 74 54 66 100 13 79 4 88 21 1 11 45 91 29 87 100 29 54 82 78",
"output": "13"
},
{
"input": "29\n91 84 26 84 9 63 52 9 65 56 90 2 36 7 67 33 91 14 65 38 53 36 81 83 85 14 33 95 51",
"output": "17"
},
{
"input": "100\n2 88 92 82 87 100 78 28 84 43 78 32 43 33 97 19 15 52 29 84 57 72 54 13 99 28 82 79 40 70 34 92 91 53 9 88 27 43 14 92 72 37 26 37 20 95 19 34 49 64 33 37 34 27 80 79 9 54 99 68 25 4 68 73 46 66 24 78 3 87 26 52 50 84 4 95 23 83 39 58 86 36 33 16 98 2 84 19 53 12 69 60 10 11 78 17 79 92 77 59",
"output": "45"
},
{
"input": "100\n2 95 45 73 9 54 20 97 57 82 88 26 18 71 25 27 75 54 31 11 58 85 69 75 72 91 76 5 25 80 45 49 4 73 8 81 81 38 5 12 53 77 7 96 90 35 28 80 73 94 19 69 96 17 94 49 69 9 32 19 5 12 46 29 26 40 59 59 6 95 82 50 72 2 45 69 12 5 72 29 39 72 23 96 81 28 28 56 68 58 37 41 30 1 90 84 15 24 96 43",
"output": "53"
},
{
"input": "100\n27 72 35 91 13 10 35 45 24 55 83 84 63 96 29 79 34 67 63 92 48 83 18 77 28 27 49 66 29 88 55 15 6 58 14 67 94 36 77 7 7 64 61 52 71 18 36 99 76 6 50 67 16 13 41 7 89 73 61 51 78 22 78 32 76 100 3 31 89 71 63 53 15 85 77 54 89 33 68 74 3 23 57 5 43 89 75 35 9 86 90 11 31 46 48 37 74 17 77 8",
"output": "40"
},
{
"input": "100\n69 98 69 88 11 49 55 8 25 91 17 81 47 26 15 73 96 71 18 42 42 61 48 14 92 78 35 72 4 27 62 75 83 79 17 16 46 80 96 90 82 54 37 69 85 21 67 70 96 10 46 63 21 59 56 92 54 88 77 30 75 45 44 29 86 100 51 11 65 69 66 56 82 63 27 1 51 51 13 10 3 55 26 85 34 16 87 72 13 100 81 71 90 95 86 50 83 55 55 54",
"output": "53"
},
{
"input": "100\n34 35 99 64 2 66 78 93 20 48 12 79 19 10 87 7 42 92 60 79 5 2 24 89 57 48 63 92 74 4 16 51 7 12 90 48 87 17 18 73 51 58 97 97 25 38 15 97 96 73 67 91 6 75 14 13 87 79 75 3 15 55 35 95 71 45 10 13 20 37 82 26 2 22 13 83 97 84 39 79 43 100 54 59 98 8 61 34 7 65 75 44 24 77 73 88 34 95 44 77",
"output": "55"
},
{
"input": "100\n15 86 3 1 51 26 74 85 37 87 64 58 10 6 57 26 30 47 85 65 24 72 50 40 12 35 91 47 91 60 47 87 95 34 80 91 26 3 36 39 14 86 28 70 51 44 28 21 72 79 57 61 16 71 100 94 57 67 36 74 24 21 89 85 25 2 97 67 76 53 76 80 97 64 35 13 8 32 21 52 62 61 67 14 74 73 66 44 55 76 24 3 43 42 99 61 36 80 38 66",
"output": "52"
},
{
"input": "100\n45 16 54 54 80 94 74 93 75 85 58 95 79 30 81 2 84 4 57 23 92 64 78 1 50 36 13 27 56 54 10 77 87 1 5 38 85 74 94 82 30 45 72 83 82 30 81 82 82 3 69 82 7 92 39 60 94 42 41 5 3 17 67 21 79 44 79 96 28 3 53 68 79 89 63 83 1 44 4 31 84 15 73 77 19 66 54 6 73 1 67 24 91 11 86 45 96 82 20 89",
"output": "51"
},
{
"input": "100\n84 23 50 32 90 71 92 43 58 70 6 82 7 55 85 19 70 89 12 26 29 56 74 30 2 27 4 39 63 67 91 81 11 33 75 10 82 88 39 43 43 80 68 35 55 67 53 62 73 65 86 74 43 51 14 48 42 92 83 57 22 33 24 99 5 27 78 96 7 28 11 15 8 38 85 67 5 92 24 96 57 59 14 95 91 4 9 18 45 33 74 83 64 85 14 51 51 94 29 2",
"output": "53"
},
{
"input": "100\n77 56 56 45 73 55 32 37 39 50 30 95 79 21 44 34 51 43 86 91 39 30 85 15 35 93 100 14 57 31 80 79 38 40 88 4 91 54 7 95 76 26 62 84 17 33 67 47 6 82 69 51 17 2 59 24 11 12 31 90 12 11 55 38 72 49 30 50 42 46 5 97 9 9 30 45 86 23 19 82 40 42 5 40 35 98 35 32 60 60 5 28 84 35 21 49 68 53 68 23",
"output": "48"
},
{
"input": "100\n78 38 79 61 45 86 83 83 86 90 74 69 2 84 73 39 2 5 20 71 24 80 54 89 58 34 77 40 39 62 2 47 28 53 97 75 88 98 94 96 33 71 44 90 47 36 19 89 87 98 90 87 5 85 34 79 82 3 42 88 89 63 35 7 89 30 40 48 12 41 56 76 83 60 80 80 39 56 77 4 72 96 30 55 57 51 7 19 11 1 66 1 91 87 11 62 95 85 79 25",
"output": "48"
},
{
"input": "100\n5 34 23 20 76 75 19 51 17 82 60 13 83 6 65 16 20 43 66 54 87 10 87 73 50 24 16 98 33 28 80 52 54 82 26 92 14 13 84 92 94 29 61 21 60 20 48 94 24 20 75 70 58 27 68 45 86 89 29 8 67 38 83 48 18 100 11 22 46 84 52 97 70 19 50 75 3 7 52 53 72 41 18 31 1 38 49 53 11 64 99 76 9 87 48 12 100 32 44 71",
"output": "58"
},
{
"input": "100\n76 89 68 78 24 72 73 95 98 72 58 15 2 5 56 32 9 65 50 70 94 31 29 54 89 52 31 93 43 56 26 35 72 95 51 55 78 70 11 92 17 5 54 94 81 31 78 95 73 91 95 37 59 9 53 48 65 55 84 8 45 97 64 37 96 34 36 53 66 17 72 48 99 23 27 18 92 84 44 73 60 78 53 29 68 99 19 39 61 40 69 6 77 12 47 29 15 4 8 45",
"output": "53"
},
{
"input": "100\n82 40 31 53 8 50 85 93 3 84 54 17 96 59 51 42 18 19 35 84 79 31 17 46 54 82 72 49 35 73 26 89 61 73 3 50 12 29 25 77 88 21 58 24 22 89 96 54 82 29 96 56 77 16 1 68 90 93 20 23 57 22 31 18 92 90 51 14 50 72 31 54 12 50 66 62 2 34 17 45 68 50 87 97 23 71 1 72 17 82 42 15 20 78 4 49 66 59 10 17",
"output": "54"
},
{
"input": "100\n32 82 82 24 39 53 48 5 29 24 9 37 91 37 91 95 1 97 84 52 12 56 93 47 22 20 14 17 40 22 79 34 24 2 69 30 69 29 3 89 21 46 60 92 39 29 18 24 49 18 40 22 60 13 77 50 39 64 50 70 99 8 66 31 90 38 20 54 7 21 5 56 41 68 69 20 54 89 69 62 9 53 43 89 81 97 15 2 52 78 89 65 16 61 59 42 56 25 32 52",
"output": "49"
},
{
"input": "100\n72 54 23 24 97 14 99 87 15 25 7 23 17 87 72 31 71 87 34 82 51 77 74 85 62 38 24 7 84 48 98 21 29 71 70 84 25 58 67 92 18 44 32 9 81 15 53 29 63 18 86 16 7 31 38 99 70 32 89 16 23 11 66 96 69 82 97 59 6 9 49 80 85 19 6 9 52 51 85 74 53 46 73 55 31 63 78 61 34 80 77 65 87 77 92 52 89 8 52 31",
"output": "44"
},
{
"input": "100\n56 88 8 19 7 15 11 54 35 50 19 57 63 72 51 43 50 19 57 90 40 100 8 92 11 96 30 32 59 65 93 47 62 3 50 41 30 50 72 83 61 46 83 60 20 46 33 1 5 18 83 22 34 16 41 95 63 63 7 59 55 95 91 29 64 60 64 81 45 45 10 9 88 37 69 85 21 82 41 76 42 34 47 78 51 83 65 100 13 22 59 76 63 1 26 86 36 94 99 74",
"output": "46"
},
{
"input": "100\n27 89 67 60 62 80 43 50 28 88 72 5 94 11 63 91 18 78 99 3 71 26 12 97 74 62 23 24 22 3 100 72 98 7 94 32 12 75 61 88 42 48 10 14 45 9 48 56 73 76 70 70 79 90 35 39 96 37 81 11 19 65 99 39 23 79 34 61 35 74 90 37 73 23 46 21 94 84 73 58 11 89 13 9 10 85 42 78 73 32 53 39 49 90 43 5 28 31 97 75",
"output": "53"
},
{
"input": "100\n33 24 97 96 1 14 99 51 13 65 67 20 46 88 42 44 20 49 5 89 98 83 15 40 74 83 58 3 10 79 34 2 69 28 37 100 55 52 14 8 44 94 97 89 6 42 11 28 30 33 55 56 20 57 52 25 75 1 87 42 62 41 37 12 54 85 95 80 42 36 94 96 28 76 54 36 4 17 26 24 62 15 17 79 84 36 92 78 74 91 96 77 54 92 81 91 62 98 37 37",
"output": "43"
},
{
"input": "100\n86 24 61 15 11 85 1 31 47 36 23 36 59 34 3 27 16 29 82 28 58 52 52 66 71 61 98 39 60 20 67 41 67 90 73 29 92 17 70 95 58 98 58 32 21 73 46 56 87 72 80 75 40 27 94 31 59 92 93 37 14 99 96 21 97 23 81 91 52 52 96 94 92 28 38 29 52 16 57 27 17 24 91 21 79 55 96 98 95 94 23 78 79 12 77 35 32 75 100 82",
"output": "51"
},
{
"input": "100\n88 85 41 37 69 21 7 69 36 5 92 26 64 75 22 46 67 20 70 22 62 66 38 24 47 49 68 30 90 31 67 86 86 82 9 51 43 45 48 42 73 44 31 94 45 60 54 66 20 87 11 94 34 32 87 66 56 28 75 39 37 90 72 93 55 72 31 42 30 71 87 61 4 12 12 81 23 61 56 98 71 32 30 33 96 63 92 16 8 78 47 91 47 54 49 3 81 82 41 5",
"output": "47"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "5\n1 1 3 2 2",
"output": "3"
}
] | 1,654,442,356 | 2,147,483,647 | Python 3 | OK | TESTS | 52 | 92 | 0 | n = int(input())
values = input().split()
a = []
all_s = 0
for i in range(n):
s = int(values[i])
a.append(s)
all_s += s
c = 0
for i in range(n):
if (all_s - a[i])%2 == 0:
c += 1
print(c) | Title: Cookies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Olga came to visit the twins Anna and Maria and saw that they have many cookies. The cookies are distributed into bags. As there are many cookies, Olga decided that it's no big deal if she steals a bag. However, she doesn't want the sisters to quarrel because of nothing when they divide the cookies. That's why Olga wants to steal a bag with cookies so that the number of cookies in the remaining bags was even, that is, so that Anna and Maria could evenly divide it into two (even 0 remaining cookies will do, just as any other even number). How many ways there are to steal exactly one cookie bag so that the total number of cookies in the remaining bags was even?
Input Specification:
The first line contains the only integer *n* (1<=≤<=*n*<=≤<=100) — the number of cookie bags Anna and Maria have. The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=100) — the number of cookies in the *i*-th bag.
Output Specification:
Print in the only line the only number — the sought number of ways. If there are no such ways print 0.
Demo Input:
['1\n1\n', '10\n1 2 2 3 4 4 4 2 2 2\n', '11\n2 2 2 2 2 2 2 2 2 2 99\n']
Demo Output:
['1\n', '8\n', '1\n']
Note:
In the first sample Olga should take the only bag so that the twins ended up with the even number of cookies.
In the second sample Olga can take any of five bags with two cookies or any of three bags with four cookies — 5 + 3 = 8 ways in total.
In the third sample, no matter which bag with two cookies Olga chooses, the twins are left with 2 * 9 + 99 = 117 cookies. Thus, Olga has only one option: to take the bag with 99 cookies. | ```python
n = int(input())
values = input().split()
a = []
all_s = 0
for i in range(n):
s = int(values[i])
a.append(s)
all_s += s
c = 0
for i in range(n):
if (all_s - a[i])%2 == 0:
c += 1
print(c)
``` | 3 | |
638 | A | Home Numbers | PROGRAMMING | 1,100 | [
"*special",
"constructive algorithms",
"math"
] | null | null | The main street of Berland is a straight line with *n* houses built along it (*n* is an even number). The houses are located at both sides of the street. The houses with odd numbers are at one side of the street and are numbered from 1 to *n*<=-<=1 in the order from the beginning of the street to the end (in the picture: from left to right). The houses with even numbers are at the other side of the street and are numbered from 2 to *n* in the order from the end of the street to its beginning (in the picture: from right to left). The corresponding houses with even and odd numbers are strictly opposite each other, that is, house 1 is opposite house *n*, house 3 is opposite house *n*<=-<=2, house 5 is opposite house *n*<=-<=4 and so on.
Vasya needs to get to house number *a* as quickly as possible. He starts driving from the beginning of the street and drives his car to house *a*. To get from the beginning of the street to houses number 1 and *n*, he spends exactly 1 second. He also spends exactly one second to drive the distance between two neighbouring houses. Vasya can park at any side of the road, so the distance between the beginning of the street at the houses that stand opposite one another should be considered the same.
Your task is: find the minimum time Vasya needs to reach house *a*. | The first line of the input contains two integers, *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100<=000) — the number of houses on the street and the number of the house that Vasya needs to reach, correspondingly. It is guaranteed that number *n* is even. | Print a single integer — the minimum time Vasya needs to get from the beginning of the street to house *a*. | [
"4 2\n",
"8 5\n"
] | [
"2\n",
"3\n"
] | In the first sample there are only four houses on the street, two houses at each side. House 2 will be the last at Vasya's right.
The second sample corresponds to picture with *n* = 8. House 5 is the one before last at Vasya's left. | 500 | [
{
"input": "4 2",
"output": "2"
},
{
"input": "8 5",
"output": "3"
},
{
"input": "2 1",
"output": "1"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "10 1",
"output": "1"
},
{
"input": "10 10",
"output": "1"
},
{
"input": "100000 100000",
"output": "1"
},
{
"input": "100000 2",
"output": "50000"
},
{
"input": "100000 3",
"output": "2"
},
{
"input": "100000 99999",
"output": "50000"
},
{
"input": "100 100",
"output": "1"
},
{
"input": "3000 34",
"output": "1484"
},
{
"input": "2000 1",
"output": "1"
},
{
"input": "100000 1",
"output": "1"
},
{
"input": "24842 1038",
"output": "11903"
},
{
"input": "1628 274",
"output": "678"
},
{
"input": "16186 337",
"output": "169"
},
{
"input": "24562 2009",
"output": "1005"
},
{
"input": "9456 3443",
"output": "1722"
},
{
"input": "5610 332",
"output": "2640"
},
{
"input": "1764 1288",
"output": "239"
},
{
"input": "28588 13902",
"output": "7344"
},
{
"input": "92480 43074",
"output": "24704"
},
{
"input": "40022 26492",
"output": "6766"
},
{
"input": "85766 64050",
"output": "10859"
},
{
"input": "67808 61809",
"output": "30905"
},
{
"input": "80124 68695",
"output": "34348"
},
{
"input": "95522 91716",
"output": "1904"
},
{
"input": "7752 2915",
"output": "1458"
},
{
"input": "5094 5058",
"output": "19"
},
{
"input": "6144 4792",
"output": "677"
},
{
"input": "34334 20793",
"output": "10397"
},
{
"input": "23538 10243",
"output": "5122"
},
{
"input": "9328 7933",
"output": "3967"
},
{
"input": "11110 9885",
"output": "4943"
},
{
"input": "26096 2778",
"output": "11660"
},
{
"input": "75062 5323",
"output": "2662"
},
{
"input": "94790 7722",
"output": "43535"
},
{
"input": "90616 32240",
"output": "29189"
},
{
"input": "96998 8992",
"output": "44004"
},
{
"input": "95130 19219",
"output": "9610"
},
{
"input": "92586 8812",
"output": "41888"
},
{
"input": "3266 3044",
"output": "112"
},
{
"input": "5026 4697",
"output": "2349"
},
{
"input": "3044 2904",
"output": "71"
},
{
"input": "6022 5396",
"output": "314"
},
{
"input": "31270 25522",
"output": "2875"
},
{
"input": "82156 75519",
"output": "37760"
},
{
"input": "34614 27913",
"output": "13957"
},
{
"input": "88024 61143",
"output": "30572"
},
{
"input": "91870 55672",
"output": "18100"
},
{
"input": "95718 4868",
"output": "45426"
},
{
"input": "99564 358",
"output": "49604"
},
{
"input": "89266 13047",
"output": "6524"
},
{
"input": "90904 16455",
"output": "8228"
},
{
"input": "94750 13761",
"output": "6881"
},
{
"input": "100000 23458",
"output": "38272"
},
{
"input": "100000 23457",
"output": "11729"
},
{
"input": "59140 24272",
"output": "17435"
},
{
"input": "9860 8516",
"output": "673"
},
{
"input": "25988 2733",
"output": "1367"
},
{
"input": "9412 5309",
"output": "2655"
},
{
"input": "25540 23601",
"output": "11801"
},
{
"input": "76260 6050",
"output": "35106"
},
{
"input": "92388 39118",
"output": "26636"
},
{
"input": "8516 5495",
"output": "2748"
},
{
"input": "91940 37847",
"output": "18924"
},
{
"input": "30518 286",
"output": "15117"
},
{
"input": "46646 19345",
"output": "9673"
}
] | 1,696,142,973 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 77 | 0 | n,a=map(int,input().split())
if a%2==0:
print(1+int((n-a)/2))
else:
print(1+int((n-a+1)/2)) | Title: Home Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The main street of Berland is a straight line with *n* houses built along it (*n* is an even number). The houses are located at both sides of the street. The houses with odd numbers are at one side of the street and are numbered from 1 to *n*<=-<=1 in the order from the beginning of the street to the end (in the picture: from left to right). The houses with even numbers are at the other side of the street and are numbered from 2 to *n* in the order from the end of the street to its beginning (in the picture: from right to left). The corresponding houses with even and odd numbers are strictly opposite each other, that is, house 1 is opposite house *n*, house 3 is opposite house *n*<=-<=2, house 5 is opposite house *n*<=-<=4 and so on.
Vasya needs to get to house number *a* as quickly as possible. He starts driving from the beginning of the street and drives his car to house *a*. To get from the beginning of the street to houses number 1 and *n*, he spends exactly 1 second. He also spends exactly one second to drive the distance between two neighbouring houses. Vasya can park at any side of the road, so the distance between the beginning of the street at the houses that stand opposite one another should be considered the same.
Your task is: find the minimum time Vasya needs to reach house *a*.
Input Specification:
The first line of the input contains two integers, *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100<=000) — the number of houses on the street and the number of the house that Vasya needs to reach, correspondingly. It is guaranteed that number *n* is even.
Output Specification:
Print a single integer — the minimum time Vasya needs to get from the beginning of the street to house *a*.
Demo Input:
['4 2\n', '8 5\n']
Demo Output:
['2\n', '3\n']
Note:
In the first sample there are only four houses on the street, two houses at each side. House 2 will be the last at Vasya's right.
The second sample corresponds to picture with *n* = 8. House 5 is the one before last at Vasya's left. | ```python
n,a=map(int,input().split())
if a%2==0:
print(1+int((n-a)/2))
else:
print(1+int((n-a+1)/2))
``` | 0 | |
577 | A | Multiplication Table | PROGRAMMING | 1,000 | [
"implementation",
"number theory"
] | null | null | Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=×<=*j*. The rows and columns are numbered starting from 1.
You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*. | The single line contains numbers *n* and *x* (1<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=109) — the size of the table and the number that we are looking for in the table. | Print a single number: the number of times *x* occurs in the table. | [
"10 5\n",
"6 12\n",
"5 13\n"
] | [
"2\n",
"4\n",
"0\n"
] | A table for the second sample test is given below. The occurrences of number 12 are marked bold. | 500 | [
{
"input": "10 5",
"output": "2"
},
{
"input": "6 12",
"output": "4"
},
{
"input": "5 13",
"output": "0"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 1",
"output": "1"
},
{
"input": "100000 1",
"output": "1"
},
{
"input": "1 1000000000",
"output": "0"
},
{
"input": "100000 1000000000",
"output": "16"
},
{
"input": "100000 362880",
"output": "154"
},
{
"input": "1 4",
"output": "0"
},
{
"input": "9 12",
"output": "4"
},
{
"input": "10 123",
"output": "0"
},
{
"input": "9551 975275379",
"output": "0"
},
{
"input": "17286 948615687",
"output": "0"
},
{
"input": "58942 936593001",
"output": "0"
},
{
"input": "50000 989460910",
"output": "4"
},
{
"input": "22741 989460910",
"output": "0"
},
{
"input": "22740 989460910",
"output": "0"
},
{
"input": "100000 989460910",
"output": "4"
},
{
"input": "100000 98280",
"output": "128"
},
{
"input": "100000 997920",
"output": "222"
},
{
"input": "100000 720720",
"output": "226"
},
{
"input": "100000 2162160",
"output": "282"
},
{
"input": "100000 4324320",
"output": "320"
},
{
"input": "100000 8648640",
"output": "348"
},
{
"input": "100000 183783600",
"output": "438"
},
{
"input": "100000 551350800",
"output": "392"
},
{
"input": "40000 551350800",
"output": "150"
},
{
"input": "20000 400000000",
"output": "1"
},
{
"input": "19999 400000000",
"output": "0"
},
{
"input": "19999 399960001",
"output": "1"
},
{
"input": "31621 999887641",
"output": "1"
},
{
"input": "31622 999887641",
"output": "1"
},
{
"input": "31620 999887641",
"output": "0"
},
{
"input": "100000 999887641",
"output": "3"
},
{
"input": "100000 25",
"output": "3"
},
{
"input": "100000 1",
"output": "1"
},
{
"input": "100000 3628800",
"output": "220"
},
{
"input": "100000 39916800",
"output": "328"
},
{
"input": "100000 479001600",
"output": "254"
},
{
"input": "4 9",
"output": "1"
},
{
"input": "2 6",
"output": "0"
},
{
"input": "20 100",
"output": "3"
},
{
"input": "10 3",
"output": "2"
},
{
"input": "4 4",
"output": "3"
},
{
"input": "2 4",
"output": "1"
},
{
"input": "89874 1",
"output": "1"
}
] | 1,651,265,608 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 3 | 30 | 0 | line2=input()
line2=line2.split()
x=int(line2[0])
y=int(line2[1])
c=0
u=int(min(x,y))
for i in range(int(u/2)):
if(y%(i+1)==0):
t=y/(i+1)
if(t<=x and (i+1)<=x):
c=c+2
print(c) | Title: Multiplication Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=×<=*j*. The rows and columns are numbered starting from 1.
You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*.
Input Specification:
The single line contains numbers *n* and *x* (1<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=109) — the size of the table and the number that we are looking for in the table.
Output Specification:
Print a single number: the number of times *x* occurs in the table.
Demo Input:
['10 5\n', '6 12\n', '5 13\n']
Demo Output:
['2\n', '4\n', '0\n']
Note:
A table for the second sample test is given below. The occurrences of number 12 are marked bold. | ```python
line2=input()
line2=line2.split()
x=int(line2[0])
y=int(line2[1])
c=0
u=int(min(x,y))
for i in range(int(u/2)):
if(y%(i+1)==0):
t=y/(i+1)
if(t<=x and (i+1)<=x):
c=c+2
print(c)
``` | 0 | |
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\n9 7\n9 1\n2 3\n5 2\n21 16\n20 19\n23 14\n27 6\n25 21\n14 1\n18 14\n7 2\n19 12\n30 27\n4 27\n24 12\n25 20\n26 22\n21 17\n11 6\n5 28\n28 24\n17 30\n2 5\n30 10\n4 21",
"output": "Kiwis\nCarrots\nGrapes\nGrapes"
},
{
"input": "39898 39898 3 1\n4567 8901\n12345 23456\n24680 35679\n29292 12121",
"output": "Grapes"
},
{
"input": "1 1 1 1\n1 1\n1 1",
"output": "Waste"
},
{
"input": "1 2 1 2\n1 1\n1 2\n1 1",
"output": "Carrots\nWaste"
},
{
"input": "1 6 3 12\n1 2\n1 4\n1 6\n1 1\n1 2\n1 6\n1 2\n1 5\n1 3\n1 4\n1 5\n1 5\n1 4\n1 6\n1 3",
"output": "Carrots\nWaste\nWaste\nWaste\nGrapes\nKiwis\nWaste\nGrapes\nGrapes\nWaste\nWaste\nKiwis"
},
{
"input": "1 10 1 3\n1 5\n1 5\n1 5\n1 5",
"output": "Waste\nWaste\nWaste"
},
{
"input": "5 5 1 3\n2 2\n2 2\n2 2\n2 2",
"output": "Waste\nWaste\nWaste"
},
{
"input": "3 3 3 5\n1 1\n2 2\n3 3\n1 1\n2 2\n2 2\n2 2\n3 3",
"output": "Waste\nWaste\nWaste\nWaste\nWaste"
},
{
"input": "1 10 2 4\n1 9\n1 5\n1 5\n1 6\n1 9\n1 10",
"output": "Waste\nKiwis\nWaste\nKiwis"
},
{
"input": "5 2 2 1\n3 2\n4 2\n5 2",
"output": "Kiwis"
}
] | 1,630,332,405 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 1 | 186 | 20,172,800 | n,m,k,t = map(int, input().split())
waste, qry = [], []
a,b = 0,0
for _ in range(k):
a, b = map(int, input().split())
waste.append(m*(a-1)+b)
for i in range(t):
a, b = map(int, input().split())
qry.append([m*(a-1)+b, i])
ans = ['']*(t)
waste.sort()
qry.sort()
d = {0: "Carrots", 1: "Kiwis", 2: "Grapes", -1: "Waste"}
i,j = 0,0
while i<t:
if j<k and qry[i][0] == waste[j]:
ans[qry[i][1]] = d[-1]
i+=1
j+=1
continue
ans[qry[i][1]] = d[(qry[i][0]-j-1)%3]
i+=1
for i in ans:
print(i)
| 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())
waste, qry = [], []
a,b = 0,0
for _ in range(k):
a, b = map(int, input().split())
waste.append(m*(a-1)+b)
for i in range(t):
a, b = map(int, input().split())
qry.append([m*(a-1)+b, i])
ans = ['']*(t)
waste.sort()
qry.sort()
d = {0: "Carrots", 1: "Kiwis", 2: "Grapes", -1: "Waste"}
i,j = 0,0
while i<t:
if j<k and qry[i][0] == waste[j]:
ans[qry[i][1]] = d[-1]
i+=1
j+=1
continue
ans[qry[i][1]] = d[(qry[i][0]-j-1)%3]
i+=1
for i in ans:
print(i)
``` | 0 |
82 | A | Double Cola | PROGRAMMING | 1,100 | [
"implementation",
"math"
] | A. Double Cola | 1 | 256 | Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon. | The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers. | Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially. | [
"1\n",
"6\n",
"1802\n"
] | [
"Sheldon\n",
"Sheldon\n",
"Penny\n"
] | none | 500 | [
{
"input": "1",
"output": "Sheldon"
},
{
"input": "6",
"output": "Sheldon"
},
{
"input": "1802",
"output": "Penny"
},
{
"input": "1",
"output": "Sheldon"
},
{
"input": "2",
"output": "Leonard"
},
{
"input": "3",
"output": "Penny"
},
{
"input": "4",
"output": "Rajesh"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "10",
"output": "Penny"
},
{
"input": "534",
"output": "Rajesh"
},
{
"input": "5033",
"output": "Howard"
},
{
"input": "10010",
"output": "Howard"
},
{
"input": "500000000",
"output": "Penny"
},
{
"input": "63",
"output": "Rajesh"
},
{
"input": "841",
"output": "Leonard"
},
{
"input": "3667",
"output": "Penny"
},
{
"input": "38614",
"output": "Howard"
},
{
"input": "282798",
"output": "Rajesh"
},
{
"input": "9266286",
"output": "Rajesh"
},
{
"input": "27385966",
"output": "Leonard"
},
{
"input": "121580142",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
},
{
"input": "300",
"output": "Howard"
},
{
"input": "1745",
"output": "Leonard"
},
{
"input": "8302",
"output": "Rajesh"
},
{
"input": "184518",
"output": "Sheldon"
},
{
"input": "1154414",
"output": "Rajesh"
},
{
"input": "28643950",
"output": "Leonard"
},
{
"input": "159222638",
"output": "Howard"
},
{
"input": "24",
"output": "Penny"
},
{
"input": "505",
"output": "Penny"
},
{
"input": "4425",
"output": "Rajesh"
},
{
"input": "12079",
"output": "Sheldon"
},
{
"input": "469726",
"output": "Penny"
},
{
"input": "3961838",
"output": "Penny"
},
{
"input": "57710446",
"output": "Leonard"
},
{
"input": "80719470",
"output": "Howard"
},
{
"input": "1000000000",
"output": "Penny"
},
{
"input": "999999999",
"output": "Penny"
},
{
"input": "999999998",
"output": "Penny"
},
{
"input": "5",
"output": "Howard"
}
] | 1,460,545,976 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 61 | 0 | n = "Sheldon,Leonard,Penny,Rajesh,Howard".split(",")
print(n[ int(input())%6]) | Title: Double Cola
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sheldon, Leonard, Penny, Rajesh and Howard are in the queue for a "Double Cola" drink vending machine; there are no other people in the queue. The first one in the queue (Sheldon) buys a can, drinks it and doubles! The resulting two Sheldons go to the end of the queue. Then the next in the queue (Leonard) buys a can, drinks it and gets to the end of the queue as two Leonards, and so on. This process continues ad infinitum.
For example, Penny drinks the third can of cola and the queue will look like this: Rajesh, Howard, Sheldon, Sheldon, Leonard, Leonard, Penny, Penny.
Write a program that will print the name of a man who will drink the *n*-th can.
Note that in the very beginning the queue looks like that: Sheldon, Leonard, Penny, Rajesh, Howard. The first person is Sheldon.
Input Specification:
The input data consist of a single integer *n* (1<=≤<=*n*<=≤<=109).
It is guaranteed that the pretests check the spelling of all the five names, that is, that they contain all the five possible answers.
Output Specification:
Print the single line — the name of the person who drinks the *n*-th can of cola. The cans are numbered starting from 1. Please note that you should spell the names like this: "Sheldon", "Leonard", "Penny", "Rajesh", "Howard" (without the quotes). In that order precisely the friends are in the queue initially.
Demo Input:
['1\n', '6\n', '1802\n']
Demo Output:
['Sheldon\n', 'Sheldon\n', 'Penny\n']
Note:
none | ```python
n = "Sheldon,Leonard,Penny,Rajesh,Howard".split(",")
print(n[ int(input())%6])
``` | 0 |
327 | A | Flipping Game | PROGRAMMING | 1,200 | [
"brute force",
"dp",
"implementation"
] | null | null | Iahub got bored, so he invented a game to be played on paper.
He writes *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices *i* and *j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) and flips all values *a**k* for which their positions are in range [*i*,<=*j*] (that is *i*<=≤<=*k*<=≤<=*j*). Flip the value of *x* means to apply operation *x*<==<=1 - *x*.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub. | The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100). In the second line of the input there are *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. It is guaranteed that each of those *n* values is either 0 or 1. | Print an integer — the maximal number of 1s that can be obtained after exactly one move. | [
"5\n1 0 0 1 0\n",
"4\n1 0 0 1\n"
] | [
"4\n",
"4\n"
] | In the first case, flip the segment from 2 to 5 (*i* = 2, *j* = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (*i* = 2, *j* = 3) will turn all numbers into 1. | 500 | [
{
"input": "5\n1 0 0 1 0",
"output": "4"
},
{
"input": "4\n1 0 0 1",
"output": "4"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "8\n1 0 0 0 1 0 0 0",
"output": "7"
},
{
"input": "18\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "18"
},
{
"input": "23\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "22"
},
{
"input": "100\n0 1 0 1 1 1 0 1 0 1 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 0 0 0 1 1 1 1 0 0 1 1 1 0 0 1 1 0 1 1 1 0 0 0 1 0 0 0 0 0 1 1 0 0 1 1 1 1 1 1 1 1 0 1 1 1 0 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1",
"output": "70"
},
{
"input": "100\n0 1 1 0 1 0 0 1 0 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 1 0 1 0 1 0 1 0 0 0 1 0 0 0 0 0 1 1 0 1 0 1 0 1 1 1 0 1 0 1 1 0 0 1 1 0 0 1 1 1 0 0 1 0 0 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 0 0 0 1 0 0 0 0 1 1 1 1",
"output": "60"
},
{
"input": "18\n0 1 0 1 0 1 0 1 0 1 1 0 1 1 0 1 1 0",
"output": "11"
},
{
"input": "25\n0 1 0 0 0 0 0 1 0 1 0 1 0 0 0 0 1 1 1 0 0 1 1 0 1",
"output": "18"
},
{
"input": "55\n0 0 1 1 0 0 0 1 0 1 1 0 1 1 1 0 1 1 1 1 1 0 0 1 0 0 1 0 1 1 0 0 1 0 1 1 0 1 1 1 1 0 1 1 0 0 0 0 1 1 0 1 1 1 1",
"output": "36"
},
{
"input": "75\n1 1 0 1 0 1 1 0 0 0 0 0 1 1 1 1 1 0 1 0 1 0 0 0 0 1 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 1 0 1 0 1 0 0 0 0 1 0 0 1 1 1 0 0 1 0 1 1 0 0 0 0 1 1 0 0 0 1 0 0 0",
"output": "44"
},
{
"input": "100\n0 0 1 0 1 0 0 1 1 0 1 1 0 1 0 1 1 0 0 0 0 0 1 0 0 1 1 0 0 0 1 0 0 1 1 0 0 1 1 1 0 0 0 0 1 0 1 1 1 0 0 1 0 1 1 1 1 1 1 1 0 1 0 1 0 0 1 0 1 1 1 0 0 0 0 1 0 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 0 0 1 1 0 1 0 1",
"output": "61"
},
{
"input": "100\n0 0 0 1 0 0 0 1 0 1 1 0 1 1 1 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 1 0 1 1 0 0 0 1 1 1 1 0 1 1 0 1 1 1 1 0 1 0 0 1 0 1 0 0 0 0 1 1 0 0 1 0 0 1 1 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 1 0 0 1 0 1 1 1 0 1 1 0",
"output": "61"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "100"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "99"
},
{
"input": "100\n0 0 1 1 1 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 0 1 1 1 1 1 1 0 1 0 0 1 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 0 0 1 0 0 1 1 0 0 0 1 1 1 0 1 1 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 1 1 1 0 1 1 1 0 1 1 1 1 1 0 0 0 0 1 0",
"output": "61"
},
{
"input": "100\n0 1 1 1 1 1 1 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 0 1 1 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 1 0 1 0 0 0 0 1 0 1 0 1 0 0 1 0 1 0 1 1 0 0 0 0 0 1 1 1 0 1 1 0 1 1 0 0 0 1 0 1 1 0 0 0 0 0 0 1",
"output": "59"
},
{
"input": "99\n1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 0 0 0 0 1 1 0 0 1 0 1 1 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 1 0 1 0 0 1 1 0 1 0 0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 0 0 1 0 0 0 0 0 1 0 1 1 1",
"output": "61"
},
{
"input": "2\n1 1",
"output": "1"
}
] | 1,674,617,031 | 2,147,483,647 | Python 3 | OK | TESTS | 21 | 154 | 0 | n=int(input())
a=list(map(int,input().split()))
summ=[]
summ.append(0)
maxsumm=0
count=1
window=1
b=a.copy()
for i in range(0,n,1):
for j in range(0,n-i,1):
b=a.copy()
for k in range(j,j+window,1):
if b[k]==1:
b[k]=0
else:
b[k]=1
summ.append(sum(b))
maxsumm=max(maxsumm,summ[count])
count+=1
window+=1
print(maxsumm) | Title: Flipping Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub got bored, so he invented a game to be played on paper.
He writes *n* integers *a*1,<=*a*2,<=...,<=*a**n*. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices *i* and *j* (1<=≤<=*i*<=≤<=*j*<=≤<=*n*) and flips all values *a**k* for which their positions are in range [*i*,<=*j*] (that is *i*<=≤<=*k*<=≤<=*j*). Flip the value of *x* means to apply operation *x*<==<=1 - *x*.
The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100). In the second line of the input there are *n* integers: *a*1,<=*a*2,<=...,<=*a**n*. It is guaranteed that each of those *n* values is either 0 or 1.
Output Specification:
Print an integer — the maximal number of 1s that can be obtained after exactly one move.
Demo Input:
['5\n1 0 0 1 0\n', '4\n1 0 0 1\n']
Demo Output:
['4\n', '4\n']
Note:
In the first case, flip the segment from 2 to 5 (*i* = 2, *j* = 5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].
In the second case, flipping only the second and the third element (*i* = 2, *j* = 3) will turn all numbers into 1. | ```python
n=int(input())
a=list(map(int,input().split()))
summ=[]
summ.append(0)
maxsumm=0
count=1
window=1
b=a.copy()
for i in range(0,n,1):
for j in range(0,n-i,1):
b=a.copy()
for k in range(j,j+window,1):
if b[k]==1:
b[k]=0
else:
b[k]=1
summ.append(sum(b))
maxsumm=max(maxsumm,summ[count])
count+=1
window+=1
print(maxsumm)
``` | 3 | |
171 | F | ucyhf | PROGRAMMING | 1,600 | [
"*special",
"brute force",
"implementation",
"number theory"
] | null | null | qd ucyhf yi q fhycu dkcruh mxeiu huluhiu yi q tyvvuhudj fhycu dkcruh. oekh jqia yi je vydt jxu djx ucyhf. | jxu ydfkj sediyiji ev q iydwbu ydjuwuh *d* (1<=≤<=*d*<=≤<=11184) — jxu edu-rqiut ydtun ev jxu ucyhf je vydt. | ekjfkj q iydwbu dkcruh. | [
"1\n"
] | [
"13\n"
] | none | 0 | [
{
"input": "1",
"output": "13"
},
{
"input": "2",
"output": "17"
},
{
"input": "3",
"output": "31"
},
{
"input": "4",
"output": "37"
},
{
"input": "5",
"output": "71"
},
{
"input": "6",
"output": "73"
},
{
"input": "7",
"output": "79"
},
{
"input": "8",
"output": "97"
},
{
"input": "9",
"output": "107"
},
{
"input": "10",
"output": "113"
},
{
"input": "8216",
"output": "768377"
},
{
"input": "119",
"output": "3359"
},
{
"input": "10618",
"output": "975193"
},
{
"input": "6692",
"output": "399731"
},
{
"input": "10962",
"output": "990511"
},
{
"input": "6848",
"output": "706463"
},
{
"input": "4859",
"output": "323149"
},
{
"input": "8653",
"output": "787433"
},
{
"input": "6826",
"output": "705533"
},
{
"input": "10526",
"output": "971513"
},
{
"input": "9819",
"output": "939487"
},
{
"input": "10844",
"output": "984341"
},
{
"input": "7779",
"output": "748169"
},
{
"input": "1340",
"output": "91009"
},
{
"input": "4020",
"output": "190871"
},
{
"input": "2279",
"output": "122867"
},
{
"input": "5581",
"output": "353057"
},
{
"input": "11107",
"output": "997001"
},
{
"input": "7397",
"output": "731053"
},
{
"input": "5273",
"output": "340573"
},
{
"input": "10476",
"output": "969481"
},
{
"input": "7161",
"output": "720611"
},
{
"input": "4168",
"output": "196193"
},
{
"input": "1438",
"output": "93887"
},
{
"input": "6327",
"output": "384913"
},
{
"input": "10107",
"output": "953077"
},
{
"input": "6399",
"output": "388313"
},
{
"input": "11182",
"output": "999853"
},
{
"input": "11183",
"output": "999931"
},
{
"input": "11184",
"output": "999983"
}
] | 1,621,654,894 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 10 | 746 | 1,228,800 | def process(n):
prime_set = set([])
primes = []
for p in range(2, 2*10**5):
prime = True
for p2 in primes:
if p2*p2 > p:
break
if p % p2==0:
prime = False
break
if prime:
prime_set.add(str(p))
primes.append(p)
emirp = []
for p in prime_set:
if p[::-1] in prime_set and p != p[::-1]:
emirp.append(int(p))
return sorted(emirp)[n-1]
print(process(int(input()))) | Title: ucyhf
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
qd ucyhf yi q fhycu dkcruh mxeiu huluhiu yi q tyvvuhudj fhycu dkcruh. oekh jqia yi je vydt jxu djx ucyhf.
Input Specification:
jxu ydfkj sediyiji ev q iydwbu ydjuwuh *d* (1<=≤<=*d*<=≤<=11184) — jxu edu-rqiut ydtun ev jxu ucyhf je vydt.
Output Specification:
ekjfkj q iydwbu dkcruh.
Demo Input:
['1\n']
Demo Output:
['13\n']
Note:
none | ```python
def process(n):
prime_set = set([])
primes = []
for p in range(2, 2*10**5):
prime = True
for p2 in primes:
if p2*p2 > p:
break
if p % p2==0:
prime = False
break
if prime:
prime_set.add(str(p))
primes.append(p)
emirp = []
for p in prime_set:
if p[::-1] in prime_set and p != p[::-1]:
emirp.append(int(p))
return sorted(emirp)[n-1]
print(process(int(input())))
``` | -1 | |
749 | A | Bachgold Problem | PROGRAMMING | 800 | [
"greedy",
"implementation",
"math",
"number theory"
] | null | null | Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*. | The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000). | The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them. | [
"5\n",
"6\n"
] | [
"2\n2 3\n",
"3\n2 2 2\n"
] | none | 500 | [
{
"input": "5",
"output": "2\n2 3"
},
{
"input": "6",
"output": "3\n2 2 2"
},
{
"input": "2",
"output": "1\n2"
},
{
"input": "3",
"output": "1\n3"
},
{
"input": "99999",
"output": "49999\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "100000",
"output": "50000\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "7",
"output": "3\n2 2 3"
},
{
"input": "4",
"output": "2\n2 2"
},
{
"input": "8",
"output": "4\n2 2 2 2"
},
{
"input": "9",
"output": "4\n2 2 2 3"
},
{
"input": "99995",
"output": "49997\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "99996",
"output": "49998\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "10",
"output": "5\n2 2 2 2 2"
},
{
"input": "11",
"output": "5\n2 2 2 2 3"
},
{
"input": "99997",
"output": "49998\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "99998",
"output": "49999\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "12",
"output": "6\n2 2 2 2 2 2"
},
{
"input": "13",
"output": "6\n2 2 2 2 2 3"
},
{
"input": "99993",
"output": "49996\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "99994",
"output": "49997\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "14",
"output": "7\n2 2 2 2 2 2 2"
},
{
"input": "15",
"output": "7\n2 2 2 2 2 2 3"
},
{
"input": "53",
"output": "26\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3"
},
{
"input": "57",
"output": "28\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3"
},
{
"input": "61",
"output": "30\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3"
},
{
"input": "774",
"output": "387\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "202",
"output": "101\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2"
},
{
"input": "530",
"output": "265\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "7166",
"output": "3583\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "9294",
"output": "4647\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "2422",
"output": "1211\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "15326",
"output": "7663\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "11454",
"output": "5727\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "14878",
"output": "7439\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
},
{
"input": "90672",
"output": "45336\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "99544",
"output": "49772\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "90472",
"output": "45236\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ..."
},
{
"input": "23",
"output": "11\n2 2 2 2 2 2 2 2 2 2 3"
},
{
"input": "93",
"output": "46\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3"
},
{
"input": "19",
"output": "9\n2 2 2 2 2 2 2 2 3"
},
{
"input": "11110",
"output": "5555\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2..."
}
] | 1,689,695,340 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 41 | 62 | 0 | n = int(input())
print(n//2)
if n%2:
print((n-2)//2*"2 ",end="")
print("3",end="")
else:
print((n-2)//2*"2 ",end="")
print("2",end="") | Title: Bachgold Problem
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bachgold problem is very easy to formulate. Given a positive integer *n* represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer *k* is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and *k*.
Input Specification:
The only line of the input contains a single integer *n* (2<=≤<=*n*<=≤<=100<=000).
Output Specification:
The first line of the output contains a single integer *k* — maximum possible number of primes in representation.
The second line should contain *k* primes with their sum equal to *n*. You can print them in any order. If there are several optimal solution, print any of them.
Demo Input:
['5\n', '6\n']
Demo Output:
['2\n2 3\n', '3\n2 2 2\n']
Note:
none | ```python
n = int(input())
print(n//2)
if n%2:
print((n-2)//2*"2 ",end="")
print("3",end="")
else:
print((n-2)//2*"2 ",end="")
print("2",end="")
``` | 3 | |
32 | B | Borze | PROGRAMMING | 800 | [
"expression parsing",
"implementation"
] | B. Borze | 2 | 256 | Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet. | The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes). | Output the decoded ternary number. It can have leading zeroes. | [
".-.--\n",
"--.\n",
"-..-.--\n"
] | [
"012",
"20",
"1012"
] | none | 1,000 | [
{
"input": ".-.--",
"output": "012"
},
{
"input": "--.",
"output": "20"
},
{
"input": "-..-.--",
"output": "1012"
},
{
"input": "---..",
"output": "210"
},
{
"input": "..--.---..",
"output": "0020210"
},
{
"input": "-.....----.",
"output": "10000220"
},
{
"input": ".",
"output": "0"
},
{
"input": "-.",
"output": "1"
},
{
"input": "--",
"output": "2"
},
{
"input": "..",
"output": "00"
},
{
"input": "--.",
"output": "20"
},
{
"input": ".--.",
"output": "020"
},
{
"input": ".-.-..",
"output": "0110"
},
{
"input": "----.-.",
"output": "2201"
},
{
"input": "-..--.-.",
"output": "10201"
},
{
"input": "..--..--.",
"output": "0020020"
},
{
"input": "-.-.---.--..-..-.-.-..-..-.--.",
"output": "112120010111010120"
},
{
"input": "---.-.-.------..-..-..-..-.-..-.--.-.-..-.-.-----..-.-.",
"output": "21112220010101011012011011221011"
},
{
"input": "-.-..--.-.-.-.-.-..-.-.-.---------.--.---..--...--.-----.-.-.-...--.-.-.---.------.--..-.--.-----.-...-..------",
"output": "11020111110111222212021020002022111100201121222020012022110010222"
},
{
"input": "-.-..-.--.---..---.-..---.-...-.-.----..-.---.-.---..-.--.---.-.-------.---.--....----.-.---.---.---.----.-----..---.-.-.-.-----.--.-------.-..",
"output": "110120210211021100112200121121012021122212120000220121212122022102111122120222110"
},
{
"input": ".-..-.-.---.-----.--.---...-.--.-.-....-..",
"output": "01011212212021001201100010"
},
{
"input": ".------.-.---..--...-..-..-.-.-.--.--.-..-.--...-.-.---.-.-.------..--..-.---..----.-..-.--.---.-.----.-.---...-.-.-.-----.-.-.---.---.-.....-.-...-----.-...-.---.-..-.-----.--...---.-.-..-.--.-.---..",
"output": "022201210200010101112020101200011211122200200121022010120211220121001112211121211000011002211001211012212000211101201210"
},
{
"input": ".-.--.---.-----.-.-----.-.-..-----..-..----..--.-.--.----..---.---..-.-.-----..-------.----..----.-..---...-----..-..-----...-..-.-.-----....---..---..-.-----...-.--...--.-.---.-.-.-.-.-...---..----.",
"output": "01202122112211102210102200201202200212101122102221220022010210022101022100101122100021021012210012000201211111100210220"
},
{
"input": "..-.-.-.---.-.-.-..-.-..-.-.---.-------.---..-----.---....-.---.--.--.-.---.---------.-..---.-.-.--..---.---.-.---.-.-..-.-..-.-.-.----.--.-....--------.-.---..----.------.-.-.--.--.-----.-----.----",
"output": "0011121111011011212221210221210001212020121222211021112002121121110110111220201000222201210220222011202022122122"
},
{
"input": "-..-------.------.-..--.-.-..--.-.-..-----..-.-.-..-..-..--.---..-----..---..-..--.-..-.-.---...-.....-------.---.-----.-...-.-...-.-.---.---.-----.--.--...-.--..-.-..-...-.-.-.-.---..---.-..-.-.-.-..",
"output": "102221222010201102011022101110101020210221021010201011210010000222121221100110011212122120200012001101001111210211011110"
},
{
"input": ".-.----.-.--..-.-.-.-..----..-.-...--.-.---.---.-------..-.--..-......--.------.--.----.--...-.--.--..-----..-.....--.--.-.-.------..--------.----------..-.---.----.---.-..--..-.....-..------.--.",
"output": "012201200111102200110020121212221012001000002022202022020001202002210100002020112220022220222220012122021102001000010222020"
},
{
"input": "------.-----.-....--.-.----.-.---.-.-..---.-.---.-----..-...-.-.---..-.-.-..-.-.-...-.-.-.----..--.------.----.-..-.--...-.-------...-.-..-.-.--.--.---..--..--------.--.-.-.---.-.-.-...----.--..-.--..",
"output": "222022110002012201211102112122101001121011101110011122002022202201012000122210011011202021020022220201121111002202001200"
},
{
"input": "-.---...----...--.--...-.--.----",
"output": "121002200020200012022"
},
{
"input": "--.--.--.---.--.-.---.-.-..-..--.-..---.-.....-..---.-----.--...-.-.-------.-.--.-.----.-..-.------.",
"output": "202020212012111010201021100001021221200011222112012201012220"
}
] | 1,676,136,073 | 2,147,483,647 | Python 3 | OK | TESTS | 30 | 92 | 0 | code=list(input())
i=0
while i<len(code):
if code[i]=='-' and code[i+1]=='.':
print('1',end='')
i+=2
elif code[i]=='-' and code[i+1]=='-':
print('2',end='')
i+=2
else:
print('0',end='')
i+=1
print('\n') | Title: Borze
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Ternary numeric notation is quite popular in Berland. To telegraph the ternary number the Borze alphabet is used. Digit 0 is transmitted as «.», 1 as «-.» and 2 as «--». You are to decode the Borze code, i.e. to find out the ternary number given its representation in Borze alphabet.
Input Specification:
The first line contains a number in Borze code. The length of the string is between 1 and 200 characters. It's guaranteed that the given string is a valid Borze code of some ternary number (this number can have leading zeroes).
Output Specification:
Output the decoded ternary number. It can have leading zeroes.
Demo Input:
['.-.--\n', '--.\n', '-..-.--\n']
Demo Output:
['012', '20', '1012']
Note:
none | ```python
code=list(input())
i=0
while i<len(code):
if code[i]=='-' and code[i+1]=='.':
print('1',end='')
i+=2
elif code[i]=='-' and code[i+1]=='-':
print('2',end='')
i+=2
else:
print('0',end='')
i+=1
print('\n')
``` | 3.977 |
294 | A | Shaass and Oskols | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots. | The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment. | On the *i*-th line of the output print the number of birds on the *i*-th wire. | [
"5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n",
"3\n2 4 1\n1\n2 2\n"
] | [
"0\n12\n5\n0\n16\n",
"3\n0\n3\n"
] | none | 500 | [
{
"input": "5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6",
"output": "0\n12\n5\n0\n16"
},
{
"input": "3\n2 4 1\n1\n2 2",
"output": "3\n0\n3"
},
{
"input": "5\n58 51 45 27 48\n5\n4 9\n5 15\n4 5\n5 8\n1 43",
"output": "0\n66\n57\n7\n0"
},
{
"input": "10\n48 53 10 28 91 56 81 2 67 52\n2\n2 40\n6 51",
"output": "87\n0\n23\n28\n141\n0\n86\n2\n67\n52"
},
{
"input": "2\n72 45\n6\n1 69\n2 41\n1 19\n2 7\n1 5\n2 1",
"output": "0\n0"
},
{
"input": "10\n95 54 36 39 98 30 19 24 14 12\n3\n9 5\n8 15\n7 5",
"output": "95\n54\n36\n39\n98\n34\n0\n28\n13\n21"
},
{
"input": "100\n95 15 25 18 64 62 23 59 70 84 50 26 87 35 75 86 0 22 77 60 66 41 21 9 75 50 25 3 69 14 39 68 64 46 59 99 2 0 21 76 90 12 61 42 6 91 36 39 47 41 93 81 66 57 70 36 68 89 52 1 19 93 67 22 76 20 8 81 98 18 100 73 61 93 75 80 53 72 40 20 2 86 33 59 27 16 11 26 55 44 47 55 94 58 2 55 82 22 9 95\n0",
"output": "95\n15\n25\n18\n64\n62\n23\n59\n70\n84\n50\n26\n87\n35\n75\n86\n0\n22\n77\n60\n66\n41\n21\n9\n75\n50\n25\n3\n69\n14\n39\n68\n64\n46\n59\n99\n2\n0\n21\n76\n90\n12\n61\n42\n6\n91\n36\n39\n47\n41\n93\n81\n66\n57\n70\n36\n68\n89\n52\n1\n19\n93\n67\n22\n76\n20\n8\n81\n98\n18\n100\n73\n61\n93\n75\n80\n53\n72\n40\n20\n2\n86\n33\n59\n27\n16\n11\n26\n55\n44\n47\n55\n94\n58\n2\n55\n82\n22\n9\n95"
},
{
"input": "1\n100\n1\n1 100",
"output": "0"
},
{
"input": "1\n100\n1\n1 1",
"output": "0"
},
{
"input": "1\n50\n1\n1 25",
"output": "0"
},
{
"input": "2\n50 0\n1\n1 1",
"output": "0\n49"
},
{
"input": "1\n10\n0",
"output": "10"
}
] | 1,688,137,082 | 2,147,483,647 | Python 3 | OK | TESTS | 31 | 92 | 0 | n = int(input())
a = [int(i) for i in input().split()]
for _ in range(int(input())):
x, y = map(int, input().split())
if x != 1:
a[x - 2] += (y - 1)
if x != n:
a[x] += (a[x - 1] - y)
a[x - 1] = 0
for i in a:
print(i)
| Title: Shaass and Oskols
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Shaass has decided to hunt some birds. There are *n* horizontal electricity wires aligned parallel to each other. Wires are numbered 1 to *n* from top to bottom. On each wire there are some oskols sitting next to each other. Oskol is the name of a delicious kind of birds in Shaass's territory. Supposed there are *a**i* oskols sitting on the *i*-th wire.
