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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
912
|
A
|
Tricky Alchemy
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are *A* yellow and *B* blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
|
The first line features two integers *A* and *B* (0<=≤<=*A*,<=*B*<=≤<=109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=109) — the respective amounts of yellow, green and blue balls to be obtained.
|
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
|
[
"4 3\n2 1 1\n",
"3 9\n1 1 3\n",
"12345678 87654321\n43043751 1000000000 53798715\n"
] |
[
"2\n",
"1\n",
"2147483648\n"
] |
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
| 500
|
[
{
"input": "4 3\n2 1 1",
"output": "2"
},
{
"input": "3 9\n1 1 3",
"output": "1"
},
{
"input": "12345678 87654321\n43043751 1000000000 53798715",
"output": "2147483648"
},
{
"input": "12 12\n3 5 2",
"output": "0"
},
{
"input": "770 1390\n170 442 311",
"output": "12"
},
{
"input": "3555165 6693472\n1499112 556941 3075290",
"output": "3089339"
},
{
"input": "0 0\n1000000000 1000000000 1000000000",
"output": "7000000000"
},
{
"input": "1 1\n0 1 0",
"output": "0"
},
{
"input": "117708228 562858833\n118004008 360437130 154015822",
"output": "738362681"
},
{
"input": "999998118 700178721\n822106746 82987112 547955384",
"output": "1753877029"
},
{
"input": "566568710 765371101\n60614022 80126928 809950465",
"output": "1744607222"
},
{
"input": "448858599 829062060\n764716760 97644201 203890025",
"output": "1178219122"
},
{
"input": "626115781 966381948\n395190569 820194184 229233367",
"output": "1525971878"
},
{
"input": "803372962 103701834\n394260597 837711458 623172928",
"output": "3426388098"
},
{
"input": "980630143 241021722\n24734406 928857659 312079781",
"output": "1624075280"
},
{
"input": "862920032 378341609\n360240924 241342224 337423122",
"output": "974174021"
},
{
"input": "40177212 515661496\n64343660 963892207 731362684",
"output": "3694721078"
},
{
"input": "217434393 579352456\n694817470 981409480 756706026",
"output": "4825785129"
},
{
"input": "394691574 716672343\n398920207 72555681 150645586",
"output": "475704521"
},
{
"input": "276981463 853992230\n29394015 90072954 839552440",
"output": "1754738044"
},
{
"input": "843552056 919184611\n341530221 423649259 101547519",
"output": "263157645"
},
{
"input": "20809236 56504497\n972004030 441166533 495487081",
"output": "4235488636"
},
{
"input": "198066417 825228166\n602477839 532312735 520830423",
"output": "2808777834"
},
{
"input": "80356306 962548053\n601547868 549830008 914769984",
"output": "4004161345"
},
{
"input": "257613487 394835231\n642087093 567347282 308709545",
"output": "2692548667"
},
{
"input": "139903376 532155119\n641157122 289897263 629020178",
"output": "3077110809"
},
{
"input": "612127849 669475006\n271630930 676010757 22959739",
"output": "682559736"
},
{
"input": "0 0\n0 0 0",
"output": "0"
},
{
"input": "1000000000 1000000000\n499999998 4 333333332",
"output": "0"
},
{
"input": "1000000000 1000000000\n1000000000 1000000000 1000000000",
"output": "5000000000"
},
{
"input": "4 3\n1 0 1",
"output": "0"
},
{
"input": "4 12\n1 2 3",
"output": "0"
},
{
"input": "4 20\n1 2 1",
"output": "0"
},
{
"input": "100 10\n2 3 4",
"output": "5"
},
{
"input": "6 0\n1 1 1",
"output": "4"
},
{
"input": "25 5\n3 3 3",
"output": "7"
},
{
"input": "48 27\n22 39 20",
"output": "107"
},
{
"input": "4 0\n1 1 1",
"output": "4"
}
] | 1,515,701,267
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 38
| 62
| 5,632,000
|
# -*- coding: utf-8 -*-
"""
Created on Thu Jan 11 23:59:16 2018
@author: Sand Boa
"""
def validator(chunk:int, i:int, j:int)->int:
pass
if __name__ == "__main__":
yellow_crys, blue_crys = list(map(int,input().split()))
yellow, green, blue = list(map(int,input().split()))
yello_summ = max(((yellow*2+green)-yellow_crys),0)
blue_summ = max(((blue*3+green )-blue_crys),0)
ans=(yello_summ+blue_summ)
print(ans)
|
Title: Tricky Alchemy
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
During the winter holidays, the demand for Christmas balls is exceptionally high. Since it's already 2018, the advances in alchemy allow easy and efficient ball creation by utilizing magic crystals.
Grisha needs to obtain some yellow, green and blue balls. It's known that to produce a yellow ball one needs two yellow crystals, green — one yellow and one blue, and for a blue ball, three blue crystals are enough.
Right now there are *A* yellow and *B* blue crystals in Grisha's disposal. Find out how many additional crystals he should acquire in order to produce the required number of balls.
Input Specification:
The first line features two integers *A* and *B* (0<=≤<=*A*,<=*B*<=≤<=109), denoting the number of yellow and blue crystals respectively at Grisha's disposal.
The next line contains three integers *x*, *y* and *z* (0<=≤<=*x*,<=*y*,<=*z*<=≤<=109) — the respective amounts of yellow, green and blue balls to be obtained.
Output Specification:
Print a single integer — the minimum number of crystals that Grisha should acquire in addition.
Demo Input:
['4 3\n2 1 1\n', '3 9\n1 1 3\n', '12345678 87654321\n43043751 1000000000 53798715\n']
Demo Output:
['2\n', '1\n', '2147483648\n']
Note:
In the first sample case, Grisha needs five yellow and four blue crystals to create two yellow balls, one green ball, and one blue ball. To do that, Grisha needs to obtain two additional crystals: one yellow and one blue.
|
```python
# -*- coding: utf-8 -*-
"""
Created on Thu Jan 11 23:59:16 2018
@author: Sand Boa
"""
def validator(chunk:int, i:int, j:int)->int:
pass
if __name__ == "__main__":
yellow_crys, blue_crys = list(map(int,input().split()))
yellow, green, blue = list(map(int,input().split()))
yello_summ = max(((yellow*2+green)-yellow_crys),0)
blue_summ = max(((blue*3+green )-blue_crys),0)
ans=(yello_summ+blue_summ)
print(ans)
```
| 3
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size *a* can climb into some car with size *b* if and only if *a*<=≤<=*b*, he or she likes it if and only if he can climb into this car and 2*a*<=≥<=*b*.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
|
You are given four integers *V*1, *V*2, *V*3, *V**m*(1<=≤<=*V**i*<=≤<=100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that *V*1<=><=*V*2<=><=*V*3.
|
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
|
[
"50 30 10 10\n",
"100 50 10 21\n"
] |
[
"50\n30\n10\n",
"-1\n"
] |
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
| 0
|
[
{
"input": "50 30 10 10",
"output": "50\n30\n10"
},
{
"input": "100 50 10 21",
"output": "-1"
},
{
"input": "100 50 19 10",
"output": "100\n50\n19"
},
{
"input": "99 50 25 49",
"output": "100\n99\n49"
},
{
"input": "3 2 1 1",
"output": "4\n3\n1"
},
{
"input": "100 99 98 100",
"output": "-1"
},
{
"input": "100 40 30 40",
"output": "-1"
},
{
"input": "100 50 19 25",
"output": "100\n51\n25"
},
{
"input": "100 50 19 30",
"output": "100\n61\n30"
},
{
"input": "49 48 25 49",
"output": "-1"
},
{
"input": "48 47 23 46",
"output": "94\n93\n46"
},
{
"input": "37 23 16 20",
"output": "42\n41\n20"
},
{
"input": "98 2 1 1",
"output": "98\n3\n1"
},
{
"input": "99 22 13 14",
"output": "99\n29\n14"
},
{
"input": "97 95 3 2",
"output": "97\n95\n3"
},
{
"input": "27 3 2 3",
"output": "-1"
},
{
"input": "13 7 6 2",
"output": "-1"
},
{
"input": "19 17 11 6",
"output": "19\n17\n11"
},
{
"input": "15 6 4 5",
"output": "15\n11\n5"
},
{
"input": "21 3 1 3",
"output": "-1"
},
{
"input": "100 99 98 97",
"output": "196\n195\n98"
},
{
"input": "99 32 13 5",
"output": "-1"
},
{
"input": "5 4 3 1",
"output": "-1"
},
{
"input": "50 20 4 8",
"output": "50\n20\n8"
},
{
"input": "60 30 10 1",
"output": "-1"
},
{
"input": "100 50 10 1",
"output": "-1"
},
{
"input": "40 30 20 10",
"output": "40\n30\n20"
},
{
"input": "50 49 5 10",
"output": "50\n49\n10"
},
{
"input": "99 98 50 97",
"output": "196\n195\n97"
},
{
"input": "50 30 10 5",
"output": "50\n30\n10"
},
{
"input": "4 3 2 1",
"output": "4\n3\n2"
},
{
"input": "100 50 3 1",
"output": "-1"
}
] | 1,514,038,606
| 706
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 2
| 46
| 5,529,600
|
v1, v2, v3, vm = map(int, input().split())
if vm > v3*2 or v3 > vm*2 or vm>=v2:
print(-1)
else:
print(2*v1, min(v1-1,2*v2), min(v2-1,2*v3))
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A family consisting of father bear, mother bear and son bear owns three cars. Father bear can climb into the largest car and he likes it. Also, mother bear can climb into the middle car and she likes it. Moreover, son bear can climb into the smallest car and he likes it. It's known that the largest car is strictly larger than the middle car, and the middle car is strictly larger than the smallest car.
Masha came to test these cars. She could climb into all cars, but she liked only the smallest car.
It's known that a character with size *a* can climb into some car with size *b* if and only if *a*<=≤<=*b*, he or she likes it if and only if he can climb into this car and 2*a*<=≥<=*b*.
You are given sizes of bears and Masha. Find out some possible integer non-negative sizes of cars.
Input Specification:
You are given four integers *V*1, *V*2, *V*3, *V**m*(1<=≤<=*V**i*<=≤<=100) — sizes of father bear, mother bear, son bear and Masha, respectively. It's guaranteed that *V*1<=><=*V*2<=><=*V*3.
Output Specification:
Output three integers — sizes of father bear's car, mother bear's car and son bear's car, respectively.
If there are multiple possible solutions, print any.
If there is no solution, print "-1" (without quotes).
Demo Input:
['50 30 10 10\n', '100 50 10 21\n']
Demo Output:
['50\n30\n10\n', '-1\n']
Note:
In first test case all conditions for cars' sizes are satisfied.
In second test case there is no answer, because Masha should be able to climb into smallest car (so size of smallest car in not less than 21), but son bear should like it, so maximum possible size of it is 20.
|
```python
v1, v2, v3, vm = map(int, input().split())
if vm > v3*2 or v3 > vm*2 or vm>=v2:
print(-1)
else:
print(2*v1, min(v1-1,2*v2), min(v2-1,2*v3))
```
| 0
|
|
743
|
B
|
Chloe and the sequence
|
PROGRAMMING
| 1,200
|
[
"binary search",
"bitmasks",
"constructive algorithms",
"implementation"
] | null | null |
Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad.
Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (*n*<=-<=1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1,<=2,<=1] after the first step, the sequence [1,<=2,<=1,<=3,<=1,<=2,<=1] after the second step.
The task is to find the value of the element with index *k* (the elements are numbered from 1) in the obtained sequence, i. e. after (*n*<=-<=1) steps.
Please help Chloe to solve the problem!
|
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=2*n*<=-<=1).
|
Print single integer — the integer at the *k*-th position in the obtained sequence.
|
[
"3 2\n",
"4 8\n"
] |
[
"2",
"4"
] |
In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2.
In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4.
| 1,000
|
[
{
"input": "3 2",
"output": "2"
},
{
"input": "4 8",
"output": "4"
},
{
"input": "5 27",
"output": "1"
},
{
"input": "7 44",
"output": "3"
},
{
"input": "15 18432",
"output": "12"
},
{
"input": "20 259676",
"output": "3"
},
{
"input": "30 671088640",
"output": "28"
},
{
"input": "38 137438953472",
"output": "38"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "3 3",
"output": "1"
},
{
"input": "4 12",
"output": "3"
},
{
"input": "8 224",
"output": "6"
},
{
"input": "13 1368",
"output": "4"
},
{
"input": "16 49152",
"output": "15"
},
{
"input": "19 1024",
"output": "11"
},
{
"input": "24 15204352",
"output": "20"
},
{
"input": "27 6586544",
"output": "5"
},
{
"input": "31 536870912",
"output": "30"
},
{
"input": "38 94489280512",
"output": "34"
},
{
"input": "42 215268308020",
"output": "3"
},
{
"input": "42 3452074721280",
"output": "21"
},
{
"input": "46 34394312982528",
"output": "15"
},
{
"input": "48 133865540681728",
"output": "39"
},
{
"input": "49 76541041985542",
"output": "2"
},
{
"input": "49 104822971826176",
"output": "34"
},
{
"input": "49 351843720888320",
"output": "47"
},
{
"input": "50 743042492421629",
"output": "1"
},
{
"input": "50 666266740436818",
"output": "2"
},
{
"input": "50 704239287953456",
"output": "5"
},
{
"input": "50 116938486513664",
"output": "17"
},
{
"input": "50 806423059496960",
"output": "37"
},
{
"input": "50 985162418487296",
"output": "48"
},
{
"input": "50 844424930131968",
"output": "49"
},
{
"input": "50 562949953421312",
"output": "50"
},
{
"input": "50 1",
"output": "1"
},
{
"input": "50 1125899906842623",
"output": "1"
},
{
"input": "50 1125899906842620",
"output": "3"
},
{
"input": "39 549755813887",
"output": "1"
},
{
"input": "50 100000000000",
"output": "12"
}
] | 1,667,049,396
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 39
| 77
| 0
|
def rec(l,h,n,k):
m=(l+h)//2
if(m>k):
return rec(l,m-1,n-1,k)
elif(m==k):
return n
else:
return rec(m+1,h,n-1,k)
n,k=map(int,input().split())
print(rec(1,(1<<n)-1,n,k))
|
Title: Chloe and the sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad.
Let's consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (*n*<=-<=1) steps. On each step we take the sequence we've got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven't used before. For example, we get the sequence [1,<=2,<=1] after the first step, the sequence [1,<=2,<=1,<=3,<=1,<=2,<=1] after the second step.
The task is to find the value of the element with index *k* (the elements are numbered from 1) in the obtained sequence, i. e. after (*n*<=-<=1) steps.
Please help Chloe to solve the problem!
Input Specification:
The only line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=50, 1<=≤<=*k*<=≤<=2*n*<=-<=1).
Output Specification:
Print single integer — the integer at the *k*-th position in the obtained sequence.
Demo Input:
['3 2\n', '4 8\n']
Demo Output:
['2', '4']
Note:
In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2.
In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4.
|
```python
def rec(l,h,n,k):
m=(l+h)//2
if(m>k):
return rec(l,m-1,n-1,k)
elif(m==k):
return n
else:
return rec(m+1,h,n-1,k)
n,k=map(int,input().split())
print(rec(1,(1<<n)-1,n,k))
```
| 3
|
|
710
|
E
|
Generate a String
|
PROGRAMMING
| 2,000
|
[
"dfs and similar",
"dp"
] | null | null |
zscoder wants to generate an input file for some programming competition problem.
His input is a string consisting of *n* letters 'a'. He is too lazy to write a generator so he will manually generate the input in a text editor.
Initially, the text editor is empty. It takes him *x* seconds to insert or delete a letter 'a' from the text file and *y* seconds to copy the contents of the entire text file, and duplicate it.
zscoder wants to find the minimum amount of time needed for him to create the input file of exactly *n* letters 'a'. Help him to determine the amount of time needed to generate the input.
|
The only line contains three integers *n*, *x* and *y* (1<=≤<=*n*<=≤<=107, 1<=≤<=*x*,<=*y*<=≤<=109) — the number of letters 'a' in the input file and the parameters from the problem statement.
|
Print the only integer *t* — the minimum amount of time needed to generate the input file.
|
[
"8 1 1\n",
"8 1 10\n"
] |
[
"4\n",
"8\n"
] |
none
| 0
|
[
{
"input": "8 1 1",
"output": "4"
},
{
"input": "8 1 10",
"output": "8"
},
{
"input": "10 62 99",
"output": "384"
},
{
"input": "88 417 591",
"output": "4623"
},
{
"input": "57 5289 8444",
"output": "60221"
},
{
"input": "382 81437847 324871127",
"output": "2519291691"
},
{
"input": "244 575154303 436759189",
"output": "5219536421"
},
{
"input": "85 902510038 553915152",
"output": "6933531064"
},
{
"input": "1926 84641582 820814219",
"output": "7184606427"
},
{
"input": "3768 561740421 232937477",
"output": "5042211408"
},
{
"input": "2313 184063453 204869248",
"output": "2969009745"
},
{
"input": "35896 278270961 253614967",
"output": "5195579310"
},
{
"input": "483867 138842067 556741142",
"output": "10712805143"
},
{
"input": "4528217 187553422 956731625",
"output": "21178755627"
},
{
"input": "10000000 1000000000 1",
"output": "8000000023"
},
{
"input": "10000000 1 100",
"output": "1757"
},
{
"input": "10000000 1 1000000000",
"output": "10000000"
},
{
"input": "10000000 1 1000",
"output": "14224"
},
{
"input": "10000000 1 10",
"output": "214"
},
{
"input": "1 1 1",
"output": "1"
},
{
"input": "10000000 998 998",
"output": "30938"
},
{
"input": "9999999 987654321 123456789",
"output": "11728395036"
},
{
"input": "9999999 1 2",
"output": "54"
},
{
"input": "10000000 1 1",
"output": "31"
},
{
"input": "11478 29358 26962",
"output": "556012"
},
{
"input": "4314870 1000000000 1",
"output": "7000000022"
},
{
"input": "7186329 608148870 290497442",
"output": "12762929866"
},
{
"input": "9917781 1 1",
"output": "35"
},
{
"input": "7789084 807239576 813643932",
"output": "25165322688"
},
{
"input": "58087 1 100000000",
"output": "58087"
},
{
"input": "9999991 2 3",
"output": "88"
}
] | 1,471,965,594
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 62
| 4,608,000
|
def gen_string():
""" This function is a solution for the problem at:
http://codeforces.com/problemset/problem/710/E
Idea:
Dynamic Programming. To calculate the shortest time to generate
a string of length n, we only have to compare the times to generate
those strings 2 ways:
- first: best way to generate string of length n-1 + append a letter
- second: best way to generate string of length n/2 + duplicate
If n is odd the second option is the only one
Complexity:
- Time: O(n)
- Space: O(n)
"""
line = input().split()
n, x, y = (int(x) for x in line)
solution = [0] * (n+1)
for i in range(1,n+1):
solution[i] = solution[i-1] + x
if i % 2 == 0:
solution[i] = min(solution[i], solution[i//2] + y)
return solution[n]
print(gen_string())
|
Title: Generate a String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
zscoder wants to generate an input file for some programming competition problem.
His input is a string consisting of *n* letters 'a'. He is too lazy to write a generator so he will manually generate the input in a text editor.
Initially, the text editor is empty. It takes him *x* seconds to insert or delete a letter 'a' from the text file and *y* seconds to copy the contents of the entire text file, and duplicate it.
zscoder wants to find the minimum amount of time needed for him to create the input file of exactly *n* letters 'a'. Help him to determine the amount of time needed to generate the input.
Input Specification:
The only line contains three integers *n*, *x* and *y* (1<=≤<=*n*<=≤<=107, 1<=≤<=*x*,<=*y*<=≤<=109) — the number of letters 'a' in the input file and the parameters from the problem statement.
Output Specification:
Print the only integer *t* — the minimum amount of time needed to generate the input file.
Demo Input:
['8 1 1\n', '8 1 10\n']
Demo Output:
['4\n', '8\n']
Note:
none
|
```python
def gen_string():
""" This function is a solution for the problem at:
http://codeforces.com/problemset/problem/710/E
Idea:
Dynamic Programming. To calculate the shortest time to generate
a string of length n, we only have to compare the times to generate
those strings 2 ways:
- first: best way to generate string of length n-1 + append a letter
- second: best way to generate string of length n/2 + duplicate
If n is odd the second option is the only one
Complexity:
- Time: O(n)
- Space: O(n)
"""
line = input().split()
n, x, y = (int(x) for x in line)
solution = [0] * (n+1)
for i in range(1,n+1):
solution[i] = solution[i-1] + x
if i % 2 == 0:
solution[i] = min(solution[i], solution[i//2] + y)
return solution[n]
print(gen_string())
```
| 0
|
|
116
|
A
|
Tram
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
|
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
|
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
|
[
"4\n0 3\n2 5\n4 2\n4 0\n"
] |
[
"6\n"
] |
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.
| 500
|
[
{
"input": "4\n0 3\n2 5\n4 2\n4 0",
"output": "6"
},
{
"input": "5\n0 4\n4 6\n6 5\n5 4\n4 0",
"output": "6"
},
{
"input": "10\n0 5\n1 7\n10 8\n5 3\n0 5\n3 3\n8 8\n0 6\n10 1\n9 0",
"output": "18"
},
{
"input": "3\n0 1\n1 1\n1 0",
"output": "1"
},
{
"input": "4\n0 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "3\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "5\n0 73\n73 189\n189 766\n766 0\n0 0",
"output": "766"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 1\n1 0",
"output": "1"
},
{
"input": "5\n0 917\n917 923\n904 992\n1000 0\n11 0",
"output": "1011"
},
{
"input": "5\n0 1\n1 2\n2 1\n1 2\n2 0",
"output": "2"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "20\n0 7\n2 1\n2 2\n5 7\n2 6\n6 10\n2 4\n0 4\n7 4\n8 0\n10 6\n2 1\n6 1\n1 7\n0 3\n8 7\n6 3\n6 3\n1 1\n3 0",
"output": "22"
},
{
"input": "5\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "10\n0 592\n258 598\n389 203\n249 836\n196 635\n478 482\n994 987\n1000 0\n769 0\n0 0",
"output": "1776"
},
{
"input": "10\n0 1\n1 0\n0 0\n0 0\n0 0\n0 1\n1 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "10\n0 926\n926 938\n938 931\n931 964\n937 989\n983 936\n908 949\n997 932\n945 988\n988 0",
"output": "1016"
},
{
"input": "10\n0 1\n1 2\n1 2\n2 2\n2 2\n2 2\n1 1\n1 1\n2 1\n2 0",
"output": "3"
},
{
"input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "10\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "50\n0 332\n332 268\n268 56\n56 711\n420 180\n160 834\n149 341\n373 777\n763 93\n994 407\n86 803\n700 132\n471 608\n429 467\n75 5\n638 305\n405 853\n316 478\n643 163\n18 131\n648 241\n241 766\n316 847\n640 380\n923 759\n789 41\n125 421\n421 9\n9 388\n388 829\n408 108\n462 856\n816 411\n518 688\n290 7\n405 912\n397 772\n396 652\n394 146\n27 648\n462 617\n514 433\n780 35\n710 705\n460 390\n194 508\n643 56\n172 469\n1000 0\n194 0",
"output": "2071"
},
{
"input": "50\n0 0\n0 1\n1 1\n0 1\n0 0\n1 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 1\n1 0\n0 1\n0 0\n1 1\n1 0\n0 1\n0 0\n1 1\n0 1\n1 0\n1 1\n1 0\n0 0\n1 1\n1 0\n0 1\n0 0\n0 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 0\n0 1\n1 0\n0 0\n0 1\n1 1\n1 1\n0 1\n0 0\n1 0\n1 0",
"output": "3"
},
{
"input": "50\n0 926\n926 971\n915 980\n920 965\n954 944\n928 952\n955 980\n916 980\n906 935\n944 913\n905 923\n912 922\n965 934\n912 900\n946 930\n931 983\n979 905\n925 969\n924 926\n910 914\n921 977\n934 979\n962 986\n942 909\n976 903\n982 982\n991 941\n954 929\n902 980\n947 983\n919 924\n917 943\n916 905\n907 913\n964 977\n984 904\n905 999\n950 970\n986 906\n993 970\n960 994\n963 983\n918 986\n980 900\n931 986\n993 997\n941 909\n907 909\n1000 0\n278 0",
"output": "1329"
},
{
"input": "2\n0 863\n863 0",
"output": "863"
},
{
"input": "50\n0 1\n1 2\n2 2\n1 1\n1 1\n1 2\n1 2\n1 1\n1 2\n1 1\n1 1\n1 2\n1 2\n1 1\n2 1\n2 2\n1 2\n2 2\n1 2\n2 1\n2 1\n2 2\n2 1\n1 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n2 2\n2 1\n1 2\n2 2\n1 2\n1 1\n1 1\n2 1\n2 1\n2 2\n2 1\n2 1\n1 2\n1 2\n1 2\n1 2\n2 0\n2 0\n2 0\n0 0",
"output": "8"
},
{
"input": "50\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "100\n0 1\n0 0\n0 0\n1 0\n0 0\n0 1\n0 1\n1 1\n0 0\n0 0\n1 1\n0 0\n1 1\n0 1\n1 1\n0 1\n1 1\n1 0\n1 0\n0 0\n1 0\n0 1\n1 0\n0 0\n0 0\n1 1\n1 1\n0 1\n0 0\n1 0\n1 1\n0 1\n1 0\n1 1\n0 1\n1 1\n1 0\n0 0\n0 0\n0 1\n0 0\n0 1\n1 1\n0 0\n1 1\n1 1\n0 0\n0 1\n1 0\n0 1\n0 0\n0 1\n0 1\n1 1\n1 1\n1 1\n0 0\n0 0\n1 1\n0 1\n0 1\n1 0\n0 0\n0 0\n1 1\n0 1\n0 1\n1 1\n1 1\n0 1\n1 1\n1 1\n0 0\n1 0\n0 1\n0 0\n0 0\n1 1\n1 1\n1 1\n1 1\n0 1\n1 0\n1 0\n1 0\n1 0\n1 0\n0 0\n1 0\n1 0\n0 0\n1 0\n0 0\n0 1\n1 0\n0 1\n1 0\n1 0\n1 0\n1 0",
"output": "11"
},
{
"input": "100\n0 2\n1 2\n2 1\n1 2\n1 2\n2 1\n2 2\n1 1\n1 1\n2 1\n1 2\n2 1\n1 2\n2 2\n2 2\n2 2\n1 2\n2 2\n2 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 2\n1 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n1 1\n2 2\n2 1\n1 2\n1 1\n1 2\n2 1\n2 2\n1 1\n2 1\n1 1\n2 1\n1 1\n1 2\n2 2\n2 2\n1 1\n2 2\n1 2\n2 1\n2 1\n1 1\n1 1\n1 2\n1 2\n1 1\n1 1\n2 1\n1 2\n1 2\n2 1\n2 2\n2 2\n2 2\n2 1\n2 2\n1 1\n1 2\n1 2\n1 1\n2 2\n2 2\n1 1\n2 1\n1 1\n1 2\n1 2\n1 2\n1 1\n1 1\n2 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 1\n2 2\n1 2\n2 0\n2 0\n2 0\n1 0",
"output": "7"
},
{
"input": "100\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 900\n900 0",
"output": "900"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1000\n1000 0",
"output": "1000"
},
{
"input": "3\n0 802\n175 188\n815 0",
"output": "815"
},
{
"input": "3\n0 910\n910 976\n976 0",
"output": "976"
},
{
"input": "3\n0 2\n2 1\n1 0",
"output": "2"
}
] | 1,685,892,870
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 62
| 0
|
import sys
a = int(input(""))
s = 0
b = [int(i) for i in sys.stdin.readline().split()]
for i in b:
s = s - b[0]
s = s + b[1]
print(end = "\n")
print(s)
|
Title: Tram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
Input Specification:
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
Output Specification:
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
Demo Input:
['4\n0 3\n2 5\n4 2\n4 0\n']
Demo Output:
['6\n']
Note:
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.
|
```python
import sys
a = int(input(""))
s = 0
b = [int(i) for i in sys.stdin.readline().split()]
for i in b:
s = s - b[0]
s = s + b[1]
print(end = "\n")
print(s)
```
| 0
|
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "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"
}
] | 1,671,528,128
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
s=input()
length=len(s)
if length<=10:
print(s)
elif length>10:
print(s[0]+str(length-2)+s[length-1])
|
Title: Way Too Long Words
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
Output Specification:
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
Demo Input:
['4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n']
Demo Output:
['word\nl10n\ni18n\np43s\n']
Note:
none
|
```python
s=input()
length=len(s)
if length<=10:
print(s)
elif length>10:
print(s[0]+str(length-2)+s[length-1])
```
| 0
|
484
|
A
|
Bits
|
PROGRAMMING
| 1,700
|
[
"bitmasks",
"constructive algorithms"
] | null | null |
Let's denote as the number of bits set ('1' bits) in the binary representation of the non-negative integer *x*.
You are given multiple queries consisting of pairs of integers *l* and *r*. For each query, find the *x*, such that *l*<=≤<=*x*<=≤<=*r*, and is maximum possible. If there are multiple such numbers find the smallest of them.
|
The first line contains integer *n* — the number of queries (1<=≤<=*n*<=≤<=10000).
Each of the following *n* lines contain two integers *l**i*,<=*r**i* — the arguments for the corresponding query (0<=≤<=*l**i*<=≤<=*r**i*<=≤<=1018).
|
For each query print the answer in a separate line.
|
[
"3\n1 2\n2 4\n1 10\n"
] |
[
"1\n3\n7\n"
] |
The binary representations of numbers from 1 to 10 are listed below:
1<sub class="lower-index">10</sub> = 1<sub class="lower-index">2</sub>
2<sub class="lower-index">10</sub> = 10<sub class="lower-index">2</sub>
3<sub class="lower-index">10</sub> = 11<sub class="lower-index">2</sub>
4<sub class="lower-index">10</sub> = 100<sub class="lower-index">2</sub>
5<sub class="lower-index">10</sub> = 101<sub class="lower-index">2</sub>
6<sub class="lower-index">10</sub> = 110<sub class="lower-index">2</sub>
7<sub class="lower-index">10</sub> = 111<sub class="lower-index">2</sub>
8<sub class="lower-index">10</sub> = 1000<sub class="lower-index">2</sub>
9<sub class="lower-index">10</sub> = 1001<sub class="lower-index">2</sub>
10<sub class="lower-index">10</sub> = 1010<sub class="lower-index">2</sub>
| 500
|
[
{
"input": "3\n1 2\n2 4\n1 10",
"output": "1\n3\n7"
},
{
"input": "55\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n2 2\n2 3\n2 4\n2 5\n2 6\n2 7\n2 8\n2 9\n2 10\n3 3\n3 4\n3 5\n3 6\n3 7\n3 8\n3 9\n3 10\n4 4\n4 5\n4 6\n4 7\n4 8\n4 9\n4 10\n5 5\n5 6\n5 7\n5 8\n5 9\n5 10\n6 6\n6 7\n6 8\n6 9\n6 10\n7 7\n7 8\n7 9\n7 10\n8 8\n8 9\n8 10\n9 9\n9 10\n10 10",
"output": "1\n1\n3\n3\n3\n3\n7\n7\n7\n7\n2\n3\n3\n3\n3\n7\n7\n7\n7\n3\n3\n3\n3\n7\n7\n7\n7\n4\n5\n5\n7\n7\n7\n7\n5\n5\n7\n7\n7\n7\n6\n7\n7\n7\n7\n7\n7\n7\n7\n8\n9\n9\n9\n9\n10"
},
{
"input": "18\n1 10\n1 100\n1 1000\n1 10000\n1 100000\n1 1000000\n1 10000000\n1 100000000\n1 1000000000\n1 10000000000\n1 100000000000\n1 1000000000000\n1 10000000000000\n1 100000000000000\n1 1000000000000000\n1 10000000000000000\n1 100000000000000000\n1 1000000000000000000",
"output": "7\n63\n511\n8191\n65535\n524287\n8388607\n67108863\n536870911\n8589934591\n68719476735\n549755813887\n8796093022207\n70368744177663\n562949953421311\n9007199254740991\n72057594037927935\n576460752303423487"
},
{
"input": "3\n0 0\n1 3\n2 4",
"output": "0\n3\n3"
},
{
"input": "17\n0 0\n0 8\n1 8\n36 39\n3 4\n3 7\n2 17\n8 12\n9 12\n10 12\n10 15\n6 14\n8 15\n9 15\n15 15\n100000000000000000 1000000000000000000\n99999999999999999 1000000000000000000",
"output": "0\n7\n7\n39\n3\n7\n15\n11\n11\n11\n15\n7\n15\n15\n15\n576460752303423487\n576460752303423487"
}
] | 1,619,460,239
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 27
| 373
| 0
|
def main():
cases = int(input())
for i in range(cases):
left, right = list(map(int, str(input()).split(" ")))
res = left
for j in range(65):
if ((1 << j) & left) == 0:
if (res | (1 << j)) > right:
break
res |= (1 << j)
print(res)
if __name__ == "__main__":
main()
|
Title: Bits
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's denote as the number of bits set ('1' bits) in the binary representation of the non-negative integer *x*.
You are given multiple queries consisting of pairs of integers *l* and *r*. For each query, find the *x*, such that *l*<=≤<=*x*<=≤<=*r*, and is maximum possible. If there are multiple such numbers find the smallest of them.
Input Specification:
The first line contains integer *n* — the number of queries (1<=≤<=*n*<=≤<=10000).
Each of the following *n* lines contain two integers *l**i*,<=*r**i* — the arguments for the corresponding query (0<=≤<=*l**i*<=≤<=*r**i*<=≤<=1018).
Output Specification:
For each query print the answer in a separate line.
Demo Input:
['3\n1 2\n2 4\n1 10\n']
Demo Output:
['1\n3\n7\n']
Note:
The binary representations of numbers from 1 to 10 are listed below:
1<sub class="lower-index">10</sub> = 1<sub class="lower-index">2</sub>
2<sub class="lower-index">10</sub> = 10<sub class="lower-index">2</sub>
3<sub class="lower-index">10</sub> = 11<sub class="lower-index">2</sub>
4<sub class="lower-index">10</sub> = 100<sub class="lower-index">2</sub>
5<sub class="lower-index">10</sub> = 101<sub class="lower-index">2</sub>
6<sub class="lower-index">10</sub> = 110<sub class="lower-index">2</sub>
7<sub class="lower-index">10</sub> = 111<sub class="lower-index">2</sub>
8<sub class="lower-index">10</sub> = 1000<sub class="lower-index">2</sub>
9<sub class="lower-index">10</sub> = 1001<sub class="lower-index">2</sub>
10<sub class="lower-index">10</sub> = 1010<sub class="lower-index">2</sub>
|
```python
def main():
cases = int(input())
for i in range(cases):
left, right = list(map(int, str(input()).split(" ")))
res = left
for j in range(65):
if ((1 << j) & left) == 0:
if (res | (1 << j)) > right:
break
res |= (1 << j)
print(res)
if __name__ == "__main__":
main()
```
| 3
|
|
296
|
A
|
Yaroslav and Permutations
|
PROGRAMMING
| 1,100
|
[
"greedy",
"math"
] | null | null |
Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.
Help Yaroslav.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements.
|
In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.
|
[
"1\n1\n",
"3\n1 1 2\n",
"4\n7 7 7 7\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
In the first sample the initial array fits well.
In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.