Sometimes Shaass shots one of the birds and the bird dies (suppose that this bird sat at the *i*-th wire). Consequently all the birds on the *i*-th wire to the left of the dead bird get scared and jump up on the wire number *i*<=-<=1, if there exists no upper wire they fly away. Also all the birds to the right of the dead bird jump down on wire number *i*<=+<=1, if there exists no such wire they fly away.
Shaass has shot *m* birds. You're given the initial number of birds on each wire, tell him how many birds are sitting on each wire after the shots.
Input Specification:
The first line of the input contains an integer *n*, (1<=≤<=*n*<=≤<=100). The next line contains a list of space-separated integers *a*1,<=*a*2,<=...,<=*a**n*, (0<=≤<=*a**i*<=≤<=100).
The third line contains an integer *m*, (0<=≤<=*m*<=≤<=100). Each of the next *m* lines contains two integers *x**i* and *y**i*. The integers mean that for the *i*-th time Shaass shoot the *y**i*-th (from left) bird on the *x**i*-th wire, (1<=≤<=*x**i*<=≤<=*n*,<=1<=≤<=*y**i*). It's guaranteed there will be at least *y**i* birds on the *x**i*-th wire at that moment.
Output Specification:
On the *i*-th line of the output print the number of birds on the *i*-th wire.
Demo Input:
['5\n10 10 10 10 10\n5\n2 5\n3 13\n2 12\n1 13\n4 6\n', '3\n2 4 1\n1\n2 2\n']
Demo Output:
['0\n12\n5\n0\n16\n', '3\n0\n3\n']
Note:
none | ```python
n = int(input())
a = [int(i) for i in input().split()]
for _ in range(int(input())):
x, y = map(int, input().split())
if x != 1:
a[x - 2] += (y - 1)
if x != n:
a[x] += (a[x - 1] - y)
a[x - 1] = 0
for i in a:
print(i)
``` | 3 | |
71 | A | Way Too Long Words | PROGRAMMING | 800 | [
"strings"
] | A. Way Too Long Words | 1 | 256 | Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. | Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. | [
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] | [
"word\nl10n\ni18n\np43s\n"
] | none | 500 | [
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,690,992,726 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | a=int(input())
lst = []
for i in range(a):
m = input()
lst.append(m)
for i in lst:
if len(m)>10:
print(m[0]+str(len(m)-2)+m[len(m)-1])
else:
print(m)
| Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none | ```python
a=int(input())
lst = []
for i in range(a):
m = input()
lst.append(m)
for i in lst:
if len(m)>10:
print(m[0]+str(len(m)-2)+m[len(m)-1])
else:
print(m)
``` | 0 |
544 | A | Set of Strings | PROGRAMMING | 1,100 | [
"implementation",
"strings"
] | null | null | You are given a string *q*. A sequence of *k* strings *s*1,<=*s*2,<=...,<=*s**k* is called beautiful, if the concatenation of these strings is string *q* (formally, *s*1<=+<=*s*2<=+<=...<=+<=*s**k*<==<=*q*) and the first characters of these strings are distinct.
Find any beautiful sequence of strings or determine that the beautiful sequence doesn't exist. | The first line contains a positive integer *k* (1<=≤<=*k*<=≤<=26) — the number of strings that should be in a beautiful sequence.
The second line contains string *q*, consisting of lowercase Latin letters. The length of the string is within range from 1 to 100, inclusive. | If such sequence doesn't exist, then print in a single line "NO" (without the quotes). Otherwise, print in the first line "YES" (without the quotes) and in the next *k* lines print the beautiful sequence of strings *s*1,<=*s*2,<=...,<=*s**k*.
If there are multiple possible answers, print any of them. | [
"1\nabca\n",
"2\naaacas\n",
"4\nabc\n"
] | [
"YES\nabca\n",
"YES\naaa\ncas\n",
"NO\n"
] | In the second sample there are two possible answers: {"*aaaca*", "*s*"} and {"*aaa*", "*cas*"}. | 500 | [
{
"input": "1\nabca",
"output": "YES\nabca"
},
{
"input": "2\naaacas",
"output": "YES\naaa\ncas"
},
{
"input": "4\nabc",
"output": "NO"
},
{
"input": "3\nnddkhkhkdndknndkhrnhddkrdhrnrrnkkdnnndndrdhnknknhnrnnkrrdhrkhkrkhnkhkhhrhdnrndnknrrhdrdrkhdrkkhkrnkk",
"output": "YES\nn\ndd\nkhkhkdndknndkhrnhddkrdhrnrrnkkdnnndndrdhnknknhnrnnkrrdhrkhkrkhnkhkhhrhdnrndnknrrhdrdrkhdrkkhkrnkk"
},
{
"input": "26\nbiibfmmfifmffbmmfmbmbmiimbmiffmffibibfbiffibibiiimbffbbfbifmiibffbmbbbfmfibmibfffibfbffmfmimbmmmfmfm",
"output": "NO"
},
{
"input": "3\nkydoybxlfeugtrbvqnrjtzshorrsrwsxkvlwyolbaadtzpmyyfllxuciia",
"output": "YES\nk\ny\ndoybxlfeugtrbvqnrjtzshorrsrwsxkvlwyolbaadtzpmyyfllxuciia"
},
{
"input": "3\nssussususskkskkskuusksuuussksukkskuksukukusssususuususkkuukssuksskusukkssuksskskuskusussusskskksksus",
"output": "YES\nss\nussususs\nkkskkskuusksuuussksukkskuksukukusssususuususkkuukssuksskusukkssuksskskuskusussusskskksksus"
},
{
"input": "5\naaaaabcdef",
"output": "YES\naaaaa\nb\nc\nd\nef"
},
{
"input": "3\niiiiiiimiriiriwmimtmwrhhxmbmhwgghhgbqhywebrblyhlxjrthoooltehrmdhqhuodjmsjwcgrfnttiitpmqvbhlafwtzyikc",
"output": "YES\niiiiiii\nmi\nriiriwmimtmwrhhxmbmhwgghhgbqhywebrblyhlxjrthoooltehrmdhqhuodjmsjwcgrfnttiitpmqvbhlafwtzyikc"
},
{
"input": "20\ngggggllglgllltgtlglttstsgtttsslhhlssghgagtlsaghhoggtfgsaahtotdodthfltdxggxislnttlanxonhnkddtigppitdh",
"output": "NO"
},
{
"input": "16\nkkkkkkyykkynkknkkonyokdndkyonokdywkwykdkdotknnwzkoywiooinkcyzyntcdnitnppnpziomyzdspomoqmomcyrrospppn",
"output": "NO"
},
{
"input": "15\nwwwgggowgwwhoohwgwghwyohhggywhyyodgwydwgggkhgyydqyggkgkpokgthqghidhworprodtcogqkwgtfiodwdurcctkmrfmh",
"output": "YES\nwww\nggg\nowgww\nhoohwgwghw\nyohhggywhyyo\ndgwydwggg\nkhgyyd\nqyggkgk\npokg\nthqgh\nidhwo\nrprodt\ncogqkwgt\nfiodwd\nurcctkmrfmh"
},
{
"input": "15\nnnnnnntnttttttqqnqqynnqqwwnnnwneenhwtyhhoqeyeqyeuthwtnhtpnphhwetjhouhwnpojvvovoswwjryrwerbwwpbvrwvjj",
"output": "YES\nnnnnnn\ntntttttt\nqqnqq\nynnqq\nwwnnnwn\neen\nhwtyhh\noqeyeqye\nuthwtnht\npnphhwet\njhouhwnpoj\nvvovo\nswwj\nryrwer\nbwwpbvrwvjj"
},
{
"input": "15\nvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv",
"output": "NO"
},
{
"input": "1\niiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiaaaaaiiiiaiaiiiiaaiaiiiaiiaiaaiaiiaiiiiiaiiiaiiiaiaiaai",
"output": "YES\niiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiaaaaaiiiiaiaiiiiaaiaiiiaiiaiaaiaiiaiiiiiaiiiaiiiaiaiaai"
},
{
"input": "26\nvvvnnsnnnpsnnswwspncvshtncwphaphmwnwkhvvhuvctvnehemowkmtzissswjaxuuvphzrmfzihamdqmmyhhijbitlipgltyy",
"output": "YES\nvvv\nnn\nsnnn\npsnns\nwwspn\ncvs\nh\ntncwph\naph\nmwnw\nkhvvh\nuvctvn\nehem\nowkmt\nz\nisssw\nja\nxuuvphz\nrm\nfziham\nd\nqmm\nyhhij\nbit\nlip\ngltyy"
},
{
"input": "26\njexzsbwaih",
"output": "NO"
},
{
"input": "1\nk",
"output": "YES\nk"
},
{
"input": "1\nzz",
"output": "YES\nzz"
},
{
"input": "3\nziw",
"output": "YES\nz\ni\nw"
},
{
"input": "26\ntjmbyqwuahlixegopkzrfndcsv",
"output": "YES\nt\nj\nm\nb\ny\nq\nw\nu\na\nh\nl\ni\nx\ne\ng\no\np\nk\nz\nr\nf\nn\nd\nc\ns\nv"
},
{
"input": "25\nvobekscyadzqwnjxruplifmthg",
"output": "YES\nv\no\nb\ne\nk\ns\nc\ny\na\nd\nz\nq\nw\nn\nj\nx\nr\nu\np\nl\ni\nf\nm\nt\nhg"
},
{
"input": "26\nlllplzkkzflzflffzznnnnfgflqlttlmtnkzlztskngyymitqagattkdllyutzimsrskpapcmuupjdopxqlnhqcscwvdtxbflefy",
"output": "YES\nlll\npl\nz\nkkz\nflzflffzz\nnnnnf\ngfl\nql\nttl\nmtnkzlzt\nskng\nyym\nitq\nagattk\ndlly\nutzims\nrskpap\ncmuup\njd\nop\nxqln\nhqcsc\nw\nvdtx\nbfl\nefy"
},
{
"input": "25\nkkrrkrkrkrsrskpskbrppdsdbgbkrbllkbswdwcchgskmkhwiidicczlscsodtjglxbmeotzxnmbjmoqgkquglaoxgcykxvbhdi",
"output": "YES\nkk\nrrkrkrkr\nsrsk\npsk\nbrpp\ndsdb\ngbkrb\nllkbs\nwdw\ncc\nhgsk\nmkhw\niidicc\nzlscs\nod\nt\njgl\nxbm\neotzx\nnmbjmo\nqgkq\nugl\naoxgc\nykx\nvbhdi"
},
{
"input": "25\nuuuuuccpucubccbupxubcbpujiliwbpqbpyiweuywaxwqasbsllwehceruytjvphytraawgbjmerfeymoayujqranlvkpkiypadr",
"output": "YES\nuuuuu\ncc\npucu\nbccbup\nxubcbpu\nj\ni\nli\nwbp\nqbp\nyiw\neuyw\naxwqa\nsbsllwe\nhce\nruy\ntj\nvphytraaw\ngbj\nmer\nfeym\noayujqra\nnlv\nkpkiypa\ndr"
},
{
"input": "26\nxxjxodrogovufvohrodliretxxyjqnrbzmicorptkjafiwmsbwml",
"output": "YES\nxx\njx\no\nd\nro\ngo\nv\nu\nfvo\nhrod\nl\nir\ne\ntxx\nyj\nq\nnr\nb\nz\nmi\ncor\npt\nkj\nafi\nwm\nsbwml"
},
{
"input": "26\npjhsxjbvkqntwmsdnrguecaofylzti",
"output": "YES\np\nj\nh\ns\nxj\nb\nv\nk\nq\nn\nt\nw\nms\ndn\nr\ng\nu\ne\nc\na\no\nf\ny\nl\nzt\ni"
},
{
"input": "25\nrrrrqqwrlqrwglrlylwhrrwyvrhvzgvqahrhgsvavtggyduayivxzgeicinlnrkapoepbsfyjjrt",
"output": "YES\nrrrr\nqq\nwr\nlqrw\nglrl\nylw\nhrrwy\nvrhv\nzgvq\nahrhg\nsvav\ntggy\nd\nuay\niv\nxzg\nei\nci\nnlnr\nka\np\noep\nbs\nfy\njjrt"
},
{
"input": "26\ncccccccaacwwaxxaacczacnnnqqwnaggzqrwagcnabxnrcvgjqjamqzgdntzanaxvjfwqlvdttuzjoxiwtkqvrydospmpeirhg",
"output": "YES\nccccccc\naac\nwwa\nxxaacc\nzac\nnnn\nqqwna\nggzq\nrwagcna\nbxnrc\nvg\njqja\nmqzg\ndn\ntzanaxvj\nfwq\nlvdtt\nuzj\nox\niwt\nkqvr\nydo\ns\npmp\neir\nhg"
},
{
"input": "4\nsssssssssssssssssssssssssslsslslsllsslssslslssllaaslsaalsasaassllasasalrlrslarlaarlsrsaslasarlr",
"output": "YES\nssssssssssssssssssssssssss\nlsslslsllsslssslslssll\naaslsaalsasaassllasasal\nrlrslarlaarlsrsaslasarlr"
},
{
"input": "26\na",
"output": "NO"
},
{
"input": "26\nab",
"output": "NO"
},
{
"input": "2\nab",
"output": "YES\na\nb"
}
] | 1,604,431,075 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 109 | 307,200 | # The first line contains a positive integer k (1 ≤ k ≤ 26) — the number of strings that should be in a beautiful sequence.
# The second line contains string q, consisting of lowercase Latin letters. The length of the string is within range from 1 to 100, inclusive.
K = int(input())
Q = input()
chars = set()
firstOcc = []
for i in range(len(Q)):
if Q[i] not in chars:
chars.add(Q[i])
firstOcc.append(i)
if len(chars) < K:
print("NO")
else:
print("YES")
index = 0
for i in range(K):
# last one
if i != K - 1:
print(Q[firstOcc[i]:firstOcc[i + 1]])
else:
print(Q[firstOcc[i]:])
| Title: Set of Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a string *q*. A sequence of *k* strings *s*1,<=*s*2,<=...,<=*s**k* is called beautiful, if the concatenation of these strings is string *q* (formally, *s*1<=+<=*s*2<=+<=...<=+<=*s**k*<==<=*q*) and the first characters of these strings are distinct.
Find any beautiful sequence of strings or determine that the beautiful sequence doesn't exist.
Input Specification:
The first line contains a positive integer *k* (1<=≤<=*k*<=≤<=26) — the number of strings that should be in a beautiful sequence.
The second line contains string *q*, consisting of lowercase Latin letters. The length of the string is within range from 1 to 100, inclusive.
Output Specification:
If such sequence doesn't exist, then print in a single line "NO" (without the quotes). Otherwise, print in the first line "YES" (without the quotes) and in the next *k* lines print the beautiful sequence of strings *s*1,<=*s*2,<=...,<=*s**k*.
If there are multiple possible answers, print any of them.
Demo Input:
['1\nabca\n', '2\naaacas\n', '4\nabc\n']
Demo Output:
['YES\nabca\n', 'YES\naaa\ncas\n', 'NO\n']
Note:
In the second sample there are two possible answers: {"*aaaca*", "*s*"} and {"*aaa*", "*cas*"}. | ```python
# The first line contains a positive integer k (1 ≤ k ≤ 26) — the number of strings that should be in a beautiful sequence.
# The second line contains string q, consisting of lowercase Latin letters. The length of the string is within range from 1 to 100, inclusive.
K = int(input())
Q = input()
chars = set()
firstOcc = []
for i in range(len(Q)):
if Q[i] not in chars:
chars.add(Q[i])
firstOcc.append(i)
if len(chars) < K:
print("NO")
else:
print("YES")
index = 0
for i in range(K):
# last one
if i != K - 1:
print(Q[firstOcc[i]:firstOcc[i + 1]])
else:
print(Q[firstOcc[i]:])
``` | 3 | |
71 | A | Way Too Long Words | PROGRAMMING | 800 | [
"strings"
] | A. Way Too Long Words | 1 | 256 | Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes. | The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters. | Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data. | [
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] | [
"word\nl10n\ni18n\np43s\n"
] | none | 500 | [
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,691,179,956 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | x = input()
while True:
if len(x) > 10:
y = (len(x)-2)
a,z = (x[0], x[-1])
print(f"{a}{y}{z}")
break
elif len(x) <= 10:
print(x)
break
pneumonoultramicroscopicsilicovolcanoconiosis
| Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none | ```python
x = input()
while True:
if len(x) > 10:
y = (len(x)-2)
a,z = (x[0], x[-1])
print(f"{a}{y}{z}")
break
elif len(x) <= 10:
print(x)
break
pneumonoultramicroscopicsilicovolcanoconiosis
``` | 0 |
357 | B | Flag Day | PROGRAMMING | 1,400 | [
"constructive algorithms",
"implementation"
] | null | null | In Berland, there is the national holiday coming — the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
- overall, there must be *m* dances;- exactly three people must take part in each dance;- each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has *n* dancers, and their number can be less than 3*m*. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the *m* dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the *n* dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances. | The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=105) and *m* (1<=≤<=*m*<=≤<=105) — the number of dancers and the number of dances, correspondingly. Then *m* lines follow, describing the dances in the order of dancing them. The *i*-th line contains three distinct integers — the numbers of the dancers that take part in the *i*-th dance. The dancers are numbered from 1 to *n*. Each dancer takes part in at least one dance. | Print *n* space-separated integers: the *i*-th number must represent the color of the *i*-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists. | [
"7 3\n1 2 3\n1 4 5\n4 6 7\n",
"9 3\n3 6 9\n2 5 8\n1 4 7\n",
"5 2\n4 1 5\n3 1 2\n"
] | [
"1 2 3 3 2 2 1 \n",
"1 1 1 2 2 2 3 3 3 \n",
"2 3 1 1 3 \n"
] | none | 1,000 | [
{
"input": "7 3\n1 2 3\n1 4 5\n4 6 7",
"output": "1 2 3 3 2 2 1 "
},
{
"input": "9 3\n3 6 9\n2 5 8\n1 4 7",
"output": "1 1 1 2 2 2 3 3 3 "
},
{
"input": "5 2\n4 1 5\n3 1 2",
"output": "2 3 1 1 3 "
},
{
"input": "14 5\n1 5 3\n13 10 11\n6 3 8\n14 9 2\n7 4 12",
"output": "1 3 3 2 2 2 1 1 2 2 3 3 1 1 "
},
{
"input": "14 6\n14 3 13\n10 14 5\n6 2 10\n7 13 9\n12 11 8\n1 4 9",
"output": "2 2 2 3 2 1 2 3 1 3 2 1 3 1 "
},
{
"input": "14 6\n11 13 10\n3 10 14\n2 7 12\n13 1 9\n5 11 4\n8 6 5",
"output": "1 1 2 2 3 2 2 1 3 3 1 3 2 1 "
},
{
"input": "13 5\n13 6 2\n13 3 8\n11 4 7\n10 9 5\n1 12 6",
"output": "3 3 3 2 3 2 3 2 2 1 1 1 1 "
},
{
"input": "14 6\n5 4 8\n5 7 12\n3 6 12\n7 11 14\n10 13 2\n10 1 9",
"output": "3 3 3 2 1 1 3 3 2 1 2 2 2 1 "
},
{
"input": "14 5\n4 13 2\n7 2 11\n6 1 5\n14 12 8\n10 3 9",
"output": "2 3 2 1 3 1 2 3 3 1 1 2 2 1 "
},
{
"input": "14 6\n2 14 5\n3 4 5\n6 13 14\n7 13 12\n8 10 11\n9 6 1",
"output": "1 1 1 2 3 3 3 1 2 2 3 2 1 2 "
},
{
"input": "14 6\n7 14 12\n6 1 12\n13 5 2\n2 3 9\n7 4 11\n5 8 10",
"output": "2 3 2 3 2 1 1 1 1 3 2 3 1 2 "
},
{
"input": "13 6\n8 7 6\n11 7 3\n13 9 3\n12 1 13\n8 10 4\n2 7 5",
"output": "3 1 3 2 3 3 2 1 2 3 1 2 1 "
},
{
"input": "13 5\n8 4 3\n1 9 5\n6 2 11\n12 10 4\n7 10 13",
"output": "1 2 3 2 3 1 3 1 2 1 3 3 2 "
},
{
"input": "20 8\n16 19 12\n13 3 5\n1 5 17\n10 19 7\n8 18 2\n3 11 14\n9 20 12\n4 15 6",
"output": "2 3 2 1 3 3 3 1 1 1 1 3 1 3 2 1 1 2 2 2 "
},
{
"input": "19 7\n10 18 14\n5 9 11\n9 17 7\n3 15 4\n6 8 12\n1 2 18\n13 16 19",
"output": "3 1 1 3 1 1 3 2 2 1 3 3 1 3 2 2 1 2 3 "
},
{
"input": "18 7\n17 4 13\n7 1 6\n16 9 13\n9 2 5\n11 12 17\n14 8 10\n3 15 18",
"output": "2 1 1 2 3 3 1 2 2 3 2 3 3 1 2 1 1 3 "
},
{
"input": "20 7\n8 5 11\n3 19 20\n16 1 17\n9 6 2\n7 18 13\n14 12 18\n10 4 15",
"output": "2 3 1 2 2 2 1 1 1 1 3 1 3 3 3 1 3 2 2 3 "
},
{
"input": "20 7\n6 11 20\n19 5 2\n15 10 12\n3 7 8\n9 1 6\n13 17 18\n14 16 4",
"output": "3 3 1 3 2 1 2 3 2 2 2 3 1 1 1 2 2 3 1 3 "
},
{
"input": "18 7\n15 5 1\n6 11 4\n14 8 17\n11 12 13\n3 8 16\n9 4 7\n2 18 10",
"output": "3 1 1 3 2 1 1 2 2 3 2 1 3 1 1 3 3 2 "
},
{
"input": "19 7\n3 10 8\n17 7 4\n1 19 18\n2 9 5\n12 11 15\n11 14 6\n13 9 16",
"output": "1 1 1 3 3 3 2 3 2 2 2 1 1 1 3 3 1 3 2 "
},
{
"input": "19 7\n18 14 4\n3 11 6\n8 10 7\n10 19 16\n17 13 15\n5 1 14\n12 9 2",
"output": "1 3 1 3 3 3 3 1 2 2 2 1 2 2 3 3 1 1 1 "
},
{
"input": "20 7\n18 7 15\n17 5 20\n9 19 12\n16 13 10\n3 6 1\n3 8 11\n4 2 14",
"output": "3 2 1 1 2 2 2 3 1 3 2 3 2 3 3 1 1 1 2 3 "
},
{
"input": "18 7\n8 4 6\n13 17 3\n9 8 12\n12 16 5\n18 2 7\n11 1 10\n5 15 14",
"output": "2 2 3 2 3 3 3 1 3 3 1 2 1 1 2 1 2 1 "
},
{
"input": "99 37\n40 10 7\n10 3 5\n10 31 37\n87 48 24\n33 47 38\n34 87 2\n2 35 28\n99 28 76\n66 51 97\n72 77 9\n18 17 67\n23 69 98\n58 89 99\n42 44 52\n65 41 80\n70 92 74\n62 88 45\n68 27 61\n6 83 95\n39 85 49\n57 75 77\n59 54 81\n56 20 82\n96 4 53\n90 7 11\n16 43 84\n19 25 59\n68 8 93\n73 94 78\n15 71 79\n26 12 50\n30 32 4\n14 22 29\n46 21 36\n60 55 86\n91 8 63\n13 1 64",
"output": "2 2 1 2 3 1 3 3 3 2 1 2 1 1 1 1 2 1 2 2 2 2 1 3 3 1 2 3 3 3 1 1 1 3 1 3 3 3 1 1 2 1 2 2 3 1 2 2 3 3 2 3 3 2 2 1 3 3 1 1 3 1 1 3 1 1 3 1 2 1 2 1 1 3 1 1 2 3 3 3 3 3 2 3 2 3 1 2 1 2 2 2 2 2 3 1 3 3 2 "
},
{
"input": "99 41\n11 70 20\n57 11 76\n52 11 64\n49 70 15\n19 61 17\n71 77 21\n77 59 39\n37 64 68\n17 84 36\n46 11 90\n35 11 14\n36 25 80\n12 43 48\n18 78 42\n82 94 15\n22 10 84\n63 86 4\n98 86 50\n92 60 9\n73 42 65\n21 5 27\n30 24 23\n7 88 49\n40 97 45\n81 56 17\n79 61 33\n13 3 77\n54 6 28\n99 58 8\n29 95 24\n89 74 32\n51 89 66\n87 91 96\n22 34 38\n1 53 72\n55 97 26\n41 16 44\n2 31 47\n83 67 91\n75 85 69\n93 47 62",
"output": "1 1 1 3 2 2 2 3 3 1 1 1 3 2 3 2 3 1 1 3 3 3 3 2 3 3 1 3 3 1 2 3 3 2 3 1 1 1 3 1 1 3 2 3 3 3 3 3 1 3 3 3 2 1 1 2 3 2 1 2 2 1 1 2 1 2 1 3 3 2 1 3 2 2 1 2 2 2 1 2 1 1 3 2 2 2 1 3 1 2 2 1 2 2 1 3 2 1 1 "
},
{
"input": "99 38\n70 56 92\n61 70 68\n18 92 91\n82 43 55\n37 5 43\n47 27 26\n64 63 40\n20 61 57\n69 80 59\n60 89 50\n33 25 86\n38 15 73\n96 85 90\n3 12 64\n95 23 48\n66 30 9\n38 99 45\n67 88 71\n74 11 81\n28 51 79\n72 92 34\n16 77 31\n65 18 94\n3 41 2\n36 42 81\n22 77 83\n44 24 52\n10 75 97\n54 21 53\n4 29 32\n58 39 98\n46 62 16\n76 5 84\n8 87 13\n6 41 14\n19 21 78\n7 49 93\n17 1 35",
"output": "2 3 2 1 1 3 1 1 3 1 2 3 3 2 2 1 1 2 1 2 2 1 2 2 2 3 2 1 2 2 3 3 1 1 3 1 3 1 2 3 1 2 2 1 2 2 1 3 2 3 2 3 3 1 3 2 1 1 3 1 3 3 2 1 1 1 1 2 1 1 3 2 3 1 2 3 2 3 3 2 3 1 3 2 2 3 2 2 2 3 1 3 3 3 1 1 3 3 3 "
},
{
"input": "98 38\n70 23 73\n73 29 86\n93 82 30\n6 29 10\n7 22 78\n55 61 87\n98 2 12\n11 5 54\n44 56 60\n89 76 50\n37 72 43\n47 41 61\n85 40 38\n48 93 20\n90 64 29\n31 68 25\n83 57 41\n51 90 3\n91 97 66\n96 95 1\n50 84 71\n53 19 5\n45 42 28\n16 17 89\n63 58 15\n26 47 39\n21 24 19\n80 74 38\n14 46 75\n88 65 36\n77 92 33\n17 59 34\n35 69 79\n13 94 39\n8 52 4\n67 27 9\n65 62 18\n81 32 49",
"output": "3 2 1 3 2 1 1 1 3 3 1 3 2 1 3 2 3 3 1 2 2 2 2 3 3 2 2 3 2 3 1 2 3 1 1 3 1 3 1 2 1 2 3 1 1 2 3 3 3 3 2 2 3 3 1 2 3 2 2 3 2 1 1 1 2 3 1 2 2 1 1 2 3 2 3 2 1 3 3 1 1 2 2 2 1 1 3 1 1 3 1 2 1 3 2 1 2 1 "
},
{
"input": "99 42\n61 66 47\n10 47 96\n68 86 67\n21 29 10\n55 44 47\n12 82 4\n45 71 55\n86 3 95\n16 99 93\n14 92 82\n12 59 20\n73 24 8\n79 72 48\n44 87 39\n87 84 97\n47 70 37\n49 77 95\n39 75 28\n75 25 5\n44 41 36\n76 86 78\n73 6 90\n8 22 58\n9 72 63\n81 42 14\n1 21 35\n91 54 15\n30 13 39\n56 89 79\n11 2 76\n19 65 52\n23 85 74\n7 38 24\n57 94 81\n43 34 60\n62 82 27\n69 64 53\n18 32 17\n24 31 88\n51 50 33\n40 80 98\n83 46 26",
"output": "2 3 1 3 3 3 3 3 1 2 2 1 2 3 3 1 3 1 1 2 3 2 1 2 1 3 3 1 1 1 1 2 3 2 1 3 1 1 3 1 1 2 1 2 2 2 3 3 1 2 1 3 3 2 1 2 2 1 3 3 1 1 3 2 2 2 3 1 1 2 3 2 1 3 2 1 2 3 1 2 1 2 1 3 2 2 1 3 3 2 1 1 3 3 3 1 2 3 2 "
}
] | 1,591,080,645 | 2,147,483,647 | Python 3 | OK | TESTS | 63 | 514 | 10,035,200 | n,m=map(int,input().split())
li=[]
for i in range(m):
li.append([int(x) for x in input().split()])
d={}
for i in range(m):
x=li[i]
use=[True]*3
l=[]
for ele in x:
if d.get(ele,0)!=0:
use[d.get(ele)-1]=False
for i in range(3):
if use[i]==True:
l.append(i)
for ele in x:
if d.get(ele,0)==0:
d[ele]=l.pop()+1
for ele in sorted(d):
print(d[ele],end=" ") | Title: Flag Day
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland, there is the national holiday coming — the Flag Day. In the honor of this event the president of the country decided to make a big dance party and asked your agency to organize it. He has several conditions:
- overall, there must be *m* dances;- exactly three people must take part in each dance;- each dance must have one dancer in white clothes, one dancer in red clothes and one dancer in blue clothes (these are the colors of the national flag of Berland).
The agency has *n* dancers, and their number can be less than 3*m*. That is, some dancers will probably have to dance in more than one dance. All of your dancers must dance on the party. However, if some dance has two or more dancers from a previous dance, then the current dance stops being spectacular. Your agency cannot allow that to happen, so each dance has at most one dancer who has danced in some previous dance.
You considered all the criteria and made the plan for the *m* dances: each dance had three dancers participating in it. Your task is to determine the clothes color for each of the *n* dancers so that the President's third condition fulfilled: each dance must have a dancer in white, a dancer in red and a dancer in blue. The dancers cannot change clothes between the dances.
Input Specification:
The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=105) and *m* (1<=≤<=*m*<=≤<=105) — the number of dancers and the number of dances, correspondingly. Then *m* lines follow, describing the dances in the order of dancing them. The *i*-th line contains three distinct integers — the numbers of the dancers that take part in the *i*-th dance. The dancers are numbered from 1 to *n*. Each dancer takes part in at least one dance.
Output Specification:
Print *n* space-separated integers: the *i*-th number must represent the color of the *i*-th dancer's clothes (1 for white, 2 for red, 3 for blue). If there are multiple valid solutions, print any of them. It is guaranteed that at least one solution exists.
Demo Input:
['7 3\n1 2 3\n1 4 5\n4 6 7\n', '9 3\n3 6 9\n2 5 8\n1 4 7\n', '5 2\n4 1 5\n3 1 2\n']
Demo Output:
['1 2 3 3 2 2 1 \n', '1 1 1 2 2 2 3 3 3 \n', '2 3 1 1 3 \n']
Note:
none | ```python
n,m=map(int,input().split())
li=[]
for i in range(m):
li.append([int(x) for x in input().split()])
d={}
for i in range(m):
x=li[i]
use=[True]*3
l=[]
for ele in x:
if d.get(ele,0)!=0:
use[d.get(ele)-1]=False
for i in range(3):
if use[i]==True:
l.append(i)
for ele in x:
if d.get(ele,0)==0:
d[ele]=l.pop()+1
for ele in sorted(d):
print(d[ele],end=" ")
``` | 3 | |
22 | A | Second Order Statistics | PROGRAMMING | 800 | [
"brute force"
] | A. Second Order Statistics | 2 | 256 | Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem. | The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value. | If the given sequence has the second order statistics, output this order statistics, otherwise output NO. | [
"4\n1 2 2 -4\n",
"5\n1 2 3 1 1\n"
] | [
"1\n",
"2\n"
] | none | 0 | [
{
"input": "4\n1 2 2 -4",
"output": "1"
},
{
"input": "5\n1 2 3 1 1",
"output": "2"
},
{
"input": "1\n28",
"output": "NO"
},
{
"input": "2\n-28 12",
"output": "12"
},
{
"input": "3\n-83 40 -80",
"output": "-80"
},
{
"input": "8\n93 77 -92 26 21 -48 53 91",
"output": "-48"
},
{
"input": "20\n-72 -9 -86 80 7 -10 40 -27 -94 92 96 56 28 -19 79 36 -3 -73 -63 -49",
"output": "-86"
},
{
"input": "49\n-74 -100 -80 23 -8 -83 -41 -20 48 17 46 -73 -55 67 85 4 40 -60 -69 -75 56 -74 -42 93 74 -95 64 -46 97 -47 55 0 -78 -34 -31 40 -63 -49 -76 48 21 -1 -49 -29 -98 -11 76 26 94",
"output": "-98"
},
{
"input": "88\n63 48 1 -53 -89 -49 64 -70 -49 71 -17 -16 76 81 -26 -50 67 -59 -56 97 2 100 14 18 -91 -80 42 92 -25 -88 59 8 -56 38 48 -71 -78 24 -14 48 -1 69 73 -76 54 16 -92 44 47 33 -34 -17 -81 21 -59 -61 53 26 10 -76 67 35 -29 70 65 -13 -29 81 80 32 74 -6 34 46 57 1 -45 -55 69 79 -58 11 -2 22 -18 -16 -89 -46",
"output": "-91"
},
{
"input": "100\n34 32 88 20 76 53 -71 -39 -98 -10 57 37 63 -3 -54 -64 -78 -82 73 20 -30 -4 22 75 51 -64 -91 29 -52 -48 83 19 18 -47 46 57 -44 95 89 89 -30 84 -83 67 58 -99 -90 -53 92 -60 -5 -56 -61 27 68 -48 52 -95 64 -48 -30 -67 66 89 14 -33 -31 -91 39 7 -94 -54 92 -96 -99 -83 -16 91 -28 -66 81 44 14 -85 -21 18 40 16 -13 -82 -33 47 -10 -40 -19 10 25 60 -34 -89",
"output": "-98"
},
{
"input": "2\n-1 -1",
"output": "NO"
},
{
"input": "3\n-2 -2 -2",
"output": "NO"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "NO"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100",
"output": "100"
},
{
"input": "10\n40 71 -85 -85 40 -85 -85 64 -85 47",
"output": "40"
},
{
"input": "23\n-90 -90 -41 -64 -64 -90 -15 10 -43 -90 -64 -64 89 -64 36 47 38 -90 -64 -90 -90 68 -90",
"output": "-64"
},
{
"input": "39\n-97 -93 -42 -93 -97 -93 56 -97 -97 -97 76 -33 -60 91 7 82 17 47 -97 -97 -93 73 -97 12 -97 -97 -97 -97 56 -92 -83 -93 -93 49 -93 -97 -97 -17 -93",
"output": "-93"
},
{
"input": "51\n-21 6 -35 -98 -86 -98 -86 -43 -65 32 -98 -40 96 -98 -98 -98 -98 -86 -86 -98 56 -86 -98 -98 -30 -98 -86 -31 -98 -86 -86 -86 -86 -30 96 -86 -86 -86 -60 25 88 -86 -86 58 31 -47 57 -86 37 44 -83",
"output": "-86"
},
{
"input": "66\n-14 -95 65 -95 -95 -97 -90 -71 -97 -97 70 -95 -95 -97 -95 -27 35 -87 -95 -5 -97 -97 87 34 -49 -95 -97 -95 -97 -95 -30 -95 -97 47 -95 -17 -97 -95 -97 -69 51 -97 -97 -95 -75 87 59 21 63 56 76 -91 98 -97 6 -97 -95 -95 -97 -73 11 -97 -35 -95 -95 -43",
"output": "-95"
},
{
"input": "77\n-67 -93 -93 -92 97 29 93 -93 -93 -5 -93 -7 60 -92 -93 44 -84 68 -92 -93 69 -92 -37 56 43 -93 35 -92 -93 19 -79 18 -92 -93 -93 -37 -93 -47 -93 -92 -92 74 67 19 40 -92 -92 -92 -92 -93 -93 -41 -93 -92 -93 -93 -92 -93 51 -80 6 -42 -92 -92 -66 -12 -92 -92 -3 93 -92 -49 -93 40 62 -92 -92",
"output": "-92"
},
{
"input": "89\n-98 40 16 -87 -98 63 -100 55 -96 -98 -21 -100 -93 26 -98 -98 -100 -89 -98 -5 -65 -28 -100 -6 -66 67 -100 -98 -98 10 -98 -98 -70 7 -98 2 -100 -100 -98 25 -100 -100 -98 23 -68 -100 -98 3 98 -100 -98 -98 -98 -98 -24 -100 -100 -9 -98 35 -100 99 -5 -98 -100 -100 37 -100 -84 57 -98 40 -47 -100 -1 -92 -76 -98 -98 -100 -100 -100 -63 30 21 -100 -100 -100 -12",
"output": "-98"
},
{
"input": "99\n10 -84 -100 -100 73 -64 -100 -94 33 -100 -100 -100 -100 71 64 24 7 -100 -32 -100 -100 77 -100 62 -12 55 45 -100 -100 -80 -100 -100 -100 -100 -100 -100 -100 -100 -100 -39 -48 -100 -34 47 -100 -100 -100 -100 -100 -77 -100 -100 -100 -100 -100 -100 -52 40 -55 -100 -44 -100 72 33 70 -100 -100 -78 -100 -3 100 -77 22 -100 95 -30 -100 10 -69 -100 -100 -100 -100 52 -39 -100 -100 -100 7 -100 -98 -66 95 -17 -100 52 -100 68 -100",
"output": "-98"
},
{
"input": "100\n-99 -98 -64 89 53 57 -99 29 -78 18 -3 -54 76 -98 -99 -98 37 -98 19 -47 89 73 -98 -91 -99 -99 -98 -48 -99 22 -99 -98 -99 -99 -98 -60 84 67 -99 -98 20 -98 88 -98 46 -98 -99 -98 -99 -71 -99 -98 -98 -39 83 95 -98 63 -98 -99 32 -98 -99 -64 57 -30 -53 -83 -4 -99 58 20 -98 -10 -99 -44 -99 -99 -99 -99 75 34 -98 -52 -98 -30 -98 -99 -98 -98 51 -99 -99 -99 -98 -99 -99 -82 -90 92",
"output": "-98"
},
{
"input": "3\n1 2 3",
"output": "2"
},
{
"input": "3\n1 3 2",
"output": "2"
},
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 3 1",
"output": "2"
},
{
"input": "3\n3 1 2",
"output": "2"
},
{
"input": "3\n3 2 1",
"output": "2"
},
{
"input": "9\n99 99 99 99 100 100 100 100 100",
"output": "100"
},
{
"input": "5\n-100 -100 -100 -100 -100",
"output": "NO"
}
] | 1,632,186,434 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 2 | 92 | 6,656,000 | n = int(input())
v = input().split(" ")
for i in range(0, n):
v[i] = int(v[i])
v = set(v)
v.discard(min(v))
print(min(v))
| Title: Second Order Statistics
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem.
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value.
Output Specification:
If the given sequence has the second order statistics, output this order statistics, otherwise output NO.
Demo Input:
['4\n1 2 2 -4\n', '5\n1 2 3 1 1\n']
Demo Output:
['1\n', '2\n']
Note:
none | ```python
n = int(input())
v = input().split(" ")
for i in range(0, n):
v[i] = int(v[i])
v = set(v)
v.discard(min(v))
print(min(v))
``` | -1 |
1,005 | B | Delete from the Left | PROGRAMMING | 900 | [
"brute force",
"implementation",
"strings"
] | null | null | You are given two strings $s$ and $t$. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by $1$. You can't choose a string if it is empty.
For example:
- by applying a move to the string "where", the result is the string "here", - by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings $s$ and $t$ equal. | The first line of the input contains $s$. In the second line of the input contains $t$. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and $2\cdot10^5$, inclusive. | Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings. | [
"test\nwest\n",
"codeforces\nyes\n",
"test\nyes\n",
"b\nab\n"
] | [
"2\n",
"9\n",
"7\n",
"1\n"
] | In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" $8$ times. As a result, the string becomes "codeforces" $\to$ "es". The move should be applied to the string "yes" once. The result is the same string "yes" $\to$ "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted. | 0 | [
{
"input": "test\nwest",
"output": "2"
},
{
"input": "codeforces\nyes",
"output": "9"
},
{
"input": "test\nyes",
"output": "7"
},
{
"input": "b\nab",
"output": "1"
},
{
"input": "z\nz",
"output": "0"
},
{
"input": "abacabadabacaba\nabacabadacaba",
"output": "18"
},
{
"input": "z\na",
"output": "2"
},
{
"input": "ama\nsama",
"output": "1"
},
{
"input": "aaabc\nbc",
"output": "3"
},
{
"input": "lxuyspmieeuyxluv\ndfwxqvavzxctblaa",
"output": "32"
},
{
"input": "ahmedaly\naly",
"output": "5"
},
{
"input": "aa\na",
"output": "1"
},
{
"input": "aa\nba",
"output": "2"
},
{
"input": "cccba\nba",
"output": "3"
},
{
"input": "a\nab",
"output": "3"
},
{
"input": "dadc\ndddc",
"output": "4"
},
{
"input": "aacaaab\nb",
"output": "6"
},
{
"input": "dddc\ncdcde",
"output": "9"
},
{
"input": "bbdab\ndab",
"output": "2"
},
{
"input": "a\naa",
"output": "1"
},
{
"input": "aa\nab",
"output": "4"
},
{
"input": "codeforces\nces",
"output": "7"
}
] | 1,689,355,257 | 2,147,483,647 | Python 3 | OK | TESTS | 28 | 93 | 819,200 | s1=input()
s2=input()
num=len(s1)+len(s2)
i=len(s1)-1
j=len(s2)-1
dl=0
while(s1[i]==s2[j]) and i>=0 and j>=0:
dl+=2
i-=1
j-=1
print(num-dl) | Title: Delete from the Left
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two strings $s$ and $t$. In a single move, you can choose any of two strings and delete the first (that is, the leftmost) character. After a move, the length of the string decreases by $1$. You can't choose a string if it is empty.
For example:
- by applying a move to the string "where", the result is the string "here", - by applying a move to the string "a", the result is an empty string "".
You are required to make two given strings equal using the fewest number of moves. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the initial strings.
Write a program that finds the minimum number of moves to make two given strings $s$ and $t$ equal.
Input Specification:
The first line of the input contains $s$. In the second line of the input contains $t$. Both strings consist only of lowercase Latin letters. The number of letters in each string is between 1 and $2\cdot10^5$, inclusive.
Output Specification:
Output the fewest number of moves required. It is possible that, in the end, both strings will be equal to the empty string, and so, are equal to each other. In this case, the answer is obviously the sum of the lengths of the given strings.
Demo Input:
['test\nwest\n', 'codeforces\nyes\n', 'test\nyes\n', 'b\nab\n']
Demo Output:
['2\n', '9\n', '7\n', '1\n']
Note:
In the first example, you should apply the move once to the first string and apply the move once to the second string. As a result, both strings will be equal to "est".
In the second example, the move should be applied to the string "codeforces" $8$ times. As a result, the string becomes "codeforces" $\to$ "es". The move should be applied to the string "yes" once. The result is the same string "yes" $\to$ "es".
In the third example, you can make the strings equal only by completely deleting them. That is, in the end, both strings will be empty.
In the fourth example, the first character of the second string should be deleted. | ```python
s1=input()
s2=input()
num=len(s1)+len(s2)
i=len(s1)-1
j=len(s2)-1
dl=0
while(s1[i]==s2[j]) and i>=0 and j>=0:
dl+=2
i-=1
j-=1
print(num-dl)
``` | 3 | |
26 | A | Almost Prime | PROGRAMMING | 900 | [
"number theory"
] | A. Almost Prime | 2 | 256 | A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and *n*, inclusive. | Input contains one integer number *n* (1<=≤<=*n*<=≤<=3000). | Output the amount of almost prime numbers between 1 and *n*, inclusive. | [
"10\n",
"21\n"
] | [
"2\n",
"8\n"
] | none | 500 | [
{
"input": "10",
"output": "2"
},
{
"input": "21",
"output": "8"
},
{
"input": "1",
"output": "0"
},
{
"input": "2",
"output": "0"
},
{
"input": "4",
"output": "0"
},
{
"input": "3",
"output": "0"
},
{
"input": "8",
"output": "1"
},
{
"input": "19",
"output": "6"
},
{
"input": "40",
"output": "19"
},
{
"input": "77",
"output": "41"
},
{
"input": "222",
"output": "125"
},
{
"input": "987",
"output": "501"
},
{
"input": "1000",
"output": "508"
},
{
"input": "2000",
"output": "958"
},
{
"input": "3000",
"output": "1375"
},
{
"input": "2999",
"output": "1375"
},
{
"input": "2998",
"output": "1375"
},
{
"input": "2997",
"output": "1374"
},
{
"input": "1429",
"output": "706"
},
{
"input": "1673",
"output": "808"
},
{
"input": "1500",
"output": "732"
},
{
"input": "500",
"output": "266"
},
{
"input": "856",
"output": "439"
}
] | 1,637,293,178 | 2,147,483,647 | Python 3 | OK | TESTS | 23 | 216 | 0 | n=int(input())
MM=n//2+1
prime=[True for i in range(MM)]
primes=[]
for i in range(2, MM):
if prime[i]:
for j in range(i+i, MM, i):
prime[j] = False
primes.append(i)
ans = 0
for i in range(6, n + 1):
cnt = 0
for p in primes:
if i % p == 0:
cnt += 1
if cnt == 2:
ans += 1
print(ans) | Title: Almost Prime
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A number is called almost prime if it has exactly two distinct prime divisors. For example, numbers 6, 18, 24 are almost prime, while 4, 8, 9, 42 are not. Find the amount of almost prime numbers which are between 1 and *n*, inclusive.
Input Specification:
Input contains one integer number *n* (1<=≤<=*n*<=≤<=3000).
Output Specification:
Output the amount of almost prime numbers between 1 and *n*, inclusive.
Demo Input:
['10\n', '21\n']
Demo Output:
['2\n', '8\n']
Note:
none | ```python
n=int(input())
MM=n//2+1
prime=[True for i in range(MM)]
primes=[]
for i in range(2, MM):
if prime[i]:
for j in range(i+i, MM, i):
prime[j] = False
primes.append(i)
ans = 0
for i in range(6, n + 1):
cnt = 0
for p in primes:
if i % p == 0:
cnt += 1
if cnt == 2:
ans += 1
print(ans)
``` | 3.946 |
554 | A | Kyoya and Photobooks | PROGRAMMING | 900 | [
"brute force",
"math",
"strings"
] | null | null | Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem. | The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=20). String *s* consists only of lowercase English letters. | Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make. | [
"a\n",
"hi\n"
] | [
"51\n",
"76\n"
] | In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets. | 250 | [
{
"input": "a",
"output": "51"
},
{
"input": "hi",
"output": "76"
},
{
"input": "y",
"output": "51"
},
{
"input": "kgan",
"output": "126"
},
{
"input": "zoabkyuvus",
"output": "276"
},
{
"input": "spyemhyznjieyhhbk",
"output": "451"
},
{
"input": "xulsyfkuizjauadjjopu",
"output": "526"
},
{
"input": "e",
"output": "51"
},
{
"input": "zv",
"output": "76"
},
{
"input": "jgv",
"output": "101"
},
{
"input": "zsfo",
"output": "126"
},
{
"input": "jselr",
"output": "151"
},
{
"input": "dwemig",
"output": "176"
},
{
"input": "mddoxsf",
"output": "201"
},
{
"input": "jgirkrmi",
"output": "226"
},
{
"input": "spkxurcum",
"output": "251"
},
{
"input": "fykkiubdkt",
"output": "276"
},
{
"input": "fznbcxsxygs",
"output": "301"
},
{
"input": "qcrvrdqcbtou",
"output": "326"
},
{
"input": "qktrbjzrqgmlr",
"output": "351"
},
{
"input": "foamodbvptlxxg",
"output": "376"
},
{
"input": "ydzpjhsidipricw",
"output": "401"
},
{
"input": "lpfpndmjfvqejdgf",
"output": "426"
},
{
"input": "ofkvparuvjtggnmab",
"output": "451"
},
{
"input": "xxncfutrtxcwdzwbgs",
"output": "476"
},
{
"input": "zovhffccflkgqncsdte",
"output": "501"
},
{
"input": "cskgsxywlvfeicoueglr",
"output": "526"
},
{
"input": "gggggggggggggggggggg",
"output": "526"
},
{
"input": "qdqdddqddqqddqddqdqd",
"output": "526"
},
{
"input": "takttttaakaaktakttkt",
"output": "526"
},
{
"input": "coccoooogogcgocccmcg",
"output": "526"
},
{
"input": "kskkskkkssksssk",
"output": "401"
},
{
"input": "lllllllllllllll",
"output": "401"
}
] | 1,577,570,688 | 2,147,483,647 | Python 3 | OK | TESTS | 33 | 109 | 307,200 | x=input()
print((len(x)+1)*26-len(x)) | Title: Kyoya and Photobooks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplicated). A photo booklet can be described as a string of lowercase letters, consisting of the photos in the booklet in order. He now wants to sell some "special edition" photobooks, each with one extra photo inserted anywhere in the book. He wants to make as many distinct photobooks as possible, so he can make more money. He asks Haruhi, how many distinct photobooks can he make by inserting one extra photo into the photobook he already has?
Please help Haruhi solve this problem.
Input Specification:
The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=20). String *s* consists only of lowercase English letters.
Output Specification:
Output a single integer equal to the number of distinct photobooks Kyoya Ootori can make.