In the third sample Yarosav can't get the array he needs.
| 500
|
[
{
"input": "1\n1",
"output": "YES"
},
{
"input": "3\n1 1 2",
"output": "YES"
},
{
"input": "4\n7 7 7 7",
"output": "NO"
},
{
"input": "4\n479 170 465 146",
"output": "YES"
},
{
"input": "5\n996 437 605 996 293",
"output": "YES"
},
{
"input": "6\n727 539 896 668 36 896",
"output": "YES"
},
{
"input": "7\n674 712 674 674 674 674 674",
"output": "NO"
},
{
"input": "8\n742 742 742 742 742 289 742 742",
"output": "NO"
},
{
"input": "9\n730 351 806 806 806 630 85 757 967",
"output": "YES"
},
{
"input": "10\n324 539 83 440 834 640 440 440 440 440",
"output": "YES"
},
{
"input": "7\n925 830 925 98 987 162 356",
"output": "YES"
},
{
"input": "68\n575 32 53 351 151 942 725 967 431 108 192 8 338 458 288 754 384 946 910 210 759 222 589 423 947 507 31 414 169 901 592 763 656 411 360 625 538 549 484 596 42 603 351 292 837 375 21 597 22 349 200 669 485 282 735 54 1000 419 939 901 789 128 468 729 894 649 484 808",
"output": "YES"
},
{
"input": "22\n618 814 515 310 617 936 452 601 250 520 557 799 304 225 9 845 610 990 703 196 486 94",
"output": "YES"
},
{
"input": "44\n459 581 449 449 449 449 449 449 449 623 449 449 449 449 449 449 449 449 889 449 203 273 329 449 449 449 449 449 449 845 882 323 22 449 449 893 449 449 449 449 449 870 449 402",
"output": "NO"
},
{
"input": "90\n424 3 586 183 286 89 427 618 758 833 933 170 155 722 190 977 330 369 693 426 556 435 550 442 513 146 61 719 754 140 424 280 997 688 530 550 438 867 950 194 196 298 417 287 106 489 283 456 735 115 702 317 672 787 264 314 356 186 54 913 809 833 946 314 757 322 559 647 983 482 145 197 223 130 162 536 451 174 467 45 660 293 440 254 25 155 511 746 650 187",
"output": "YES"
},
{
"input": "14\n959 203 478 315 788 788 373 834 488 519 774 764 193 103",
"output": "YES"
},
{
"input": "81\n544 528 528 528 528 4 506 528 32 528 528 528 528 528 528 528 528 975 528 528 528 528 528 528 528 528 528 528 528 528 528 20 528 528 528 528 528 528 528 528 852 528 528 120 528 528 61 11 528 528 528 228 528 165 883 528 488 475 628 528 528 528 528 528 528 597 528 528 528 528 528 528 528 528 528 528 528 412 528 521 925",
"output": "NO"
},
{
"input": "89\n354 356 352 355 355 355 352 354 354 352 355 356 355 352 354 356 354 355 355 354 353 352 352 355 355 356 352 352 353 356 352 353 354 352 355 352 353 353 353 354 353 354 354 353 356 353 353 354 354 354 354 353 352 353 355 356 356 352 356 354 353 352 355 354 356 356 356 354 354 356 354 355 354 355 353 352 354 355 352 355 355 354 356 353 353 352 356 352 353",
"output": "YES"
},
{
"input": "71\n284 284 285 285 285 284 285 284 284 285 284 285 284 284 285 284 285 285 285 285 284 284 285 285 284 284 284 285 284 285 284 285 285 284 284 284 285 284 284 285 285 285 284 284 285 284 285 285 284 285 285 284 285 284 284 284 285 285 284 285 284 285 285 285 285 284 284 285 285 284 285",
"output": "NO"
},
{
"input": "28\n602 216 214 825 814 760 814 28 76 814 814 288 814 814 222 707 11 490 814 543 914 705 814 751 976 814 814 99",
"output": "YES"
},
{
"input": "48\n546 547 914 263 986 945 914 914 509 871 324 914 153 571 914 914 914 528 970 566 544 914 914 914 410 914 914 589 609 222 914 889 691 844 621 68 914 36 914 39 630 749 914 258 945 914 727 26",
"output": "YES"
},
{
"input": "56\n516 76 516 197 516 427 174 516 706 813 94 37 516 815 516 516 937 483 16 516 842 516 638 691 516 635 516 516 453 263 516 516 635 257 125 214 29 81 516 51 362 516 677 516 903 516 949 654 221 924 516 879 516 516 972 516",
"output": "YES"
},
{
"input": "46\n314 723 314 314 314 235 314 314 314 314 270 314 59 972 314 216 816 40 314 314 314 314 314 314 314 381 314 314 314 314 314 314 314 789 314 957 114 942 314 314 29 314 314 72 314 314",
"output": "NO"
},
{
"input": "72\n169 169 169 599 694 81 250 529 865 406 817 169 667 169 965 169 169 663 65 169 903 169 942 763 169 807 169 603 169 169 13 169 169 810 169 291 169 169 169 169 169 169 169 713 169 440 169 169 169 169 169 480 169 169 867 169 169 169 169 169 169 169 169 393 169 169 459 169 99 169 601 800",
"output": "NO"
},
{
"input": "100\n317 316 317 316 317 316 317 316 317 316 316 317 317 316 317 316 316 316 317 316 317 317 316 317 316 316 316 316 316 316 317 316 317 317 317 317 317 317 316 316 316 317 316 317 316 317 316 317 317 316 317 316 317 317 316 317 316 317 316 317 316 316 316 317 317 317 317 317 316 317 317 316 316 316 316 317 317 316 317 316 316 316 316 316 316 317 316 316 317 317 317 317 317 317 317 317 317 316 316 317",
"output": "NO"
},
{
"input": "100\n510 510 510 162 969 32 510 511 510 510 911 183 496 875 903 461 510 510 123 578 510 510 510 510 510 755 510 673 510 510 763 510 510 909 510 435 487 959 807 510 368 788 557 448 284 332 510 949 510 510 777 112 857 926 487 510 510 510 678 510 510 197 829 427 698 704 409 509 510 238 314 851 510 651 510 455 682 510 714 635 973 510 443 878 510 510 510 591 510 24 596 510 43 183 510 510 671 652 214 784",
"output": "YES"
},
{
"input": "100\n476 477 474 476 476 475 473 476 474 475 473 477 476 476 474 476 474 475 476 477 473 473 473 474 474 476 473 473 476 476 475 476 473 474 473 473 477 475 475 475 476 475 477 477 477 476 475 475 475 473 476 477 475 476 477 473 474 477 473 475 476 476 474 477 476 474 473 477 473 475 477 473 476 474 477 473 475 477 473 476 476 475 476 475 474 473 477 473 475 473 477 473 473 474 475 473 477 476 477 474",
"output": "YES"
},
{
"input": "100\n498 498 498 498 498 499 498 499 499 499 498 498 498 498 499 498 499 499 498 499 498 498 498 499 499 499 498 498 499 499 498 498 498 499 498 499 498 498 498 499 498 499 498 498 498 498 499 498 498 499 498 498 499 498 499 499 498 499 499 499 498 498 498 498 499 498 499 498 499 499 499 499 498 498 499 499 498 499 499 498 498 499 499 498 498 499 499 499 498 498 499 498 498 498 499 499 499 498 498 499",
"output": "NO"
},
{
"input": "100\n858 53 816 816 816 816 816 816 816 181 816 816 816 816 579 879 816 948 171 816 816 150 866 816 816 816 897 816 816 816 816 816 816 706 816 539 816 816 816 816 816 816 423 487 816 615 254 816 816 816 816 83 816 816 816 816 816 816 816 816 816 816 816 136 775 999 816 816 816 644 816 816 816 816 927 816 802 816 856 816 816 816 816 816 816 816 816 816 816 700 816 816 816 816 982 477 816 891 806 816",
"output": "NO"
},
{
"input": "100\n167 169 169 167 169 169 167 167 167 167 168 166 170 170 169 170 170 170 169 168 166 167 170 169 167 169 168 169 166 170 166 167 170 166 166 167 169 166 166 169 166 167 168 168 170 167 168 166 168 170 167 168 167 169 169 166 168 167 170 168 167 169 168 169 166 168 168 169 169 166 170 168 167 169 170 168 167 169 168 167 168 168 166 169 170 170 166 166 167 170 167 168 167 167 169 169 166 166 169 167",
"output": "YES"
},
{
"input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "NO"
},
{
"input": "99\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "NO"
},
{
"input": "100\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 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": "YES"
},
{
"input": "99\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 1000 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": "YES"
},
{
"input": "2\n1 1",
"output": "NO"
},
{
"input": "1\n1000",
"output": "YES"
},
{
"input": "12\n2 2 4 4 4 4 6 6 6 6 6 6",
"output": "YES"
}
] | 1,629,025,089
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 154
| 6,758,400
|
n = int(input())
a = input().split(' ')
s = [int(i) for i in a]
max_num = max(s,key=s.count)
max_count = s.count(max_num)
if max_count>((n//2)+(n%2)):
print("NO")
else:
print("YES")
|
Title: Yaroslav and Permutations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Yaroslav has an array that consists of *n* integers. In one second Yaroslav can swap two neighboring array elements. Now Yaroslav is wondering if he can obtain an array where any two neighboring elements would be distinct in a finite time.
Help Yaroslav.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=100) — the number of elements in the array. The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=1000) — the array elements.
Output Specification:
In the single line print "YES" (without the quotes) if Yaroslav can obtain the array he needs, and "NO" (without the quotes) otherwise.
Demo Input:
['1\n1\n', '3\n1 1 2\n', '4\n7 7 7 7\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
In the first sample the initial array fits well.
In the second sample Yaroslav can get array: 1, 2, 1. He can swap the last and the second last elements to obtain it.
In the third sample Yarosav can't get the array he needs.
|
```python
n = int(input())
a = input().split(' ')
s = [int(i) for i in a]
max_num = max(s,key=s.count)
max_count = s.count(max_num)
if max_count>((n//2)+(n%2)):
print("NO")
else:
print("YES")
```
| 3
|
|
228
|
A
|
Is your horseshoe on the other hoof?
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
|
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
|
Print a single integer — the minimum number of horseshoes Valera needs to buy.
|
[
"1 7 3 3\n",
"7 7 7 7\n"
] |
[
"1\n",
"3\n"
] |
none
| 500
|
[
{
"input": "1 7 3 3",
"output": "1"
},
{
"input": "7 7 7 7",
"output": "3"
},
{
"input": "81170865 673572653 756938629 995577259",
"output": "0"
},
{
"input": "3491663 217797045 522540872 715355328",
"output": "0"
},
{
"input": "251590420 586975278 916631563 586975278",
"output": "1"
},
{
"input": "259504825 377489979 588153796 377489979",
"output": "1"
},
{
"input": "652588203 931100304 931100304 652588203",
"output": "2"
},
{
"input": "391958720 651507265 391958720 651507265",
"output": "2"
},
{
"input": "90793237 90793237 90793237 90793237",
"output": "3"
},
{
"input": "551651653 551651653 551651653 551651653",
"output": "3"
},
{
"input": "156630260 609654355 668943582 973622757",
"output": "0"
},
{
"input": "17061017 110313588 434481173 796661222",
"output": "0"
},
{
"input": "24975422 256716298 337790533 690960249",
"output": "0"
},
{
"input": "255635360 732742923 798648949 883146723",
"output": "0"
},
{
"input": "133315691 265159773 734556507 265159773",
"output": "1"
},
{
"input": "28442865 741657755 978106882 978106882",
"output": "1"
},
{
"input": "131245479 174845575 497483467 131245479",
"output": "1"
},
{
"input": "139159884 616215581 958341883 616215581",
"output": "1"
},
{
"input": "147784432 947653080 947653080 947653080",
"output": "2"
},
{
"input": "94055790 756126496 756126496 94055790",
"output": "2"
},
{
"input": "240458500 511952208 240458500 511952208",
"output": "2"
},
{
"input": "681828506 972810624 972810624 681828506",
"output": "2"
},
{
"input": "454961014 454961014 454961014 454961014",
"output": "3"
},
{
"input": "915819430 915819430 915819430 915819430",
"output": "3"
},
{
"input": "671645142 671645142 671645142 671645142",
"output": "3"
},
{
"input": "132503558 132503558 132503558 132503558",
"output": "3"
},
{
"input": "5 5 999999 6",
"output": "1"
},
{
"input": "1 1 2 5",
"output": "1"
},
{
"input": "2 1 2 3",
"output": "1"
},
{
"input": "1 1 3 5",
"output": "1"
},
{
"input": "1 1 3 3",
"output": "2"
},
{
"input": "2 2 2 1",
"output": "2"
},
{
"input": "3 1 1 1",
"output": "2"
},
{
"input": "1 2 2 2",
"output": "2"
}
] | 1,691,559,870
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 62
| 0
|
def minimum_horseshoes(c):
distinct_colors = len(set(c))
return max(0, 4 - distinct_colors)
c = list(map(int, input().split()))
res = minimum_horseshoes_to_buy(c)
print(res)
|
Title: Is your horseshoe on the other hoof?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera the Horse is going to the party with friends. He has been following the fashion trends for a while, and he knows that it is very popular to wear all horseshoes of different color. Valera has got four horseshoes left from the last year, but maybe some of them have the same color. In this case he needs to go to the store and buy some few more horseshoes, not to lose face in front of his stylish comrades.
Fortunately, the store sells horseshoes of all colors under the sun and Valera has enough money to buy any four of them. However, in order to save the money, he would like to spend as little money as possible, so you need to help Valera and determine what is the minimum number of horseshoes he needs to buy to wear four horseshoes of different colors to a party.
Input Specification:
The first line contains four space-separated integers *s*1,<=*s*2,<=*s*3,<=*s*4 (1<=≤<=*s*1,<=*s*2,<=*s*3,<=*s*4<=≤<=109) — the colors of horseshoes Valera has.
Consider all possible colors indexed with integers.
Output Specification:
Print a single integer — the minimum number of horseshoes Valera needs to buy.
Demo Input:
['1 7 3 3\n', '7 7 7 7\n']
Demo Output:
['1\n', '3\n']
Note:
none
|
```python
def minimum_horseshoes(c):
distinct_colors = len(set(c))
return max(0, 4 - distinct_colors)
c = list(map(int, input().split()))
res = minimum_horseshoes_to_buy(c)
print(res)
```
| -1
|
|
1,009
|
B
|
Minimum Ternary String
|
PROGRAMMING
| 1,400
|
[
"greedy",
"implementation"
] | null | null |
You are given a ternary string (it is a string which consists only of characters '0', '1' and '2').
You can swap any two adjacent (consecutive) characters '0' and '1' (i.e. replace "01" with "10" or vice versa) or any two adjacent (consecutive) characters '1' and '2' (i.e. replace "12" with "21" or vice versa).
For example, for string "010210" we can perform the following moves:
- "010210" $\rightarrow$ "100210"; - "010210" $\rightarrow$ "001210"; - "010210" $\rightarrow$ "010120"; - "010210" $\rightarrow$ "010201".
Note than you cannot swap "02" $\rightarrow$ "20" and vice versa. You cannot perform any other operations with the given string excluding described above.
You task is to obtain the minimum possible (lexicographically) string by using these swaps arbitrary number of times (possibly, zero).
String $a$ is lexicographically less than string $b$ (if strings $a$ and $b$ have the same length) if there exists some position $i$ ($1 \le i \le |a|$, where $|s|$ is the length of the string $s$) such that for every $j < i$ holds $a_j = b_j$, and $a_i < b_i$.
|
The first line of the input contains the string $s$ consisting only of characters '0', '1' and '2', its length is between $1$ and $10^5$ (inclusive).
|
Print a single string — the minimum possible (lexicographically) string you can obtain by using the swaps described above arbitrary number of times (possibly, zero).
|
[
"100210\n",
"11222121\n",
"20\n"
] |
[
"001120\n",
"11112222\n",
"20\n"
] |
none
| 0
|
[
{
"input": "100210",
"output": "001120"
},
{
"input": "11222121",
"output": "11112222"
},
{
"input": "20",
"output": "20"
},
{
"input": "1002",
"output": "0012"
},
{
"input": "10",
"output": "01"
},
{
"input": "000021",
"output": "000012"
},
{
"input": "021",
"output": "012"
},
{
"input": "2",
"output": "2"
},
{
"input": "201",
"output": "120"
},
{
"input": "2112120",
"output": "1112220"
},
{
"input": "102",
"output": "012"
},
{
"input": "202",
"output": "202"
},
{
"input": "220201",
"output": "122020"
},
{
"input": "12",
"output": "12"
},
{
"input": "100022202",
"output": "000122202"
},
{
"input": "01",
"output": "01"
},
{
"input": "1",
"output": "1"
}
] | 1,635,719,120
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 5
| 780
| 5,427,200
|
x = input()
O = 0
z = ""
y = []
for i in x :
if i == "1" :
O +=1
else :
y.append(i)
n = len(x)
i = -1
while i+1 < n and y[i+1] == "0" :
i+=1
for k in range(O):
y.insert(k+1+i,"1")
for i in y :
z+=i
print(z)
|
Title: Minimum Ternary String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a ternary string (it is a string which consists only of characters '0', '1' and '2').
You can swap any two adjacent (consecutive) characters '0' and '1' (i.e. replace "01" with "10" or vice versa) or any two adjacent (consecutive) characters '1' and '2' (i.e. replace "12" with "21" or vice versa).
For example, for string "010210" we can perform the following moves:
- "010210" $\rightarrow$ "100210"; - "010210" $\rightarrow$ "001210"; - "010210" $\rightarrow$ "010120"; - "010210" $\rightarrow$ "010201".
Note than you cannot swap "02" $\rightarrow$ "20" and vice versa. You cannot perform any other operations with the given string excluding described above.
You task is to obtain the minimum possible (lexicographically) string by using these swaps arbitrary number of times (possibly, zero).
String $a$ is lexicographically less than string $b$ (if strings $a$ and $b$ have the same length) if there exists some position $i$ ($1 \le i \le |a|$, where $|s|$ is the length of the string $s$) such that for every $j < i$ holds $a_j = b_j$, and $a_i < b_i$.
Input Specification:
The first line of the input contains the string $s$ consisting only of characters '0', '1' and '2', its length is between $1$ and $10^5$ (inclusive).
Output Specification:
Print a single string — the minimum possible (lexicographically) string you can obtain by using the swaps described above arbitrary number of times (possibly, zero).
Demo Input:
['100210\n', '11222121\n', '20\n']
Demo Output:
['001120\n', '11112222\n', '20\n']
Note:
none
|
```python
x = input()
O = 0
z = ""
y = []
for i in x :
if i == "1" :
O +=1
else :
y.append(i)
n = len(x)
i = -1
while i+1 < n and y[i+1] == "0" :
i+=1
for k in range(O):
y.insert(k+1+i,"1")
for i in y :
z+=i
print(z)
```
| -1
|
|
139
|
A
|
Petr and Book
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book.
|
The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero.
|
Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
|
[
"100\n15 20 20 15 10 30 45\n",
"2\n1 0 0 0 0 0 0\n"
] |
[
"6\n",
"1\n"
] |
Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book.
| 500
|
[
{
"input": "100\n15 20 20 15 10 30 45",
"output": "6"
},
{
"input": "2\n1 0 0 0 0 0 0",
"output": "1"
},
{
"input": "100\n100 200 100 200 300 400 500",
"output": "1"
},
{
"input": "3\n1 1 1 1 1 1 1",
"output": "3"
},
{
"input": "1\n1 1 1 1 1 1 1",
"output": "1"
},
{
"input": "20\n5 3 7 2 1 6 4",
"output": "6"
},
{
"input": "10\n5 1 1 1 1 1 5",
"output": "6"
},
{
"input": "50\n10 1 10 1 10 1 10",
"output": "1"
},
{
"input": "77\n11 11 11 11 11 11 10",
"output": "1"
},
{
"input": "1\n1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "1000\n100 100 100 100 100 100 100",
"output": "3"
},
{
"input": "999\n10 20 10 20 30 20 10",
"output": "3"
},
{
"input": "433\n109 58 77 10 39 125 15",
"output": "7"
},
{
"input": "1\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "5\n1 0 1 0 1 0 1",
"output": "1"
},
{
"input": "997\n1 1 0 0 1 0 1",
"output": "1"
},
{
"input": "1000\n1 1 1 1 1 1 1",
"output": "6"
},
{
"input": "1000\n1000 1000 1000 1000 1000 1000 1000",
"output": "1"
},
{
"input": "1000\n1 0 0 0 0 0 0",
"output": "1"
},
{
"input": "1000\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "1000\n1 0 0 1 0 0 1",
"output": "1"
},
{
"input": "509\n105 23 98 0 7 0 155",
"output": "2"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "7"
},
{
"input": "2\n1 1 0 0 0 0 0",
"output": "2"
},
{
"input": "1\n0 0 0 0 0 1 0",
"output": "6"
},
{
"input": "10\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "5\n0 0 0 0 0 6 0",
"output": "6"
},
{
"input": "3\n0 1 0 0 0 0 0",
"output": "2"
},
{
"input": "10\n0 0 0 0 0 0 10",
"output": "7"
},
{
"input": "28\n1 2 3 4 5 6 7",
"output": "7"
},
{
"input": "100\n5 5 5 5 5 5 5",
"output": "6"
},
{
"input": "4\n1 0 0 0 0 0 1",
"output": "7"
},
{
"input": "2\n0 0 0 0 0 0 1",
"output": "7"
},
{
"input": "7\n0 0 0 0 0 0 7",
"output": "7"
},
{
"input": "7\n2 1 1 1 1 1 0",
"output": "6"
},
{
"input": "2\n0 0 1 1 0 0 0",
"output": "4"
},
{
"input": "6\n1 1 1 1 1 1 0",
"output": "6"
},
{
"input": "5\n1 1 1 0 0 1 1",
"output": "7"
},
{
"input": "100\n10 20 30 10 10 10 10",
"output": "7"
},
{
"input": "1\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "70\n10 10 10 10 10 10 10",
"output": "7"
},
{
"input": "22\n1 2 3 4 5 6 10",
"output": "7"
},
{
"input": "5\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "2\n0 0 0 1 0 0 0",
"output": "4"
},
{
"input": "6\n1 0 0 0 0 0 2",
"output": "7"
},
{
"input": "10\n1 2 2 1 2 1 1",
"output": "7"
},
{
"input": "5\n0 0 0 0 0 0 10",
"output": "7"
},
{
"input": "4\n0 1 1 0 0 0 0",
"output": "3"
},
{
"input": "100\n0 0 0 0 0 1 0",
"output": "6"
}
] | 1,596,281,083
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 218
| 20,172,800
|
n = int(input())
l1 = [int(x) for x in input().split()]
weekly = n%sum(l1)
n = 7
l1.reverse()
while weekly>=0:
weekly-=l1.pop()
n+=1
print(n%7)
|
Title: Petr and Book
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One Sunday Petr went to a bookshop and bought a new book on sports programming. The book had exactly *n* pages.
Petr decided to start reading it starting from the next day, that is, from Monday. Petr's got a very tight schedule and for each day of the week he knows how many pages he will be able to read on that day. Some days are so busy that Petr will have no time to read whatsoever. However, we know that he will be able to read at least one page a week.
Assuming that Petr will not skip days and will read as much as he can every day, determine on which day of the week he will read the last page of the book.
Input Specification:
The first input line contains the single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of pages in the book.
The second line contains seven non-negative space-separated integers that do not exceed 1000 — those integers represent how many pages Petr can read on Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday correspondingly. It is guaranteed that at least one of those numbers is larger than zero.
Output Specification:
Print a single number — the number of the day of the week, when Petr will finish reading the book. The days of the week are numbered starting with one in the natural order: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
Demo Input:
['100\n15 20 20 15 10 30 45\n', '2\n1 0 0 0 0 0 0\n']
Demo Output:
['6\n', '1\n']
Note:
Note to the first sample:
By the end of Monday and therefore, by the beginning of Tuesday Petr has 85 pages left. He has 65 pages left by Wednesday, 45 by Thursday, 30 by Friday, 20 by Saturday and on Saturday Petr finishes reading the book (and he also has time to read 10 pages of something else).
Note to the second sample:
On Monday of the first week Petr will read the first page. On Monday of the second week Petr will read the second page and will finish reading the book.
|
```python
n = int(input())
l1 = [int(x) for x in input().split()]
weekly = n%sum(l1)
n = 7
l1.reverse()
while weekly>=0:
weekly-=l1.pop()
n+=1
print(n%7)
```
| 0
|
|
1,003
|
D
|
Coins and Queries
|
PROGRAMMING
| 1,600
|
[
"greedy"
] | null | null |
Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. It is guaranteed that all the values are integer powers of $2$ (i.e. $a_i = 2^d$ for some non-negative integer number $d$).
Polycarp wants to know answers on $q$ queries. The $j$-th query is described as integer number $b_j$. The answer to the query is the minimum number of coins that is necessary to obtain the value $b_j$ using some subset of coins (Polycarp can use only coins he has). If Polycarp can't obtain the value $b_j$, the answer to the $j$-th query is -1.
The queries are independent (the answer on the query doesn't affect Polycarp's coins).
|
The first line of the input contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of coins and the number of queries.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ — values of coins ($1 \le a_i \le 2 \cdot 10^9$). It is guaranteed that all $a_i$ are integer powers of $2$ (i.e. $a_i = 2^d$ for some non-negative integer number $d$).
The next $q$ lines contain one integer each. The $j$-th line contains one integer $b_j$ — the value of the $j$-th query ($1 \le b_j \le 10^9$).
|
Print $q$ integers $ans_j$. The $j$-th integer must be equal to the answer on the $j$-th query. If Polycarp can't obtain the value $b_j$ the answer to the $j$-th query is -1.
|
[
"5 4\n2 4 8 2 4\n8\n5\n14\n10\n"
] |
[
"1\n-1\n3\n2\n"
] |
none
| 0
|
[
{
"input": "5 4\n2 4 8 2 4\n8\n5\n14\n10",
"output": "1\n-1\n3\n2"
},
{
"input": "3 3\n1 1 1\n1\n2\n3",
"output": "1\n2\n3"
},
{
"input": "4 1\n2 4 16 32\n14",
"output": "-1"
},
{
"input": "1 10\n8\n1\n2\n3\n4\n5\n6\n7\n8\n9\n16",
"output": "-1\n-1\n-1\n-1\n-1\n-1\n-1\n1\n-1\n-1"
},
{
"input": "1 10\n4\n1\n2\n3\n4\n5\n6\n7\n8\n9\n16",
"output": "-1\n-1\n-1\n1\n-1\n-1\n-1\n-1\n-1\n-1"
},
{
"input": "1 10\n2\n1\n2\n3\n4\n5\n6\n7\n8\n9\n16",
"output": "-1\n1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1"
},
{
"input": "1 10\n1\n1\n2\n3\n4\n5\n6\n7\n8\n9\n16",
"output": "1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1\n-1"
}
] | 1,676,146,238
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 3
| 451
| 23,552,000
|
import sys
input = sys.stdin.readline
from math import log2
n, q = map(int, input().split())
d = [0]*31
w = list(map(int, input().split()))
for i in w:
d[int(log2(i))] += 1
x = [0]*31
for i in range(31):
if d[i] > 0:
x[i] = 1
else:
a, b = 0, 2
for j in range(i-1, -1, -1):
if d[j] >= b:
x[i] = a+b
break
else:
b -= d[j]
a += d[j]
b *= 2
else:
x[i] = 0
for _ in range(q):
a = bin(int(input()))[2:][::-1]
c = 0
for i in range(len(a)):
if a[i] == '1':
if x[i] == 0:
print(-1)
break
c += x[i]
else:
print(c)
|
Title: Coins and Queries
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has $n$ coins, the value of the $i$-th coin is $a_i$. It is guaranteed that all the values are integer powers of $2$ (i.e. $a_i = 2^d$ for some non-negative integer number $d$).
Polycarp wants to know answers on $q$ queries. The $j$-th query is described as integer number $b_j$. The answer to the query is the minimum number of coins that is necessary to obtain the value $b_j$ using some subset of coins (Polycarp can use only coins he has). If Polycarp can't obtain the value $b_j$, the answer to the $j$-th query is -1.
The queries are independent (the answer on the query doesn't affect Polycarp's coins).
Input Specification:
The first line of the input contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of coins and the number of queries.
The second line of the input contains $n$ integers $a_1, a_2, \dots, a_n$ — values of coins ($1 \le a_i \le 2 \cdot 10^9$). It is guaranteed that all $a_i$ are integer powers of $2$ (i.e. $a_i = 2^d$ for some non-negative integer number $d$).
The next $q$ lines contain one integer each. The $j$-th line contains one integer $b_j$ — the value of the $j$-th query ($1 \le b_j \le 10^9$).
Output Specification:
Print $q$ integers $ans_j$. The $j$-th integer must be equal to the answer on the $j$-th query. If Polycarp can't obtain the value $b_j$ the answer to the $j$-th query is -1.
Demo Input:
['5 4\n2 4 8 2 4\n8\n5\n14\n10\n']
Demo Output:
['1\n-1\n3\n2\n']
Note:
none
|
```python
import sys
input = sys.stdin.readline
from math import log2
n, q = map(int, input().split())
d = [0]*31
w = list(map(int, input().split()))
for i in w:
d[int(log2(i))] += 1
x = [0]*31
for i in range(31):
if d[i] > 0:
x[i] = 1
else:
a, b = 0, 2
for j in range(i-1, -1, -1):
if d[j] >= b:
x[i] = a+b
break
else:
b -= d[j]
a += d[j]
b *= 2
else:
x[i] = 0
for _ in range(q):
a = bin(int(input()))[2:][::-1]
c = 0
for i in range(len(a)):
if a[i] == '1':
if x[i] == 0:
print(-1)
break
c += x[i]
else:
print(c)
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Bike is interested in permutations. A permutation of length *n* is an integer sequence such that each integer from 0 to (*n*<=-<=1) appears exactly once in it. For example, [0,<=2,<=1] is a permutation of length 3 while both [0,<=2,<=2] and [1,<=2,<=3] is not.
A permutation triple of permutations of length *n* (*a*,<=*b*,<=*c*) is called a Lucky Permutation Triple if and only if . The sign *a**i* denotes the *i*-th element of permutation *a*. The modular equality described above denotes that the remainders after dividing *a**i*<=+<=*b**i* by *n* and dividing *c**i* by *n* are equal.
Now, he has an integer *n* and wants to find a Lucky Permutation Triple. Could you please help him?
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105).
|
If no Lucky Permutation Triple of length *n* exists print -1.
Otherwise, you need to print three lines. Each line contains *n* space-seperated integers. The first line must contain permutation *a*, the second line — permutation *b*, the third — permutation *c*.
If there are multiple solutions, print any of them.
|
[
"5\n",
"2\n"
] |
[
"1 4 3 2 0\n1 0 2 4 3\n2 4 0 1 3\n",
"-1\n"
] |
In Sample 1, the permutation triple ([1, 4, 3, 2, 0], [1, 0, 2, 4, 3], [2, 4, 0, 1, 3]) is Lucky Permutation Triple, as following holds:
- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/a6bf1b9b57809dbec5021f65f89616f259587c07.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/48cc13134296b68f459f69d78e0240859aaec702.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ac44412de7b46833e90348a6b3298f9796e3977c.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3825b0bb758208dda2ead1c5224c05d89ad9ab55.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0a72e2da40048a507839927a211267ac01c9bf89.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In Sample 2, you can easily notice that no lucky permutation triple exists.
| 0
|
[
{
"input": "5",
"output": "1 4 3 2 0\n1 0 2 4 3\n2 4 0 1 3"
},
{
"input": "2",
"output": "-1"
},
{
"input": "8",
"output": "-1"
},
{
"input": "9",
"output": "0 1 2 3 4 5 6 7 8 \n0 1 2 3 4 5 6 7 8 \n0 2 4 6 8 1 3 5 7 "
},
{
"input": "2",
"output": "-1"
},
{
"input": "77",
"output": "0 1 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 \n0 1 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 \n0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 4..."
},
{
"input": "6",
"output": "-1"
},
{
"input": "87",
"output": "0 1 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 \n0 1 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 \n0 2 4..."
},
{
"input": "72",
"output": "-1"
},
{
"input": "1",
"output": "0 \n0 \n0 "
},
{
"input": "23",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 \n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 \n0 2 4 6 8 10 12 14 16 18 20 22 1 3 5 7 9 11 13 15 17 19 21 "
},
{
"input": "52",
"output": "-1"
},
{
"input": "32",
"output": "-1"
},
{
"input": "25",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 \n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 \n0 2 4 6 8 10 12 14 16 18 20 22 24 1 3 5 7 9 11 13 15 17 19 21 23 "
},
{
"input": "54",
"output": "-1"
},
{
"input": "39",
"output": "0 1 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 \n0 1 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 \n0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 "
},
{
"input": "20",
"output": "-1"
},
{
"input": "53",
"output": "0 1 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 \n0 1 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 \n0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 "
},
{
"input": "34",
"output": "-1"
},
{
"input": "23",
"output": "0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 \n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 \n0 2 4 6 8 10 12 14 16 18 20 22 1 3 5 7 9 11 13 15 17 19 21 "
},
{
"input": "37123",
"output": "0 1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "41904",
"output": "-1"
},
{
"input": "46684",
"output": "-1"
},
{
"input": "67817",
"output": "0 1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "72598",
"output": "-1"
},
{
"input": "85891",
"output": "0 1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "74320",
"output": "-1"
},
{
"input": "11805",
"output": "0 1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "16586",
"output": "-1"
},
{
"input": "5014",
"output": "-1"
},
{
"input": "73268",
"output": "-1"
},
{
"input": "61697",
"output": "0 1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "99182",
"output": "-1"
},
{
"input": "79771",
"output": "0 1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "68199",
"output": "0 1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "5684",
"output": "-1"
},
{
"input": "10465",
"output": "0 1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "31598",
"output": "-1"
},
{
"input": "36379",
"output": "0 1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "16968",
"output": "-1"
},
{
"input": "93061",
"output": "0 1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "73650",
"output": "-1"
},
{
"input": "94783",
"output": "0 1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "99564",
"output": "-1"
},
{
"input": "37049",
"output": "0 1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "25478",
"output": "-1"
},
{
"input": "30259",
"output": "0 1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "43551",
"output": "0 1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "31980",
"output": "-1"
},
{
"input": "69465",
"output": "0 1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "1",
"output": "0 \n0 \n0 "
},
{
"input": "100000",
"output": "-1"
},
{
"input": "99999",
"output": "0 1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 1..."
},
{
"input": "99998",
"output": "-1"
}
] | 1,586,754,859
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 54
| 842
| 10,547,200
|
n = int(input())
if n%2 == 0:
print(-1)
else:
[print(i,end=' ') if i < n else print() for i in range(n+1)]
[print(i,end=' ') if i < n else print() for i in range(n+1)]
[print((i+i)%n,end=' ') if i < n else print() for i in range(n+1)]
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Bike is interested in permutations. A permutation of length *n* is an integer sequence such that each integer from 0 to (*n*<=-<=1) appears exactly once in it. For example, [0,<=2,<=1] is a permutation of length 3 while both [0,<=2,<=2] and [1,<=2,<=3] is not.
A permutation triple of permutations of length *n* (*a*,<=*b*,<=*c*) is called a Lucky Permutation Triple if and only if . The sign *a**i* denotes the *i*-th element of permutation *a*. The modular equality described above denotes that the remainders after dividing *a**i*<=+<=*b**i* by *n* and dividing *c**i* by *n* are equal.
Now, he has an integer *n* and wants to find a Lucky Permutation Triple. Could you please help him?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105).
Output Specification:
If no Lucky Permutation Triple of length *n* exists print -1.
Otherwise, you need to print three lines. Each line contains *n* space-seperated integers. The first line must contain permutation *a*, the second line — permutation *b*, the third — permutation *c*.
If there are multiple solutions, print any of them.
Demo Input:
['5\n', '2\n']
Demo Output:
['1 4 3 2 0\n1 0 2 4 3\n2 4 0 1 3\n', '-1\n']
Note:
In Sample 1, the permutation triple ([1, 4, 3, 2, 0], [1, 0, 2, 4, 3], [2, 4, 0, 1, 3]) is Lucky Permutation Triple, as following holds:
- <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/a6bf1b9b57809dbec5021f65f89616f259587c07.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/48cc13134296b68f459f69d78e0240859aaec702.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/ac44412de7b46833e90348a6b3298f9796e3977c.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/3825b0bb758208dda2ead1c5224c05d89ad9ab55.png" style="max-width: 100.0%;max-height: 100.0%;"/>; - <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/0a72e2da40048a507839927a211267ac01c9bf89.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In Sample 2, you can easily notice that no lucky permutation triple exists.
|
```python
n = int(input())
if n%2 == 0:
print(-1)
else:
[print(i,end=' ') if i < n else print() for i in range(n+1)]
[print(i,end=' ') if i < n else print() for i in range(n+1)]
[print((i+i)%n,end=' ') if i < n else print() for i in range(n+1)]
```
| 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"
}
] | 1,507,146,490
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 46
| 0
|
NOT = int(input())
treesIndexes = [[], []]
trees = {}
start = 0
end = 1
numberOfApplesCollected = 0
for x in range(NOT):
temp = input().split(" ")
trees[temp[0]] = temp[1]
if int(temp[0]) > 0:
treesIndexes[1].append(temp[0])
continue
treesIndexes[0].append(temp[1])
if len(treesIndexes[1]) > len(treesIndexes[0]):
start = 1
end = 0
treesIndexes[0].sort()
treesIndexes[1].sort()
while len(treesIndexes[start]) > 0:
numberOfApplesCollected += int(trees[treesIndexes[start][start - 1]])
del treesIndexes[start][start - 1]
if len(trees[end]) != 0:
numberOfApplesCollected += int(trees[treesIndexes[end][end - 1]])
del treesIndexes[end][end - 1]
else:
break
print(numberOfApplesCollected)
|
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
NOT = int(input())
treesIndexes = [[], []]
trees = {}
start = 0
end = 1
numberOfApplesCollected = 0
for x in range(NOT):
temp = input().split(" ")
trees[temp[0]] = temp[1]
if int(temp[0]) > 0:
treesIndexes[1].append(temp[0])
continue
treesIndexes[0].append(temp[1])
if len(treesIndexes[1]) > len(treesIndexes[0]):
start = 1
end = 0
treesIndexes[0].sort()
treesIndexes[1].sort()
while len(treesIndexes[start]) > 0:
numberOfApplesCollected += int(trees[treesIndexes[start][start - 1]])
del treesIndexes[start][start - 1]
if len(trees[end]) != 0:
numberOfApplesCollected += int(trees[treesIndexes[end][end - 1]])
del treesIndexes[end][end - 1]
else:
break
print(numberOfApplesCollected)
```
| -1
|
|
483
|
A
|
Counterexample
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation",
"math",
"number theory"
] | null | null |
Your friend has recently learned about coprime numbers. A pair of numbers {*a*,<=*b*} is called coprime if the maximum number that divides both *a* and *b* is equal to one.
Your friend often comes up with different statements. He has recently supposed that if the pair (*a*,<=*b*) is coprime and the pair (*b*,<=*c*) is coprime, then the pair (*a*,<=*c*) is coprime.
You want to find a counterexample for your friend's statement. Therefore, your task is to find three distinct numbers (*a*,<=*b*,<=*c*), for which the statement is false, and the numbers meet the condition *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*.
More specifically, you need to find three numbers (*a*,<=*b*,<=*c*), such that *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*, pairs (*a*,<=*b*) and (*b*,<=*c*) are coprime, and pair (*a*,<=*c*) is not coprime.
|
The single line contains two positive space-separated integers *l*, *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018; *r*<=-<=*l*<=≤<=50).
|
Print three positive space-separated integers *a*, *b*, *c* — three distinct numbers (*a*,<=*b*,<=*c*) that form the counterexample. If there are several solutions, you are allowed to print any of them. The numbers must be printed in ascending order.
If the counterexample does not exist, print the single number -1.
|
[
"2 4\n",
"10 11\n",
"900000000000000009 900000000000000029\n"
] |
[
"2 3 4\n",
"-1\n",
"900000000000000009 900000000000000010 900000000000000021\n"
] |
In the first sample pair (2, 4) is not coprime and pairs (2, 3) and (3, 4) are.
In the second sample you cannot form a group of three distinct integers, so the answer is -1.