Demo Input:
['a\n', 'hi\n']
Demo Output:
['51\n', '76\n']
Note:
In the first case, we can make 'ab','ac',...,'az','ba','ca',...,'za', and 'aa', producing a total of 51 distinct photo booklets. | ```python
x=input()
print((len(x)+1)*26-len(x))
``` | 3 | |
50 | A | Domino piling | PROGRAMMING | 800 | [
"greedy",
"math"
] | A. Domino piling | 2 | 256 | You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions. | In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16). | Output one number — the maximal number of dominoes, which can be placed. | [
"2 4\n",
"3 3\n"
] | [
"4\n",
"4\n"
] | none | 500 | [
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,634,677,755 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 92 | 0 | m, n = tuple(map(int, input().split()))
area = m * n
if area % 2 == 0:
print(int(area/2))
else:
print(int((area-1)/2)) | Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
m, n = tuple(map(int, input().split()))
area = m * n
if area % 2 == 0:
print(int(area/2))
else:
print(int((area-1)/2))
``` | 3.977 |
155 | A | I_love_\%username\% | PROGRAMMING | 800 | [
"brute force"
] | null | null | Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him. | The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000. | Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests. | [
"5\n100 50 200 150 200\n",
"10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n"
] | [
"2\n",
"4\n"
] | In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing. | 500 | [
{
"input": "5\n100 50 200 150 200",
"output": "2"
},
{
"input": "10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242",
"output": "4"
},
{
"input": "1\n6",
"output": "0"
},
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "5\n100 36 53 7 81",
"output": "2"
},
{
"input": "5\n7 36 53 81 100",
"output": "4"
},
{
"input": "5\n100 81 53 36 7",
"output": "4"
},
{
"input": "10\n8 6 3 4 9 10 7 7 1 3",
"output": "5"
},
{
"input": "10\n1627 1675 1488 1390 1812 1137 1746 1324 1952 1862",
"output": "6"
},
{
"input": "10\n1 3 3 4 6 7 7 8 9 10",
"output": "7"
},
{
"input": "10\n1952 1862 1812 1746 1675 1627 1488 1390 1324 1137",
"output": "9"
},
{
"input": "25\n1448 4549 2310 2725 2091 3509 1565 2475 2232 3989 4231 779 2967 2702 608 3739 721 1552 2767 530 3114 665 1940 48 4198",
"output": "5"
},
{
"input": "33\n1097 1132 1091 1104 1049 1038 1023 1080 1104 1029 1035 1061 1049 1060 1088 1106 1105 1087 1063 1076 1054 1103 1047 1041 1028 1120 1126 1063 1117 1110 1044 1093 1101",
"output": "5"
},
{
"input": "34\n821 5536 2491 6074 7216 9885 764 1603 778 8736 8987 771 617 1587 8943 7922 439 7367 4115 8886 7878 6899 8811 5752 3184 3401 9760 9400 8995 4681 1323 6637 6554 6498",
"output": "7"
},
{
"input": "68\n6764 6877 6950 6768 6839 6755 6726 6778 6699 6805 6777 6985 6821 6801 6791 6805 6940 6761 6677 6999 6911 6699 6959 6933 6903 6843 6972 6717 6997 6756 6789 6668 6735 6852 6735 6880 6723 6834 6810 6694 6780 6679 6698 6857 6826 6896 6979 6968 6957 6988 6960 6700 6919 6892 6984 6685 6813 6678 6715 6857 6976 6902 6780 6686 6777 6686 6842 6679",
"output": "9"
},
{
"input": "60\n9000 9014 9034 9081 9131 9162 9174 9199 9202 9220 9221 9223 9229 9235 9251 9260 9268 9269 9270 9298 9307 9309 9313 9323 9386 9399 9407 9495 9497 9529 9531 9544 9614 9615 9627 9627 9643 9654 9656 9657 9685 9699 9701 9736 9745 9758 9799 9827 9843 9845 9854 9854 9885 9891 9896 9913 9942 9963 9986 9992",
"output": "57"
},
{
"input": "100\n7 61 12 52 41 16 34 99 30 44 48 89 31 54 21 1 48 52 61 15 35 87 21 76 64 92 44 81 16 93 84 92 32 15 68 76 53 39 26 4 11 26 7 4 99 99 61 65 55 85 65 67 47 39 2 74 63 49 98 87 5 94 22 30 25 42 31 84 49 23 89 60 16 26 92 27 9 57 75 61 94 35 83 47 99 100 63 24 91 88 79 10 15 45 22 64 3 11 89 83",
"output": "4"
},
{
"input": "100\n9999 9999 9999 9998 9998 9998 9997 9996 9996 9995 9993 9993 9991 9990 9989 9986 9984 9984 9983 9981 9981 9980 9980 9980 9979 9977 9977 9977 9977 9977 9976 9976 9975 9975 9973 9972 9972 9972 9972 9971 9969 9969 9969 9968 9967 9965 9962 9962 9962 9962 9960 9958 9958 9955 9954 9954 9954 9954 9950 9949 9949 9947 9943 9943 9940 9940 9939 9937 9937 9937 9936 9935 9932 9932 9931 9929 9927 9927 9926 9923 9919 9919 9918 9918 9918 9917 9917 9916 9916 9915 9914 9911 9910 9907 9907 9906 9905 9905 9904 9901",
"output": "57"
},
{
"input": "100\n158 159 157 156 155 154 160 153 161 152 162 163 151 164 150 165 149 148 147 166 146 167 145 144 143 142 168 141 169 170 140 139 138 137 171 136 135 134 172 173 174 133 175 132 177 178 131 179 180 130 129 181 128 127 182 126 183 125 124 184 123 122 121 120 119 118 117 185 186 187 188 189 116 190 115 191 192 193 114 113 112 111 110 109 108 194 195 107 106 105 196 197 104 198 199 103 102 200 101 100",
"output": "99"
},
{
"input": "2\n0 10000",
"output": "1"
},
{
"input": "2\n5 5",
"output": "0"
},
{
"input": "2\n1 1",
"output": "0"
},
{
"input": "2\n10 10",
"output": "0"
},
{
"input": "1\n0",
"output": "0"
}
] | 1,684,860,775 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 7 | 92 | 0 | t=int(input())
s=list(map(int,input().split()))
c=0
v=0
for i in range(len(s)-1):
if s[i+1]>s[i]:
c+=1
if c==0:
for i in range(len(s)-1):
if s[i+1]<s[i]:
v+=1
print(v)
else:
print(c) | Title: I_love_\%username\%
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya adores sport programming. He can't write programs but he loves to watch the contests' progress. Vasya even has a favorite coder and Vasya pays special attention to him.
One day Vasya decided to collect the results of all contests where his favorite coder participated and track the progress of his coolness. For each contest where this coder participated, he wrote out a single non-negative number — the number of points his favorite coder earned in the contest. Vasya wrote out the points for the contest in the order, in which the contests run (naturally, no two contests ran simultaneously).
Vasya considers a coder's performance in a contest amazing in two situations: he can break either his best or his worst performance record. First, it is amazing if during the contest the coder earns strictly more points that he earned on each past contest. Second, it is amazing if during the contest the coder earns strictly less points that he earned on each past contest. A coder's first contest isn't considered amazing. Now he wants to count the number of amazing performances the coder had throughout his whole history of participating in contests. But the list of earned points turned out long and Vasya can't code... That's why he asks you to help him.
Input Specification:
The first line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of contests where the coder participated.
The next line contains *n* space-separated non-negative integer numbers — they are the points which the coder has earned. The points are given in the chronological order. All points do not exceed 10000.
Output Specification:
Print the single number — the number of amazing performances the coder has had during his whole history of participating in the contests.
Demo Input:
['5\n100 50 200 150 200\n', '10\n4664 6496 5814 7010 5762 5736 6944 4850 3698 7242\n']
Demo Output:
['2\n', '4\n']
Note:
In the first sample the performances number 2 and 3 are amazing.
In the second sample the performances number 2, 4, 9 and 10 are amazing. | ```python
t=int(input())
s=list(map(int,input().split()))
c=0
v=0
for i in range(len(s)-1):
if s[i+1]>s[i]:
c+=1
if c==0:
for i in range(len(s)-1):
if s[i+1]<s[i]:
v+=1
print(v)
else:
print(c)
``` | 0 | |
549 | G | Happy Line | PROGRAMMING | 2,200 | [
"constructive algorithms",
"greedy",
"sortings"
] | null | null | Do you like summer? Residents of Berland do. They especially love eating ice cream in the hot summer. So this summer day a large queue of *n* Berland residents lined up in front of the ice cream stall. We know that each of them has a certain amount of berland dollars with them. The residents of Berland are nice people, so each person agrees to swap places with the person right behind him for just 1 dollar. More formally, if person *a* stands just behind person *b*, then person *a* can pay person *b* 1 dollar, then *a* and *b* get swapped. Of course, if person *a* has zero dollars, he can not swap places with person *b*.
Residents of Berland are strange people. In particular, they get upset when there is someone with a strictly smaller sum of money in the line in front of them.
Can you help the residents of Berland form such order in the line so that they were all happy? A happy resident is the one who stands first in the line or the one in front of who another resident stands with not less number of dollars. Note that the people of Berland are people of honor and they agree to swap places only in the manner described above. | The first line contains integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of residents who stand in the line.
The second line contains *n* space-separated integers *a**i* (0<=≤<=*a**i*<=≤<=109), where *a**i* is the number of Berland dollars of a man standing on the *i*-th position in the line. The positions are numbered starting from the end of the line. | If it is impossible to make all the residents happy, print ":(" without the quotes. Otherwise, print in the single line *n* space-separated integers, the *i*-th of them must be equal to the number of money of the person on position *i* in the new line. If there are multiple answers, print any of them. | [
"2\n11 8\n",
"5\n10 9 7 10 6\n",
"3\n12 3 3\n"
] | [
"9 10 ",
":(\n",
"4 4 10 "
] | In the first sample two residents should swap places, after that the first resident has 10 dollars and he is at the head of the line and the second resident will have 9 coins and he will be at the end of the line.
In the second sample it is impossible to achieve the desired result.
In the third sample the first person can swap with the second one, then they will have the following numbers of dollars: 4 11 3, then the second person (in the new line) swaps with the third one, and the resulting numbers of dollars will equal to: 4 4 10. In this line everybody will be happy. | 1,500 | [
{
"input": "2\n11 8",
"output": "9 10 "
},
{
"input": "5\n10 9 7 10 6",
"output": ":("
},
{
"input": "3\n12 3 3",
"output": "4 4 10 "
},
{
"input": "4\n7 3 9 10",
"output": "4 6 9 10 "
},
{
"input": "1\n1",
"output": "1 "
},
{
"input": "5\n15 5 8 6 3",
"output": "6 6 7 7 11 "
},
{
"input": "1\n1000000000",
"output": "1000000000 "
},
{
"input": "2\n2 1",
"output": ":("
},
{
"input": "3\n11 1 9",
"output": ":("
},
{
"input": "1\n0",
"output": "0 "
},
{
"input": "20\n38 32 21 44 26 24 15 32 34 40 31 33 33 13 26 28 12 10 14 18",
"output": "21 22 24 24 24 24 24 25 25 28 28 28 28 28 28 28 28 28 29 30 "
},
{
"input": "20\n43 38 20 41 16 37 27 29 19 17 24 19 28 8 14 32 13 21 32 16",
"output": "20 20 20 23 23 23 23 23 25 25 25 25 26 26 26 27 27 27 29 31 "
},
{
"input": "20\n44 50 41 18 28 31 21 38 12 20 28 15 12 29 16 31 34 24 19 15",
"output": "20 20 22 23 23 24 24 25 26 27 27 27 29 29 29 29 29 29 32 32 "
},
{
"input": "10\n920480900 920480899 920480898 920480897 920480896 920480895 920480894 920480893 920480892 920480891",
"output": ":("
},
{
"input": "10\n536259132 536259132 536259132 536259132 536259132 536259132 536259132 536259132 536259132 536259132",
"output": "536259132 536259132 536259132 536259132 536259132 536259132 536259132 536259132 536259132 536259132 "
},
{
"input": "10\n876584065 876584063 876584061 876584059 876584057 876584055 876584053 876584051 876584049 876584047",
"output": "876584056 876584056 876584056 876584056 876584056 876584056 876584056 876584056 876584056 876584056 "
},
{
"input": "10\n528402489 528402486 528402483 528402480 528402477 528402474 528402471 528402468 528402465 528402462",
"output": "528402471 528402472 528402473 528402474 528402475 528402476 528402477 528402478 528402479 528402480 "
},
{
"input": "10\n383593860 383593860 383593860 383593860 383593860 383593858 383593856 383593854 383593852 383593850",
"output": ":("
},
{
"input": "10\n198447418 198447416 198447414 198447412 198447410 198447410 198447410 198447410 198447410 198447410",
"output": ":("
},
{
"input": "10\n520230360 520230361 520230362 520230363 520230364 520230365 520230366 520230367 520230368 520230369",
"output": "520230360 520230361 520230362 520230363 520230364 520230365 520230366 520230367 520230368 520230369 "
}
] | 1,433,602,239 | 6,639 | Python 3 | TIME_LIMIT_EXCEEDED | TESTS | 18 | 1,000 | 14,233,600 | n = int(input())
s = [int(i) for i in input().split()]
for i in range(n-1, 0, -1):
if s[i] < s[i-1]:
u = 0
for j in range(i-1, n-1):
if s[j]>s[j+1]:
s[j], s[j+1] = s[j+1]+1, s[j]-1
if s != sorted(s):
print(":(")
else:
print(*s)
| Title: Happy Line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Do you like summer? Residents of Berland do. They especially love eating ice cream in the hot summer. So this summer day a large queue of *n* Berland residents lined up in front of the ice cream stall. We know that each of them has a certain amount of berland dollars with them. The residents of Berland are nice people, so each person agrees to swap places with the person right behind him for just 1 dollar. More formally, if person *a* stands just behind person *b*, then person *a* can pay person *b* 1 dollar, then *a* and *b* get swapped. Of course, if person *a* has zero dollars, he can not swap places with person *b*.
Residents of Berland are strange people. In particular, they get upset when there is someone with a strictly smaller sum of money in the line in front of them.
Can you help the residents of Berland form such order in the line so that they were all happy? A happy resident is the one who stands first in the line or the one in front of who another resident stands with not less number of dollars. Note that the people of Berland are people of honor and they agree to swap places only in the manner described above.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of residents who stand in the line.
The second line contains *n* space-separated integers *a**i* (0<=≤<=*a**i*<=≤<=109), where *a**i* is the number of Berland dollars of a man standing on the *i*-th position in the line. The positions are numbered starting from the end of the line.
Output Specification:
If it is impossible to make all the residents happy, print ":(" without the quotes. Otherwise, print in the single line *n* space-separated integers, the *i*-th of them must be equal to the number of money of the person on position *i* in the new line. If there are multiple answers, print any of them.
Demo Input:
['2\n11 8\n', '5\n10 9 7 10 6\n', '3\n12 3 3\n']
Demo Output:
['9 10 ', ':(\n', '4 4 10 ']
Note:
In the first sample two residents should swap places, after that the first resident has 10 dollars and he is at the head of the line and the second resident will have 9 coins and he will be at the end of the line.
In the second sample it is impossible to achieve the desired result.
In the third sample the first person can swap with the second one, then they will have the following numbers of dollars: 4 11 3, then the second person (in the new line) swaps with the third one, and the resulting numbers of dollars will equal to: 4 4 10. In this line everybody will be happy. | ```python
n = int(input())
s = [int(i) for i in input().split()]
for i in range(n-1, 0, -1):
if s[i] < s[i-1]:
u = 0
for j in range(i-1, n-1):
if s[j]>s[j+1]:
s[j], s[j+1] = s[j+1]+1, s[j]-1
if s != sorted(s):
print(":(")
else:
print(*s)
``` | 0 | |
740 | A | Alyona and copybooks | PROGRAMMING | 1,300 | [
"brute force",
"implementation"
] | null | null | Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for *a* rubles, a pack of two copybooks for *b* rubles, and a pack of three copybooks for *c* rubles. Alyona already has *n* copybooks.
What is the minimum amount of rubles she should pay to buy such number of copybooks *k* that *n*<=+<=*k* is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase. | The only line contains 4 integers *n*, *a*, *b*, *c* (1<=≤<=*n*,<=*a*,<=*b*,<=*c*<=≤<=109). | Print the minimum amount of rubles she should pay to buy such number of copybooks *k* that *n*<=+<=*k* is divisible by 4. | [
"1 1 3 4\n",
"6 2 1 1\n",
"4 4 4 4\n",
"999999999 1000000000 1000000000 1000000000\n"
] | [
"3\n",
"1\n",
"0\n",
"1000000000\n"
] | In the first example Alyona can buy 3 packs of 1 copybook for 3*a* = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.
In the second example Alyuna can buy a pack of 2 copybooks for *b* = 1 ruble. She will have 8 copybooks in total.
In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.
In the fourth example Alyona should buy one pack of one copybook. | 500 | [
{
"input": "1 1 3 4",
"output": "3"
},
{
"input": "6 2 1 1",
"output": "1"
},
{
"input": "4 4 4 4",
"output": "0"
},
{
"input": "999999999 1000000000 1000000000 1000000000",
"output": "1000000000"
},
{
"input": "1016 3 2 1",
"output": "0"
},
{
"input": "17 100 100 1",
"output": "1"
},
{
"input": "17 2 3 100",
"output": "5"
},
{
"input": "18 1 3 3",
"output": "2"
},
{
"input": "19 1 1 1",
"output": "1"
},
{
"input": "999999997 999999990 1000000000 1000000000",
"output": "1000000000"
},
{
"input": "999999998 1000000000 999999990 1000000000",
"output": "999999990"
},
{
"input": "634074578 336470888 481199252 167959139",
"output": "335918278"
},
{
"input": "999999999 1000000000 1000000000 999999990",
"output": "1000000000"
},
{
"input": "804928248 75475634 54748096 641009859",
"output": "0"
},
{
"input": "535590429 374288891 923264237 524125987",
"output": "524125987"
},
{
"input": "561219907 673102149 496813081 702209411",
"output": "673102149"
},
{
"input": "291882089 412106895 365329221 585325539",
"output": "585325539"
},
{
"input": "757703054 5887448 643910770 58376259",
"output": "11774896"
},
{
"input": "783332532 449924898 72235422 941492387",
"output": "0"
},
{
"input": "513994713 43705451 940751563 824608515",
"output": "131116353"
},
{
"input": "539624191 782710197 514300407 2691939",
"output": "8075817"
},
{
"input": "983359971 640274071 598196518 802030518",
"output": "640274071"
},
{
"input": "8989449 379278816 26521171 685146646",
"output": "405799987"
},
{
"input": "34618927 678092074 895037311 863230070",
"output": "678092074"
},
{
"input": "205472596 417096820 468586155 41313494",
"output": "0"
},
{
"input": "19 5 1 2",
"output": "3"
},
{
"input": "17 1 2 2",
"output": "2"
},
{
"input": "18 3 3 1",
"output": "2"
},
{
"input": "19 4 3 1",
"output": "3"
},
{
"input": "936134778 715910077 747167704 219396918",
"output": "438793836"
},
{
"input": "961764255 454914823 615683844 102513046",
"output": "307539138"
},
{
"input": "692426437 48695377 189232688 985629174",
"output": "146086131"
},
{
"input": "863280107 347508634 912524637 458679894",
"output": "347508634"
},
{
"input": "593942288 86513380 486073481 341796022",
"output": "0"
},
{
"input": "914539062 680293934 764655030 519879446",
"output": "764655030"
},
{
"input": "552472140 509061481 586588704 452405440",
"output": "0"
},
{
"input": "723325809 807874739 160137548 335521569",
"output": "335521569"
},
{
"input": "748955287 546879484 733686393 808572289",
"output": "546879484"
},
{
"input": "774584765 845692742 162011045 691688417",
"output": "691688417"
},
{
"input": "505246946 439473295 30527185 869771841",
"output": "30527185"
},
{
"input": "676100616 178478041 604076030 752887969",
"output": "0"
},
{
"input": "701730093 477291299 177624874 930971393",
"output": "654916173"
},
{
"input": "432392275 216296044 751173719 109054817",
"output": "216296044"
},
{
"input": "458021753 810076598 324722563 992170945",
"output": "992170945"
},
{
"input": "188683934 254114048 48014511 170254369",
"output": "48014511"
},
{
"input": "561775796 937657403 280013594 248004555",
"output": "0"
},
{
"input": "1000000000 1000000000 1000000000 1000000000",
"output": "0"
},
{
"input": "3 10000 10000 3",
"output": "9"
},
{
"input": "3 12 3 4",
"output": "7"
},
{
"input": "3 10000 10000 1",
"output": "3"
},
{
"input": "3 1000 1000 1",
"output": "3"
},
{
"input": "3 10 10 1",
"output": "3"
},
{
"input": "3 100 100 1",
"output": "3"
},
{
"input": "3 100000 10000 1",
"output": "3"
},
{
"input": "7 10 2 3",
"output": "5"
},
{
"input": "3 1000 1000 2",
"output": "6"
},
{
"input": "1 100000 1 100000",
"output": "100000"
},
{
"input": "7 4 3 1",
"output": "3"
},
{
"input": "3 1000 1000 3",
"output": "9"
},
{
"input": "3 1000 1 1",
"output": "2"
},
{
"input": "3 10 1 1",
"output": "2"
},
{
"input": "3 100000 1 1",
"output": "2"
},
{
"input": "3 100 1 1",
"output": "2"
},
{
"input": "3 100000 100000 1",
"output": "3"
},
{
"input": "3 1000 1 100",
"output": "101"
},
{
"input": "3 1000000000 1 1000000000",
"output": "1000000000"
},
{
"input": "3 1000 1 10",
"output": "11"
},
{
"input": "3 200 1 100",
"output": "101"
},
{
"input": "7 4 1 1",
"output": "2"
},
{
"input": "7 4 12 1",
"output": "3"
},
{
"input": "3 9 1 1",
"output": "2"
},
{
"input": "3 10000000 1000000 1",
"output": "3"
},
{
"input": "7 1000 1000 1",
"output": "3"
},
{
"input": "3 10000 1 30",
"output": "31"
},
{
"input": "3 1000 1 2",
"output": "3"
},
{
"input": "7 12 6 1",
"output": "3"
},
{
"input": "3 100000 1 1000",
"output": "1001"
},
{
"input": "7 1000 1000 3",
"output": "9"
},
{
"input": "3 4 3 1",
"output": "3"
},
{
"input": "3 3000000 1 100000",
"output": "100001"
},
{
"input": "3 3 1 1",
"output": "2"
},
{
"input": "3 10 1 5",
"output": "6"
},
{
"input": "3 2000 2000 1",
"output": "3"
},
{
"input": "3 10000000 10000000 1",
"output": "3"
},
{
"input": "3 5 1 1",
"output": "2"
},
{
"input": "3 100 1 33",
"output": "34"
},
{
"input": "7 9 2 7",
"output": "9"
},
{
"input": "4448 2 3 6",
"output": "0"
},
{
"input": "2228 1 6 3",
"output": "0"
}
] | 1,479,919,171 | 271 | Python 3 | WRONG_ANSWER | PRETESTS | 6 | 61 | 0 | n, a, b, c = map(int,input().split())
ost = (-n) % 4
if ost == 0:
print(0)
elif ost == 1:
print(min(a, 3 * c))
elif ost == 2:
print(min(2 * a, b, 2 * c))
else:
print(min(3 * a, c)) | Title: Alyona and copybooks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for *a* rubles, a pack of two copybooks for *b* rubles, and a pack of three copybooks for *c* rubles. Alyona already has *n* copybooks.
What is the minimum amount of rubles she should pay to buy such number of copybooks *k* that *n*<=+<=*k* is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.
Input Specification:
The only line contains 4 integers *n*, *a*, *b*, *c* (1<=≤<=*n*,<=*a*,<=*b*,<=*c*<=≤<=109).
Output Specification:
Print the minimum amount of rubles she should pay to buy such number of copybooks *k* that *n*<=+<=*k* is divisible by 4.
Demo Input:
['1 1 3 4\n', '6 2 1 1\n', '4 4 4 4\n', '999999999 1000000000 1000000000 1000000000\n']
Demo Output:
['3\n', '1\n', '0\n', '1000000000\n']
Note:
In the first example Alyona can buy 3 packs of 1 copybook for 3*a* = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.
In the second example Alyuna can buy a pack of 2 copybooks for *b* = 1 ruble. She will have 8 copybooks in total.
In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.
In the fourth example Alyona should buy one pack of one copybook. | ```python
n, a, b, c = map(int,input().split())
ost = (-n) % 4
if ost == 0:
print(0)
elif ost == 1:
print(min(a, 3 * c))
elif ost == 2:
print(min(2 * a, b, 2 * c))
else:
print(min(3 * a, c))
``` | 0 | |
931 | A | Friends Meeting | PROGRAMMING | 800 | [
"brute force",
"greedy",
"implementation",
"math"
] | null | null | Two friends are on the coordinate axis *Ox* in points with integer coordinates. One of them is in the point *x*1<==<=*a*, another one is in the point *x*2<==<=*b*.
Each of the friends can move by one along the line in any direction unlimited number of times. When a friend moves, the tiredness of a friend changes according to the following rules: the first move increases the tiredness by 1, the second move increases the tiredness by 2, the third — by 3 and so on. For example, if a friend moves first to the left, then to the right (returning to the same point), and then again to the left his tiredness becomes equal to 1<=+<=2<=+<=3<==<=6.
The friends want to meet in a integer point. Determine the minimum total tiredness they should gain, if they meet in the same point. | The first line contains a single integer *a* (1<=≤<=*a*<=≤<=1000) — the initial position of the first friend.
The second line contains a single integer *b* (1<=≤<=*b*<=≤<=1000) — the initial position of the second friend.
It is guaranteed that *a*<=≠<=*b*. | Print the minimum possible total tiredness if the friends meet in the same point. | [
"3\n4\n",
"101\n99\n",
"5\n10\n"
] | [
"1\n",
"2\n",
"9\n"
] | In the first example the first friend should move by one to the right (then the meeting happens at point 4), or the second friend should move by one to the left (then the meeting happens at point 3). In both cases, the total tiredness becomes 1.
In the second example the first friend should move by one to the left, and the second friend should move by one to the right. Then they meet in the point 100, and the total tiredness becomes 1 + 1 = 2.
In the third example one of the optimal ways is the following. The first friend should move three times to the right, and the second friend — two times to the left. Thus the friends meet in the point 8, and the total tiredness becomes 1 + 2 + 3 + 1 + 2 = 9. | 500 | [
{
"input": "3\n4",
"output": "1"
},
{
"input": "101\n99",
"output": "2"
},
{
"input": "5\n10",
"output": "9"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "1\n1000",
"output": "250000"
},
{
"input": "999\n1000",
"output": "1"
},
{
"input": "1000\n999",
"output": "1"
},
{
"input": "1000\n1",
"output": "250000"
},
{
"input": "2\n1",
"output": "1"
},
{
"input": "2\n999",
"output": "249001"
},
{
"input": "2\n998",
"output": "248502"
},
{
"input": "999\n2",
"output": "249001"
},
{
"input": "998\n2",
"output": "248502"
},
{
"input": "2\n1000",
"output": "249500"
},
{
"input": "1000\n2",
"output": "249500"
},
{
"input": "1\n999",
"output": "249500"
},
{
"input": "999\n1",
"output": "249500"
},
{
"input": "188\n762",
"output": "82656"
},
{
"input": "596\n777",
"output": "8281"
},
{
"input": "773\n70",
"output": "123904"
},
{
"input": "825\n729",
"output": "2352"
},
{
"input": "944\n348",
"output": "89102"
},
{
"input": "352\n445",
"output": "2209"
},
{
"input": "529\n656",
"output": "4096"
},
{
"input": "19\n315",
"output": "22052"
},
{
"input": "138\n370",
"output": "13572"
},
{
"input": "546\n593",
"output": "576"
},
{
"input": "285\n242",
"output": "484"
},
{
"input": "773\n901",
"output": "4160"
},
{
"input": "892\n520",
"output": "34782"
},
{
"input": "864\n179",
"output": "117649"
},
{
"input": "479\n470",
"output": "25"
},
{
"input": "967\n487",
"output": "57840"
},
{
"input": "648\n106",
"output": "73712"
},
{
"input": "58\n765",
"output": "125316"
},
{
"input": "235\n56",
"output": "8100"
},
{
"input": "285\n153",
"output": "4422"
},
{
"input": "943\n13",
"output": "216690"
},
{
"input": "675\n541",
"output": "4556"
},
{
"input": "4\n912",
"output": "206570"
}
] | 1,643,022,670 | 2,147,483,647 | PyPy 3 | OK | TESTS | 40 | 92 | 0 | a = int(input())
b = int(input())
diff = abs (a-b)
if diff % 2 == 0:
steps = int(diff / 2)
tiredness = 0
for i in range (1, steps+1):
tiredness +=i
print (tiredness*2)
else:
steps = int(diff // 2)
tiredness = 0
for i in range (1, steps+1):
tiredness +=i
print (tiredness*2 + (steps+1)) | Title: Friends Meeting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two friends are on the coordinate axis *Ox* in points with integer coordinates. One of them is in the point *x*1<==<=*a*, another one is in the point *x*2<==<=*b*.
Each of the friends can move by one along the line in any direction unlimited number of times. When a friend moves, the tiredness of a friend changes according to the following rules: the first move increases the tiredness by 1, the second move increases the tiredness by 2, the third — by 3 and so on. For example, if a friend moves first to the left, then to the right (returning to the same point), and then again to the left his tiredness becomes equal to 1<=+<=2<=+<=3<==<=6.
The friends want to meet in a integer point. Determine the minimum total tiredness they should gain, if they meet in the same point.
Input Specification:
The first line contains a single integer *a* (1<=≤<=*a*<=≤<=1000) — the initial position of the first friend.
The second line contains a single integer *b* (1<=≤<=*b*<=≤<=1000) — the initial position of the second friend.
It is guaranteed that *a*<=≠<=*b*.
Output Specification:
Print the minimum possible total tiredness if the friends meet in the same point.
Demo Input:
['3\n4\n', '101\n99\n', '5\n10\n']
Demo Output:
['1\n', '2\n', '9\n']
Note:
In the first example the first friend should move by one to the right (then the meeting happens at point 4), or the second friend should move by one to the left (then the meeting happens at point 3). In both cases, the total tiredness becomes 1.
In the second example the first friend should move by one to the left, and the second friend should move by one to the right. Then they meet in the point 100, and the total tiredness becomes 1 + 1 = 2.
In the third example one of the optimal ways is the following. The first friend should move three times to the right, and the second friend — two times to the left. Thus the friends meet in the point 8, and the total tiredness becomes 1 + 2 + 3 + 1 + 2 = 9. | ```python
a = int(input())
b = int(input())
diff = abs (a-b)
if diff % 2 == 0:
steps = int(diff / 2)
tiredness = 0
for i in range (1, steps+1):
tiredness +=i
print (tiredness*2)
else:
steps = int(diff // 2)
tiredness = 0
for i in range (1, steps+1):
tiredness +=i
print (tiredness*2 + (steps+1))
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn! | The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length. | Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise. | [
"aaba\nabaa\n",
"aabb\nabab\n"
] | [
"YES\n",
"NO\n"
] | In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa". | 0 | [
{
"input": "aaba\nabaa",
"output": "YES"
},
{
"input": "aabb\nabab",
"output": "NO"
},
{
"input": "a\na",
"output": "YES"
},
{
"input": "a\nb",
"output": "NO"
},
{
"input": "ab\nab",
"output": "YES"
},
{
"input": "ab\nba",
"output": "YES"
},
{
"input": "ab\nbb",
"output": "NO"
},
{
"input": "zzaa\naazz",
"output": "YES"
},
{
"input": "azza\nzaaz",
"output": "YES"
},
{
"input": "abc\nabc",
"output": "YES"
},
{
"input": "abc\nacb",
"output": "NO"
},
{
"input": "azzz\nzzaz",
"output": "YES"
},
{
"input": "abcd\ndcab",
"output": "YES"
},
{
"input": "abcd\ncdab",
"output": "YES"
},
{
"input": "abcd\ndcba",
"output": "YES"
},
{
"input": "abcd\nacbd",
"output": "NO"
},
{
"input": "oloaxgddgujq\noloaxgujqddg",
"output": "YES"
},
{
"input": "uwzwdxfmosmqatyv\ndxfmzwwusomqvyta",
"output": "YES"
},
{
"input": "hagnzomowtledfdotnll\nledfdotnllomowthagnz",
"output": "YES"
},
{
"input": "snyaydaeobufdg\nsnyaydaeobufdg",
"output": "YES"
},
{
"input": "baaaaa\nabaaaa",
"output": "NO"
},
{
"input": "hhiisug\nmzdjwju",
"output": "NO"
},
{
"input": "bbbabbabaaab\naaaabbabbbbb",
"output": "NO"
},
{
"input": "bbaaab\naababb",
"output": "NO"
},
{
"input": "aabbaaaa\naaaaabab",
"output": "NO"
},
{
"input": "aab\naba",
"output": "NO"
},
{
"input": "abcddd\nbacddd",
"output": "NO"
},
{
"input": "qgiufelsfhanx\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "aabaababaaba\naababaaababa",
"output": "NO"
},
{
"input": "nocdqzdriyyil\naaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "zdmctxl\nkojqhgw",
"output": "NO"
},
{
"input": "yhwepqwyhwepqwyhwepqweahnqtueahnqtueahnqtuyhwepqwyhwepqwyhwepqwyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtueahnqtueahnqtueahnqtueahnqtu\neahnqtueahnqtueahnqtuyhwepqweahnqtuyhwepqwyhwepqweahnqtuyhwepqweahnqtuyhwepqweahnqtueahnqtuyhwepqweahnqtueahnqtuyhwepqwyhwepqwyhwepqw",
"output": "NO"
},
{
"input": "abc\nbac",
"output": "NO"
},
{
"input": "ottceez\npcstdvz",
"output": "NO"
}
] | 1,650,575,530 | 2,147,483,647 | PyPy 3-64 | COMPILATION_ERROR | TESTS | 0 | 0 | 0 | #include <bits/stdc++.h>
#include <iostream>
using namespace std;
map<pair<size_t, size_t>, size_t> pair_to_id;
map<pair<size_t, size_t>, size_t> seg_id1, seg_id2;
size_t last_id;
vector<size_t> a1, a2;
void calc_ids1(size_t l, size_t r) {
if (l < r) {
size_t m = (l + r) / 2;
calc_ids1(l, m);
calc_ids1(m + 1, r);
pair<size_t, size_t> cur = {seg_id1[{l, m}], seg_id1[{m + 1, r}]};
if (cur.first > cur.second) {
swap(cur.first, cur.second);
}
if (!pair_to_id.contains(cur)) {
pair_to_id[cur] = last_id++;
}
seg_id1[{l, r}] = pair_to_id[cur];
}
}
void calc_ids2(size_t l, size_t r) {
if (l < r) {
size_t m = (l + r) / 2;
calc_ids2(l, m);
calc_ids2(m + 1, r);
pair<size_t, size_t> cur = {seg_id2[{l, m}], seg_id2[{m + 1, r}]};
if (cur.first > cur.second) {
swap(cur.first, cur.second);
}
if (!pair_to_id.contains(cur)) {
pair_to_id[cur] = last_id++;
}
seg_id2[{l, r}] = pair_to_id[cur];
}
}
int main() {
ios::sync_with_stdio(false);
string s1, s2;
cin >> s1 >> s2;
size_t n = s1.size();
size_t k = n;
while (k % 2 == 0) {
k /= 2;
}
vector<string> subs;
for (size_t i = 0; i < n; i += k) {
subs.push_back(s1.substr(i, k));
subs.push_back(s2.substr(i, k));
}
sort(subs.begin(), subs.end());
for (size_t i = 0; i < n; i += k) {
size_t j = i / k;
a1.push_back(lower_bound(subs.begin(), subs.end(), s1.substr(i, k)) - subs.begin());
seg_id1[{j, j}] = a1.back();
a2.push_back(lower_bound(subs.begin(), subs.end(), s2.substr(i, k)) - subs.begin());
seg_id2[{j, j}] = a2.back();
}
last_id = n;
calc_ids1(0, a1.size() - 1);
calc_ids2(0, a2.size() - 1);
std::cout << (seg_id1[{0, a1.size() - 1}] == seg_id2[{0, a2.size() - 1}] ? "YES\n" : "NO\n");
}
| Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today on a lecture about strings Gerald learned a new definition of string equivalency. Two strings *a* and *b* of equal length are called equivalent in one of the two cases:
1. They are equal. 1. If we split string *a* into two halves of the same size *a*1 and *a*2, and string *b* into two halves of the same size *b*1 and *b*2, then one of the following is correct: *a*1 is equivalent to *b*1, and *a*2 is equivalent to *b*2 1. *a*1 is equivalent to *b*2, and *a*2 is equivalent to *b*1
As a home task, the teacher gave two strings to his students and asked to determine if they are equivalent.
Gerald has already completed this home task. Now it's your turn!
Input Specification:
The first two lines of the input contain two strings given by the teacher. Each of them has the length from 1 to 200<=000 and consists of lowercase English letters. The strings have the same length.
Output Specification:
Print "YES" (without the quotes), if these two strings are equivalent, and "NO" (without the quotes) otherwise.
Demo Input:
['aaba\nabaa\n', 'aabb\nabab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample you should split the first string into strings "aa" and "ba", the second one — into strings "ab" and "aa". "aa" is equivalent to "aa"; "ab" is equivalent to "ba" as "ab" = "a" + "b", "ba" = "b" + "a".
In the second sample the first string can be splitted into strings "aa" and "bb", that are equivalent only to themselves. That's why string "aabb" is equivalent only to itself and to string "bbaa". | ```python
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
map<pair<size_t, size_t>, size_t> pair_to_id;
map<pair<size_t, size_t>, size_t> seg_id1, seg_id2;
size_t last_id;
vector<size_t> a1, a2;
void calc_ids1(size_t l, size_t r) {
if (l < r) {
size_t m = (l + r) / 2;
calc_ids1(l, m);
calc_ids1(m + 1, r);
pair<size_t, size_t> cur = {seg_id1[{l, m}], seg_id1[{m + 1, r}]};
if (cur.first > cur.second) {
swap(cur.first, cur.second);
}
if (!pair_to_id.contains(cur)) {
pair_to_id[cur] = last_id++;
}
seg_id1[{l, r}] = pair_to_id[cur];
}
}
void calc_ids2(size_t l, size_t r) {
if (l < r) {
size_t m = (l + r) / 2;
calc_ids2(l, m);
calc_ids2(m + 1, r);
pair<size_t, size_t> cur = {seg_id2[{l, m}], seg_id2[{m + 1, r}]};
if (cur.first > cur.second) {
swap(cur.first, cur.second);
}
if (!pair_to_id.contains(cur)) {
pair_to_id[cur] = last_id++;
}
seg_id2[{l, r}] = pair_to_id[cur];
}
}
int main() {
ios::sync_with_stdio(false);
string s1, s2;
cin >> s1 >> s2;
size_t n = s1.size();
size_t k = n;
while (k % 2 == 0) {
k /= 2;
}
vector<string> subs;
for (size_t i = 0; i < n; i += k) {
subs.push_back(s1.substr(i, k));
subs.push_back(s2.substr(i, k));
}
sort(subs.begin(), subs.end());
for (size_t i = 0; i < n; i += k) {
size_t j = i / k;
a1.push_back(lower_bound(subs.begin(), subs.end(), s1.substr(i, k)) - subs.begin());
seg_id1[{j, j}] = a1.back();
a2.push_back(lower_bound(subs.begin(), subs.end(), s2.substr(i, k)) - subs.begin());
seg_id2[{j, j}] = a2.back();
}
last_id = n;
calc_ids1(0, a1.size() - 1);
calc_ids2(0, a2.size() - 1);
std::cout << (seg_id1[{0, a1.size() - 1}] == seg_id2[{0, a2.size() - 1}] ? "YES\n" : "NO\n");
}
``` | -1 | |
0 | none | none | none | 0 | [
"none"
] | null | null | The last stage of Football World Cup is played using the play-off system.
There are *n* teams left in this stage, they are enumerated from 1 to *n*. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids *a* and *b* can meet. | The only line contains three integers *n*, *a* and *b* (2<=≤<=*n*<=≤<=256, 1<=≤<=*a*,<=*b*<=≤<=*n*) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that *n* is such that in each round an even number of team advance, and that *a* and *b* are not equal. | In the only line print "Final!" (without quotes), if teams *a* and *b* can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams *a* and *b* can meet. The round are enumerated from 1. | [
"4 1 2\n",
"8 2 6\n",
"8 7 5\n"
] | [
"1\n",
"Final!\n",
"2\n"
] | In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round. | 0 | [
{
"input": "4 1 2",
"output": "1"
},
{
"input": "8 2 6",
"output": "Final!"
},
{
"input": "8 7 5",
"output": "2"
},
{
"input": "128 30 98",
"output": "Final!"
},
{
"input": "256 128 256",
"output": "Final!"
},
{
"input": "256 2 127",
"output": "7"
},
{
"input": "2 1 2",
"output": "Final!"
},
{
"input": "2 2 1",
"output": "Final!"
},
{
"input": "4 1 3",
"output": "Final!"
},
{
"input": "4 1 4",
"output": "Final!"
},
{
"input": "4 2 1",
"output": "1"
},
{
"input": "4 2 3",
"output": "Final!"
},
{
"input": "4 2 4",
"output": "Final!"
},
{
"input": "4 3 1",
"output": "Final!"
},
{
"input": "4 3 2",
"output": "Final!"
},
{
"input": "4 3 4",
"output": "1"
},
{
"input": "4 4 1",
"output": "Final!"
},
{
"input": "4 4 2",
"output": "Final!"
},
{
"input": "4 4 3",
"output": "1"
},
{
"input": "8 8 7",
"output": "1"
},
{
"input": "8 8 5",
"output": "2"
},
{
"input": "8 8 1",
"output": "Final!"
},
{
"input": "16 4 3",
"output": "1"
},
{
"input": "16 2 4",
"output": "2"
},
{
"input": "16 14 11",
"output": "3"
},
{
"input": "16 3 11",
"output": "Final!"
},
{
"input": "32 10 9",
"output": "1"
},
{
"input": "32 25 28",
"output": "2"
},
{
"input": "32 22 18",
"output": "3"
},
{
"input": "32 17 25",
"output": "4"
},
{
"input": "32 18 3",
"output": "Final!"
},
{
"input": "64 40 39",
"output": "1"
},
{
"input": "64 60 58",
"output": "2"
},
{
"input": "64 34 37",
"output": "3"
},
{
"input": "64 26 24",
"output": "4"
},
{
"input": "64 50 43",
"output": "5"
},
{
"input": "64 17 42",
"output": "Final!"
},
{
"input": "128 116 115",
"output": "1"
},
{
"input": "128 35 33",
"output": "2"
},
{
"input": "128 61 59",
"output": "3"
},
{
"input": "128 116 123",
"output": "4"
},
{
"input": "128 17 15",
"output": "5"
},
{
"input": "128 124 77",
"output": "6"
},
{
"input": "128 4 80",
"output": "Final!"
},
{
"input": "256 224 223",
"output": "1"
},
{
"input": "256 24 22",
"output": "2"
},
{
"input": "256 199 196",
"output": "3"
},
{
"input": "256 148 159",
"output": "4"
},
{
"input": "256 178 166",
"output": "5"
},
{
"input": "256 75 97",
"output": "6"
},
{
"input": "256 185 200",
"output": "7"
},
{
"input": "256 3 238",
"output": "Final!"
},
{
"input": "256 128 129",
"output": "Final!"
},
{
"input": "256 255 129",
"output": "7"
},
{
"input": "256 255 128",
"output": "Final!"
},
{
"input": "256 129 256",
"output": "7"
},
{
"input": "128 98 69",
"output": "6"
},
{
"input": "128 47 83",
"output": "Final!"
},
{
"input": "16 2 3",
"output": "2"
},
{
"input": "64 32 30",
"output": "2"
},
{
"input": "8 4 5",
"output": "Final!"
},
{
"input": "8 7 8",
"output": "1"
},
{
"input": "8 2 3",
"output": "2"
},
{
"input": "8 2 5",
"output": "Final!"
}
] | 1,689,849,960 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 3 | 46 | 0 | from math import log
n, a, b = map(int, input().split())
rasn = abs(b - a)
if (log(rasn, 2) + 1 == log(n, 2)):
print("Final!")
else:
print(int(log(rasn, 2) + 1)) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The last stage of Football World Cup is played using the play-off system.
There are *n* teams left in this stage, they are enumerated from 1 to *n*. Several rounds are held, in each round the remaining teams are sorted in the order of their ids, then the first in this order plays with the second, the third — with the fourth, the fifth — with the sixth, and so on. It is guaranteed that in each round there is even number of teams. The winner of each game advances to the next round, the loser is eliminated from the tournament, there are no draws. In the last round there is the only game with two remaining teams: the round is called the Final, the winner is called the champion, and the tournament is over.
Arkady wants his two favorite teams to play in the Final. Unfortunately, the team ids are already determined, and it may happen that it is impossible for teams to meet in the Final, because they are to meet in some earlier stage, if they are strong enough. Determine, in which round the teams with ids *a* and *b* can meet.
Input Specification:
The only line contains three integers *n*, *a* and *b* (2<=≤<=*n*<=≤<=256, 1<=≤<=*a*,<=*b*<=≤<=*n*) — the total number of teams, and the ids of the teams that Arkady is interested in.
It is guaranteed that *n* is such that in each round an even number of team advance, and that *a* and *b* are not equal.
Output Specification:
In the only line print "Final!" (without quotes), if teams *a* and *b* can meet in the Final.
Otherwise, print a single integer — the number of the round in which teams *a* and *b* can meet. The round are enumerated from 1.
Demo Input:
['4 1 2\n', '8 2 6\n', '8 7 5\n']
Demo Output:
['1\n', 'Final!\n', '2\n']
Note:
In the first example teams 1 and 2 meet in the first round.
In the second example teams 2 and 6 can only meet in the third round, which is the Final, if they win all their opponents in earlier rounds.
In the third example the teams with ids 7 and 5 can meet in the second round, if they win their opponents in the first round. | ```python
from math import log
n, a, b = map(int, input().split())
rasn = abs(b - a)
if (log(rasn, 2) + 1 == log(n, 2)):
print("Final!")
else:
print(int(log(rasn, 2) + 1))
``` | 0 | |
99 | A | Help Far Away Kingdom | PROGRAMMING | 800 | [
"strings"
] | A. Help Far Away Kingdom | 2 | 256 | In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Kingdom healers had already had three appointments with the merchants who were asked to sell, say, exactly 0.273549107 beer barrels. To deal with the problem somehow, the King issued an order obliging rounding up all numbers to the closest integer to simplify calculations. Specifically, the order went like this:
- If a number's integer part does not end with digit 9 and its fractional part is strictly less than 0.5, then the rounded up number coincides with the number’s integer part. - If a number's integer part does not end with digit 9 and its fractional part is not less than 0.5, the rounded up number is obtained if we add 1 to the last digit of the number’s integer part.- If the number’s integer part ends with digit 9, to round up the numbers one should go to Vasilisa the Wise. In the whole Kingdom she is the only one who can perform the tricky operation of carrying into the next position.
Merchants found the algorithm very sophisticated and they asked you (the ACMers) to help them. Can you write a program that would perform the rounding according to the King’s order? | The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empty set of decimal digits). The number's length does not exceed 1000 characters, including the dot. There are no other characters in the input data. | If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes). | [
"0.0\n",
"1.49\n",
"1.50\n",
"2.71828182845904523536\n",
"3.14159265358979323846\n",
"12345678901234567890.1\n",
"123456789123456789.999\n"
] | [
"0",
"1",
"2",
"3",
"3",
"12345678901234567890",
"GOTO Vasilisa."
] | none | 500 | [
{
"input": "0.0",
"output": "0"
},
{
"input": "1.49",
"output": "1"
},
{
"input": "1.50",
"output": "2"
},
{
"input": "2.71828182845904523536",
"output": "3"
},
{
"input": "3.14159265358979323846",
"output": "3"
},
{
"input": "12345678901234567890.1",
"output": "12345678901234567890"
},
{
"input": "123456789123456789.999",
"output": "GOTO Vasilisa."
},
{
"input": "12345678901234567890.9",
"output": "12345678901234567891"
},
{
"input": "123456789123456788.999",
"output": "123456789123456789"
},
{
"input": "9.000",
"output": "GOTO Vasilisa."
},
{
"input": "0.1",
"output": "0"
},
{
"input": "0.2",
"output": "0"
},
{
"input": "0.3",
"output": "0"
},
{
"input": "0.4",
"output": "0"
},
{
"input": "0.5",
"output": "1"
},
{
"input": "0.6",
"output": "1"
},
{
"input": "0.7",
"output": "1"
},
{
"input": "0.8",
"output": "1"
},
{
"input": "0.9",
"output": "1"
},
{
"input": "1.0",
"output": "1"
},
{
"input": "1.1",
"output": "1"
},
{
"input": "1.2",
"output": "1"
},
{
"input": "1.3",
"output": "1"
},
{
"input": "1.4",
"output": "1"
},
{
"input": "1.5",
"output": "2"
},
{
"input": "1.6",
"output": "2"
},
{
"input": "1.7",
"output": "2"
},
{
"input": "1.8",
"output": "2"
},
{
"input": "1.9",
"output": "2"
},
{
"input": "2.0",
"output": "2"
},
{
"input": "2.1",
"output": "2"
},
{
"input": "2.2",
"output": "2"
},
{
"input": "2.3",
"output": "2"
},
{
"input": "2.4",
"output": "2"
},
{
"input": "2.5",
"output": "3"
},
{
"input": "2.6",
"output": "3"
},
{
"input": "2.7",
"output": "3"
},
{
"input": "2.8",
"output": "3"
},
{
"input": "2.9",
"output": "3"
},
{
"input": "3.0",
"output": "3"
},
{
"input": "3.1",
"output": "3"
},
{
"input": "3.2",
"output": "3"
},
{
"input": "3.3",
"output": "3"
},
{
"input": "3.4",
"output": "3"
},
{
"input": "3.5",
"output": "4"
},
{
"input": "3.6",
"output": "4"
},
{
"input": "3.7",
"output": "4"
},
{
"input": "3.8",
"output": "4"
},
{
"input": "3.9",
"output": "4"
},
{
"input": "4.0",
"output": "4"
},
{
"input": "4.1",
"output": "4"
},
{
"input": "4.2",
"output": "4"
},
{
"input": "4.3",
"output": "4"
},
{
"input": "4.4",
"output": "4"
},
{
"input": "4.5",
"output": "5"
},
{
"input": "4.6",
"output": "5"
},
{
"input": "4.7",
"output": "5"
},
{
"input": "4.8",
"output": "5"
},
{
"input": "4.9",
"output": "5"
},
{
"input": "5.0",
"output": "5"
},
{
"input": "5.1",
"output": "5"
},
{
"input": "5.2",
"output": "5"
},
{
"input": "5.3",
"output": "5"
},
{
"input": "5.4",
"output": "5"
},
{
"input": "5.5",
"output": "6"
},
{
"input": "5.6",
"output": "6"
},
{
"input": "5.7",
"output": "6"
},
{
"input": "5.8",
"output": "6"
},
{
"input": "5.9",
"output": "6"
},
{
"input": "6.0",
"output": "6"
},
{
"input": "6.1",
"output": "6"
},
{
"input": "6.2",
"output": "6"
},
{
"input": "6.3",
"output": "6"
},
{
"input": "6.4",
"output": "6"
},
{
"input": "6.5",
"output": "7"
},
{
"input": "6.6",
"output": "7"
},
{
"input": "6.7",
"output": "7"
},
{
"input": "6.8",
"output": "7"
},
{
"input": "6.9",
"output": "7"
},
{
"input": "7.0",
"output": "7"
},
{
"input": "7.1",
"output": "7"
},
{
"input": "7.2",
"output": "7"
},
{
"input": "7.3",
"output": "7"
},
{
"input": "7.4",
"output": "7"
},
{
"input": "7.5",
"output": "8"
},
{
"input": "7.6",
"output": "8"
},
{
"input": "7.7",
"output": "8"
},
{
"input": "7.8",
"output": "8"
},
{
"input": "7.9",
"output": "8"
},
{
"input": "8.0",
"output": "8"
},
{
"input": "8.1",
"output": "8"
},
{
"input": "8.2",
"output": "8"
},
{
"input": "8.3",
"output": "8"
},
{
"input": "8.4",
"output": "8"
},
{
"input": "8.5",
"output": "9"
},
{
"input": "8.6",
"output": "9"
},
{
"input": "8.7",
"output": "9"
},
{
"input": "8.8",
"output": "9"
},
{
"input": "8.9",
"output": "9"
},
{
"input": "9.0",
"output": "GOTO Vasilisa."
},
{
"input": "9.1",
"output": "GOTO Vasilisa."
},
{
"input": "9.2",
"output": "GOTO Vasilisa."
},
{
"input": "9.3",
"output": "GOTO Vasilisa."
},
{
"input": "9.4",
"output": "GOTO Vasilisa."
},
{
"input": "9.5",
"output": "GOTO Vasilisa."
},
{
"input": "9.6",
"output": "GOTO Vasilisa."
},
{
"input": "9.7",
"output": "GOTO Vasilisa."
},
{
"input": "9.8",
"output": "GOTO Vasilisa."
},
{
"input": "9.9",
"output": "GOTO Vasilisa."
},
{
"input": "609942239104813108618306232517836377583566292129955473517174437591594761209877970062547641606473593416245554763832875919009472288995880898848455284062760160557686724163817329189799336769669146848904803188614226720978399787805489531837751080926098.1664915772983166314490532653577560222779830866949001942720729759794777105570672781798092416748052690224813237139640723361527601154465287615917169132637313918577673651098507390501962",
"output": "609942239104813108618306232517836377583566292129955473517174437591594761209877970062547641606473593416245554763832875919009472288995880898848455284062760160557686724163817329189799336769669146848904803188614226720978399787805489531837751080926098"
},
{
"input": "7002108534951820589946967018226114921984364117669853212254634761258884835434844673935047882480101006606512119541798298905598015607366335061012709906661245805358900665571472645463994925687210711492820804158354236327017974683658305043146543214454877759341394.20211856263503281388748282682120712214711232598021393495443628276945042110862480888110959179019986486690931930108026302665438087068150666835901617457150158918705186964935221768346957536540345814875615118637945520917367155931078965",
"output": "7002108534951820589946967018226114921984364117669853212254634761258884835434844673935047882480101006606512119541798298905598015607366335061012709906661245805358900665571472645463994925687210711492820804158354236327017974683658305043146543214454877759341394"
},
{
"input": "1950583094879039694852660558765931995628486712128191844305265555887022812284005463780616067.5000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1950583094879039694852660558765931995628486712128191844305265555887022812284005463780616068"
},
{
"input": "718130341896330596635811874410345440628950330.500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "718130341896330596635811874410345440628950331"
},
{
"input": "927925904158088313481229162503626281882161630091489367140850985555900173018122871746924067186432044676083646964286435457446768031295712712803570690846298544912543439221596866052681116386179629036945370280722.500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "927925904158088313481229162503626281882161630091489367140850985555900173018122871746924067186432044676083646964286435457446768031295712712803570690846298544912543439221596866052681116386179629036945370280723"
},
{
"input": "68289614863244584294178637364598054554769889.500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "GOTO Vasilisa."