In the third sample it is easy to see that numbers 900000000000000009 and 900000000000000021 are divisible by three.
| 500
|
[
{
"input": "2 4",
"output": "2 3 4"
},
{
"input": "10 11",
"output": "-1"
},
{
"input": "900000000000000009 900000000000000029",
"output": "900000000000000009 900000000000000010 900000000000000021"
},
{
"input": "640097987171091791 640097987171091835",
"output": "640097987171091792 640097987171091793 640097987171091794"
},
{
"input": "19534350415104721 19534350415104725",
"output": "19534350415104722 19534350415104723 19534350415104724"
},
{
"input": "933700505788726243 933700505788726280",
"output": "933700505788726244 933700505788726245 933700505788726246"
},
{
"input": "1 3",
"output": "-1"
},
{
"input": "1 4",
"output": "2 3 4"
},
{
"input": "1 1",
"output": "-1"
},
{
"input": "266540997167959130 266540997167959164",
"output": "266540997167959130 266540997167959131 266540997167959132"
},
{
"input": "267367244641009850 267367244641009899",
"output": "267367244641009850 267367244641009851 267367244641009852"
},
{
"input": "268193483524125978 268193483524125993",
"output": "268193483524125978 268193483524125979 268193483524125980"
},
{
"input": "269019726702209402 269019726702209432",
"output": "269019726702209402 269019726702209403 269019726702209404"
},
{
"input": "269845965585325530 269845965585325576",
"output": "269845965585325530 269845965585325531 269845965585325532"
},
{
"input": "270672213058376250 270672213058376260",
"output": "270672213058376250 270672213058376251 270672213058376252"
},
{
"input": "271498451941492378 271498451941492378",
"output": "-1"
},
{
"input": "272324690824608506 272324690824608523",
"output": "272324690824608506 272324690824608507 272324690824608508"
},
{
"input": "273150934002691930 273150934002691962",
"output": "273150934002691930 273150934002691931 273150934002691932"
},
{
"input": "996517375802030516 996517375802030524",
"output": "996517375802030516 996517375802030517 996517375802030518"
},
{
"input": "997343614685146644 997343614685146694",
"output": "997343614685146644 997343614685146645 997343614685146646"
},
{
"input": "998169857863230068 998169857863230083",
"output": "998169857863230068 998169857863230069 998169857863230070"
},
{
"input": "998996101041313492 998996101041313522",
"output": "998996101041313492 998996101041313493 998996101041313494"
},
{
"input": "999822344219396916 999822344219396961",
"output": "999822344219396916 999822344219396917 999822344219396918"
},
{
"input": "648583102513043 648583102513053",
"output": "648583102513044 648583102513045 648583102513046"
},
{
"input": "266540997167959130 266540997167959131",
"output": "-1"
},
{
"input": "267367244641009850 267367244641009850",
"output": "-1"
},
{
"input": "268193483524125978 268193483524125979",
"output": "-1"
},
{
"input": "269019726702209402 269019726702209402",
"output": "-1"
},
{
"input": "269845965585325530 269845965585325530",
"output": "-1"
},
{
"input": "270672213058376250 270672213058376254",
"output": "270672213058376250 270672213058376251 270672213058376252"
},
{
"input": "271498451941492378 271498451941492379",
"output": "-1"
},
{
"input": "272324690824608506 272324690824608508",
"output": "272324690824608506 272324690824608507 272324690824608508"
},
{
"input": "273150934002691930 273150934002691931",
"output": "-1"
},
{
"input": "996517375802030516 996517375802030518",
"output": "996517375802030516 996517375802030517 996517375802030518"
},
{
"input": "997343614685146644 997343614685146644",
"output": "-1"
},
{
"input": "2147483647 2147483649",
"output": "-1"
},
{
"input": "3 5",
"output": "-1"
},
{
"input": "1 7",
"output": "2 3 4"
},
{
"input": "9 12",
"output": "9 11 12"
},
{
"input": "4 4",
"output": "-1"
},
{
"input": "11 13",
"output": "-1"
},
{
"input": "2 2",
"output": "-1"
}
] | 1,691,943,804
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 78
| 0
|
import math
l,r = map(int , input().split())
ans= []
ans2=[]
def iscoprime(x,y):
if math.gcd(x,y) == 1 :
return True
else:
return False
for i in range(l,r+1):
if r-l ==1:
ans2 = -1
break
for j in range(l,r+1):
if iscoprime(i,j):
if sorted([i,j]) not in ans:
ans.append([i,j])
def iscoprimepair(lst):
a = lst[0]
b = lst[1]
c = lst[2]
if math.gcd(a,b) == 1 and math.gcd(b,c) ==1 and math.gcd(a,c) == 1 :
return True
else:
return False
for pair in ans:
for anotherpair in ans:
if pair[1] == anotherpair[0]:
if not iscoprimepair(list(set(pair+anotherpair))):
ans2 = list(set(pair+anotherpair))
if isinstance(ans2 , list):
print(" ".join(str(e) for e in ans2))
else:
print(ans2)
|
Title: Counterexample
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend has recently learned about coprime numbers. A pair of numbers {*a*,<=*b*} is called coprime if the maximum number that divides both *a* and *b* is equal to one.
Your friend often comes up with different statements. He has recently supposed that if the pair (*a*,<=*b*) is coprime and the pair (*b*,<=*c*) is coprime, then the pair (*a*,<=*c*) is coprime.
You want to find a counterexample for your friend's statement. Therefore, your task is to find three distinct numbers (*a*,<=*b*,<=*c*), for which the statement is false, and the numbers meet the condition *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*.
More specifically, you need to find three numbers (*a*,<=*b*,<=*c*), such that *l*<=≤<=*a*<=<<=*b*<=<<=*c*<=≤<=*r*, pairs (*a*,<=*b*) and (*b*,<=*c*) are coprime, and pair (*a*,<=*c*) is not coprime.
Input Specification:
The single line contains two positive space-separated integers *l*, *r* (1<=≤<=*l*<=≤<=*r*<=≤<=1018; *r*<=-<=*l*<=≤<=50).
Output Specification:
Print three positive space-separated integers *a*, *b*, *c* — three distinct numbers (*a*,<=*b*,<=*c*) that form the counterexample. If there are several solutions, you are allowed to print any of them. The numbers must be printed in ascending order.
If the counterexample does not exist, print the single number -1.
Demo Input:
['2 4\n', '10 11\n', '900000000000000009 900000000000000029\n']
Demo Output:
['2 3 4\n', '-1\n', '900000000000000009 900000000000000010 900000000000000021\n']
Note:
In the first sample pair (2, 4) is not coprime and pairs (2, 3) and (3, 4) are.
In the second sample you cannot form a group of three distinct integers, so the answer is -1.
In the third sample it is easy to see that numbers 900000000000000009 and 900000000000000021 are divisible by three.
|
```python
import math
l,r = map(int , input().split())
ans= []
ans2=[]
def iscoprime(x,y):
if math.gcd(x,y) == 1 :
return True
else:
return False
for i in range(l,r+1):
if r-l ==1:
ans2 = -1
break
for j in range(l,r+1):
if iscoprime(i,j):
if sorted([i,j]) not in ans:
ans.append([i,j])
def iscoprimepair(lst):
a = lst[0]
b = lst[1]
c = lst[2]
if math.gcd(a,b) == 1 and math.gcd(b,c) ==1 and math.gcd(a,c) == 1 :
return True
else:
return False
for pair in ans:
for anotherpair in ans:
if pair[1] == anotherpair[0]:
if not iscoprimepair(list(set(pair+anotherpair))):
ans2 = list(set(pair+anotherpair))
if isinstance(ans2 , list):
print(" ".join(str(e) for e in ans2))
else:
print(ans2)
```
| 0
|
|
9
|
A
|
Die Roll
|
PROGRAMMING
| 800
|
[
"math",
"probabilities"
] |
A. Die Roll
|
1
|
64
|
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
|
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
|
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
|
[
"4 2\n"
] |
[
"1/2\n"
] |
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
| 0
|
[
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
"output": "1/6"
},
{
"input": "2 1",
"output": "5/6"
},
{
"input": "2 2",
"output": "5/6"
},
{
"input": "2 3",
"output": "2/3"
},
{
"input": "2 4",
"output": "1/2"
},
{
"input": "2 5",
"output": "1/3"
},
{
"input": "2 6",
"output": "1/6"
},
{
"input": "3 1",
"output": "2/3"
},
{
"input": "3 2",
"output": "2/3"
},
{
"input": "3 3",
"output": "2/3"
},
{
"input": "3 4",
"output": "1/2"
},
{
"input": "3 5",
"output": "1/3"
},
{
"input": "3 6",
"output": "1/6"
},
{
"input": "4 1",
"output": "1/2"
},
{
"input": "4 3",
"output": "1/2"
},
{
"input": "4 4",
"output": "1/2"
},
{
"input": "4 5",
"output": "1/3"
},
{
"input": "4 6",
"output": "1/6"
},
{
"input": "5 1",
"output": "1/3"
},
{
"input": "5 2",
"output": "1/3"
},
{
"input": "5 3",
"output": "1/3"
},
{
"input": "5 4",
"output": "1/3"
},
{
"input": "5 5",
"output": "1/3"
},
{
"input": "5 6",
"output": "1/6"
},
{
"input": "6 1",
"output": "1/6"
},
{
"input": "6 2",
"output": "1/6"
},
{
"input": "6 3",
"output": "1/6"
},
{
"input": "6 4",
"output": "1/6"
},
{
"input": "6 5",
"output": "1/6"
},
{
"input": "6 6",
"output": "1/6"
}
] | 1,512,556,342
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 92
| 5,529,600
|
a,b=map(int,input().split())
c=max(a,b)-1
def nsd(a,b):
ost=a%b
while ost!=0:
a=b
b=ost
ost=a%b
return b
n=nsd(c,6)
c=c//n
d=6//n
print(str(c)+'/'+str(d))
|
Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
|
```python
a,b=map(int,input().split())
c=max(a,b)-1
def nsd(a,b):
ost=a%b
while ost!=0:
a=b
b=ost
ost=a%b
return b
n=nsd(c,6)
c=c//n
d=6//n
print(str(c)+'/'+str(d))
```
| 0
|
940
|
A
|
Points on the line
|
PROGRAMMING
| 1,200
|
[
"brute force",
"greedy",
"sortings"
] | null | null |
We've got no test cases. A big olympiad is coming up. But the problemsetters' number one priority should be adding another problem to the round.
The diameter of a multiset of points on the line is the largest distance between two points from this set. For example, the diameter of the multiset {1,<=3,<=2,<=1} is 2.
Diameter of multiset consisting of one point is 0.
You are given *n* points on the line. What is the minimum number of points you have to remove, so that the diameter of the multiset of the remaining points will not exceed *d*?
|
The first line contains two integers *n* and *d* (1<=≤<=*n*<=≤<=100,<=0<=≤<=*d*<=≤<=100) — the amount of points and the maximum allowed diameter respectively.
The second line contains *n* space separated integers (1<=≤<=*x**i*<=≤<=100) — the coordinates of the points.
|
Output a single integer — the minimum number of points you have to remove.
|
[
"3 1\n2 1 4\n",
"3 0\n7 7 7\n",
"6 3\n1 3 4 6 9 10\n"
] |
[
"1\n",
"0\n",
"3\n"
] |
In the first test case the optimal strategy is to remove the point with coordinate 4. The remaining points will have coordinates 1 and 2, so the diameter will be equal to 2 - 1 = 1.
In the second test case the diameter is equal to 0, so its is unnecessary to remove any points.
In the third test case the optimal strategy is to remove points with coordinates 1, 9 and 10. The remaining points will have coordinates 3, 4 and 6, so the diameter will be equal to 6 - 3 = 3.
| 500
|
[
{
"input": "3 1\n2 1 4",
"output": "1"
},
{
"input": "3 0\n7 7 7",
"output": "0"
},
{
"input": "6 3\n1 3 4 6 9 10",
"output": "3"
},
{
"input": "11 5\n10 11 12 13 14 15 16 17 18 19 20",
"output": "5"
},
{
"input": "1 100\n1",
"output": "0"
},
{
"input": "100 10\n22 75 26 45 72 81 47 29 97 2 75 25 82 84 17 56 32 2 28 37 57 39 18 11 79 6 40 68 68 16 40 63 93 49 91 10 55 68 31 80 57 18 34 28 76 55 21 80 22 45 11 67 67 74 91 4 35 34 65 80 21 95 1 52 25 31 2 53 96 22 89 99 7 66 32 2 68 33 75 92 84 10 94 28 54 12 9 80 43 21 51 92 20 97 7 25 67 17 38 100",
"output": "84"
},
{
"input": "100 70\n22 75 26 45 72 81 47 29 97 2 75 25 82 84 17 56 32 2 28 37 57 39 18 11 79 6 40 68 68 16 40 63 93 49 91 10 55 68 31 80 57 18 34 28 76 55 21 80 22 45 11 67 67 74 91 4 35 34 65 80 21 95 1 52 25 31 2 53 96 22 89 99 7 66 32 2 68 33 75 92 84 10 94 28 54 12 9 80 43 21 51 92 20 97 7 25 67 17 38 100",
"output": "27"
},
{
"input": "1 10\n25",
"output": "0"
},
{
"input": "70 80\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",
"output": "0"
},
{
"input": "3 1\n25 26 27",
"output": "1"
},
{
"input": "100 5\n51 56 52 60 52 53 52 60 56 54 55 50 53 51 57 53 52 54 54 52 51 55 50 56 60 51 58 50 60 59 50 54 60 55 55 57 54 59 59 55 55 52 56 57 59 54 53 57 52 50 50 55 59 54 54 56 51 58 52 51 56 56 58 56 54 54 57 52 51 58 56 57 54 59 58 53 50 52 50 60 57 51 54 59 54 54 52 55 53 55 51 53 52 54 51 56 55 53 58 56",
"output": "34"
},
{
"input": "100 11\n44 89 57 64 94 96 73 96 55 52 91 73 73 93 51 62 63 85 43 75 60 78 98 55 80 84 65 75 61 88 62 71 53 57 94 85 60 96 66 96 61 72 97 64 51 44 63 82 67 86 60 57 74 85 57 79 61 94 86 78 84 56 60 75 91 91 92 62 89 85 79 57 76 97 65 56 46 78 51 69 50 52 85 80 76 71 81 51 90 71 77 60 63 62 84 59 79 84 69 81",
"output": "70"
},
{
"input": "100 0\n22 75 26 45 72 81 47 29 97 2 75 25 82 84 17 56 32 2 28 37 57 39 18 11 79 6 40 68 68 16 40 63 93 49 91 10 55 68 31 80 57 18 34 28 76 55 21 80 22 45 11 67 67 74 91 4 35 34 65 80 21 95 1 52 25 31 2 53 96 22 89 99 7 66 32 2 68 33 75 92 84 10 94 28 54 12 9 80 43 21 51 92 20 97 7 25 67 17 38 100",
"output": "96"
},
{
"input": "100 100\n22 75 26 45 72 81 47 29 97 2 75 25 82 84 17 56 32 2 28 37 57 39 18 11 79 6 40 68 68 16 40 63 93 49 91 10 55 68 31 80 57 18 34 28 76 55 21 80 22 45 11 67 67 74 91 4 35 34 65 80 21 95 1 52 25 31 2 53 96 22 89 99 7 66 32 2 68 33 75 92 84 10 94 28 54 12 9 80 43 21 51 92 20 97 7 25 67 17 38 100",
"output": "0"
},
{
"input": "76 32\n50 53 69 58 55 39 40 42 40 55 58 73 55 72 75 44 45 55 46 60 60 42 41 64 77 39 68 51 61 49 38 41 56 57 64 43 78 36 39 63 40 66 52 76 39 68 39 73 40 68 54 60 35 67 69 52 58 52 38 63 69 38 69 60 73 64 65 41 59 55 37 57 40 34 35 35",
"output": "13"
},
{
"input": "100 1\n22 75 26 45 72 81 47 29 97 2 75 25 82 84 17 56 32 2 28 37 57 39 18 11 79 6 40 68 68 16 40 63 93 49 91 10 55 68 31 80 57 18 34 28 76 55 21 80 22 45 11 67 67 74 91 4 35 34 65 80 21 95 1 52 25 31 2 53 96 22 89 99 7 66 32 2 68 33 75 92 84 10 94 28 54 12 9 80 43 21 51 92 20 97 7 25 67 17 38 100",
"output": "93"
},
{
"input": "100 5\n22 75 26 45 72 81 47 29 97 2 75 25 82 84 17 56 32 2 28 37 57 39 18 11 79 6 40 68 68 16 40 63 93 49 91 10 55 68 31 80 57 18 34 28 76 55 21 80 22 45 11 67 67 74 91 4 35 34 65 80 21 95 1 52 25 31 2 53 96 22 89 99 7 66 32 2 68 33 75 92 84 10 94 28 54 12 9 80 43 21 51 92 20 97 7 25 67 17 38 100",
"output": "89"
},
{
"input": "98 64\n2 29 36 55 58 15 25 33 7 16 61 1 4 24 63 26 36 16 16 3 57 39 56 7 11 24 20 12 22 10 56 5 11 39 61 52 27 54 21 6 61 36 40 52 54 5 15 52 58 23 45 39 65 16 27 40 13 64 47 24 51 29 9 18 49 49 8 47 2 64 7 63 49 10 20 26 34 3 45 66 8 46 16 32 16 38 3 6 15 17 35 48 36 5 57 29 61 15",
"output": "1"
},
{
"input": "100 56\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": "43"
},
{
"input": "100 0\n14 13 14 13 14 13 13 13 13 14 13 13 14 14 13 14 14 14 14 13 13 13 14 13 13 14 14 14 14 14 14 13 13 13 13 14 13 14 13 14 13 14 14 14 14 13 13 14 14 13 13 13 13 14 13 14 13 14 13 14 13 13 13 14 13 13 14 13 14 14 13 13 13 14 14 14 14 13 13 14 14 14 14 14 14 14 13 14 13 13 13 14 14 13 13 13 13 13 14 14",
"output": "50"
},
{
"input": "100 0\n14 17 18 22 19 18 19 21 19 19 22 22 19 21 24 23 24 19 25 24 24 21 20 13 26 18 17 15 25 13 17 20 20 21 13 22 27 15 18 27 19 15 16 25 18 17 18 22 19 17 18 24 14 16 18 16 22 16 17 27 18 17 18 24 22 13 14 20 23 19 16 21 19 13 14 14 25 15 27 24 26 22 16 20 16 14 21 27 15 23 23 24 27 14 24 17 19 24 15 27",
"output": "89"
},
{
"input": "100 100\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": "0"
},
{
"input": "1 100\n22",
"output": "0"
},
{
"input": "1 0\n22",
"output": "0"
},
{
"input": "1 99\n99",
"output": "0"
},
{
"input": "1 5\n6",
"output": "0"
},
{
"input": "3 1\n10 20 30",
"output": "2"
},
{
"input": "3 0\n1 2 3",
"output": "2"
},
{
"input": "3 2\n1 50 99",
"output": "2"
},
{
"input": "7 4\n1 3 4 9 10 11 12",
"output": "3"
},
{
"input": "2 5\n67 23",
"output": "1"
},
{
"input": "4 2\n1 4 7 9",
"output": "2"
},
{
"input": "2 0\n1 2",
"output": "1"
},
{
"input": "8 1\n3 3 3 5 5 5 5 5",
"output": "3"
},
{
"input": "5 1\n3 5 5 5 6",
"output": "1"
}
] | 1,519,480,059
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 61
| 6,348,800
|
#import sys
#sys.stdin = open('in.txt', 'r')
n, d = list(map(int, input().split()))
a = list(map(int, input().split()))
sorted(a)
ans = 10000
for i in range(0,n):
for j in range(i+1,n):
if abs(a[j]-a[i]) <= d:
ans = min(ans,n-(j+1)+i)
if ans == 10000:
print(n-1)
else:
print(ans)
|
Title: Points on the line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We've got no test cases. A big olympiad is coming up. But the problemsetters' number one priority should be adding another problem to the round.
The diameter of a multiset of points on the line is the largest distance between two points from this set. For example, the diameter of the multiset {1,<=3,<=2,<=1} is 2.
Diameter of multiset consisting of one point is 0.
You are given *n* points on the line. What is the minimum number of points you have to remove, so that the diameter of the multiset of the remaining points will not exceed *d*?
Input Specification:
The first line contains two integers *n* and *d* (1<=≤<=*n*<=≤<=100,<=0<=≤<=*d*<=≤<=100) — the amount of points and the maximum allowed diameter respectively.
The second line contains *n* space separated integers (1<=≤<=*x**i*<=≤<=100) — the coordinates of the points.
Output Specification:
Output a single integer — the minimum number of points you have to remove.
Demo Input:
['3 1\n2 1 4\n', '3 0\n7 7 7\n', '6 3\n1 3 4 6 9 10\n']
Demo Output:
['1\n', '0\n', '3\n']
Note:
In the first test case the optimal strategy is to remove the point with coordinate 4. The remaining points will have coordinates 1 and 2, so the diameter will be equal to 2 - 1 = 1.
In the second test case the diameter is equal to 0, so its is unnecessary to remove any points.
In the third test case the optimal strategy is to remove points with coordinates 1, 9 and 10. The remaining points will have coordinates 3, 4 and 6, so the diameter will be equal to 6 - 3 = 3.
|
```python
#import sys
#sys.stdin = open('in.txt', 'r')
n, d = list(map(int, input().split()))
a = list(map(int, input().split()))
sorted(a)
ans = 10000
for i in range(0,n):
for j in range(i+1,n):
if abs(a[j]-a[i]) <= d:
ans = min(ans,n-(j+1)+i)
if ans == 10000:
print(n-1)
else:
print(ans)
```
| 0
|
|
758
|
A
|
Holiday Of Equality
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.
|
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen.
|
In the only line print the integer *S* — the minimum number of burles which are had to spend.
|
[
"5\n0 1 2 3 4\n",
"5\n1 1 0 1 1\n",
"3\n1 3 1\n",
"1\n12\n"
] |
[
"10",
"1",
"4",
"0"
] |
In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles.
| 500
|
[
{
"input": "5\n0 1 2 3 4",
"output": "10"
},
{
"input": "5\n1 1 0 1 1",
"output": "1"
},
{
"input": "3\n1 3 1",
"output": "4"
},
{
"input": "1\n12",
"output": "0"
},
{
"input": "3\n1 2 3",
"output": "3"
},
{
"input": "14\n52518 718438 358883 462189 853171 592966 225788 46977 814826 295697 676256 561479 56545 764281",
"output": "5464380"
},
{
"input": "21\n842556 216391 427181 626688 775504 168309 851038 448402 880826 73697 593338 519033 135115 20128 424606 939484 846242 756907 377058 241543 29353",
"output": "9535765"
},
{
"input": "3\n1 3 2",
"output": "3"
},
{
"input": "3\n2 1 3",
"output": "3"
},
{
"input": "3\n2 3 1",
"output": "3"
},
{
"input": "3\n3 1 2",
"output": "3"
},
{
"input": "3\n3 2 1",
"output": "3"
},
{
"input": "1\n228503",
"output": "0"
},
{
"input": "2\n32576 550340",
"output": "517764"
},
{
"input": "3\n910648 542843 537125",
"output": "741328"
},
{
"input": "4\n751720 572344 569387 893618",
"output": "787403"
},
{
"input": "6\n433864 631347 597596 794426 713555 231193",
"output": "1364575"
},
{
"input": "9\n31078 645168 695751 126111 375934 150495 838412 434477 993107",
"output": "4647430"
},
{
"input": "30\n315421 772664 560686 654312 151528 356749 351486 707462 820089 226682 546700 136028 824236 842130 578079 337807 665903 764100 617900 822937 992759 591749 651310 742085 767695 695442 17967 515106 81059 186025",
"output": "13488674"
},
{
"input": "45\n908719 394261 815134 419990 926993 383792 772842 277695 527137 655356 684956 695716 273062 550324 106247 399133 442382 33076 462920 294674 846052 817752 421365 474141 290471 358990 109812 74492 543281 169434 919692 786809 24028 197184 310029 801476 699355 429672 51343 374128 776726 850380 293868 981569 550763",
"output": "21993384"
},
{
"input": "56\n100728 972537 13846 385421 756708 184642 259487 319707 376662 221694 675284 972837 499419 13846 38267 289898 901299 831197 954715 197515 514102 910423 127555 883934 362472 870788 538802 741008 973434 448124 391526 363321 947321 544618 68006 782313 955075 741981 815027 723297 585059 718114 700739 413489 454091 736144 308999 98065 3716 347323 9635 289003 986510 607065 60236 273351",
"output": "26984185"
},
{
"input": "70\n644488 5444 150441 714420 602059 335330 510670 196555 546346 740011 509449 850947 692874 524857 750434 952985 223744 374727 896124 753037 367352 679050 560202 172728 569291 778616 332116 286927 843598 372698 244906 498046 900681 709791 420904 724593 864493 813094 791377 39998 296710 625656 403891 579231 706693 984045 16901 574259 562265 761104 930361 256045 124461 538980 573508 372148 988722 108592 784354 55302 232524 277205 782251 299943 436488 743389 324618 742543 266915 99642",
"output": "32816391"
},
{
"input": "1\n0",
"output": "0"
},
{
"input": "1\n1000000",
"output": "0"
},
{
"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": "0"
},
{
"input": "100\n1000000 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": "99000000"
},
{
"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 1000000",
"output": "99000000"
},
{
"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 1000000 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": "99000000"
},
{
"input": "3\n0 0 0",
"output": "0"
},
{
"input": "50\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",
"output": "0"
},
{
"input": "3\n5 0 0",
"output": "10"
},
{
"input": "5\n2 10 0 0 0",
"output": "38"
}
] | 1,685,541,634
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 41
| 46
| 0
|
n = int(input())
a = list(map(int, input().split()))
a.sort()
y = a[n-1]
c = 0
for i in range(n):
if a[i] < y :
x = abs(y-a[i])
c = c + x
print(c)
|
Title: Holiday Of Equality
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.
Totally in Berland there are *n* citizens, the welfare of each of them is estimated as the integer in *a**i* burles (burle is the currency in Berland).
You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.
Input Specification:
The first line contains the integer *n* (1<=≤<=*n*<=≤<=100) — the number of citizens in the kingdom.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n*, where *a**i* (0<=≤<=*a**i*<=≤<=106) — the welfare of the *i*-th citizen.
Output Specification:
In the only line print the integer *S* — the minimum number of burles which are had to spend.
Demo Input:
['5\n0 1 2 3 4\n', '5\n1 1 0 1 1\n', '3\n1 3 1\n', '1\n12\n']
Demo Output:
['10', '1', '4', '0']
Note:
In the first example if we add to the first citizen 4 burles, to the second 3, to the third 2 and to the fourth 1, then the welfare of all citizens will equal 4.
In the second example it is enough to give one burle to the third citizen.
In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal 3.
In the fourth example it is possible to give nothing to everyone because all citizens have 12 burles.
|
```python
n = int(input())
a = list(map(int, input().split()))
a.sort()
y = a[n-1]
c = 0
for i in range(n):
if a[i] < y :
x = abs(y-a[i])
c = c + x
print(c)
```
| 3
|
|
1,000
|
B
|
Light It Up
|
PROGRAMMING
| 1,500
|
[
"greedy"
] | null | null |
Recently, you bought a brand new smart lamp with programming features. At first, you set up a schedule to the lamp. Every day it will turn power on at moment $0$ and turn power off at moment $M$. Moreover, the lamp allows you to set a program of switching its state (states are "lights on" and "lights off"). Unfortunately, some program is already installed into the lamp.
The lamp allows only good programs. Good program can be represented as a non-empty array $a$, where $0 < a_1 < a_2 < \dots < a_{|a|} < M$. All $a_i$ must be integers. Of course, preinstalled program is a good program.
The lamp follows program $a$ in next manner: at moment $0$ turns power and light on. Then at moment $a_i$ the lamp flips its state to opposite (if it was lit, it turns off, and vice versa). The state of the lamp flips instantly: for example, if you turn the light off at moment $1$ and then do nothing, the total time when the lamp is lit will be $1$. Finally, at moment $M$ the lamp is turning its power off regardless of its state.
Since you are not among those people who read instructions, and you don't understand the language it's written in, you realize (after some testing) the only possible way to alter the preinstalled program. You can insert at most one element into the program $a$, so it still should be a good program after alteration. Insertion can be done between any pair of consecutive elements of $a$, or even at the begining or at the end of $a$.
Find such a way to alter the program that the total time when the lamp is lit is maximum possible. Maybe you should leave program untouched. If the lamp is lit from $x$ till moment $y$, then its lit for $y - x$ units of time. Segments of time when the lamp is lit are summed up.
|
First line contains two space separated integers $n$ and $M$ ($1 \le n \le 10^5$, $2 \le M \le 10^9$) — the length of program $a$ and the moment when power turns off.
Second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($0 < a_1 < a_2 < \dots < a_n < M$) — initially installed program $a$.
|
Print the only integer — maximum possible total time when the lamp is lit.
|
[
"3 10\n4 6 7\n",
"2 12\n1 10\n",
"2 7\n3 4\n"
] |
[
"8\n",
"9\n",
"6\n"
] |
In the first example, one of possible optimal solutions is to insert value $x = 3$ before $a_1$, so program will be $[3, 4, 6, 7]$ and time of lamp being lit equals $(3 - 0) + (6 - 4) + (10 - 7) = 8$. Other possible solution is to insert $x = 5$ in appropriate place.
In the second example, there is only one optimal solution: to insert $x = 2$ between $a_1$ and $a_2$. Program will become $[1, 2, 10]$, and answer will be $(1 - 0) + (10 - 2) = 9$.
In the third example, optimal answer is to leave program untouched, so answer will be $(3 - 0) + (7 - 4) = 6$.
| 0
|
[
{
"input": "3 10\n4 6 7",
"output": "8"
},
{
"input": "2 12\n1 10",
"output": "9"
},
{
"input": "2 7\n3 4",
"output": "6"
},
{
"input": "1 2\n1",
"output": "1"
},
{
"input": "5 10\n1 3 5 6 8",
"output": "6"
},
{
"input": "7 1000000000\n1 10001 10011 20011 20021 40021 40031",
"output": "999999969"
},
{
"input": "7 1000000000\n3 10001 10011 20011 20021 40021 40031",
"output": "999999969"
},
{
"input": "1 10\n1",
"output": "9"
},
{
"input": "1 10000000\n1",
"output": "9999999"
},
{
"input": "1 8\n1",
"output": "7"
},
{
"input": "7 17\n1 5 9 10 11 14 16",
"output": "9"
},
{
"input": "4 17\n1 5 9 10",
"output": "12"
},
{
"input": "5 12\n1 2 3 4 5",
"output": "9"
},
{
"input": "2 1000000000\n100 111",
"output": "999999989"
},
{
"input": "3 90591\n90579 90580 90581",
"output": "90589"
},
{
"input": "5 16\n1 2 3 4 5",
"output": "13"
},
{
"input": "6 981231233\n1 6 15 30 130 1875",
"output": "981229468"
},
{
"input": "1 3\n1",
"output": "2"
},
{
"input": "3 12\n4 7 11",
"output": "8"
},
{
"input": "1 100000\n1",
"output": "99999"
},
{
"input": "2 5\n1 3",
"output": "3"
},
{
"input": "1 1000000000\n1",
"output": "999999999"
},
{
"input": "1 1000000000\n999999999",
"output": "999999999"
},
{
"input": "3 4\n1 2 3",
"output": "2"
},
{
"input": "3 5\n1 2 3",
"output": "3"
},
{
"input": "1 4\n2",
"output": "3"
},
{
"input": "3 5\n1 3 4",
"output": "3"
},
{
"input": "9 20\n5 9 11 12 14 15 16 17 19",
"output": "12"
},
{
"input": "2 4\n1 2",
"output": "3"
},
{
"input": "1 10\n2",
"output": "9"
},
{
"input": "2 6\n2 4",
"output": "4"
},
{
"input": "2 4\n1 3",
"output": "2"
},
{
"input": "3 6\n1 2 4",
"output": "4"
},
{
"input": "7 9863\n65 96 97 98 101 112 1115",
"output": "9819"
}
] | 1,666,186,134
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 30
| 0
|
n,M = map(int,input().split())
a = list(map(int,input().split()))
from copy import copy
if n%2 == 0:
res = [M+sum(a[0:n-1:2])-sum(a[1:n:2])]
for i in range(n):
if a[i]-1 != a[i-1] and a[i]-1 != 0 :
b = copy(a)
b.insert(i,a[i]-1)
res.append(sum(b[0:n+1:2])-sum(b[1:n:2]))
if a[i]+1 != a[i+1] and a[i]+1 != 0 :
b = copy(a)
b.insert(i+1,a[i]+1)
res.append(sum(b[0:n+1:2])-sum(b[1:n:2]))
else:
res = [sum(a[0:n:2])-sum(a[1:n-1:2])]
for i in range(n):
if a[i]-1 != a[i-1] and a[i]-1 != 0 :
b = copy(a)
b.insert(i,a[i]-1)
res.append(M+sum(b[0:n:2])-sum(b[1:n+1:2]))
if a[i]+1 != a[i+1] and a[i]+1 != 0 :
b = copy(a)
b.insert(i+1,a[i]+1)
res.append(M+sum(b[0:n:2])-sum(b[1:n+1:2]))
print(max(res))
|
Title: Light It Up
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, you bought a brand new smart lamp with programming features. At first, you set up a schedule to the lamp. Every day it will turn power on at moment $0$ and turn power off at moment $M$. Moreover, the lamp allows you to set a program of switching its state (states are "lights on" and "lights off"). Unfortunately, some program is already installed into the lamp.
The lamp allows only good programs. Good program can be represented as a non-empty array $a$, where $0 < a_1 < a_2 < \dots < a_{|a|} < M$. All $a_i$ must be integers. Of course, preinstalled program is a good program.
The lamp follows program $a$ in next manner: at moment $0$ turns power and light on. Then at moment $a_i$ the lamp flips its state to opposite (if it was lit, it turns off, and vice versa). The state of the lamp flips instantly: for example, if you turn the light off at moment $1$ and then do nothing, the total time when the lamp is lit will be $1$. Finally, at moment $M$ the lamp is turning its power off regardless of its state.
Since you are not among those people who read instructions, and you don't understand the language it's written in, you realize (after some testing) the only possible way to alter the preinstalled program. You can insert at most one element into the program $a$, so it still should be a good program after alteration. Insertion can be done between any pair of consecutive elements of $a$, or even at the begining or at the end of $a$.
Find such a way to alter the program that the total time when the lamp is lit is maximum possible. Maybe you should leave program untouched. If the lamp is lit from $x$ till moment $y$, then its lit for $y - x$ units of time. Segments of time when the lamp is lit are summed up.
Input Specification:
First line contains two space separated integers $n$ and $M$ ($1 \le n \le 10^5$, $2 \le M \le 10^9$) — the length of program $a$ and the moment when power turns off.
Second line contains $n$ space separated integers $a_1, a_2, \dots, a_n$ ($0 < a_1 < a_2 < \dots < a_n < M$) — initially installed program $a$.
Output Specification:
Print the only integer — maximum possible total time when the lamp is lit.
Demo Input:
['3 10\n4 6 7\n', '2 12\n1 10\n', '2 7\n3 4\n']
Demo Output:
['8\n', '9\n', '6\n']
Note:
In the first example, one of possible optimal solutions is to insert value $x = 3$ before $a_1$, so program will be $[3, 4, 6, 7]$ and time of lamp being lit equals $(3 - 0) + (6 - 4) + (10 - 7) = 8$. Other possible solution is to insert $x = 5$ in appropriate place.
In the second example, there is only one optimal solution: to insert $x = 2$ between $a_1$ and $a_2$. Program will become $[1, 2, 10]$, and answer will be $(1 - 0) + (10 - 2) = 9$.
In the third example, optimal answer is to leave program untouched, so answer will be $(3 - 0) + (7 - 4) = 6$.
|
```python
n,M = map(int,input().split())
a = list(map(int,input().split()))
from copy import copy
if n%2 == 0:
res = [M+sum(a[0:n-1:2])-sum(a[1:n:2])]
for i in range(n):
if a[i]-1 != a[i-1] and a[i]-1 != 0 :
b = copy(a)
b.insert(i,a[i]-1)
res.append(sum(b[0:n+1:2])-sum(b[1:n:2]))
if a[i]+1 != a[i+1] and a[i]+1 != 0 :
b = copy(a)
b.insert(i+1,a[i]+1)
res.append(sum(b[0:n+1:2])-sum(b[1:n:2]))
else:
res = [sum(a[0:n:2])-sum(a[1:n-1:2])]
for i in range(n):
if a[i]-1 != a[i-1] and a[i]-1 != 0 :
b = copy(a)
b.insert(i,a[i]-1)
res.append(M+sum(b[0:n:2])-sum(b[1:n+1:2]))
if a[i]+1 != a[i+1] and a[i]+1 != 0 :
b = copy(a)
b.insert(i+1,a[i]+1)
res.append(M+sum(b[0:n:2])-sum(b[1:n+1:2]))
print(max(res))
```
| -1
|
|
822
|
A
|
I'm bored with life
|
PROGRAMMING
| 800
|
[
"implementation",
"math",
"number theory"
] | null | null |
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
|
The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12).
|
Print a single integer denoting the greatest common divisor of integers *A*! and *B*!.
|
[
"4 3\n"
] |
[
"6\n"
] |
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.
| 500
|
[
{
"input": "4 3",
"output": "6"
},
{
"input": "10 399603090",
"output": "3628800"
},
{
"input": "6 973151934",
"output": "720"
},
{
"input": "2 841668075",
"output": "2"
},
{
"input": "7 415216919",
"output": "5040"
},
{
"input": "3 283733059",
"output": "6"
},
{
"input": "11 562314608",
"output": "39916800"
},
{
"input": "3 990639260",
"output": "6"
},
{
"input": "11 859155400",
"output": "39916800"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "5 3",
"output": "6"
},
{
"input": "1 4",
"output": "1"
},
{
"input": "5 4",
"output": "24"
},
{
"input": "1 12",
"output": "1"
},
{
"input": "9 7",
"output": "5040"
},
{
"input": "2 3",
"output": "2"
},
{
"input": "6 11",
"output": "720"
},
{
"input": "6 7",
"output": "720"
},
{
"input": "11 11",
"output": "39916800"
},
{
"input": "4 999832660",
"output": "24"
},
{
"input": "7 999228288",
"output": "5040"
},
{
"input": "11 999257105",
"output": "39916800"
},
{
"input": "11 999286606",
"output": "39916800"
},
{
"input": "3 999279109",
"output": "6"
},
{
"input": "999632727 11",
"output": "39916800"
},
{
"input": "999625230 7",
"output": "5040"
},
{
"input": "999617047 3",
"output": "6"
},
{
"input": "999646548 7",
"output": "5040"
},
{
"input": "999639051 3",
"output": "6"
},
{
"input": "12 12",
"output": "479001600"
},
{
"input": "12 1",
"output": "1"
},
{
"input": "1213 5",
"output": "120"
},
{
"input": "8 9",
"output": "40320"
},
{
"input": "12 9",
"output": "362880"
},
{
"input": "12 1000000000",
"output": "479001600"
},
{
"input": "1000000000 1",
"output": "1"
},
{
"input": "12 13",
"output": "479001600"
},
{
"input": "2 29845",
"output": "2"
},
{
"input": "10 21",
"output": "3628800"
},
{
"input": "12 20",
"output": "479001600"
},
{
"input": "15 12",
"output": "479001600"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1000000000 12",
"output": "479001600"
},
{
"input": "11 30",
"output": "39916800"
},
{
"input": "17 12",
"output": "479001600"
},
{
"input": "4 19",
"output": "24"
},
{
"input": "12 15",
"output": "479001600"
},
{
"input": "20 6",
"output": "720"
},
{
"input": "10 20",
"output": "3628800"
},
{
"input": "10 10",
"output": "3628800"
},
{
"input": "22 12",
"output": "479001600"
},
{
"input": "20 12",
"output": "479001600"
},
{
"input": "12 23",
"output": "479001600"
},
{
"input": "12 22",
"output": "479001600"
},
{
"input": "18 3",
"output": "6"
},
{
"input": "14 10",
"output": "3628800"
},
{
"input": "14 12",
"output": "479001600"
},
{
"input": "8 3",
"output": "6"
},
{
"input": "5 5",
"output": "120"
}
] | 1,667,731,774
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 59
| 62
| 0
|
numbers = tuple (map (int, input ().split (' ')))
result = 1
for i in range (1, min (numbers) + 1):
result *= i
print (result)
|
Title: I'm bored with life
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for the period of university studies. Consequently Noora had to leave Vičkopolis and move to Pavlopolis. Thus Leha was left completely alone in a quiet town Vičkopolis. He almost even fell into a depression from boredom!
Leha came up with a task for himself to relax a little. He chooses two integers *A* and *B* and then calculates the greatest common divisor of integers "*A* factorial" and "*B* factorial". Formally the hacker wants to find out GCD(*A*!,<=*B*!). It's well known that the factorial of an integer *x* is a product of all positive integers less than or equal to *x*. Thus *x*!<==<=1·2·3·...·(*x*<=-<=1)·*x*. For example 4!<==<=1·2·3·4<==<=24. Recall that GCD(*x*,<=*y*) is the largest positive integer *q* that divides (without a remainder) both *x* and *y*.
Leha has learned how to solve this task very effective. You are able to cope with it not worse, aren't you?
Input Specification:
The first and single line contains two integers *A* and *B* (1<=≤<=*A*,<=*B*<=≤<=109,<=*min*(*A*,<=*B*)<=≤<=12).
Output Specification:
Print a single integer denoting the greatest common divisor of integers *A*! and *B*!.
Demo Input:
['4 3\n']
Demo Output:
['6\n']
Note:
Consider the sample.
4! = 1·2·3·4 = 24. 3! = 1·2·3 = 6. The greatest common divisor of integers 24 and 6 is exactly 6.
|
```python
numbers = tuple (map (int, input ().split (' ')))
result = 1
for i in range (1, min (numbers) + 1):
result *= i
print (result)
```
| 3
|
|
43
|
A
|
Football
|
PROGRAMMING
| 1,000
|
[
"strings"
] |
A. Football
|
2
|
256
|
One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
|
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
|
[
"1\nABC\n",
"5\nA\nABA\nABA\nA\nA\n"
] |
[
"ABC\n",
"A\n"
] |
none
| 500
|
[
{
"input": "1\nABC",
"output": "ABC"
},
{
"input": "5\nA\nABA\nABA\nA\nA",
"output": "A"
},
{
"input": "2\nXTSJEP\nXTSJEP",
"output": "XTSJEP"
},
{
"input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ",
"output": "XZYDJAEDZ"
},
{
"input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD",
"output": "QCCYXL"
},
{
"input": "3\nAZID\nEERWBC\nEERWBC",
"output": "EERWBC"
},
{
"input": "3\nHNCGYL\nHNCGYL\nHNCGYL",
"output": "HNCGYL"
},
{
"input": "4\nZZWZTG\nZZWZTG\nZZWZTG\nZZWZTG",
"output": "ZZWZTG"
},
{
"input": "4\nA\nA\nKUDLJMXCSE\nA",
"output": "A"
},
{
"input": "5\nPHBTW\nPHBTW\nPHBTW\nPHBTW\nPHBTW",
"output": "PHBTW"
},
{
"input": "5\nPKUZYTFYWN\nPKUZYTFYWN\nSTC\nPKUZYTFYWN\nPKUZYTFYWN",
"output": "PKUZYTFYWN"
},
{
"input": "5\nHH\nHH\nNTQWPA\nNTQWPA\nHH",
"output": "HH"
},
{
"input": "10\nW\nW\nW\nW\nW\nD\nW\nD\nD\nW",
"output": "W"
},
{
"input": "19\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nXBCP\nXBCP\nXBCP\nTGACNIH\nXBCP\nXBCP\nTGACNIH\nTGACNIH\nXBCP",
"output": "XBCP"
},
{
"input": "33\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nOWQWCKLLF\nOWQWCKLLF\nPYPAS\nPYPAS\nOWQWCKLLF\nPYPAS\nPYPAS",
"output": "PYPAS"
},
{
"input": "51\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC\nNC",
"output": "NC"
},
{
"input": "89\nH\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nVOCI\nVOCI\nH\nH\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nVOCI\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI\nH\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nVOCI\nH\nH\nH\nH\nH\nVOCI\nH\nVOCI\nH\nVOCI\nVOCI",
"output": "VOCI"
},
{
"input": "100\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nM\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA\nHA",
"output": "HA"
},
{
"input": "100\nG\nG\nS\nS\nG\nG\nS\nS\nG\nS\nS\nS\nG\nS\nG\nG\nS\nG\nS\nS\nG\nS\nS\nS\nS\nS\nG\nS\nG\nS\nS\nG\nG\nG\nS\nS\nS\nS\nG\nS\nS\nG\nG\nG\nG\nG\nS\nG\nG\nS\nS\nS\nS\nS\nG\nG\nS\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nS\nS\nG\nS\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nG\nS\nS\nG\nS\nS\nS\nS\nG\nG\nG\nS\nG\nG\nG\nS",
"output": "G"
},
{
"input": "100\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nOBH\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL\nWL",
"output": "WL"
}
] | 1,670,953,150
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 2
| 60
| 0
|
x=int(input())
y = []
count1 = 0
count2 = 0
for i in range(x):
y.append(input())
teams = set(y)
team1 = teams.pop()
if x>1:
team2 = teams.pop()
for i in range(x):
if y[i] == team1:
count1+=1
else:
count2 +=1
if count1 > count2:
print(str(team1))
else:
print(str(team2))
|
Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
Output Specification:
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
Demo Input:
['1\nABC\n', '5\nA\nABA\nABA\nA\nA\n']
Demo Output:
['ABC\n', 'A\n']
Note:
none
|
```python
x=int(input())
y = []
count1 = 0
count2 = 0
for i in range(x):
y.append(input())
teams = set(y)
team1 = teams.pop()
if x>1:
team2 = teams.pop()
for i in range(x):
if y[i] == team1:
count1+=1
else:
count2 +=1
if count1 > count2:
print(str(team1))
else:
print(str(team2))
```
| -1
|
761
|
B
|
Dasha and friends
|
PROGRAMMING
| 1,300
|
[
"brute force",
"implementation",
"math"
] | null | null |
Running with barriers on the circle track is very popular in the country where Dasha lives, so no wonder that on her way to classes she saw the following situation:
The track is the circle with length *L*, in distinct points of which there are *n* barriers. Athlete always run the track in counterclockwise direction if you look on him from above. All barriers are located at integer distance from each other along the track.
Her friends the parrot Kefa and the leopard Sasha participated in competitions and each of them ran one lap. Each of the friends started from some integral point on the track. Both friends wrote the distance from their start along the track to each of the *n* barriers. Thus, each of them wrote *n* integers in the ascending order, each of them was between 0 and *L*<=-<=1, inclusively.
There are several tracks in the country, all of them have same length and same number of barriers, but the positions of the barriers can differ among different tracks. Now Dasha is interested if it is possible that Kefa and Sasha ran the same track or they participated on different tracks.
Write the program which will check that Kefa's and Sasha's tracks coincide (it means that one can be obtained from the other by changing the start position). Note that they always run the track in one direction — counterclockwise, if you look on a track from above.
|
The first line contains two integers *n* and *L* (1<=≤<=*n*<=≤<=50, *n*<=≤<=*L*<=≤<=100) — the number of barriers on a track and its length.
The second line contains *n* distinct integers in the ascending order — the distance from Kefa's start to each barrier in the order of its appearance. All integers are in the range from 0 to *L*<=-<=1 inclusively.
The second line contains *n* distinct integers in the ascending order — the distance from Sasha's start to each barrier in the order of its overcoming. All integers are in the range from 0 to *L*<=-<=1 inclusively.
|
Print "YES" (without quotes), if Kefa and Sasha ran the coinciding tracks (it means that the position of all barriers coincides, if they start running from the same points on the track). Otherwise print "NO" (without quotes).
|
[
"3 8\n2 4 6\n1 5 7\n",
"4 9\n2 3 5 8\n0 1 3 6\n",
"2 4\n1 3\n1 2\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
The first test is analyzed in the statement.
| 1,000
|
[
{
"input": "3 8\n2 4 6\n1 5 7",
"output": "YES"
},
{
"input": "4 9\n2 3 5 8\n0 1 3 6",
"output": "YES"
},
{
"input": "2 4\n1 3\n1 2",
"output": "NO"
},
{
"input": "5 9\n0 2 5 6 7\n1 3 6 7 8",
"output": "YES"
},
{
"input": "5 60\n7 26 27 40 59\n14 22 41 42 55",
"output": "YES"
},
{
"input": "20 29\n0 1 2 4 5 8 9 12 14 15 17 19 20 21 22 23 25 26 27 28\n0 2 4 5 6 7 8 10 11 12 13 14 15 16 18 19 22 23 26 28",
"output": "YES"
},
{
"input": "35 41\n0 1 2 3 4 5 6 7 9 10 11 12 13 14 18 19 20 21 22 23 24 25 26 28 30 31 32 33 34 35 36 37 38 39 40\n0 1 2 3 4 5 7 8 9 10 11 12 16 17 18 19 20 21 22 23 24 26 28 29 30 31 32 33 34 35 36 37 38 39 40",
"output": "YES"
},
{
"input": "40 63\n0 2 3 4 5 6 9 10 12 15 17 19 23 25 26 27 28 29 30 31 33 34 36 37 38 39 40 43 45 49 50 52 53 54 55 57 58 60 61 62\n1 2 3 4 5 8 10 14 15 17 18 19 20 22 23 25 26 27 28 30 31 32 33 34 37 38 40 43 46 47 51 53 54 55 56 57 58 59 61 62",
"output": "NO"
},
{
"input": "50 97\n1 2 3 4 6 9 10 11 12 13 14 21 22 23 24 25 28 29 30 31 32 33 34 36 37 40 41 45 53 56 59 64 65 69 70 71 72 73 74 77 81 84 85 86 87 89 91 92 95 96\n0 1 2 3 6 10 13 14 15 16 18 20 21 24 25 27 28 29 30 33 35 36 37 38 39 40 47 48 49 50 51 54 55 56 57 58 59 60 62 63 66 67 71 79 82 85 90 91 95 96",
"output": "NO"
},
{
"input": "50 100\n0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98\n1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99",
"output": "YES"
},
{
"input": "1 2\n0\n0",
"output": "YES"
},
{
"input": "1 2\n0\n1",
"output": "YES"
},
{
"input": "1 2\n1\n0",
"output": "YES"
},
{
"input": "1 2\n1\n1",
"output": "YES"
},
{
"input": "1 1\n0\n0",
"output": "YES"
},
{
"input": "5 12\n2 3 4 8 10\n2 3 4 8 10",
"output": "YES"
},
{
"input": "1 18\n3\n10",
"output": "YES"
},
{
"input": "1 75\n65\n8",
"output": "YES"
},
{
"input": "2 16\n4 13\n2 11",
"output": "YES"
},
{
"input": "2 95\n45 59\n3 84",
"output": "YES"
},
{
"input": "3 53\n29 43 50\n29 43 50",
"output": "YES"
},
{
"input": "3 60\n39 46 51\n43 50 55",
"output": "YES"
},
{
"input": "4 4\n0 1 2 3\n0 1 2 3",
"output": "YES"
},
{
"input": "4 93\n45 48 50 90\n20 68 71 73",
"output": "YES"
},
{
"input": "6 18\n0 3 8 11 15 16\n2 7 10 14 15 17",
"output": "YES"
},
{
"input": "6 87\n0 1 21 31 34 66\n11 12 32 42 45 77",
"output": "YES"
},
{
"input": "7 26\n0 3 9 13 14 19 20\n4 7 13 17 18 23 24",
"output": "YES"
},
{
"input": "7 81\n0 12 19 24 25 35 59\n1 8 13 14 24 48 70",
"output": "YES"
},
{
"input": "8 20\n0 1 2 3 5 6 14 15\n1 2 10 11 16 17 18 19",
"output": "YES"
},
{
"input": "8 94\n0 8 11 27 38 54 57 89\n1 33 38 46 49 65 76 92",
"output": "YES"
},
{
"input": "9 18\n1 3 6 8 11 12 13 16 17\n0 2 5 6 7 10 11 13 15",
"output": "YES"
},
{
"input": "9 90\n10 11 27 33 34 55 63 84 87\n9 12 25 26 42 48 49 70 78",
"output": "YES"
},
{
"input": "10 42\n4 9 10 14 15 16 19 33 36 40\n0 14 17 21 27 32 33 37 38 39",
"output": "YES"
},
{
"input": "10 73\n4 5 15 19 20 25 28 42 57 58\n3 4 9 12 26 41 42 61 62 72",
"output": "YES"
},
{
"input": "11 11\n0 1 2 3 4 5 6 7 8 9 10\n0 1 2 3 4 5 6 7 8 9 10",
"output": "YES"
},
{
"input": "11 57\n1 4 27 30 31 35 37 41 50 52 56\n22 25 26 30 32 36 45 47 51 53 56",
"output": "YES"
},
{
"input": "12 73\n5 9 11 20 25 36 40 41 44 48 56 60\n12 16 18 27 32 43 47 48 51 55 63 67",
"output": "YES"
},
{
"input": "12 95\n1 37 42 46 56 58 59 62 64 71 76 80\n2 18 54 59 63 73 75 76 79 81 88 93",
"output": "YES"
},
{
"input": "13 29\n2 5 6 9 12 17 18 19 20 21 22 24 27\n0 3 6 11 12 13 14 15 16 18 21 25 28",
"output": "YES"
},
{
"input": "13 90\n9 18 23 30 31 36 39 44 58 59 74 82 87\n1 6 18 27 32 39 40 45 48 53 67 68 83",
"output": "YES"
},
{
"input": "14 29\n1 2 3 4 5 7 9 12 13 20 21 22 23 24\n0 3 4 11 12 13 14 15 21 22 23 24 25 27",
"output": "YES"
},
{
"input": "14 94\n7 8 9 21 34 35 36 37 38 43 46 52 84 93\n2 3 4 16 29 30 31 32 33 38 41 47 79 88",
"output": "YES"
},
{
"input": "15 19\n1 2 3 4 5 6 7 8 9 10 11 13 14 16 17\n0 1 2 3 4 5 6 7 8 9 10 12 13 15 16",
"output": "YES"
},
{
"input": "15 27\n2 3 4 5 6 7 8 9 10 11 12 14 17 24 26\n2 3 4 5 6 7 8 9 10 11 12 14 17 24 26",
"output": "YES"
},
{
"input": "16 28\n3 5 6 7 9 10 11 12 13 14 17 19 20 25 26 27\n0 5 6 7 11 13 14 15 17 18 19 20 21 22 25 27",
"output": "YES"
},
{
"input": "16 93\n5 6 10 11 13 14 41 43 46 61 63 70 74 79 83 92\n0 9 15 16 20 21 23 24 51 53 56 71 73 80 84 89",
"output": "YES"
},
{
"input": "17 49\n2 5 11 12 16 18 19 21 22 24 36 37 38 39 40 44 47\n1 7 8 12 14 15 17 18 20 32 33 34 35 36 40 43 47",
"output": "YES"
},
{
"input": "17 86\n16 17 25 33 39 41 50 51 54 56 66 70 72 73 77 80 85\n3 9 11 20 21 24 26 36 40 42 43 47 50 55 72 73 81",
"output": "YES"
},
{
"input": "18 20\n0 1 2 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19\n0 1 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19",
"output": "YES"
},
{
"input": "18 82\n0 5 10 13 14 16 21 28 29 30 44 46 61 64 69 71 77 78\n0 5 8 9 11 16 23 24 25 39 41 56 59 64 66 72 73 77",
"output": "YES"
},
{
"input": "19 25\n0 1 2 3 5 7 9 10 12 13 16 17 18 19 20 21 22 23 24\n0 3 4 5 6 7 8 9 10 11 12 13 14 15 17 19 21 22 24",
"output": "YES"
},
{
"input": "19 91\n5 17 18 20 22 25 26 31 32 33 43 47 54 61 62 64 77 80 87\n4 5 6 16 20 27 34 35 37 50 53 60 69 81 82 84 86 89 90",
"output": "YES"
},
{
"input": "20 53\n2 6 8 9 16 17 20 21 22 23 25 26 35 36 38 39 44 46 47 50\n4 5 8 9 10 11 13 14 23 24 26 27 32 34 35 38 43 47 49 50",
"output": "YES"
},
{
"input": "21 44\n0 1 3 4 6 7 8 9 10 11 12 15 17 18 21 22 27 29 34 36 42\n1 7 9 10 12 13 15 16 17 18 19 20 21 24 26 27 30 31 36 38 43",
"output": "YES"
},
{
"input": "21 94\n3 5 6 8 9 15 16 20 28 31 35 39 49 50 53 61 71 82 85 89 90\n6 17 20 24 25 32 34 35 37 38 44 45 49 57 60 64 68 78 79 82 90",
"output": "YES"
},
{
"input": "22 24\n0 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 22 23\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 21 22 23",
"output": "YES"
},
{
"input": "22 85\n3 5 7 14 18 21 25 32 38 41 53 58 61 62 66 70 71 73 75 76 79 83\n3 6 18 23 26 27 31 35 36 38 40 41 44 48 53 55 57 64 68 71 75 82",
"output": "YES"
},
{
"input": "23 38\n0 2 4 5 7 8 12 13 14 16 17 18 21 22 24 27 28 30 31 32 35 36 37\n0 1 2 3 5 7 8 10 11 15 16 17 19 20 21 24 25 27 30 31 33 34 35",
"output": "YES"
},
{
"input": "23 93\n1 3 5 10 19 22 26 27 30 35 39 53 55 60 66 67 75 76 77 80 82 89 90\n9 11 16 22 23 31 32 33 36 38 45 46 50 52 54 59 68 71 75 76 79 84 88",
"output": "YES"
},
{
"input": "24 37\n1 4 5 6 8 11 12 13 15 16 17 19 20 21 23 26 27 28 30 31 33 34 35 36\n0 3 4 5 7 8 10 11 12 13 15 18 19 20 22 25 26 27 29 30 31 33 34 35",
"output": "YES"
},
{
"input": "24 94\n9 10 13 14 16 18 19 22 24 29 32 35 48 55 57 63 64 69 72 77 78 85 90 92\n1 7 8 13 16 21 22 29 34 36 47 48 51 52 54 56 57 60 62 67 70 73 86 93",
"output": "YES"
},
{
"input": "25 45\n0 1 2 4 6 7 8 9 13 14 17 19 21 22 23 25 28 29 30 31 34 36 38 39 42\n1 3 4 5 7 10 11 12 13 16 18 20 21 24 27 28 29 31 33 34 35 36 40 41 44",
"output": "YES"
},
{
"input": "25 72\n1 2 6 8 9 11 15 18 19 20 26 29 31 33 34 40 41 43 45 48 58 60 68 69 71\n0 6 9 11 13 14 20 21 23 25 28 38 40 48 49 51 53 54 58 60 61 63 67 70 71",
"output": "YES"
},
{
"input": "26 47\n0 2 5 7 8 9 10 12 13 14 20 22 23 25 27 29 31 32 33 35 36 37 38 42 44 45\n0 2 4 6 8 9 10 12 13 14 15 19 21 22 24 26 29 31 32 33 34 36 37 38 44 46",
"output": "YES"
},
{
"input": "26 99\n0 1 13 20 21 22 25 26 27 28 32 39 44 47 56 58 60 62 71 81 83 87 89 93 94 98\n6 8 12 14 18 19 23 24 25 37 44 45 46 49 50 51 52 56 63 68 71 80 82 84 86 95",
"output": "YES"
},
{
"input": "27 35\n0 2 3 4 5 6 7 8 10 11 12 13 14 15 16 17 19 20 21 23 26 27 29 30 31 32 33\n0 1 2 3 5 7 8 9 10 11 12 13 15 16 17 18 19 20 21 22 24 25 26 28 31 32 34",
"output": "YES"
},
{
"input": "27 51\n1 2 4 7 8 11 13 17 20 21 23 24 25 28 29 30 34 35 37 38 40 43 45 46 47 48 50\n0 1 2 4 6 7 9 12 13 16 18 22 25 26 28 29 30 33 34 35 39 40 42 43 45 48 50",
"output": "YES"
},
{
"input": "28 38\n1 4 5 7 8 9 10 11 12 14 15 16 18 19 20 21 22 23 24 25 28 29 30 32 33 35 36 37\n0 1 2 3 4 5 6 9 10 11 13 14 16 17 18 20 23 24 26 27 28 29 30 31 33 34 35 37",
"output": "YES"
},
{
"input": "28 67\n0 1 2 3 6 9 10 15 18 22 24 25 30 35 36 38 39 47 48 49 51 53 55 56 58 62 63 64\n4 7 11 13 14 19 24 25 27 28 36 37 38 40 42 44 45 47 51 52 53 56 57 58 59 62 65 66",
"output": "YES"
},
{
"input": "29 29\n0 1 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\n0 1 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",
"output": "YES"
},
{
"input": "29 93\n1 2 11 13 18 21 27 28 30 38 41 42 46 54 55 56 60 61 63 64 66 69 71 72 77 81 83 89 90\n2 10 11 12 16 17 19 20 22 25 27 28 33 37 39 45 46 50 51 60 62 67 70 76 77 79 87 90 91",
"output": "YES"
},
{
"input": "30 63\n0 2 3 5 6 7 8 10 13 18 19 21 22 23 26 32 35 37 38 39 40 41 43 44 49 51 53 54 58 61\n0 2 3 5 6 7 8 10 13 18 19 21 22 23 26 32 35 37 38 39 40 41 43 44 49 51 53 54 58 61",
"output": "YES"
},
{
"input": "30 91\n1 2 3 7 8 9 13 16 17 19 27 29 38 45 47 52 53 55 61 62 66 77 78 79 80 81 82 84 88 89\n3 4 5 9 12 13 15 23 25 34 41 43 48 49 51 57 58 62 73 74 75 76 77 78 80 84 85 88 89 90",
"output": "YES"
},
{
"input": "31 39\n0 1 2 3 4 5 6 7 8 10 11 13 14 17 18 20 21 23 24 25 27 28 29 30 31 33 34 35 36 37 38\n0 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 18 19 21 22 25 26 28 29 31 32 33 35 36 37 38",
"output": "YES"
},
{
"input": "31 95\n9 12 14 15 21 23 26 28 30 36 37 42 47 51 54 56 59 62 64 65 66 70 72 74 75 79 82 85 87 91 93\n0 2 3 7 10 13 15 19 21 32 35 37 38 44 46 49 51 53 59 60 65 70 74 77 79 82 85 87 88 89 93",
"output": "YES"
},
{
"input": "32 61\n0 2 3 5 7 10 13 14 15 18 19 20 21 22 23 24 26 32 33 34 36 38 43 46 47 51 54 55 56 57 58 59\n1 2 4 6 9 12 13 14 17 18 19 20 21 22 23 25 31 32 33 35 37 42 45 46 50 53 54 55 56 57 58 60",
"output": "YES"
},
{
"input": "32 86\n5 7 9 10 13 17 18 19 25 26 28 32 33 37 38 43 45 47 50 53 57 58 60 69 73 74 75 77 80 82 83 85\n7 11 12 13 15 18 20 21 23 29 31 33 34 37 41 42 43 49 50 52 56 57 61 62 67 69 71 74 77 81 82 84",
"output": "YES"
},
{
"input": "33 44\n0 1 2 3 5 9 10 11 12 13 14 15 17 18 20 21 22 23 24 25 26 27 28 30 31 32 35 36 38 39 41 42 43\n0 2 3 4 7 8 10 11 13 14 15 16 17 18 19 21 25 26 27 28 29 30 31 33 34 36 37 38 39 40 41 42 43",
"output": "YES"
},
{
"input": "33 73\n3 6 7 8 9 10 11 13 14 15 17 19 22 23 26 27 28 31 33 34 35 37 42 44 48 52 54 57 62 63 64 67 68\n2 3 4 7 8 16 19 20 21 22 23 24 26 27 28 30 32 35 36 39 40 41 44 46 47 48 50 55 57 61 65 67 70",
"output": "YES"
},
{
"input": "34 52\n1 2 3 4 5 6 8 9 10 12 13 14 15 16 17 19 21 24 26 27 28 29 31 33 35 36 37 39 40 45 46 49 50 51\n0 1 2 3 4 6 7 8 10 11 12 13 14 15 17 19 22 24 25 26 27 29 31 33 34 35 37 38 43 44 47 48 49 51",
"output": "YES"
},
{
"input": "34 68\n0 7 9 10 11 14 15 16 20 21 22 24 26 32 34 35 37 38 40 41 42 43 44 45 47 50 53 55 57 58 59 62 64 65\n0 1 2 3 5 8 11 13 15 16 17 20 22 23 26 33 35 36 37 40 41 42 46 47 48 50 52 58 60 61 63 64 66 67",
"output": "YES"
},
{
"input": "35 90\n4 5 7 8 10 11 12 13 14 22 27 29 31 33 34 38 46 49 52 53 54 55 56 57 60 61 64 69 77 81 83 86 87 88 89\n4 7 10 11 12 13 14 15 18 19 22 27 35 39 41 44 45 46 47 52 53 55 56 58 59 60 61 62 70 75 77 79 81 82 86",
"output": "YES"
},
{
"input": "36 43\n1 2 3 4 6 7 8 9 10 11 14 16 17 18 19 20 21 22 23 24 25 26 27 29 30 31 32 33 34 35 36 37 38 39 40 42\n0 1 2 3 4 5 6 8 9 10 11 12 13 14 15 16 17 18 19 21 23 24 25 26 28 29 30 31 32 33 36 38 39 40 41 42",
"output": "YES"
},
{
"input": "36 84\n1 3 6 13 15 16 17 18 19 21 23 26 29 33 38 40 42 45 49 50 53 54 57 58 60 61 64 65 67 70 73 76 78 79 81 83\n0 2 5 8 12 17 19 21 24 28 29 32 33 36 37 39 40 43 44 46 49 52 55 57 58 60 62 64 66 69 76 78 79 80 81 82",
"output": "YES"
},
{
"input": "37 46\n0 1 3 6 7 8 9 10 12 13 14 16 17 19 20 21 22 23 24 25 26 27 28 29 30 31 33 34 35 36 37 39 40 41 42 43 44\n0 3 4 5 6 7 9 10 11 13 14 16 17 18 19 20 21 22 23 24 25 26 27 28 30 31 32 33 34 36 37 38 39 40 41 43 44",
"output": "YES"
},
{
"input": "37 97\n0 5 10 11 12 15 16 18 19 25 28 29 34 35 36 37 38 40 46 47 48 49 55 58 60 61 62 64 65 70 76 77 80 82 88 94 96\n1 7 13 15 16 21 26 27 28 31 32 34 35 41 44 45 50 51 52 53 54 56 62 63 64 65 71 74 76 77 78 80 81 86 92 93 96",
"output": "YES"
},
{
"input": "38 58\n1 2 3 4 5 8 9 11 12 13 15 16 17 22 23 24 25 26 27 29 30 31 32 33 34 36 37 40 41 43 46 47 48 52 53 55 56 57\n1 2 3 5 6 7 8 9 12 13 15 16 17 19 20 21 26 27 28 29 30 31 33 34 35 36 37 38 40 41 44 45 47 50 51 52 56 57",
"output": "YES"
},
{
"input": "38 92\n1 2 3 5 6 7 12 14 15 16 17 18 20 22 29 31 33 34 38 41 43 49 54 55 57 58 61 63 66 67 69 73 75 76 82 85 88 90\n1 3 4 10 13 16 18 21 22 23 25 26 27 32 34 35 36 37 38 40 42 49 51 53 54 58 61 63 69 74 75 77 78 81 83 86 87 89",
"output": "YES"
},
{
"input": "39 59\n0 1 2 3 5 6 7 8 9 10 11 12 13 15 16 17 19 24 25 28 29 31 32 33 35 37 38 40 41 42 43 45 46 47 49 50 53 55 56\n0 1 3 4 5 6 8 9 10 12 13 16 18 19 22 23 24 25 27 28 29 30 31 32 33 34 35 37 38 39 41 46 47 50 51 53 54 55 57",
"output": "YES"
},
{
"input": "39 67\n1 3 5 7 8 16 18 20 21 23 24 25 27 28 29 31 32 34 36 38 40 43 44 46 47 48 49 50 52 53 54 55 58 59 61 62 63 64 66\n0 1 2 4 6 8 10 12 13 21 23 25 26 28 29 30 32 33 34 36 37 39 41 43 45 48 49 51 52 53 54 55 57 58 59 60 63 64 66",
"output": "YES"
},
{
"input": "40 63\n0 2 3 4 5 6 9 10 12 15 18 19 23 25 26 27 28 29 30 31 33 34 36 37 38 39 40 43 45 49 50 52 53 54 55 57 58 60 61 62\n1 2 3 4 5 8 10 14 15 17 18 19 20 22 23 25 26 27 28 30 31 32 33 34 37 38 40 43 46 47 51 53 54 55 56 57 58 59 61 62",
"output": "YES"
},
{
"input": "40 96\n5 11 12 13 14 16 17 18 19 24 30 31 32 33 37 42 46 50 53 54 55 58 60 61 64 67 68 69 70 72 75 76 77 81 84 85 89 91 92 93\n2 7 11 15 18 19 20 23 25 26 29 32 33 34 35 37 40 41 42 46 49 50 54 56 57 58 66 72 73 74 75 77 78 79 80 85 91 92 93 94",
"output": "YES"
},
{
"input": "41 67\n0 2 3 5 8 10 11 12 13 14 15 19 20 21 22 26 29 30 31 32 34 35 37 38 40 41 44 45 46 47 49 51 52 53 54 56 57 58 59 63 66\n2 3 4 5 9 12 13 14 15 17 18 20 21 23 24 27 28 29 30 32 34 35 36 37 39 40 41 42 46 49 50 52 53 55 58 60 61 62 63 64 65",
"output": "YES"
},
{
"input": "41 72\n0 3 4 6 7 8 9 12 13 14 16 21 23 24 25 26 27 29 31 32 33 34 35 38 40 41 45 47 49 50 51 52 56 57 58 59 61 62 65 66 69\n0 1 4 5 6 8 13 15 16 17 18 19 21 23 24 25 26 27 30 32 33 37 39 41 42 43 44 48 49 50 51 53 54 57 58 61 64 67 68 70 71",
"output": "YES"
},
{
"input": "42 48\n0 1 2 3 4 7 8 9 10 11 12 13 15 16 17 18 19 20 21 22 23 24 26 27 28 29 30 32 33 34 35 36 37 38 40 41 42 43 44 45 46 47\n0 1 2 3 4 5 6 8 9 10 11 12 14 15 16 17 18 19 20 22 23 24 25 26 27 28 29 30 31 32 33 34 37 38 39 40 41 42 43 45 46 47",
"output": "YES"
},
{
"input": "42 81\n0 1 3 6 7 8 11 13 17 18 19 21 22 24 29 30 31 32 34 35 38 44 46 48 49 50 51 52 53 55 59 61 62 63 65 66 67 69 70 72 77 80\n0 1 3 4 6 11 12 13 14 16 17 20 26 28 30 31 32 33 34 35 37 41 43 44 45 47 48 49 51 52 54 59 62 63 64 66 69 70 71 74 76 80",
"output": "YES"
},
{
"input": "43 55\n0 1 2 3 4 5 6 7 8 12 14 15 17 18 19 20 21 22 23 26 27 28 29 31 32 33 35 36 37 38 40 42 43 44 45 46 47 48 49 50 51 53 54\n1 2 4 5 6 7 8 9 10 13 14 15 16 18 19 20 22 23 24 25 27 29 30 31 32 33 34 35 36 37 38 40 41 42 43 44 45 46 47 48 49 50 54",
"output": "YES"
},
{
"input": "43 81\n2 3 4 5 6 7 9 10 12 13 18 19 20 21 23 26 27 29 30 32 34 38 39 43 46 47 48 50 51 52 54 55 58 62 64 67 69 70 71 72 73 75 80\n0 3 5 6 7 8 9 11 16 19 20 21 22 23 24 26 27 29 30 35 36 37 38 40 43 44 46 47 49 51 55 56 60 63 64 65 67 68 69 71 72 75 79",
"output": "YES"
},
{
"input": "44 54\n0 1 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 21 22 23 24 25 26 27 28 29 31 33 34 35 36 37 39 40 41 43 44 47 49 50 52 53\n0 1 2 3 4 5 6 7 8 10 12 13 14 15 16 18 19 20 22 23 26 28 29 31 32 33 34 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52",
"output": "YES"
},
{
"input": "44 93\n1 5 6 7 8 10 14 17 19 21 25 26 27 30 33 34 35 36 38 41 45 48 49 51 53 55 57 60 66 67 69 70 73 76 78 79 80 81 82 83 85 87 88 90\n0 2 4 8 9 10 13 16 17 18 19 21 24 28 31 32 34 36 38 40 43 49 50 52 53 56 59 61 62 63 64 65 66 68 70 71 73 77 81 82 83 84 86 90",
"output": "YES"
},
{
"input": "45 47\n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 43 44 45 46\n0 1 2 3 4 5 6 7 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 33 34 35 36 37 38 39 40 41 42 43 44 45 46",
"output": "YES"
},
{
"input": "45 71\n0 2 3 7 8 11 12 13 14 15 16 17 20 21 22 23 24 26 28 30 32 37 39 41 42 43 44 45 47 48 50 52 54 55 56 57 58 59 60 61 62 64 66 68 70\n0 1 2 3 4 7 8 9 10 11 13 15 17 19 24 26 28 29 30 31 32 34 35 37 39 41 42 43 44 45 46 47 48 49 51 53 55 57 58 60 61 65 66 69 70",
"output": "YES"
},
{
"input": "46 46\n0 1 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\n0 1 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",
"output": "YES"
},
{
"input": "46 93\n0 1 2 6 13 16 17 18 19 21 27 29 32 34 37 38 39 40 41 44 45 49 50 52 54 56 57 61 64 65 66 67 69 71 73 75 77 78 79 83 85 87 88 90 91 92\n0 2 4 5 7 8 9 10 11 12 16 23 26 27 28 29 31 37 39 42 44 47 48 49 50 51 54 55 59 60 62 64 66 67 71 74 75 76 77 79 81 83 85 87 88 89",
"output": "YES"
},
{
"input": "47 49\n0 1 2 3 4 5 6 7 9 10 11 12 13 14 15 16 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\n0 1 2 3 4 5 6 7 8 9 10 11 13 14 15 16 17 18 19 20 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",
"output": "YES"
},
{
"input": "47 94\n0 1 3 4 5 7 8 9 14 18 19 26 30 33 34 35 37 40 42 45 46 49 50 51 52 53 55 56 60 61 62 63 64 65 66 69 71 73 75 79 84 86 87 88 90 92 93\n1 2 3 4 6 7 8 10 11 12 17 21 22 29 33 36 37 38 40 43 45 48 49 52 53 54 55 56 58 59 63 64 65 66 67 68 69 72 74 76 78 82 87 89 90 91 93",
"output": "YES"
},
{
"input": "48 65\n0 1 2 4 5 6 7 8 9 10 11 12 15 16 17 20 22 24 25 26 27 28 30 32 33 34 35 37 38 39 44 45 46 47 48 50 51 52 53 54 55 56 57 58 59 61 62 63\n0 1 4 6 8 9 10 11 12 14 16 17 18 19 21 22 23 28 29 30 31 32 34 35 36 37 38 39 40 41 42 43 45 46 47 49 50 51 53 54 55 56 57 58 59 60 61 64",
"output": "YES"
},
{
"input": "48 90\n1 3 4 5 8 9 11 13 14 15 16 18 20 21 24 26 29 30 31 33 34 36 37 38 39 40 42 43 44 46 47 48 51 52 55 58 59 61 62 63 65 66 68 78 79 81 82 89\n0 3 4 6 8 9 10 11 13 15 16 19 21 24 25 26 28 29 31 32 33 34 35 37 38 39 41 42 43 46 47 50 53 54 56 57 58 60 61 63 73 74 76 77 84 86 88 89",
"output": "YES"
},
{
"input": "49 60\n0 1 2 5 7 8 9 10 11 12 13 14 15 16 17 19 20 21 23 25 26 27 28 29 30 31 32 33 34 36 38 39 40 41 42 43 44 46 47 48 49 50 51 52 53 54 55 58 59\n0 1 2 3 4 5 6 7 8 10 11 12 14 16 17 18 19 20 21 22 23 24 25 27 29 30 31 32 33 34 35 37 38 39 40 41 42 43 44 45 46 49 50 51 52 53 56 58 59",
"output": "YES"
},
{
"input": "49 97\n0 1 2 3 6 8 11 14 19 23 26 29 32 34 35 37 39 41 43 44 45 46 51 53 63 64 65 66 67 70 71 72 73 76 77 78 79 81 83 84 86 87 90 91 92 93 94 95 96\n0 3 4 5 6 7 8 9 10 11 12 13 16 18 21 24 29 33 36 39 42 44 45 47 49 51 53 54 55 56 61 63 73 74 75 76 77 80 81 82 83 86 87 88 89 91 93 94 96",
"output": "YES"
},
{
"input": "50 58\n0 1 2 3 5 6 7 8 10 11 12 13 14 15 16 17 18 19 21 22 23 24 25 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 49 50 54 55 56 57\n0 1 3 4 5 6 7 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 31 32 36 37 38 39 40 41 42 43 45 46 47 48 50 51 52 53 54 55 56 57",
"output": "YES"
},
{
"input": "50 97\n1 2 3 4 7 9 10 11 12 13 14 21 22 23 24 25 28 29 30 31 32 33 34 36 37 40 41 45 53 56 59 64 65 69 70 71 72 73 74 77 81 84 85 86 87 89 91 92 95 96\n0 1 2 3 6 10 13 14 15 16 18 20 21 24 25 27 28 29 30 33 35 36 37 38 39 40 47 48 49 50 51 54 55 56 57 58 59 60 62 63 66 67 71 79 82 85 90 91 95 96",
"output": "YES"
},
{
"input": "40 96\n5 11 12 13 14 16 17 18 19 24 30 31 32 33 37 42 46 50 53 54 55 58 60 61 64 67 68 69 70 72 75 76 77 81 84 85 88 91 92 93\n2 7 11 15 18 19 20 23 25 26 29 32 33 34 35 37 40 41 42 46 49 50 54 56 57 58 66 72 73 74 75 77 78 79 80 85 91 92 93 94",
"output": "NO"
},
{
"input": "41 67\n0 2 3 5 8 10 11 12 13 14 15 19 20 21 22 25 29 30 31 32 34 35 37 38 40 41 44 45 46 47 49 51 52 53 54 56 57 58 59 63 66\n2 3 4 5 9 12 13 14 15 17 18 20 21 23 24 27 28 29 30 32 34 35 36 37 39 40 41 42 46 49 50 52 53 55 58 60 61 62 63 64 65",
"output": "NO"
},
{
"input": "41 72\n0 3 4 6 7 8 9 12 13 14 16 21 23 24 25 26 27 28 31 32 33 34 35 38 40 41 45 47 49 50 51 52 56 57 58 59 61 62 65 66 69\n0 1 4 5 6 8 13 15 16 17 18 19 21 23 24 25 26 27 30 32 33 37 39 41 42 43 44 48 49 50 51 53 54 57 58 61 64 67 68 70 71",
"output": "NO"
},
{
"input": "42 48\n0 1 2 3 4 7 8 9 10 11 12 13 15 16 17 18 19 20 21 22 23 24 25 27 28 29 30 32 33 34 35 36 37 38 40 41 42 43 44 45 46 47\n0 1 2 3 4 5 6 8 9 10 11 12 14 15 16 17 18 19 20 22 23 24 25 26 27 28 29 30 31 32 33 34 37 38 39 40 41 42 43 45 46 47",
"output": "NO"
},
{
"input": "42 81\n0 1 3 6 7 8 11 13 17 18 19 20 22 24 29 30 31 32 34 35 38 44 46 48 49 50 51 52 53 55 59 61 62 63 65 66 67 69 70 72 77 80\n0 1 3 4 6 11 12 13 14 16 17 20 26 28 30 31 32 33 34 35 37 41 43 44 45 47 48 49 51 52 54 59 62 63 64 66 69 70 71 74 76 80",
"output": "NO"
},
{
"input": "43 55\n0 1 2 3 4 5 6 7 8 12 14 15 17 18 19 20 21 22 23 26 27 28 29 31 32 33 34 36 37 38 40 42 43 44 45 46 47 48 49 50 51 53 54\n1 2 4 5 6 7 8 9 10 13 14 15 16 18 19 20 22 23 24 25 27 29 30 31 32 33 34 35 36 37 38 40 41 42 43 44 45 46 47 48 49 50 54",
"output": "NO"
},
{
"input": "43 81\n2 3 4 5 6 7 9 10 12 13 17 19 20 21 23 26 27 29 30 32 34 38 39 43 46 47 48 50 51 52 54 55 58 62 64 67 69 70 71 72 73 75 80\n0 3 5 6 7 8 9 11 16 19 20 21 22 23 24 26 27 29 30 35 36 37 38 40 43 44 46 47 49 51 55 56 60 63 64 65 67 68 69 71 72 75 79",
"output": "NO"
},
{
"input": "44 54\n0 1 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 21 22 23 24 25 26 27 28 29 31 33 34 35 36 37 38 40 41 43 44 47 49 50 52 53\n0 1 2 3 4 5 6 7 8 10 12 13 14 15 16 18 19 20 22 23 26 28 29 31 32 33 34 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52",
"output": "NO"
},
{
"input": "44 93\n1 5 6 7 8 10 14 17 19 21 25 26 27 30 33 34 35 36 38 41 45 48 49 51 53 55 57 60 66 67 69 70 73 76 78 79 80 81 82 83 84 87 88 90\n0 2 4 8 9 10 13 16 17 18 19 21 24 28 31 32 34 36 38 40 43 49 50 52 53 56 59 61 62 63 64 65 66 68 70 71 73 77 81 82 83 84 86 90",
"output": "NO"
},
{
"input": "45 47\n0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 44 45 46\n0 1 2 3 4 5 6 7 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 33 34 35 36 37 38 39 40 41 42 43 44 45 46",
"output": "YES"
},
{
"input": "45 71\n0 2 3 7 8 11 12 13 14 15 16 17 20 21 22 23 24 26 28 30 32 37 39 40 42 43 44 45 47 48 50 52 54 55 56 57 58 59 60 61 62 64 66 68 70\n0 1 2 3 4 7 8 9 10 11 13 15 17 19 24 26 28 29 30 31 32 34 35 37 39 41 42 43 44 45 46 47 48 49 51 53 55 57 58 60 61 65 66 69 70",
"output": "NO"
},
{
"input": "46 46\n0 1 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\n0 1 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",
"output": "YES"
},
{
"input": "46 93\n0 1 2 6 13 16 17 18 19 21 27 29 32 34 37 38 39 40 41 44 45 49 50 52 54 56 57 61 64 65 66 67 69 71 73 75 77 78 79 83 85 86 88 90 91 92\n0 2 4 5 7 8 9 10 11 12 16 23 26 27 28 29 31 37 39 42 44 47 48 49 50 51 54 55 59 60 62 64 66 67 71 74 75 76 77 79 81 83 85 87 88 89",
"output": "NO"
},
{
"input": "47 49\n0 1 2 3 4 5 6 7 9 10 11 12 13 14 15 16 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\n0 1 2 3 4 5 6 7 8 9 10 11 13 14 15 16 17 18 19 20 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",
"output": "YES"
},
{
"input": "47 94\n0 1 3 4 5 7 8 9 14 18 19 26 30 33 34 35 37 40 42 44 46 49 50 51 52 53 55 56 60 61 62 63 64 65 66 69 71 73 75 79 84 86 87 88 90 92 93\n1 2 3 4 6 7 8 10 11 12 17 21 22 29 33 36 37 38 40 43 45 48 49 52 53 54 55 56 58 59 63 64 65 66 67 68 69 72 74 76 78 82 87 89 90 91 93",
"output": "NO"
},
{
"input": "48 65\n0 1 2 4 5 6 7 8 9 10 11 12 15 16 17 20 21 24 25 26 27 28 30 32 33 34 35 37 38 39 44 45 46 47 48 50 51 52 53 54 55 56 57 58 59 61 62 63\n0 1 4 6 8 9 10 11 12 14 16 17 18 19 21 22 23 28 29 30 31 32 34 35 36 37 38 39 40 41 42 43 45 46 47 49 50 51 53 54 55 56 57 58 59 60 61 64",
"output": "NO"
},
{
"input": "48 90\n1 3 4 5 8 9 11 13 14 15 16 17 20 21 24 26 29 30 31 33 34 36 37 38 39 40 42 43 44 46 47 48 51 52 55 58 59 61 62 63 65 66 68 78 79 81 82 89\n0 3 4 6 8 9 10 11 13 15 16 19 21 24 25 26 28 29 31 32 33 34 35 37 38 39 41 42 43 46 47 50 53 54 56 57 58 60 61 63 73 74 76 77 84 86 88 89",
"output": "NO"
},
{
"input": "49 60\n0 1 2 5 7 8 9 10 11 12 13 14 15 16 17 18 20 21 23 25 26 27 28 29 30 31 32 33 34 36 38 39 40 41 42 43 44 46 47 48 49 50 51 52 53 54 55 58 59\n0 1 2 3 4 5 6 7 8 10 11 12 14 16 17 18 19 20 21 22 23 24 25 27 29 30 31 32 33 34 35 37 38 39 40 41 42 43 44 45 46 49 50 51 52 53 56 58 59",
"output": "NO"
},
{
"input": "49 97\n0 1 2 3 5 8 11 14 19 23 26 29 32 34 35 37 39 41 43 44 45 46 51 53 63 64 65 66 67 70 71 72 73 76 77 78 79 81 83 84 86 87 90 91 92 93 94 95 96\n0 3 4 5 6 7 8 9 10 11 12 13 16 18 21 24 29 33 36 39 42 44 45 47 49 51 53 54 55 56 61 63 73 74 75 76 77 80 81 82 83 86 87 88 89 91 93 94 96",
"output": "NO"
},
{
"input": "50 58\n0 1 2 3 5 6 7 8 10 11 12 13 14 15 16 17 18 19 21 22 23 24 25 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 50 54 55 56 57\n0 1 3 4 5 6 7 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 31 32 36 37 38 39 40 41 42 43 45 46 47 48 50 51 52 53 54 55 56 57",
"output": "NO"
},
{
"input": "5 10\n0 1 3 5 7\n0 1 2 4 7",
"output": "NO"
},
{
"input": "5 8\n0 2 4 6 7\n0 2 3 5 7",
"output": "NO"
}
] | 1,539,154,319
| 2,759
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 4
| 280
| 7,475,200
|
from sys import stdin, stdout
import cProfile, math
from collections import Counter
printHeap = str()
test = False
memory_constrained = False
def display(string_to_print):
stdout.write(str(string_to_print) + "\n")
def test_print(output):
if test:
stdout.write(str(output) + "\n")
def display_list(list1, sep=" "):
stdout.write(sep.join(map(str, list1)) + "\n")
def get_int():
return int(stdin.readline().strip())
def get_tuple():
return map(int, stdin.readline().split())
def get_list():
return list(map(int, stdin.readline().split()))
memory = dict()
def clear_cache():
global memory
memory = dict()
def cached_fn(fn, *args):
global memory
if args in memory:
return memory[args]
else:
result = fn(*args)
memory[args] = result
return result
# ----------------------------------------------------------------------------------- MAIN PROGRAM
def main():
n, l = get_tuple()
l1 = get_list()
l2 = get_list()
d1,d2 = list(),[]
for i,ele in enumerate(l1):
if (i+1!=len(l1)):
d1.append(l1[i+1]-ele)
d1.append(8-l1[-1]+l1[0])
for i,ele in enumerate(l2):
if (i+1!=len(l2)):
d2.append(l2[i+1]-ele)
d2.append(8-l2[-1]+l2[0])
d1.sort()
d2.sort()
if (d1==d2 or (len(l1)==1 and len(l2)==1)):
display("YES")
else: display("NO")
# --------------------------------------------------------------------------------------------- END
cProfile.run('main()') if test else main()
|
Title: Dasha and friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Running with barriers on the circle track is very popular in the country where Dasha lives, so no wonder that on her way to classes she saw the following situation:
The track is the circle with length *L*, in distinct points of which there are *n* barriers. Athlete always run the track in counterclockwise direction if you look on him from above. All barriers are located at integer distance from each other along the track.