},
{
"input": "7536521504744364134984603189602839063535643888645969434165019366202558753840519.4999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999",
"output": "GOTO Vasilisa."
},
{
"input": "646188694587964249318078225173.5000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "646188694587964249318078225174"
},
{
"input": "7017472758334494297677068672222822550374695787171163207025095950296957261530.50",
"output": "7017472758334494297677068672222822550374695787171163207025095950296957261531"
},
{
"input": "259085737066615534998640212505663524594409165063310128108448186246980628179842202905722595400477937071746695941939306735605849342959111887834258250883469840846714848774368.50000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "259085737066615534998640212505663524594409165063310128108448186246980628179842202905722595400477937071746695941939306735605849342959111887834258250883469840846714848774369"
},
{
"input": "682500858233333594535201113441004740771119672961581796618069185960107115823662126812159957094407454522028503739299.5000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "GOTO Vasilisa."
},
{
"input": "41203422675619090661099806687619.49999999999999999999999999999999999999999999999999999999999999999999999999999999999999999",
"output": "GOTO Vasilisa."
}
] | 1,636,318,566 | 2,147,483,647 | Python 3 | OK | TESTS | 150 | 124 | 4,505,600 | def rounder(s):
st = ""
tochkaindex1 = s.index(".")+2
for i in range(tochkaindex1):
st+=s[i]
return st
def integer(s):
sum = 0
st = ""
while s[sum]!=".":
st+=str(s[sum])
sum+=1
return st
a = str(input())
if str(integer(a))[-1] == "9":
print("GOTO Vasilisa.")
elif int(str(rounder(a))[-1])>=5 and str(integer(a))[-1] != "9":
print(int(str(integer(a)))+1)
else:
print(int(str(integer(a))))
| Title: Help Far Away Kingdom
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
In a far away kingdom lived the King, the Prince, the Shoemaker, the Dressmaker and many other citizens. They lived happily until great trouble came into the Kingdom. The ACMers settled there.
Most damage those strange creatures inflicted upon the kingdom was that they loved high precision numbers. As a result, the Kingdom healers had already had three appointments with the merchants who were asked to sell, say, exactly 0.273549107 beer barrels. To deal with the problem somehow, the King issued an order obliging rounding up all numbers to the closest integer to simplify calculations. Specifically, the order went like this:
- If a number's integer part does not end with digit 9 and its fractional part is strictly less than 0.5, then the rounded up number coincides with the number’s integer part. - If a number's integer part does not end with digit 9 and its fractional part is not less than 0.5, the rounded up number is obtained if we add 1 to the last digit of the number’s integer part.- If the number’s integer part ends with digit 9, to round up the numbers one should go to Vasilisa the Wise. In the whole Kingdom she is the only one who can perform the tricky operation of carrying into the next position.
Merchants found the algorithm very sophisticated and they asked you (the ACMers) to help them. Can you write a program that would perform the rounding according to the King’s order?
Input Specification:
The first line contains a single number to round up — the integer part (a non-empty set of decimal digits that do not start with 0 — with the exception of a case when the set consists of a single digit — in this case 0 can go first), then follows character «.» (a dot), and then follows the fractional part (any non-empty set of decimal digits). The number's length does not exceed 1000 characters, including the dot. There are no other characters in the input data.
Output Specification:
If the last number of the integer part is not equal to 9, print the rounded-up number without leading zeroes. Otherwise, print the message "GOTO Vasilisa." (without the quotes).
Demo Input:
['0.0\n', '1.49\n', '1.50\n', '2.71828182845904523536\n', '3.14159265358979323846\n', '12345678901234567890.1\n', '123456789123456789.999\n']
Demo Output:
['0', '1', '2', '3', '3', '12345678901234567890', 'GOTO Vasilisa.']
Note:
none | ```python
def rounder(s):
st = ""
tochkaindex1 = s.index(".")+2
for i in range(tochkaindex1):
st+=s[i]
return st
def integer(s):
sum = 0
st = ""
while s[sum]!=".":
st+=str(s[sum])
sum+=1
return st
a = str(input())
if str(integer(a))[-1] == "9":
print("GOTO Vasilisa.")
elif int(str(rounder(a))[-1])>=5 and str(integer(a))[-1] != "9":
print(int(str(integer(a)))+1)
else:
print(int(str(integer(a))))
``` | 3.960608 |
0 | none | none | none | 0 | [
"none"
] | null | null | Поликарп мечтает стать программистом и фанатеет от степеней двойки. Среди двух чисел ему больше нравится то, которое делится на большую степень числа 2.
По заданной последовательности целых положительных чисел *a*1,<=*a*2,<=...,<=*a**n* требуется найти *r* — максимальную степень числа 2, на которую делится хотя бы одно из чисел последовательности. Кроме того, требуется вывести количество чисел *a**i*, которые делятся на *r*. | В первой строке записано целое число *n* (1<=≤<=*n*<=≤<=100) — длина последовательности *a*.
Во второй строке записана последовательность целых чисел *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109). | Выведите два числа:
- *r* — максимальную степень двойки, на которую делится хотя бы одно из чисел заданной последовательности, - количество элементов последовательности, которые делятся на *r*. | [
"5\n80 7 16 4 48\n",
"4\n21 5 3 33\n"
] | [
"16 3\n",
"1 4\n"
] | В первом тестовом примере максимальная степень двойки, на которую делится хотя бы одно число, равна 16 = 2<sup class="upper-index">4</sup>, на неё делятся числа 80, 16 и 48.
Во втором тестовом примере все четыре числа нечётные, поэтому делятся только на 1 = 2<sup class="upper-index">0</sup>. Это и будет максимальной степенью двойки для данного примера. | 0 | [
{
"input": "5\n80 7 16 4 48",
"output": "16 3"
},
{
"input": "4\n21 5 3 33",
"output": "1 4"
},
{
"input": "10\n8 112 52 86 93 102 24 24 100 826791168",
"output": "256 1"
},
{
"input": "3\n458297759 18 104",
"output": "8 1"
},
{
"input": "7\n12 14 40 8 74 104 11",
"output": "8 3"
},
{
"input": "11\n35 16 664311776 46 48 52 63 82 84 80 23",
"output": "32 1"
},
{
"input": "7\n67 68 58 24 96 73 72",
"output": "32 1"
},
{
"input": "8\n48 112 40 8 112 14 80 36",
"output": "16 4"
},
{
"input": "10\n14 6 68 8 84 949689614 91 26 80 56",
"output": "16 1"
},
{
"input": "4\n39 56 939117699 56",
"output": "8 2"
},
{
"input": "5\n90 18 56 64 32",
"output": "64 1"
},
{
"input": "9\n64 95 32 64 96 80 100 96 66",
"output": "64 2"
},
{
"input": "20\n105407881 735510073 587127085 111067442 126807503 250859170 778634763 919694130 592496831 462347734 532487590 475786023 951527598 183635985 612791353 447723541 409812454 900700354 801564406 532793851",
"output": "2 9"
},
{
"input": "20\n850632510 530517796 700510265 454664263 131947796 418444926 921278498 251889644 705327498 892480283 884422799 479219117 399278535 80826412 496934492 448261193 39033930 49426174 621130971 808191947",
"output": "4 5"
},
{
"input": "20\n780355354 620754888 193377552 463211662 46248927 312489308 472238901 823707535 138518748 267363170 19751630 193171944 411443343 858525221 458019868 490268043 7864848 218005780 744553112 83590041",
"output": "16 2"
},
{
"input": "20\n29023024 579267278 217400978 121454376 235087976 154574217 708760940 84623652 195299056 329204104 527952531 822521791 513319036 285749488 292843688 389260660 498981613 835987320 444201058 251639011",
"output": "16 3"
},
{
"input": "20\n267784376 576420580 392773522 296581728 508523192 812838532 920098710 624114448 194991560 850559568 29915376 785467756 490019770 524237000 871021232 970867040 769417893 210139479 445850586 333230268",
"output": "32 2"
},
{
"input": "20\n860654784 630481952 430211228 13468621 33985780 279050728 782571295 83521731 818343376 508318323 550168944 763113524 152970477 502262855 934672824 712697136 451447464 732781790 71573907 50381000",
"output": "64 1"
},
{
"input": "20\n673865536 152236510 957204496 401364096 969402746 287701920 768559538 642049008 736330680 179648832 480094052 225156558 957671104 726304328 612058916 257008256 173639040 673864512 431405191 454360662",
"output": "128 3"
},
{
"input": "20\n706678380 597303020 176804438 146220776 485004772 799346560 692789954 737954674 398118372 231976240 957701828 556811840 74342144 966291136 893909760 745234360 44276827 878935416 975182148 322390872",
"output": "256 2"
},
{
"input": "20\n442107036 883530112 852749824 997931232 902004480 838557324 186049792 827163136 3843737 603467472 383038751 548720704 843680384 906511492 591629504 41722624 79778650 839163077 880599104 456148480",
"output": "512 3"
},
{
"input": "20\n667815852 318176276 693849088 724201296 188710200 39249152 929966576 651876056 580647856 575425536 367972188 647585808 833274694 578646160 593232968 747635620 973200384 608104976 754724885 832141532",
"output": "1024 3"
},
{
"input": "20\n448394296 216942008 573160113 728121900 769966592 164290016 721604576 970539238 338262776 947927236 587084928 648622584 194610176 435895128 896641600 70371299 323855936 292543040 28980004 787518144",
"output": "1024 1"
},
{
"input": "20\n269609216 130664082 366702720 254341120 817371149 791314720 886888448 933572608 411407552 86828928 280842240 259838684 821718144 131427072 316135424 189065544 173073728 20176393 508466777 927373184",
"output": "4096 3"
},
{
"input": "20\n620004352 728068096 230808280 347805952 153777664 828290048 941633792 681387488 689396208 283672752 130113536 124222464 425923944 365087488 68677632 957876224 86529928 278224896 516674048 203400656",
"output": "8192 2"
},
{
"input": "20\n957116416 938908864 254662656 28720000 829892752 344974528 22716709 493757015 729003570 868597760 675246081 648372096 233462945 949382272 600301600 979810000 695847936 383948336 388551600 125714432",
"output": "16384 2"
},
{
"input": "20\n793801200 316289782 968725504 765722788 172605440 945717248 613921792 54457344 725727714 598109120 390593416 438184064 245627755 91785071 855031808 778218454 34154240 686966990 736207232 674856960",
"output": "32768 2"
},
{
"input": "20\n356744192 260087808 498705408 60572928 360008038 968932864 66422016 929599488 973047264 426826855 483623936 826974208 487705600 787624960 951492608 343212032 661494459 244741040 409686016 20327511",
"output": "65536 1"
},
{
"input": "20\n775136472 720896 585826304 479121408 116085124 608963940 286154752 103731391 192445952 745342784 887373824 351469568 201183616 579608192 26928128 861796540 579858432 30678450 359436288 476635136",
"output": "65536 3"
},
{
"input": "20\n962265088 792592384 175088192 998003136 645410688 697606715 918541862 551100416 247733780 770750336 264468484 185550848 201588736 989953440 799341904 355386616 611975168 792208864 272619608 239038496",
"output": "262144 1"
},
{
"input": "20\n420610048 361879464 11984896 648474908 567803904 811903488 777519104 677117952 794312704 362283008 946274304 824280862 716753756 318224096 114032640 470286336 153747456 698417152 410143376 577402584",
"output": "524288 4"
},
{
"input": "20\n916422656 61941502 755804160 993386496 371458048 361240704 791150592 218434752 532807680 517277964 158990336 877143936 263469056 339735363 438097920 842006528 200088380 651153211 932184064 443051520",
"output": "1048576 2"
},
{
"input": "20\n855113728 457249360 705319632 368396992 886571008 261505024 647304078 73518537 749228480 194262008 148995424 860356608 950009856 649068544 430006272 363153728 105360192 37648488 426766453 565502131",
"output": "2097152 1"
},
{
"input": "20\n283119671 29874944 739246080 972336073 468233952 926107648 665047794 354091008 141526171 529537472 937811232 970178560 948105794 433304784 186908672 912261120 858259456 972472320 4889883 76907904",
"output": "2097152 1"
},
{
"input": "20\n386375302 77596672 411041792 683671552 667767296 625737728 947733007 612950256 129715876 813694976 782385152 164455808 647131408 385566720 65880960 96010240 484900864 385339335 533303296 660648928",
"output": "8388608 2"
},
{
"input": "20\n786432000 123007744 351391826 216719360 391303168 920693440 652456192 715653120 915090432 26559797 680345413 346423296 401288334 437510144 122224640 652606928 182370304 683278336 656773980 316229632",
"output": "2097152 1"
},
{
"input": "20\n850523026 993052212 100663296 830207504 281863570 426977070 803471360 796327936 369098752 842702848 526319616 785973248 501219328 801245229 853162496 689842176 292886257 859104512 631656960 512295035",
"output": "33554432 2"
},
{
"input": "20\n694157312 375128064 494927872 199980576 823836320 358259440 950409408 215242336 126689280 697139200 471849008 854435840 935337515 589698048 157286400 161334770 738197504 594549920 416464896 690872320",
"output": "67108864 1"
},
{
"input": "20\n785580032 987226112 398458880 437531712 335544320 676369664 533876736 181598976 117440512 924319744 469762048 443505526 949904673 710590464 533015579 17793024 781975552 803666112 973078528 866337472",
"output": "67108864 2"
},
{
"input": "20\n558266065 720866235 285275468 139607080 150235150 855638016 815792128 358744064 620756992 438077440 73404848 892534480 939524096 195793792 838860800 364010680 931135488 254242133 374493422 578846720",
"output": "134217728 1"
},
{
"input": "20\n909362176 291766272 557318144 348389376 426770432 526614528 213516288 932012606 344981504 138412032 6291456 354007477 536870912 557842432 536870912 668205056 398003707 609566463 893207232 83886080",
"output": "536870912 2"
},
{
"input": "1\n167959139",
"output": "1 1"
},
{
"input": "2\n641009859 54748095",
"output": "1 2"
},
{
"input": "10\n1000000 1 1000000 1 1000000 1 1000000 1 1000000 1",
"output": "64 5"
},
{
"input": "5\n1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "512 5"
},
{
"input": "8\n1 1 1 1 1 1 1 1",
"output": "1 8"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100",
"output": "64 1"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "1 100"
},
{
"input": "100\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "2 100"
},
{
"input": "100\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3",
"output": "1 100"
},
{
"input": "1\n536870912",
"output": "536870912 1"
}
] | 1,458,976,228 | 628 | Python 3 | OK | TESTS | 172 | 77 | 4,608,000 | n = int(input())
a = list(map(int, input().split()))
ans = 1
ansc = 0
for x in a:
i = 0
while (x % (2 ** i) == 0):
i += 1
i -= 1
if (2 ** i == ans):
ansc += 1
if (2 ** i > ans):
ans = 2 ** i
ansc = 1
print(str(ans) + ' ' + str(ansc)) | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Поликарп мечтает стать программистом и фанатеет от степеней двойки. Среди двух чисел ему больше нравится то, которое делится на большую степень числа 2.
По заданной последовательности целых положительных чисел *a*1,<=*a*2,<=...,<=*a**n* требуется найти *r* — максимальную степень числа 2, на которую делится хотя бы одно из чисел последовательности. Кроме того, требуется вывести количество чисел *a**i*, которые делятся на *r*.
Input Specification:
В первой строке записано целое число *n* (1<=≤<=*n*<=≤<=100) — длина последовательности *a*.
Во второй строке записана последовательность целых чисел *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Выведите два числа:
- *r* — максимальную степень двойки, на которую делится хотя бы одно из чисел заданной последовательности, - количество элементов последовательности, которые делятся на *r*.
Demo Input:
['5\n80 7 16 4 48\n', '4\n21 5 3 33\n']
Demo Output:
['16 3\n', '1 4\n']
Note:
В первом тестовом примере максимальная степень двойки, на которую делится хотя бы одно число, равна 16 = 2<sup class="upper-index">4</sup>, на неё делятся числа 80, 16 и 48.
Во втором тестовом примере все четыре числа нечётные, поэтому делятся только на 1 = 2<sup class="upper-index">0</sup>. Это и будет максимальной степенью двойки для данного примера. | ```python
n = int(input())
a = list(map(int, input().split()))
ans = 1
ansc = 0
for x in a:
i = 0
while (x % (2 ** i) == 0):
i += 1
i -= 1
if (2 ** i == ans):
ansc += 1
if (2 ** i > ans):
ans = 2 ** i
ansc = 1
print(str(ans) + ' ' + str(ansc))
``` | 3 | |
0 | none | none | none | 0 | [
"none"
] | null | null | Sereja has an *n*<=×<=*m* rectangular table *a*, each cell of the table contains a zero or a number one. Sereja wants his table to meet the following requirement: each connected component of the same values forms a rectangle with sides parallel to the sides of the table. Rectangles should be filled with cells, that is, if a component form a rectangle of size *h*<=×<=*w*, then the component must contain exactly *hw* cells.
A connected component of the same values is a set of cells of the table that meet the following conditions:
- every two cells of the set have the same value; - the cells of the set form a connected region on the table (two cells are connected if they are adjacent in some row or some column of the table); - it is impossible to add any cell to the set unless we violate the two previous conditions.
Can Sereja change the values of at most *k* cells of the table so that the table met the described requirement? What minimum number of table cells should he change in this case? | The first line contains integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=100; 1<=≤<=*k*<=≤<=10). Next *n* lines describe the table *a*: the *i*-th of them contains *m* integers *a**i*1,<=*a**i*2,<=...,<=*a**im* (0<=≤<=*a**i*,<=*j*<=≤<=1) — the values in the cells of the *i*-th row. | Print -1, if it is impossible to meet the requirement. Otherwise, print the minimum number of cells which should be changed. | [
"5 5 2\n1 1 1 1 1\n1 1 1 1 1\n1 1 0 1 1\n1 1 1 1 1\n1 1 1 1 1\n",
"3 4 1\n1 0 0 0\n0 1 1 1\n1 1 1 0\n",
"3 4 1\n1 0 0 1\n0 1 1 0\n1 0 0 1\n"
] | [
"1\n",
"-1\n",
"0\n"
] | none | 0 | [] | 1,691,657,769 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 46 | 0 | print("_RANDOM_GUESS_1691657769.6756732")# 1691657769.6756897 | Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja has an *n*<=×<=*m* rectangular table *a*, each cell of the table contains a zero or a number one. Sereja wants his table to meet the following requirement: each connected component of the same values forms a rectangle with sides parallel to the sides of the table. Rectangles should be filled with cells, that is, if a component form a rectangle of size *h*<=×<=*w*, then the component must contain exactly *hw* cells.
A connected component of the same values is a set of cells of the table that meet the following conditions:
- every two cells of the set have the same value; - the cells of the set form a connected region on the table (two cells are connected if they are adjacent in some row or some column of the table); - it is impossible to add any cell to the set unless we violate the two previous conditions.
Can Sereja change the values of at most *k* cells of the table so that the table met the described requirement? What minimum number of table cells should he change in this case?
Input Specification:
The first line contains integers *n*, *m* and *k* (1<=≤<=*n*,<=*m*<=≤<=100; 1<=≤<=*k*<=≤<=10). Next *n* lines describe the table *a*: the *i*-th of them contains *m* integers *a**i*1,<=*a**i*2,<=...,<=*a**im* (0<=≤<=*a**i*,<=*j*<=≤<=1) — the values in the cells of the *i*-th row.
Output Specification:
Print -1, if it is impossible to meet the requirement. Otherwise, print the minimum number of cells which should be changed.
Demo Input:
['5 5 2\n1 1 1 1 1\n1 1 1 1 1\n1 1 0 1 1\n1 1 1 1 1\n1 1 1 1 1\n', '3 4 1\n1 0 0 0\n0 1 1 1\n1 1 1 0\n', '3 4 1\n1 0 0 1\n0 1 1 0\n1 0 0 1\n']
Demo Output:
['1\n', '-1\n', '0\n']
Note:
none | ```python
print("_RANDOM_GUESS_1691657769.6756732")# 1691657769.6756897
``` | 0 | |
96 | A | Football | PROGRAMMING | 900 | [
"implementation",
"strings"
] | A. Football | 2 | 256 | Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not. | The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field. | Print "YES" if the situation is dangerous. Otherwise, print "NO". | [
"001001\n",
"1000000001\n"
] | [
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "001001",
"output": "NO"
},
{
"input": "1000000001",
"output": "YES"
},
{
"input": "00100110111111101",
"output": "YES"
},
{
"input": "11110111111111111",
"output": "YES"
},
{
"input": "01",
"output": "NO"
},
{
"input": "10100101",
"output": "NO"
},
{
"input": "1010010100000000010",
"output": "YES"
},
{
"input": "101010101",
"output": "NO"
},
{
"input": "000000000100000000000110101100000",
"output": "YES"
},
{
"input": "100001000000110101100000",
"output": "NO"
},
{
"input": "100001000011010110000",
"output": "NO"
},
{
"input": "010",
"output": "NO"
},
{
"input": "10101011111111111111111111111100",
"output": "YES"
},
{
"input": "1001101100",
"output": "NO"
},
{
"input": "1001101010",
"output": "NO"
},
{
"input": "1111100111",
"output": "NO"
},
{
"input": "00110110001110001111",
"output": "NO"
},
{
"input": "11110001001111110001",
"output": "NO"
},
{
"input": "10001111001011111101",
"output": "NO"
},
{
"input": "10000010100000001000110001010100001001001010011",
"output": "YES"
},
{
"input": "01111011111010111100101100001011001010111110000010",
"output": "NO"
},
{
"input": "00100000100100101110011001011011101110110110010100",
"output": "NO"
},
{
"input": "10110100110001001011110101110010100010000000000100101010111110111110100011",
"output": "YES"
},
{
"input": "00011101010101111001011011001101101011111101000010100000111000011100101011",
"output": "NO"
},
{
"input": "01110000110100110101110100111000101101011101011110110100100111100001110111",
"output": "NO"
},
{
"input": "11110110011000100111100111101101011111110100010101011011111101110110110111",
"output": "YES"
},
{
"input": "100100010101110010001011001110100011100010011110100101100011010001001010001001101111001100",
"output": "NO"
},
{
"input": "111110010001011010010011111100110110001111000010100011011100111101111101110010101111011110000001010",
"output": "NO"
},
{
"input": "111110111100010100000100001010111011101011000111011011011010110010100010000101011111000011010011110",
"output": "NO"
},
{
"input": "1011110110111010110111111010010010100011111011110000011000110010011110111010110100011010100010111000",
"output": "NO"
},
{
"input": "0010100111100010110110000011100111110100111110001010000100111111111010111100101101010101001011010110",
"output": "YES"
},
{
"input": "1111010100010100101011101100101101110011000010100010000001111100010011100101010001101111000001011000",
"output": "NO"
},
{
"input": "0101100011001110001110100111100011010101011000000000110110010010111100101111010111100011101100100101",
"output": "YES"
},
{
"input": "0001101110011101110000000010011111101001101111100001001010110000110001100000010001111011011110001101",
"output": "YES"
},
{
"input": "1000010000100000100010000100001000010000100001000010000100001000010000100001000010000100001000010000",
"output": "NO"
},
{
"input": "1000000000100000000010000000001000000000100000000010000000001000000000100000000010000000001000000000",
"output": "YES"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "YES"
},
{
"input": "1111111111111111111111111111111111111111011111111111111111111111111111111111111111111111111111111111",
"output": "YES"
},
{
"input": "10100101000",
"output": "NO"
},
{
"input": "11110111011101",
"output": "NO"
},
{
"input": "10000000",
"output": "YES"
},
{
"input": "00000001",
"output": "YES"
},
{
"input": "01111111",
"output": "YES"
},
{
"input": "11111110",
"output": "YES"
}
] | 1,666,448,436 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 2 | 92 | 0 | string = input()
temp = None
counter = int()
for letter in string:
x = int()
if temp == None:
temp = letter
else:
if letter == temp:
counter += 1
else:
temp = letter
counter = 0
if counter == 7:
print('YES')
break
if counter < 7:
print('NO')
| Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
Input Specification:
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
Output Specification:
Print "YES" if the situation is dangerous. Otherwise, print "NO".
Demo Input:
['001001\n', '1000000001\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none | ```python
string = input()
temp = None
counter = int()
for letter in string:
x = int()
if temp == None:
temp = letter
else:
if letter == temp:
counter += 1
else:
temp = letter
counter = 0
if counter == 7:
print('YES')
break
if counter < 7:
print('NO')
``` | 0 |
253 | A | Boys and Girls | PROGRAMMING | 1,100 | [
"greedy"
] | null | null | There are *n* boys and *m* girls studying in the class. They should stand in a line so that boys and girls alternated there as much as possible. Let's assume that positions in the line are indexed from left to right by numbers from 1 to *n*<=+<=*m*. Then the number of integers *i* (1<=≤<=*i*<=<<=*n*<=+<=*m*) such that positions with indexes *i* and *i*<=+<=1 contain children of different genders (position *i* has a girl and position *i*<=+<=1 has a boy or vice versa) must be as large as possible.
Help the children and tell them how to form the line. | The single line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), separated by a space. | Print a line of *n*<=+<=*m* characters. Print on the *i*-th position of the line character "B", if the *i*-th position of your arrangement should have a boy and "G", if it should have a girl.
Of course, the number of characters "B" should equal *n* and the number of characters "G" should equal *m*. If there are multiple optimal solutions, print any of them. | [
"3 3\n",
"4 2\n"
] | [
"GBGBGB\n",
"BGBGBB\n"
] | In the first sample another possible answer is BGBGBG.
In the second sample answer BBGBGB is also optimal. | 500 | [
{
"input": "3 3",
"output": "GBGBGB"
},
{
"input": "4 2",
"output": "BGBGBB"
},
{
"input": "5 5",
"output": "GBGBGBGBGB"
},
{
"input": "6 4",
"output": "BGBGBGBGBB"
},
{
"input": "100 1",
"output": "BGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "76 48",
"output": "BGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "100 90",
"output": "BGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBBBBBBBBBB"
},
{
"input": "90 100",
"output": "GBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGGGGGGGGGG"
},
{
"input": "1 98",
"output": "GBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "1 100",
"output": "GBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "56 98",
"output": "GBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "89 89",
"output": "GBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGB"
},
{
"input": "18 94",
"output": "GBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "84 27",
"output": "BGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "1 1",
"output": "GB"
},
{
"input": "1 2",
"output": "GBG"
},
{
"input": "2 1",
"output": "BGB"
},
{
"input": "1 34",
"output": "GBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "46 2",
"output": "BGBGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "99 3",
"output": "BGBGBGBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB"
},
{
"input": "10 100",
"output": "GBGBGBGBGBGBGBGBGBGBGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG"
},
{
"input": "100 100",
"output": "GBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGBGB"
},
{
"input": "1 4",
"output": "GBGGG"
}
] | 1,647,640,461 | 2,147,483,647 | Python 3 | OK | TESTS | 23 | 92 | 0 | INPUT = open('input.txt', 'r')
OUTPUT = open('output.txt', 'w')
n, m = map(int, INPUT.read().split())
if n>m:
OUTPUT.write(m*"BG"+(n-m)*"B")
else:
OUTPUT.write(n*"GB"+(m-n)*"G")
| Title: Boys and Girls
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* boys and *m* girls studying in the class. They should stand in a line so that boys and girls alternated there as much as possible. Let's assume that positions in the line are indexed from left to right by numbers from 1 to *n*<=+<=*m*. Then the number of integers *i* (1<=≤<=*i*<=<<=*n*<=+<=*m*) such that positions with indexes *i* and *i*<=+<=1 contain children of different genders (position *i* has a girl and position *i*<=+<=1 has a boy or vice versa) must be as large as possible.
Help the children and tell them how to form the line.
Input Specification:
The single line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100), separated by a space.
Output Specification:
Print a line of *n*<=+<=*m* characters. Print on the *i*-th position of the line character "B", if the *i*-th position of your arrangement should have a boy and "G", if it should have a girl.
Of course, the number of characters "B" should equal *n* and the number of characters "G" should equal *m*. If there are multiple optimal solutions, print any of them.
Demo Input:
['3 3\n', '4 2\n']
Demo Output:
['GBGBGB\n', 'BGBGBB\n']
Note:
In the first sample another possible answer is BGBGBG.
In the second sample answer BBGBGB is also optimal. | ```python
INPUT = open('input.txt', 'r')
OUTPUT = open('output.txt', 'w')
n, m = map(int, INPUT.read().split())
if n>m:
OUTPUT.write(m*"BG"+(n-m)*"B")
else:
OUTPUT.write(n*"GB"+(m-n)*"G")
``` | 3 | |
520 | A | Pangram | PROGRAMMING | 800 | [
"implementation",
"strings"
] | null | null | A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.
You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string.
The second line contains the string. The string consists only of uppercase and lowercase Latin letters. | Output "YES", if the string is a pangram and "NO" otherwise. | [
"12\ntoosmallword\n",
"35\nTheQuickBrownFoxJumpsOverTheLazyDog\n"
] | [
"NO\n",
"YES\n"
] | none | 500 | [
{
"input": "12\ntoosmallword",
"output": "NO"
},
{
"input": "35\nTheQuickBrownFoxJumpsOverTheLazyDog",
"output": "YES"
},
{
"input": "1\na",
"output": "NO"
},
{
"input": "26\nqwertyuiopasdfghjklzxcvbnm",
"output": "YES"
},
{
"input": "26\nABCDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "YES"
},
{
"input": "48\nthereisasyetinsufficientdataforameaningfulanswer",
"output": "NO"
},
{
"input": "30\nToBeOrNotToBeThatIsTheQuestion",
"output": "NO"
},
{
"input": "30\njackdawslovemybigsphinxofquarz",
"output": "NO"
},
{
"input": "31\nTHEFIVEBOXINGWIZARDSJUMPQUICKLY",
"output": "YES"
},
{
"input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "NO"
},
{
"input": "26\nMGJYIZDKsbhpVeNFlquRTcWoAx",
"output": "YES"
},
{
"input": "26\nfWMOhAPsbIVtyUEZrGNQXDklCJ",
"output": "YES"
},
{
"input": "26\nngPMVFSThiRCwLEuyOAbKxQzDJ",
"output": "YES"
},
{
"input": "25\nnxYTzLFwzNolAumjgcAboyxAj",
"output": "NO"
},
{
"input": "26\npRWdodGdxUESvcScPGbUoooZsC",
"output": "NO"
},
{
"input": "66\nBovdMlDzTaqKllZILFVfxbLGsRnzmtVVTmqiIDTYrossLEPlmsPrkUYtWEsGHVOnFj",
"output": "NO"
},
{
"input": "100\nmKtsiDRJypUieHIkvJaMFkwaKxcCIbBszZQLIyPpCDCjhNpAnYFngLjRpnKWpKWtGnwoSteeZXuFHWQxxxOpFlNeYTwKocsXuCoa",
"output": "YES"
},
{
"input": "26\nEoqxUbsLjPytUHMiFnvcGWZdRK",
"output": "NO"
},
{
"input": "26\nvCUFRKElZOnjmXGylWQaHDiPst",
"output": "NO"
},
{
"input": "26\nWtrPuaHdXLKJMsnvQfgOiJZBEY",
"output": "NO"
},
{
"input": "26\npGiFluRteQwkaVoPszJyNBChxM",
"output": "NO"
},
{
"input": "26\ncTUpqjPmANrdbzSFhlWIoKxgVY",
"output": "NO"
},
{
"input": "26\nLndjgvAEuICHKxPwqYztosrmBN",
"output": "NO"
},
{
"input": "26\nMdaXJrCipnOZLykfqHWEStevbU",
"output": "NO"
},
{
"input": "26\nEjDWsVxfKTqGXRnUMOLYcIzPba",
"output": "NO"
},
{
"input": "26\nxKwzRMpunYaqsdfaBgJcVElTHo",
"output": "NO"
},
{
"input": "26\nnRYUQsTwCPLZkgshfEXvBdoiMa",
"output": "NO"
},
{
"input": "26\nHNCQPfJutyAlDGsvRxZWMEbIdO",
"output": "NO"
},
{
"input": "26\nDaHJIpvKznQcmUyWsTGObXRFDe",
"output": "NO"
},
{
"input": "26\nkqvAnFAiRhzlJbtyuWedXSPcOG",
"output": "NO"
},
{
"input": "26\nhlrvgdwsIOyjcmUZXtAKEqoBpF",
"output": "NO"
},
{
"input": "26\njLfXXiMhBTcAwQVReGnpKzdsYu",
"output": "NO"
},
{
"input": "26\nlNMcVuwItjxRBGAekjhyDsQOzf",
"output": "NO"
},
{
"input": "26\nRkSwbNoYldUGtAZvpFMcxhIJFE",
"output": "NO"
},
{
"input": "26\nDqspXZJTuONYieKgaHLMBwfVSC",
"output": "NO"
},
{
"input": "26\necOyUkqNljFHRVXtIpWabGMLDz",
"output": "NO"
},
{
"input": "26\nEKAvqZhBnPmVCDRlgWJfOusxYI",
"output": "NO"
},
{
"input": "26\naLbgqeYchKdMrsZxIPFvTOWNjA",
"output": "NO"
},
{
"input": "26\nxfpBLsndiqtacOCHGmeWUjRkYz",
"output": "NO"
},
{
"input": "26\nXsbRKtqleZPNIVCdfUhyagAomJ",
"output": "NO"
},
{
"input": "26\nAmVtbrwquEthZcjKPLiyDgSoNF",
"output": "NO"
},
{
"input": "26\nOhvXDcwqAUmSEPRZGnjFLiKtNB",
"output": "NO"
},
{
"input": "26\nEKWJqCFLRmstxVBdYuinpbhaOg",
"output": "NO"
},
{
"input": "26\nmnbvcxxlkjhgfdsapoiuytrewq",
"output": "NO"
},
{
"input": "26\naAbcdefghijklmnopqrstuvwxy",
"output": "NO"
},
{
"input": "30\nABCDEFGHTYRIOPLabcdefghtyriopl",
"output": "NO"
},
{
"input": "25\nabcdefghijklmnopqrstuvwxy",
"output": "NO"
},
{
"input": "26\nabcdefhijklmnopqrstVxyzABC",
"output": "NO"
},
{
"input": "25\nqwertyuiopasdfghjklxcvbnm",
"output": "NO"
},
{
"input": "34\nTheQuickBrownFoxJumpsOverTheLayDog",
"output": "NO"
},
{
"input": "26\nabcdefghigklmnopqrstuvwxyz",
"output": "NO"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxyA",
"output": "NO"
},
{
"input": "50\nqazwsxedcrfvtgbyhnujmikolQWERTYUIOASDFGHJKLZXCVBNM",
"output": "NO"
},
{
"input": "35\nTheQuickBrownFoxJumpsOverTheLasyDog",
"output": "NO"
},
{
"input": "25\nbcdefghijklmnopqrstuvwxyz",
"output": "NO"
},
{
"input": "38\nAbCdEfGhIjKlMnOpQrStVwXyZzzzzzzaaaaaaa",
"output": "NO"
},
{
"input": "26\nabcdefghiklmnopqrstvxyzABC",
"output": "NO"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxzZ",
"output": "NO"
},
{
"input": "50\nabcdefghijklmnopqrstuvwxyABCDEFGHIJKLMNOPQRSTUVWXY",
"output": "NO"
}
] | 1,694,187,266 | 2,147,483,647 | Python 3 | OK | TESTS | 59 | 46 | 0 | len_word=int(input())
word=input().lower()
count_simvols=set()
answer=[]
for i in word:
count_simvols.add(i)
for i in count_simvols:
answer.append(i)
if len(count_simvols)==26:
print('YES')
else:
print('NO') | Title: Pangram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.
You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of characters in the string.
The second line contains the string. The string consists only of uppercase and lowercase Latin letters.
Output Specification:
Output "YES", if the string is a pangram and "NO" otherwise.
Demo Input:
['12\ntoosmallword\n', '35\nTheQuickBrownFoxJumpsOverTheLazyDog\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none | ```python
len_word=int(input())
word=input().lower()
count_simvols=set()
answer=[]
for i in word:
count_simvols.add(i)
for i in count_simvols:
answer.append(i)
if len(count_simvols)==26:
print('YES')
else:
print('NO')
``` | 3 | |
219 | A | k-String | PROGRAMMING | 1,000 | [
"implementation",
"strings"
] | null | null | A string is called a *k*-string if it can be represented as *k* concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1-string, a 2-string and a 4-string, but it is not a 3-string, a 5-string, or a 6-string and so on. Obviously any string is a 1-string.
You are given a string *s*, consisting of lowercase English letters and a positive integer *k*. Your task is to reorder the letters in the string *s* in such a way that the resulting string is a *k*-string. | The first input line contains integer *k* (1<=≤<=*k*<=≤<=1000). The second line contains *s*, all characters in *s* are lowercase English letters. The string length *s* satisfies the inequality 1<=≤<=|*s*|<=≤<=1000, where |*s*| is the length of string *s*. | Rearrange the letters in string *s* in such a way that the result is a *k*-string. Print the result on a single output line. If there are multiple solutions, print any of them.
If the solution doesn't exist, print "-1" (without quotes). | [
"2\naazz\n",
"3\nabcabcabz\n"
] | [
"azaz\n",
"-1\n"
] | none | 500 | [
{
"input": "2\naazz",
"output": "azaz"
},
{
"input": "3\nabcabcabz",
"output": "-1"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "2\nabba",
"output": "abab"
},
{
"input": "2\naaab",
"output": "-1"
},
{
"input": "7\nabacaba",
"output": "-1"
},
{
"input": "5\naaaaa",
"output": "aaaaa"
},
{
"input": "3\naabaaaaabb",
"output": "-1"
},
{
"input": "2\naaab",
"output": "-1"
},
{
"input": "2\nbabac",
"output": "-1"
},
{
"input": "3\nbbbccc",
"output": "bcbcbc"
},
{
"input": "2\naa",
"output": "aa"
},
{
"input": "250\ncececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececece",
"output": "cececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececececece"
},
{
"input": "15\nabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaaabaabbbcababaaa",
"output": "aaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbcaaaaaaaabbbbbbc"
},
{
"input": "1\naaa",
"output": "aaa"
},
{
"input": "1\naabaab",
"output": "aaaabb"
},
{
"input": "2\naabbbbccccccdddddddd",
"output": "abbcccddddabbcccdddd"
},
{
"input": "3\naaaaaaaaacccdddddd",
"output": "aaacddaaacddaaacdd"
},
{
"input": "3\naaaaaaaaacccbbbbbb",
"output": "aaabbcaaabbcaaabbc"
},
{
"input": "2\naazzzz",
"output": "azzazz"
},
{
"input": "2\naaaa",
"output": "aaaa"
},
{
"input": "2\naaaazz",
"output": "aazaaz"
},
{
"input": "2\naaaabb",
"output": "aabaab"
},
{
"input": "2\naabbbb",
"output": "abbabb"
},
{
"input": "2\naaaazzzz",
"output": "aazzaazz"
},
{
"input": "2\naaazz",
"output": "-1"
},
{
"input": "2\nbbbbaa",
"output": "abbabb"
},
{
"input": "2\naaaaaabbbb",
"output": "aaabbaaabb"
},
{
"input": "2\naaaaaazz",
"output": "aaazaaaz"
},
{
"input": "2\naaaabbbb",
"output": "aabbaabb"
},
{
"input": "2\naabaab",
"output": "aabaab"
},
{
"input": "1\naaaaa",
"output": "aaaaa"
},
{
"input": "2\nbbaaaa",
"output": "aabaab"
},
{
"input": "2\nacaccc",
"output": "accacc"
}
] | 1,649,352,818 | 2,147,483,647 | Python 3 | OK | TESTS | 42 | 92 | 0 | k = int(input())
s = input()
h = {}
for c in s:
if c in h:
h[c] += 1
else:
h[c] = 1
check = 0
for i in h.keys():
if h[i] % k == 0:
h[i] = h[i]//k
else:
check = -1
break
if check == -1:
print(-1)
else:
s = ""
for i in h.keys():
s += i*h[i]
print ( s*k ) | Title: k-String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A string is called a *k*-string if it can be represented as *k* concatenated copies of some string. For example, the string "aabaabaabaab" is at the same time a 1-string, a 2-string and a 4-string, but it is not a 3-string, a 5-string, or a 6-string and so on. Obviously any string is a 1-string.
You are given a string *s*, consisting of lowercase English letters and a positive integer *k*. Your task is to reorder the letters in the string *s* in such a way that the resulting string is a *k*-string.
Input Specification:
The first input line contains integer *k* (1<=≤<=*k*<=≤<=1000). The second line contains *s*, all characters in *s* are lowercase English letters. The string length *s* satisfies the inequality 1<=≤<=|*s*|<=≤<=1000, where |*s*| is the length of string *s*.
Output Specification:
Rearrange the letters in string *s* in such a way that the result is a *k*-string. Print the result on a single output line. If there are multiple solutions, print any of them.
If the solution doesn't exist, print "-1" (without quotes).
Demo Input:
['2\naazz\n', '3\nabcabcabz\n']
Demo Output:
['azaz\n', '-1\n']
Note:
none | ```python
k = int(input())
s = input()
h = {}
for c in s:
if c in h:
h[c] += 1
else:
h[c] = 1
check = 0
for i in h.keys():
if h[i] % k == 0:
h[i] = h[i]//k
else:
check = -1
break
if check == -1:
print(-1)
else:
s = ""
for i in h.keys():
s += i*h[i]
print ( s*k )
``` | 3 | |
1 | B | Spreadsheet | PROGRAMMING | 1,600 | [
"implementation",
"math"
] | B. Spreadsheets | 10 | 64 | In the popular spreadsheets systems (for example, in Excel) the following numeration of columns is used. The first column has number A, the second — number B, etc. till column 26 that is marked by Z. Then there are two-letter numbers: column 27 has number AA, 28 — AB, column 52 is marked by AZ. After ZZ there follow three-letter numbers, etc.
The rows are marked by integer numbers starting with 1. The cell name is the concatenation of the column and the row numbers. For example, BC23 is the name for the cell that is in column 55, row 23.
Sometimes another numeration system is used: RXCY, where X and Y are integer numbers, showing the column and the row numbers respectfully. For instance, R23C55 is the cell from the previous example.
Your task is to write a program that reads the given sequence of cell coordinates and produce each item written according to the rules of another numeration system. | The first line of the input contains integer number *n* (1<=≤<=*n*<=≤<=105), the number of coordinates in the test. Then there follow *n* lines, each of them contains coordinates. All the coordinates are correct, there are no cells with the column and/or the row numbers larger than 106 . | Write *n* lines, each line should contain a cell coordinates in the other numeration system. | [
"2\nR23C55\nBC23\n"
] | [
"BC23\nR23C55\n"
] | none | 0 | [
{
"input": "2\nR23C55\nBC23",
"output": "BC23\nR23C55"
},
{
"input": "1\nA1",
"output": "R1C1"
},
{
"input": "5\nR8C3\nD1\nR7C2\nR8C9\nR8C9",
"output": "C8\nR1C4\nB7\nI8\nI8"
},
{
"input": "4\nR4C25\nR90C35\nAP55\nX83",
"output": "Y4\nAI90\nR55C42\nR83C24"
},
{
"input": "10\nR50C12\nR23C47\nY96\nR44C13\nR19C21\nR95C73\nBK12\nR51C74\nAY34\nR63C25",
"output": "L50\nAU23\nR96C25\nM44\nU19\nBU95\nR12C63\nBV51\nR34C51\nY63"
}
] | 1,649,320,537 | 2,137 | Python 3 | WRONG_ANSWER | TESTS | 5 | 92 | 409,600 | import string
n = int(input())
for _ in range(n):
s = input()
last_isnumeric = False
arr = [[s[0]]]
for i in range(1, len(s)):
c = s[i]
isnumeric = c.isnumeric()
if isnumeric:
if last_isnumeric:
arr[-1].append(c)
else:
arr.append([c])
else:
if last_isnumeric:
arr.append([c])
else:
arr[-1].append(c)
last_isnumeric = isnumeric
if len(arr) == 2:
row = ''.join(arr[1])
col = 0
for c in arr[0]:
col = col * 26 + (ord(c) - 64)
print(f'R{row}C{col}')
else:
row = ''.join(arr[1])
col_int = int(''.join(arr[3]))
col = []
while col_int > 0:
col.append(col_int % 26)
col_int //= 26
col = [string.ascii_uppercase[c-1] for c in col]
col.reverse()
col = ''.join(col)
print(f'{col}{row}')
| Title: Spreadsheet
Time Limit: 10 seconds
Memory Limit: 64 megabytes
Problem Description:
In the popular spreadsheets systems (for example, in Excel) the following numeration of columns is used. The first column has number A, the second — number B, etc. till column 26 that is marked by Z. Then there are two-letter numbers: column 27 has number AA, 28 — AB, column 52 is marked by AZ. After ZZ there follow three-letter numbers, etc.
The rows are marked by integer numbers starting with 1. The cell name is the concatenation of the column and the row numbers. For example, BC23 is the name for the cell that is in column 55, row 23.
Sometimes another numeration system is used: RXCY, where X and Y are integer numbers, showing the column and the row numbers respectfully. For instance, R23C55 is the cell from the previous example.
Your task is to write a program that reads the given sequence of cell coordinates and produce each item written according to the rules of another numeration system.
Input Specification:
The first line of the input contains integer number *n* (1<=≤<=*n*<=≤<=105), the number of coordinates in the test. Then there follow *n* lines, each of them contains coordinates. All the coordinates are correct, there are no cells with the column and/or the row numbers larger than 106 .
Output Specification:
Write *n* lines, each line should contain a cell coordinates in the other numeration system.
Demo Input:
['2\nR23C55\nBC23\n']
Demo Output:
['BC23\nR23C55\n']
Note:
none | ```python
import string
n = int(input())
for _ in range(n):
s = input()
last_isnumeric = False
arr = [[s[0]]]
for i in range(1, len(s)):
c = s[i]
isnumeric = c.isnumeric()
if isnumeric:
if last_isnumeric:
arr[-1].append(c)
else:
arr.append([c])
else:
if last_isnumeric:
arr.append([c])
else:
arr[-1].append(c)
last_isnumeric = isnumeric
if len(arr) == 2:
row = ''.join(arr[1])
col = 0
for c in arr[0]:
col = col * 26 + (ord(c) - 64)
print(f'R{row}C{col}')
else:
row = ''.join(arr[1])
col_int = int(''.join(arr[3]))
col = []
while col_int > 0:
col.append(col_int % 26)
col_int //= 26
col = [string.ascii_uppercase[c-1] for c in col]
col.reverse()
col = ''.join(col)
print(f'{col}{row}')
``` | 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",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,633,544,102 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 218 | 22,323,200 | m = int(input())
n = int(input())
board = (n*m)/2
print(int(board))
| Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
m = int(input())
n = int(input())
board = (n*m)/2
print(int(board))
``` | -1 |
707 | A | Brain's Photos | PROGRAMMING | 800 | [
"implementation"
] | null | null | Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored. | The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'. | Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line. | [
"2 2\nC M\nY Y\n",
"3 2\nW W\nW W\nB B\n",
"1 1\nW\n"
] | [
"#Color",
"#Black&White",
"#Black&White"
] | none | 500 | [
{
"input": "2 2\nC M\nY Y",
"output": "#Color"
},
{
"input": "3 2\nW W\nW W\nB B",
"output": "#Black&White"
},
{
"input": "1 1\nW",
"output": "#Black&White"
},
{
"input": "2 3\nW W W\nB G Y",
"output": "#Color"
},
{
"input": "1 1\nW",
"output": "#Black&White"
},
{
"input": "5 5\nW G B Y M\nG B Y M C\nB Y M C W\nY M C W G\nM C W G B",
"output": "#Color"
},
{
"input": "1 6\nC M Y W G B",
"output": "#Color"
},
{
"input": "1 3\nW G B",
"output": "#Black&White"
},
{
"input": "1 1\nW",
"output": "#Black&White"
},
{
"input": "5 5\nW G B W G\nG B W G B\nB W G B W\nW G B W G\nG B W G B",
"output": "#Black&White"
},
{
"input": "2 3\nW W W\nB G C",
"output": "#Color"
},
{
"input": "2 3\nW W W\nB G M",
"output": "#Color"
},
{
"input": "3 3\nC B W\nB Y M\nB B W",
"output": "#Color"
},
{
"input": "1 3\nW C W",
"output": "#Color"
},
{
"input": "3 3\nB W B\nB C W\nB W W",
"output": "#Color"
},
{
"input": "1 2\nW Y",
"output": "#Color"
},
{
"input": "1 1\nG",
"output": "#Black&White"
},
{
"input": "1 3\nB G W",
"output": "#Black&White"
},
{
"input": "2 2\nW W\nB C",
"output": "#Color"
},
{
"input": "1 1\nM",
"output": "#Color"
},
{
"input": "1 2\nW C",
"output": "#Color"
},
{
"input": "2 3\nW W M\nW W M",
"output": "#Color"
},
{
"input": "3 2\nW W\nW W\nB C",
"output": "#Color"
},
{
"input": "2 3\nW W C\nW W W",
"output": "#Color"
},
{
"input": "1 3\nG G G",
"output": "#Black&White"
},
{
"input": "1 1\nC",
"output": "#Color"
},
{
"input": "1 2\nC W",
"output": "#Color"
},
{
"input": "1 3\nW W C",
"output": "#Color"
},
{
"input": "2 2\nW B\nB G",
"output": "#Black&White"
},
{
"input": "2 2\nB B\nY Y",
"output": "#Color"
},
{
"input": "2 2\nG G\nC C",
"output": "#Color"
},
{
"input": "1 2\nB C",
"output": "#Color"
},
{
"input": "1 4\nG G G C",
"output": "#Color"
},
{
"input": "1 1\nB",
"output": "#Black&White"
},
{
"input": "2 1\nY\nB",
"output": "#Color"
}
] | 1,687,864,763 | 2,147,483,647 | Python 3 | OK | TESTS | 50 | 46 | 0 | n, m = map(int, input().split())
for i in range(n):
a = list(input().split())
for j in range(m):
if a[j] == 'C' or a[j] == 'Y' or a[j] == 'M':
print('#Color')
quit()
print('#Black&White') | Title: Brain's Photos
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.
As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).
Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!
As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.
Photo can be represented as a matrix sized *n*<=×<=*m*, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:
- 'C' (cyan)- 'M' (magenta)- 'Y' (yellow)- 'W' (white)- 'G' (grey)- 'B' (black)
The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored.
Input Specification:
The first line of the input contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100) — the number of photo pixel matrix rows and columns respectively.
Then *n* lines describing matrix rows follow. Each of them contains *m* space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'.
Output Specification:
Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line.