Her friends the parrot Kefa and the leopard Sasha participated in competitions and each of them ran one lap. Each of the friends started from some integral point on the track. Both friends wrote the distance from their start along the track to each of the *n* barriers. Thus, each of them wrote *n* integers in the ascending order, each of them was between 0 and *L*<=-<=1, inclusively.
There are several tracks in the country, all of them have same length and same number of barriers, but the positions of the barriers can differ among different tracks. Now Dasha is interested if it is possible that Kefa and Sasha ran the same track or they participated on different tracks.
Write the program which will check that Kefa's and Sasha's tracks coincide (it means that one can be obtained from the other by changing the start position). Note that they always run the track in one direction — counterclockwise, if you look on a track from above.
Input Specification:
The first line contains two integers *n* and *L* (1<=≤<=*n*<=≤<=50, *n*<=≤<=*L*<=≤<=100) — the number of barriers on a track and its length.
The second line contains *n* distinct integers in the ascending order — the distance from Kefa's start to each barrier in the order of its appearance. All integers are in the range from 0 to *L*<=-<=1 inclusively.
The second line contains *n* distinct integers in the ascending order — the distance from Sasha's start to each barrier in the order of its overcoming. All integers are in the range from 0 to *L*<=-<=1 inclusively.
Output Specification:
Print "YES" (without quotes), if Kefa and Sasha ran the coinciding tracks (it means that the position of all barriers coincides, if they start running from the same points on the track). Otherwise print "NO" (without quotes).
Demo Input:
['3 8\n2 4 6\n1 5 7\n', '4 9\n2 3 5 8\n0 1 3 6\n', '2 4\n1 3\n1 2\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
The first test is analyzed in the statement.
|
```python
from sys import stdin, stdout
import cProfile, math
from collections import Counter
printHeap = str()
test = False
memory_constrained = False
def display(string_to_print):
stdout.write(str(string_to_print) + "\n")
def test_print(output):
if test:
stdout.write(str(output) + "\n")
def display_list(list1, sep=" "):
stdout.write(sep.join(map(str, list1)) + "\n")
def get_int():
return int(stdin.readline().strip())
def get_tuple():
return map(int, stdin.readline().split())
def get_list():
return list(map(int, stdin.readline().split()))
memory = dict()
def clear_cache():
global memory
memory = dict()
def cached_fn(fn, *args):
global memory
if args in memory:
return memory[args]
else:
result = fn(*args)
memory[args] = result
return result
# ----------------------------------------------------------------------------------- MAIN PROGRAM
def main():
n, l = get_tuple()
l1 = get_list()
l2 = get_list()
d1,d2 = list(),[]
for i,ele in enumerate(l1):
if (i+1!=len(l1)):
d1.append(l1[i+1]-ele)
d1.append(8-l1[-1]+l1[0])
for i,ele in enumerate(l2):
if (i+1!=len(l2)):
d2.append(l2[i+1]-ele)
d2.append(8-l2[-1]+l2[0])
d1.sort()
d2.sort()
if (d1==d2 or (len(l1)==1 and len(l2)==1)):
display("YES")
else: display("NO")
# --------------------------------------------------------------------------------------------- END
cProfile.run('main()') if test else main()
```
| 0
|
|
281
|
A
|
Word Capitalization
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] | null | null |
Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged.
|
A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.
|
Output the given word after capitalization.
|
[
"ApPLe\n",
"konjac\n"
] |
[
"ApPLe\n",
"Konjac\n"
] |
none
| 500
|
[
{
"input": "ApPLe",
"output": "ApPLe"
},
{
"input": "konjac",
"output": "Konjac"
},
{
"input": "a",
"output": "A"
},
{
"input": "A",
"output": "A"
},
{
"input": "z",
"output": "Z"
},
{
"input": "ABACABA",
"output": "ABACABA"
},
{
"input": "xYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX",
"output": "XYaPxPxHxGePfGtQySlNrLxSjDtNnTaRaEpAhPaQpWnDzMqGgRgEwJxGiBdZnMtHxFbObCaGiCeZkUqIgBhHtNvAqAlHpMnQhNeQbMyZrCdElVwHtKrPpJjIaHuIlYwHaRkAkUpPlOhNlBtXwDsKzPyHrPiUwNlXtTaPuMwTqYtJySgFoXvLiHbQwMjSvXsQfKhVlOxGdQkWjBhEyQvBjPoFkThNeRhTuIzFjInJtEfPjOlOsJpJuLgLzFnZmKvFgFrNsOnVqFcNiMfCqTpKnVyLwNqFiTySpWeTdFnWuTwDkRjVxNyQvTrOoEiExYiFaIrLoFmJfZcDkHuWjYfCeEqCvEsZiWnJaEmFbMjDvYwEeJeGcKbVbChGsIzNlExHzHiTlHcSaKxLuZxX"
},
{
"input": "rZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO",
"output": "RZhIcQlXpNcPgXrOjTiOlMoTgXgIhCfMwZfWoFzGhEkQlOoMjIuShPlZfWkNnMyQfYdUhVgQuSmYoElEtZpDyHtOxXgCpWbZqSbYnPqBcNqRtPgCnJnAyIvNsAhRbNeVlMwZyRyJnFgIsCnSbOdLvUyIeOzQvRpMoMoHfNhHwKvTcHuYnYySfPmAiNwAiWdZnWlLvGfBbRbRrCrBqIgIdWkWiBsNyYkKdNxZdGaToSsDnXpRaGrKxBpQsCzBdQgZzBkGeHgGxNrIyQlSzWsTmSnZwOcHqQpNcQvJlPvKaPiQaMaYsQjUeCqQdCjPgUbDmWiJmNiXgExLqOcCtSwSePnUxIuZfIfBeWbEiVbXnUsPwWyAiXyRbZgKwOqFfCtQuKxEmVeRlAkOeXkO"
},
{
"input": "hDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD",
"output": "HDgZlUmLhYbLkLcNcKeOwJwTePbOvLaRvNzQbSbLsPeHqLhUqWtUbNdQfQqFfXeJqJwWuOrFnDdZiPxIkDyVmHbHvXfIlFqSgAcSyWbOlSlRuPhWdEpEzEeLnXwCtWuVcHaUeRgCiYsIvOaIgDnFuDbRnMoCmPrZfLeFpSjQaTfHgZwZvAzDuSeNwSoWuJvLqKqAuUxFaCxFfRcEjEsJpOfCtDiVrBqNsNwPuGoRgPzRpLpYnNyQxKaNnDnYiJrCrVcHlOxPiPcDbEgKfLwBjLhKcNeMgJhJmOiJvPfOaPaEuGqWvRbErKrIpDkEoQnKwJnTlStLyNsHyOjZfKoIjXwUvRrWpSyYhRpQdLqGmErAiNcGqAqIrTeTiMuPmCrEkHdBrLyCxPtYpRqD"
},
{
"input": "qUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW",
"output": "QUdLgGrJeGmIzIeZrCjUtBpYfRvNdXdRpGsThIsEmJjTiMqEwRxBeBaSxEuWrNvExKePjPnXhPzBpWnHiDhTvZhBuIjDnZpTcEkCvRkAcTmMuXhGgErWgFyGyToOyVwYlCuQpTfJkVdWmFyBqQhJjYtXrBbFdHzDlGsFbHmHbFgXgFhIyDhZyEqEiEwNxSeByBwLiVeSnCxIdHbGjOjJrZeVkOzGeMmQrJkVyGhDtCzOlPeAzGrBlWwEnAdUfVaIjNrRyJjCnHkUvFuKuKeKbLzSbEmUcXtVkZzXzKlOrPgQiDmCcCvIyAdBwOeUuLbRmScNcWxIkOkJuIsBxTrIqXhDzLcYdVtPgZdZfAxTmUtByGiTsJkSySjXdJvEwNmSmNoWsChPdAzJrBoW"
},
{
"input": "kHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL",
"output": "KHbApGoBcLmIwUlXkVgUmWzYeLoDbGaOkWbIuXoRwMfKuOoMzAoXrBoTvYxGrMbRjDuRxAbGsTnErIiHnHoLeRnTbFiRfDdOkNlWiAcOsChLdLqFqXlDpDoDtPxXqAmSvYgPvOcCpOlWtOjYwFkGkHuCaHwZcFdOfHjBmIxTeSiHkWjXyFcCtOlSuJsZkDxUgPeZkJwMmNpErUlBcGuMlJwKkWnOzFeFiSiPsEvMmQiCsYeHlLuHoMgBjFoZkXlObDkSoQcVyReTmRsFzRhTuIvCeBqVsQdQyTyZjStGrTyDcEcAgTgMiIcVkLbZbGvWeHtXwEqWkXfTcPyHhHjYwIeVxLyVmHmMkUsGiHmNnQuMsXaFyPpVqNrBhOiWmNkBbQuHvQdOjPjKiZcL"
},
{
"input": "aHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC",
"output": "AHmRbLgNuWkLxLnWvUbYwTeZeYiOlLhTuOvKfLnVmCiPcMkSgVrYjZiLuRjCiXhAnVzVcTlVeJdBvPdDfFvHkTuIhCdBjEsXbVmGcLrPfNvRdFsZkSdNpYsJeIhIcNqSoLkOjUlYlDmXsOxPbQtIoUxFjGnRtBhFaJvBeEzHsAtVoQbAfYjJqReBiKeUwRqYrUjPjBoHkOkPzDwEwUgTxQxAvKzUpMhKyOhPmEhYhItQwPeKsKaKlUhGuMcTtSwFtXfJsDsFlTtOjVvVfGtBtFlQyIcBaMsPaJlPqUcUvLmReZiFbXxVtRhTzJkLkAjVqTyVuFeKlTyQgUzMsXjOxQnVfTaWmThEnEoIhZeZdStBkKeLpAhJnFoJvQyGwDiStLjEwGfZwBuWsEfC"
},
{
"input": "sLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN",
"output": "SLlZkDiDmEdNaXuUuJwHqYvRtOdGfTiTpEpAoSqAbJaChOiCvHgSwZwEuPkMmXiLcKdXqSsEyViEbZpZsHeZpTuXoGcRmOiQfBfApPjDqSqElWeSeOhUyWjLyNoRuYeGfGwNqUsQoTyVvWeNgNdZfDxGwGfLsDjIdInSqDlMuNvFaHbScZkTlVwNcJpEjMaPaOtFgJjBjOcLlLmDnQrShIrJhOcUmPnZhTxNeClQsZaEaVaReLyQpLwEqJpUwYhLiRzCzKfOoFeTiXzPiNbOsZaZaLgCiNnMkBcFwGgAwPeNyTxJcCtBgXcToKlWaWcBaIvBpNxPeClQlWeQqRyEtAkJdBtSrFdDvAbUlKyLdCuTtXxFvRcKnYnWzVdYqDeCmOqPxUaFjQdTdCtN"
},
{
"input": "iRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE",
"output": "IRuStKvVhJdJbQwRoIuLiVdTpKaOqKfYlYwAzIpPtUwUtMeKyCaOlXmVrKwWeImYmVuXdLkRlHwFxKqZbZtTzNgOzDbGqTfZnKmUzAcIjDcEmQgYyFbEfWzRpKvCkDmAqDiIiRcLvMxWaJqCgYqXgIcLdNaZlBnXtJyKaMnEaWfXfXwTbDnAiYnWqKbAtDpYdUbZrCzWgRnHzYxFgCdDbOkAgTqBuLqMeStHcDxGnVhSgMzVeTaZoTfLjMxQfRuPcFqVlRyYdHyOdJsDoCeWrUuJyIiAqHwHyVpEeEoMaJwAoUfPtBeJqGhMaHiBjKwAlXoZpUsDhHgMxBkVbLcEvNtJbGnPsUwAvXrAkTlXwYvEnOpNeWyIkRnEnTrIyAcLkRgMyYcKrGiDaAyE"
},
{
"input": "cRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP",
"output": "CRtJkOxHzUbJcDdHzJtLbVmSoWuHoTkVrPqQaVmXeBrHxJbQfNrQbAaMrEhVdQnPxNyCjErKxPoEdWkVrBbDeNmEgBxYiBtWdAfHiLuSwIxJuHpSkAxPoYdNkGoLySsNhUmGoZhDzAfWhJdPlJzQkZbOnMtTkClIoCqOlIcJcMlGjUyOiEmHdYfIcPtTgQhLlLcPqQjAnQnUzHpCaQsCnYgQsBcJrQwBnWsIwFfSfGuYgTzQmShFpKqEeRlRkVfMuZbUsDoFoPrNuNwTtJqFkRiXxPvKyElDzLoUnIwAaBaOiNxMpEvPzSpGpFhMtGhGdJrFnZmNiMcUfMtBnDuUnXqDcMsNyGoLwLeNnLfRsIwRfBtXkHrFcPsLdXaAoYaDzYnZuQeVcZrElWmP"
},
{
"input": "wVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG",
"output": "WVaCsGxZrBbFnTbKsCoYlAvUkIpBaYpYmJkMlPwCaFvUkDxAiJgIqWsFqZlFvTtAnGzEwXbYiBdFfFxRiDoUkLmRfAwOlKeOlKgXdUnVqLkTuXtNdQpBpXtLvZxWoBeNePyHcWmZyRiUkPlRqYiQdGeXwOhHbCqVjDcEvJmBkRwWnMqPjXpUsIyXqGjHsEsDwZiFpIbTkQaUlUeFxMwJzSaHdHnDhLaLdTuYgFuJsEcMmDvXyPjKsSeBaRwNtPuOuBtNeOhQdVgKzPzOdYtPjPfDzQzHoWcYjFbSvRgGdGsCmGnQsErToBkCwGeQaCbBpYkLhHxTbUvRnJpZtXjKrHdRiUmUbSlJyGaLnWsCrJbBnSjFaZrIzIrThCmGhQcMsTtOxCuUcRaEyPaG"
},
{
"input": "kEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV",
"output": "KEiLxLmPjGzNoGkJdBlAfXhThYhMsHmZoZbGyCvNiUoLoZdAxUbGyQiEfXvPzZzJrPbEcMpHsMjIkRrVvDvQtHuKmXvGpQtXbPzJpFjJdUgWcPdFxLjLtXgVpEiFhImHnKkGiWnZbJqRjCyEwHsNbYfYfTyBaEuKlCtWnOqHmIgGrFmQiYrBnLiFcGuZxXlMfEuVoCxPkVrQvZoIpEhKsYtXrPxLcSfQqXsWaDgVlOnAzUvAhOhMrJfGtWcOwQfRjPmGhDyAeXrNqBvEiDfCiIvWxPjTwPlXpVsMjVjUnCkXgBuWnZaDyJpWkCfBrWnHxMhJgItHdRqNrQaEeRjAuUwRkUdRhEeGlSqVqGmOjNcUhFfXjCmWzBrGvIuZpRyWkWiLyUwFpYjNmNfV"
},
{
"input": "eIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI",
"output": "EIhDoLmDeReKqXsHcVgFxUqNfScAiQnFrTlCgSuTtXiYvBxKaPaGvUeYfSgHqEaWcHxKpFaSlCxGqAmNeFcIzFcZsBiVoZhUjXaDaIcKoBzYdIlEnKfScRqSkYpPtVsVhXsBwUsUfAqRoCkBxWbHgDiCkRtPvUwVgDjOzObYwNiQwXlGnAqEkHdSqLgUkOdZiWaHqQnOhUnDhIzCiQtVcJlGoRfLuVlFjWqSuMsLgLwOdZvKtWdRuRqDoBoInKqPbJdXpIqLtFlMlDaWgSiKbFpCxOnQeNeQzXeKsBzIjCyPxCmBnYuHzQoYxZgGzSgGtZiTeQmUeWlNzZeKiJbQmEjIiDhPeSyZlNdHpZnIkPdJzSeJpPiXxToKyBjJfPwNzZpWzIzGySqPxLtI"
},
{
"input": "uOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX",
"output": "UOoQzIeTwYeKpJtGoUdNiXbPgEwVsZkAnJcArHxIpEnEhZwQhZvAiOuLeMkVqLeDsAyKeYgFxGmRoLaRsZjAeXgNfYhBkHeDrHdPuTuYhKmDlAvYzYxCdYgYfVaYlGeVqTeSfBxQePbQrKsTaIkGzMjFrQlJuYaMxWpQkLdEcDsIiMnHnDtThRvAcKyGwBsHqKdXpJfIeTeZtYjFbMeUoXoXzGrShTwSwBpQlKeDrZdCjRqNtXoTsIzBkWbMsObTtDvYaPhUeLeHqHeMpZmTaCcIqXzAmGnPfNdDaFhOqWqDrWuFiBpRjZrQmAdViOuMbFfRyXyWfHgRkGpPnDrEqQcEmHcKpEvWlBrOtJbUaXbThJaSxCbVoGvTmHvZrHvXpCvLaYbRiHzYuQyX"
},
{
"input": "lZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ",
"output": "LZqBqKeGvNdSeYuWxRiVnFtYbKuJwQtUcKnVtQhAlOeUzMaAuTaEnDdPfDcNyHgEoBmYjZyFePeJrRiKyAzFnBfAuGiUyLrIeLrNhBeBdVcEeKgCcBrQzDsPwGcNnZvTsEaYmFfMeOmMdNuZbUtDoQoNcGwDqEkEjIdQaPwAxJbXeNxOgKgXoEbZiIsVkRrNpNyAkLeHkNfEpLuQvEcMbIoGaDzXbEtNsLgGfOkZaFiUsOvEjVeCaMcZqMzKeAdXxJsVeCrZaFpJtZxInQxFaSmGgSsVyGeLlFgFqTpIbAvPkIfJrVcJeBxSdEvPyVwIjHpYrLrKqLnAmCuGmPoZrSbOtGaLaTmBmSuUyAmAsRiMqOtRjJhPhAfXaJnTpLbFqPmJgFcBxImTqIiJ"
},
{
"input": "P",
"output": "P"
},
{
"input": "Xyzzy",
"output": "Xyzzy"
},
{
"input": "Zzz",
"output": "Zzz"
},
{
"input": "Zp",
"output": "Zp"
}
] | 1,698,048,352
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 92
| 0
|
s=input()
a=s[0]
c=a.upper()+s[1:]
print(c)
|
Title: Word Capitalization
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Capitalization is writing a word with its first letter as a capital letter. Your task is to capitalize the given word.
Note, that during capitalization all the letters except the first one remains unchanged.
Input Specification:
A single line contains a non-empty word. This word consists of lowercase and uppercase English letters. The length of the word will not exceed 103.
Output Specification:
Output the given word after capitalization.
Demo Input:
['ApPLe\n', 'konjac\n']
Demo Output:
['ApPLe\n', 'Konjac\n']
Note:
none
|
```python
s=input()
a=s[0]
c=a.upper()+s[1:]
print(c)
```
| 3
|
|
559
|
B
|
Equivalent Strings
|
PROGRAMMING
| 1,700
|
[
"divide and conquer",
"hashing",
"sortings",
"strings"
] | 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".
| 1,000
|
[
{
"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"
}
] | 1,548,678,658
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 93
| 0
|
def f(s):
if(len(s) % 2 != 0):
return s
sorted([f(s[:len(s)//2]), f(s[len(s)//2 :])])
return s
if(f(input()) == f(input())):
print("YES")
else:
print("NO")
|
Title: Equivalent Strings
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
def f(s):
if(len(s) % 2 != 0):
return s
sorted([f(s[:len(s)//2]), f(s[len(s)//2 :])])
return s
if(f(input()) == f(input())):
print("YES")
else:
print("NO")
```
| 0
|
|
230
|
B
|
T-primes
|
PROGRAMMING
| 1,300
|
[
"binary search",
"implementation",
"math",
"number theory"
] | null | null |
We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors.
You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not.
|
The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.
|
Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't.
|
[
"3\n4 5 6\n"
] |
[
"YES\nNO\nNO\n"
] |
The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".
| 500
|
[
{
"input": "3\n4 5 6",
"output": "YES\nNO\nNO"
},
{
"input": "2\n48 49",
"output": "NO\nYES"
},
{
"input": "10\n10 9 8 7 6 5 4 3 2 1",
"output": "NO\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nNO\nNO"
},
{
"input": "1\n36",
"output": "NO"
},
{
"input": "1\n999966000289",
"output": "YES"
},
{
"input": "1\n999993399999",
"output": "NO"
},
{
"input": "9\n111 121 131 111 121 131 111 121 131",
"output": "NO\nYES\nNO\nNO\nYES\nNO\nNO\nYES\nNO"
},
{
"input": "1\n1",
"output": "NO"
},
{
"input": "1\n10",
"output": "NO"
},
{
"input": "1\n976197352729",
"output": "NO"
},
{
"input": "1\n1000000000000",
"output": "NO"
},
{
"input": "1\n9",
"output": "YES"
},
{
"input": "6\n549755813888 847288609443 762939453125 678223072849 285311670611 137858491849",
"output": "NO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "3\n223092870 6469693230 200560490130",
"output": "NO\nNO\nNO"
},
{
"input": "2\n81 25",
"output": "NO\nYES"
},
{
"input": "1\n16",
"output": "NO"
},
{
"input": "22\n1 2 3 4 5 6 7 8 9 10 12752041 64 121 144 27550356289 124 24657 23756 135153365 25235235235 42351351 81",
"output": "NO\nNO\nNO\nYES\nNO\nNO\nNO\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nYES\nNO\nNO\nNO\nNO\nNO\nNO\nNO"
},
{
"input": "1\n225",
"output": "NO"
},
{
"input": "1\n1521",
"output": "NO"
},
{
"input": "1\n81",
"output": "NO"
}
] | 1,696,943,101
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 15
| 2,000
| 14,336,000
|
import math
def ifprime():
prime_list=[True]*1000000
prime_list[0],prime_list[1]=False,False
for i in range(2,1000000):
if prime_list[i]:
for j in range(i*i,1000000,i):
prime_list[j]=False
return prime_list
n = int(input())
l = list(map(int, input().split()))
for _ in l:
if _ == 4:
print('YES')
elif _ %2 == 0 or _<4:
print('NO')
else:
a = math.isqrt(_)
if a ** 2 != _:
print('NO')
else:
print('YES' if ifprime()[a] else 'NO')
|
Title: T-primes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
We know that prime numbers are positive integers that have exactly two distinct positive divisors. Similarly, we'll call a positive integer *t* Т-prime, if *t* has exactly three distinct positive divisors.
You are given an array of *n* positive integers. For each of them determine whether it is Т-prime or not.
Input Specification:
The first line contains a single positive integer, *n* (1<=≤<=*n*<=≤<=105), showing how many numbers are in the array. The next line contains *n* space-separated integers *x**i* (1<=≤<=*x**i*<=≤<=1012).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is advised to use the cin, cout streams or the %I64d specifier.
Output Specification:
Print *n* lines: the *i*-th line should contain "YES" (without the quotes), if number *x**i* is Т-prime, and "NO" (without the quotes), if it isn't.
Demo Input:
['3\n4 5 6\n']
Demo Output:
['YES\nNO\nNO\n']
Note:
The given test has three numbers. The first number 4 has exactly three divisors — 1, 2 and 4, thus the answer for this number is "YES". The second number 5 has two divisors (1 and 5), and the third number 6 has four divisors (1, 2, 3, 6), hence the answer for them is "NO".
|
```python
import math
def ifprime():
prime_list=[True]*1000000
prime_list[0],prime_list[1]=False,False
for i in range(2,1000000):
if prime_list[i]:
for j in range(i*i,1000000,i):
prime_list[j]=False
return prime_list
n = int(input())
l = list(map(int, input().split()))
for _ in l:
if _ == 4:
print('YES')
elif _ %2 == 0 or _<4:
print('NO')
else:
a = math.isqrt(_)
if a ** 2 != _:
print('NO')
else:
print('YES' if ifprime()[a] else 'NO')
```
| 0
|
|
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"
}
] | 1,604,569,424
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 248
| 0
|
#!/usr/bin/env python
# coding: utf-8
# In[75]:
masukan1 = input()
masukan2 = input()
balik = masukan1[::-1]
if masukan2 == balik:
print ('YES')
else:
print ('NO')
# In[ ]:
|
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
#!/usr/bin/env python
# coding: utf-8
# In[75]:
masukan1 = input()
masukan2 = input()
balik = masukan1[::-1]
if masukan2 == balik:
print ('YES')
else:
print ('NO')
# In[ ]:
```
| 3.938
|
424
|
B
|
Megacity
|
PROGRAMMING
| 1,200
|
[
"binary search",
"greedy",
"implementation",
"sortings"
] | null | null |
The administration of the Tomsk Region firmly believes that it's time to become a megacity (that is, get population of one million). Instead of improving the demographic situation, they decided to achieve its goal by expanding the boundaries of the city.
The city of Tomsk can be represented as point on the plane with coordinates (0; 0). The city is surrounded with *n* other locations, the *i*-th one has coordinates (*x**i*, *y**i*) with the population of *k**i* people. You can widen the city boundaries to a circle of radius *r*. In such case all locations inside the circle and on its border are included into the city.
Your goal is to write a program that will determine the minimum radius *r*, to which is necessary to expand the boundaries of Tomsk, so that it becomes a megacity.
|
The first line of the input contains two integers *n* and *s* (1<=≤<=*n*<=≤<=103; 1<=≤<=*s*<=<<=106) — the number of locatons around Tomsk city and the population of the city. Then *n* lines follow. The *i*-th line contains three integers — the *x**i* and *y**i* coordinate values of the *i*-th location and the number *k**i* of people in it (1<=≤<=*k**i*<=<<=106). Each coordinate is an integer and doesn't exceed 104 in its absolute value.
It is guaranteed that no two locations are at the same point and no location is at point (0; 0).
|
In the output, print "-1" (without the quotes), if Tomsk won't be able to become a megacity. Otherwise, in the first line print a single real number — the minimum radius of the circle that the city needs to expand to in order to become a megacity.
The answer is considered correct if the absolute or relative error don't exceed 10<=-<=6.
|
[
"4 999998\n1 1 1\n2 2 1\n3 3 1\n2 -2 1\n",
"4 999998\n1 1 2\n2 2 1\n3 3 1\n2 -2 1\n",
"2 1\n1 1 999997\n2 2 1\n"
] |
[
"2.8284271\n",
"1.4142136\n",
"-1"
] |
none
| 1,000
|
[
{
"input": "4 999998\n1 1 1\n2 2 1\n3 3 1\n2 -2 1",
"output": "2.8284271"
},
{
"input": "4 999998\n1 1 2\n2 2 1\n3 3 1\n2 -2 1",
"output": "1.4142136"
},
{
"input": "2 1\n1 1 999997\n2 2 1",
"output": "-1"
},
{
"input": "4 999998\n3 3 10\n-3 3 10\n3 -3 10\n-3 -3 10",
"output": "4.2426407"
},
{
"input": "15 95473\n-9 6 199715\n0 -8 110607\n0 2 6621\n-3 -2 59894\n-10 -8 175440\n-2 0 25814\n10 -4 68131\n7 1 9971\n6 7 821\n6 5 20208\n6 2 68468\n0 7 37427\n1 -3 13337\n-10 7 113041\n-6 -2 44028",
"output": "12.8062485"
},
{
"input": "20 93350\n13 -28 486\n26 -26 48487\n5 -23 143368\n-23 -25 10371\n-2 -7 75193\n0 -8 3\n-6 -11 5015\n-19 -18 315278\n28 -15 45801\n21 8 4590\n-4 -28 12926\n-16 17 9405\n-28 -23 222092\n1 -10 1857\n14 -28 35170\n-4 -22 22036\n-2 -10 1260\n-1 12 375745\n-19 -24 38845\n10 -25 9256",
"output": "26.1725047"
},
{
"input": "30 505231\n-18 16 88130\n-10 16 15693\n16 -32 660\n-27 17 19042\n30 -37 6680\n36 19 299674\n-45 21 3300\n11 27 76\n-49 -34 28649\n-1 11 31401\n25 42 20858\n-40 6 455660\n-29 43 105001\n-38 10 6042\n19 -45 65551\n20 -9 148533\n-5 -24 393442\n-43 2 8577\n-39 18 97059\n12 28 39189\n35 23 28178\n40 -34 51687\n23 41 219028\n21 -44 927\n47 8 13206\n33 41 97342\n10 18 24895\n0 12 288\n0 -44 1065\n-25 43 44231",
"output": "24.5153013"
},
{
"input": "2 500000\n936 1000 500000\n961 976 500000",
"output": "1369.7065379"
},
{
"input": "10 764008\n959 32 23049\n-513 797 38979\n-603 -838 24916\n598 -430 25414\n-280 -624 18714\n330 891 21296\n-347 -68 27466\n650 -842 30125\n-314 889 35394\n275 969 5711",
"output": "1063.7029661"
},
{
"input": "30 295830\n1 -4 24773\n4 3 26175\n-2 -3 14789\n2 -1 46618\n-2 -2 52997\n-3 0 517\n-2 0 18173\n-4 -3 54465\n2 4 63579\n4 -4 41821\n2 2 11018\n0 4 42856\n0 -1 51885\n-3 4 57137\n3 0 4688\n0 2 60137\n-4 4 33484\n-1 3 66196\n3 -1 53634\n0 -2 41630\n-2 1 54606\n2 -2 2978\n2 -3 23733\n1 -2 35248\n-3 -3 15124\n-2 -4 26518\n4 0 28151\n4 -1 18348\n3 3 16914\n-4 2 26013",
"output": "4.4721360"
},
{
"input": "10 511500\n-5129 -3858 76954\n1296 1130 36126\n1219 6732 102003\n-8026 -178 4150\n-3261 1342 105429\n7965 -3013 62561\n5607 8963 53539\n-9044 -3999 16509\n1406 4103 115667\n-3716 2522 110626",
"output": "6841.4753526"
},
{
"input": "20 39342\n2 0 36476\n-3 1 136925\n1 3 31234\n0 -3 23785\n-1 3 77700\n-3 -1 50490\n-1 -3 13965\n-3 2 121093\n3 0 118933\n-3 0 125552\n-3 3 54779\n-2 0 96250\n1 2 142643\n2 2 23848\n0 2 29845\n0 -2 80462\n-1 1 91852\n-1 2 26526\n0 -1 136272\n1 1 108999",
"output": "3.0000000"
},
{
"input": "2 1\n1 0 1\n0 1 999999",
"output": "1.0000000"
},
{
"input": "2 999997\n1 1 1\n1 2 1",
"output": "-1"
}
] | 1,642,670,567
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 54
| 124
| 2,560,000
|
import math
def main_function():
n, s = [int(i) for i in input().split(" ")]
r_s = []
current_r = 0
for i in range(n):
x, y, k = [int(u) for u in input().split(" ")]
r_s.append([math.sqrt(x ** 2 + y ** 2), k])
r_s.sort(key=lambda x:x[0])
for i in r_s:
r, k = i
if s >= 1000000:
print(current_r)
break
current_r = r
s += k
if s >= 1000000:
print(current_r)
break
if s < 1000000:
print(-1)
main_function()
#main_function()
|
Title: Megacity
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The administration of the Tomsk Region firmly believes that it's time to become a megacity (that is, get population of one million). Instead of improving the demographic situation, they decided to achieve its goal by expanding the boundaries of the city.