Demo Input:
['2 2\nC M\nY Y\n', '3 2\nW W\nW W\nB B\n', '1 1\nW\n']
Demo Output:
['#Color', '#Black&White', '#Black&White']
Note:
none | ```python
n, m = map(int, input().split())
for i in range(n):
a = list(input().split())
for j in range(m):
if a[j] == 'C' or a[j] == 'Y' or a[j] == 'M':
print('#Color')
quit()
print('#Black&White')
``` | 3 | |
59 | A | Word | PROGRAMMING | 800 | [
"implementation",
"strings"
] | A. Word | 2 | 256 | Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word. | The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100. | Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one. | [
"HoUse\n",
"ViP\n",
"maTRIx\n"
] | [
"house\n",
"VIP\n",
"matrix\n"
] | none | 500 | [
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "chnenu"
},
{
"input": "ERPZGrodyu",
"output": "erpzgrodyu"
},
{
"input": "KSXBXWpebh",
"output": "KSXBXWPEBH"
},
{
"input": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv",
"output": "qvxpqullmcbegsdskddortcvxyqlbvxmmkhevovnezubvpvnrcajpxraeaxizgaowtfkzywvhnbgzsxbhkaipcmoumtikkiyyaiv"
},
{
"input": "Amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd",
"output": "amnhaxtaopjzrkqlbroiyipitndczpunwygstmzevgyjdzyanxkdqnvgkikfabwouwkkbzuiuvgvxgpizsvqsbwepktpdrgdkmfd"
},
{
"input": "ISAGFJFARYFBLOPQDSHWGMCNKMFTLVFUGNJEWGWNBLXUIATXEkqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv",
"output": "isagfjfaryfblopqdshwgmcnkmftlvfugnjewgwnblxuiatxekqiettmmjgydwcpafqrppdsrrrtguinqbgmzzfqwonkpgpcwenv"
},
{
"input": "XHRPXZEGHSOCJPICUIXSKFUZUPYTSGJSDIYBCMNMNBPNDBXLXBzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg",
"output": "xhrpxzeghsocjpicuixskfuzupytsgjsdiybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsyvncqmfhautvxudqdhgg"
},
{
"input": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGAdkcetqjljtmttlonpekcovdzebzdkzggwfsxhapmjkdbuceak",
"output": "RJIQZMJCIMSNDBOHBRAWIENODSALETAKGKPYUFGVEFGCBRENZGADKCETQJLJTMTTLONPEKCOVDZEBZDKZGGWFSXHAPMJKDBUCEAK"
},
{
"input": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFw",
"output": "DWLWOBHNMMGTFOLFAECKBRNNGLYLYDXTGTVRLMEESZOIUATZZZXUFUZDLSJXMEVRTESSFBWLNZZCLCQWEVNNUCXYVHNGNXHCBDFW"
},
{
"input": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB",
"output": "NYCNHJWGBOCOTSPETKKHVWFGAQYNHOVJWJHCIEFOUQZXOYUIEQDZALFKTEHTVDBVJMEUBJUBCMNVPWGDPNCHQHZJRCHYRFPVIGUB"
},
{
"input": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge",
"output": "igxoixiecetohtgjgbqzvlaobkhstejxdklghowtvwunnnvauriohuspsdmpzckprwajyxldoyckgjivjpmbfqtszmtocovxwge"
},
{
"input": "Ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw",
"output": "ykkekrsqolzryiwsmdlnbmfautxxxauoojrddvwklgnlyrfcvhorrzbmtcrvpaypqhcffdqhwziipyyskcmztjprjqvmzzqhqnw"
},
{
"input": "YQOMLKYAORUQQUCQZCDYMIVDHGWZFFRMUVTAWCHERFPMNRYRIkgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks",
"output": "yqomlkyaoruqqucqzcdymivdhgwzffrmuvtawcherfpmnryrikgqrciokgajamehmcxgerpudvsqyonjonsxgbnefftzmygncks"
},
{
"input": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJcuusigdqfkumewjtdyitveeiaybwrhomrwmpdipjwiuxfnwuz",
"output": "CDOZDPBVVVHNBJVBYHEOXWFLJKRWJCAJMIFCOZWWYFKVWOGTVJCUUSIGDQFKUMEWJTDYITVEEIAYBWRHOMRWMPDIPJWIUXFNWUZ"
},
{
"input": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWuckzcpxosodcjaaakvlxpbiigsiauviilylnnqlyucziihqg",
"output": "WHIUVEXHVOOIJIDVJVPQUBJMEVPMPDKQWJKFBZSGSKUXMIPPMJWUCKZCPXOSODCJAAAKVLXPBIIGSIAUVIILYLNNQLYUCZIIHQG"
},
{
"input": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO",
"output": "VGHUNFOXKETUYMZDJNGTAOIOANYXSGYNFOGOFFLDAWEUKYFOZXCJTCAFXZYLQZERYZLRSQXYQGAPCSUDPMEYTNCTTTMFAGVDWBO"
},
{
"input": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDd",
"output": "EDUANHCQDOCJHFONTTSWBUJSTTIXBIXMAIUFSGFLXWAYIURTVAVZPYQDLAWIGCLMPHXCEFCJPFAAHXVNGQUFNHADAIUAZIDMHDD"
},
{
"input": "EBWOVSNLYTWWXrnovgugogtjahnmatomubebyxakas",
"output": "ebwovsnlytwwxrnovgugogtjahnmatomubebyxakas"
},
{
"input": "AORNNDKTRLRVGDPXJKXFTPFpopuzrlqumrxssskvbm",
"output": "AORNNDKTRLRVGDPXJKXFTPFPOPUZRLQUMRXSSSKVBM"
},
{
"input": "DBBZJCWQEVGYHEEUHJAWRHBDPsnplijbwtdibqrrheebfxqcvn",
"output": "dbbzjcwqevgyheeuhjawrhbdpsnplijbwtdibqrrheebfxqcvn"
},
{
"input": "t",
"output": "t"
},
{
"input": "N",
"output": "N"
},
{
"input": "kv",
"output": "kv"
},
{
"input": "Ur",
"output": "ur"
},
{
"input": "CN",
"output": "CN"
}
] | 1,636,207,056 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 62 | 4,198,400 | s = input("")
count_upper = 0
count_lower = 0
for i in s:
if s.isupper():
count_upper += 1
elif s.islower():
count_lower += 1
if count_upper > count_lower:
print(s.upper())
elif count_lower >= count_upper:
print(s.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
s = input("")
count_upper = 0
count_lower = 0
for i in s:
if s.isupper():
count_upper += 1
elif s.islower():
count_lower += 1
if count_upper > count_lower:
print(s.upper())
elif count_lower >= count_upper:
print(s.lower())
``` | 0 |
842 | C | Ilya And The Tree | PROGRAMMING | 2,000 | [
"dfs and similar",
"graphs",
"math",
"number theory",
"trees"
] | null | null | Ilya is very fond of graphs, especially trees. During his last trip to the forest Ilya found a very interesting tree rooted at vertex 1. There is an integer number written on each vertex of the tree; the number written on vertex *i* is equal to *a**i*.
Ilya believes that the beauty of the vertex *x* is the greatest common divisor of all numbers written on the vertices on the path from the root to *x*, including this vertex itself. In addition, Ilya can change the number in one arbitrary vertex to 0 or leave all vertices unchanged. Now for each vertex Ilya wants to know the maximum possible beauty it can have.
For each vertex the answer must be considered independently.
The beauty of the root equals to number written on it. | First line contains one integer number *n* — the number of vertices in tree (1<=≤<=*n*<=≤<=2·105).
Next line contains *n* integer numbers *a**i* (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*a**i*<=≤<=2·105).
Each of next *n*<=-<=1 lines contains two integer numbers *x* and *y* (1<=≤<=*x*,<=*y*<=≤<=*n*, *x*<=≠<=*y*), which means that there is an edge (*x*,<=*y*) in the tree. | Output *n* numbers separated by spaces, where *i*-th number equals to maximum possible beauty of vertex *i*. | [
"2\n6 2\n1 2\n",
"3\n6 2 3\n1 2\n1 3\n",
"1\n10\n"
] | [
"6 6 \n",
"6 6 6 \n",
"10 \n"
] | none | 1,500 | [
{
"input": "2\n6 2\n1 2",
"output": "6 6 "
},
{
"input": "3\n6 2 3\n1 2\n1 3",
"output": "6 6 6 "
},
{
"input": "1\n10",
"output": "10 "
},
{
"input": "10\n2 3 4 5 6 7 8 9 10 11\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n4 8\n8 9\n9 10",
"output": "2 3 2 1 1 1 1 1 1 1 "
},
{
"input": "4\n6 2 3 2\n1 2\n2 3\n3 4",
"output": "6 6 3 2 "
}
] | 1,624,951,915 | 2,147,483,647 | PyPy 3 | TIME_LIMIT_EXCEEDED | TESTS | 14 | 2,000 | 53,145,600 | import os
import sys
from io import BytesIO, IOBase
from collections import deque
from math import gcd
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
n = int(input())
A = list(map(int,input().split()))
G = [[] for _ in range(n)]
for _ in range(n-1):
x,y = map(int,input().split())
G[x-1].append(y-1)
G[y-1].append(x-1)
par = [-1]*n
lvl = [-1]*n
q = deque([0])
lvl[0] = 1
while q:
x = q.popleft()
for y in G[x]:
if lvl[y] == -1:
par[y] = x
lvl[y] = 1 + lvl[x]
q.append(y)
# suppose that A[0] = 0
ans = [0]*n
q = deque([])
for x in G[0]:
ans[x] = A[x]
for y in G[x]:
if y != 0:
q.append(y)
while q:
x = q.popleft()
ans[x] = gcd(ans[par[x]], A[x])
for y in G[x]:
if y != par[x]:
q.append(y)
# suppose that A[0] != 0
DIVISOR = set()
d = 1
while d*d <= A[0]:
if A[0] % d == 0:
DIVISOR.add(d)
DIVISOR.add(A[0] // d)
d += 1
def update(d):
cnt = [0]*n
cnt[0] = 1
q = deque(G[0])
while q:
x = q.popleft()
cnt[x] = cnt[par[x]] + (A[x]%d == 0)
if cnt[x] >= lvl[x] - 1:
ans[x] = max(d, ans[x])
for y in G[x]:
if y != par[x]:
q.append(y)
for d in DIVISOR:
update(d)
ans[0] = A[0]
print(*ans)
| Title: Ilya And The Tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ilya is very fond of graphs, especially trees. During his last trip to the forest Ilya found a very interesting tree rooted at vertex 1. There is an integer number written on each vertex of the tree; the number written on vertex *i* is equal to *a**i*.
Ilya believes that the beauty of the vertex *x* is the greatest common divisor of all numbers written on the vertices on the path from the root to *x*, including this vertex itself. In addition, Ilya can change the number in one arbitrary vertex to 0 or leave all vertices unchanged. Now for each vertex Ilya wants to know the maximum possible beauty it can have.
For each vertex the answer must be considered independently.
The beauty of the root equals to number written on it.
Input Specification:
First line contains one integer number *n* — the number of vertices in tree (1<=≤<=*n*<=≤<=2·105).
Next line contains *n* integer numbers *a**i* (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*a**i*<=≤<=2·105).
Each of next *n*<=-<=1 lines contains two integer numbers *x* and *y* (1<=≤<=*x*,<=*y*<=≤<=*n*, *x*<=≠<=*y*), which means that there is an edge (*x*,<=*y*) in the tree.
Output Specification:
Output *n* numbers separated by spaces, where *i*-th number equals to maximum possible beauty of vertex *i*.
Demo Input:
['2\n6 2\n1 2\n', '3\n6 2 3\n1 2\n1 3\n', '1\n10\n']
Demo Output:
['6 6 \n', '6 6 6 \n', '10 \n']
Note:
none | ```python
import os
import sys
from io import BytesIO, IOBase
from collections import deque
from math import gcd
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
n = int(input())
A = list(map(int,input().split()))
G = [[] for _ in range(n)]
for _ in range(n-1):
x,y = map(int,input().split())
G[x-1].append(y-1)
G[y-1].append(x-1)
par = [-1]*n
lvl = [-1]*n
q = deque([0])
lvl[0] = 1
while q:
x = q.popleft()
for y in G[x]:
if lvl[y] == -1:
par[y] = x
lvl[y] = 1 + lvl[x]
q.append(y)
# suppose that A[0] = 0
ans = [0]*n
q = deque([])
for x in G[0]:
ans[x] = A[x]
for y in G[x]:
if y != 0:
q.append(y)
while q:
x = q.popleft()
ans[x] = gcd(ans[par[x]], A[x])
for y in G[x]:
if y != par[x]:
q.append(y)
# suppose that A[0] != 0
DIVISOR = set()
d = 1
while d*d <= A[0]:
if A[0] % d == 0:
DIVISOR.add(d)
DIVISOR.add(A[0] // d)
d += 1
def update(d):
cnt = [0]*n
cnt[0] = 1
q = deque(G[0])
while q:
x = q.popleft()
cnt[x] = cnt[par[x]] + (A[x]%d == 0)
if cnt[x] >= lvl[x] - 1:
ans[x] = max(d, ans[x])
for y in G[x]:
if y != par[x]:
q.append(y)
for d in DIVISOR:
update(d)
ans[0] = A[0]
print(*ans)
``` | 0 | |
282 | A | Bit++ | PROGRAMMING | 800 | [
"implementation"
] | null | null | The classic programming language of Bitland is Bit++. This language is so peculiar and complicated.
The language is that peculiar as it has exactly one variable, called *x*. Also, there are two operations:
- Operation ++ increases the value of variable *x* by 1. - Operation -- decreases the value of variable *x* by 1.
A statement in language Bit++ is a sequence, consisting of exactly one operation and one variable *x*. The statement is written without spaces, that is, it can only contain characters "+", "-", "X". Executing a statement means applying the operation it contains.
A programme in Bit++ is a sequence of statements, each of them needs to be executed. Executing a programme means executing all the statements it contains.
You're given a programme in language Bit++. The initial value of *x* is 0. Execute the programme and find its final value (the value of the variable when this programme is executed). | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=150) — the number of statements in the programme.
Next *n* lines contain a statement each. Each statement contains exactly one operation (++ or --) and exactly one variable *x* (denoted as letter «X»). Thus, there are no empty statements. The operation and the variable can be written in any order. | Print a single integer — the final value of *x*. | [
"1\n++X\n",
"2\nX++\n--X\n"
] | [
"1\n",
"0\n"
] | none | 500 | [
{
"input": "1\n++X",
"output": "1"
},
{
"input": "2\nX++\n--X",
"output": "0"
},
{
"input": "3\n++X\n++X\n++X",
"output": "3"
},
{
"input": "2\n--X\n--X",
"output": "-2"
},
{
"input": "5\n++X\n--X\n++X\n--X\n--X",
"output": "-1"
},
{
"input": "28\nX--\n++X\nX++\nX++\nX++\n--X\n--X\nX++\nX--\n++X\nX++\n--X\nX--\nX++\nX--\n++X\n++X\nX++\nX++\nX++\nX++\n--X\n++X\n--X\n--X\n--X\n--X\nX++",
"output": "4"
},
{
"input": "94\nX++\nX++\n++X\n++X\nX--\n--X\nX++\n--X\nX++\n++X\nX++\n++X\n--X\n--X\n++X\nX++\n--X\nX--\nX--\n--X\nX--\nX--\n--X\n++X\n--X\nX--\nX--\nX++\n++X\n--X\nX--\n++X\n--X\n--X\nX--\nX--\nX++\nX++\nX--\nX++\nX--\nX--\nX--\n--X\nX--\nX--\nX--\nX++\n++X\nX--\n++X\nX++\n--X\n--X\n--X\n--X\n++X\nX--\n--X\n--X\n++X\nX--\nX--\nX++\n++X\nX++\n++X\n--X\n--X\nX--\n++X\nX--\nX--\n++X\n++X\n++X\n++X\nX++\n++X\n--X\nX++\n--X\n--X\n++X\n--X\nX++\n++X\nX++\n--X\nX--\nX--\n--X\n++X\nX++",
"output": "-10"
},
{
"input": "56\n--X\nX--\n--X\n--X\nX--\nX--\n--X\nX++\n++X\n--X\nX++\nX--\n--X\n++X\n--X\nX--\nX--\n++X\nX--\nX--\n--X\n++X\n--X\n++X\n--X\nX++\n++X\nX++\n--X\n++X\nX++\nX++\n--X\nX++\nX--\n--X\nX--\n--X\nX++\n++X\n--X\n++X\nX++\nX--\n--X\n--X\n++X\nX--\nX--\n--X\nX--\n--X\nX++\n--X\n++X\n--X",
"output": "-14"
},
{
"input": "59\nX--\n--X\nX++\n++X\nX--\n--X\n--X\n++X\n++X\n++X\n++X\nX++\n++X\n++X\nX++\n--X\nX--\nX++\n++X\n--X\nX++\n--X\n++X\nX++\n--X\n--X\nX++\nX++\n--X\nX++\nX++\nX++\nX--\nX--\n--X\nX++\nX--\nX--\n++X\nX--\nX++\n--X\nX++\nX--\nX--\nX--\nX--\n++X\n--X\nX++\nX++\nX--\nX++\n++X\nX--\nX++\nX--\nX--\n++X",
"output": "3"
},
{
"input": "87\n--X\n++X\n--X\nX++\n--X\nX--\n--X\n++X\nX--\n++X\n--X\n--X\nX++\n--X\nX--\nX++\n++X\n--X\n++X\n++X\n--X\n++X\n--X\nX--\n++X\n++X\nX--\nX++\nX++\n--X\n--X\n++X\nX--\n--X\n++X\n--X\nX++\n--X\n--X\nX--\n++X\n++X\n--X\nX--\nX--\nX--\nX--\nX--\nX++\n--X\n++X\n--X\nX++\n++X\nX++\n++X\n--X\nX++\n++X\nX--\n--X\nX++\n++X\nX++\nX++\n--X\n--X\n++X\n--X\nX++\nX++\n++X\nX++\nX++\nX++\nX++\n--X\n--X\n--X\n--X\n--X\n--X\n--X\nX--\n--X\n++X\n++X",
"output": "-5"
},
{
"input": "101\nX++\nX++\nX++\n++X\n--X\nX--\nX++\nX--\nX--\n--X\n--X\n++X\nX++\n++X\n++X\nX--\n--X\n++X\nX++\nX--\n++X\n--X\n--X\n--X\n++X\n--X\n++X\nX++\nX++\n++X\n--X\nX++\nX--\nX++\n++X\n++X\nX--\nX--\nX--\nX++\nX++\nX--\nX--\nX++\n++X\n++X\n++X\n--X\n--X\n++X\nX--\nX--\n--X\n++X\nX--\n++X\nX++\n++X\nX--\nX--\n--X\n++X\n--X\n++X\n++X\n--X\nX++\n++X\nX--\n++X\nX--\n++X\nX++\nX--\n++X\nX++\n--X\nX++\nX++\n++X\n--X\n++X\n--X\nX++\n--X\nX--\n--X\n++X\n++X\n++X\n--X\nX--\nX--\nX--\nX--\n--X\n--X\n--X\n++X\n--X\n--X",
"output": "1"
},
{
"input": "63\n--X\nX--\n++X\n--X\n++X\nX++\n--X\n--X\nX++\n--X\n--X\nX++\nX--\nX--\n--X\n++X\nX--\nX--\nX++\n++X\nX++\nX++\n--X\n--X\n++X\nX--\nX--\nX--\n++X\nX++\nX--\n--X\nX--\n++X\n++X\nX++\n++X\nX++\nX++\n--X\nX--\n++X\nX--\n--X\nX--\nX--\nX--\n++X\n++X\n++X\n++X\nX++\nX++\n++X\n--X\n--X\n++X\n++X\n++X\nX--\n++X\n++X\nX--",
"output": "1"
},
{
"input": "45\n--X\n++X\nX--\n++X\n++X\nX++\n--X\n--X\n--X\n--X\n--X\n--X\n--X\nX++\n++X\nX--\n++X\n++X\nX--\nX++\nX--\n--X\nX--\n++X\n++X\n--X\n--X\nX--\nX--\n--X\n++X\nX--\n--X\n++X\n++X\n--X\n--X\nX--\n++X\n++X\nX++\nX++\n++X\n++X\nX++",
"output": "-3"
},
{
"input": "21\n++X\nX++\n--X\nX--\nX++\n++X\n--X\nX--\nX++\nX--\nX--\nX--\nX++\n++X\nX++\n++X\n--X\nX--\n--X\nX++\n++X",
"output": "1"
},
{
"input": "100\n--X\n++X\nX++\n++X\nX--\n++X\nX--\nX++\n--X\nX++\nX--\nX--\nX--\n++X\nX--\nX++\nX++\n++X\nX++\nX++\nX++\nX++\n++X\nX++\n++X\nX--\n--X\n++X\nX--\n--X\n++X\n++X\nX--\nX++\nX++\nX++\n++X\n--X\n++X\nX++\nX--\n++X\n++X\n--X\n++X\nX--\nX--\nX--\nX++\nX--\nX--\nX++\nX++\n--X\nX++\nX++\n--X\nX--\n--X\n++X\n--X\n++X\n++X\nX--\n--X\n++X\n++X\n--X\n--X\n++X\nX++\nX--\nX++\nX--\nX++\nX++\n--X\nX--\nX--\n++X\nX--\n--X\n--X\nX++\n--X\n--X\nX--\nX--\n++X\n++X\nX--\n++X\nX++\n--X\n++X\n++X\nX++\n--X\n--X\nX++",
"output": "8"
},
{
"input": "17\nX++\nX++\n++X\n--X\n--X\n++X\n++X\n--X\nX--\nX++\nX--\n--X\n--X\nX--\n++X\nX--\nX++",
"output": "-1"
},
{
"input": "77\n++X\nX++\n--X\nX--\n--X\n--X\nX--\nX++\nX--\nX++\nX--\n++X\n--X\n--X\n--X\n--X\n++X\nX--\nX++\nX--\n--X\nX--\n--X\nX--\n++X\n--X\n++X\n++X\nX++\nX++\nX--\n--X\nX--\nX--\nX++\n--X\n--X\n++X\nX--\nX--\n++X\nX++\nX--\n++X\n--X\nX++\nX--\n++X\n++X\n++X\nX--\nX--\nX--\n--X\n++X\n++X\n++X\nX++\n--X\n--X\n++X\n--X\nX--\nX++\n++X\nX++\n++X\nX--\nX++\nX++\n--X\nX++\nX++\nX++\n--X\nX++\nX--",
"output": "-5"
},
{
"input": "21\nX--\n++X\n--X\nX--\n++X\nX--\n++X\nX--\n--X\n++X\nX++\n++X\nX++\n++X\nX--\n--X\nX++\nX++\nX--\n++X\nX--",
"output": "1"
},
{
"input": "1\nX--",
"output": "-1"
}
] | 1,699,550,707 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 36 | 62 | 0 | def execute_programme(n, l):
x = 0
for statement in l:
if "++" in statement:
x += 1
else:
x -= 1
return x
n = int(input())
l = [input() for _ in range(n)]
result = execute_programme(n, l)
print(result)
| Title: Bit++
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The classic programming language of Bitland is Bit++. This language is so peculiar and complicated.
The language is that peculiar as it has exactly one variable, called *x*. Also, there are two operations:
- Operation ++ increases the value of variable *x* by 1. - Operation -- decreases the value of variable *x* by 1.
A statement in language Bit++ is a sequence, consisting of exactly one operation and one variable *x*. The statement is written without spaces, that is, it can only contain characters "+", "-", "X". Executing a statement means applying the operation it contains.
A programme in Bit++ is a sequence of statements, each of them needs to be executed. Executing a programme means executing all the statements it contains.
You're given a programme in language Bit++. The initial value of *x* is 0. Execute the programme and find its final value (the value of the variable when this programme is executed).
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=150) — the number of statements in the programme.
Next *n* lines contain a statement each. Each statement contains exactly one operation (++ or --) and exactly one variable *x* (denoted as letter «X»). Thus, there are no empty statements. The operation and the variable can be written in any order.
Output Specification:
Print a single integer — the final value of *x*.
Demo Input:
['1\n++X\n', '2\nX++\n--X\n']
Demo Output:
['1\n', '0\n']
Note:
none | ```python
def execute_programme(n, l):
x = 0
for statement in l:
if "++" in statement:
x += 1
else:
x -= 1
return x
n = int(input())
l = [input() for _ in range(n)]
result = execute_programme(n, l)
print(result)
``` | 3 | |
909 | A | Generate Login | PROGRAMMING | 1,000 | [
"brute force",
"greedy",
"sortings"
] | null | null | The preferred way to generate user login in Polygon is to concatenate a prefix of the user's first name and a prefix of their last name, in that order. Each prefix must be non-empty, and any of the prefixes can be the full name. Typically there are multiple possible logins for each person.
You are given the first and the last name of a user. Return the alphabetically earliest login they can get (regardless of other potential Polygon users).
As a reminder, a prefix of a string *s* is its substring which occurs at the beginning of *s*: "a", "ab", "abc" etc. are prefixes of string "{abcdef}" but "b" and 'bc" are not. A string *a* is alphabetically earlier than a string *b*, if *a* is a prefix of *b*, or *a* and *b* coincide up to some position, and then *a* has a letter that is alphabetically earlier than the corresponding letter in *b*: "a" and "ab" are alphabetically earlier than "ac" but "b" and "ba" are alphabetically later than "ac". | The input consists of a single line containing two space-separated strings: the first and the last names. Each character of each string is a lowercase English letter. The length of each string is between 1 and 10, inclusive. | Output a single string — alphabetically earliest possible login formed from these names. The output should be given in lowercase as well. | [
"harry potter\n",
"tom riddle\n"
] | [
"hap\n",
"tomr\n"
] | none | 500 | [
{
"input": "harry potter",
"output": "hap"
},
{
"input": "tom riddle",
"output": "tomr"
},
{
"input": "a qdpinbmcrf",
"output": "aq"
},
{
"input": "wixjzniiub ssdfodfgap",
"output": "wis"
},
{
"input": "z z",
"output": "zz"
},
{
"input": "ertuyivhfg v",
"output": "ertuv"
},
{
"input": "asdfghjkli ware",
"output": "asdfghjkliw"
},
{
"input": "udggmyop ze",
"output": "udggmyopz"
},
{
"input": "fapkdme rtzxovx",
"output": "fapkdmer"
},
{
"input": "mybiqxmnqq l",
"output": "ml"
},
{
"input": "dtbqya fyyymv",
"output": "df"
},
{
"input": "fyclu zokbxiahao",
"output": "fycluz"
},
{
"input": "qngatnviv rdych",
"output": "qngar"
},
{
"input": "ttvnhrnng lqkfulhrn",
"output": "tl"
},
{
"input": "fya fgx",
"output": "ff"
},
{
"input": "nuis zvjjqlre",
"output": "nuisz"
},
{
"input": "ly qtsmze",
"output": "lq"
},
{
"input": "d kgfpjsurfw",
"output": "dk"
},
{
"input": "lwli ewrpu",
"output": "le"
},
{
"input": "rr wldsfubcs",
"output": "rrw"
},
{
"input": "h qart",
"output": "hq"
},
{
"input": "vugvblnzx kqdwdulm",
"output": "vk"
},
{
"input": "xohesmku ef",
"output": "xe"
},
{
"input": "twvvsl wtcyawv",
"output": "tw"
},
{
"input": "obljndajv q",
"output": "obljndajq"
},
{
"input": "jjxwj kxccwx",
"output": "jjk"
},
{
"input": "sk fftzmv",
"output": "sf"
},
{
"input": "cgpegngs aufzxkyyrw",
"output": "ca"
},
{
"input": "reyjzjdvq skuch",
"output": "res"
},
{
"input": "ardaae mxgdulijf",
"output": "am"
},
{
"input": "bgopsdfji uaps",
"output": "bgopsdfjiu"
},
{
"input": "amolfed pun",
"output": "amolfedp"
},
{
"input": "badkiln yort",
"output": "badkilny"
},
{
"input": "aaaaaaaaaz york",
"output": "aaaaaaaaay"
},
{
"input": "bbbbcbbbbd c",
"output": "bbbbc"
},
{
"input": "aa ab",
"output": "aa"
},
{
"input": "ab b",
"output": "ab"
},
{
"input": "aaaaa ab",
"output": "aa"
},
{
"input": "aa a",
"output": "aa"
},
{
"input": "aba b",
"output": "ab"
},
{
"input": "aaaaaaa aaaaaa",
"output": "aa"
},
{
"input": "a a",
"output": "aa"
},
{
"input": "a aa",
"output": "aa"
},
{
"input": "a b",
"output": "ab"
},
{
"input": "b a",
"output": "ba"
},
{
"input": "z a",
"output": "za"
},
{
"input": "aaa a",
"output": "aa"
},
{
"input": "aa aa",
"output": "aa"
},
{
"input": "a aaa",
"output": "aa"
},
{
"input": "aaaaaaaaaa aaaaaaaaaa",
"output": "aa"
},
{
"input": "aaaaaaaaaa a",
"output": "aa"
},
{
"input": "a aaaaaaaaaa",
"output": "aa"
},
{
"input": "zzaa b",
"output": "zb"
},
{
"input": "ca cf",
"output": "cac"
},
{
"input": "abhi ia",
"output": "abhi"
},
{
"input": "aaaa aaaab",
"output": "aa"
},
{
"input": "aar raa",
"output": "aar"
},
{
"input": "harry hotter",
"output": "hah"
},
{
"input": "aaaaaaa a",
"output": "aa"
},
{
"input": "apple pie",
"output": "ap"
},
{
"input": "aaa aaa",
"output": "aa"
},
{
"input": "kabc buba",
"output": "kab"
},
{
"input": "asd ss",
"output": "as"
},
{
"input": "bbb b",
"output": "bb"
}
] | 1,514,393,459 | 959 | Python 3 | OK | TESTS | 64 | 62 | 5,632,000 | x=input().split()
a=x[0][0]
for i in range(1, len(x[0])):
if(x[0][i]<x[1][0]):
a+=x[0][i]
else:
break
a+=x[1][0]
print(a)
| Title: Generate Login
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The preferred way to generate user login in Polygon is to concatenate a prefix of the user's first name and a prefix of their last name, in that order. Each prefix must be non-empty, and any of the prefixes can be the full name. Typically there are multiple possible logins for each person.
You are given the first and the last name of a user. Return the alphabetically earliest login they can get (regardless of other potential Polygon users).
As a reminder, a prefix of a string *s* is its substring which occurs at the beginning of *s*: "a", "ab", "abc" etc. are prefixes of string "{abcdef}" but "b" and 'bc" are not. A string *a* is alphabetically earlier than a string *b*, if *a* is a prefix of *b*, or *a* and *b* coincide up to some position, and then *a* has a letter that is alphabetically earlier than the corresponding letter in *b*: "a" and "ab" are alphabetically earlier than "ac" but "b" and "ba" are alphabetically later than "ac".
Input Specification:
The input consists of a single line containing two space-separated strings: the first and the last names. Each character of each string is a lowercase English letter. The length of each string is between 1 and 10, inclusive.
Output Specification:
Output a single string — alphabetically earliest possible login formed from these names. The output should be given in lowercase as well.
Demo Input:
['harry potter\n', 'tom riddle\n']
Demo Output:
['hap\n', 'tomr\n']
Note:
none | ```python
x=input().split()
a=x[0][0]
for i in range(1, len(x[0])):
if(x[0][i]<x[1][0]):
a+=x[0][i]
else:
break
a+=x[1][0]
print(a)
``` | 3 | |
527 | A | Playing with Paper | PROGRAMMING | 1,100 | [
"implementation",
"math"
] | null | null | One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular *a* mm <=×<= *b* mm sheet of paper (*a*<=><=*b*). Usually the first step in making an origami is making a square piece of paper from the rectangular sheet by folding the sheet along the bisector of the right angle, and cutting the excess part.
After making a paper ship from the square piece, Vasya looked on the remaining (*a*<=-<=*b*) mm <=×<= *b* mm strip of paper. He got the idea to use this strip of paper in the same way to make an origami, and then use the remainder (if it exists) and so on. At the moment when he is left with a square piece of paper, he will make the last ship from it and stop.
Can you determine how many ships Vasya will make during the lesson? | The first line of the input contains two integers *a*, *b* (1<=≤<=*b*<=<<=*a*<=≤<=1012) — the sizes of the original sheet of paper. | Print a single integer — the number of ships that Vasya will make. | [
"2 1\n",
"10 7\n",
"1000000000000 1\n"
] | [
"2\n",
"6\n",
"1000000000000\n"
] | Pictures to the first and second sample test. | 500 | [
{
"input": "2 1",
"output": "2"
},
{
"input": "10 7",
"output": "6"
},
{
"input": "1000000000000 1",
"output": "1000000000000"
},
{
"input": "3 1",
"output": "3"
},
{
"input": "4 1",
"output": "4"
},
{
"input": "3 2",
"output": "3"
},
{
"input": "4 2",
"output": "2"
},
{
"input": "1000 700",
"output": "6"
},
{
"input": "959986566087 524054155168",
"output": "90"
},
{
"input": "4 3",
"output": "4"
},
{
"input": "7 6",
"output": "7"
},
{
"input": "1000 999",
"output": "1000"
},
{
"input": "1000 998",
"output": "500"
},
{
"input": "1000 997",
"output": "336"
},
{
"input": "42 1",
"output": "42"
},
{
"input": "1000 1",
"output": "1000"
},
{
"input": "8 5",
"output": "5"
},
{
"input": "13 8",
"output": "6"
},
{
"input": "987 610",
"output": "15"
},
{
"input": "442 42",
"output": "22"
},
{
"input": "754 466",
"output": "13"
},
{
"input": "1000000000000 999999999999",
"output": "1000000000000"
},
{
"input": "1000000000000 999999999998",
"output": "500000000000"
},
{
"input": "941 14",
"output": "74"
},
{
"input": "998 2",
"output": "499"
},
{
"input": "1000 42",
"output": "32"
},
{
"input": "1000 17",
"output": "66"
},
{
"input": "5 1",
"output": "5"
},
{
"input": "5 2",
"output": "4"
},
{
"input": "5 3",
"output": "4"
},
{
"input": "5 4",
"output": "5"
},
{
"input": "293 210",
"output": "17"
},
{
"input": "787878787878 424242424242",
"output": "8"
},
{
"input": "956722026041 591286729879",
"output": "58"
},
{
"input": "956722026041 365435296162",
"output": "58"
},
{
"input": "628625247282 464807889701",
"output": "102"
},
{
"input": "695928431619 424778620208",
"output": "167"
},
{
"input": "1000000000000 42",
"output": "23809523821"
},
{
"input": "987654345678 23",
"output": "42941493300"
},
{
"input": "10000000001 2",
"output": "5000000002"
},
{
"input": "1000000000000 2",
"output": "500000000000"
},
{
"input": "1000000000000 3",
"output": "333333333336"
},
{
"input": "100000000000 3",
"output": "33333333336"
},
{
"input": "100000000000 23",
"output": "4347826109"
},
{
"input": "999999999997 7",
"output": "142857142861"
},
{
"input": "8589934592 4294967296",
"output": "2"
}
] | 1,467,631,611 | 2,147,483,647 | Python 3 | OK | TESTS | 46 | 92 | 0 | # Problem 527A
# Playing with Paper
dim = list(map(int, input().split(" ")))
length = dim[0]
breadth = dim[1]
no_of_ships = 0
while (length - breadth) > 0:
no_of_ships += length // breadth
length -= (breadth * (length // breadth))
if (length > 0) and (length < breadth):
temp = breadth
breadth = length
length = temp
print(no_of_ships)
| Title: Playing with Paper
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular *a* mm <=×<= *b* mm sheet of paper (*a*<=><=*b*). Usually the first step in making an origami is making a square piece of paper from the rectangular sheet by folding the sheet along the bisector of the right angle, and cutting the excess part.
After making a paper ship from the square piece, Vasya looked on the remaining (*a*<=-<=*b*) mm <=×<= *b* mm strip of paper. He got the idea to use this strip of paper in the same way to make an origami, and then use the remainder (if it exists) and so on. At the moment when he is left with a square piece of paper, he will make the last ship from it and stop.
Can you determine how many ships Vasya will make during the lesson?
Input Specification:
The first line of the input contains two integers *a*, *b* (1<=≤<=*b*<=<<=*a*<=≤<=1012) — the sizes of the original sheet of paper.
Output Specification:
Print a single integer — the number of ships that Vasya will make.
Demo Input:
['2 1\n', '10 7\n', '1000000000000 1\n']
Demo Output:
['2\n', '6\n', '1000000000000\n']
Note:
Pictures to the first and second sample test. | ```python
# Problem 527A
# Playing with Paper
dim = list(map(int, input().split(" ")))
length = dim[0]
breadth = dim[1]
no_of_ships = 0
while (length - breadth) > 0:
no_of_ships += length // breadth
length -= (breadth * (length // breadth))
if (length > 0) and (length < breadth):
temp = breadth
breadth = length
length = temp
print(no_of_ships)
``` | 3 | |
61 | B | Hard Work | PROGRAMMING | 1,300 | [
"strings"
] | B. Hard Work | 2 | 256 | After the contest in comparing numbers, Shapur's teacher found out that he is a real genius and that no one could possibly do the calculations faster than him even using a super computer!
Some days before the contest, the teacher took a very simple-looking exam and all his *n* students took part in the exam. The teacher gave them 3 strings and asked them to concatenate them. Concatenating strings means to put them in some arbitrary order one after the other. For example from concatenating Alireza and Amir we can get to AlirezaAmir or AmirAlireza depending on the order of concatenation.
Unfortunately enough, the teacher forgot to ask students to concatenate their strings in a pre-defined order so each student did it the way he/she liked.
Now the teacher knows that Shapur is such a fast-calculating genius boy and asks him to correct the students' papers.
Shapur is not good at doing such a time-taking task. He rather likes to finish up with it as soon as possible and take his time to solve 3-SAT in polynomial time. Moreover, the teacher has given some advice that Shapur has to follow. Here's what the teacher said:
- As I expect you know, the strings I gave to my students (including you) contained only lowercase and uppercase Persian Mikhi-Script letters. These letters are too much like Latin letters, so to make your task much harder I converted all the initial strings and all of the students' answers to Latin. - As latin alphabet has much less characters than Mikhi-Script, I added three odd-looking characters to the answers, these include "-", ";" and "_". These characters are my own invention of course! And I call them Signs. - The length of all initial strings was less than or equal to 100 and the lengths of my students' answers are less than or equal to 600 - My son, not all students are genius as you are. It is quite possible that they make minor mistakes changing case of some characters. For example they may write ALiReZaAmIR instead of AlirezaAmir. Don't be picky and ignore these mistakes. - Those signs which I previously talked to you about are not important. You can ignore them, since many students are in the mood for adding extra signs or forgetting about a sign. So something like Iran;;-- is the same as --;IRAN - You should indicate for any of my students if his answer was right or wrong. Do this by writing "WA" for Wrong answer or "ACC" for a correct answer. - I should remind you that none of the strings (initial strings or answers) are empty. - Finally, do these as soon as possible. You have less than 2 hours to complete this. | The first three lines contain a string each. These are the initial strings. They consists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). All the initial strings have length from 1 to 100, inclusively.
In the fourth line there is a single integer *n* (0<=≤<=*n*<=≤<=1000), the number of students.
Next *n* lines contain a student's answer each. It is guaranteed that the answer meets what the teacher said. Each answer iconsists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). Length is from 1 to 600, inclusively. | For each student write in a different line. Print "WA" if his answer is wrong or "ACC" if his answer is OK. | [
"Iran_\nPersian;\nW_o;n;d;e;r;f;u;l;\n7\nWonderfulPersianIran\nwonderful_PersIAN_IRAN;;_\nWONDERFUL___IRAN__PERSIAN__;;\nIra__Persiann__Wonderful\nWonder;;fulPersian___;I;r;a;n;\n__________IranPersianWonderful__________\nPersianIran_is_Wonderful\n",
"Shapur;;\nis___\na_genius\n3\nShapur__a_is___geniUs\nis___shapur___a__Genius;\nShapur;;is;;a;;geni;;us;;\n"
] | [
"ACC\nACC\nACC\nWA\nACC\nACC\nWA\n",
"WA\nACC\nACC\n"
] | none | 1,000 | [
{
"input": "Iran_\nPersian;\nW_o;n;d;e;r;f;u;l;\n7\nWonderfulPersianIran\nwonderful_PersIAN_IRAN;;_\nWONDERFUL___IRAN__PERSIAN__;;\nIra__Persiann__Wonderful\nWonder;;fulPersian___;I;r;a;n;\n__________IranPersianWonderful__________\nPersianIran_is_Wonderful",
"output": "ACC\nACC\nACC\nWA\nACC\nACC\nWA"
},
{
"input": "Shapur;;\nis___\na_genius\n3\nShapur__a_is___geniUs\nis___shapur___a__Genius;\nShapur;;is;;a;;geni;;us;;",
"output": "WA\nACC\nACC"
},
{
"input": "rr\nrrx\nab\n1\nrabrrrx",
"output": "WA"
},
{
"input": "AB\nBC\nCD\n1\nABCDZZ",
"output": "WA"
},
{
"input": "aa\naaa\nz\n1\naazaaa",
"output": "ACC"
},
{
"input": "aa\naaa\nz\n1\naaazaa",
"output": "ACC"
},
{
"input": "as\nav\nax\n1\n-------",
"output": "WA"
},
{
"input": "a\nab\nb\n1\nabcd",
"output": "WA"
},
{
"input": "c\naba\ncc\n2\nccabac\nabcacc",
"output": "ACC\nWA"
},
{
"input": "ab\na\nb\n1\nabcd",
"output": "WA"
},
{
"input": "ACB\nTB\nAC\n1\nATBACBC",
"output": "WA"
},
{
"input": "cc\naba\ncc\n1\nccabaxx",
"output": "WA"
}
] | 1,567,981,338 | 1,038 | PyPy 3 | WRONG_ANSWER | TESTS | 32 | 187 | 3,276,800 | strings = [input().replace(';', '').replace('_', '').replace('-', '').lower() for _ in range(3)]
givens = list()
for _ in range(int(input())):
givens.append(input().replace(';', '').replace('_', '').replace('-', '').lower())
for given in givens:
if all([string in given for string in strings]) and len(''.join(strings)) == len(given):
print('ACC')
else:
print('WA')
| Title: Hard Work
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
After the contest in comparing numbers, Shapur's teacher found out that he is a real genius and that no one could possibly do the calculations faster than him even using a super computer!
Some days before the contest, the teacher took a very simple-looking exam and all his *n* students took part in the exam. The teacher gave them 3 strings and asked them to concatenate them. Concatenating strings means to put them in some arbitrary order one after the other. For example from concatenating Alireza and Amir we can get to AlirezaAmir or AmirAlireza depending on the order of concatenation.
Unfortunately enough, the teacher forgot to ask students to concatenate their strings in a pre-defined order so each student did it the way he/she liked.
Now the teacher knows that Shapur is such a fast-calculating genius boy and asks him to correct the students' papers.
Shapur is not good at doing such a time-taking task. He rather likes to finish up with it as soon as possible and take his time to solve 3-SAT in polynomial time. Moreover, the teacher has given some advice that Shapur has to follow. Here's what the teacher said:
- As I expect you know, the strings I gave to my students (including you) contained only lowercase and uppercase Persian Mikhi-Script letters. These letters are too much like Latin letters, so to make your task much harder I converted all the initial strings and all of the students' answers to Latin. - As latin alphabet has much less characters than Mikhi-Script, I added three odd-looking characters to the answers, these include "-", ";" and "_". These characters are my own invention of course! And I call them Signs. - The length of all initial strings was less than or equal to 100 and the lengths of my students' answers are less than or equal to 600 - My son, not all students are genius as you are. It is quite possible that they make minor mistakes changing case of some characters. For example they may write ALiReZaAmIR instead of AlirezaAmir. Don't be picky and ignore these mistakes. - Those signs which I previously talked to you about are not important. You can ignore them, since many students are in the mood for adding extra signs or forgetting about a sign. So something like Iran;;-- is the same as --;IRAN - You should indicate for any of my students if his answer was right or wrong. Do this by writing "WA" for Wrong answer or "ACC" for a correct answer. - I should remind you that none of the strings (initial strings or answers) are empty. - Finally, do these as soon as possible. You have less than 2 hours to complete this.
Input Specification:
The first three lines contain a string each. These are the initial strings. They consists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). All the initial strings have length from 1 to 100, inclusively.
In the fourth line there is a single integer *n* (0<=≤<=*n*<=≤<=1000), the number of students.
Next *n* lines contain a student's answer each. It is guaranteed that the answer meets what the teacher said. Each answer iconsists only of lowercase and uppercase Latin letters and signs ("-", ";" and "_"). Length is from 1 to 600, inclusively.
Output Specification:
For each student write in a different line. Print "WA" if his answer is wrong or "ACC" if his answer is OK.
Demo Input:
['Iran_\nPersian;\nW_o;n;d;e;r;f;u;l;\n7\nWonderfulPersianIran\nwonderful_PersIAN_IRAN;;_\nWONDERFUL___IRAN__PERSIAN__;;\nIra__Persiann__Wonderful\nWonder;;fulPersian___;I;r;a;n;\n__________IranPersianWonderful__________\nPersianIran_is_Wonderful\n', 'Shapur;;\nis___\na_genius\n3\nShapur__a_is___geniUs\nis___shapur___a__Genius;\nShapur;;is;;a;;geni;;us;;\n']
Demo Output:
['ACC\nACC\nACC\nWA\nACC\nACC\nWA\n', 'WA\nACC\nACC\n']
Note:
none | ```python
strings = [input().replace(';', '').replace('_', '').replace('-', '').lower() for _ in range(3)]
givens = list()
for _ in range(int(input())):
givens.append(input().replace(';', '').replace('_', '').replace('-', '').lower())
for given in givens:
if all([string in given for string in strings]) and len(''.join(strings)) == len(given):
print('ACC')
else:
print('WA')
``` | 0 |
522 | A | Reposts | PROGRAMMING | 1,200 | [
"*special",
"dfs and similar",
"dp",
"graphs",
"trees"
] | null | null | One day Polycarp published a funny picture in a social network making a poll about the color of his handle. Many of his friends started reposting Polycarp's joke to their news feed. Some of them reposted the reposts and so on.
These events are given as a sequence of strings "name1 reposted name2", where name1 is the name of the person who reposted the joke, and name2 is the name of the person from whose news feed the joke was reposted. It is guaranteed that for each string "name1 reposted name2" user "name1" didn't have the joke in his feed yet, and "name2" already had it in his feed by the moment of repost. Polycarp was registered as "Polycarp" and initially the joke was only in his feed.
Polycarp measures the popularity of the joke as the length of the largest repost chain. Print the popularity of Polycarp's joke. | The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=200) — the number of reposts. Next follow the reposts in the order they were made. Each of them is written on a single line and looks as "name1 reposted name2". All the names in the input consist of lowercase or uppercase English letters and/or digits and have lengths from 2 to 24 characters, inclusive.
We know that the user names are case-insensitive, that is, two names that only differ in the letter case correspond to the same social network user. | Print a single integer — the maximum length of a repost chain. | [
"5\ntourist reposted Polycarp\nPetr reposted Tourist\nWJMZBMR reposted Petr\nsdya reposted wjmzbmr\nvepifanov reposted sdya\n",
"6\nMike reposted Polycarp\nMax reposted Polycarp\nEveryOne reposted Polycarp\n111 reposted Polycarp\nVkCup reposted Polycarp\nCodeforces reposted Polycarp\n",
"1\nSoMeStRaNgEgUe reposted PoLyCaRp\n"
] | [
"6\n",
"2\n",
"2\n"
] | none | 500 | [
{
"input": "5\ntourist reposted Polycarp\nPetr reposted Tourist\nWJMZBMR reposted Petr\nsdya reposted wjmzbmr\nvepifanov reposted sdya",
"output": "6"
},
{
"input": "6\nMike reposted Polycarp\nMax reposted Polycarp\nEveryOne reposted Polycarp\n111 reposted Polycarp\nVkCup reposted Polycarp\nCodeforces reposted Polycarp",
"output": "2"
},
{
"input": "1\nSoMeStRaNgEgUe reposted PoLyCaRp",
"output": "2"
},
{
"input": "1\niuNtwVf reposted POlYcarP",
"output": "2"
},
{
"input": "10\ncs reposted poLYCaRp\nAFIkDrY7Of4V7Mq reposted CS\nsoBiwyN7KOvoFUfbhux reposted aFikDry7Of4v7MQ\nvb6LbwA reposted sObIWYN7KOvoFufBHUx\nDtWKIcVwIHgj4Rcv reposted vb6lbwa\nkt reposted DTwKicvwihgJ4rCV\n75K reposted kT\njKzyxx1 reposted 75K\nuoS reposted jkZyXX1\npZJskHTCIqE3YyZ5ME reposted uoS",
"output": "11"
},
{
"input": "10\nvxrUpCXvx8Isq reposted pOLYcaRP\nICb1 reposted vXRUpCxvX8ISq\nJFMt4b8jZE7iF2m8by7y2 reposted Icb1\nqkG6ZkMIf9QRrBFQU reposted ICb1\nnawsNfcR2palIMnmKZ reposted pOlYcaRP\nKksyH reposted jFMT4b8JzE7If2M8by7y2\nwJtWwQS5FvzN0h8CxrYyL reposted NawsNfcR2paLIMnmKz\nDpBcBPYAcTXEdhldI6tPl reposted NaWSnFCr2pALiMnmkZ\nlEnwTVnlwdQg2vaIRQry reposted kKSYh\nQUVFgwllaWO reposted Wjtwwqs5FVzN0H8cxRyyl",
"output": "6"
},
{
"input": "10\nkkuLGEiHv reposted POLYcArp\n3oX1AoUqyw1eR3nCADY9hLwd reposted kkuLGeIHV\nwf97dqq5bx1dPIchCoT reposted 3OX1AOuQYW1eR3ncAdY9hLwD\nWANr8h reposted Wf97dQQ5bx1dpIcHcoT\n3Fb736lkljZK2LtSbfL reposted wANR8h\n6nq9xLOn reposted 3fB736lKlJZk2LtSbFL\nWL reposted 3Fb736lKLjZk2LTSbfl\ndvxn4Xtc6SBcvKf1 reposted wF97DQq5bX1dPiChCOt\nMCcPLIMISqxDzrj reposted 6nQ9XLOn\nxsQL4Z2Iu reposted MCcpLiMiSqxdzrj",
"output": "9"
},
{
"input": "10\nsMA4 reposted pOLyCARP\nlq3 reposted pOlycARp\nEa16LSFTQxLJnE reposted polYcARp\nkvZVZhJwXcWsnC7NA1DV2WvS reposted polYCArp\nEYqqlrjRwddI reposted pOlyCArP\nsPqQCA67Y6PBBbcaV3EhooO reposted ea16LSFTqxLJne\njjPnneZdF6WLZ3v reposted Ea16LSFTqxLjNe\nWEoi6UpnfBUx79 reposted ea16LSFtqXljNe\nqi4yra reposted eYqqlRJrWDDI\ncw7E1UCSUD reposted eYqqLRJRwDdI",
"output": "3"
}
] | 1,627,747,489 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 46 | 6,963,200 | n = int(input())
a = []
for j in range(0, n):
a.append(input())
if n != 1:
s = 1
pol = ''
q = 0
for i in range(0,n):
c, b, d = a[i].split()
if d == 'Polycarp' or pol.find(str(c)) != 0 or pol.find(capitalize(str(c))) != 0:
if d == 'Polycarp' and q < 1:
pol += c + ' '
s += 1
q += 1
elif d == 'Polycarp' and q >= 1:
pol = ''
s = 2
else:
pol += c + ' '
s += 1
else:
pol = ''
s = 1
print(s)
else:
print(2)
| Title: Reposts
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Polycarp published a funny picture in a social network making a poll about the color of his handle. Many of his friends started reposting Polycarp's joke to their news feed. Some of them reposted the reposts and so on.
These events are given as a sequence of strings "name1 reposted name2", where name1 is the name of the person who reposted the joke, and name2 is the name of the person from whose news feed the joke was reposted. It is guaranteed that for each string "name1 reposted name2" user "name1" didn't have the joke in his feed yet, and "name2" already had it in his feed by the moment of repost. Polycarp was registered as "Polycarp" and initially the joke was only in his feed.
Polycarp measures the popularity of the joke as the length of the largest repost chain. Print the popularity of Polycarp's joke.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=200) — the number of reposts. Next follow the reposts in the order they were made. Each of them is written on a single line and looks as "name1 reposted name2". All the names in the input consist of lowercase or uppercase English letters and/or digits and have lengths from 2 to 24 characters, inclusive.
We know that the user names are case-insensitive, that is, two names that only differ in the letter case correspond to the same social network user.
Output Specification:
Print a single integer — the maximum length of a repost chain.