The city of Tomsk can be represented as point on the plane with coordinates (0; 0). The city is surrounded with *n* other locations, the *i*-th one has coordinates (*x**i*, *y**i*) with the population of *k**i* people. You can widen the city boundaries to a circle of radius *r*. In such case all locations inside the circle and on its border are included into the city.
Your goal is to write a program that will determine the minimum radius *r*, to which is necessary to expand the boundaries of Tomsk, so that it becomes a megacity.
Input Specification:
The first line of the input contains two integers *n* and *s* (1<=≤<=*n*<=≤<=103; 1<=≤<=*s*<=<<=106) — the number of locatons around Tomsk city and the population of the city. Then *n* lines follow. The *i*-th line contains three integers — the *x**i* and *y**i* coordinate values of the *i*-th location and the number *k**i* of people in it (1<=≤<=*k**i*<=<<=106). Each coordinate is an integer and doesn't exceed 104 in its absolute value.
It is guaranteed that no two locations are at the same point and no location is at point (0; 0).
Output Specification:
In the output, print "-1" (without the quotes), if Tomsk won't be able to become a megacity. Otherwise, in the first line print a single real number — the minimum radius of the circle that the city needs to expand to in order to become a megacity.
The answer is considered correct if the absolute or relative error don't exceed 10<=-<=6.
Demo Input:
['4 999998\n1 1 1\n2 2 1\n3 3 1\n2 -2 1\n', '4 999998\n1 1 2\n2 2 1\n3 3 1\n2 -2 1\n', '2 1\n1 1 999997\n2 2 1\n']
Demo Output:
['2.8284271\n', '1.4142136\n', '-1']
Note:
none
|
```python
import math
def main_function():
n, s = [int(i) for i in input().split(" ")]
r_s = []
current_r = 0
for i in range(n):
x, y, k = [int(u) for u in input().split(" ")]
r_s.append([math.sqrt(x ** 2 + y ** 2), k])
r_s.sort(key=lambda x:x[0])
for i in r_s:
r, k = i
if s >= 1000000:
print(current_r)
break
current_r = r
s += k
if s >= 1000000:
print(current_r)
break
if s < 1000000:
print(-1)
main_function()
#main_function()
```
| 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"
}
] | 1,541,855,517
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
n = int(input())
a = list(map(int, input().split()))
2n = 0
2n1 = 0
cnt = 0
cnt2 = 0
for i in a:
if (i%2)==0:
cnt2+=1
2n = i
else:
cnt+=1
2n1 = i
if cnt2>1:
print(2n)
else:
print(2n1)
|
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()))
2n = 0
2n1 = 0
cnt = 0
cnt2 = 0
for i in a:
if (i%2)==0:
cnt2+=1
2n = i
else:
cnt+=1
2n1 = i
if cnt2>1:
print(2n)
else:
print(2n1)
```
| -1
|
14
|
B
|
Young Photographer
|
PROGRAMMING
| 1,000
|
[
"implementation"
] |
B. Young Photographer
|
2
|
64
|
Among other things, Bob is keen on photography. Especially he likes to take pictures of sportsmen. That was the reason why he placed himself in position *x*0 of a long straight racetrack and got ready to take pictures. But the problem was that not all the runners passed him. The total amount of sportsmen, training at that racetrack, equals *n*. And each of them regularly runs distances within a particular segment of the racetrack, which is the same for each sportsman. For example, the first sportsman runs from position *a*1 to position *b*1, the second — from *a*2 to *b*2
What is the minimum distance that Bob should move to have a chance to take pictures of each sportsman? Bob can take a picture of a sportsman, if he stands within the segment that this sportsman covers on the racetrack.
|
The first line of the input file contains integers *n* and *x*0 (1<=≤<=*n*<=≤<=100; 0<=≤<=*x*0<=≤<=1000). The following *n* lines contain pairs of integers *a**i*,<=*b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000; *a**i*<=≠<=*b**i*).
|
Output the required minimum distance in the same units as the positions on the racetrack. If there is no such a position, output -1.
|
[
"3 3\n0 7\n14 2\n4 6\n"
] |
[
"1\n"
] |
none
| 0
|
[
{
"input": "3 3\n0 7\n14 2\n4 6",
"output": "1"
},
{
"input": "1 1\n0 10",
"output": "0"
},
{
"input": "2 2\n1 2\n3 2",
"output": "0"
},
{
"input": "3 2\n1 2\n2 3\n3 4",
"output": "-1"
},
{
"input": "2 4\n10 4\n1 5",
"output": "0"
},
{
"input": "1 10\n1 9",
"output": "1"
},
{
"input": "1 10\n123 12",
"output": "2"
},
{
"input": "1 17\n10 17",
"output": "0"
},
{
"input": "1 22\n22 33",
"output": "0"
},
{
"input": "1 3\n1 2",
"output": "1"
},
{
"input": "2 5\n0 3\n2 1",
"output": "3"
},
{
"input": "3 3\n7 3\n6 4\n3 7",
"output": "1"
},
{
"input": "4 9\n8 6\n11 5\n5 11\n8 3",
"output": "1"
},
{
"input": "2 4\n1 4\n4 0",
"output": "0"
},
{
"input": "3 7\n5 8\n7 5\n4 7",
"output": "0"
},
{
"input": "4 7\n8 2\n5 7\n8 2\n5 8",
"output": "0"
},
{
"input": "2 3\n4 1\n4 1",
"output": "0"
},
{
"input": "3 8\n7 2\n3 7\n5 2",
"output": "3"
},
{
"input": "4 0\n9 1\n8 1\n8 4\n4 5",
"output": "4"
},
{
"input": "4 7\n2 5\n3 6\n3 5\n7 4",
"output": "2"
},
{
"input": "10 16\n4 18\n6 19\n22 1\n23 0\n1 22\n9 22\n4 19\n0 14\n6 14\n0 16",
"output": "2"
},
{
"input": "20 1\n35 8\n40 6\n49 5\n48 18\n46 16\n45 16\n44 10\n16 44\n8 46\n2 45\n38 3\n42 1\n13 35\n35 18\n12 33\n32 11\n31 3\n50 20\n47 6\n38 2",
"output": "19"
},
{
"input": "30 43\n17 72\n75 26\n23 69\n83 30\n15 82\n4 67\n83 27\n33 62\n26 83\n70 26\n69 25\n16 67\n77 26\n66 33\n7 88\n70 9\n10 79\n76 9\n30 77\n77 28\n21 68\n81 14\n13 72\n88 15\n60 29\n87 28\n16 58\n6 58\n71 9\n83 18",
"output": "0"
},
{
"input": "40 69\n29 109\n28 87\n52 106\n101 34\n32 92\n91 60\n90 47\n62 102\n33 72\n27 87\n45 78\n103 37\n94 33\n56 98\n38 79\n31 83\n105 53\n47 89\n50 83\n93 62\n96 49\n47 75\n89 47\n89 61\n93 54\n46 100\n110 41\n103 28\n101 57\n100 62\n71 37\n65 80\n86 28\n73 42\n96 44\n33 111\n98 39\n87 55\n108 65\n31 101",
"output": "0"
},
{
"input": "50 77\n95 55\n113 33\n101 17\n109 56\n117 7\n77 12\n14 84\n57 101\n96 28\n108 22\n105 12\n17 114\n51 115\n18 112\n104 25\n50 115\n14 111\n55 113\n124 20\n101 37\n18 121\n41 90\n77 41\n117 16\n8 83\n92 45\n48 86\n16 84\n13 98\n40 107\n14 94\n23 111\n36 121\n50 100\n35 90\n103 37\n96 51\n109 15\n13 117\n117 42\n112 45\n88 36\n51 121\n127 49\n112 15\n9 95\n122 46\n126 40\n57 93\n56 88",
"output": "0"
},
{
"input": "5 12\n2 7\n7 5\n3 10\n11 3\n2 11",
"output": "5"
},
{
"input": "15 15\n12 37\n40 4\n38 8\n5 36\n11 31\n21 33\n9 37\n4 38\n8 33\n5 39\n7 39\n38 16\n16 41\n38 9\n5 32",
"output": "6"
},
{
"input": "25 40\n66 26\n56 19\n64 38\n64 23\n25 49\n51 26\n67 20\n65 35\n33 66\n28 63\n27 57\n40 56\n59 26\n35 56\n39 67\n30 63\n69 22\n21 63\n67 22\n20 66\n26 65\n64 26\n44 57\n57 41\n35 50",
"output": "4"
},
{
"input": "50 77\n24 119\n43 119\n102 22\n117 30\n127 54\n93 19\n120 9\n118 27\n98 16\n17 105\n22 127\n109 52\n115 40\n11 121\n12 120\n113 30\n13 108\n33 124\n31 116\n112 39\n37 108\n127 28\n127 39\n120 29\n19 114\n103 18\n106 16\n24 121\n93 10\n36 112\n104 40\n39 100\n36 97\n83 9\n14 114\n126 12\n85 47\n25 84\n105 29\n35 113\n102 19\n8 110\n111 28\n94 12\n11 115\n40 124\n39 85\n47 93\n94 31\n17 121",
"output": "0"
},
{
"input": "1 21\n973 373",
"output": "352"
},
{
"input": "2 212\n831 551\n810 753",
"output": "541"
},
{
"input": "3 404\n690 728\n820 260\n186 402",
"output": "-1"
},
{
"input": "4 906\n548 906\n830 457\n228 638\n464 167",
"output": "-1"
},
{
"input": "5 97\n97 393\n840 965\n269 183\n596 49\n975 62",
"output": "-1"
},
{
"input": "3 183\n416 335\n773 648\n434 198",
"output": "-1"
},
{
"input": "3 868\n251 927\n862 464\n157 756",
"output": "112"
},
{
"input": "3 242\n397 208\n951 279\n570 622",
"output": "-1"
},
{
"input": "3 618\n543 800\n38 94\n293 179",
"output": "-1"
},
{
"input": "3 993\n378 81\n127 911\n16 737",
"output": "615"
},
{
"input": "5 12\n11 1\n9 6\n1 11\n3 8\n874 842",
"output": "-1"
},
{
"input": "15 16\n11 40\n5 32\n5 31\n36 10\n34 9\n43 6\n28 6\n34 8\n43 15\n9 28\n14 34\n34 6\n7 31\n31 14\n68 478",
"output": "-1"
},
{
"input": "25 57\n47 31\n64 21\n43 56\n47 19\n70 27\n28 61\n41 61\n39 45\n46 21\n55 35\n70 22\n22 69\n30 67\n55 42\n37 58\n50 28\n57 42\n35 48\n68 40\n38 50\n62 20\n31 52\n38 70\n64 35\n666 393",
"output": "-1"
},
{
"input": "50 118\n83 55\n101 33\n89 17\n97 56\n105 7\n65 12\n14 72\n57 89\n84 28\n96 22\n93 12\n17 102\n51 103\n18 100\n92 25\n50 103\n14 99\n55 101\n112 20\n89 37\n18 109\n41 78\n65 41\n105 16\n8 71\n80 45\n48 74\n16 72\n13 86\n40 95\n14 82\n23 99\n36 109\n50 88\n35 78\n91 37\n84 51\n97 15\n13 105\n105 42\n100 45\n76 36\n51 109\n115 49\n100 15\n9 83\n110 46\n114 40\n57 81\n528 348",
"output": "-1"
},
{
"input": "1 21\n0 1000",
"output": "0"
}
] | 1,646,214,780
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 60
| 0
|
inp=list(map(int,input().split()))
n=inp[0]
x0=inp[1]
pos=[]
a=[]
b=[]
shifts=10000
for x in range(n):
inp=list(map(int,input().split()))
a.append(inp[0])
b.append(inp[1])
for x in range(max(max(a)+1,max(b)+1)):
niceposition=True
for y in range(n):
if(x<min(a[y],b[y]) or x>max(a[y],b[y])):
niceposition=False
break
if(niceposition):
shifts=min(shifts,abs(x-x0))
print(shifts)
|
Title: Young Photographer
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
Among other things, Bob is keen on photography. Especially he likes to take pictures of sportsmen. That was the reason why he placed himself in position *x*0 of a long straight racetrack and got ready to take pictures. But the problem was that not all the runners passed him. The total amount of sportsmen, training at that racetrack, equals *n*. And each of them regularly runs distances within a particular segment of the racetrack, which is the same for each sportsman. For example, the first sportsman runs from position *a*1 to position *b*1, the second — from *a*2 to *b*2
What is the minimum distance that Bob should move to have a chance to take pictures of each sportsman? Bob can take a picture of a sportsman, if he stands within the segment that this sportsman covers on the racetrack.
Input Specification:
The first line of the input file contains integers *n* and *x*0 (1<=≤<=*n*<=≤<=100; 0<=≤<=*x*0<=≤<=1000). The following *n* lines contain pairs of integers *a**i*,<=*b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000; *a**i*<=≠<=*b**i*).
Output Specification:
Output the required minimum distance in the same units as the positions on the racetrack. If there is no such a position, output -1.
Demo Input:
['3 3\n0 7\n14 2\n4 6\n']
Demo Output:
['1\n']
Note:
none
|
```python
inp=list(map(int,input().split()))
n=inp[0]
x0=inp[1]
pos=[]
a=[]
b=[]
shifts=10000
for x in range(n):
inp=list(map(int,input().split()))
a.append(inp[0])
b.append(inp[1])
for x in range(max(max(a)+1,max(b)+1)):
niceposition=True
for y in range(n):
if(x<min(a[y],b[y]) or x>max(a[y],b[y])):
niceposition=False
break
if(niceposition):
shifts=min(shifts,abs(x-x0))
print(shifts)
```
| 0
|
148
|
A
|
Insomnia cure
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation",
"math"
] | null | null |
«One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine.
However, just counting dragons was boring as well, so she entertained herself at best she could. Tonight she imagined that all dragons were here to steal her, and she was fighting them off. Every *k*-th dragon got punched in the face with a frying pan. Every *l*-th dragon got his tail shut into the balcony door. Every *m*-th dragon got his paws trampled with sharp heels. Finally, she threatened every *n*-th dragon to call her mom, and he withdrew in panic.
How many imaginary dragons suffered moral or physical damage tonight, if the princess counted a total of *d* dragons?
|
Input data contains integer numbers *k*,<=*l*,<=*m*,<=*n* and *d*, each number in a separate line (1<=≤<=*k*,<=*l*,<=*m*,<=*n*<=≤<=10, 1<=≤<=*d*<=≤<=105).
|
Output the number of damaged dragons.
|
[
"1\n2\n3\n4\n12\n",
"2\n3\n4\n5\n24\n"
] |
[
"12\n",
"17\n"
] |
In the first case every first dragon got punched with a frying pan. Some of the dragons suffered from other reasons as well, but the pan alone would be enough.
In the second case dragons 1, 7, 11, 13, 17, 19 and 23 escaped unharmed.
| 1,000
|
[
{
"input": "1\n2\n3\n4\n12",
"output": "12"
},
{
"input": "2\n3\n4\n5\n24",
"output": "17"
},
{
"input": "1\n1\n1\n1\n100000",
"output": "100000"
},
{
"input": "10\n9\n8\n7\n6",
"output": "0"
},
{
"input": "8\n4\n4\n3\n65437",
"output": "32718"
},
{
"input": "8\n4\n1\n10\n59392",
"output": "59392"
},
{
"input": "4\n1\n8\n7\n44835",
"output": "44835"
},
{
"input": "6\n1\n7\n2\n62982",
"output": "62982"
},
{
"input": "2\n7\n4\n9\n56937",
"output": "35246"
},
{
"input": "2\n9\n8\n1\n75083",
"output": "75083"
},
{
"input": "8\n7\n7\n6\n69038",
"output": "24656"
},
{
"input": "4\n4\n2\n3\n54481",
"output": "36320"
},
{
"input": "6\n4\n9\n8\n72628",
"output": "28244"
},
{
"input": "9\n7\n8\n10\n42357",
"output": "16540"
},
{
"input": "5\n6\n4\n3\n60504",
"output": "36302"
},
{
"input": "7\n2\n3\n8\n21754",
"output": "15539"
},
{
"input": "1\n2\n10\n4\n39901",
"output": "39901"
},
{
"input": "3\n4\n7\n1\n58048",
"output": "58048"
},
{
"input": "9\n10\n4\n6\n52003",
"output": "21956"
},
{
"input": "5\n10\n9\n3\n70149",
"output": "32736"
},
{
"input": "5\n5\n5\n10\n55592",
"output": "11118"
},
{
"input": "1\n5\n2\n6\n49547",
"output": "49547"
},
{
"input": "3\n7\n7\n7\n84046",
"output": "36019"
},
{
"input": "10\n2\n1\n8\n63537",
"output": "63537"
},
{
"input": "7\n2\n10\n5\n81684",
"output": "53678"
},
{
"input": "7\n1\n6\n8\n99831",
"output": "99831"
},
{
"input": "3\n9\n3\n3\n61082",
"output": "20360"
},
{
"input": "5\n5\n8\n9\n79228",
"output": "29931"
},
{
"input": "1\n5\n5\n6\n89535",
"output": "89535"
},
{
"input": "5\n5\n2\n1\n7682",
"output": "7682"
},
{
"input": "7\n4\n8\n8\n25829",
"output": "9224"
},
{
"input": "3\n4\n3\n3\n87079",
"output": "43539"
},
{
"input": "5\n8\n4\n9\n5226",
"output": "2438"
},
{
"input": "4\n9\n6\n6\n1070",
"output": "415"
},
{
"input": "10\n10\n10\n10\n100000",
"output": "10000"
},
{
"input": "1\n1\n1\n1\n1",
"output": "1"
},
{
"input": "10\n10\n10\n10\n1",
"output": "0"
},
{
"input": "10\n10\n1\n10\n10",
"output": "10"
},
{
"input": "10\n10\n1\n10\n100",
"output": "100"
},
{
"input": "2\n2\n2\n2\n1",
"output": "0"
}
] | 1,688,864,894
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 40
| 654
| 22,220,800
|
# Import helpful pages
from logging.handlers import WatchedFileHandler
import queue
from re import A
from sys import stdin, stdout
from collections.abc import MutableSet, Sequence, Set
from collections import defaultdict
from tkinter.messagebox import NO
from typing import List
import copy
import math
import itertools
import heapq
import time
import bisect
from itertools import count, cycle
from unicodedata import name
from zlib import Z_BEST_COMPRESSION
# Helper Objects
class SortedSet(MutableSet, Sequence):
def __init__(self, iterable=None, key=None, init_set=set()):
self._key = key
self._set = init_set
self._len = 0
self._load = self.DEFAULT_LOAD_FACTOR
self._lists = []
self._maxes = []
self._index = []
self._offset = 0
if iterable is not None:
self._update(iterable)
@classmethod
def _fromset(cls, values, key=None):
return SortedSet(key=key, init_set=values)
def __contains__(self, value):
return value in self._set
def __len__(self):
return len(self._set)
def __iter__(self):
return iter(self._list)
def clear(self):
self._set.clear()
#self._list.clear()
def count(self, value):
return 1 if value in self._set else 0
class SegNode(object):
def __init__(self, start, end):
self.left = None
self.right = None
self.start = start
self.end = end
self.payload = set()
def is_leaf(self) -> bool:
return not self.left and not self.right
def get_midpoint(self) -> int:
return (self.start + self.end) // 2
def split(self) -> bool:
if self.is_leaf() and self.start != self.end:
midpoint = self.get_midpoint()
self.left = SegNode(self.start, midpoint)
self.right = SegNode(midpoint + 1, self.end)
return True
return False
def has_non_match(self, color):
len_check = 1 if color in self.payload else 0
return len(self.payload) > len_check
class SegArray(object):
def __init__(self, min_val:int, max_val:int):
"""Initialization."""
self.root = SegNode(min_val, max_val)
def update(self, start: int, end: int, payload:int, root:SegNode=None) -> int:
"""Update payload range"""
if not root:
root = self.root
if root.is_leaf() and not (root.start == start and root.end == end):
root.split()
if root.is_leaf():
root.payload.add(payload)
return root.payload
if end <= root.get_midpoint():
self.update(start, end, root=root.left)
elif start > root.get_midpoint():
self.update(start, end, root=root.right)
else:
self.update(start, root.left.end, root=root.left)
self.update(root.right.start, end, root=root.right)
# Propogate the changes after recursive call returns
root.payload.add(payload)
return root.payload
def get_in_range(self, start: int, end:int, root:SegNode=None) -> int:
"""set of elements nums[i..j]"""
if not root:
root = self.root
#If the range exactly matches the root, we already have the sum
if root.start == start and root.end == end:
return root.total
#If end of the range is less than the mid, the entire interval lies
#in the left subtree
if end <= root.get_midpoint():
return self.get_in_range(start, end, root=root.left)
#If start of the interval is greater than mid, the entire inteval lies
#in the right subtree
elif start >= root.get_midpoint() + 1:
return self.get_in_range(start, end, root=root.right)
#Otherwise, the interval is split. So we calculate the sum recursively,
#by splitting the interval
else:
left = self.get_in_range(start, root.get_midpoint(), root=root.left)
right = self.get_in_range(root.get_midpoint()+1, end, root=root.right)
return left.union(right)
def closest_non_match(self, start:int, end:int, payload:int, root:SegNode=None) -> int:
if not root:
root = self.root
# If there are no other payloads then no match found
if not root.has_non_match():
return -1
if root.is_leaf():
# Leaf and in range
if start <= root.end and root.end <= end:
return 0
if start <= root.start and root.start <= end:
return 0
# Leaf and out of range
min_dist = min(abs(root.start - end), abs(root.end - start))
return min_dist
# sub range
if end <= root.get_midpoint():
if root.left.has_non_match(payload):
return self.closest_non_match(start, end, payload, root.left)
return self.closest_non_match(start, end, payload, root.right)
elif start > root.get_midpoint():
if root.right.has_non_match(payload):
return self.closest_non_match(start, end, payload, root.right)
return self.closest_non_match(start, end, payload, root.left)
elif root.left.has_non_match(payload) and root.right.has_non_match(payload):
left = self.closest_non_match(start, root.get_midpoint(), payload, root.left)
left = self.closest_non_match(root.get_midpoint() + 1, end, payload, root.right)
elif root.left.has_non_match(payload):
pass
# if range is to the left of current node; left is closer if possible
if end < root.start:
next = root.right
if root.left.has_non_match(payload):
next = root.left
return self.closest_non_match(start, end, payload, next)
elif start > root.end:
next = root.left
if root.right.has_non_match(payload):
next = root.right
return self.closest_non_match(start, end, payload, next)
if start <= root.start and end <= root.end:
needed_len = 2 if payload in root.payload else 1
if len(root.payload) > needed_len:
return 0
return -1
if end <= root.get_midpoint():
pass
class IntervalNode(object):
def __init__(self, l, r):
self.l = l
self.r = r
self.split = (r - l) // 2 + l
self.left = None
self.right = None
self.count = 0
def split(self, val):
if self.has_child():
if val < self.split:
self.left.split(val)
else:
self.right.split(val)
else:
self.split = val
self.left = IntervalNode(self.l, self.split)
self.left.count = self.count
self.right = IntervalNode(self.split + 1, self.r)
self.right.count = self.count
def in_range(self, val):
return self.l <= val and self.r >= val
def add_range(self, l, r):
l = max(l, self.l)
r = min(r, self.r)
if l == self.l and r == self.r:
self.count += 1
if self.has_child():
self.left.add_range(l, r)
self.right.add_range(l, r)
else:
if l <= self.split:
if not self.has_child() and self.l < self.r:
self.left = IntervalNode(self.l, self.split)
self.left.count = self.count
self.right = IntervalNode(self.split + 1, self.r)
self.right.count = self.count
self.left.add_range(l, r)
if r > self.split:
if not self.has_child() and self.l < self.r:
self.left = IntervalNode(self.l, self.split)
self.left.count = self.count
self.right = IntervalNode(self.split + 1, self.r)
self.right.count = self.count
self.right.add_range(l, r)
def has_child(self):
return self.left and self.right and self.l < self.r
def get_count(self, li):
if self.has_child():
self.left.get_count(li)
self.right.get_count(li)
else:
li.append((self.count, self.l, self.r))
# Helper methods for execution
def get_int_dd(d=0):
def def_value():
return d
return defaultdict(def_value)
def get_counts(li):
dd = get_int_dd()
for i in li:
dd[i] += 1
return dd
def get_li_dd():
def def_value():
return []
return defaultdict(def_value)
def get_set_dd():
def def_value():
return set()
return defaultdict(def_value)
def get_index_map(li, w_offset=False):
dd = get_li_dd()
for i,n in enumerate(li):
dd[n].append(i + (1 if w_offset else 0))
return dd
def to_int_list(str):
str.strip()
return [int(i) for i in str.split()]
def to_char_list(str):
str.strip()
return [str[i] for i in range(len(str))]
# INPUT METHODS
def read_line():
return stdin.readline().strip()
def read_int():
return int(read_line())
def read_ints():
return [int(i) for i in read_line().split()]
def read_str_list(read_n=True, n=1):
if read_n:
n = read_int()
return [read_line() for i in range(n)]
# OUTPUT METHODS
def write_out(value):
stdout.write(str(value))
def write_line(value):
stdout.write(str(value) + '\n')
COMMENTS_ON = True
def comment(value):
if COMMENTS_ON:
stdout.write(str(value) + '\n')
def execute_on_input(calc_ans, read_case=read_line, t=None, write_output=write_line):
if not t:
t = read_int()
for eval in [calc_ans(read_case()) for input in range(t)]:
if eval is not None:
write_output(eval)
primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999, 2003, 2011, 2017, 2027, 2029, 2039, 2053, 2063, 2069, 2081, 2083, 2087, 2089, 2099, 2111, 2113, 2129, 2131, 2137, 2141, 2143, 2153, 2161, 2179, 2203, 2207, 2213, 2221, 2237, 2239, 2243, 2251, 2267, 2269, 2273, 2281, 2287, 2293, 2297, 2309, 2311, 2333, 2339, 2341, 2347, 2351, 2357, 2371, 2377, 2381, 2383, 2389, 2393, 2399, 2411, 2417, 2423, 2437, 2441, 2447, 2459, 2467, 2473, 2477, 2503, 2521, 2531, 2539, 2543, 2549, 2551, 2557, 2579, 2591, 2593, 2609, 2617, 2621, 2633, 2647, 2657, 2659, 2663, 2671, 2677, 2683, 2687, 2689, 2693, 2699, 2707, 2711, 2713, 2719, 2729, 2731, 2741, 2749, 2753, 2767, 2777, 2789, 2791, 2797, 2801, 2803, 2819, 2833, 2837, 2843, 2851, 2857, 2861, 2879, 2887, 2897, 2903, 2909, 2917, 2927, 2939, 2953, 2957, 2963, 2969, 2971, 2999, 3001, 3011, 3019, 3023, 3037, 3041, 3049, 3061, 3067, 3079, 3083, 3089, 3109, 3119, 3121, 3137, 3163, 3167, 3169, 3181, 3187, 3191]
fibonacci = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, 2971215073, 4807526976, 7778742049]
def qA():
def read():
return [read_int() for _ in range(5)]
def calc(input):
k, l, m, n, d = input
return sum([1 if (i % k == 0) or (i % l == 0) or (i % m == 0) or (i % n == 0) else 0 for i in range(1, d+1)])
return max([
a*b*c,
a+b+c,
(a+b)*c,
a*(b+c)
])
# Execute on each input line.
execute_on_input(calc, read_case=read, t=1)
def qB():
def read():
return read_int(), read_ints()
def calc(input):
return input
# Execute on each input line.
execute_on_input(calc, read_case=read)
def qTemplate():
def read():
return read_int(), read_ints()
def calc(input):
return input
# Execute on each input line.
execute_on_input(calc, read_case=read)
# Call the methods
if __name__ == "__main__":
qA()
|
Title: Insomnia cure
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
«One dragon. Two dragon. Three dragon», — the princess was counting. She had trouble falling asleep, and she got bored of counting lambs when she was nine.
However, just counting dragons was boring as well, so she entertained herself at best she could. Tonight she imagined that all dragons were here to steal her, and she was fighting them off. Every *k*-th dragon got punched in the face with a frying pan. Every *l*-th dragon got his tail shut into the balcony door. Every *m*-th dragon got his paws trampled with sharp heels. Finally, she threatened every *n*-th dragon to call her mom, and he withdrew in panic.
How many imaginary dragons suffered moral or physical damage tonight, if the princess counted a total of *d* dragons?
Input Specification:
Input data contains integer numbers *k*,<=*l*,<=*m*,<=*n* and *d*, each number in a separate line (1<=≤<=*k*,<=*l*,<=*m*,<=*n*<=≤<=10, 1<=≤<=*d*<=≤<=105).
Output Specification:
Output the number of damaged dragons.
Demo Input:
['1\n2\n3\n4\n12\n', '2\n3\n4\n5\n24\n']
Demo Output:
['12\n', '17\n']
Note:
In the first case every first dragon got punched with a frying pan. Some of the dragons suffered from other reasons as well, but the pan alone would be enough.
In the second case dragons 1, 7, 11, 13, 17, 19 and 23 escaped unharmed.
|
```python
# Import helpful pages
from logging.handlers import WatchedFileHandler
import queue
from re import A
from sys import stdin, stdout
from collections.abc import MutableSet, Sequence, Set
from collections import defaultdict
from tkinter.messagebox import NO
from typing import List
import copy
import math
import itertools
import heapq
import time
import bisect
from itertools import count, cycle
from unicodedata import name
from zlib import Z_BEST_COMPRESSION
# Helper Objects
class SortedSet(MutableSet, Sequence):
def __init__(self, iterable=None, key=None, init_set=set()):
self._key = key
self._set = init_set
self._len = 0
self._load = self.DEFAULT_LOAD_FACTOR
self._lists = []
self._maxes = []
self._index = []
self._offset = 0
if iterable is not None:
self._update(iterable)
@classmethod
def _fromset(cls, values, key=None):
return SortedSet(key=key, init_set=values)
def __contains__(self, value):
return value in self._set
def __len__(self):
return len(self._set)
def __iter__(self):
return iter(self._list)
def clear(self):
self._set.clear()
#self._list.clear()
def count(self, value):
return 1 if value in self._set else 0
class SegNode(object):
def __init__(self, start, end):
self.left = None
self.right = None
self.start = start
self.end = end
self.payload = set()
def is_leaf(self) -> bool:
return not self.left and not self.right
def get_midpoint(self) -> int:
return (self.start + self.end) // 2
def split(self) -> bool:
if self.is_leaf() and self.start != self.end:
midpoint = self.get_midpoint()
self.left = SegNode(self.start, midpoint)
self.right = SegNode(midpoint + 1, self.end)
return True
return False
def has_non_match(self, color):
len_check = 1 if color in self.payload else 0
return len(self.payload) > len_check
class SegArray(object):
def __init__(self, min_val:int, max_val:int):
"""Initialization."""
self.root = SegNode(min_val, max_val)
def update(self, start: int, end: int, payload:int, root:SegNode=None) -> int:
"""Update payload range"""
if not root:
root = self.root
if root.is_leaf() and not (root.start == start and root.end == end):
root.split()
if root.is_leaf():
root.payload.add(payload)
return root.payload
if end <= root.get_midpoint():
self.update(start, end, root=root.left)
elif start > root.get_midpoint():
self.update(start, end, root=root.right)
else:
self.update(start, root.left.end, root=root.left)
self.update(root.right.start, end, root=root.right)
# Propogate the changes after recursive call returns
root.payload.add(payload)
return root.payload
def get_in_range(self, start: int, end:int, root:SegNode=None) -> int:
"""set of elements nums[i..j]"""
if not root:
root = self.root
#If the range exactly matches the root, we already have the sum
if root.start == start and root.end == end:
return root.total
#If end of the range is less than the mid, the entire interval lies
#in the left subtree
if end <= root.get_midpoint():
return self.get_in_range(start, end, root=root.left)
#If start of the interval is greater than mid, the entire inteval lies
#in the right subtree
elif start >= root.get_midpoint() + 1:
return self.get_in_range(start, end, root=root.right)
#Otherwise, the interval is split. So we calculate the sum recursively,
#by splitting the interval
else:
left = self.get_in_range(start, root.get_midpoint(), root=root.left)
right = self.get_in_range(root.get_midpoint()+1, end, root=root.right)
return left.union(right)
def closest_non_match(self, start:int, end:int, payload:int, root:SegNode=None) -> int:
if not root:
root = self.root
# If there are no other payloads then no match found
if not root.has_non_match():
return -1
if root.is_leaf():
# Leaf and in range
if start <= root.end and root.end <= end:
return 0
if start <= root.start and root.start <= end:
return 0
# Leaf and out of range
min_dist = min(abs(root.start - end), abs(root.end - start))
return min_dist
# sub range
if end <= root.get_midpoint():
if root.left.has_non_match(payload):
return self.closest_non_match(start, end, payload, root.left)
return self.closest_non_match(start, end, payload, root.right)
elif start > root.get_midpoint():
if root.right.has_non_match(payload):
return self.closest_non_match(start, end, payload, root.right)
return self.closest_non_match(start, end, payload, root.left)
elif root.left.has_non_match(payload) and root.right.has_non_match(payload):
left = self.closest_non_match(start, root.get_midpoint(), payload, root.left)
left = self.closest_non_match(root.get_midpoint() + 1, end, payload, root.right)
elif root.left.has_non_match(payload):
pass
# if range is to the left of current node; left is closer if possible
if end < root.start:
next = root.right
if root.left.has_non_match(payload):
next = root.left
return self.closest_non_match(start, end, payload, next)
elif start > root.end:
next = root.left
if root.right.has_non_match(payload):
next = root.right
return self.closest_non_match(start, end, payload, next)
if start <= root.start and end <= root.end:
needed_len = 2 if payload in root.payload else 1
if len(root.payload) > needed_len:
return 0
return -1
if end <= root.get_midpoint():
pass
class IntervalNode(object):
def __init__(self, l, r):
self.l = l
self.r = r
self.split = (r - l) // 2 + l
self.left = None
self.right = None
self.count = 0
def split(self, val):
if self.has_child():
if val < self.split:
self.left.split(val)
else:
self.right.split(val)
else:
self.split = val
self.left = IntervalNode(self.l, self.split)
self.left.count = self.count
self.right = IntervalNode(self.split + 1, self.r)
self.right.count = self.count
def in_range(self, val):
return self.l <= val and self.r >= val
def add_range(self, l, r):
l = max(l, self.l)
r = min(r, self.r)
if l == self.l and r == self.r:
self.count += 1
if self.has_child():
self.left.add_range(l, r)
self.right.add_range(l, r)
else:
if l <= self.split:
if not self.has_child() and self.l < self.r:
self.left = IntervalNode(self.l, self.split)
self.left.count = self.count
self.right = IntervalNode(self.split + 1, self.r)
self.right.count = self.count
self.left.add_range(l, r)
if r > self.split:
if not self.has_child() and self.l < self.r:
self.left = IntervalNode(self.l, self.split)
self.left.count = self.count
self.right = IntervalNode(self.split + 1, self.r)
self.right.count = self.count
self.right.add_range(l, r)
def has_child(self):
return self.left and self.right and self.l < self.r
def get_count(self, li):
if self.has_child():
self.left.get_count(li)
self.right.get_count(li)
else:
li.append((self.count, self.l, self.r))
# Helper methods for execution
def get_int_dd(d=0):
def def_value():
return d
return defaultdict(def_value)
def get_counts(li):
dd = get_int_dd()
for i in li:
dd[i] += 1
return dd
def get_li_dd():
def def_value():
return []
return defaultdict(def_value)
def get_set_dd():
def def_value():
return set()
return defaultdict(def_value)
def get_index_map(li, w_offset=False):
dd = get_li_dd()
for i,n in enumerate(li):
dd[n].append(i + (1 if w_offset else 0))
return dd
def to_int_list(str):
str.strip()
return [int(i) for i in str.split()]
def to_char_list(str):
str.strip()
return [str[i] for i in range(len(str))]
# INPUT METHODS
def read_line():
return stdin.readline().strip()
def read_int():
return int(read_line())
def read_ints():
return [int(i) for i in read_line().split()]
def read_str_list(read_n=True, n=1):
if read_n:
n = read_int()
return [read_line() for i in range(n)]
# OUTPUT METHODS
def write_out(value):
stdout.write(str(value))
def write_line(value):
stdout.write(str(value) + '\n')
COMMENTS_ON = True
def comment(value):
if COMMENTS_ON:
stdout.write(str(value) + '\n')
def execute_on_input(calc_ans, read_case=read_line, t=None, write_output=write_line):
if not t:
t = read_int()
for eval in [calc_ans(read_case()) for input in range(t)]:
if eval is not None:
write_output(eval)
primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999, 2003, 2011, 2017, 2027, 2029, 2039, 2053, 2063, 2069, 2081, 2083, 2087, 2089, 2099, 2111, 2113, 2129, 2131, 2137, 2141, 2143, 2153, 2161, 2179, 2203, 2207, 2213, 2221, 2237, 2239, 2243, 2251, 2267, 2269, 2273, 2281, 2287, 2293, 2297, 2309, 2311, 2333, 2339, 2341, 2347, 2351, 2357, 2371, 2377, 2381, 2383, 2389, 2393, 2399, 2411, 2417, 2423, 2437, 2441, 2447, 2459, 2467, 2473, 2477, 2503, 2521, 2531, 2539, 2543, 2549, 2551, 2557, 2579, 2591, 2593, 2609, 2617, 2621, 2633, 2647, 2657, 2659, 2663, 2671, 2677, 2683, 2687, 2689, 2693, 2699, 2707, 2711, 2713, 2719, 2729, 2731, 2741, 2749, 2753, 2767, 2777, 2789, 2791, 2797, 2801, 2803, 2819, 2833, 2837, 2843, 2851, 2857, 2861, 2879, 2887, 2897, 2903, 2909, 2917, 2927, 2939, 2953, 2957, 2963, 2969, 2971, 2999, 3001, 3011, 3019, 3023, 3037, 3041, 3049, 3061, 3067, 3079, 3083, 3089, 3109, 3119, 3121, 3137, 3163, 3167, 3169, 3181, 3187, 3191]
fibonacci = [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, 2971215073, 4807526976, 7778742049]
def qA():
def read():
return [read_int() for _ in range(5)]
def calc(input):
k, l, m, n, d = input
return sum([1 if (i % k == 0) or (i % l == 0) or (i % m == 0) or (i % n == 0) else 0 for i in range(1, d+1)])
return max([
a*b*c,
a+b+c,
(a+b)*c,
a*(b+c)
])
# Execute on each input line.
execute_on_input(calc, read_case=read, t=1)
def qB():
def read():
return read_int(), read_ints()
def calc(input):
return input
# Execute on each input line.
execute_on_input(calc, read_case=read)
def qTemplate():
def read():
return read_int(), read_ints()
def calc(input):
return input
# Execute on each input line.
execute_on_input(calc, read_case=read)
# Call the methods
if __name__ == "__main__":
qA()
```
| 3
|
|
560
|
A
|
Currency System in Geraldion
|
PROGRAMMING
| 1,000
|
[
"implementation",
"sortings"
] | null | null |
A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the system is not perfect and sometimes it happens that Geraldionians cannot express a certain sum of money with any set of banknotes. Of course, they can use any number of banknotes of each value. Such sum is called unfortunate. Gerald wondered: what is the minimum unfortunate sum?
|
The first line contains number *n* (1<=≤<=*n*<=≤<=1000) — the number of values of the banknotes that used in Geraldion.