Demo Input:
['5\ntourist reposted Polycarp\nPetr reposted Tourist\nWJMZBMR reposted Petr\nsdya reposted wjmzbmr\nvepifanov reposted sdya\n', '6\nMike reposted Polycarp\nMax reposted Polycarp\nEveryOne reposted Polycarp\n111 reposted Polycarp\nVkCup reposted Polycarp\nCodeforces reposted Polycarp\n', '1\nSoMeStRaNgEgUe reposted PoLyCaRp\n']
Demo Output:
['6\n', '2\n', '2\n']
Note:
none | ```python
n = int(input())
a = []
for j in range(0, n):
a.append(input())
if n != 1:
s = 1
pol = ''
q = 0
for i in range(0,n):
c, b, d = a[i].split()
if d == 'Polycarp' or pol.find(str(c)) != 0 or pol.find(capitalize(str(c))) != 0:
if d == 'Polycarp' and q < 1:
pol += c + ' '
s += 1
q += 1
elif d == 'Polycarp' and q >= 1:
pol = ''
s = 2
else:
pol += c + ' '
s += 1
else:
pol = ''
s = 1
print(s)
else:
print(2)
``` | 0 | |
49 | A | Sleuth | PROGRAMMING | 800 | [
"implementation"
] | A. Sleuth | 2 | 256 | Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z | The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter. | Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters. | [
"Is it a melon?\n",
"Is it an apple?\n",
"Is it a banana ?\n",
"Is it an apple and a banana simultaneouSLY?\n"
] | [
"NO\n",
"YES\n",
"YES\n",
"YES\n"
] | none | 500 | [
{
"input": "Is it a melon?",
"output": "NO"
},
{
"input": "Is it an apple?",
"output": "YES"
},
{
"input": " Is it a banana ?",
"output": "YES"
},
{
"input": "Is it an apple and a banana simultaneouSLY?",
"output": "YES"
},
{
"input": "oHtSbDwzHb?",
"output": "NO"
},
{
"input": "sZecYdUvZHrXx?",
"output": "NO"
},
{
"input": "uMtXK?",
"output": "NO"
},
{
"input": "U?",
"output": "YES"
},
{
"input": "aqFDkCUKeHMyvZFcAyWlMUSQTFomtaWjoKLVyxLCw vcufPBFbaljOuHWiDCROYTcmbgzbaqHXKPOYEbuEtRqqoxBbOETCsQzhw?",
"output": "NO"
},
{
"input": "dJcNqQiFXzcbsj fItCpBLyXOnrSBPebwyFHlxUJHqCUzzCmcAvMiKL NunwOXnKeIxUZmBVwiCUfPkjRAkTPbkYCmwRRnDSLaz?",
"output": "NO"
},
{
"input": "gxzXbdcAQMuFKuuiPohtMgeypr wpDIoDSyOYTdvylcg SoEBZjnMHHYZGEqKgCgBeTbyTwyGuPZxkxsnSczotBdYyfcQsOVDVC?",
"output": "NO"
},
{
"input": "FQXBisXaJFMiHFQlXjixBDMaQuIbyqSBKGsBfTmBKCjszlGVZxEOqYYqRTUkGpSDDAoOXyXcQbHcPaegeOUBNeSD JiKOdECPOF?",
"output": "NO"
},
{
"input": "YhCuZnrWUBEed?",
"output": "NO"
},
{
"input": "hh?",
"output": "NO"
},
{
"input": "whU?",
"output": "YES"
},
{
"input": "fgwg?",
"output": "NO"
},
{
"input": "GlEmEPKrYcOnBNJUIFjszWUyVdvWw DGDjoCMtRJUburkPToCyDrOtMr?",
"output": "NO"
},
{
"input": "n?",
"output": "NO"
},
{
"input": "BueDOlxgzeNlxrzRrMbKiQdmGujEKmGxclvaPpTuHmTqBp?",
"output": "NO"
},
{
"input": "iehvZNQXDGCuVmJPOEysLyUryTdfaIxIuTzTadDbqRQGoCLXkxnyfWSGoLXebNnQQNTqAQJebbyYvHOfpUnXeWdjx?",
"output": "NO"
},
{
"input": " J ?",
"output": "NO"
},
{
"input": " j ?",
"output": "NO"
},
{
"input": " o ?",
"output": "YES"
},
{
"input": " T ?",
"output": "NO"
},
{
"input": " q ?",
"output": "NO"
},
{
"input": " j ?",
"output": "NO"
},
{
"input": " c ?",
"output": "NO"
},
{
"input": " B ?",
"output": "NO"
},
{
"input": "LuhxDHVwMPTtUIUMIQTuQETgXCOQPsfdFlyHvpfOVedjUTpGLAZGOHloIjJJtOLAlHPivzA?",
"output": "YES"
},
{
"input": "wmztmzFfwbGyOmNHENUFMTsFEMWYA?",
"output": "YES"
},
{
"input": "wGsfZCSwN PEUhNUrLfABrxA?",
"output": "YES"
},
{
"input": "mCDHENXjYbgMdBimAdPnewaHfpGWowjWrVAdvWczjw iDcUbyzMsmsnwbviiKiAyGVA?",
"output": "YES"
},
{
"input": "ARIWnwqFqxsQXsXXzHqvFjxOCttAGPUzDtWzsenPYdNXuFOIUGYZsLLK IaoxiyjBBRThoelwdPTkuCQfcBLUEJpCPIrVZlvUWA?",
"output": "YES"
},
{
"input": " PslvVpgpN BXkMFBEVXsyZFIQbBEFxGkYTeXKrOdcmhbiTUatYRUoYAayrchqbksswIlfIjerZPqptvCGnMUhyrQSvwltRhFzA?",
"output": "YES"
},
{
"input": "HpBkttwSjBXDmyleGiRWNUMPaAIE uzTrp KJDzaUiCdsMYOoWKHoUhWUoecCPmACymMUUbGav UMRpCytPETwNFAObZJA?",
"output": "YES"
}
] | 1,544,094,669 | 2,147,483,647 | Python 3 | OK | TESTS | 35 | 248 | 0 | s = input().upper()
gl = 'AEIOUY'
sogl = 'BCDFGHJKLMNPQRSTVWXZ'
for i in s[::-1]:
if i in gl:
print('YES')
break
elif i in sogl:
print('NO')
break
| Title: Sleuth
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya plays the sleuth with his friends. The rules of the game are as follows: those who play for the first time, that is Vasya is the sleuth, he should investigate a "crime" and find out what is happening. He can ask any questions whatsoever that can be answered with "Yes" or "No". All the rest agree beforehand to answer the questions like that: if the question’s last letter is a vowel, they answer "Yes" and if the last letter is a consonant, they answer "No". Of course, the sleuth knows nothing about it and his task is to understand that.
Unfortunately, Vasya is not very smart. After 5 hours of endless stupid questions everybody except Vasya got bored. That’s why Vasya’s friends ask you to write a program that would give answers instead of them.
The English alphabet vowels are: A, E, I, O, U, Y
The English alphabet consonants are: B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z
Input Specification:
The single line contains a question represented by a non-empty line consisting of large and small Latin letters, spaces and a question mark. The line length does not exceed 100. It is guaranteed that the question mark occurs exactly once in the line — as the last symbol and that the line contains at least one letter.
Output Specification:
Print answer for the question in a single line: YES if the answer is "Yes", NO if the answer is "No".
Remember that in the reply to the question the last letter, not the last character counts. I. e. the spaces and the question mark do not count as letters.
Demo Input:
['Is it a melon?\n', 'Is it an apple?\n', 'Is it a banana ?\n', 'Is it an apple and a banana simultaneouSLY?\n']
Demo Output:
['NO\n', 'YES\n', 'YES\n', 'YES\n']
Note:
none | ```python
s = input().upper()
gl = 'AEIOUY'
sogl = 'BCDFGHJKLMNPQRSTVWXZ'
for i in s[::-1]:
if i in gl:
print('YES')
break
elif i in sogl:
print('NO')
break
``` | 3.938 |
832 | B | Petya and Exam | PROGRAMMING | 1,600 | [
"implementation",
"strings"
] | null | null | It's hard times now. Today Petya needs to score 100 points on Informatics exam. The tasks seem easy to Petya, but he thinks he lacks time to finish them all, so he asks you to help with one..
There is a glob pattern in the statements (a string consisting of lowercase English letters, characters "?" and "*"). It is known that character "*" occurs no more than once in the pattern.
Also, *n* query strings are given, it is required to determine for each of them if the pattern matches it or not.
Everything seemed easy to Petya, but then he discovered that the special pattern characters differ from their usual meaning.
A pattern matches a string if it is possible to replace each character "?" with one good lowercase English letter, and the character "*" (if there is one) with any, including empty, string of bad lowercase English letters, so that the resulting string is the same as the given string.
The good letters are given to Petya. All the others are bad. | The first line contains a string with length from 1 to 26 consisting of distinct lowercase English letters. These letters are good letters, all the others are bad.
The second line contains the pattern — a string *s* of lowercase English letters, characters "?" and "*" (1<=≤<=|*s*|<=≤<=105). It is guaranteed that character "*" occurs in *s* no more than once.
The third line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of query strings.
*n* lines follow, each of them contains single non-empty string consisting of lowercase English letters — a query string.
It is guaranteed that the total length of all query strings is not greater than 105. | Print *n* lines: in the *i*-th of them print "YES" if the pattern matches the *i*-th query string, and "NO" otherwise.
You can choose the case (lower or upper) for each letter arbitrary. | [
"ab\na?a\n2\naaa\naab\n",
"abc\na?a?a*\n4\nabacaba\nabaca\napapa\naaaaax\n"
] | [
"YES\nNO\n",
"NO\nYES\nNO\nYES\n"
] | In the first example we can replace "?" with good letters "a" and "b", so we can see that the answer for the first query is "YES", and the answer for the second query is "NO", because we can't match the third letter.
Explanation of the second example.
- The first query: "NO", because character "*" can be replaced with a string of bad letters only, but the only way to match the query string is to replace it with the string "ba", in which both letters are good. - The second query: "YES", because characters "?" can be replaced with corresponding good letters, and character "*" can be replaced with empty string, and the strings will coincide. - The third query: "NO", because characters "?" can't be replaced with bad letters. - The fourth query: "YES", because characters "?" can be replaced with good letters "a", and character "*" can be replaced with a string of bad letters "x". | 1,000 | [
{
"input": "ab\na?a\n2\naaa\naab",
"output": "YES\nNO"
},
{
"input": "abc\na?a?a*\n4\nabacaba\nabaca\napapa\naaaaax",
"output": "NO\nYES\nNO\nYES"
},
{
"input": "s\nc*?cb\n26\nbbaa\nb\ncc\ncbaab\nacacc\nca\na\nc\ncb\nabb\nba\nb\nba\ncac\nccccb\nccb\nbbbc\nabbcb\na\nbc\nc\na\nabb\nca\ncacb\nac",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "o\n*\n28\nbac\nbcc\ncbcb\ncaabc\ncb\nacab\ncbccb\ncbccc\nc\nbbaa\ncaaaa\nbbc\nba\nc\ncacbc\ncbab\naa\nac\nacc\na\nac\nbac\naaac\nba\nabbbb\nbbcc\nbaacb\naabaa",
"output": "YES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES\nYES"
},
{
"input": "u\n*b??c\n23\na\nbcbcc\nacb\na\nbacaa\nbb\nb\nbcba\ncbbcc\nb\nabbb\nbcacb\nabcb\ncbca\nb\ncba\ncabcb\nbc\ncc\naaacc\nccac\ncc\nccbcb",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "g\nc?*\n58\nb\ncaac\nbbc\nabb\ncaccc\ncb\naba\nbcaa\ncca\ncbbcb\ncac\nbbaca\nbcba\nbba\nabbab\nccc\nc\nbcb\naac\nbcbbc\nbca\nc\ncbb\nccabb\naaccc\nccaa\nc\nc\nbcca\naa\nccb\ncb\ncbcb\ncc\nab\ncccc\nbbbab\nbab\na\nc\ncbba\nbbacb\naa\nb\nbaab\nacabb\nbcbab\ncbbcb\nbc\ncccba\naa\ncccca\ncacc\naacbb\na\nc\nab\nccca",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "g\nbc*a\n40\nbabac\nccbb\ncacbc\nc\na\naba\nbc\na\nba\nbbcca\nccbac\na\nc\nbabc\ncccbc\nab\nabca\nccb\nacbbb\nb\nbbac\naa\nb\nca\nbc\naaba\nbaaaa\nbcc\nab\na\naba\nb\nc\nba\nbc\nca\nbb\nc\nc\nca",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "g\n*cc?\n93\nac\ncaab\nacaca\ncccc\nbcc\nbab\nbc\nc\nc\nbbaa\nb\ncc\ncb\naa\nabcbb\nbccc\nc\ncbcbc\nac\nca\nbcba\nbb\nbab\nba\nb\nbbba\nbabbc\nbacab\nbc\na\ncbccc\nbbaac\ncbab\ncab\ncc\ncbbcb\nc\nc\ncbaa\nca\nbacab\nc\nbcac\nbbbc\nc\nac\nccab\nccccb\ncccab\nc\nacb\nac\nbccba\nca\nbbbbc\naaca\naa\na\nbabac\nbb\nc\ncac\naca\naacb\naacbb\na\nacaab\ncbb\nbcc\ncb\nbcbaa\ncca\nb\nbaac\nbcca\nc\ncbb\nac\nc\naccc\naac\nbcbc\nabc\nbacab\nb\na\na\nbbacc\ncb\na\nccac\nb\nbbc",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "c\n*\n83\nbbc\ncacb\nbcbc\naca\nba\nc\nccac\nab\nab\nbacba\nbb\nc\nbcc\nc\ncbc\ncbbb\nac\nb\nacbcb\nbccc\ncccb\nb\na\nca\nc\nccaa\naa\ncacb\nccc\na\nccc\nababb\nbab\ncaa\nbaa\na\ncc\ncbbbc\naaaa\nabbab\naabac\nbcbab\nbcb\nacaa\nbcb\na\ncca\na\nbacc\nacacb\nc\nc\ncba\nbcaca\na\ncaac\na\nb\na\nccc\naabca\nbbab\nb\nac\nbabc\nc\nac\nba\nbbcb\nc\naaab\ncab\nacb\nbba\nbbcba\nc\na\naccbb\naaccc\nac\nbaa\nbaabb\nabca",
"output": "NO\nNO\nNO\nNO\nYES\nNO\nNO\nYES\nYES\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nYES\nYES\nNO\nNO\nNO\nYES\nNO\nNO\nYES\nNO\nYES\nYES\nNO\nYES\nYES\nNO\nNO\nYES\nYES\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nYES\nYES\nYES\nNO\nNO\nYES\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nYES\nNO\nNO\nYES\nNO\nNO\nYES\nNO\nNO\nNO\nYES\nYES\nNO"
},
{
"input": "s\n*cb\n70\nab\nccb\naaab\nb\nab\ncba\na\nbbaca\nac\nccacb\nbaabb\naaab\nccca\ncb\nba\nbccac\nc\ncc\ncbbbb\ncab\nabbb\ncbb\naabc\ncac\nacb\na\nc\nc\ncbbbb\nbaaca\ncbcc\nbc\naa\nabcb\nacbbc\nbaaa\naa\ncc\ncc\nb\nb\nbcba\ncbacc\nbcb\ncaabc\nacaac\ncb\ncba\ncbaaa\nbcaaa\naccbb\naccac\nca\nacaa\ncc\nc\nb\nbac\nb\nbab\nb\ncca\naacc\nacb\nccc\nbc\nb\naab\naaca\naac",
"output": "NO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "k\n*b\n70\ncbc\nb\ncca\nacbc\nca\nab\nc\nbbb\nbaa\nbbb\nac\nbaacc\nbaabc\naac\na\nba\nb\nc\nc\nba\ncacbb\nabb\nbc\nabcb\nca\na\nbbbbb\ncca\nccacc\ncbaca\nba\ncbcca\ncb\nc\nbbbba\ncca\nabaac\na\nac\nc\nccbc\nbcac\nbcb\na\nc\nabbca\nbaacb\ncc\nacba\nc\nbcc\ncbba\nccba\na\na\ncbb\ncba\nb\naaaac\ncb\nbaacb\nab\nc\ncbbcb\nbab\nac\nca\nc\nac\nb",
"output": "NO\nYES\nNO\nNO\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nYES\nYES\nNO\nYES\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nYES\nNO\nYES\nYES\nYES\nNO\nYES\nYES\nNO\nNO\nNO\nNO\nYES"
},
{
"input": "l\na*\n40\nacbb\naba\nb\naab\nbb\nbbab\ncaba\naab\naaab\nacac\nacbaa\nbca\nac\nbb\na\nba\naaa\nbc\nbba\ncca\naacab\na\nc\nca\naacaa\nbaac\nbb\nc\nba\nc\nbab\nb\na\ncabaa\nccacc\ncbbab\nbaaca\ncabb\naaccc\nbcbac",
"output": "YES\nYES\nNO\nYES\nNO\nNO\nNO\nYES\nYES\nYES\nYES\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nNO\nNO\nYES\nYES\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nYES\nNO"
},
{
"input": "u\ncba*\n26\ncaa\ncccb\nbc\nbacb\nca\nccaaa\nb\naaca\nba\ncacc\ncccac\nabba\nbabc\na\nac\nca\nbbba\na\naa\naaabb\nb\nc\nbba\nbbba\nacaa\nba",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "cba\n?*cbc\n88\ncccca\ncbc\nb\nbcb\naaa\ncaac\nbacb\nacbb\na\nab\ncbcca\nbccc\nabcc\naca\nba\nbbac\nacc\ncba\nbcba\nbc\naa\nab\ncaba\ncccab\ncba\ncbcc\nba\ncacbb\nabcc\na\nc\nbac\nccaba\nb\nac\nbbb\nac\nccaca\na\nba\nacbcc\nbbc\nacbc\nbbabc\nccbb\nb\nacaa\na\nba\nacb\na\nab\naa\nbbbb\naabb\nbcbc\nb\nca\nb\nccab\nab\nc\nb\naabab\nc\ncbbbc\nacbbb\nbacaa\nbcccc\ncbac\nc\nac\nb\nca\ncbb\nccbc\nc\nc\nbcb\nc\nbaaba\nc\nbac\nb\nba\ncb\ncc\nbaaca",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "a\naa\n1\naaa",
"output": "NO"
},
{
"input": "a\naaa\n1\naaaa",
"output": "NO"
},
{
"input": "a\naa*aa\n1\naaa",
"output": "NO"
},
{
"input": "a\nbb*bb\n1\nbbbbbbbbbbbbbbbb",
"output": "YES"
},
{
"input": "a\na*\n1\nabbbbbbb",
"output": "YES"
},
{
"input": "a\na?a\n1\naaab",
"output": "NO"
},
{
"input": "xy\ncab*aba\n1\ncaba",
"output": "NO"
},
{
"input": "a\n*\n4\nb\na\nab\nba",
"output": "YES\nNO\nNO\nNO"
},
{
"input": "abc\na?a?*a\n3\nababxa\nababca\nababa",
"output": "YES\nNO\nYES"
},
{
"input": "abc\n??a*df?c\n6\nabadfcc\naaaadfac\nbbagthfac\nacadddfac\ndaagdffc\naaaadfcc",
"output": "YES\nNO\nNO\nYES\nNO\nNO"
},
{
"input": "abc\nabc*a\n1\nabckka",
"output": "YES"
},
{
"input": "b\n*a\n1\naba",
"output": "NO"
},
{
"input": "a\nabc*g\n1\nabcdefg",
"output": "YES"
},
{
"input": "a\nab\n1\na",
"output": "NO"
},
{
"input": "abcdefghijklmnopqrstuvwxyz\n*a\n1\na",
"output": "YES"
},
{
"input": "as\naba*aba\n1\naba",
"output": "NO"
},
{
"input": "ab\naweerrtab\n4\naw\naweerrtabwqeqrw\naweerrtabxcvxcbcxbdsfdsfewrewrqweq\naweerrtabaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "NO\nNO\nNO\nNO"
},
{
"input": "a\na\n1\nab",
"output": "NO"
},
{
"input": "a\na*b\n1\nabb",
"output": "YES"
},
{
"input": "a\nb*a\n1\nbbadd",
"output": "NO"
},
{
"input": "a\naaaa\n1\naaa",
"output": "NO"
},
{
"input": "z\n*abcd\n1\nggggggg",
"output": "NO"
},
{
"input": "abc\n*??\n1\nqqqqqqqqab",
"output": "YES"
},
{
"input": "b\naa\n1\na",
"output": "NO"
},
{
"input": "ab\na*pa\n1\nappppa",
"output": "YES"
},
{
"input": "a\nbbb\n1\nbbbbb",
"output": "NO"
},
{
"input": "ab\nabcd?\n1\nabcd",
"output": "NO"
},
{
"input": "c\na\n1\nab",
"output": "NO"
}
] | 1,633,625,929 | 2,147,483,647 | PyPy 3 | WRONG_ANSWER | TESTS | 11 | 358 | 26,112,000 | def coolf(b):
coolp = -1
for i in range(n):
coolp+=1
if(coolp >= len(b)):
if(a[i] == '*' and i + 1 >= n):
print("YES")
else:
print('NO')
return
if(a[i] == b[coolp]):
continue
elif(a[i] == '?' and b[coolp] in gud):
continue
elif(a[i] == '*'):
while(coolp < len(b) and not b[coolp] in gud):
coolp+=1
if(coolp + 1 >= len(b) and i + 1 < n):
print('NO')
return
continue
else:
print("NO")
return
if(coolp + 1 < len(b)):
print("NO")
else:
print("YES")
return
gud = input()
a = input()
n = len(a)
for _ in range(int(input())):
#b = input()
coolf(input()) | Title: Petya and Exam
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's hard times now. Today Petya needs to score 100 points on Informatics exam. The tasks seem easy to Petya, but he thinks he lacks time to finish them all, so he asks you to help with one..
There is a glob pattern in the statements (a string consisting of lowercase English letters, characters "?" and "*"). It is known that character "*" occurs no more than once in the pattern.
Also, *n* query strings are given, it is required to determine for each of them if the pattern matches it or not.
Everything seemed easy to Petya, but then he discovered that the special pattern characters differ from their usual meaning.
A pattern matches a string if it is possible to replace each character "?" with one good lowercase English letter, and the character "*" (if there is one) with any, including empty, string of bad lowercase English letters, so that the resulting string is the same as the given string.
The good letters are given to Petya. All the others are bad.
Input Specification:
The first line contains a string with length from 1 to 26 consisting of distinct lowercase English letters. These letters are good letters, all the others are bad.
The second line contains the pattern — a string *s* of lowercase English letters, characters "?" and "*" (1<=≤<=|*s*|<=≤<=105). It is guaranteed that character "*" occurs in *s* no more than once.
The third line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of query strings.
*n* lines follow, each of them contains single non-empty string consisting of lowercase English letters — a query string.
It is guaranteed that the total length of all query strings is not greater than 105.
Output Specification:
Print *n* lines: in the *i*-th of them print "YES" if the pattern matches the *i*-th query string, and "NO" otherwise.
You can choose the case (lower or upper) for each letter arbitrary.
Demo Input:
['ab\na?a\n2\naaa\naab\n', 'abc\na?a?a*\n4\nabacaba\nabaca\napapa\naaaaax\n']
Demo Output:
['YES\nNO\n', 'NO\nYES\nNO\nYES\n']
Note:
In the first example we can replace "?" with good letters "a" and "b", so we can see that the answer for the first query is "YES", and the answer for the second query is "NO", because we can't match the third letter.
Explanation of the second example.
- The first query: "NO", because character "*" can be replaced with a string of bad letters only, but the only way to match the query string is to replace it with the string "ba", in which both letters are good. - The second query: "YES", because characters "?" can be replaced with corresponding good letters, and character "*" can be replaced with empty string, and the strings will coincide. - The third query: "NO", because characters "?" can't be replaced with bad letters. - The fourth query: "YES", because characters "?" can be replaced with good letters "a", and character "*" can be replaced with a string of bad letters "x". | ```python
def coolf(b):
coolp = -1
for i in range(n):
coolp+=1
if(coolp >= len(b)):
if(a[i] == '*' and i + 1 >= n):
print("YES")
else:
print('NO')
return
if(a[i] == b[coolp]):
continue
elif(a[i] == '?' and b[coolp] in gud):
continue
elif(a[i] == '*'):
while(coolp < len(b) and not b[coolp] in gud):
coolp+=1
if(coolp + 1 >= len(b) and i + 1 < n):
print('NO')
return
continue
else:
print("NO")
return
if(coolp + 1 < len(b)):
print("NO")
else:
print("YES")
return
gud = input()
a = input()
n = len(a)
for _ in range(int(input())):
#b = input()
coolf(input())
``` | 0 | |
681 | C | Heap Operations | PROGRAMMING | 1,600 | [
"constructive algorithms",
"data structures",
"greedy"
] | null | null | Petya has recently learned data structure named "Binary heap".
The heap he is now operating with allows the following operations:
- put the given number into the heap; - get the value of the minimum element in the heap; - extract the minimum element from the heap;
Thus, at any moment of time the heap contains several integers (possibly none), some of them might be equal.
In order to better learn this data structure Petya took an empty heap and applied some operations above to it. Also, he carefully wrote down all the operations and their results to his event log, following the format:
- insert *x* — put the element with value *x* in the heap; - getMin *x* — the value of the minimum element contained in the heap was equal to *x*; - removeMin — the minimum element was extracted from the heap (only one instance, if there were many).
All the operations were correct, i.e. there was at least one element in the heap each time getMin or removeMin operations were applied.
While Petya was away for a lunch, his little brother Vova came to the room, took away some of the pages from Petya's log and used them to make paper boats.
Now Vova is worried, if he made Petya's sequence of operations inconsistent. For example, if one apply operations one-by-one in the order they are written in the event log, results of getMin operations might differ from the results recorded by Petya, and some of getMin or removeMin operations may be incorrect, as the heap is empty at the moment they are applied.
Now Vova wants to add some new operation records to the event log in order to make the resulting sequence of operations correct. That is, the result of each getMin operation is equal to the result in the record, and the heap is non-empty when getMin ad removeMin are applied. Vova wants to complete this as fast as possible, as the Petya may get back at any moment. He asks you to add the least possible number of operation records to the current log. Note that arbitrary number of operations may be added at the beginning, between any two other operations, or at the end of the log. | The first line of the input contains the only integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of the records left in Petya's journal.
Each of the following *n* lines describe the records in the current log in the order they are applied. Format described in the statement is used. All numbers in the input are integers not exceeding 109 by their absolute value. | The first line of the output should contain a single integer *m* — the minimum possible number of records in the modified sequence of operations.
Next *m* lines should contain the corrected sequence of records following the format of the input (described in the statement), one per line and in the order they are applied. All the numbers in the output should be integers not exceeding 109 by their absolute value.
Note that the input sequence of operations must be the subsequence of the output sequence.
It's guaranteed that there exists the correct answer consisting of no more than 1<=000<=000 operations. | [
"2\ninsert 3\ngetMin 4\n",
"4\ninsert 1\ninsert 1\nremoveMin\ngetMin 2\n"
] | [
"4\ninsert 3\nremoveMin\ninsert 4\ngetMin 4\n",
"6\ninsert 1\ninsert 1\nremoveMin\nremoveMin\ninsert 2\ngetMin 2\n"
] | In the first sample, after number 3 is inserted into the heap, the minimum number is 3. To make the result of the first getMin equal to 4 one should firstly remove number 3 from the heap and then add number 4 into the heap.
In the second sample case number 1 is inserted two times, so should be similarly removed twice. | 1,500 | [
{
"input": "2\ninsert 3\ngetMin 4",
"output": "4\ninsert 3\nremoveMin\ninsert 4\ngetMin 4"
},
{
"input": "4\ninsert 1\ninsert 1\nremoveMin\ngetMin 2",
"output": "6\ninsert 1\ninsert 1\nremoveMin\nremoveMin\ninsert 2\ngetMin 2"
},
{
"input": "1\ninsert 1",
"output": "1\ninsert 1"
},
{
"input": "1\ngetMin 31",
"output": "2\ninsert 31\ngetMin 31"
},
{
"input": "1\nremoveMin",
"output": "2\ninsert 0\nremoveMin"
},
{
"input": "2\ninsert 2\ngetMin 2",
"output": "2\ninsert 2\ngetMin 2"
},
{
"input": "2\ninsert 31\nremoveMin",
"output": "2\ninsert 31\nremoveMin"
},
{
"input": "2\ngetMin 31\nremoveMin",
"output": "3\ninsert 31\ngetMin 31\nremoveMin"
},
{
"input": "2\nremoveMin\ngetMin 31",
"output": "4\ninsert 0\nremoveMin\ninsert 31\ngetMin 31"
},
{
"input": "8\ninsert 219147240\nremoveMin\ngetMin 923854124\nremoveMin\ngetMin -876779400\nremoveMin\ninsert 387686853\ngetMin 749998368",
"output": "12\ninsert 219147240\nremoveMin\ninsert 923854124\ngetMin 923854124\nremoveMin\ninsert -876779400\ngetMin -876779400\nremoveMin\ninsert 387686853\nremoveMin\ninsert 749998368\ngetMin 749998368"
},
{
"input": "2\nremoveMin\ninsert 450653162",
"output": "3\ninsert 0\nremoveMin\ninsert 450653162"
},
{
"input": "6\ninsert -799688192\ngetMin 491561656\nremoveMin\ninsert -805250162\ninsert -945439443\nremoveMin",
"output": "8\ninsert -799688192\nremoveMin\ninsert 491561656\ngetMin 491561656\nremoveMin\ninsert -805250162\ninsert -945439443\nremoveMin"
},
{
"input": "30\ninsert 62350949\ngetMin -928976719\nremoveMin\ngetMin 766590157\ngetMin -276914351\ninsert 858958907\ngetMin -794653029\ngetMin 505812710\ngetMin -181182543\ninsert -805198995\nremoveMin\ninsert -200361579\nremoveMin\ninsert 988531216\ninsert -474257426\ninsert 579296921\nremoveMin\ninsert -410043658\ngetMin 716684155\nremoveMin\ngetMin -850837161\ngetMin 368670814\ninsert 579000842\nremoveMin\ngetMin -169833018\ninsert 313148949\nremoveMin\nremoveMin\ngetMin 228901059\ngetMin 599172503",
"output": "52\ninsert 62350949\ninsert -928976719\ngetMin -928976719\nremoveMin\nremoveMin\ninsert 766590157\ngetMin 766590157\ninsert -276914351\ngetMin -276914351\ninsert 858958907\ninsert -794653029\ngetMin -794653029\nremoveMin\nremoveMin\ninsert 505812710\ngetMin 505812710\ninsert -181182543\ngetMin -181182543\ninsert -805198995\nremoveMin\ninsert -200361579\nremoveMin\ninsert 988531216\ninsert -474257426\ninsert 579296921\nremoveMin\ninsert -410043658\nremoveMin\nremoveMin\nremoveMin\nremoveMin\ninsert 71668415..."
},
{
"input": "9\ninsert 3\ninsert 4\ninsert 5\nremoveMin\ngetMin 3\nremoveMin\ngetMin 4\nremoveMin\ngetMin 5",
"output": "10\ninsert 3\ninsert 4\ninsert 5\nremoveMin\ninsert 3\ngetMin 3\nremoveMin\ngetMin 4\nremoveMin\ngetMin 5"
},
{
"input": "9\ninsert 3\ninsert 4\ninsert 5\nremoveMin\ngetMin 5\nremoveMin\ngetMin 4\nremoveMin\ngetMin 3",
"output": "12\ninsert 3\ninsert 4\ninsert 5\nremoveMin\nremoveMin\ngetMin 5\nremoveMin\ninsert 4\ngetMin 4\nremoveMin\ninsert 3\ngetMin 3"
},
{
"input": "1\ngetMin 0",
"output": "2\ninsert 0\ngetMin 0"
},
{
"input": "3\ninsert 3\ninsert 4\ngetMin 4",
"output": "4\ninsert 3\ninsert 4\nremoveMin\ngetMin 4"
},
{
"input": "3\ninsert 1\ninsert 2\ngetMin 2",
"output": "4\ninsert 1\ninsert 2\nremoveMin\ngetMin 2"
},
{
"input": "3\ninsert -1\ninsert 0\ngetMin 0",
"output": "4\ninsert -1\ninsert 0\nremoveMin\ngetMin 0"
},
{
"input": "2\ninsert 0\ngetMin 0",
"output": "2\ninsert 0\ngetMin 0"
},
{
"input": "1\ninsert -1",
"output": "1\ninsert -1"
},
{
"input": "3\ninsert 1\ninsert 0\ngetMin 1",
"output": "4\ninsert 1\ninsert 0\nremoveMin\ngetMin 1"
}
] | 1,683,345,623 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 9 | 326 | 11,468,800 | import heapq
lists= []
heapq.heapify(lists)
result = []
n = int(input())
for _ in range(n):
get = input()
if get[0] == "r":
if lists:
heapq.heappop(lists)
else:
result.append("insert 0")
elif get[0] == "i":
splitted = get.split()
heapq.heappush(lists , int(splitted[1]) )
else:
ele = get.split()
ele = int(ele[1])
while lists and lists[0] < ele:
#we have to pop till we make it eqaul since no ele should exist
heapq.heappop(lists)
result.append("removeMin")
if not lists:
heapq.heappush(lists , ele)
result.append("insert " + str(ele))
result.append(get)
print(len(result))
for i in result:
print(i)
| Title: Heap Operations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya has recently learned data structure named "Binary heap".
The heap he is now operating with allows the following operations:
- put the given number into the heap; - get the value of the minimum element in the heap; - extract the minimum element from the heap;
Thus, at any moment of time the heap contains several integers (possibly none), some of them might be equal.
In order to better learn this data structure Petya took an empty heap and applied some operations above to it. Also, he carefully wrote down all the operations and their results to his event log, following the format:
- insert *x* — put the element with value *x* in the heap; - getMin *x* — the value of the minimum element contained in the heap was equal to *x*; - removeMin — the minimum element was extracted from the heap (only one instance, if there were many).
All the operations were correct, i.e. there was at least one element in the heap each time getMin or removeMin operations were applied.
While Petya was away for a lunch, his little brother Vova came to the room, took away some of the pages from Petya's log and used them to make paper boats.
Now Vova is worried, if he made Petya's sequence of operations inconsistent. For example, if one apply operations one-by-one in the order they are written in the event log, results of getMin operations might differ from the results recorded by Petya, and some of getMin or removeMin operations may be incorrect, as the heap is empty at the moment they are applied.
Now Vova wants to add some new operation records to the event log in order to make the resulting sequence of operations correct. That is, the result of each getMin operation is equal to the result in the record, and the heap is non-empty when getMin ad removeMin are applied. Vova wants to complete this as fast as possible, as the Petya may get back at any moment. He asks you to add the least possible number of operation records to the current log. Note that arbitrary number of operations may be added at the beginning, between any two other operations, or at the end of the log.
Input Specification:
The first line of the input contains the only integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of the records left in Petya's journal.
Each of the following *n* lines describe the records in the current log in the order they are applied. Format described in the statement is used. All numbers in the input are integers not exceeding 109 by their absolute value.
Output Specification:
The first line of the output should contain a single integer *m* — the minimum possible number of records in the modified sequence of operations.
Next *m* lines should contain the corrected sequence of records following the format of the input (described in the statement), one per line and in the order they are applied. All the numbers in the output should be integers not exceeding 109 by their absolute value.
Note that the input sequence of operations must be the subsequence of the output sequence.
It's guaranteed that there exists the correct answer consisting of no more than 1<=000<=000 operations.
Demo Input:
['2\ninsert 3\ngetMin 4\n', '4\ninsert 1\ninsert 1\nremoveMin\ngetMin 2\n']
Demo Output:
['4\ninsert 3\nremoveMin\ninsert 4\ngetMin 4\n', '6\ninsert 1\ninsert 1\nremoveMin\nremoveMin\ninsert 2\ngetMin 2\n']
Note:
In the first sample, after number 3 is inserted into the heap, the minimum number is 3. To make the result of the first getMin equal to 4 one should firstly remove number 3 from the heap and then add number 4 into the heap.
In the second sample case number 1 is inserted two times, so should be similarly removed twice. | ```python
import heapq
lists= []
heapq.heapify(lists)
result = []
n = int(input())
for _ in range(n):
get = input()
if get[0] == "r":
if lists:
heapq.heappop(lists)
else:
result.append("insert 0")
elif get[0] == "i":
splitted = get.split()
heapq.heappush(lists , int(splitted[1]) )
else:
ele = get.split()
ele = int(ele[1])
while lists and lists[0] < ele:
#we have to pop till we make it eqaul since no ele should exist
heapq.heappop(lists)
result.append("removeMin")
if not lists:
heapq.heappush(lists , ele)
result.append("insert " + str(ele))
result.append(get)
print(len(result))
for i in result:
print(i)
``` | 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",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,667,472,284 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 62 | 0 | n = int(input())
m = int(input())
print(int(m * n / 2)) | Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none | ```python
n = int(input())
m = int(input())
print(int(m * n / 2))
``` | -1 |
732 | A | Buy a Shovel | PROGRAMMING | 800 | [
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null | Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel. | The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels. | Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change. | [
"117 3\n",
"237 7\n",
"15 2\n"
] | [
"9\n",
"1\n",
"2\n"
] | In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change. | 500 | [
{
"input": "117 3",
"output": "9"
},
{
"input": "237 7",
"output": "1"
},
{
"input": "15 2",
"output": "2"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 9",
"output": "9"
},
{
"input": "1000 3",
"output": "1"
},
{
"input": "1000 1",
"output": "1"
},
{
"input": "1000 9",
"output": "1"
},
{
"input": "1 2",
"output": "2"
},
{
"input": "999 9",
"output": "1"
},
{
"input": "999 8",
"output": "2"
},
{
"input": "105 6",
"output": "2"
},
{
"input": "403 9",
"output": "3"
},
{
"input": "546 4",
"output": "4"
},
{
"input": "228 9",
"output": "5"
},
{
"input": "57 2",
"output": "6"
},
{
"input": "437 9",
"output": "7"
},
{
"input": "997 6",
"output": "8"
},
{
"input": "109 1",
"output": "9"
},
{
"input": "998 9",
"output": "5"
},
{
"input": "4 2",
"output": "3"
},
{
"input": "9 3",
"output": "7"
},
{
"input": "8 2",
"output": "4"
},
{
"input": "1 3",
"output": "3"
},
{
"input": "1 4",
"output": "4"
},
{
"input": "1 5",
"output": "5"
},
{
"input": "1 6",
"output": "6"
},
{
"input": "1 7",
"output": "7"
},
{
"input": "1 8",
"output": "8"
},
{
"input": "100 3",
"output": "1"
},
{
"input": "1000 2",
"output": "1"
},
{
"input": "1000 4",
"output": "1"
},
{
"input": "1000 5",
"output": "1"
},
{
"input": "1000 6",
"output": "1"
},
{
"input": "1000 7",
"output": "1"
},
{
"input": "1000 8",
"output": "1"
},
{
"input": "23 4",
"output": "8"
},
{
"input": "33 1",
"output": "7"
},
{
"input": "33 2",
"output": "4"
},
{
"input": "666 5",
"output": "5"
},
{
"input": "2 3",
"output": "5"
},
{
"input": "5 5",
"output": "1"
},
{
"input": "3 6",
"output": "2"
},
{
"input": "12 4",
"output": "2"
},
{
"input": "15 5",
"output": "1"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "25 5",
"output": "1"
},
{
"input": "2 9",
"output": "5"
},
{
"input": "6 7",
"output": "5"
},
{
"input": "8 9",
"output": "5"
},
{
"input": "2 7",
"output": "5"
},
{
"input": "4 7",
"output": "5"
},
{
"input": "2 1",
"output": "5"
},
{
"input": "261 1",
"output": "1"
}
] | 1,687,252,705 | 2,147,483,647 | Python 3 | OK | TESTS | 54 | 46 | 0 | a = input().split(" ")
b = a[0]
c = int(b[-1])
r = int(a[1])
output = 10
for i in range(1,10):
if int(str(c*i)[-1]) == 0:
output = i
break
if int(str(c*i)[-1]) == r:
output = i
break
print(output) | Title: Buy a Shovel
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp urgently needs a shovel! He comes to the shop and chooses an appropriate one. The shovel that Policarp chooses is sold for *k* burles. Assume that there is an unlimited number of such shovels in the shop.
In his pocket Polycarp has an unlimited number of "10-burle coins" and exactly one coin of *r* burles (1<=≤<=*r*<=≤<=9).
What is the minimum number of shovels Polycarp has to buy so that he can pay for the purchase without any change? It is obvious that he can pay for 10 shovels without any change (by paying the requied amount of 10-burle coins and not using the coin of *r* burles). But perhaps he can buy fewer shovels and pay without any change. Note that Polycarp should buy at least one shovel.
Input Specification:
The single line of input contains two integers *k* and *r* (1<=≤<=*k*<=≤<=1000, 1<=≤<=*r*<=≤<=9) — the price of one shovel and the denomination of the coin in Polycarp's pocket that is different from "10-burle coins".
Remember that he has an unlimited number of coins in the denomination of 10, that is, Polycarp has enough money to buy any number of shovels.
Output Specification:
Print the required minimum number of shovels Polycarp has to buy so that he can pay for them without any change.
Demo Input:
['117 3\n', '237 7\n', '15 2\n']
Demo Output:
['9\n', '1\n', '2\n']
Note:
In the first example Polycarp can buy 9 shovels and pay 9·117 = 1053 burles. Indeed, he can pay this sum by using 10-burle coins and one 3-burle coin. He can't buy fewer shovels without any change.
In the second example it is enough for Polycarp to buy one shovel.
In the third example Polycarp should buy two shovels and pay 2·15 = 30 burles. It is obvious that he can pay this sum without any change. | ```python
a = input().split(" ")
b = a[0]
c = int(b[-1])
r = int(a[1])
output = 10
for i in range(1,10):
if int(str(c*i)[-1]) == 0:
output = i
break
if int(str(c*i)[-1]) == r:
output = i
break
print(output)
``` | 3 | |
245 | B | Internet Address | PROGRAMMING | 1,100 | [
"implementation",
"strings"
] | null | null | Vasya is an active Internet user. One day he came across an Internet resource he liked, so he wrote its address in the notebook. We know that the address of the written resource has format:
where:
- <protocol> can equal either "http" (without the quotes) or "ftp" (without the quotes), - <domain> is a non-empty string, consisting of lowercase English letters, - the /<context> part may not be present. If it is present, then <context> is a non-empty string, consisting of lowercase English letters.
If string <context> isn't present in the address, then the additional character "/" isn't written. Thus, the address has either two characters "/" (the ones that go before the domain), or three (an extra one in front of the context).
When the boy came home, he found out that the address he wrote in his notebook had no punctuation marks. Vasya must have been in a lot of hurry and didn't write characters ":", "/", ".".
Help Vasya to restore the possible address of the recorded Internet resource. | The first line contains a non-empty string that Vasya wrote out in his notebook. This line consists of lowercase English letters only.
It is guaranteed that the given string contains at most 50 letters. It is guaranteed that the given string can be obtained from some correct Internet resource address, described above. | Print a single line — the address of the Internet resource that Vasya liked. If there are several addresses that meet the problem limitations, you are allowed to print any of them. | [
"httpsunrux\n",
"ftphttprururu\n"
] | [
"http://sun.ru/x\n",
"ftp://http.ru/ruru\n"
] | In the second sample there are two more possible answers: "ftp://httpruru.ru" and "ftp://httpru.ru/ru". | 0 | [
{
"input": "httpsunrux",
"output": "http://sun.ru/x"
},
{
"input": "ftphttprururu",
"output": "ftp://http.ru/ruru"
},
{
"input": "httpuururrururruruurururrrrrurrurrurruruuruuu",
"output": "http://uu.ru/rrururruruurururrrrrurrurrurruruuruuu"
},
{
"input": "httpabuaruauabbaruru",
"output": "http://abua.ru/auabbaruru"
},
{
"input": "httpuurrruurruuruuruuurrrurururuurruuuuuuruurr",
"output": "http://uurr.ru/urruuruuruuurrrurururuurruuuuuuruurr"
},
{
"input": "httpruhhphhhpuhruruhhpruhhphruhhru",
"output": "http://ruhhphhhpuh.ru/ruhhpruhhphruhhru"
},
{
"input": "httpftprftprutprururftruruftptp",
"output": "http://ftprftp.ru/tprururftruruftptp"
},
{
"input": "httpfttpftpfttftpftpftppfrurururu",
"output": "http://fttpftpfttftpftpftppf.ru/rururu"
},
{
"input": "httpruhttttpruhttprupruhttpruhtturuhttphtruuru",
"output": "http://ruhttttp.ru/httprupruhttpruhtturuhttphtruuru"
},
{
"input": "httpsjkazaaghasjkasjkabruru",
"output": "http://sjkazaaghasjkasjkab.ru/ru"
},
{
"input": "httpftphttptphttphrururuhpftphtpftphtpftphtptpft",
"output": "http://ftphttptphttph.ru/ruruhpftphtpftphtpftphtptpft"
},
{
"input": "httpppppru",
"output": "http://pppp.ru"
},
{
"input": "ftprrurururrurururuurrururruuru",
"output": "ftp://r.ru/rururrurururuurrururruuru"
},
{
"input": "ftpabaruru",
"output": "ftp://aba.ru/ru"
},
{
"input": "ftpruurruurururururuuruuur",
"output": "ftp://ruur.ru/urururururuuruuur"
},
{
"input": "ftphhphruhhpruhhpuhhpuruhhphruhhruhhpuhhru",
"output": "ftp://hhph.ru/hhpruhhpuhhpuruhhphruhhruhhpuhhru"
},
{
"input": "ftparua",
"output": "ftp://a.ru/a"
},
{
"input": "httpzru",
"output": "http://z.ru"
},
{
"input": "httprrur",
"output": "http://r.ru/r"
},
{
"input": "ftprru",
"output": "ftp://r.ru"
}
] | 1,590,385,318 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 186 | 307,200 | s = input()
flag = 0
if(s[0]=='f'):
flag=1
ind = s.find('ru')
if((flag == 1 and ind==3) or (flag==0 and ind ==4)):
ind = s[ind+2:].find('ru')
if(flag==1):
print( s[:3] + '://' + s[3:ind] + '.' + s[ind:ind+2] + ('/' + s[ind+2:] if(len(s)>=ind+2) else ''))
else:
print( s[:4] + '://' + s[4:ind] + '.' + s[ind:ind+2] + ('/' + s[ind+2:] if(len(s)>=ind+2) else '')) | Title: Internet Address
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya is an active Internet user. One day he came across an Internet resource he liked, so he wrote its address in the notebook. We know that the address of the written resource has format:
where:
- <protocol> can equal either "http" (without the quotes) or "ftp" (without the quotes), - <domain> is a non-empty string, consisting of lowercase English letters, - the /<context> part may not be present. If it is present, then <context> is a non-empty string, consisting of lowercase English letters.
If string <context> isn't present in the address, then the additional character "/" isn't written. Thus, the address has either two characters "/" (the ones that go before the domain), or three (an extra one in front of the context).
When the boy came home, he found out that the address he wrote in his notebook had no punctuation marks. Vasya must have been in a lot of hurry and didn't write characters ":", "/", ".".
Help Vasya to restore the possible address of the recorded Internet resource.
Input Specification:
The first line contains a non-empty string that Vasya wrote out in his notebook. This line consists of lowercase English letters only.
It is guaranteed that the given string contains at most 50 letters. It is guaranteed that the given string can be obtained from some correct Internet resource address, described above.
Output Specification:
Print a single line — the address of the Internet resource that Vasya liked. If there are several addresses that meet the problem limitations, you are allowed to print any of them.
Demo Input:
['httpsunrux\n', 'ftphttprururu\n']
Demo Output:
['http://sun.ru/x\n', 'ftp://http.ru/ruru\n']
Note:
In the second sample there are two more possible answers: "ftp://httpruru.ru" and "ftp://httpru.ru/ru". | ```python
s = input()
flag = 0
if(s[0]=='f'):
flag=1
ind = s.find('ru')
if((flag == 1 and ind==3) or (flag==0 and ind ==4)):
ind = s[ind+2:].find('ru')
if(flag==1):
print( s[:3] + '://' + s[3:ind] + '.' + s[ind:ind+2] + ('/' + s[ind+2:] if(len(s)>=ind+2) else ''))
else:
print( s[:4] + '://' + s[4:ind] + '.' + s[ind:ind+2] + ('/' + s[ind+2:] if(len(s)>=ind+2) else ''))
``` | 0 | |
761 | A | Dasha and Stairs | PROGRAMMING | 1,000 | [
"brute force",
"constructive algorithms",
"implementation",
"math"
] | null | null | On her way to programming school tiger Dasha faced her first test — a huge staircase!
The steps were numbered from one to infinity. As we know, tigers are very fond of all striped things, it is possible that it has something to do with their color. So on some interval of her way she calculated two values — the number of steps with even and odd numbers.
You need to check whether there is an interval of steps from the *l*-th to the *r*-th (1<=≤<=*l*<=≤<=*r*), for which values that Dasha has found are correct. | In the only line you are given two integers *a*, *b* (0<=≤<=*a*,<=*b*<=≤<=100) — the number of even and odd steps, accordingly. | In the only line print "YES", if the interval of steps described above exists, and "NO" otherwise. | [
"2 3\n",
"3 1\n"
] | [
"YES\n",
"NO\n"
] | In the first example one of suitable intervals is from 1 to 5. The interval contains two even steps — 2 and 4, and three odd: 1, 3 and 5. | 500 | [
{
"input": "2 3",
"output": "YES"
},
{
"input": "3 1",
"output": "NO"
},
{
"input": "5 4",
"output": "YES"
},
{
"input": "9 9",
"output": "YES"
},
{
"input": "85 95",
"output": "NO"
},
{
"input": "0 1",
"output": "YES"
},
{
"input": "89 25",
"output": "NO"
},
{
"input": "74 73",
"output": "YES"
},
{
"input": "62 39",
"output": "NO"
},
{
"input": "57 57",
"output": "YES"
},
{
"input": "100 99",
"output": "YES"
},
{
"input": "0 0",
"output": "NO"
},
{
"input": "98 100",
"output": "NO"
},
{
"input": "99 100",
"output": "YES"
},
{
"input": "1 0",
"output": "YES"
},
{
"input": "100 0",
"output": "NO"
},
{
"input": "0 100",
"output": "NO"
},
{
"input": "100 98",
"output": "NO"
},
{
"input": "100 100",
"output": "YES"
},
{
"input": "0 5",
"output": "NO"
},
{
"input": "2 2",
"output": "YES"
}
] | 1,595,754,022 | 2,147,483,647 | PyPy 3 | OK | TESTS | 21 | 155 | 20,172,800 | l1 = [int(x) for x in input().split()]
if l1[0] or l1[1]:
if max(l1[0],l1[1])-min(l1[0],l1[1])>1:
print("NO")
else:
print("YES")
else:
print("NO") | Title: Dasha and Stairs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On her way to programming school tiger Dasha faced her first test — a huge staircase!
The steps were numbered from one to infinity. As we know, tigers are very fond of all striped things, it is possible that it has something to do with their color. So on some interval of her way she calculated two values — the number of steps with even and odd numbers.
You need to check whether there is an interval of steps from the *l*-th to the *r*-th (1<=≤<=*l*<=≤<=*r*), for which values that Dasha has found are correct.
Input Specification:
In the only line you are given two integers *a*, *b* (0<=≤<=*a*,<=*b*<=≤<=100) — the number of even and odd steps, accordingly.
Output Specification:
In the only line print "YES", if the interval of steps described above exists, and "NO" otherwise.
Demo Input:
['2 3\n', '3 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example one of suitable intervals is from 1 to 5. The interval contains two even steps — 2 and 4, and three odd: 1, 3 and 5. | ```python
l1 = [int(x) for x in input().split()]
if l1[0] or l1[1]:
if max(l1[0],l1[1])-min(l1[0],l1[1])>1:
print("NO")
else:
print("YES")
else:
print("NO")
``` | 3 | |
282 | A | Bit++ | PROGRAMMING | 800 | [
"implementation"
] | null | null | The classic programming language of Bitland is Bit++. This language is so peculiar and complicated.
The language is that peculiar as it has exactly one variable, called *x*. Also, there are two operations:
- Operation ++ increases the value of variable *x* by 1. - Operation -- decreases the value of variable *x* by 1.
A statement in language Bit++ is a sequence, consisting of exactly one operation and one variable *x*. The statement is written without spaces, that is, it can only contain characters "+", "-", "X". Executing a statement means applying the operation it contains.
A programme in Bit++ is a sequence of statements, each of them needs to be executed. Executing a programme means executing all the statements it contains.
You're given a programme in language Bit++. The initial value of *x* is 0. Execute the programme and find its final value (the value of the variable when this programme is executed). | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=150) — the number of statements in the programme.
Next *n* lines contain a statement each. Each statement contains exactly one operation (++ or --) and exactly one variable *x* (denoted as letter «X»). Thus, there are no empty statements. The operation and the variable can be written in any order. | Print a single integer — the final value of *x*. | [
"1\n++X\n",
"2\nX++\n--X\n"
] | [
"1\n",
"0\n"
] | none | 500 | [
{
"input": "1\n++X",
"output": "1"
},
{
"input": "2\nX++\n--X",
"output": "0"
},
{
"input": "3\n++X\n++X\n++X",
"output": "3"
},
{
"input": "2\n--X\n--X",
"output": "-2"
},
{
"input": "5\n++X\n--X\n++X\n--X\n--X",
"output": "-1"
},
{
"input": "28\nX--\n++X\nX++\nX++\nX++\n--X\n--X\nX++\nX--\n++X\nX++\n--X\nX--\nX++\nX--\n++X\n++X\nX++\nX++\nX++\nX++\n--X\n++X\n--X\n--X\n--X\n--X\nX++",
"output": "4"
},
{
"input": "94\nX++\nX++\n++X\n++X\nX--\n--X\nX++\n--X\nX++\n++X\nX++\n++X\n--X\n--X\n++X\nX++\n--X\nX--\nX--\n--X\nX--\nX--\n--X\n++X\n--X\nX--\nX--\nX++\n++X\n--X\nX--\n++X\n--X\n--X\nX--\nX--\nX++\nX++\nX--\nX++\nX--\nX--\nX--\n--X\nX--\nX--\nX--\nX++\n++X\nX--\n++X\nX++\n--X\n--X\n--X\n--X\n++X\nX--\n--X\n--X\n++X\nX--\nX--\nX++\n++X\nX++\n++X\n--X\n--X\nX--\n++X\nX--\nX--\n++X\n++X\n++X\n++X\nX++\n++X\n--X\nX++\n--X\n--X\n++X\n--X\nX++\n++X\nX++\n--X\nX--\nX--\n--X\n++X\nX++",
"output": "-10"
},
{
"input": "56\n--X\nX--\n--X\n--X\nX--\nX--\n--X\nX++\n++X\n--X\nX++\nX--\n--X\n++X\n--X\nX--\nX--\n++X\nX--\nX--\n--X\n++X\n--X\n++X\n--X\nX++\n++X\nX++\n--X\n++X\nX++\nX++\n--X\nX++\nX--\n--X\nX--\n--X\nX++\n++X\n--X\n++X\nX++\nX--\n--X\n--X\n++X\nX--\nX--\n--X\nX--\n--X\nX++\n--X\n++X\n--X",
"output": "-14"
},
{
"input": "59\nX--\n--X\nX++\n++X\nX--\n--X\n--X\n++X\n++X\n++X\n++X\nX++\n++X\n++X\nX++\n--X\nX--\nX++\n++X\n--X\nX++\n--X\n++X\nX++\n--X\n--X\nX++\nX++\n--X\nX++\nX++\nX++\nX--\nX--\n--X\nX++\nX--\nX--\n++X\nX--\nX++\n--X\nX++\nX--\nX--\nX--\nX--\n++X\n--X\nX++\nX++\nX--\nX++\n++X\nX--\nX++\nX--\nX--\n++X",
"output": "3"
},
{
"input": "87\n--X\n++X\n--X\nX++\n--X\nX--\n--X\n++X\nX--\n++X\n--X\n--X\nX++\n--X\nX--\nX++\n++X\n--X\n++X\n++X\n--X\n++X\n--X\nX--\n++X\n++X\nX--\nX++\nX++\n--X\n--X\n++X\nX--\n--X\n++X\n--X\nX++\n--X\n--X\nX--\n++X\n++X\n--X\nX--\nX--\nX--\nX--\nX--\nX++\n--X\n++X\n--X\nX++\n++X\nX++\n++X\n--X\nX++\n++X\nX--\n--X\nX++\n++X\nX++\nX++\n--X\n--X\n++X\n--X\nX++\nX++\n++X\nX++\nX++\nX++\nX++\n--X\n--X\n--X\n--X\n--X\n--X\n--X\nX--\n--X\n++X\n++X",
"output": "-5"
},
{
"input": "101\nX++\nX++\nX++\n++X\n--X\nX--\nX++\nX--\nX--\n--X\n--X\n++X\nX++\n++X\n++X\nX--\n--X\n++X\nX++\nX--\n++X\n--X\n--X\n--X\n++X\n--X\n++X\nX++\nX++\n++X\n--X\nX++\nX--\nX++\n++X\n++X\nX--\nX--\nX--\nX++\nX++\nX--\nX--\nX++\n++X\n++X\n++X\n--X\n--X\n++X\nX--\nX--\n--X\n++X\nX--\n++X\nX++\n++X\nX--\nX--\n--X\n++X\n--X\n++X\n++X\n--X\nX++\n++X\nX--\n++X\nX--\n++X\nX++\nX--\n++X\nX++\n--X\nX++\nX++\n++X\n--X\n++X\n--X\nX++\n--X\nX--\n--X\n++X\n++X\n++X\n--X\nX--\nX--\nX--\nX--\n--X\n--X\n--X\n++X\n--X\n--X",
"output": "1"
},
{
"input": "63\n--X\nX--\n++X\n--X\n++X\nX++\n--X\n--X\nX++\n--X\n--X\nX++\nX--\nX--\n--X\n++X\nX--\nX--\nX++\n++X\nX++\nX++\n--X\n--X\n++X\nX--\nX--\nX--\n++X\nX++\nX--\n--X\nX--\n++X\n++X\nX++\n++X\nX++\nX++\n--X\nX--\n++X\nX--\n--X\nX--\nX--\nX--\n++X\n++X\n++X\n++X\nX++\nX++\n++X\n--X\n--X\n++X\n++X\n++X\nX--\n++X\n++X\nX--",
"output": "1"
},
{
"input": "45\n--X\n++X\nX--\n++X\n++X\nX++\n--X\n--X\n--X\n--X\n--X\n--X\n--X\nX++\n++X\nX--\n++X\n++X\nX--\nX++\nX--\n--X\nX--\n++X\n++X\n--X\n--X\nX--\nX--\n--X\n++X\nX--\n--X\n++X\n++X\n--X\n--X\nX--\n++X\n++X\nX++\nX++\n++X\n++X\nX++",
"output": "-3"
},
{
"input": "21\n++X\nX++\n--X\nX--\nX++\n++X\n--X\nX--\nX++\nX--\nX--\nX--\nX++\n++X\nX++\n++X\n--X\nX--\n--X\nX++\n++X",
"output": "1"
},
{
"input": "100\n--X\n++X\nX++\n++X\nX--\n++X\nX--\nX++\n--X\nX++\nX--\nX--\nX--\n++X\nX--\nX++\nX++\n++X\nX++\nX++\nX++\nX++\n++X\nX++\n++X\nX--\n--X\n++X\nX--\n--X\n++X\n++X\nX--\nX++\nX++\nX++\n++X\n--X\n++X\nX++\nX--\n++X\n++X\n--X\n++X\nX--\nX--\nX--\nX++\nX--\nX--\nX++\nX++\n--X\nX++\nX++\n--X\nX--\n--X\n++X\n--X\n++X\n++X\nX--\n--X\n++X\n++X\n--X\n--X\n++X\nX++\nX--\nX++\nX--\nX++\nX++\n--X\nX--\nX--\n++X\nX--\n--X\n--X\nX++\n--X\n--X\nX--\nX--\n++X\n++X\nX--\n++X\nX++\n--X\n++X\n++X\nX++\n--X\n--X\nX++",
"output": "8"
},
{
"input": "17\nX++\nX++\n++X\n--X\n--X\n++X\n++X\n--X\nX--\nX++\nX--\n--X\n--X\nX--\n++X\nX--\nX++",
"output": "-1"
},
{
"input": "77\n++X\nX++\n--X\nX--\n--X\n--X\nX--\nX++\nX--\nX++\nX--\n++X\n--X\n--X\n--X\n--X\n++X\nX--\nX++\nX--\n--X\nX--\n--X\nX--\n++X\n--X\n++X\n++X\nX++\nX++\nX--\n--X\nX--\nX--\nX++\n--X\n--X\n++X\nX--\nX--\n++X\nX++\nX--\n++X\n--X\nX++\nX--\n++X\n++X\n++X\nX--\nX--\nX--\n--X\n++X\n++X\n++X\nX++\n--X\n--X\n++X\n--X\nX--\nX++\n++X\nX++\n++X\nX--\nX++\nX++\n--X\nX++\nX++\nX++\n--X\nX++\nX--",
"output": "-5"
},
{
"input": "21\nX--\n++X\n--X\nX--\n++X\nX--\n++X\nX--\n--X\n++X\nX++\n++X\nX++\n++X\nX--\n--X\nX++\nX++\nX--\n++X\nX--",
"output": "1"
},
{
"input": "1\nX--",
"output": "-1"
}
] | 1,697,368,698 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 15 | 0 | n = int(input())
x = 0
for i in range(0,n):
p = input("")
if p == "++x":
x += 1
elif p == "--x":
x -= 1
print(x) | Title: Bit++
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The classic programming language of Bitland is Bit++. This language is so peculiar and complicated.
The language is that peculiar as it has exactly one variable, called *x*. Also, there are two operations:
- Operation ++ increases the value of variable *x* by 1. - Operation -- decreases the value of variable *x* by 1.
A statement in language Bit++ is a sequence, consisting of exactly one operation and one variable *x*. The statement is written without spaces, that is, it can only contain characters "+", "-", "X". Executing a statement means applying the operation it contains.
A programme in Bit++ is a sequence of statements, each of them needs to be executed. Executing a programme means executing all the statements it contains.
You're given a programme in language Bit++. The initial value of *x* is 0. Execute the programme and find its final value (the value of the variable when this programme is executed).
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=150) — the number of statements in the programme.
Next *n* lines contain a statement each. Each statement contains exactly one operation (++ or --) and exactly one variable *x* (denoted as letter «X»). Thus, there are no empty statements. The operation and the variable can be written in any order.
Output Specification:
Print a single integer — the final value of *x*.
Demo Input:
['1\n++X\n', '2\nX++\n--X\n']
Demo Output:
['1\n', '0\n']
Note:
none | ```python
n = int(input())
x = 0
for i in range(0,n):
p = input("")
if p == "++x":
x += 1
elif p == "--x":
x -= 1
print(x)
``` | 0 | |
670 | A | Holidays | PROGRAMMING | 900 | [
"brute force",
"constructive algorithms",
"greedy",
"math"
] | null | null | On the planet Mars a year lasts exactly *n* days (there are no leap years on Mars). But Martians have the same weeks as earthlings — 5 work days and then 2 days off. Your task is to determine the minimum possible and the maximum possible number of days off per year on Mars. | The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=1<=000<=000) — the number of days in a year on Mars. | Print two integers — the minimum possible and the maximum possible number of days off per year on Mars. | [
"14\n",
"2\n"
] | [
"4 4\n",
"0 2\n"
] | In the first sample there are 14 days in a year on Mars, and therefore independently of the day a year starts with there will be exactly 4 days off .
In the second sample there are only 2 days in a year on Mars, and they can both be either work days or days off. | 500 | [
{
"input": "14",
"output": "4 4"
},
{
"input": "2",
"output": "0 2"
},
{
"input": "1",
"output": "0 1"
},
{
"input": "3",
"output": "0 2"
},
{
"input": "4",
"output": "0 2"
},
{
"input": "5",
"output": "0 2"
},
{
"input": "6",
"output": "1 2"
},
{
"input": "7",
"output": "2 2"
},
{
"input": "8",
"output": "2 3"
},
{
"input": "9",
"output": "2 4"
},
{
"input": "10",
"output": "2 4"
},
{
"input": "11",
"output": "2 4"
},
{
"input": "12",
"output": "2 4"
},
{
"input": "13",
"output": "3 4"
},
{
"input": "1000000",
"output": "285714 285715"
},
{
"input": "16",
"output": "4 6"
},
{
"input": "17",
"output": "4 6"
},
{
"input": "18",
"output": "4 6"
},
{
"input": "19",
"output": "4 6"
},
{
"input": "20",
"output": "5 6"
},
{
"input": "21",
"output": "6 6"
},
{
"input": "22",
"output": "6 7"
},
{
"input": "23",
"output": "6 8"
},
{
"input": "24",
"output": "6 8"
},
{
"input": "25",
"output": "6 8"
},
{
"input": "26",
"output": "6 8"
},
{
"input": "27",
"output": "7 8"
},
{
"input": "28",
"output": "8 8"
},
{
"input": "29",
"output": "8 9"
},
{
"input": "30",
"output": "8 10"
},
{
"input": "100",
"output": "28 30"
},
{
"input": "99",
"output": "28 29"
},
{
"input": "98",
"output": "28 28"
},
{
"input": "97",
"output": "27 28"
},
{
"input": "96",
"output": "26 28"
},
{
"input": "95",
"output": "26 28"
},
{
"input": "94",
"output": "26 28"
},
{
"input": "93",
"output": "26 28"
},
{
"input": "92",
"output": "26 27"
},
{
"input": "91",
"output": "26 26"
},
{
"input": "90",
"output": "25 26"
},
{
"input": "89",
"output": "24 26"
},
{
"input": "88",
"output": "24 26"
},
{
"input": "87",
"output": "24 26"
},
{
"input": "86",
"output": "24 26"
},
{
"input": "85",
"output": "24 25"
},
{
"input": "84",
"output": "24 24"
},
{
"input": "83",
"output": "23 24"
},
{
"input": "82",
"output": "22 24"
},
{
"input": "81",
"output": "22 24"
},
{
"input": "80",
"output": "22 24"
},
{
"input": "1000",
"output": "285 286"
},
{
"input": "999",
"output": "284 286"
},
{
"input": "998",
"output": "284 286"
},
{
"input": "997",
"output": "284 286"
},
{
"input": "996",
"output": "284 286"
},
{
"input": "995",
"output": "284 285"
},
{
"input": "994",
"output": "284 284"
},
{
"input": "993",
"output": "283 284"
},
{
"input": "992",
"output": "282 284"
},
{
"input": "991",
"output": "282 284"
},
{
"input": "990",
"output": "282 284"
},
{
"input": "989",
"output": "282 284"
},
{
"input": "988",
"output": "282 283"
},
{
"input": "987",
"output": "282 282"
},
{
"input": "986",
"output": "281 282"
},
{
"input": "985",
"output": "280 282"
},
{
"input": "984",
"output": "280 282"
},
{
"input": "983",
"output": "280 282"
},
{
"input": "982",
"output": "280 282"
},
{
"input": "981",
"output": "280 281"
},
{
"input": "980",
"output": "280 280"
},
{
"input": "10000",
"output": "2856 2858"
},
{
"input": "9999",
"output": "2856 2858"
},
{
"input": "9998",
"output": "2856 2858"
},
{
"input": "9997",
"output": "2856 2857"
},
{
"input": "9996",
"output": "2856 2856"
},
{
"input": "9995",
"output": "2855 2856"
},
{
"input": "9994",
"output": "2854 2856"
},
{
"input": "9993",
"output": "2854 2856"
},
{
"input": "9992",
"output": "2854 2856"
},
{
"input": "9991",
"output": "2854 2856"
},
{
"input": "9990",
"output": "2854 2855"
},
{
"input": "9989",
"output": "2854 2854"
},
{
"input": "9988",
"output": "2853 2854"
},
{
"input": "9987",
"output": "2852 2854"
},
{
"input": "9986",
"output": "2852 2854"
},
{
"input": "9985",
"output": "2852 2854"
},
{
"input": "9984",
"output": "2852 2854"
},
{
"input": "9983",
"output": "2852 2853"
},
{
"input": "9982",
"output": "2852 2852"
},
{
"input": "9981",
"output": "2851 2852"
},
{
"input": "9980",
"output": "2850 2852"
},
{
"input": "100000",
"output": "28570 28572"
},
{
"input": "99999",
"output": "28570 28572"
},
{
"input": "99998",
"output": "28570 28572"
},
{
"input": "99997",
"output": "28570 28572"
},
{
"input": "99996",
"output": "28570 28571"
},
{
"input": "99995",
"output": "28570 28570"
},
{
"input": "99994",
"output": "28569 28570"
},
{
"input": "99993",
"output": "28568 28570"
},
{
"input": "99992",
"output": "28568 28570"
},
{
"input": "99991",
"output": "28568 28570"
},
{
"input": "99990",
"output": "28568 28570"
},
{
"input": "99989",
"output": "28568 28569"
},
{
"input": "99988",
"output": "28568 28568"
},
{
"input": "99987",
"output": "28567 28568"
},
{
"input": "99986",
"output": "28566 28568"
},
{
"input": "99985",
"output": "28566 28568"
},
{
"input": "99984",
"output": "28566 28568"
},
{
"input": "99983",
"output": "28566 28568"
},
{
"input": "99982",
"output": "28566 28567"
},
{
"input": "99981",
"output": "28566 28566"
},
{
"input": "99980",
"output": "28565 28566"
},
{
"input": "999999",
"output": "285714 285714"
},
{
"input": "999998",
"output": "285713 285714"
},
{
"input": "999997",
"output": "285712 285714"
},
{
"input": "999996",
"output": "285712 285714"
},
{
"input": "999995",
"output": "285712 285714"
},
{
"input": "999994",
"output": "285712 285714"
},
{
"input": "999993",
"output": "285712 285713"
},
{
"input": "999992",
"output": "285712 285712"
},
{
"input": "999991",
"output": "285711 285712"
},
{
"input": "999990",
"output": "285710 285712"
},
{
"input": "999989",
"output": "285710 285712"
},
{
"input": "999988",
"output": "285710 285712"
},
{
"input": "999987",
"output": "285710 285712"
},
{
"input": "999986",
"output": "285710 285711"
},
{
"input": "999985",
"output": "285710 285710"
},
{
"input": "999984",
"output": "285709 285710"
},
{
"input": "999983",
"output": "285708 285710"
},
{
"input": "999982",
"output": "285708 285710"
},
{
"input": "999981",
"output": "285708 285710"
},
{
"input": "999980",
"output": "285708 285710"
},
{
"input": "234123",
"output": "66892 66893"
},
{
"input": "234122",
"output": "66892 66892"
},
{
"input": "234121",
"output": "66891 66892"
},
{
"input": "234120",
"output": "66890 66892"
},
{
"input": "234119",
"output": "66890 66892"
},
{
"input": "234118",
"output": "66890 66892"
},
{
"input": "234117",
"output": "66890 66892"
},
{
"input": "234116",
"output": "66890 66891"
},
{
"input": "234115",
"output": "66890 66890"
},
{
"input": "234114",
"output": "66889 66890"
},
{
"input": "234113",
"output": "66888 66890"
},
{
"input": "234112",
"output": "66888 66890"
},
{
"input": "234111",
"output": "66888 66890"
},
{
"input": "234110",
"output": "66888 66890"
},
{
"input": "234109",
"output": "66888 66889"
},
{
"input": "234108",
"output": "66888 66888"
},
{
"input": "234107",
"output": "66887 66888"
},
{
"input": "234106",
"output": "66886 66888"
},
{
"input": "234105",
"output": "66886 66888"
},
{
"input": "234104",
"output": "66886 66888"
},
{
"input": "234103",
"output": "66886 66888"
},
{
"input": "868531",
"output": "248151 248152"
},
{
"input": "868530",
"output": "248150 248152"
},
{
"input": "868529",
"output": "248150 248152"
},
{
"input": "868528",
"output": "248150 248152"
},
{
"input": "868527",
"output": "248150 248152"
},
{
"input": "868526",
"output": "248150 248151"
},
{
"input": "868525",
"output": "248150 248150"
},
{
"input": "868524",
"output": "248149 248150"
},
{
"input": "868523",
"output": "248148 248150"
},
{
"input": "868522",
"output": "248148 248150"
},
{
"input": "868521",
"output": "248148 248150"
},
{
"input": "868520",
"output": "248148 248150"
},
{
"input": "868519",
"output": "248148 248149"
},
{
"input": "868518",
"output": "248148 248148"
},
{
"input": "868517",
"output": "248147 248148"
},
{
"input": "868516",
"output": "248146 248148"
},
{
"input": "868515",
"output": "248146 248148"
},
{
"input": "868514",
"output": "248146 248148"
},
{
"input": "868513",
"output": "248146 248148"
},
{
"input": "868512",
"output": "248146 248147"
},
{
"input": "868511",
"output": "248146 248146"
},
{
"input": "123413",
"output": "35260 35262"
},
{
"input": "123412",
"output": "35260 35262"
},
{
"input": "123411",
"output": "35260 35261"
},
{
"input": "123410",
"output": "35260 35260"
},
{
"input": "123409",
"output": "35259 35260"
},
{
"input": "123408",
"output": "35258 35260"
},
{
"input": "123407",
"output": "35258 35260"
},
{
"input": "123406",
"output": "35258 35260"
},
{
"input": "123405",
"output": "35258 35260"
},
{
"input": "123404",
"output": "35258 35259"
},
{
"input": "123403",
"output": "35258 35258"
},
{
"input": "123402",
"output": "35257 35258"
},
{
"input": "123401",
"output": "35256 35258"
},
{
"input": "123400",
"output": "35256 35258"
},
{
"input": "123399",
"output": "35256 35258"
},
{
"input": "123398",
"output": "35256 35258"
},
{
"input": "123397",
"output": "35256 35257"
},
{
"input": "123396",
"output": "35256 35256"
},
{
"input": "123395",
"output": "35255 35256"
},
{
"input": "123394",
"output": "35254 35256"
},
{
"input": "123393",
"output": "35254 35256"
},
{
"input": "15",
"output": "4 5"
}
] | 1,647,765,343 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 31 | 0 | n=int(input())
mi=n//7
ma=n//7
mi*=2
ma*=2
if n%7==1:
ma+=1
elif n%7>=2:
ma+=2
print(mi,ma)
| Title: Holidays
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
On the planet Mars a year lasts exactly *n* days (there are no leap years on Mars). But Martians have the same weeks as earthlings — 5 work days and then 2 days off. Your task is to determine the minimum possible and the maximum possible number of days off per year on Mars.
Input Specification:
The first line of the input contains a positive integer *n* (1<=≤<=*n*<=≤<=1<=000<=000) — the number of days in a year on Mars.
Output Specification:
Print two integers — the minimum possible and the maximum possible number of days off per year on Mars.
Demo Input:
['14\n', '2\n']
Demo Output:
['4 4\n', '0 2\n']
Note:
In the first sample there are 14 days in a year on Mars, and therefore independently of the day a year starts with there will be exactly 4 days off .
In the second sample there are only 2 days in a year on Mars, and they can both be either work days or days off. | ```python
n=int(input())
mi=n//7
ma=n//7
mi*=2
ma*=2
if n%7==1:
ma+=1
elif n%7>=2:
ma+=2
print(mi,ma)
``` | 0 | |
189 | A | Cut Ribbon | PROGRAMMING | 1,300 | [
"brute force",
"dp"
] | null | null | Polycarpus has a ribbon, its length is *n*. He wants to cut the ribbon in a way that fulfils the following two conditions:
- After the cutting each ribbon piece should have length *a*, *b* or *c*. - After the cutting the number of ribbon pieces should be maximum.
Help Polycarpus and find the number of ribbon pieces after the required cutting. | The first line contains four space-separated integers *n*, *a*, *b* and *c* (1<=≤<=*n*,<=*a*,<=*b*,<=*c*<=≤<=4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers *a*, *b* and *c* can coincide. | Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists. | [
"5 5 3 2\n",
"7 5 5 2\n"
] | [
"2\n",
"2\n"
] | In the first example Polycarpus can cut the ribbon in such way: the first piece has length 2, the second piece has length 3.
In the second example Polycarpus can cut the ribbon in such way: the first piece has length 5, the second piece has length 2. | 500 | [
{
"input": "5 5 3 2",
"output": "2"
},
{
"input": "7 5 5 2",
"output": "2"
},
{
"input": "4 4 4 4",
"output": "1"
},
{
"input": "1 1 1 1",
"output": "1"
},
{
"input": "4000 1 2 3",
"output": "4000"
},
{
"input": "4000 3 4 5",
"output": "1333"
},
{
"input": "10 3 4 5",
"output": "3"
},
{
"input": "100 23 15 50",
"output": "2"
},
{
"input": "3119 3515 1021 7",
"output": "11"
},
{
"input": "918 102 1327 1733",
"output": "9"
},
{
"input": "3164 42 430 1309",
"output": "15"
},
{
"input": "3043 317 1141 2438",
"output": "7"
},
{
"input": "26 1 772 2683",
"output": "26"
},
{
"input": "370 2 1 15",
"output": "370"
},
{
"input": "734 12 6 2",
"output": "367"
},
{
"input": "418 18 14 17",
"output": "29"
},
{
"input": "18 16 28 9",
"output": "2"
},
{
"input": "14 6 2 17",
"output": "7"
},
{
"input": "29 27 18 2",
"output": "2"
},
{
"input": "29 12 7 10",
"output": "3"
},
{
"input": "27 23 4 3",
"output": "9"
},
{
"input": "5 14 5 2",
"output": "1"
},
{
"input": "5 17 26 5",
"output": "1"
},
{
"input": "9 1 10 3",
"output": "9"
},
{
"input": "2 19 15 1",
"output": "2"
},
{
"input": "4 6 4 9",
"output": "1"
},
{
"input": "10 6 2 9",
"output": "5"
},
{
"input": "2 2 9 6",
"output": "1"
},
{
"input": "6 2 4 1",
"output": "6"
},
{
"input": "27 24 5 27",
"output": "1"
},
{
"input": "2683 83 26 2709",
"output": "101"
},
{
"input": "728 412 789 158",
"output": "3"
},
{
"input": "3964 4 2916 176",
"output": "991"
},
{
"input": "3399 2035 2 3334",
"output": "683"
},
{
"input": "3455 244 3301 3",
"output": "991"
},
{
"input": "595 2263 3625 1",
"output": "595"
},
{
"input": "4000 1 1 1",
"output": "4000"
},
{
"input": "3999 2 2 3999",
"output": "1"
},
{
"input": "25 6 8 11",
"output": "3"
},
{
"input": "4000 500 1000 2000",
"output": "8"
},
{
"input": "53 10 11 23",
"output": "5"
},
{
"input": "100 100 1 1",
"output": "100"
},
{
"input": "17 3 4 10",
"output": "5"
},
{
"input": "413 101 102 105",
"output": "4"
},
{
"input": "490 4 49 50",
"output": "111"
},
{
"input": "3999 2 3 3",
"output": "1999"
},
{
"input": "8 3 8 4",
"output": "2"
},
{
"input": "5 1 3 3",
"output": "5"
},
{
"input": "100 3 17 22",
"output": "27"
},
{
"input": "4000 2 3 4",
"output": "2000"
},
{
"input": "4000 3 3 5",
"output": "1332"
},
{
"input": "13 4 6 7",
"output": "2"
},
{
"input": "4000 5 2 2",
"output": "2000"
},
{
"input": "3999 2 2 3",
"output": "1999"
},
{
"input": "4000 33 7 3333",
"output": "564"
},
{
"input": "60 33 20 9",
"output": "4"
},
{
"input": "100 9 11 99",
"output": "10"
},
{
"input": "2009 6 8 9",
"output": "334"
}
] | 1,686,372,537 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 4 | 171 | 1,536,000 | def max_ribbon_pieces(n, a, b, c):
return ans
n, a, b, c = map(int, input().split())
# Calculate the maximum number of ribbon pieces
ans = 0
for i in range(n + 1):
for j in range(n + 1):
rem = n - i * a - j * b
if rem % c == 0:
k = rem // c
ans = max(ans, i + j + k)
print(ans)
| Title: Cut Ribbon
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus has a ribbon, its length is *n*. He wants to cut the ribbon in a way that fulfils the following two conditions:
- After the cutting each ribbon piece should have length *a*, *b* or *c*. - After the cutting the number of ribbon pieces should be maximum.
Help Polycarpus and find the number of ribbon pieces after the required cutting.
Input Specification:
The first line contains four space-separated integers *n*, *a*, *b* and *c* (1<=≤<=*n*,<=*a*,<=*b*,<=*c*<=≤<=4000) — the length of the original ribbon and the acceptable lengths of the ribbon pieces after the cutting, correspondingly. The numbers *a*, *b* and *c* can coincide.
Output Specification:
Print a single number — the maximum possible number of ribbon pieces. It is guaranteed that at least one correct ribbon cutting exists.
Demo Input:
['5 5 3 2\n', '7 5 5 2\n']
Demo Output:
['2\n', '2\n']
Note:
In the first example Polycarpus can cut the ribbon in such way: the first piece has length 2, the second piece has length 3.
In the second example Polycarpus can cut the ribbon in such way: the first piece has length 5, the second piece has length 2. | ```python
def max_ribbon_pieces(n, a, b, c):
return ans
n, a, b, c = map(int, input().split())
# Calculate the maximum number of ribbon pieces
ans = 0
for i in range(n + 1):
for j in range(n + 1):
rem = n - i * a - j * b
if rem % c == 0:
k = rem // c
ans = max(ans, i + j + k)
print(ans)
``` | 0 | |
527 | C | Glass Carving | PROGRAMMING | 1,500 | [
"binary search",
"data structures",
"implementation"
] | null | null | Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular *w* mm <=×<= *h* mm sheet of glass, a diamond glass cutter and lots of enthusiasm. What he lacks is understanding of what to carve and how.
In order not to waste time, he decided to practice the technique of carving. To do this, he makes vertical and horizontal cuts through the entire sheet. This process results in making smaller rectangular fragments of glass. Leonid does not move the newly made glass fragments. In particular, a cut divides each fragment of glass that it goes through into smaller fragments.
After each cut Leonid tries to determine what area the largest of the currently available glass fragments has. Since there appear more and more fragments, this question takes him more and more time and distracts him from the fascinating process.
Leonid offers to divide the labor — he will cut glass, and you will calculate the area of the maximum fragment after each cut. Do you agree? | The first line contains three integers *w*,<=*h*,<=*n* (2<=≤<=*w*,<=*h*<=≤<=200<=000, 1<=≤<=*n*<=≤<=200<=000).
Next *n* lines contain the descriptions of the cuts. Each description has the form *H* *y* or *V* *x*. In the first case Leonid makes the horizontal cut at the distance *y* millimeters (1<=≤<=*y*<=≤<=*h*<=-<=1) from the lower edge of the original sheet of glass. In the second case Leonid makes a vertical cut at distance *x* (1<=≤<=*x*<=≤<=*w*<=-<=1) millimeters from the left edge of the original sheet of glass. It is guaranteed that Leonid won't make two identical cuts. | After each cut print on a single line the area of the maximum available glass fragment in mm2. | [
"4 3 4\nH 2\nV 2\nV 3\nV 1\n",
"7 6 5\nH 4\nV 3\nV 5\nH 2\nV 1\n"
] | [
"8\n4\n4\n2\n",
"28\n16\n12\n6\n4\n"
] | Picture for the first sample test: | 1,500 | [
{
"input": "4 3 4\nH 2\nV 2\nV 3\nV 1",
"output": "8\n4\n4\n2"
},
{
"input": "7 6 5\nH 4\nV 3\nV 5\nH 2\nV 1",
"output": "28\n16\n12\n6\n4"
},
{
"input": "2 2 1\nV 1",
"output": "2"
},
{
"input": "2 2 1\nH 1",
"output": "2"
},
{
"input": "2 2 2\nV 1\nH 1",
"output": "2\n1"
},
{
"input": "2 2 2\nH 1\nV 1",
"output": "2\n1"
},
{
"input": "10 10 10\nV 6\nH 8\nV 4\nV 8\nH 2\nH 5\nV 9\nH 7\nH 3\nV 7",
"output": "60\n48\n32\n32\n24\n12\n12\n12\n8\n8"
},
{
"input": "5 15 10\nH 8\nH 9\nV 1\nH 2\nH 6\nH 4\nH 1\nV 2\nH 13\nV 3",
"output": "40\n40\n32\n24\n24\n24\n24\n18\n12\n8"
},
{
"input": "15 5 10\nV 13\nV 10\nV 3\nH 2\nV 9\nV 7\nV 2\nH 1\nV 4\nH 3",
"output": "65\n50\n35\n21\n18\n12\n12\n12\n9\n6"
},
{
"input": "2 3 1\nH 1",
"output": "4"
},
{
"input": "200000 200000 1\nH 1",
"output": "39999800000"
},
{
"input": "2 4 1\nH 2",
"output": "4"
}
] | 1,585,267,018 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 61 | 307,200 | w, h, n = map(int, input().split())
h_lengths = [h]
w_lengths = [w]
h_lengths_record = [h]
w_lengths_record = [w]
def get_index_split(arr, start, stop, dist):
mid = (start + stop) // 2
if mid == 0:
return 0
if dist > arr[mid - 1] and dist < arr[mid]:
return mid
return get_index_split(arr, start, mid, dist)
def get_new_lengths(h_lengths, h_lengths_record, index):
new_h_lengths = [-1 for j in range(len(h_lengths) + 1)]
new_h_lengths_record = [-1 for j in range(len(h_lengths) + 1)]
for j in range(index):
new_h_lengths[j] = h_lengths[j]
new_h_lengths_record[j] = h_lengths_record[j]
if index > 0:
new_h_lengths[index] = dist - h_lengths_record[index - 1]
else:
new_h_lengths[index] = dist
new_h_lengths[index + 1] = h_lengths_record[index] - dist
new_h_lengths_record[index + 1] = h_lengths_record[index]
if index > 0:
new_h_lengths_record[index] = h_lengths_record[index - 1] + dist
else:
new_h_lengths_record[index] = dist
for j in range(index + 2, len(new_h_lengths)):
new_h_lengths[j] = h_lengths[j - 1]
new_h_lengths_record[j] = h_lengths_record[j - 1]
return new_h_lengths, new_h_lengths_record
for i in range(n):
cut, dist = input().split()
dist = int(dist)
if cut == 'H':
index = get_index_split(h_lengths_record, 0, len(h_lengths_record), dist)
h_lengths, h_lengths_record = get_new_lengths(h_lengths, h_lengths_record, index)
if cut == 'V':
index = get_index_split(w_lengths_record, 0, len(w_lengths_record), dist)
w_lengths, w_lengths_record = get_new_lengths(w_lengths, w_lengths_record, index)
print(index, h_lengths, w_lengths)
print(max(h_lengths) * max(w_lengths))
| Title: Glass Carving
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Leonid wants to become a glass carver (the person who creates beautiful artworks by cutting the glass). He already has a rectangular *w* mm <=×<= *h* mm sheet of glass, a diamond glass cutter and lots of enthusiasm. What he lacks is understanding of what to carve and how.
In order not to waste time, he decided to practice the technique of carving. To do this, he makes vertical and horizontal cuts through the entire sheet. This process results in making smaller rectangular fragments of glass. Leonid does not move the newly made glass fragments. In particular, a cut divides each fragment of glass that it goes through into smaller fragments.
After each cut Leonid tries to determine what area the largest of the currently available glass fragments has. Since there appear more and more fragments, this question takes him more and more time and distracts him from the fascinating process.
Leonid offers to divide the labor — he will cut glass, and you will calculate the area of the maximum fragment after each cut. Do you agree?
Input Specification:
The first line contains three integers *w*,<=*h*,<=*n* (2<=≤<=*w*,<=*h*<=≤<=200<=000, 1<=≤<=*n*<=≤<=200<=000).
Next *n* lines contain the descriptions of the cuts. Each description has the form *H* *y* or *V* *x*. In the first case Leonid makes the horizontal cut at the distance *y* millimeters (1<=≤<=*y*<=≤<=*h*<=-<=1) from the lower edge of the original sheet of glass. In the second case Leonid makes a vertical cut at distance *x* (1<=≤<=*x*<=≤<=*w*<=-<=1) millimeters from the left edge of the original sheet of glass. It is guaranteed that Leonid won't make two identical cuts.
Output Specification:
After each cut print on a single line the area of the maximum available glass fragment in mm2.
Demo Input:
['4 3 4\nH 2\nV 2\nV 3\nV 1\n', '7 6 5\nH 4\nV 3\nV 5\nH 2\nV 1\n']
Demo Output:
['8\n4\n4\n2\n', '28\n16\n12\n6\n4\n']
Note:
Picture for the first sample test: | ```python
w, h, n = map(int, input().split())
h_lengths = [h]
w_lengths = [w]
h_lengths_record = [h]
w_lengths_record = [w]
def get_index_split(arr, start, stop, dist):
mid = (start + stop) // 2
if mid == 0:
return 0
if dist > arr[mid - 1] and dist < arr[mid]:
return mid
return get_index_split(arr, start, mid, dist)
def get_new_lengths(h_lengths, h_lengths_record, index):
new_h_lengths = [-1 for j in range(len(h_lengths) + 1)]
new_h_lengths_record = [-1 for j in range(len(h_lengths) + 1)]
for j in range(index):
new_h_lengths[j] = h_lengths[j]
new_h_lengths_record[j] = h_lengths_record[j]
if index > 0:
new_h_lengths[index] = dist - h_lengths_record[index - 1]
else:
new_h_lengths[index] = dist
new_h_lengths[index + 1] = h_lengths_record[index] - dist
new_h_lengths_record[index + 1] = h_lengths_record[index]
if index > 0:
new_h_lengths_record[index] = h_lengths_record[index - 1] + dist
else:
new_h_lengths_record[index] = dist
for j in range(index + 2, len(new_h_lengths)):
new_h_lengths[j] = h_lengths[j - 1]
new_h_lengths_record[j] = h_lengths_record[j - 1]
return new_h_lengths, new_h_lengths_record
for i in range(n):
cut, dist = input().split()
dist = int(dist)
if cut == 'H':
index = get_index_split(h_lengths_record, 0, len(h_lengths_record), dist)
h_lengths, h_lengths_record = get_new_lengths(h_lengths, h_lengths_record, index)
if cut == 'V':
index = get_index_split(w_lengths_record, 0, len(w_lengths_record), dist)
w_lengths, w_lengths_record = get_new_lengths(w_lengths, w_lengths_record, index)
print(index, h_lengths, w_lengths)
print(max(h_lengths) * max(w_lengths))
``` | 0 | |
61 | A | Ultra-Fast Mathematician | PROGRAMMING | 800 | [
"implementation"
] | A. Ultra-Fast Mathematician | 2 | 256 | Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate. | There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100. | Write one line — the corresponding answer. Do not omit the leading 0s. | [
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] | [
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] | none | 500 | [
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,511,259,443 | 2,147,483,647 | Python 3 | OK | TESTS | 102 | 62 | 0 | fstLine = input()
scndLine = input()
arr = []
for i,j in zip(fstLine,scndLine):
arr.append('01'[i != j])
print(''.join(arr))
'''
Sample Input:
1010100
0100101
Sample Output:
1110001
'''
| Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none | ```python
fstLine = input()
scndLine = input()
arr = []
for i,j in zip(fstLine,scndLine):
arr.append('01'[i != j])
print(''.join(arr))
'''
Sample Input:
1010100
0100101
Sample Output:
1110001
'''
``` | 3.9845 |
991 | C | Candies | PROGRAMMING | 1,500 | [
"binary search",
"implementation"
] | null | null | After passing a test, Vasya got himself a box of $n$ candies. He decided to eat an equal amount of candies each morning until there are no more candies. However, Petya also noticed the box and decided to get some candies for himself.
This means the process of eating candies is the following: in the beginning Vasya chooses a single integer $k$, same for all days. After that, in the morning he eats $k$ candies from the box (if there are less than $k$ candies in the box, he eats them all), then in the evening Petya eats $10\%$ of the candies remaining in the box. If there are still candies left in the box, the process repeats — next day Vasya eats $k$ candies again, and Petya — $10\%$ of the candies left in a box, and so on.
If the amount of candies in the box is not divisible by $10$, Petya rounds the amount he takes from the box down. For example, if there were $97$ candies in the box, Petya would eat only $9$ of them. In particular, if there are less than $10$ candies in a box, Petya won't eat any at all.
Your task is to find out the minimal amount of $k$ that can be chosen by Vasya so that he would eat at least half of the $n$ candies he initially got. Note that the number $k$ must be integer. | The first line contains a single integer $n$ ($1 \leq n \leq 10^{18}$) — the initial amount of candies in the box. | Output a single integer — the minimal amount of $k$ that would allow Vasya to eat at least half of candies he got. | [
"68\n"
] | [
"3\n"
] | In the sample, the amount of candies, with $k=3$, would change in the following way (Vasya eats first):
$68 \to 65 \to 59 \to 56 \to 51 \to 48 \to 44 \to 41 \\ \to 37 \to 34 \to 31 \to 28 \to 26 \to 23 \to 21 \to 18 \to 17 \to 14 \\ \to 13 \to 10 \to 9 \to 6 \to 6 \to 3 \to 3 \to 0$.
In total, Vasya would eat $39$ candies, while Petya — $29$. | 1,250 | [
{
"input": "68",
"output": "3"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "1"
},
{
"input": "42",
"output": "1"
},
{
"input": "43",
"output": "2"
},
{
"input": "756",
"output": "29"
},
{
"input": "999999972",
"output": "39259423"
},
{
"input": "999999973",
"output": "39259424"
},
{
"input": "1000000000000000000",
"output": "39259424579862572"
},
{
"input": "6",
"output": "1"
},
{
"input": "3",
"output": "1"
},
{
"input": "4",
"output": "1"
},
{
"input": "5",
"output": "1"
},
{
"input": "66",
"output": "2"
},
{
"input": "67",
"output": "3"
},
{
"input": "1000",
"output": "39"
},
{
"input": "10000",
"output": "392"
},
{
"input": "100500",
"output": "3945"
},
{
"input": "1000000",
"output": "39259"
},
{
"input": "10000000",
"output": "392594"
},
{
"input": "100000000",
"output": "3925942"
},
{
"input": "123456789",
"output": "4846842"
},
{
"input": "543212345",
"output": "21326204"
},
{
"input": "505050505",
"output": "19827992"
},
{
"input": "777777777",
"output": "30535108"
},
{
"input": "888888871",
"output": "34897266"
},
{
"input": "1000000000",
"output": "39259424"
},
{
"input": "999999999999999973",
"output": "39259424579862572"
},
{
"input": "999999999999999998",
"output": "39259424579862572"
},
{
"input": "999999999999999999",
"output": "39259424579862573"
},
{
"input": "100000000000000000",
"output": "3925942457986257"
},
{
"input": "540776028375043656",
"output": "21230555700587649"
},
{
"input": "210364830044445976",
"output": "8258802179385535"
},
{
"input": "297107279239074256",
"output": "11664260821414605"
},
{
"input": "773524766411950187",
"output": "30368137227605772"
},
{
"input": "228684941775227220",
"output": "8978039224174797"
},
{
"input": "878782039723446310",
"output": "34500477210660436"
},
{
"input": "615090701338187389",
"output": "24148106998961343"
},
{
"input": "325990422297859188",
"output": "12798196397960353"
},
{
"input": "255163492355051023",
"output": "10017571883647466"
},
{
"input": "276392003308849171",
"output": "10850991008380891"
},
{
"input": "601",
"output": "23"
},
{
"input": "983",
"output": "38"
},
{
"input": "729",
"output": "29"
},
{
"input": "70",
"output": "3"
},
{
"input": "703",
"output": "28"
},
{
"input": "257",
"output": "10"
},
{
"input": "526",
"output": "20"
},
{
"input": "466",
"output": "18"
},
{
"input": "738",
"output": "29"
},
{
"input": "116",
"output": "5"
},
{
"input": "888888888888888887",
"output": "34897266293211176"
},
{
"input": "888888888888888888",
"output": "34897266293211176"
},
{
"input": "888888888888888889",
"output": "34897266293211176"
},
{
"input": "999999999999999969",
"output": "39259424579862571"
},
{
"input": "999999999999999970",
"output": "39259424579862571"
},
{
"input": "999999999999999971",
"output": "39259424579862572"
},
{
"input": "999999999999999943",
"output": "39259424579862571"
},
{
"input": "999999999999999944",
"output": "39259424579862570"
},
{
"input": "999999999999999945",
"output": "39259424579862571"
},
{
"input": "999999999999999917",
"output": "39259424579862570"
},
{
"input": "999999999999999918",
"output": "39259424579862569"
},
{
"input": "999999999999999919",
"output": "39259424579862570"
},
{
"input": "99999999999999957",
"output": "3925942457986255"
},
{
"input": "99999999999999958",
"output": "3925942457986255"
},
{
"input": "99999999999999959",
"output": "3925942457986256"
},
{
"input": "888888888888888853",
"output": "34897266293211174"
},
{
"input": "888888888888888854",
"output": "34897266293211174"
},
{
"input": "888888888888888855",
"output": "34897266293211175"
}
] | 1,583,245,520 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 1 | 62 | 0 | n = int(input().strip())
k = 0
for i in range(1,int(n/2)):
s = n
a = 0
b = 0
while s > 0:
s -= i
a += i
if s//10 > 0:
b += s//10
s -= s//10
if a >= n / 2 and b < n/2:
k = i
print(k)
exit(0) | Title: Candies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After passing a test, Vasya got himself a box of $n$ candies. He decided to eat an equal amount of candies each morning until there are no more candies. However, Petya also noticed the box and decided to get some candies for himself.
This means the process of eating candies is the following: in the beginning Vasya chooses a single integer $k$, same for all days. After that, in the morning he eats $k$ candies from the box (if there are less than $k$ candies in the box, he eats them all), then in the evening Petya eats $10\%$ of the candies remaining in the box. If there are still candies left in the box, the process repeats — next day Vasya eats $k$ candies again, and Petya — $10\%$ of the candies left in a box, and so on.
If the amount of candies in the box is not divisible by $10$, Petya rounds the amount he takes from the box down. For example, if there were $97$ candies in the box, Petya would eat only $9$ of them. In particular, if there are less than $10$ candies in a box, Petya won't eat any at all.
Your task is to find out the minimal amount of $k$ that can be chosen by Vasya so that he would eat at least half of the $n$ candies he initially got. Note that the number $k$ must be integer.
Input Specification:
The first line contains a single integer $n$ ($1 \leq n \leq 10^{18}$) — the initial amount of candies in the box.
Output Specification:
Output a single integer — the minimal amount of $k$ that would allow Vasya to eat at least half of candies he got.
Demo Input:
['68\n']
Demo Output:
['3\n']
Note:
In the sample, the amount of candies, with $k=3$, would change in the following way (Vasya eats first):
$68 \to 65 \to 59 \to 56 \to 51 \to 48 \to 44 \to 41 \\ \to 37 \to 34 \to 31 \to 28 \to 26 \to 23 \to 21 \to 18 \to 17 \to 14 \\ \to 13 \to 10 \to 9 \to 6 \to 6 \to 3 \to 3 \to 0$.
In total, Vasya would eat $39$ candies, while Petya — $29$. | ```python
n = int(input().strip())
k = 0
for i in range(1,int(n/2)):
s = n
a = 0
b = 0
while s > 0:
s -= i
a += i
if s//10 > 0:
b += s//10
s -= s//10
if a >= n / 2 and b < n/2:
k = i
print(k)
exit(0)
``` | 0 | |
69 | A | Young Physicist | PROGRAMMING | 1,000 | [
"implementation",
"math"
] | A. Young Physicist | 2 | 256 | A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces. | The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100). | Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not. | [
"3\n4 1 7\n-2 4 -1\n1 -5 -3\n",
"3\n3 -1 7\n-5 2 -4\n2 -1 -3\n"
] | [
"NO",
"YES"
] | none | 500 | [
{
"input": "3\n4 1 7\n-2 4 -1\n1 -5 -3",
"output": "NO"
},
{
"input": "3\n3 -1 7\n-5 2 -4\n2 -1 -3",
"output": "YES"
},
{
"input": "10\n21 32 -46\n43 -35 21\n42 2 -50\n22 40 20\n-27 -9 38\n-4 1 1\n-40 6 -31\n-13 -2 34\n-21 34 -12\n-32 -29 41",
"output": "NO"
},
{
"input": "10\n25 -33 43\n-27 -42 28\n-35 -20 19\n41 -42 -1\n49 -39 -4\n-49 -22 7\n-19 29 41\n8 -27 -43\n8 34 9\n-11 -3 33",
"output": "NO"
},
{
"input": "10\n-6 21 18\n20 -11 -8\n37 -11 41\n-5 8 33\n29 23 32\n30 -33 -11\n39 -49 -36\n28 34 -49\n22 29 -34\n-18 -6 7",
"output": "NO"
},
{
"input": "10\n47 -2 -27\n0 26 -14\n5 -12 33\n2 18 3\n45 -30 -49\n4 -18 8\n-46 -44 -41\n-22 -10 -40\n-35 -21 26\n33 20 38",
"output": "NO"
},
{
"input": "13\n-3 -36 -46\n-11 -50 37\n42 -11 -15\n9 42 44\n-29 -12 24\n3 9 -40\n-35 13 50\n14 43 18\n-13 8 24\n-48 -15 10\n50 9 -50\n21 0 -50\n0 0 -6",
"output": "YES"
},
{
"input": "14\n43 23 17\n4 17 44\n5 -5 -16\n-43 -7 -6\n47 -48 12\n50 47 -45\n2 14 43\n37 -30 15\n4 -17 -11\n17 9 -45\n-50 -3 -8\n-50 0 0\n-50 0 0\n-16 0 0",
"output": "YES"
},
{
"input": "13\n29 49 -11\n38 -11 -20\n25 1 -40\n-11 28 11\n23 -19 1\n45 -41 -17\n-3 0 -19\n-13 -33 49\n-30 0 28\n34 17 45\n-50 9 -27\n-50 0 0\n-37 0 0",
"output": "YES"
},
{
"input": "12\n3 28 -35\n-32 -44 -17\n9 -25 -6\n-42 -22 20\n-19 15 38\n-21 38 48\n-1 -37 -28\n-10 -13 -50\n-5 21 29\n34 28 50\n50 11 -49\n34 0 0",
"output": "YES"
},
{
"input": "37\n-64 -79 26\n-22 59 93\n-5 39 -12\n77 -9 76\n55 -86 57\n83 100 -97\n-70 94 84\n-14 46 -94\n26 72 35\n14 78 -62\n17 82 92\n-57 11 91\n23 15 92\n-80 -1 1\n12 39 18\n-23 -99 -75\n-34 50 19\n-39 84 -7\n45 -30 -39\n-60 49 37\n45 -16 -72\n33 -51 -56\n-48 28 5\n97 91 88\n45 -82 -11\n-21 -15 -90\n-53 73 -26\n-74 85 -90\n-40 23 38\n100 -13 49\n32 -100 -100\n0 -100 -70\n0 -100 0\n0 -100 0\n0 -100 0\n0 -100 0\n0 -37 0",
"output": "YES"
},
{
"input": "4\n68 3 100\n68 21 -100\n-100 -24 0\n-36 0 0",
"output": "YES"
},
{
"input": "33\n-1 -46 -12\n45 -16 -21\n-11 45 -21\n-60 -42 -93\n-22 -45 93\n37 96 85\n-76 26 83\n-4 9 55\n7 -52 -9\n66 8 -85\n-100 -54 11\n-29 59 74\n-24 12 2\n-56 81 85\n-92 69 -52\n-26 -97 91\n54 59 -51\n58 21 -57\n7 68 56\n-47 -20 -51\n-59 77 -13\n-85 27 91\n79 60 -56\n66 -80 5\n21 -99 42\n-31 -29 98\n66 93 76\n-49 45 61\n100 -100 -100\n100 -100 -100\n66 -75 -100\n0 0 -100\n0 0 -87",
"output": "YES"
},
{
"input": "3\n1 2 3\n3 2 1\n0 0 0",
"output": "NO"
},
{
"input": "2\n5 -23 12\n0 0 0",
"output": "NO"
},
{
"input": "1\n0 0 0",
"output": "YES"
},
{
"input": "1\n1 -2 0",
"output": "NO"
},
{
"input": "2\n-23 77 -86\n23 -77 86",
"output": "YES"
},
{
"input": "26\n86 7 20\n-57 -64 39\n-45 6 -93\n-44 -21 100\n-11 -49 21\n73 -71 -80\n-2 -89 56\n-65 -2 7\n5 14 84\n57 41 13\n-12 69 54\n40 -25 27\n-17 -59 0\n64 -91 -30\n-53 9 42\n-54 -8 14\n-35 82 27\n-48 -59 -80\n88 70 79\n94 57 97\n44 63 25\n84 -90 -40\n-100 100 -100\n-92 100 -100\n0 10 -100\n0 0 -82",
"output": "YES"
},
{
"input": "42\n11 27 92\n-18 -56 -57\n1 71 81\n33 -92 30\n82 83 49\n-87 -61 -1\n-49 45 49\n73 26 15\n-22 22 -77\n29 -93 87\n-68 44 -90\n-4 -84 20\n85 67 -6\n-39 26 77\n-28 -64 20\n65 -97 24\n-72 -39 51\n35 -75 -91\n39 -44 -8\n-25 -27 -57\n91 8 -46\n-98 -94 56\n94 -60 59\n-9 -95 18\n-53 -37 98\n-8 -94 -84\n-52 55 60\n15 -14 37\n65 -43 -25\n94 12 66\n-8 -19 -83\n29 81 -78\n-58 57 33\n24 86 -84\n-53 32 -88\n-14 7 3\n89 97 -53\n-5 -28 -91\n-100 100 -6\n-84 100 0\n0 100 0\n0 70 0",
"output": "YES"
},
{
"input": "3\n96 49 -12\n2 -66 28\n-98 17 -16",
"output": "YES"
},
{
"input": "5\n70 -46 86\n-100 94 24\n-27 63 -63\n57 -100 -47\n0 -11 0",
"output": "YES"
},
{
"input": "18\n-86 -28 70\n-31 -89 42\n31 -48 -55\n95 -17 -43\n24 -95 -85\n-21 -14 31\n68 -18 81\n13 31 60\n-15 28 99\n-42 15 9\n28 -61 -62\n-16 71 29\n-28 75 -48\n-77 -67 36\n-100 83 89\n100 100 -100\n57 34 -100\n0 0 -53",
"output": "YES"
},
{
"input": "44\n52 -54 -29\n-82 -5 -94\n-54 43 43\n91 16 71\n7 80 -91\n3 15 29\n-99 -6 -77\n-3 -77 -64\n73 67 34\n25 -10 -18\n-29 91 63\n-72 86 -16\n-68 85 -81\n-3 36 44\n-74 -14 -80\n34 -96 -97\n-76 -78 -33\n-24 44 -58\n98 12 77\n95 -63 -6\n-51 3 -90\n-92 -10 72\n7 3 -68\n57 -53 71\n29 57 -48\n35 -60 10\n79 -70 -61\n-20 77 55\n-86 -15 -35\n84 -88 -18\n100 -42 77\n-20 46 8\n-41 -43 -65\n38 -98 -23\n-100 65 45\n-7 -91 -63\n46 88 -85\n48 59 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 100\n0 0 1",
"output": "YES"
},
{
"input": "18\n-14 -64 -91\n-8 -66 -86\n-23 92 -40\n6 -3 -53\n57 41 78\n-79 42 -22\n-88 -17 45\n4 -45 44\n83 -18 -25\n34 86 -92\n75 -30 12\n44 99 11\n-67 -13 72\n22 83 -56\n-37 71 72\n-9 -100 100\n0 -100 31\n0 -58 0",
"output": "YES"
},
{
"input": "23\n-70 37 78\n42 84 6\n28 -94 -24\n-49 76 95\n-67 18 84\n-53 78 -5\n65 -63 -64\n-66 -64 -69\n81 -93 95\n10 52 -79\n-89 -61 -64\n-64 -47 43\n-81 -35 55\n80 82 73\n-60 -81 -18\n49 -10 -19\n-58 70 12\n-24 -15 -93\n98 -93 -54\n-28 -75 11\n100 100 -63\n100 100 0\n56 34 0",
"output": "YES"
},
{
"input": "3\n0 2 -2\n1 -1 3\n-3 0 0",
"output": "NO"
}
] | 1,693,230,649 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 92 | 0 | T = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
C = list(map(int, input().split()))
g = A[0]+B[0]+C[0]
f = A[1]+B[1]+C[1]
e = A[2]+B[2]+C[2]
if(g==0 and f==0 and e==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
T = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
C = list(map(int, input().split()))
g = A[0]+B[0]+C[0]
f = A[1]+B[1]+C[1]
e = A[2]+B[2]+C[2]
if(g==0 and f==0 and e==0):
print("YES")
else:
print("NO")
``` | 0 |
976 | A | Minimum Binary Number | PROGRAMMING | 800 | [
"implementation"
] | null | null | String can be called correct if it consists of characters "0" and "1" and there are no redundant leading zeroes. Here are some examples: "0", "10", "1001".
You are given a correct string *s*.
You can perform two different operations on this string:
1. swap any pair of adjacent characters (for example, "101" "110"); 1. replace "11" with "1" (for example, "110" "10").
Let *val*(*s*) be such a number that *s* is its binary representation.
Correct string *a* is less than some other correct string *b* iff *val*(*a*)<=<<=*val*(*b*).