The second line contains *n* distinct space-separated numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the values of the banknotes.
|
Print a single line — the minimum unfortunate sum. If there are no unfortunate sums, print <=-<=1.
|
[
"5\n1 2 3 4 5\n"
] |
[
"-1\n"
] |
none
| 500
|
[
{
"input": "5\n1 2 3 4 5",
"output": "-1"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "10\n371054 506438 397130 1 766759 208409 769264 549213 641270 771837",
"output": "-1"
},
{
"input": "10\n635370 154890 909382 220996 276501 716105 538714 140162 171960 271264",
"output": "1"
},
{
"input": "50\n110876 835020 859879 999908 712969 788264 287153 921820 330355 499311 209594 484829 296329 940051 174081 931503 1 780512 390075 97866 124255 950067 697612 244256 782385 789882 37608 82153 399889 598867 416717 377988 535636 511221 792568 683271 131077 290194 496712 330720 587436 563481 645817 942562 654093 980561 382937 48293 582608 116156",
"output": "-1"
},
{
"input": "50\n474421 421097 217233 156339 27075 733996 281778 863492 184707 956857 288561 70997 393786 337382 663642 131184 637 273801 799870 295017 392338 842567 161819 297705 102013 930684 375703 838048 154915 138503 629056 256591 893619 19263 787927 684541 320265 841090 421423 490879 394582 493952 619247 633202 612928 50907 276653 407819 489945 153173",
"output": "1"
},
{
"input": "1\n1",
"output": "-1"
},
{
"input": "1\n1000000",
"output": "1"
},
{
"input": "2\n3 2",
"output": "1"
},
{
"input": "2\n2 3",
"output": "1"
}
] | 1,464,792,086
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 16
| 124
| 23,244,800
|
input()
print(1-2*('1' in input().split()))
|
Title: Currency System in Geraldion
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the system is not perfect and sometimes it happens that Geraldionians cannot express a certain sum of money with any set of banknotes. Of course, they can use any number of banknotes of each value. Such sum is called unfortunate. Gerald wondered: what is the minimum unfortunate sum?
Input Specification:
The first line contains number *n* (1<=≤<=*n*<=≤<=1000) — the number of values of the banknotes that used in Geraldion.
The second line contains *n* distinct space-separated numbers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=106) — the values of the banknotes.
Output Specification:
Print a single line — the minimum unfortunate sum. If there are no unfortunate sums, print <=-<=1.
Demo Input:
['5\n1 2 3 4 5\n']
Demo Output:
['-1\n']
Note:
none
|
```python
input()
print(1-2*('1' in input().split()))
```
| 3
|
|
976
|
E
|
Well played!
|
PROGRAMMING
| 2,100
|
[
"greedy",
"sortings"
] | null | null |
Recently Max has got himself into popular CCG "BrainStone". As "BrainStone" is a pretty intellectual game, Max has to solve numerous hard problems during the gameplay. Here is one of them:
Max owns *n* creatures, *i*-th of them can be described with two numbers — its health *hp**i* and its damage *dmg**i*. Max also has two types of spells in stock:
1. Doubles health of the creature (*hp**i* := *hp**i*·2); 1. Assigns value of health of the creature to its damage (*dmg**i* := *hp**i*).
Spell of first type can be used no more than *a* times in total, of the second type — no more than *b* times in total. Spell can be used on a certain creature multiple times. Spells can be used in arbitrary order. It isn't necessary to use all the spells.
Max is really busy preparing for his final exams, so he asks you to determine what is the maximal total damage of all creatures he can achieve if he uses spells in most optimal way.
|
The first line contains three integers *n*, *a*, *b* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*a*<=≤<=20, 0<=≤<=*b*<=≤<=2·105) — the number of creatures, spells of the first type and spells of the second type, respectively.
The *i*-th of the next *n* lines contain two number *hp**i* and *dmg**i* (1<=≤<=*hp**i*,<=*dmg**i*<=≤<=109) — description of the *i*-th creature.
|
Print single integer — maximum total damage creatures can deal.
|
[
"2 1 1\n10 15\n6 1\n",
"3 0 3\n10 8\n7 11\n5 2\n"
] |
[
"27\n",
"26\n"
] |
In the first example Max should use the spell of the first type on the second creature, then the spell of the second type on the same creature. Then total damage will be equal to 15 + 6·2 = 27.
In the second example Max should use the spell of the second type on the first creature, then the spell of the second type on the third creature. Total damage will be equal to 10 + 11 + 5 = 26.
| 0
|
[
{
"input": "2 1 1\n10 15\n6 1",
"output": "27"
},
{
"input": "3 0 3\n10 8\n7 11\n5 2",
"output": "26"
},
{
"input": "1 0 0\n2 1",
"output": "1"
},
{
"input": "1 0 200000\n1 2",
"output": "2"
},
{
"input": "7 5 7\n29 25\n84 28\n34 34\n14 76\n85 9\n40 57\n99 88",
"output": "3533"
},
{
"input": "7 6 7\n11 75\n61 90\n22 14\n100 36\n29 48\n69 52\n16 3",
"output": "6720"
},
{
"input": "7 8 7\n88 29\n30 44\n14 1\n83 95\n73 88\n10 42\n29 26",
"output": "22840"
},
{
"input": "12 7 7\n78 189\n614 271\n981 510\n37 762\n803 106\n78 369\n787 54\n768 159\n238 111\n107 54\n207 72\n485 593",
"output": "130952"
},
{
"input": "12 20 4\n852 935\n583 820\n969 197\n219 918\n547 842\n615 163\n704 377\n326 482\n183 833\n884 994\n886 581\n909 450",
"output": "1016078777"
},
{
"input": "2 13 2\n208637 682633\n393097 724045",
"output": "3220933257"
},
{
"input": "1 0 200000\n42 1",
"output": "42"
},
{
"input": "1 6 200000\n42 1",
"output": "2688"
},
{
"input": "1 0 200000\n1 42",
"output": "42"
},
{
"input": "1 6 200000\n1 42",
"output": "64"
},
{
"input": "3 1 1\n10 9\n8 6\n7 5",
"output": "31"
},
{
"input": "1 1 0\n10 1",
"output": "1"
},
{
"input": "1 1 0\n3 4",
"output": "4"
},
{
"input": "3 20 0\n1 5\n5 1\n5 1",
"output": "7"
},
{
"input": "2 5 1\n10 1\n20 20",
"output": "641"
},
{
"input": "3 20 0\n3 2\n4 3\n5 4",
"output": "9"
},
{
"input": "2 1 0\n10 15\n6 1",
"output": "16"
},
{
"input": "5 10 0\n20 1\n22 1\n30 1\n30 5\n40 6",
"output": "14"
},
{
"input": "1 20 0\n1 5",
"output": "5"
},
{
"input": "2 3 14\n28 5\n32 47",
"output": "284"
},
{
"input": "3 1 2\n20 10\n5 1\n25 25",
"output": "71"
},
{
"input": "2 3 3\n28 5\n32 47",
"output": "284"
},
{
"input": "2 2 1\n10 15\n6 1",
"output": "41"
},
{
"input": "2 1 2\n20 1\n22 23",
"output": "64"
},
{
"input": "10 7 2\n8 6\n5 5\n3 7\n7 7\n3 8\n6 1\n10 9\n4 6\n9 5\n7 9",
"output": "1339"
},
{
"input": "3 8 1\n6 6\n7 9\n2 5",
"output": "1803"
},
{
"input": "10 4 4\n5 5\n8 1\n10 10\n3 1\n7 10\n1 7\n8 7\n5 9\n3 3\n1 1",
"output": "214"
},
{
"input": "4 8 3\n1 6\n10 10\n4 8\n9 4",
"output": "2583"
},
{
"input": "8 18 1\n8 6\n6 8\n1 7\n7 2\n5 1\n10 5\n8 3\n9 3",
"output": "2621470"
},
{
"input": "2 11 1\n1 4\n1 5",
"output": "2053"
},
{
"input": "2 19 2\n9 3\n7 2",
"output": "4718599"
},
{
"input": "5 13 0\n4 4\n8 10\n1 8\n3 9\n4 6",
"output": "37"
},
{
"input": "5 8 0\n10 7\n6 6\n6 5\n7 9\n10 7",
"output": "34"
},
{
"input": "5 20 2\n1 10\n7 8\n10 1\n6 5\n2 1",
"output": "10485785"
},
{
"input": "2 1 0\n5 6\n8 8",
"output": "14"
},
{
"input": "7 3 5\n2 6\n5 9\n5 5\n4 10\n5 7\n7 8\n3 10",
"output": "103"
},
{
"input": "10 9 0\n620118469 704168608\n528098892 341451371\n15150469 449838744\n960504540 722185004\n271837337 344050133\n940943201 419522619\n85569623 788965215\n161962866 563795701\n943389281 445744350\n610994199 473866838",
"output": "5253588583"
},
{
"input": "10 11 1\n7 3\n9 4\n1 5\n10 3\n6 1\n10 7\n8 5\n7 6\n1 4\n9 9",
"output": "20524"
},
{
"input": "2 1 200000\n44 42\n1000 1001",
"output": "2044"
},
{
"input": "5 12 2\n5 9\n8 9\n4 1\n2 5\n1 8",
"output": "32794"
},
{
"input": "4 8 1\n9 9\n7 6\n2 4\n6 10",
"output": "2324"
},
{
"input": "2 1 1\n292725479 742903381\n239450691 307356865",
"output": "1221804763"
}
] | 1,657,728,156
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 1
| 77
| 5,734,400
|
# Python3 program of Quick Select
# Standard partition process of QuickSort().
# It considers the last element as pivot
# and moves all smaller element to left of
# it and greater elements to right
def partition(arr, l, r):
x = arr[r]
i = l
for j in range(l, r):
if arr[j] <= x:
arr[i], arr[j] = arr[j], arr[i]
i += 1
arr[i], arr[r] = arr[r], arr[i]
return i
# finds the kth position (of the sorted array)
# in a given unsorted array i.e this function
# can be used to find both kth largest and
# kth smallest element in the array.
# ASSUMPTION: all elements in arr[] are distinct
def kthSmallest(arr, l, r, k):
# if k is smaller than number of
# elements in array
if (k > 0 and k <= r - l + 1):
# Partition the array around last
# element and get position of pivot
# element in sorted array
index = partition(arr, l, r)
# if position is same as k
if (index - l == k - 1):
return arr[index],index
# If position is more, recur
# for left subarray
if (index - l > k - 1):
return kthSmallest(arr, l, index - 1, k)
# Else recur for right subarray
return kthSmallest(arr, index + 1, r,
k - index + l - 1)
print("Index out of bound")
from sys import stdin,exit
#Parsea una línea
def parser():
return map(int, stdin.readline().split())
#Recibiendo los valores de n, a y b
n,a,b=parser()
#Máximo de hechizos de tipo 2 que tiene sentido utilizar
b=min(b,n)
#Inicializando las listas de vida y daño
hp=[]
dmg=[]
#Recibiendo los parámetros de las criaturas
for creature in range(n):
h,d=parser()
hp.append(h)
dmg.append(d)
#Si no hay hechizos de tipo 2 solo se pueden usar los daños de las criaturas
if b==0:
print(sum(dmg))
exit()
arr=[hp[i]-dmg[i] for i in range(n)]
minnor_difference,chosen_creature=kthSmallest(arr,0,len(arr)-1,b+1)
#Inicializando el daño total
max_damage=0
#Recorriendo las criaturas
#Si es una de las primeras b criaturas ver si es mejor utilizar un hechizo de tipo 2 o no
for i in range(n):
if hp[i]-dmg[i] > minnor_difference:
max_damage+=max(hp[i],dmg[i])
elif hp[i]-dmg[i] == minnor_difference:
if i > chosen_creature:
max_damage+=dmg[i]
else:
max_damage+=max(hp[i],dmg[i])
else:
max_damage+=dmg[i]
#Guardar el valor auxiliar de la solución óptima del primer subproblema
aux=max_damage
#Restar lo aportado por la b-ésima criatura y sumar su daño
change=max(hp[chosen_creature],dmg[chosen_creature])-dmg[chosen_creature]
#Recorriendo las criaturas para ver a cuál aplicar los hechizos tipo 1
#Si pertenece a las primeras b criaturas restar daño aportado anteriormente y sumar el daño que causará con los hechizos de tipo 1
#Quedarse con el máximo valor entre lo antes calculado y lo calculado actualmente
for i in range(n):
if hp[i]-dmg[i] >= minnor_difference:
max_damage=max(max_damage,aux-max(hp[i],dmg[i])+hp[i]*2**a)
else:
max_damage=max(max_damage,aux-change-dmg[i]+hp[i]*2**a)
#Imprimir el máximo daño total
print(max_damage)
|
Title: Well played!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently Max has got himself into popular CCG "BrainStone". As "BrainStone" is a pretty intellectual game, Max has to solve numerous hard problems during the gameplay. Here is one of them:
Max owns *n* creatures, *i*-th of them can be described with two numbers — its health *hp**i* and its damage *dmg**i*. Max also has two types of spells in stock:
1. Doubles health of the creature (*hp**i* := *hp**i*·2); 1. Assigns value of health of the creature to its damage (*dmg**i* := *hp**i*).
Spell of first type can be used no more than *a* times in total, of the second type — no more than *b* times in total. Spell can be used on a certain creature multiple times. Spells can be used in arbitrary order. It isn't necessary to use all the spells.
Max is really busy preparing for his final exams, so he asks you to determine what is the maximal total damage of all creatures he can achieve if he uses spells in most optimal way.
Input Specification:
The first line contains three integers *n*, *a*, *b* (1<=≤<=*n*<=≤<=2·105, 0<=≤<=*a*<=≤<=20, 0<=≤<=*b*<=≤<=2·105) — the number of creatures, spells of the first type and spells of the second type, respectively.
The *i*-th of the next *n* lines contain two number *hp**i* and *dmg**i* (1<=≤<=*hp**i*,<=*dmg**i*<=≤<=109) — description of the *i*-th creature.
Output Specification:
Print single integer — maximum total damage creatures can deal.
Demo Input:
['2 1 1\n10 15\n6 1\n', '3 0 3\n10 8\n7 11\n5 2\n']
Demo Output:
['27\n', '26\n']
Note:
In the first example Max should use the spell of the first type on the second creature, then the spell of the second type on the same creature. Then total damage will be equal to 15 + 6·2 = 27.
In the second example Max should use the spell of the second type on the first creature, then the spell of the second type on the third creature. Total damage will be equal to 10 + 11 + 5 = 26.
|
```python
# Python3 program of Quick Select
# Standard partition process of QuickSort().
# It considers the last element as pivot
# and moves all smaller element to left of
# it and greater elements to right
def partition(arr, l, r):
x = arr[r]
i = l
for j in range(l, r):
if arr[j] <= x:
arr[i], arr[j] = arr[j], arr[i]
i += 1
arr[i], arr[r] = arr[r], arr[i]
return i
# finds the kth position (of the sorted array)
# in a given unsorted array i.e this function
# can be used to find both kth largest and
# kth smallest element in the array.
# ASSUMPTION: all elements in arr[] are distinct
def kthSmallest(arr, l, r, k):
# if k is smaller than number of
# elements in array
if (k > 0 and k <= r - l + 1):
# Partition the array around last
# element and get position of pivot
# element in sorted array
index = partition(arr, l, r)
# if position is same as k
if (index - l == k - 1):
return arr[index],index
# If position is more, recur
# for left subarray
if (index - l > k - 1):
return kthSmallest(arr, l, index - 1, k)
# Else recur for right subarray
return kthSmallest(arr, index + 1, r,
k - index + l - 1)
print("Index out of bound")
from sys import stdin,exit
#Parsea una línea
def parser():
return map(int, stdin.readline().split())
#Recibiendo los valores de n, a y b
n,a,b=parser()
#Máximo de hechizos de tipo 2 que tiene sentido utilizar
b=min(b,n)
#Inicializando las listas de vida y daño
hp=[]
dmg=[]
#Recibiendo los parámetros de las criaturas
for creature in range(n):
h,d=parser()
hp.append(h)
dmg.append(d)
#Si no hay hechizos de tipo 2 solo se pueden usar los daños de las criaturas
if b==0:
print(sum(dmg))
exit()
arr=[hp[i]-dmg[i] for i in range(n)]
minnor_difference,chosen_creature=kthSmallest(arr,0,len(arr)-1,b+1)
#Inicializando el daño total
max_damage=0
#Recorriendo las criaturas
#Si es una de las primeras b criaturas ver si es mejor utilizar un hechizo de tipo 2 o no
for i in range(n):
if hp[i]-dmg[i] > minnor_difference:
max_damage+=max(hp[i],dmg[i])
elif hp[i]-dmg[i] == minnor_difference:
if i > chosen_creature:
max_damage+=dmg[i]
else:
max_damage+=max(hp[i],dmg[i])
else:
max_damage+=dmg[i]
#Guardar el valor auxiliar de la solución óptima del primer subproblema
aux=max_damage
#Restar lo aportado por la b-ésima criatura y sumar su daño
change=max(hp[chosen_creature],dmg[chosen_creature])-dmg[chosen_creature]
#Recorriendo las criaturas para ver a cuál aplicar los hechizos tipo 1
#Si pertenece a las primeras b criaturas restar daño aportado anteriormente y sumar el daño que causará con los hechizos de tipo 1
#Quedarse con el máximo valor entre lo antes calculado y lo calculado actualmente
for i in range(n):
if hp[i]-dmg[i] >= minnor_difference:
max_damage=max(max_damage,aux-max(hp[i],dmg[i])+hp[i]*2**a)
else:
max_damage=max(max_damage,aux-change-dmg[i]+hp[i]*2**a)
#Imprimir el máximo daño total
print(max_damage)
```
| -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"
}
] | 1,643,936,268
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 52
| 186
| 0
|
#For fast I/O
import sys
input = lambda: sys.stdin.readline().strip()
n = int(input())
l = [int(i) for i in input().split()]
odd = 0
for i in l:
if i%2 == 1:
odd += 1
if odd%2 == 1:
print(odd)
else:
print(n-odd)
|
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
#For fast I/O
import sys
input = lambda: sys.stdin.readline().strip()
n = int(input())
l = [int(i) for i in input().split()]
odd = 0
for i in l:
if i%2 == 1:
odd += 1
if odd%2 == 1:
print(odd)
else:
print(n-odd)
```
| 3
|
|
447
|
B
|
DZY Loves Strings
|
PROGRAMMING
| 1,000
|
[
"greedy",
"implementation"
] | null | null |
DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter *c* DZY knows its value *w**c*. For each special string *s*<==<=*s*1*s*2... *s*|*s*| (|*s*| is the length of the string) he represents its value with a function *f*(*s*), where
Now DZY has a string *s*. He wants to insert *k* lowercase letters into this string in order to get the largest possible value of the resulting string. Can you help him calculate the largest possible value he could get?
|
The first line contains a single string *s* (1<=≤<=|*s*|<=≤<=103).
The second line contains a single integer *k* (0<=≤<=*k*<=≤<=103).
The third line contains twenty-six integers from *w**a* to *w**z*. Each such number is non-negative and doesn't exceed 1000.
|
Print a single integer — the largest possible value of the resulting string DZY could get.
|
[
"abc\n3\n1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n"
] |
[
"41\n"
] |
In the test sample DZY can obtain "abcbbc", *value* = 1·1 + 2·2 + 3·2 + 4·2 + 5·2 + 6·2 = 41.
| 1,000
|
[
{
"input": "abc\n3\n1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "41"
},
{
"input": "mmzhr\n3\n443 497 867 471 195 670 453 413 579 466 553 881 847 642 269 996 666 702 487 209 257 741 974 133 519 453",
"output": "29978"
},
{
"input": "ajeeseerqnpaujubmajpibxrccazaawetywxmifzehojf\n23\n359 813 772 413 733 654 33 87 890 433 395 311 801 852 376 148 914 420 636 695 583 733 664 394 407 314",
"output": "1762894"
},
{
"input": "uahngxejpomhbsebcxvelfsojbaouynnlsogjyvktpwwtcyddkcdqcqs\n34\n530 709 150 660 947 830 487 142 208 276 885 542 138 214 76 184 273 753 30 195 722 236 82 691 572 585",
"output": "2960349"
},
{
"input": "xnzeqmouqyzvblcidmhbkqmtusszuczadpooslqxegldanwopilmdwzbczvrwgnwaireykwpugvpnpafbxlyggkgawghysufuegvmzvpgcqyjkoadcreaguzepbendwnowsuekxxivkziibxvxfoilofxcgnxvfefyezfhevfvtetsuhwtyxdlkccdkvqjl\n282\n170 117 627 886 751 147 414 187 150 960 410 70 576 681 641 729 798 877 611 108 772 643 683 166 305 933",
"output": "99140444"
},
{
"input": "pplkqmluhfympkjfjnfdkwrkpumgdmbkfbbldpepicbbmdgafttpopzdxsevlqbtywzkoxyviglbbxsohycbdqksrhlumsldiwzjmednbkcjishkiekfrchzuztkcxnvuykhuenqojrmzaxlaoxnljnvqgnabtmcftisaazzgbmubmpsorygyusmeonrhrgphnfhlaxrvyhuxsnnezjxmdoklpquzpvjbxgbywppmegzxknhfzyygrmejleesoqfwheulmqhonqaukyuejtwxskjldplripyihbfpookxkuehiwqthbfafyrgmykuxglpplozycgydyecqkgfjljfqvigqhuxssqqtfanwszduwbsoytnrtgc\n464\n838 95 473 955 690 84 436 19 179 437 674 626 377 365 781 4 733 776 462 203 119 256 381 668 855 686",
"output": "301124161"
},
{
"input": "qkautnuilwlhjsldfcuwhiqtgtoihifszlyvfaygrnivzgvwthkrzzdtfjcirrjjlrmjtbjlzmjeqmuffsjorjyggzefwgvmblvotvzffnwjhqxorpowzdcnfksdibezdtfjjxfozaghieksbmowrbeehuxlesmvqjsphlvauxiijm\n98\n121 622 0 691 616 959 838 161 581 862 876 830 267 812 598 106 337 73 588 323 999 17 522 399 657 495",
"output": "30125295"
},
{
"input": "tghyxqfmhz\n8\n191 893 426 203 780 326 148 259 182 140 847 636 778 97 167 773 219 891 758 993 695 603 223 779 368 165",
"output": "136422"
},
{
"input": "nyawbfjxnxjiyhwkydaruozobpphgjqdpfdqzezcsoyvurnapu\n30\n65 682 543 533 990 148 815 821 315 916 632 771 332 513 472 864 12 73 548 687 660 572 507 192 226 348",
"output": "2578628"
},
{
"input": "pylrnkrbcjgoytvdnhmlvnkknijkdgdhworlvtwuonrkhrilkewcnofodaumgvnsisxooswgrgtvdeauyxhkipfoxrrtysuepjcf\n60\n894 206 704 179 272 337 413 828 119 182 330 46 440 102 250 191 242 539 678 783 843 431 612 567 33 338",
"output": "9168707"
},
{
"input": "vhjnkrxbyhjhnjrxvwxmhxwoxttbtqosfxtcuvhfjlkyfspeypthsdkkwnqdpxdlnxsgtzvkrgqosgfjrwetqbxgoarkjhrjbspzgblsapifltkfxbfdbxqwoohlgyzijmiwnpmveybyzvasoctxsmgjehpyysmqblwnmkappbecklqjfmxhlyceordroflnposohfplrvijxbwvqdtvzhobtrumiujnyrfbwthvciinuveoizkccelxtaveiiagryqnyvsgfnipnavrtmdqlcnldepocbpzmqnarkdvykds\n276\n364 244 798 82 582 9 309 950 286 547 892 371 569 159 705 975 740 845 655 179 130 993 255 552 882 657",
"output": "144901921"
},
{
"input": "gsaddmezrnttfalbwlqbnedumvikplfosw\n12\n290 850 872 361 483 895 152 118 974 619 701 154 899 285 328 712 669 984 407 340 851 775 324 892 554 860",
"output": "809931"
},
{
"input": "a\n0\n5 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": "5"
},
{
"input": "lol\n3\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",
"output": "21"
}
] | 1,557,519,232
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 24
| 155
| 716,800
|
import sys
import math
import collections
from pprint import pprint
s = input()
n = len(s)
k = int(input())
val = list(map(int, input().split()))
ans = 0
for i in range(n):
ans += (i + 1) * val[ord(s[i]) - ord('a')]
mx = max(val)
for i in range(n, n + k):
ans += (i + 1) * mx
print(ans)
|
Title: DZY Loves Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
DZY loves collecting special strings which only contain lowercase letters. For each lowercase letter *c* DZY knows its value *w**c*. For each special string *s*<==<=*s*1*s*2... *s*|*s*| (|*s*| is the length of the string) he represents its value with a function *f*(*s*), where
Now DZY has a string *s*. He wants to insert *k* lowercase letters into this string in order to get the largest possible value of the resulting string. Can you help him calculate the largest possible value he could get?
Input Specification:
The first line contains a single string *s* (1<=≤<=|*s*|<=≤<=103).
The second line contains a single integer *k* (0<=≤<=*k*<=≤<=103).
The third line contains twenty-six integers from *w**a* to *w**z*. Each such number is non-negative and doesn't exceed 1000.
Output Specification:
Print a single integer — the largest possible value of the resulting string DZY could get.
Demo Input:
['abc\n3\n1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n']
Demo Output:
['41\n']
Note:
In the test sample DZY can obtain "abcbbc", *value* = 1·1 + 2·2 + 3·2 + 4·2 + 5·2 + 6·2 = 41.
|
```python
import sys
import math
import collections
from pprint import pprint
s = input()
n = len(s)
k = int(input())
val = list(map(int, input().split()))
ans = 0
for i in range(n):
ans += (i + 1) * val[ord(s[i]) - ord('a')]
mx = max(val)
for i in range(n, n + k):
ans += (i + 1) * mx
print(ans)
```
| 3
|
|
430
|
A
|
Points and Segments (easy)
|
PROGRAMMING
| 1,600
|
[
"constructive algorithms",
"sortings"
] | null | null |
Iahub isn't well prepared on geometry problems, but he heard that this year there will be a lot of geometry problems on the IOI selection camp. Scared, Iahub locked himself in the basement and started thinking of new problems of this kind. One of them is the following.
Iahub wants to draw *n* distinct points and *m* segments on the *OX* axis. He can draw each point with either red or blue. The drawing is good if and only if the following requirement is met: for each segment [*l**i*,<=*r**i*] consider all the red points belong to it (*r**i* points), and all the blue points belong to it (*b**i* points); each segment *i* should satisfy the inequality |*r**i*<=-<=*b**i*|<=≤<=1.
Iahub thinks that point *x* belongs to segment [*l*,<=*r*], if inequality *l*<=≤<=*x*<=≤<=*r* holds.
Iahub gives to you all coordinates of points and segments. Please, help him to find any good drawing.
|
The first line of input contains two integers: *n* (1<=≤<=*n*<=≤<=100) and *m* (1<=≤<=*m*<=≤<=100). The next line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=100) — the coordinates of the points. The following *m* lines contain the descriptions of the *m* segments. Each line contains two integers *l**i* and *r**i* (0<=≤<=*l**i*<=≤<=*r**i*<=≤<=100) — the borders of the *i*-th segment.
It's guaranteed that all the points are distinct.
|
If there is no good drawing for a given test, output a single integer -1. Otherwise output *n* integers, each integer must be 0 or 1. The *i*-th number denotes the color of the *i*-th point (0 is red, and 1 is blue).
If there are multiple good drawings you can output any of them.
|
[
"3 3\n3 7 14\n1 5\n6 10\n11 15\n",
"3 4\n1 2 3\n1 2\n2 3\n5 6\n2 2\n"
] |
[
"0 0 0",
"1 0 1 "
] |
none
| 500
|
[
{
"input": "3 3\n3 7 14\n1 5\n6 10\n11 15",
"output": "0 0 0"
},
{
"input": "3 4\n1 2 3\n1 2\n2 3\n5 6\n2 2",
"output": "1 0 1 "
},
{
"input": "10 10\n3 4 2 6 1 9 0 5 8 7\n5 7\n2 6\n0 1\n5 6\n3 4\n2 5\n2 10\n4 6\n3 6\n3 7",
"output": "0 1 1 1 0 0 1 0 1 0 "
},
{
"input": "3 3\n50 51 52\n1 5\n6 10\n11 15",
"output": "1 0 1 "
},
{
"input": "3 1\n1 2 3\n2 3",
"output": "1 0 1 "
}
] | 1,405,077,696
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 12
| 62
| 0
|
n, m = map(int, input().split())
t = list(enumerate(list(map(int, input().split()))))
t.sort(key=lambda x: x[1])
p = ['0' for i in range(n)]
for i, x in t[::2]:
p[i] = '1'
print(' '.join(p))
|
Title: Points and Segments (easy)
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Iahub isn't well prepared on geometry problems, but he heard that this year there will be a lot of geometry problems on the IOI selection camp. Scared, Iahub locked himself in the basement and started thinking of new problems of this kind. One of them is the following.
Iahub wants to draw *n* distinct points and *m* segments on the *OX* axis. He can draw each point with either red or blue. The drawing is good if and only if the following requirement is met: for each segment [*l**i*,<=*r**i*] consider all the red points belong to it (*r**i* points), and all the blue points belong to it (*b**i* points); each segment *i* should satisfy the inequality |*r**i*<=-<=*b**i*|<=≤<=1.
Iahub thinks that point *x* belongs to segment [*l*,<=*r*], if inequality *l*<=≤<=*x*<=≤<=*r* holds.
Iahub gives to you all coordinates of points and segments. Please, help him to find any good drawing.
Input Specification:
The first line of input contains two integers: *n* (1<=≤<=*n*<=≤<=100) and *m* (1<=≤<=*m*<=≤<=100). The next line contains *n* space-separated integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=100) — the coordinates of the points. The following *m* lines contain the descriptions of the *m* segments. Each line contains two integers *l**i* and *r**i* (0<=≤<=*l**i*<=≤<=*r**i*<=≤<=100) — the borders of the *i*-th segment.
It's guaranteed that all the points are distinct.
Output Specification:
If there is no good drawing for a given test, output a single integer -1. Otherwise output *n* integers, each integer must be 0 or 1. The *i*-th number denotes the color of the *i*-th point (0 is red, and 1 is blue).
If there are multiple good drawings you can output any of them.
Demo Input:
['3 3\n3 7 14\n1 5\n6 10\n11 15\n', '3 4\n1 2 3\n1 2\n2 3\n5 6\n2 2\n']
Demo Output:
['0 0 0', '1 0 1 ']
Note:
none
|
```python
n, m = map(int, input().split())
t = list(enumerate(list(map(int, input().split()))))
t.sort(key=lambda x: x[1])
p = ['0' for i in range(n)]
for i, x in t[::2]:
p[i] = '1'
print(' '.join(p))
```
| 3
|
|
672
|
B
|
Different is Good
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"implementation",
"strings"
] | null | null |
A wise man told Kerem "Different is good" once, so Kerem wants all things in his life to be different.
Kerem recently got a string *s* consisting of lowercase English letters. Since Kerem likes it when things are different, he wants all substrings of his string *s* to be distinct. Substring is a string formed by some number of consecutive characters of the string. For example, string "aba" has substrings "" (empty substring), "a", "b", "a", "ab", "ba", "aba".
If string *s* has at least two equal substrings then Kerem will change characters at some positions to some other lowercase English letters. Changing characters is a very tiring job, so Kerem want to perform as few changes as possible.
Your task is to find the minimum number of changes needed to make all the substrings of the given string distinct, or determine that it is impossible.
|
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the length of the string *s*.
The second line contains the string *s* of length *n* consisting of only lowercase English letters.
|
If it's impossible to change the string *s* such that all its substring are distinct print -1. Otherwise print the minimum required number of changes.
|
[
"2\naa\n",
"4\nkoko\n",
"5\nmurat\n"
] |
[
"1\n",
"2\n",
"0\n"
] |
In the first sample one of the possible solutions is to change the first character to 'b'.
In the second sample, one may change the first character to 'a' and second character to 'b', so the string becomes "abko".
| 1,000
|
[
{
"input": "2\naa",
"output": "1"
},
{
"input": "4\nkoko",
"output": "2"
},
{
"input": "5\nmurat",
"output": "0"
},
{
"input": "6\nacbead",
"output": "1"
},
{
"input": "7\ncdaadad",
"output": "4"
},
{
"input": "25\npeoaicnbisdocqofsqdpgobpn",
"output": "12"
},
{
"input": "25\ntcqpchnqskqjacruoaqilgebu",
"output": "7"
},
{
"input": "13\naebaecedabbee",
"output": "8"
},
{
"input": "27\naaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "-1"
},
{
"input": "10\nbababbdaee",
"output": "6"
},
{
"input": "11\ndbadcdbdbca",
"output": "7"
},
{
"input": "12\nacceaabddaaa",
"output": "7"
},
{
"input": "13\nabddfbfaeecfa",
"output": "7"
},
{
"input": "14\neeceecacdbcbbb",
"output": "9"
},
{
"input": "15\ndcbceaaggabaheb",
"output": "8"
},
{
"input": "16\nhgiegfbadgcicbhd",
"output": "7"
},
{
"input": "17\nabhfibbdddfghgfdi",
"output": "10"
},
{
"input": "26\nbbbbbabbaababaaabaaababbaa",
"output": "24"
},
{
"input": "26\nahnxdnbfbcrirerssyzydihuee",
"output": "11"
},
{
"input": "26\nhwqeqhkpxwulbsiwmnlfyhgknc",
"output": "8"
},
{
"input": "26\nrvxmulriorilidecqwmfaemifj",
"output": "10"
},
{
"input": "26\naowpmreooavnmamogdoopuisge",
"output": "12"
},
{
"input": "26\ninimevtuefhvuefirdehmmfudh",
"output": "15"
},
{
"input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "25"
},
{
"input": "27\nqdcfjtblgglnilgassirrjekcjt",
"output": "-1"
},
{
"input": "27\nabcdefghijklmnopqrstuvwxyza",
"output": "-1"
},
{
"input": "26\nqwertyuiopasdfghjklzxcvbnm",
"output": "0"
},
{
"input": "5\nzzzzz",
"output": "4"
},
{
"input": "27\naaaaaaaaaaaaaaaaabaaaaaaaaa",
"output": "-1"
},
{
"input": "1\nq",
"output": "0"
},
{
"input": "27\nqwertyuioplkjhgfdsazxcvbnmm",
"output": "-1"
},
{
"input": "9\nxxxyyyzzz",
"output": "6"
},
{
"input": "45\naaabbbcccdddeeefffgghhiijjkkkkkkkkkkkkkkkkkkk",
"output": "-1"
},
{
"input": "27\nqwertyuiopasdfghjklzxcvbnmm",
"output": "-1"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxyz",
"output": "0"
},
{
"input": "26\nabcdefghijklmnopqrstuvwxya",
"output": "1"
},
{
"input": "27\nabcdefghijklmnopqrstuvwxyzz",
"output": "-1"
},
{
"input": "26\naaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "25"
},
{
"input": "26\nqwertyuioplkjhgfdsazxcvbnm",
"output": "0"
},
{
"input": "10\nzzzzzzzzzz",
"output": "9"
},
{
"input": "1\na",
"output": "0"
},
{
"input": "30\nabcdefghtyabcdefghtyabcdefghty",
"output": "-1"
}
] | 1,541,772,448
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 7
| 108
| 204,800
|
a=int(input())
b=input()
print(a-len(set(b)))
|
Title: Different is Good
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A wise man told Kerem "Different is good" once, so Kerem wants all things in his life to be different.
Kerem recently got a string *s* consisting of lowercase English letters. Since Kerem likes it when things are different, he wants all substrings of his string *s* to be distinct. Substring is a string formed by some number of consecutive characters of the string. For example, string "aba" has substrings "" (empty substring), "a", "b", "a", "ab", "ba", "aba".
If string *s* has at least two equal substrings then Kerem will change characters at some positions to some other lowercase English letters. Changing characters is a very tiring job, so Kerem want to perform as few changes as possible.
Your task is to find the minimum number of changes needed to make all the substrings of the given string distinct, or determine that it is impossible.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100<=000) — the length of the string *s*.
The second line contains the string *s* of length *n* consisting of only lowercase English letters.
Output Specification:
If it's impossible to change the string *s* such that all its substring are distinct print -1. Otherwise print the minimum required number of changes.
Demo Input:
['2\naa\n', '4\nkoko\n', '5\nmurat\n']
Demo Output:
['1\n', '2\n', '0\n']
Note:
In the first sample one of the possible solutions is to change the first character to 'b'.
In the second sample, one may change the first character to 'a' and second character to 'b', so the string becomes "abko".
|
```python
a=int(input())
b=input()
print(a-len(set(b)))
```
| 0
|
|
998
|
A
|
Balloons
|
PROGRAMMING
| 1,000
|
[
"constructive algorithms",
"implementation"
] | null | null |
There are quite a lot of ways to have fun with inflatable balloons. For example, you can fill them with water and see what happens.
Grigory and Andrew have the same opinion. So, once upon a time, they went to the shop and bought $n$ packets with inflatable balloons, where $i$-th of them has exactly $a_i$ balloons inside.
They want to divide the balloons among themselves. In addition, there are several conditions to hold:
- Do not rip the packets (both Grigory and Andrew should get unbroken packets); - Distribute all packets (every packet should be given to someone); - Give both Grigory and Andrew at least one packet; - To provide more fun, the total number of balloons in Grigory's packets should not be equal to the total number of balloons in Andrew's packets.
Help them to divide the balloons or determine that it's impossible under these conditions.
|
The first line of input contains a single integer $n$ ($1 \le n \le 10$) — the number of packets with balloons.
The second line contains $n$ integers: $a_1$, $a_2$, $\ldots$, $a_n$ ($1 \le a_i \le 1000$) — the number of balloons inside the corresponding packet.
|
If it's impossible to divide the balloons satisfying the conditions above, print $-1$.
Otherwise, print an integer $k$ — the number of packets to give to Grigory followed by $k$ distinct integers from $1$ to $n$ — the indices of those. The order of packets doesn't matter.
If there are multiple ways to divide balloons, output any of them.
|
[
"3\n1 2 1\n",
"2\n5 5\n",
"1\n10\n"
] |
[
"2\n1 2\n",
"-1\n",
"-1\n"
] |
In the first test Grigory gets $3$ balloons in total while Andrey gets $1$.
In the second test there's only one way to divide the packets which leads to equal numbers of balloons.