Your task is to find the minimum correct string that you can obtain from the given one using the operations described above. You can use these operations any number of times in any order (or even use no operations at all). | The first line contains integer number *n* (1<=≤<=*n*<=≤<=100) — the length of string *s*.
The second line contains the string *s* consisting of characters "0" and "1". It is guaranteed that the string *s* is correct. | Print one string — the minimum correct string that you can obtain from the given one. | [
"4\n1001\n",
"1\n1\n"
] | [
"100\n",
"1\n"
] | In the first example you can obtain the answer by the following sequence of operations: "1001" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1010" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1100" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "100".
In the second example you can't obtain smaller answer no matter what operations you use. | 0 | [
{
"input": "4\n1001",
"output": "100"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "100\n1110111100001111011111111010110011111111011110000111101101011100110110001011000000101010110101011100",
"output": "1000000000000000000000000000000000000000"
},
{
"input": "100\n1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111",
"output": "1"
},
{
"input": "100\n1111111111111111111111111111111111111111111111111111111110111111111111111111111111111111111111111111",
"output": "10"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "8\n10101010",
"output": "10000"
},
{
"input": "2\n10",
"output": "10"
},
{
"input": "3\n111",
"output": "1"
},
{
"input": "5\n11100",
"output": "100"
},
{
"input": "2\n11",
"output": "1"
},
{
"input": "3\n110",
"output": "10"
},
{
"input": "50\n10010010000000000000000000000000000000001000000000",
"output": "10000000000000000000000000000000000000000000000"
}
] | 1,640,790,679 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 6 | 31 | 0 | n = int(input())
s = input()
count1 = 0
for i in s:
if(i=='0'):
count1+=1
print('1'+'0'*count1) | Title: Minimum Binary Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
String can be called correct if it consists of characters "0" and "1" and there are no redundant leading zeroes. Here are some examples: "0", "10", "1001".
You are given a correct string *s*.
You can perform two different operations on this string:
1. swap any pair of adjacent characters (for example, "101" "110"); 1. replace "11" with "1" (for example, "110" "10").
Let *val*(*s*) be such a number that *s* is its binary representation.
Correct string *a* is less than some other correct string *b* iff *val*(*a*)<=<<=*val*(*b*).
Your task is to find the minimum correct string that you can obtain from the given one using the operations described above. You can use these operations any number of times in any order (or even use no operations at all).
Input Specification:
The first line contains integer number *n* (1<=≤<=*n*<=≤<=100) — the length of string *s*.
The second line contains the string *s* consisting of characters "0" and "1". It is guaranteed that the string *s* is correct.
Output Specification:
Print one string — the minimum correct string that you can obtain from the given one.
Demo Input:
['4\n1001\n', '1\n1\n']
Demo Output:
['100\n', '1\n']
Note:
In the first example you can obtain the answer by the following sequence of operations: "1001" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1010" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "1100" <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> "100".
In the second example you can't obtain smaller answer no matter what operations you use. | ```python
n = int(input())
s = input()
count1 = 0
for i in s:
if(i=='0'):
count1+=1
print('1'+'0'*count1)
``` | 0 | |
500 | A | New Year Transportation | PROGRAMMING | 1,000 | [
"dfs and similar",
"graphs",
"implementation"
] | null | null | New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system. | The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World. | If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO". | [
"8 4\n1 2 1 2 1 2 1\n",
"8 5\n1 2 1 2 1 1 1\n"
] | [
"YES\n",
"NO\n"
] | In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | 500 | [
{
"input": "8 4\n1 2 1 2 1 2 1",
"output": "YES"
},
{
"input": "8 5\n1 2 1 2 1 1 1",
"output": "NO"
},
{
"input": "20 19\n13 16 7 6 12 1 5 7 8 6 5 7 5 5 3 3 2 2 1",
"output": "YES"
},
{
"input": "50 49\n11 7 1 41 26 36 19 16 38 14 36 35 37 27 20 27 3 6 21 2 27 11 18 17 19 16 22 8 8 9 1 7 5 12 5 6 13 6 11 2 6 3 1 5 1 1 2 2 1",
"output": "YES"
},
{
"input": "120 104\n41 15 95 85 34 11 25 42 65 39 77 80 74 17 66 73 21 14 36 63 63 79 45 24 65 7 63 80 51 21 2 19 78 28 71 2 15 23 17 68 62 18 54 39 43 70 3 46 34 23 41 65 32 10 13 18 10 3 16 48 54 18 57 28 3 24 44 50 15 2 20 22 45 44 3 29 2 27 11 2 12 25 25 31 1 2 32 4 11 30 13 16 26 21 1 13 21 8 15 5 18 13 5 15 3 8 13 6 5 1 9 7 1 2 4 1 1 2 1",
"output": "NO"
},
{
"input": "10 3\n8 3 5 4 2 3 2 2 1",
"output": "NO"
},
{
"input": "10 9\n8 3 5 4 2 3 2 2 1",
"output": "YES"
},
{
"input": "3 2\n1 1",
"output": "YES"
},
{
"input": "3 2\n2 1",
"output": "NO"
},
{
"input": "4 2\n2 1 1",
"output": "NO"
},
{
"input": "4 4\n2 2 1",
"output": "YES"
},
{
"input": "8 8\n1 2 1 2 1 2 1",
"output": "YES"
},
{
"input": "3 3\n1 1",
"output": "YES"
},
{
"input": "8 8\n1 2 1 2 1 1 1",
"output": "YES"
},
{
"input": "3 3\n2 1",
"output": "YES"
},
{
"input": "4 4\n1 1 1",
"output": "YES"
},
{
"input": "8 8\n1 1 1 1 1 1 1",
"output": "YES"
},
{
"input": "5 5\n1 1 1 1",
"output": "YES"
}
] | 1,682,617,631 | 2,147,483,647 | PyPy 3-64 | OK | TESTS | 34 | 78 | 4,505,600 | n,t=map(int,input().split())
a=list(map(int,input().split()))
k=0
e=0
for i in range(n):
try:
k+=a[k]
except:
break
if k+1==t:
e=1
break
print('YES' if e==1 else 'NO') | Title: New Year Transportation
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
New Year is coming in Line World! In this world, there are *n* cells numbered by integers from 1 to *n*, as a 1<=×<=*n* board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.
So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of *n*<=-<=1 positive integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1. For every integer *i* where 1<=≤<=*i*<=≤<=*n*<=-<=1 the condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* holds. Next, he made *n*<=-<=1 portals, numbered by integers from 1 to *n*<=-<=1. The *i*-th (1<=≤<=*i*<=≤<=*n*<=-<=1) portal connects cell *i* and cell (*i*<=+<=*a**i*), and one can travel from cell *i* to cell (*i*<=+<=*a**i*) using the *i*-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (*i*<=+<=*a**i*) to cell *i* using the *i*-th portal. It is easy to see that because of condition 1<=≤<=*a**i*<=≤<=*n*<=-<=*i* one can't leave the Line World using portals.
Currently, I am standing at cell 1, and I want to go to cell *t*. However, I don't know whether it is possible to go there. Please determine whether I can go to cell *t* by only using the construted transportation system.
Input Specification:
The first line contains two space-separated integers *n* (3<=≤<=*n*<=≤<=3<=×<=104) and *t* (2<=≤<=*t*<=≤<=*n*) — the number of cells, and the index of the cell which I want to go to.
The second line contains *n*<=-<=1 space-separated integers *a*1,<=*a*2,<=...,<=*a**n*<=-<=1 (1<=≤<=*a**i*<=≤<=*n*<=-<=*i*). It is guaranteed, that using the given transportation system, one cannot leave the Line World.
Output Specification:
If I can go to cell *t* using the transportation system, print "YES". Otherwise, print "NO".
Demo Input:
['8 4\n1 2 1 2 1 2 1\n', '8 5\n1 2 1 2 1 1 1\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.
In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit. | ```python
n,t=map(int,input().split())
a=list(map(int,input().split()))
k=0
e=0
for i in range(n):
try:
k+=a[k]
except:
break
if k+1==t:
e=1
break
print('YES' if e==1 else 'NO')
``` | 3 | |
486 | A | Calculating Function | PROGRAMMING | 800 | [
"implementation",
"math"
] | null | null | For a positive integer *n* let's define a function *f*:
*f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n*
Your task is to calculate *f*(*n*) for a given integer *n*. | The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015). | Print *f*(*n*) in a single line. | [
"4\n",
"5\n"
] | [
"2\n",
"-3\n"
] | *f*(4) = - 1 + 2 - 3 + 4 = 2
*f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3 | 500 | [
{
"input": "4",
"output": "2"
},
{
"input": "5",
"output": "-3"
},
{
"input": "1000000000",
"output": "500000000"
},
{
"input": "1000000001",
"output": "-500000001"
},
{
"input": "1000000000000000",
"output": "500000000000000"
},
{
"input": "100",
"output": "50"
},
{
"input": "101",
"output": "-51"
},
{
"input": "102",
"output": "51"
},
{
"input": "103",
"output": "-52"
},
{
"input": "104",
"output": "52"
},
{
"input": "105",
"output": "-53"
},
{
"input": "106",
"output": "53"
},
{
"input": "107",
"output": "-54"
},
{
"input": "108",
"output": "54"
},
{
"input": "109",
"output": "-55"
},
{
"input": "208170109961052",
"output": "104085054980526"
},
{
"input": "46017661651072",
"output": "23008830825536"
},
{
"input": "4018154546667",
"output": "-2009077273334"
},
{
"input": "288565475053",
"output": "-144282737527"
},
{
"input": "3052460231",
"output": "-1526230116"
},
{
"input": "29906716",
"output": "14953358"
},
{
"input": "87897701693326",
"output": "43948850846663"
},
{
"input": "8240",
"output": "4120"
},
{
"input": "577935",
"output": "-288968"
},
{
"input": "62",
"output": "31"
},
{
"input": "1",
"output": "-1"
},
{
"input": "2",
"output": "1"
},
{
"input": "9999999999999",
"output": "-5000000000000"
},
{
"input": "1000000000000",
"output": "500000000000"
},
{
"input": "99999999999999",
"output": "-50000000000000"
},
{
"input": "999999999999999",
"output": "-500000000000000"
},
{
"input": "42191359342",
"output": "21095679671"
},
{
"input": "100000000000000",
"output": "50000000000000"
},
{
"input": "145645214654154",
"output": "72822607327077"
},
{
"input": "4294967296",
"output": "2147483648"
},
{
"input": "3037000499",
"output": "-1518500250"
},
{
"input": "10000000000001",
"output": "-5000000000001"
},
{
"input": "100000017040846",
"output": "50000008520423"
},
{
"input": "98979894985999",
"output": "-49489947493000"
}
] | 1,699,068,592 | 2,147,483,647 | PyPy 3-64 | WRONG_ANSWER | TESTS | 0 | 31 | 0 | n = int(input())
k = 2*(n//2)
somapar = k*(k+1)
somaimpar = (k//2)**2
soma = somapar - somaimpar
if n%2 != 0:
soma -= n
print(soma) | Title: Calculating Function
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
For a positive integer *n* let's define a function *f*:
*f*(*n*)<==<=<=-<=1<=+<=2<=-<=3<=+<=..<=+<=(<=-<=1)*n**n*
Your task is to calculate *f*(*n*) for a given integer *n*.
Input Specification:
The single line contains the positive integer *n* (1<=≤<=*n*<=≤<=1015).
Output Specification:
Print *f*(*n*) in a single line.
Demo Input:
['4\n', '5\n']
Demo Output:
['2\n', '-3\n']
Note:
*f*(4) = - 1 + 2 - 3 + 4 = 2
*f*(5) = - 1 + 2 - 3 + 4 - 5 = - 3 | ```python
n = int(input())
k = 2*(n//2)
somapar = k*(k+1)
somaimpar = (k//2)**2
soma = somapar - somaimpar
if n%2 != 0:
soma -= n
print(soma)
``` | 0 | |
638 | A | Home Numbers | PROGRAMMING | 1,100 | [
"*special",
"constructive algorithms",
"math"
] | null | null | The main street of Berland is a straight line with *n* houses built along it (*n* is an even number). The houses are located at both sides of the street. The houses with odd numbers are at one side of the street and are numbered from 1 to *n*<=-<=1 in the order from the beginning of the street to the end (in the picture: from left to right). The houses with even numbers are at the other side of the street and are numbered from 2 to *n* in the order from the end of the street to its beginning (in the picture: from right to left). The corresponding houses with even and odd numbers are strictly opposite each other, that is, house 1 is opposite house *n*, house 3 is opposite house *n*<=-<=2, house 5 is opposite house *n*<=-<=4 and so on.
Vasya needs to get to house number *a* as quickly as possible. He starts driving from the beginning of the street and drives his car to house *a*. To get from the beginning of the street to houses number 1 and *n*, he spends exactly 1 second. He also spends exactly one second to drive the distance between two neighbouring houses. Vasya can park at any side of the road, so the distance between the beginning of the street at the houses that stand opposite one another should be considered the same.
Your task is: find the minimum time Vasya needs to reach house *a*. | The first line of the input contains two integers, *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100<=000) — the number of houses on the street and the number of the house that Vasya needs to reach, correspondingly. It is guaranteed that number *n* is even. | Print a single integer — the minimum time Vasya needs to get from the beginning of the street to house *a*. | [
"4 2\n",
"8 5\n"
] | [
"2\n",
"3\n"
] | In the first sample there are only four houses on the street, two houses at each side. House 2 will be the last at Vasya's right.
The second sample corresponds to picture with *n* = 8. House 5 is the one before last at Vasya's left. | 500 | [
{
"input": "4 2",
"output": "2"
},
{
"input": "8 5",
"output": "3"
},
{
"input": "2 1",
"output": "1"
},
{
"input": "2 2",
"output": "1"
},
{
"input": "10 1",
"output": "1"
},
{
"input": "10 10",
"output": "1"
},
{
"input": "100000 100000",
"output": "1"
},
{
"input": "100000 2",
"output": "50000"
},
{
"input": "100000 3",
"output": "2"
},
{
"input": "100000 99999",
"output": "50000"
},
{
"input": "100 100",
"output": "1"
},
{
"input": "3000 34",
"output": "1484"
},
{
"input": "2000 1",
"output": "1"
},
{
"input": "100000 1",
"output": "1"
},
{
"input": "24842 1038",
"output": "11903"
},
{
"input": "1628 274",
"output": "678"
},
{
"input": "16186 337",
"output": "169"
},
{
"input": "24562 2009",
"output": "1005"
},
{
"input": "9456 3443",
"output": "1722"
},
{
"input": "5610 332",
"output": "2640"
},
{
"input": "1764 1288",
"output": "239"
},
{
"input": "28588 13902",
"output": "7344"
},
{
"input": "92480 43074",
"output": "24704"
},
{
"input": "40022 26492",
"output": "6766"
},
{
"input": "85766 64050",
"output": "10859"
},
{
"input": "67808 61809",
"output": "30905"
},
{
"input": "80124 68695",
"output": "34348"
},
{
"input": "95522 91716",
"output": "1904"
},
{
"input": "7752 2915",
"output": "1458"
},
{
"input": "5094 5058",
"output": "19"
},
{
"input": "6144 4792",
"output": "677"
},
{
"input": "34334 20793",
"output": "10397"
},
{
"input": "23538 10243",
"output": "5122"
},
{
"input": "9328 7933",
"output": "3967"
},
{
"input": "11110 9885",
"output": "4943"
},
{
"input": "26096 2778",
"output": "11660"
},
{
"input": "75062 5323",
"output": "2662"
},
{
"input": "94790 7722",
"output": "43535"
},
{
"input": "90616 32240",
"output": "29189"
},
{
"input": "96998 8992",
"output": "44004"
},
{
"input": "95130 19219",
"output": "9610"
},
{
"input": "92586 8812",
"output": "41888"
},
{
"input": "3266 3044",
"output": "112"
},
{
"input": "5026 4697",
"output": "2349"
},
{
"input": "3044 2904",
"output": "71"
},
{
"input": "6022 5396",
"output": "314"
},
{
"input": "31270 25522",
"output": "2875"
},
{
"input": "82156 75519",
"output": "37760"
},
{
"input": "34614 27913",
"output": "13957"
},
{
"input": "88024 61143",
"output": "30572"
},
{
"input": "91870 55672",
"output": "18100"
},
{
"input": "95718 4868",
"output": "45426"
},
{
"input": "99564 358",
"output": "49604"
},
{
"input": "89266 13047",
"output": "6524"
},
{
"input": "90904 16455",
"output": "8228"
},
{
"input": "94750 13761",
"output": "6881"
},
{
"input": "100000 23458",
"output": "38272"
},
{
"input": "100000 23457",
"output": "11729"
},
{
"input": "59140 24272",
"output": "17435"
},
{
"input": "9860 8516",
"output": "673"
},
{
"input": "25988 2733",
"output": "1367"
},
{
"input": "9412 5309",
"output": "2655"
},
{
"input": "25540 23601",
"output": "11801"
},
{
"input": "76260 6050",
"output": "35106"
},
{
"input": "92388 39118",
"output": "26636"
},
{
"input": "8516 5495",
"output": "2748"
},
{
"input": "91940 37847",
"output": "18924"
},
{
"input": "30518 286",
"output": "15117"
},
{
"input": "46646 19345",
"output": "9673"
}
] | 1,458,549,011 | 73,811 | Python 3 | OK | TESTS | 69 | 62 | 4,812,800 | n, a = map(int, input().split())
if a % 2 == 0:
print((n-a)//2+1)
else:
print(a//2+1) | Title: Home Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The main street of Berland is a straight line with *n* houses built along it (*n* is an even number). The houses are located at both sides of the street. The houses with odd numbers are at one side of the street and are numbered from 1 to *n*<=-<=1 in the order from the beginning of the street to the end (in the picture: from left to right). The houses with even numbers are at the other side of the street and are numbered from 2 to *n* in the order from the end of the street to its beginning (in the picture: from right to left). The corresponding houses with even and odd numbers are strictly opposite each other, that is, house 1 is opposite house *n*, house 3 is opposite house *n*<=-<=2, house 5 is opposite house *n*<=-<=4 and so on.
Vasya needs to get to house number *a* as quickly as possible. He starts driving from the beginning of the street and drives his car to house *a*. To get from the beginning of the street to houses number 1 and *n*, he spends exactly 1 second. He also spends exactly one second to drive the distance between two neighbouring houses. Vasya can park at any side of the road, so the distance between the beginning of the street at the houses that stand opposite one another should be considered the same.
Your task is: find the minimum time Vasya needs to reach house *a*.
Input Specification:
The first line of the input contains two integers, *n* and *a* (1<=≤<=*a*<=≤<=*n*<=≤<=100<=000) — the number of houses on the street and the number of the house that Vasya needs to reach, correspondingly. It is guaranteed that number *n* is even.
Output Specification:
Print a single integer — the minimum time Vasya needs to get from the beginning of the street to house *a*.
Demo Input:
['4 2\n', '8 5\n']
Demo Output:
['2\n', '3\n']
Note:
In the first sample there are only four houses on the street, two houses at each side. House 2 will be the last at Vasya's right.
The second sample corresponds to picture with *n* = 8. House 5 is the one before last at Vasya's left. | ```python
n, a = map(int, input().split())
if a % 2 == 0:
print((n-a)//2+1)
else:
print(a//2+1)
``` | 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": "tymbzjyqhymedasloqbq",
"output": "NO"
},
{
"input": "yehluhlkwo",
"output": "NO"
},
{
"input": "hatlevhhalrohairnolsvocafgueelrqmlqlleello",
"output": "YES"
},
{
"input": "hhhtehdbllnhwmbyhvelqqyoulretpbfokflhlhreeflxeftelziclrwllrpflflbdtotvlqgoaoqldlroovbfsq",
"output": "YES"
},
{
"input": "rzlvihhghnelqtwlexmvdjjrliqllolhyewgozkuovaiezgcilelqapuoeglnwmnlftxxiigzczlouooi",
"output": "YES"
},
{
"input": "pfhhwctyqdlkrwhebfqfelhyebwllhemtrmeblgrynmvyhioesqklclocxmlffuormljszllpoo",
"output": "YES"
},
{
"input": "lqllcolohwflhfhlnaow",
"output": "NO"
},
{
"input": "heheeellollvoo",
"output": "YES"
},
{
"input": "hellooo",
"output": "YES"
},
{
"input": "o",
"output": "NO"
},
{
"input": "hhqhzeclohlehljlhtesllylrolmomvuhcxsobtsckogdv",
"output": "YES"
},
{
"input": "yoegfuzhqsihygnhpnukluutocvvwuldiighpogsifealtgkfzqbwtmgghmythcxflebrkctlldlkzlagovwlstsghbouk",
"output": "YES"
},
{
"input": "uatqtgbvrnywfacwursctpagasnhydvmlinrcnqrry",
"output": "NO"
},
{
"input": "tndtbldbllnrwmbyhvqaqqyoudrstpbfokfoclnraefuxtftmgzicorwisrpfnfpbdtatvwqgyalqtdtrjqvbfsq",
"output": "NO"
},
{
"input": "rzlvirhgemelnzdawzpaoqtxmqucnahvqnwldklrmjiiyageraijfivigvozgwngiulttxxgzczptusoi",
"output": "YES"
},
{
"input": "kgyelmchocojsnaqdsyeqgnllytbqietpdlgknwwumqkxrexgdcnwoldicwzwofpmuesjuxzrasscvyuqwspm",
"output": "YES"
},
{
"input": "pnyvrcotjvgynbeldnxieghfltmexttuxzyac",
"output": "NO"
},
{
"input": "dtwhbqoumejligbenxvzhjlhosqojetcqsynlzyhfaevbdpekgbtjrbhlltbceobcok",
"output": "YES"
},
{
"input": "crrfpfftjwhhikwzeedrlwzblckkteseofjuxjrktcjfsylmlsvogvrcxbxtffujqshslemnixoeezivksouefeqlhhokwbqjz",
"output": "YES"
},
{
"input": "jhfbndhyzdvhbvhmhmefqllujdflwdpjbehedlsqfdsqlyelwjtyloxwsvasrbqosblzbowlqjmyeilcvotdlaouxhdpoeloaovb",
"output": "YES"
},
{
"input": "hwlghueoemiqtjhhpashjsouyegdlvoyzeunlroypoprnhlyiwiuxrghekaylndhrhllllwhbebezoglydcvykllotrlaqtvmlla",
"output": "YES"
},
{
"input": "wshiaunnqnqxodholbipwhhjmyeblhgpeleblklpzwhdunmpqkbuzloetmwwxmeltkrcomulxauzlwmlklldjodozxryghsnwgcz",
"output": "YES"
},
{
"input": "shvksednttggehroewuiptvvxtrzgidravtnjwuqrlnnkxbplctzkckinpkgjopjfoxdbojtcvsuvablcbkrzajrlhgobkcxeqti",
"output": "YES"
},
{
"input": "hyyhddqhxhekehkwfhlnlsihzefwchzerevcjtokefplholrbvxlltdlafjxrfhleglrvlolojoqaolagtbeyogxlbgfolllslli",
"output": "YES"
},
{
"input": "iaagrdhhelxpdegueiulflquvlzidoprzkehthkldaytrjwhyhprwjxlltinxvuilxohqgjqcvkkdcuoliekcnlilwgqlnlzouoo",
"output": "YES"
},
{
"input": "wfluaeseldgxyvxpwuhkptdmlflnlhktwxiabsvkolsquymrmhzczzoybvlilhmvsuunggvgxzgyyffk",
"output": "NO"
},
{
"input": "loee",
"output": "NO"
},
{
"input": "izdipwylefqmjbuoznfglgcfvedeouflkeehxbasaqmiooppfsqlhprospqgxvzrcpwlfdddngoqvpwvggsnvvxhmjoslwjudjlo",
"output": "NO"
},
{
"input": "pplynflbjlseflaegzqpdxswgunhnlglqhycrbcwdfhclmcrwekkzsoosmfblljeenwfjyifcvxihhhbyhmhlkvtskeglaovjnsi",
"output": "YES"
},
{
"input": "xqjqmenkodmlhzyzmmvofdngktygbbxbzpluzcohohmalkoeuwfikblltaaigv",
"output": "NO"
},
{
"input": "hello",
"output": "YES"
},
{
"input": "oohell",
"output": "NO"
},
{
"input": "hell",
"output": "NO"
},
{
"input": "eloellohoelo",
"output": "NO"
},
{
"input": "helllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllo",
"output": "YES"
}
] | 1,668,331,919 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 0 | 30 | 0 | s = input()
hello = 'hello'
index = 0
found = None
for i in hello:
found = False
for j in range(index, len(s)):
print(f'{j}now')
if s[j] == i:
index = j+1
print(index)
found = True
break
if not found:
print('NO')
break
if found:
print('YES')
| Title: Chat room
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya has recently learned to type and log on to the Internet. He immediately entered a chat room and decided to say hello to everybody. Vasya typed the word *s*. It is considered that Vasya managed to say hello if several letters can be deleted from the typed word so that it resulted in the word "hello". For example, if Vasya types the word "ahhellllloou", it will be considered that he said hello, and if he types "hlelo", it will be considered that Vasya got misunderstood and he didn't manage to say hello. Determine whether Vasya managed to say hello by the given word *s*.
Input Specification:
The first and only line contains the word *s*, which Vasya typed. This word consisits of small Latin letters, its length is no less that 1 and no more than 100 letters.
Output Specification:
If Vasya managed to say hello, print "YES", otherwise print "NO".
Demo Input:
['ahhellllloou\n', 'hlelo\n']
Demo Output:
['YES\n', 'NO\n']
Note:
none | ```python
s = input()
hello = 'hello'
index = 0
found = None
for i in hello:
found = False
for j in range(index, len(s)):
print(f'{j}now')
if s[j] == i:
index = j+1
print(index)
found = True
break
if not found:
print('NO')
break
if found:
print('YES')
``` | 0 |
235 | A | LCM Challenge | PROGRAMMING | 1,600 | [
"number theory"
] | null | null | Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than *n*. Can you help me to find the maximum possible least common multiple of these three integers? | The first line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the *n* mentioned in the statement. | Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than *n*. | [
"9\n",
"7\n"
] | [
"504\n",
"210\n"
] | The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get. | 500 | [
{
"input": "9",
"output": "504"
},
{
"input": "7",
"output": "210"
},
{
"input": "1",
"output": "1"
},
{
"input": "5",
"output": "60"
},
{
"input": "6",
"output": "60"
},
{
"input": "33",
"output": "32736"
},
{
"input": "21",
"output": "7980"
},
{
"input": "2",
"output": "2"
},
{
"input": "41",
"output": "63960"
},
{
"input": "29",
"output": "21924"
},
{
"input": "117",
"output": "1560780"
},
{
"input": "149",
"output": "3241644"
},
{
"input": "733",
"output": "392222436"
},
{
"input": "925",
"output": "788888100"
},
{
"input": "509",
"output": "131096004"
},
{
"input": "829",
"output": "567662724"
},
{
"input": "117",
"output": "1560780"
},
{
"input": "605",
"output": "220348260"
},
{
"input": "245",
"output": "14526540"
},
{
"input": "925",
"output": "788888100"
},
{
"input": "213",
"output": "9527916"
},
{
"input": "53",
"output": "140556"
},
{
"input": "341",
"output": "39303660"
},
{
"input": "21",
"output": "7980"
},
{
"input": "605",
"output": "220348260"
},
{
"input": "149",
"output": "3241644"
},
{
"input": "733",
"output": "392222436"
},
{
"input": "117",
"output": "1560780"
},
{
"input": "53",
"output": "140556"
},
{
"input": "245",
"output": "14526540"
},
{
"input": "829",
"output": "567662724"
},
{
"input": "924",
"output": "783776526"
},
{
"input": "508",
"output": "130065780"
},
{
"input": "700",
"output": "341042100"
},
{
"input": "636",
"output": "254839470"
},
{
"input": "20",
"output": "6460"
},
{
"input": "604",
"output": "218891412"
},
{
"input": "796",
"output": "501826260"
},
{
"input": "732",
"output": "389016270"
},
{
"input": "412",
"output": "69256788"
},
{
"input": "700",
"output": "341042100"
},
{
"input": "244",
"output": "14289372"
},
{
"input": "828",
"output": "563559150"
},
{
"input": "508",
"output": "130065780"
},
{
"input": "796",
"output": "501826260"
},
{
"input": "636",
"output": "254839470"
},
{
"input": "924",
"output": "783776526"
},
{
"input": "245",
"output": "14526540"
},
{
"input": "828",
"output": "563559150"
},
{
"input": "21",
"output": "7980"
},
{
"input": "605",
"output": "220348260"
},
{
"input": "636",
"output": "254839470"
},
{
"input": "924",
"output": "783776526"
},
{
"input": "116",
"output": "1507420"
},
{
"input": "700",
"output": "341042100"
},
{
"input": "732",
"output": "389016270"
},
{
"input": "20",
"output": "6460"
},
{
"input": "508",
"output": "130065780"
},
{
"input": "148",
"output": "3154620"
},
{
"input": "828",
"output": "563559150"
},
{
"input": "763116",
"output": "444394078546562430"
},
{
"input": "756604",
"output": "433115377058855412"
},
{
"input": "447244",
"output": "89460162932862372"
},
{
"input": "372636",
"output": "51742503205363470"
},
{
"input": "546924",
"output": "163597318076822526"
},
{
"input": "540412",
"output": "157823524476316788"
},
{
"input": "714700",
"output": "365063922340784100"
},
{
"input": "520731",
"output": "141201007712496270"
},
{
"input": "695019",
"output": "335728459024850814"
},
{
"input": "688507",
"output": "326379736779169710"
},
{
"input": "862795",
"output": "642275489615199390"
},
{
"input": "668827",
"output": "299184742915995150"
},
{
"input": "810411",
"output": "532248411551110590"
},
{
"input": "836603",
"output": "585540171302562606"
},
{
"input": "978187",
"output": "935975171582120670"
},
{
"input": "816923",
"output": "545182335484592526"
},
{
"input": "958507",
"output": "880611813728059710"
},
{
"input": "984699",
"output": "954792870629291694"
},
{
"input": "642635",
"output": "265393998349453470"
},
{
"input": "296604",
"output": "26092892528622606"
},
{
"input": "1000000",
"output": "999996000003000000"
},
{
"input": "8",
"output": "280"
},
{
"input": "3",
"output": "6"
},
{
"input": "4",
"output": "12"
},
{
"input": "30",
"output": "21924"
},
{
"input": "18",
"output": "4080"
},
{
"input": "12",
"output": "990"
}
] | 1,590,299,382 | 2,147,483,647 | PyPy 3 | OK | TESTS | 87 | 374 | 1,740,800 | import math
from functools import reduce
def lcm(a, b):
g = math.gcd(a, b)
return a // g * b // g * g
def solve(n):
return max(find_from_n(n), find_from_n(n - 1))
def find_from_n(n):
if n == 0:
return 1
a = [n]
for i in range(n - 1, 0, -1):
if len(a) >= 3:
break
if all(math.gcd(i, x) == 1 for x in a):
a.append(i)
if len(a) >= 3 and a[-1] == 1:
a[-1] = a[1] - 1
return reduce(lambda x, y: lcm(x, y), a, a[0])
# assert solve(924) == 783776526
#
# assert solve(7) == 210
# assert solve(1) == 1
# assert solve(5) == 60
# assert solve(6) == 60
# assert solve(33) == 32736
# assert solve(21) == 7980
# assert solve(2) == 2
# assert solve(41) == 63960
# assert solve(29) == 21924
# assert solve(117) == 1560780
# assert solve(149) == 3241644
# assert solve(733) == 392222436
# assert solve(925) == 788888100
# assert solve(509) == 131096004
# assert solve(829) == 567662724
# assert solve(117) == 1560780
# assert solve(605) == 220348260
n = int(input())
r = max(solve(n), solve(n - 1))
print(r)
| Title: LCM Challenge
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it.
But I also don't want to use many numbers, so I'll choose three positive integers (they don't have to be distinct) which are not greater than *n*. Can you help me to find the maximum possible least common multiple of these three integers?
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=106) — the *n* mentioned in the statement.
Output Specification:
Print a single integer — the maximum possible LCM of three not necessarily distinct positive integers that are not greater than *n*.
Demo Input:
['9\n', '7\n']
Demo Output:
['504\n', '210\n']
Note:
The least common multiple of some positive integers is the least positive integer which is multiple for each of them.
The result may become very large, 32-bit integer won't be enough. So using 64-bit integers is recommended.
For the last example, we can chose numbers 7, 6, 5 and the LCM of them is 7·6·5 = 210. It is the maximum value we can get. | ```python
import math
from functools import reduce
def lcm(a, b):
g = math.gcd(a, b)
return a // g * b // g * g
def solve(n):
return max(find_from_n(n), find_from_n(n - 1))
def find_from_n(n):
if n == 0:
return 1
a = [n]
for i in range(n - 1, 0, -1):
if len(a) >= 3:
break
if all(math.gcd(i, x) == 1 for x in a):
a.append(i)
if len(a) >= 3 and a[-1] == 1:
a[-1] = a[1] - 1
return reduce(lambda x, y: lcm(x, y), a, a[0])
# assert solve(924) == 783776526
#
# assert solve(7) == 210
# assert solve(1) == 1
# assert solve(5) == 60
# assert solve(6) == 60
# assert solve(33) == 32736
# assert solve(21) == 7980
# assert solve(2) == 2
# assert solve(41) == 63960
# assert solve(29) == 21924
# assert solve(117) == 1560780
# assert solve(149) == 3241644
# assert solve(733) == 392222436
# assert solve(925) == 788888100
# assert solve(509) == 131096004
# assert solve(829) == 567662724
# assert solve(117) == 1560780
# assert solve(605) == 220348260
n = int(input())
r = max(solve(n), solve(n - 1))
print(r)
``` | 3 | |
258 | A | Little Elephant and Bits | PROGRAMMING | 1,100 | [
"greedy",
"math"
] | null | null | The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper.
To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes).
The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation. | The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits. | In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem. | [
"101\n",
"110010\n"
] | [
"11\n",
"11010\n"
] | In the first sample the best strategy is to delete the second digit. That results in number 11<sub class="lower-index">2</sub> = 3<sub class="lower-index">10</sub>.
In the second sample the best strategy is to delete the third or fourth digits — that results in number 11010<sub class="lower-index">2</sub> = 26<sub class="lower-index">10</sub>. | 500 | [
{
"input": "101",
"output": "11"
},
{
"input": "110010",
"output": "11010"
},
{
"input": "10000",
"output": "1000"
},
{
"input": "1111111110",
"output": "111111111"
},
{
"input": "10100101011110101",
"output": "1100101011110101"
},
{
"input": "111010010111",
"output": "11110010111"
},
{
"input": "11110111011100000000",
"output": "1111111011100000000"
},
{
"input": "11110010010100001110110101110011110110100111101",
"output": "1111010010100001110110101110011110110100111101"
},
{
"input": "1001011111010010100111111",
"output": "101011111010010100111111"
},
{
"input": "1111111111",
"output": "111111111"
},
{
"input": "1111111111111111111100111101001110110111111000001111110101001101001110011000001011001111111000110101",
"output": "111111111111111111110111101001110110111111000001111110101001101001110011000001011001111111000110101"
},
{
"input": "11010110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100",
"output": "1110110000100100101111110111001001010011000011011000010010100111010101000111010011101101111110001111000101000001100011101110100"
},
{
"input": "11111111111111111111111110110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011",
"output": "1111111111111111111111111110111001101100111010010101101101001011100011011000111010011110010101100010001011101011010010100001000011100001101101001100010100001001010010100100001111110100110011000101100001111111011010111001011111110111101000100101001001011"
},
{
"input": "11100010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011",
"output": "1110010010010000110101101101100111111001010001101101001001111010110010111001011010000001100110101000101111000001111101111110010000010101110011110101101010110001100011101111011100010011101100111110010111111100110101000000111101000000000110100100101111101000110101010101101001110001110000101011010101100011100100111100010001011010010001100011111110010011010011000111000100111100010110100011010010101011011011111110100001110000011011"
},
{
"input": "11",
"output": "1"
},
{
"input": "111",
"output": "11"
},
{
"input": "111111",
"output": "11111"
},
{
"input": "11111",
"output": "1111"
},
{
"input": "1111",
"output": "111"
}
] | 1,692,454,390 | 2,147,483,647 | PyPy 3 | RUNTIME_ERROR | TESTS | 0 | 154 | 2,048,000 | t = int(input())
while(t>0):
s = input()
l = list(s)
a = 0
for i in range(len(s)-2,-1,-1):
if l[i] == '0':
l.remove(l[i])
a+=1
break
if a == 1:
print(''.join(l))
else:
if l[-1] == '0':
l.remove('0')
else:
l.remove('1')
print(''.join(l))
t-=1 | Title: Little Elephant and Bits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant has an integer *a*, written in the binary notation. He wants to write this number on a piece of paper.
To make sure that the number *a* fits on the piece of paper, the Little Elephant ought to delete exactly one any digit from number *a* in the binary record. At that a new number appears. It consists of the remaining binary digits, written in the corresponding order (possible, with leading zeroes).
The Little Elephant wants the number he is going to write on the paper to be as large as possible. Help him find the maximum number that he can obtain after deleting exactly one binary digit and print it in the binary notation.
Input Specification:
The single line contains integer *a*, written in the binary notation without leading zeroes. This number contains more than 1 and at most 105 digits.
Output Specification:
In the single line print the number that is written without leading zeroes in the binary notation — the answer to the problem.
Demo Input:
['101\n', '110010\n']
Demo Output:
['11\n', '11010\n']
Note:
In the first sample the best strategy is to delete the second digit. That results in number 11<sub class="lower-index">2</sub> = 3<sub class="lower-index">10</sub>.
In the second sample the best strategy is to delete the third or fourth digits — that results in number 11010<sub class="lower-index">2</sub> = 26<sub class="lower-index">10</sub>. | ```python
t = int(input())
while(t>0):
s = input()
l = list(s)
a = 0
for i in range(len(s)-2,-1,-1):
if l[i] == '0':
l.remove(l[i])
a+=1
break
if a == 1:
print(''.join(l))
else:
if l[-1] == '0':
l.remove('0')
else:
l.remove('1')
print(''.join(l))
t-=1
``` | -1 | |
6 | A | Triangle | PROGRAMMING | 900 | [
"brute force",
"geometry"
] | A. Triangle | 2 | 64 | Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him. | The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks. | Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length. | [
"4 2 1 3\n",
"7 2 2 4\n",
"3 5 9 1\n"
] | [
"TRIANGLE\n",
"SEGMENT\n",
"IMPOSSIBLE\n"
] | none | 0 | [
{
"input": "4 2 1 3",
"output": "TRIANGLE"
},
{
"input": "7 2 2 4",
"output": "SEGMENT"
},
{
"input": "3 5 9 1",
"output": "IMPOSSIBLE"
},
{
"input": "3 1 5 1",
"output": "IMPOSSIBLE"
},
{
"input": "10 10 10 10",
"output": "TRIANGLE"
},
{
"input": "11 5 6 11",
"output": "TRIANGLE"
},
{
"input": "1 1 1 1",
"output": "TRIANGLE"
},
{
"input": "10 20 30 40",
"output": "TRIANGLE"
},
{
"input": "45 25 5 15",
"output": "IMPOSSIBLE"
},
{
"input": "20 5 8 13",
"output": "TRIANGLE"
},
{
"input": "10 30 7 20",
"output": "SEGMENT"
},
{
"input": "3 2 3 2",
"output": "TRIANGLE"
},
{
"input": "70 10 100 30",
"output": "SEGMENT"
},
{
"input": "4 8 16 2",
"output": "IMPOSSIBLE"
},
{
"input": "3 3 3 10",
"output": "TRIANGLE"
},
{
"input": "1 5 5 5",
"output": "TRIANGLE"
},
{
"input": "13 25 12 1",
"output": "SEGMENT"
},
{
"input": "10 100 7 3",
"output": "SEGMENT"
},
{
"input": "50 1 50 100",
"output": "TRIANGLE"
},
{
"input": "50 1 100 49",
"output": "SEGMENT"
},
{
"input": "49 51 100 1",
"output": "SEGMENT"
},
{
"input": "5 11 2 25",
"output": "IMPOSSIBLE"
},
{
"input": "91 50 9 40",
"output": "IMPOSSIBLE"
},
{
"input": "27 53 7 97",
"output": "IMPOSSIBLE"
},
{
"input": "51 90 24 8",
"output": "IMPOSSIBLE"
},
{
"input": "3 5 1 1",
"output": "IMPOSSIBLE"
},
{
"input": "13 49 69 15",
"output": "IMPOSSIBLE"
},
{
"input": "16 99 9 35",
"output": "IMPOSSIBLE"
},
{
"input": "27 6 18 53",
"output": "IMPOSSIBLE"
},
{
"input": "57 88 17 8",
"output": "IMPOSSIBLE"
},
{
"input": "95 20 21 43",
"output": "IMPOSSIBLE"
},
{
"input": "6 19 32 61",
"output": "IMPOSSIBLE"
},
{
"input": "100 21 30 65",
"output": "IMPOSSIBLE"
},
{
"input": "85 16 61 9",
"output": "IMPOSSIBLE"
},
{
"input": "5 6 19 82",
"output": "IMPOSSIBLE"
},
{
"input": "1 5 1 3",
"output": "IMPOSSIBLE"
},
{
"input": "65 10 36 17",
"output": "IMPOSSIBLE"
},
{
"input": "81 64 9 7",
"output": "IMPOSSIBLE"
},
{
"input": "11 30 79 43",
"output": "IMPOSSIBLE"
},
{
"input": "1 1 5 3",
"output": "IMPOSSIBLE"
},
{
"input": "21 94 61 31",
"output": "IMPOSSIBLE"
},
{
"input": "49 24 9 74",
"output": "IMPOSSIBLE"
},
{
"input": "11 19 5 77",
"output": "IMPOSSIBLE"
},
{
"input": "52 10 19 71",
"output": "SEGMENT"
},
{
"input": "2 3 7 10",
"output": "SEGMENT"
},
{
"input": "1 2 6 3",
"output": "SEGMENT"
},
{
"input": "2 6 1 8",
"output": "SEGMENT"
},
{
"input": "1 2 4 1",
"output": "SEGMENT"
},
{
"input": "4 10 6 2",
"output": "SEGMENT"
},
{
"input": "2 10 7 3",
"output": "SEGMENT"
},
{
"input": "5 2 3 9",
"output": "SEGMENT"
},
{
"input": "6 1 4 10",
"output": "SEGMENT"
},
{
"input": "10 6 4 1",
"output": "SEGMENT"
},
{
"input": "3 2 9 1",
"output": "SEGMENT"
},
{
"input": "22 80 29 7",
"output": "SEGMENT"
},
{
"input": "2 6 3 9",
"output": "SEGMENT"
},
{
"input": "3 1 2 1",
"output": "SEGMENT"
},
{
"input": "3 4 7 1",
"output": "SEGMENT"
},
{
"input": "8 4 3 1",
"output": "SEGMENT"
},
{
"input": "2 8 3 5",
"output": "SEGMENT"
},
{
"input": "4 1 2 1",
"output": "SEGMENT"
},
{
"input": "8 1 3 2",
"output": "SEGMENT"
},
{
"input": "6 2 1 8",
"output": "SEGMENT"
},
{
"input": "3 3 3 6",
"output": "TRIANGLE"
},
{
"input": "3 6 3 3",
"output": "TRIANGLE"
},
{
"input": "4 10 4 4",
"output": "TRIANGLE"
},
{
"input": "1 1 2 1",
"output": "TRIANGLE"
},
{
"input": "3 3 3 6",
"output": "TRIANGLE"
},
{
"input": "5 4 5 5",
"output": "TRIANGLE"
},
{
"input": "8 7 8 8",
"output": "TRIANGLE"
},
{
"input": "3 3 3 1",
"output": "TRIANGLE"
},
{
"input": "1 1 6 6",
"output": "TRIANGLE"
},
{
"input": "1 9 1 9",
"output": "TRIANGLE"
},
{
"input": "7 2 2 7",
"output": "TRIANGLE"
},
{
"input": "7 2 3 2",
"output": "TRIANGLE"
},
{
"input": "4 4 10 10",
"output": "TRIANGLE"
},
{
"input": "7 7 10 7",
"output": "TRIANGLE"
},
{
"input": "4 4 4 5",
"output": "TRIANGLE"
},
{
"input": "1 10 9 2",
"output": "TRIANGLE"
},
{
"input": "1 8 2 7",
"output": "TRIANGLE"
},
{
"input": "4 3 2 8",
"output": "TRIANGLE"
},
{
"input": "5 9 5 3",
"output": "TRIANGLE"
},
{
"input": "4 10 3 5",
"output": "TRIANGLE"
}
] | 1,512,168,153 | 2,147,483,647 | Python 3 | WRONG_ANSWER | TESTS | 5 | 124 | 5,529,600 | def check():
n = input()
n = n.split()
for i in range(len(n)):
n[i] = int(n[i])
for i in range(len(n)):
for k in range(i + 1, len(n)):
for p in range(k + 1, len(n)):
sides = [ n[i], n[k], n[p]]
for side in sides:
other_sides = (sum(sides)-side)
if side > other_sides:
break
elif side == other_sides:
return 'SEGMENT'
else:
return 'TRIANGLE'
return 'IMPOSSIBLE'
print (check())
| Title: Triangle
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Johnny has a younger sister Anne, who is very clever and smart. As she came home from the kindergarten, she told his brother about the task that her kindergartener asked her to solve. The task was just to construct a triangle out of four sticks of different colours. Naturally, one of the sticks is extra. It is not allowed to break the sticks or use their partial length. Anne has perfectly solved this task, now she is asking Johnny to do the same.
The boy answered that he would cope with it without any difficulty. However, after a while he found out that different tricky things can occur. It can happen that it is impossible to construct a triangle of a positive area, but it is possible to construct a degenerate triangle. It can be so, that it is impossible to construct a degenerate triangle even. As Johnny is very lazy, he does not want to consider such a big amount of cases, he asks you to help him.
Input Specification:
The first line of the input contains four space-separated positive integer numbers not exceeding 100 — lengthes of the sticks.
Output Specification:
Output TRIANGLE if it is possible to construct a non-degenerate triangle. Output SEGMENT if the first case cannot take place and it is possible to construct a degenerate triangle. Output IMPOSSIBLE if it is impossible to construct any triangle. Remember that you are to use three sticks. It is not allowed to break the sticks or use their partial length.
Demo Input:
['4 2 1 3\n', '7 2 2 4\n', '3 5 9 1\n']
Demo Output:
['TRIANGLE\n', 'SEGMENT\n', 'IMPOSSIBLE\n']
Note:
none | ```python
def check():
n = input()
n = n.split()
for i in range(len(n)):
n[i] = int(n[i])
for i in range(len(n)):
for k in range(i + 1, len(n)):
for p in range(k + 1, len(n)):
sides = [ n[i], n[k], n[p]]
for side in sides:
other_sides = (sum(sides)-side)
if side > other_sides:
break
elif side == other_sides:
return 'SEGMENT'
else:
return 'TRIANGLE'
return 'IMPOSSIBLE'
print (check())
``` | 0 |
937 | A | Olympiad | PROGRAMMING | 800 | [
"implementation",
"sortings"
] | null | null | The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points.
As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria:
- At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded.
Determine the number of ways to choose a subset of participants that will receive the diplomas. | The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants.
The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores.
It's guaranteed that at least one participant has non-zero score. | Print a single integer — the desired number of ways. | [
"4\n1 3 3 2\n",
"3\n1 1 1\n",
"4\n42 0 0 42\n"
] | [
"3\n",
"1\n",
"1\n"
] | There are three ways to choose a subset in sample case one.
1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma!
The only option in sample case two is to award everyone.
Note that in sample case three participants with zero scores cannot get anything. | 500 | [
{
"input": "4\n1 3 3 2",
"output": "3"
},
{
"input": "3\n1 1 1",
"output": "1"
},
{
"input": "4\n42 0 0 42",
"output": "1"
},
{
"input": "10\n1 0 1 0 1 0 0 0 0 1",
"output": "1"
},
{
"input": "10\n572 471 540 163 50 30 561 510 43 200",
"output": "10"
},
{
"input": "100\n122 575 426 445 172 81 247 429 97 202 175 325 382 384 417 356 132 502 328 537 57 339 518 211 479 306 140 168 268 16 140 263 593 249 391 310 555 468 231 180 157 18 334 328 276 155 21 280 322 545 111 267 467 274 291 304 235 34 365 180 21 95 501 552 325 331 302 353 296 22 289 399 7 466 32 302 568 333 75 192 284 10 94 128 154 512 9 480 243 521 551 492 420 197 207 125 367 117 438 600",
"output": "94"
},
{
"input": "100\n600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600 600",
"output": "1"
},
{
"input": "78\n5 4 13 2 5 6 2 10 10 1 2 6 7 9 6 3 5 7 1 10 2 2 7 0 2 11 11 3 1 13 3 10 6 2 0 3 0 5 0 1 4 11 1 1 7 0 12 7 5 12 0 2 12 9 8 3 4 3 4 11 4 10 2 3 10 12 5 6 1 11 2 0 8 7 9 1 3 12",
"output": "13"
},
{
"input": "34\n220 387 408 343 184 447 197 307 337 414 251 319 426 322 347 242 208 412 188 185 241 235 216 259 331 372 322 284 444 384 214 297 389 391",
"output": "33"
},
{
"input": "100\n1 2 1 0 3 0 2 0 0 1 2 0 1 3 0 3 3 1 3 0 0 2 1 2 2 1 3 3 3 3 3 2 0 0 2 1 2 3 2 3 0 1 1 3 3 2 0 3 1 0 2 2 2 1 2 3 2 1 0 3 0 2 0 3 0 2 1 0 3 1 0 2 2 1 3 1 3 0 2 3 3 1 1 3 1 3 0 3 2 0 2 3 3 0 2 0 2 0 1 3",
"output": "3"
},
{
"input": "100\n572 471 540 163 50 30 561 510 43 200 213 387 500 424 113 487 357 333 294 337 435 202 447 494 485 465 161 344 470 559 104 356 393 207 224 213 511 514 60 386 149 216 392 229 429 173 165 401 395 150 127 579 344 390 529 296 225 425 318 79 465 447 177 110 367 212 459 270 41 500 277 567 125 436 178 9 214 342 203 112 144 24 79 155 495 556 40 549 463 281 241 316 2 246 1 396 510 293 332 55",
"output": "93"
},
{
"input": "99\n5 4 13 2 5 6 2 10 10 1 2 6 7 9 6 3 5 7 1 10 2 2 7 0 2 11 11 3 1 13 3 10 6 2 0 3 0 5 0 1 4 11 1 1 7 0 12 7 5 12 0 2 12 9 8 3 4 3 4 11 4 10 2 3 10 12 5 6 1 11 2 0 8 7 9 1 3 12 2 3 9 3 7 13 7 13 0 11 8 12 2 5 9 4 0 6 6 2 13",
"output": "13"
},
{
"input": "99\n1 0 1 0 1 0 0 0 0 1 0 0 1 1 0 1 1 1 1 0 0 0 1 0 0 1 1 1 1 1 1 0 0 0 0 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 0 1 0 1 1 0 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 0 1 0 0 0 1 1 0 0 0 0 0 1",
"output": "1"
},
{
"input": "99\n21 74 25 44 71 80 46 28 96 1 74 24 81 83 16 55 31 1 27 36 56 38 17 10 78 5 39 67 67 15 39 62 92 48 90 9 54 67 30 79 56 17 33 27 75 54 20 79 21 44 10 66 66 73 90 3 34 33 64 79 20 94 0 51 24 30 1 52 95 21 88 98 6 65 31 1 67 32 74 91 83 9 93 27 53 11 8 79 42 20 50 91 19 96 6 24 66 16 37",
"output": "61"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "2\n0 600",
"output": "1"
},
{
"input": "4\n1 1 1 2",
"output": "2"
},
{
"input": "4\n0 0 1 2",
"output": "2"
},
{
"input": "1\n5",
"output": "1"
},
{
"input": "2\n0 5",
"output": "1"
},
{
"input": "5\n1 0 0 1 2",
"output": "2"
}
] | 1,621,540,247 | 2,147,483,647 | Python 3 | RUNTIME_ERROR | TESTS | 0 | 46 | 102,400 | from functools import reduce
n=int(input())
k=[int(k) for k in input().split()]
s=set(k)
s.remove(0)
print(len(s)) | Title: Olympiad
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The recent All-Berland Olympiad in Informatics featured *n* participants with each scoring a certain amount of points.
As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria:
- At least one participant should get a diploma. - None of those with score equal to zero should get awarded. - When someone is awarded, all participants with score not less than his score should also be awarded.
Determine the number of ways to choose a subset of participants that will receive the diplomas.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100) — the number of participants.
The next line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=600) — participants' scores.
It's guaranteed that at least one participant has non-zero score.
Output Specification:
Print a single integer — the desired number of ways.
Demo Input:
['4\n1 3 3 2\n', '3\n1 1 1\n', '4\n42 0 0 42\n']
Demo Output:
['3\n', '1\n', '1\n']
Note:
There are three ways to choose a subset in sample case one.
1. Only participants with 3 points will get diplomas. 1. Participants with 2 or 3 points will get diplomas. 1. Everyone will get a diploma!
The only option in sample case two is to award everyone.
Note that in sample case three participants with zero scores cannot get anything. | ```python
from functools import reduce
n=int(input())
k=[int(k) for k in input().split()]
s=set(k)
s.remove(0)
print(len(s))
``` | -1 |
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