In the third test one of the boys won't get a packet at all.
| 500
|
[
{
"input": "3\n1 2 1",
"output": "1\n1"
},
{
"input": "2\n5 5",
"output": "-1"
},
{
"input": "1\n10",
"output": "-1"
},
{
"input": "1\n1",
"output": "-1"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "1\n1"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 9",
"output": "1\n1"
},
{
"input": "10\n26 723 970 13 422 968 875 329 234 983",
"output": "1\n4"
},
{
"input": "3\n3 2 1",
"output": "1\n3"
},
{
"input": "10\n1000 1000 1000 1000 1000 1000 1000 1000 1000 1000",
"output": "1\n1"
},
{
"input": "10\n1 9 7 6 2 4 7 8 1 3",
"output": "1\n1"
},
{
"input": "2\n9 6",
"output": "1\n2"
},
{
"input": "2\n89 7",
"output": "1\n2"
},
{
"input": "2\n101 807",
"output": "1\n1"
},
{
"input": "5\n8 7 4 8 3",
"output": "1\n5"
},
{
"input": "5\n55 62 70 100 90",
"output": "1\n1"
},
{
"input": "5\n850 840 521 42 169",
"output": "1\n4"
},
{
"input": "6\n7 1 4 1 6 1",
"output": "1\n2"
},
{
"input": "6\n36 80 38 88 79 69",
"output": "1\n1"
},
{
"input": "6\n108 318 583 10 344 396",
"output": "1\n4"
},
{
"input": "9\n10 9 10 10 8 3 5 10 2",
"output": "1\n9"
},
{
"input": "9\n90 31 28 63 57 57 27 62 42",
"output": "1\n7"
},
{
"input": "9\n665 646 152 829 190 64 555 536 321",
"output": "1\n6"
},
{
"input": "10\n99 62 10 47 53 9 83 33 15 24",
"output": "1\n6"
},
{
"input": "4\n600 200 100 300",
"output": "1\n3"
},
{
"input": "2\n4 5",
"output": "1\n1"
},
{
"input": "2\n5 12",
"output": "1\n1"
},
{
"input": "2\n1 2",
"output": "1\n1"
},
{
"input": "3\n1 1 2",
"output": "1\n1"
},
{
"input": "2\n3 2",
"output": "1\n2"
},
{
"input": "3\n1 4 5",
"output": "1\n1"
},
{
"input": "4\n5 5 5 5",
"output": "1\n1"
},
{
"input": "1\n5",
"output": "-1"
},
{
"input": "3\n5 5 5",
"output": "1\n1"
},
{
"input": "5\n5 5 5 5 5",
"output": "1\n1"
},
{
"input": "4\n2 7 10 1",
"output": "1\n4"
},
{
"input": "3\n1 1 1",
"output": "1\n1"
},
{
"input": "4\n8 4 2 2",
"output": "1\n3"
},
{
"input": "2\n3 4",
"output": "1\n1"
},
{
"input": "4\n1 1 3 1",
"output": "1\n1"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "1\n1"
},
{
"input": "2\n18 19",
"output": "1\n1"
},
{
"input": "2\n17 18",
"output": "1\n1"
},
{
"input": "2\n1 3",
"output": "1\n1"
},
{
"input": "4\n5 5 4 4",
"output": "1\n3"
},
{
"input": "2\n10 11",
"output": "1\n1"
},
{
"input": "2\n10 20",
"output": "1\n1"
},
{
"input": "2\n2 1",
"output": "1\n2"
},
{
"input": "4\n2 3 2 3",
"output": "1\n1"
},
{
"input": "2\n5 6",
"output": "1\n1"
}
] | 1,531,462,714
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 124
| 0
|
def main():
n = int(input())
l = list(map(int, input().split()))
if n == 1 or n == 2 and l[0] == l[1]:
print(-1)
return
s = sum(l)
for i, x in enumerate(l, 1):
if x * 2 != s:
print(1, i, sep='\n')
return
if __name__ == '__main__':
main()
|
Title: Balloons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are quite a lot of ways to have fun with inflatable balloons. For example, you can fill them with water and see what happens.
Grigory and Andrew have the same opinion. So, once upon a time, they went to the shop and bought $n$ packets with inflatable balloons, where $i$-th of them has exactly $a_i$ balloons inside.
They want to divide the balloons among themselves. In addition, there are several conditions to hold:
- Do not rip the packets (both Grigory and Andrew should get unbroken packets); - Distribute all packets (every packet should be given to someone); - Give both Grigory and Andrew at least one packet; - To provide more fun, the total number of balloons in Grigory's packets should not be equal to the total number of balloons in Andrew's packets.
Help them to divide the balloons or determine that it's impossible under these conditions.
Input Specification:
The first line of input contains a single integer $n$ ($1 \le n \le 10$) — the number of packets with balloons.
The second line contains $n$ integers: $a_1$, $a_2$, $\ldots$, $a_n$ ($1 \le a_i \le 1000$) — the number of balloons inside the corresponding packet.
Output Specification:
If it's impossible to divide the balloons satisfying the conditions above, print $-1$.
Otherwise, print an integer $k$ — the number of packets to give to Grigory followed by $k$ distinct integers from $1$ to $n$ — the indices of those. The order of packets doesn't matter.
If there are multiple ways to divide balloons, output any of them.
Demo Input:
['3\n1 2 1\n', '2\n5 5\n', '1\n10\n']
Demo Output:
['2\n1 2\n', '-1\n', '-1\n']
Note:
In the first test Grigory gets $3$ balloons in total while Andrey gets $1$.
In the second test there's only one way to divide the packets which leads to equal numbers of balloons.
In the third test one of the boys won't get a packet at all.
|
```python
def main():
n = int(input())
l = list(map(int, input().split()))
if n == 1 or n == 2 and l[0] == l[1]:
print(-1)
return
s = sum(l)
for i, x in enumerate(l, 1):
if x * 2 != s:
print(1, i, sep='\n')
return
if __name__ == '__main__':
main()
```
| 3
|
|
411
|
A
|
Password Check
|
PROGRAMMING
| 800
|
[
"*special",
"implementation"
] | null | null |
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check.
Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions:
- the password length is at least 5 characters; - the password contains at least one large English letter; - the password contains at least one small English letter; - the password contains at least one digit.
You are given a password. Please implement the automatic check of its complexity for company Q.
|
The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_".
|
If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes).
|
[
"abacaba\n",
"X12345\n",
"CONTEST_is_STARTED!!11\n"
] |
[
"Too weak\n",
"Too weak\n",
"Correct\n"
] |
none
| 0
|
[
{
"input": "abacaba",
"output": "Too weak"
},
{
"input": "X12345",
"output": "Too weak"
},
{
"input": "CONTEST_is_STARTED!!11",
"output": "Correct"
},
{
"input": "1zA__",
"output": "Correct"
},
{
"input": "1zA_",
"output": "Too weak"
},
{
"input": "zA___",
"output": "Too weak"
},
{
"input": "1A___",
"output": "Too weak"
},
{
"input": "z1___",
"output": "Too weak"
},
{
"input": "0",
"output": "Too weak"
},
{
"input": "_",
"output": "Too weak"
},
{
"input": "a",
"output": "Too weak"
},
{
"input": "D",
"output": "Too weak"
},
{
"input": "_",
"output": "Too weak"
},
{
"input": "?",
"output": "Too weak"
},
{
"input": "?",
"output": "Too weak"
},
{
"input": "._,.!.,...?_,!.",
"output": "Too weak"
},
{
"input": "!_?_,?,?.,.,_!!!.!,.__,?!!,_!,?_,!??,?!..._!?_,?_!,?_.,._,,_.,.",
"output": "Too weak"
},
{
"input": "?..!.,,?,__.,...????_???__!,?...?.,,,,___!,.!,_,,_,??!_?_,!!?_!_??.?,.!!?_?_.,!",
"output": "Too weak"
},
{
"input": "XZX",
"output": "Too weak"
},
{
"input": "R",
"output": "Too weak"
},
{
"input": "H.FZ",
"output": "Too weak"
},
{
"input": "KSHMICWPK,LSBM_JVZ!IPDYDG_GOPCHXFJTKJBIFY,FPHMY,CB?PZEAG..,X,.GFHPIDBB,IQ?MZ",
"output": "Too weak"
},
{
"input": "EFHI,,Y?HMMUI,,FJGAY?FYPBJQMYM!DZHLFCTFWT?JOPDW,S_!OR?ATT?RWFBMAAKUHIDMHSD?LCZQY!UD_CGYGBAIRDPICYS",
"output": "Too weak"
},
{
"input": "T,NDMUYCCXH_L_FJHMCCAGX_XSCPGOUZSY?D?CNDSYRITYS,VAT!PJVKNTBMXGGRYKACLYU.RJQ_?UWKXYIDE_AE",
"output": "Too weak"
},
{
"input": "y",
"output": "Too weak"
},
{
"input": "qgw",
"output": "Too weak"
},
{
"input": "g",
"output": "Too weak"
},
{
"input": "loaray",
"output": "Too weak"
},
{
"input": "d_iymyvxolmjayhwpedocopqwmy.oalrdg!_n?.lrxpamhygps?kkzxydsbcaihfs.j?eu!oszjsy.vzu?!vs.bprz_j",
"output": "Too weak"
},
{
"input": "txguglvclyillwnono",
"output": "Too weak"
},
{
"input": "FwX",
"output": "Too weak"
},
{
"input": "Zi",
"output": "Too weak"
},
{
"input": "PodE",
"output": "Too weak"
},
{
"input": "SdoOuJ?nj_wJyf",
"output": "Too weak"
},
{
"input": "MhnfZjsUyXYw?f?ubKA",
"output": "Too weak"
},
{
"input": "CpWxDVzwHfYFfoXNtXMFuAZr",
"output": "Too weak"
},
{
"input": "9.,0",
"output": "Too weak"
},
{
"input": "5,8",
"output": "Too weak"
},
{
"input": "7",
"output": "Too weak"
},
{
"input": "34__39_02!,!,82!129!2!566",
"output": "Too weak"
},
{
"input": "96156027.65935663!_87!,44,..7914_!0_1,.4!!62!.8350!17_282!!9.2584,!!7__51.526.7",
"output": "Too weak"
},
{
"input": "90328_",
"output": "Too weak"
},
{
"input": "B9",
"output": "Too weak"
},
{
"input": "P1H",
"output": "Too weak"
},
{
"input": "J2",
"output": "Too weak"
},
{
"input": "M6BCAKW!85OSYX1D?.53KDXP42F",
"output": "Too weak"
},
{
"input": "C672F429Y8X6XU7S,.K9111UD3232YXT81S4!729ER7DZ.J7U1R_7VG6.FQO,LDH",
"output": "Too weak"
},
{
"input": "W2PI__!.O91H8OFY6AB__R30L9XOU8800?ZUD84L5KT99818NFNE35V.8LJJ5P2MM.B6B",
"output": "Too weak"
},
{
"input": "z1",
"output": "Too weak"
},
{
"input": "p1j",
"output": "Too weak"
},
{
"input": "j9",
"output": "Too weak"
},
{
"input": "v8eycoylzv0qkix5mfs_nhkn6k!?ovrk9!b69zy!4frc?k",
"output": "Too weak"
},
{
"input": "l4!m_44kpw8.jg!?oh,?y5oraw1tg7_x1.osl0!ny?_aihzhtt0e2!mr92tnk0es!1f,9he40_usa6c50l",
"output": "Too weak"
},
{
"input": "d4r!ak.igzhnu!boghwd6jl",
"output": "Too weak"
},
{
"input": "It0",
"output": "Too weak"
},
{
"input": "Yb1x",
"output": "Too weak"
},
{
"input": "Qf7",
"output": "Too weak"
},
{
"input": "Vu7jQU8.!FvHBYTsDp6AphaGfnEmySP9te",
"output": "Correct"
},
{
"input": "Ka4hGE,vkvNQbNolnfwp",
"output": "Correct"
},
{
"input": "Ee9oluD?amNItsjeQVtOjwj4w_ALCRh7F3eaZah",
"output": "Correct"
},
{
"input": "Um3Fj?QLhNuRE_Gx0cjMLOkGCm",
"output": "Correct"
},
{
"input": "Oq2LYmV9HmlaW",
"output": "Correct"
},
{
"input": "Cq7r3Wrb.lDb_0wsf7!ruUUGSf08RkxD?VsBEDdyE?SHK73TFFy0f8gmcATqGafgTv8OOg8or2HyMPIPiQ2Hsx8q5rn3_WZe",
"output": "Correct"
},
{
"input": "Wx4p1fOrEMDlQpTlIx0p.1cnFD7BnX2K8?_dNLh4cQBx_Zqsv83BnL5hGKNcBE9g3QB,!fmSvgBeQ_qiH7",
"output": "Correct"
},
{
"input": "k673,",
"output": "Too weak"
},
{
"input": "LzuYQ",
"output": "Too weak"
},
{
"input": "Pasq!",
"output": "Too weak"
},
{
"input": "x5hve",
"output": "Too weak"
},
{
"input": "b27fk",
"output": "Too weak"
},
{
"input": "h6y1l",
"output": "Too weak"
},
{
"input": "i9nij",
"output": "Too weak"
},
{
"input": "Gf5Q6",
"output": "Correct"
},
{
"input": "Uf24o",
"output": "Correct"
},
{
"input": "Oj9vu",
"output": "Correct"
},
{
"input": "c7jqaudcqmv8o7zvb5x_gp6zcgl6nwr7tz5or!28.tj8s1m2.wxz5a4id03!rq07?662vy.7.p5?vk2f2mc7ag8q3861rgd0rmbr",
"output": "Too weak"
},
{
"input": "i6a.,8jb,n0kv4.1!7h?p.96pnhhgy6cl7dg7e4o6o384ys3z.t71kkq,,w,oqi4?u,,m5!rzu6wym_4hm,ohjy!.vvksl?pt,,1",
"output": "Too weak"
},
{
"input": "M10V_MN_1K8YX2LA!89EYV7!5V9?,.IDHDP6JEC.OGLY.180LMZ6KW3Z5E17IT94ZNHS!79GN09Q6LH0,F3AYNKP?KM,QP_?XRD6",
"output": "Too weak"
},
{
"input": "Hi7zYuVXCPhaho68YgCMzzgLILM6toQTJq8akMqqrnUn6ZCD36iA1yVVpvlsIiMpCu!1QZd4ycIrQ5Kcrhk5k0jTrwdAAEEP_T2f",
"output": "Correct"
},
{
"input": "Bk2Q38vDSW5JqYu.077iYC.9YoiPc!Dh6FJWOVze6?YXiFjPNa4F1RG?154m9mY2jQobBnbxM,cDV8l1UX1?v?p.tTYIyJO!NYmE",
"output": "Correct"
},
{
"input": "Ro1HcZ.piN,JRR88DLh,WtW!pbFM076?wCSbqfK7N2s5zUySFBtzk7HV,BxHXR0zALAr016z5jvvB.WUdEcKgYFav5TygwHQC..C",
"output": "Correct"
},
{
"input": "!?.,_",
"output": "Too weak"
}
] | 1,591,638,230
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 81
| 109
| 307,200
|
i=input()
a,b,c=0,0,0
l=["!","?",".",",","_"]
if len(i)<5:
print("Too weak")
else:
for x in i:
if x==" " or x in l:
continue
elif 65<=ord(x)<=90:
a+=1
elif 97<=ord(x)<=122:
b+=1
elif 48<=ord(x)<=57:
c+=1
print(["Too weak","Correct"][a>0 and b>0 and c>0])
|
Title: Password Check
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check.
Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions:
- the password length is at least 5 characters; - the password contains at least one large English letter; - the password contains at least one small English letter; - the password contains at least one digit.
You are given a password. Please implement the automatic check of its complexity for company Q.
Input Specification:
The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_".
Output Specification:
If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes).
Demo Input:
['abacaba\n', 'X12345\n', 'CONTEST_is_STARTED!!11\n']
Demo Output:
['Too weak\n', 'Too weak\n', 'Correct\n']
Note:
none
|
```python
i=input()
a,b,c=0,0,0
l=["!","?",".",",","_"]
if len(i)<5:
print("Too weak")
else:
for x in i:
if x==" " or x in l:
continue
elif 65<=ord(x)<=90:
a+=1
elif 97<=ord(x)<=122:
b+=1
elif 48<=ord(x)<=57:
c+=1
print(["Too weak","Correct"][a>0 and b>0 and c>0])
```
| 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"
}
] | 1,698,943,064
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 8
| 61
| 0
|
from math import ceil
from math import floor
n = int(input())
ppl = ['Sheldon', 'Leonard', 'Penny', 'Rajesh', 'Howard']
while n > 5 :
n /=5
if ceil(n) - n >= 0.5:
ans = floor(n)
else:
ans = ceil(n)
for i in range(len(ppl)):
if ans-1 == i:
print(ppl[i])
|
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
from math import ceil
from math import floor
n = int(input())
ppl = ['Sheldon', 'Leonard', 'Penny', 'Rajesh', 'Howard']
while n > 5 :
n /=5
if ceil(n) - n >= 0.5:
ans = floor(n)
else:
ans = ceil(n)
for i in range(len(ppl)):
if ans-1 == i:
print(ppl[i])
```
| 0
|
789
|
B
|
Masha and geometric depression
|
PROGRAMMING
| 1,700
|
[
"brute force",
"implementation",
"math"
] | null | null |
Masha really loves algebra. On the last lesson, her strict teacher Dvastan gave she new exercise.
You are given geometric progression *b* defined by two integers *b*1 and *q*. Remind that a geometric progression is a sequence of integers *b*1,<=*b*2,<=*b*3,<=..., where for each *i*<=><=1 the respective term satisfies the condition *b**i*<==<=*b**i*<=-<=1·*q*, where *q* is called the common ratio of the progression. Progressions in Uzhlyandia are unusual: both *b*1 and *q* can equal 0. Also, Dvastan gave Masha *m* "bad" integers *a*1,<=*a*2,<=...,<=*a**m*, and an integer *l*.
Masha writes all progression terms one by one onto the board (including repetitive) while condition |*b**i*|<=≤<=*l* is satisfied (|*x*| means absolute value of *x*). There is an exception: if a term equals one of the "bad" integers, Masha skips it (doesn't write onto the board) and moves forward to the next term.
But the lesson is going to end soon, so Masha has to calculate how many integers will be written on the board. In order not to get into depression, Masha asked you for help: help her calculate how many numbers she will write, or print "inf" in case she needs to write infinitely many integers.
|
The first line of input contains four integers *b*1, *q*, *l*, *m* (-109<=≤<=*b*1,<=*q*<=≤<=109, 1<=≤<=*l*<=≤<=109, 1<=≤<=*m*<=≤<=105) — the initial term and the common ratio of progression, absolute value of maximal number that can be written on the board and the number of "bad" integers, respectively.
The second line contains *m* distinct integers *a*1,<=*a*2,<=...,<=*a**m* (-109<=≤<=*a**i*<=≤<=109) — numbers that will never be written on the board.
|
Print the only integer, meaning the number of progression terms that will be written on the board if it is finite, or "inf" (without quotes) otherwise.
|
[
"3 2 30 4\n6 14 25 48\n",
"123 1 2143435 4\n123 11 -5453 141245\n",
"123 1 2143435 4\n54343 -13 6 124\n"
] |
[
"3",
"0",
"inf"
] |
In the first sample case, Masha will write integers 3, 12, 24. Progression term 6 will be skipped because it is a "bad" integer. Terms bigger than 24 won't be written because they exceed *l* by absolute value.
In the second case, Masha won't write any number because all terms are equal 123 and this is a "bad" integer.
In the third case, Masha will write infinitely integers 123.
| 1,000
|
[
{
"input": "3 2 30 4\n6 14 25 48",
"output": "3"
},
{
"input": "123 1 2143435 4\n123 11 -5453 141245",
"output": "0"
},
{
"input": "123 1 2143435 4\n54343 -13 6 124",
"output": "inf"
},
{
"input": "3 2 25 2\n379195692 -69874783",
"output": "4"
},
{
"input": "3 2 30 3\n-691070108 -934106649 -220744807",
"output": "4"
},
{
"input": "3 3 104 17\n9 -73896485 -290898562 5254410 409659728 -916522518 -435516126 94354167 262981034 -375897180 -80186684 -173062070 -288705544 -699097793 -11447747 320434295 503414250",
"output": "3"
},
{
"input": "-1000000000 -1000000000 1 1\n232512888",
"output": "0"
},
{
"input": "11 0 228 5\n-1 0 1 5 -11245",
"output": "1"
},
{
"input": "11 0 228 5\n-1 -17 1 5 -11245",
"output": "inf"
},
{
"input": "0 0 2143435 5\n-1 -153 1 5 -11245",
"output": "inf"
},
{
"input": "123 0 2143435 4\n5433 0 123 -645",
"output": "0"
},
{
"input": "123 -1 2143435 5\n-123 0 12 5 -11245",
"output": "inf"
},
{
"input": "123 0 21 4\n543453 -123 6 1424",
"output": "0"
},
{
"input": "3 2 115 16\n24 48 12 96 3 720031148 -367712651 -838596957 558177735 -963046495 -313322487 -465018432 -618984128 -607173835 144854086 178041956",
"output": "1"
},
{
"input": "-3 0 92055 36\n-92974174 -486557474 -663622151 695596393 177960746 -563227474 -364263320 -676254242 -614140218 71456762 -764104225 705056581 -106398436 332755134 -199942822 -732751692 658942664 677739866 886535704 183687802 -784248291 -22550621 -938674499 637055091 -704750213 780395802 778342470 -999059668 -794361783 796469192 215667969 354336794 -60195289 -885080928 -290279020 201221317",
"output": "inf"
},
{
"input": "0 -3 2143435 5\n-1 0 1 5 -11245",
"output": "0"
},
{
"input": "123 -1 2143435 5\n-123 0 123 -5453 141245",
"output": "0"
},
{
"input": "123 0 2143435 4\n5433 0 -123 -645",
"output": "1"
},
{
"input": "11 0 2 5\n-1 0 1 5 -11245",
"output": "0"
},
{
"input": "2 2 4 1\n2",
"output": "1"
},
{
"input": "1 -2 1000000000 1\n0",
"output": "30"
},
{
"input": "0 8 10 1\n5",
"output": "inf"
},
{
"input": "-1000 0 10 1\n5",
"output": "0"
},
{
"input": "0 2 2143435 4\n54343 -13 6 124",
"output": "inf"
},
{
"input": "0 8 5 1\n9",
"output": "inf"
},
{
"input": "-10 1 5 1\n100",
"output": "0"
},
{
"input": "123 -1 2143435 4\n54343 -13 6 123",
"output": "inf"
},
{
"input": "-5 -1 10 1\n-5",
"output": "inf"
},
{
"input": "2 0 1 1\n2",
"output": "0"
},
{
"input": "0 5 8 1\n10",
"output": "inf"
},
{
"input": "0 5 100 2\n34 56",
"output": "inf"
},
{
"input": "15 -1 15 4\n15 -15 1 2",
"output": "0"
},
{
"input": "10 -1 2 1\n1",
"output": "0"
},
{
"input": "2 0 2 1\n2",
"output": "inf"
},
{
"input": "4 0 4 1\n0",
"output": "1"
},
{
"input": "10 10 10 1\n123",
"output": "1"
},
{
"input": "2 2 4 1\n3",
"output": "2"
},
{
"input": "0 1 1 1\n0",
"output": "0"
},
{
"input": "3 2 30 1\n3",
"output": "3"
},
{
"input": "1000000000 100000 1000000000 4\n5433 13 6 0",
"output": "1"
},
{
"input": "-2 0 1 1\n1",
"output": "0"
},
{
"input": "2 -1 10 1\n2",
"output": "inf"
},
{
"input": "1 -1 2 1\n1",
"output": "inf"
},
{
"input": "0 10 10 1\n2",
"output": "inf"
},
{
"input": "0 35 2 1\n3",
"output": "inf"
},
{
"input": "3 1 3 1\n5",
"output": "inf"
},
{
"input": "3 2 3 4\n6 14 25 48",
"output": "1"
},
{
"input": "0 69 12 1\n1",
"output": "inf"
},
{
"input": "100 0 100000 1\n100",
"output": "inf"
},
{
"input": "0 4 1000 3\n5 6 7",
"output": "inf"
},
{
"input": "0 2 100 1\n5",
"output": "inf"
},
{
"input": "3 2 24 4\n6 14 25 48",
"output": "3"
},
{
"input": "0 4 1 1\n2",
"output": "inf"
},
{
"input": "1 5 10000 1\n125",
"output": "5"
},
{
"input": "2 -1 1 1\n1",
"output": "0"
},
{
"input": "0 3 100 1\n5",
"output": "inf"
},
{
"input": "0 3 3 1\n1",
"output": "inf"
},
{
"input": "0 2 5 1\n1",
"output": "inf"
},
{
"input": "5 -1 100 1\n5",
"output": "inf"
},
{
"input": "-20 0 10 1\n0",
"output": "0"
},
{
"input": "3 0 1 1\n3",
"output": "0"
},
{
"input": "2 -1 3 1\n2",
"output": "inf"
},
{
"input": "1 1 1000000000 1\n100",
"output": "inf"
},
{
"input": "5 -1 3 1\n0",
"output": "0"
},
{
"input": "0 5 10 1\n2",
"output": "inf"
},
{
"input": "123 0 125 1\n123",
"output": "inf"
},
{
"input": "2 -1 100 1\n2",
"output": "inf"
},
{
"input": "5 2 100 1\n5",
"output": "4"
},
{
"input": "-5 0 1 1\n1",
"output": "0"
},
{
"input": "-3 0 1 1\n-3",
"output": "0"
},
{
"input": "2 -2 10 1\n1",
"output": "3"
},
{
"input": "0 2 30 4\n6 14 25 48",
"output": "inf"
},
{
"input": "1 -1 1 1\n1",
"output": "inf"
},
{
"input": "2 -1 6 1\n2",
"output": "inf"
},
{
"input": "-3 1 100 1\n-3",
"output": "0"
},
{
"input": "1 0 2 1\n1",
"output": "inf"
},
{
"input": "1000000000 999999998 1000000000 1\n0",
"output": "1"
},
{
"input": "1 0 2143435 4\n1 -123 -5453 141245",
"output": "inf"
},
{
"input": "-1000 0 100 1\n-1000",
"output": "0"
},
{
"input": "100 10 2 1\n100",
"output": "0"
},
{
"input": "-3 1 100 1\n3",
"output": "inf"
},
{
"input": "123 -1 10000 1\n123",
"output": "inf"
},
{
"input": "1 -1 2143435 4\n1 -123 -5453 141245",
"output": "inf"
},
{
"input": "5 1 5 5\n1 2 3 4 0",
"output": "inf"
},
{
"input": "-100 -1 1 1\n1",
"output": "0"
},
{
"input": "10 -1 3 2\n10 8",
"output": "0"
},
{
"input": "-10 0 5 1\n0",
"output": "0"
},
{
"input": "3 0 3 1\n0",
"output": "1"
},
{
"input": "2 0 2 1\n-1",
"output": "inf"
},
{
"input": "5 0 20 1\n5",
"output": "inf"
},
{
"input": "-4 1 1 1\n0",
"output": "0"
},
{
"input": "11 0 1111 1\n11",
"output": "inf"
},
{
"input": "2 0 3 1\n2",
"output": "inf"
},
{
"input": "-1 -1 2143435 4\n-1 -123 -5453 141245",
"output": "inf"
},
{
"input": "-100 0 50 1\n0",
"output": "0"
},
{
"input": "5 1 2 1\n2",
"output": "0"
},
{
"input": "3 0 3 1\n4",
"output": "inf"
},
{
"input": "0 23 3 1\n3",
"output": "inf"
},
{
"input": "-1000 0 100 1\n2",
"output": "0"
},
{
"input": "1 -1 10 1\n1",
"output": "inf"
}
] | 1,490,807,147
| 3,647
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 0
| 46
| 4,608,000
|
import math
b ,q , l ,m = map(int, input().split(' '))
a = list(map(int,input().split(' ')))
di = {}
for i in range(m):
di[a[i]] = 1
if q == 1 :
if b in di:
print(0)
elif b not in di:
print("inf")
elif q == 0:
if b in di and 0 in di:
print(0)
elif b in di and 0 not in di:
print(1)
else:
print("inf")
elif q == -1:
if b in di and -b in di:
print(0)
elif b in di or -b in di:
print(1)
else:
print(0)
else:
temp = b
count = 0
while(abs(temp)<=l):
if temp in di:
count = count
print(temp)
else:
count +=1
temp = temp*q
print(count)
|
Title: Masha and geometric depression
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Masha really loves algebra. On the last lesson, her strict teacher Dvastan gave she new exercise.
You are given geometric progression *b* defined by two integers *b*1 and *q*. Remind that a geometric progression is a sequence of integers *b*1,<=*b*2,<=*b*3,<=..., where for each *i*<=><=1 the respective term satisfies the condition *b**i*<==<=*b**i*<=-<=1·*q*, where *q* is called the common ratio of the progression. Progressions in Uzhlyandia are unusual: both *b*1 and *q* can equal 0. Also, Dvastan gave Masha *m* "bad" integers *a*1,<=*a*2,<=...,<=*a**m*, and an integer *l*.
Masha writes all progression terms one by one onto the board (including repetitive) while condition |*b**i*|<=≤<=*l* is satisfied (|*x*| means absolute value of *x*). There is an exception: if a term equals one of the "bad" integers, Masha skips it (doesn't write onto the board) and moves forward to the next term.
But the lesson is going to end soon, so Masha has to calculate how many integers will be written on the board. In order not to get into depression, Masha asked you for help: help her calculate how many numbers she will write, or print "inf" in case she needs to write infinitely many integers.
Input Specification:
The first line of input contains four integers *b*1, *q*, *l*, *m* (-109<=≤<=*b*1,<=*q*<=≤<=109, 1<=≤<=*l*<=≤<=109, 1<=≤<=*m*<=≤<=105) — the initial term and the common ratio of progression, absolute value of maximal number that can be written on the board and the number of "bad" integers, respectively.
The second line contains *m* distinct integers *a*1,<=*a*2,<=...,<=*a**m* (-109<=≤<=*a**i*<=≤<=109) — numbers that will never be written on the board.
Output Specification:
Print the only integer, meaning the number of progression terms that will be written on the board if it is finite, or "inf" (without quotes) otherwise.
Demo Input:
['3 2 30 4\n6 14 25 48\n', '123 1 2143435 4\n123 11 -5453 141245\n', '123 1 2143435 4\n54343 -13 6 124\n']
Demo Output:
['3', '0', 'inf']
Note:
In the first sample case, Masha will write integers 3, 12, 24. Progression term 6 will be skipped because it is a "bad" integer. Terms bigger than 24 won't be written because they exceed *l* by absolute value.
In the second case, Masha won't write any number because all terms are equal 123 and this is a "bad" integer.
In the third case, Masha will write infinitely integers 123.
|
```python
import math
b ,q , l ,m = map(int, input().split(' '))
a = list(map(int,input().split(' ')))
di = {}
for i in range(m):
di[a[i]] = 1
if q == 1 :
if b in di:
print(0)
elif b not in di:
print("inf")
elif q == 0:
if b in di and 0 in di:
print(0)
elif b in di and 0 not in di:
print(1)
else:
print("inf")
elif q == -1:
if b in di and -b in di:
print(0)
elif b in di or -b in di:
print(1)
else:
print(0)
else:
temp = b
count = 0
while(abs(temp)<=l):
if temp in di:
count = count
print(temp)
else:
count +=1
temp = temp*q
print(count)
```
| 0
|
|
614
|
A
|
Link/Cut Tree
|
PROGRAMMING
| 1,500
|
[
"brute force",
"implementation"
] | null | null |
Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, which is based on Splay trees. Specifically, he is now studying the *expose* procedure.
Unfortunately, Rostislav is unable to understand the definition of this procedure, so he decided to ask programmer Serezha to help him. Serezha agreed to help if Rostislav solves a simple task (and if he doesn't, then why would he need Splay trees anyway?)
Given integers *l*, *r* and *k*, you need to print all powers of number *k* within range from *l* to *r* inclusive. However, Rostislav doesn't want to spent time doing this, as he got interested in playing a network game called Agar with Gleb. Help him!
|
The first line of the input contains three space-separated integers *l*, *r* and *k* (1<=≤<=*l*<=≤<=*r*<=≤<=1018, 2<=≤<=*k*<=≤<=109).
|
Print all powers of number *k*, that lie within range from *l* to *r* in the increasing order. If there are no such numbers, print "-1" (without the quotes).
|
[
"1 10 2\n",
"2 4 5\n"
] |
[
"1 2 4 8 ",
"-1"
] |
Note to the first sample: numbers 2<sup class="upper-index">0</sup> = 1, 2<sup class="upper-index">1</sup> = 2, 2<sup class="upper-index">2</sup> = 4, 2<sup class="upper-index">3</sup> = 8 lie within the specified range. The number 2<sup class="upper-index">4</sup> = 16 is greater then 10, thus it shouldn't be printed.
| 500
|
[
{
"input": "1 10 2",
"output": "1 2 4 8 "
},
{
"input": "2 4 5",
"output": "-1"
},
{
"input": "18102 43332383920 28554",
"output": "28554 815330916 "
},
{
"input": "19562 31702689720 17701",
"output": "313325401 "
},
{
"input": "11729 55221128400 313",
"output": "97969 30664297 9597924961 "
},
{
"input": "5482 100347128000 342",
"output": "116964 40001688 13680577296 "
},
{
"input": "3680 37745933600 10",
"output": "10000 100000 1000000 10000000 100000000 1000000000 10000000000 "
},
{
"input": "17098 191120104800 43",
"output": "79507 3418801 147008443 6321363049 "
},
{
"input": "10462 418807699200 2",
"output": "16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368 68719476736 137438953472 274877906944 "
},
{
"input": "30061 641846400000 3",
"output": "59049 177147 531441 1594323 4782969 14348907 43046721 129140163 387420489 1162261467 3486784401 10460353203 31381059609 94143178827 282429536481 "
},
{
"input": "1 1000000000000000000 2",
"output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368 68719476736 137438953472 274877906944 549755813888 1099511627776 2199023255552 4398046511104 8796093022208 17592186044416 35184372088832 70368744177664 140737488355328 281474976710656 562949953421312 1125899906842624 2251799813685248 4503599627370496 900719925474099..."
},
{
"input": "32 2498039712000 4",
"output": "64 256 1024 4096 16384 65536 262144 1048576 4194304 16777216 67108864 268435456 1073741824 4294967296 17179869184 68719476736 274877906944 1099511627776 "
},
{
"input": "1 2576683920000 2",
"output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368 68719476736 137438953472 274877906944 549755813888 1099511627776 2199023255552 "
},
{
"input": "5 25 5",
"output": "5 25 "
},
{
"input": "1 90 90",
"output": "1 90 "
},
{
"input": "95 2200128528000 68",
"output": "4624 314432 21381376 1453933568 98867482624 "
},
{
"input": "64 426314644000 53",
"output": "2809 148877 7890481 418195493 22164361129 "
},
{
"input": "198765 198765 198765",
"output": "198765 "
},
{
"input": "42 2845016496000 12",
"output": "144 1728 20736 248832 2985984 35831808 429981696 5159780352 61917364224 743008370688 "
},
{
"input": "6 6 3",
"output": "-1"
},
{
"input": "1 10 11",
"output": "1 "
},
{
"input": "2 10 11",
"output": "-1"
},
{
"input": "87 160 41",
"output": "-1"
},
{
"input": "237171123124584251 923523399718980912 7150",
"output": "-1"
},
{
"input": "101021572000739548 453766043506276015 8898",
"output": "-1"
},
{
"input": "366070689449360724 928290634811046396 8230",
"output": "-1"
},
{
"input": "438133886369772308 942612870269666780 7193",
"output": "-1"
},
{
"input": "10 10 10",
"output": "10 "
},
{
"input": "16 16 256",
"output": "-1"
},
{
"input": "1 1000000000000000000 1000000000",
"output": "1 1000000000 1000000000000000000 "
},
{
"input": "1000000000000000000 1000000000000000000 1000000000",
"output": "1000000000000000000 "
},
{
"input": "1000000000 1000000000000000000 1000000000",
"output": "1000000000 1000000000000000000 "
},
{
"input": "1 1 4",
"output": "1 "
},
{
"input": "1 999999999999999999 1000000000",
"output": "1 1000000000 "
},
{
"input": "1 1000000000000000000 999999990",
"output": "1 999999990 999999980000000100 "
},
{
"input": "1 1000000000000000000 999999984",
"output": "1 999999984 999999968000000256 "
},
{
"input": "1 1000000000000000000 324325",
"output": "1 324325 105186705625 34114678301828125 "
},
{
"input": "1 1000000000000000000 999999523",
"output": "1 999999523 999999046000227529 "
},
{
"input": "1 243 3",
"output": "1 3 9 27 81 243 "
},
{
"input": "62769392426654367 567152589733560993 688813",
"output": "326816522793383797 "
},
{
"input": "1 1000000000000000000 690852001",
"output": "1 690852001 477276487285704001 "
},
{
"input": "1 1000000000000000000 918745157",
"output": "1 918745157 844092663510954649 "
},
{
"input": "1 1000000000000000000 131299843",
"output": "1 131299843 17239648771824649 "
},
{
"input": "2861381721051425 2861381721051425 1234",
"output": "-1"
}
] | 1,503,503,234
| 2,834
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 62
| 0
|
import sys
arr = list(map(int,input().split()))
l = arr[0]
r = arr[1]
k = arr[2]
if l == 1:
start = 1
elif l % k == 0:
start = k**(l//k)
else:
start = k**((l // k ) + 1)
if start < l or start > r:
print('-1')
else:
while(start <= r):
print(start,'',end='')
start = start*k
|
Title: Link/Cut Tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, which is based on Splay trees. Specifically, he is now studying the *expose* procedure.
Unfortunately, Rostislav is unable to understand the definition of this procedure, so he decided to ask programmer Serezha to help him. Serezha agreed to help if Rostislav solves a simple task (and if he doesn't, then why would he need Splay trees anyway?)
Given integers *l*, *r* and *k*, you need to print all powers of number *k* within range from *l* to *r* inclusive. However, Rostislav doesn't want to spent time doing this, as he got interested in playing a network game called Agar with Gleb. Help him!
Input Specification:
The first line of the input contains three space-separated integers *l*, *r* and *k* (1<=≤<=*l*<=≤<=*r*<=≤<=1018, 2<=≤<=*k*<=≤<=109).
Output Specification:
Print all powers of number *k*, that lie within range from *l* to *r* in the increasing order. If there are no such numbers, print "-1" (without the quotes).
Demo Input:
['1 10 2\n', '2 4 5\n']
Demo Output:
['1 2 4 8 ', '-1']
Note:
Note to the first sample: numbers 2<sup class="upper-index">0</sup> = 1, 2<sup class="upper-index">1</sup> = 2, 2<sup class="upper-index">2</sup> = 4, 2<sup class="upper-index">3</sup> = 8 lie within the specified range. The number 2<sup class="upper-index">4</sup> = 16 is greater then 10, thus it shouldn't be printed.
|
```python
import sys
arr = list(map(int,input().split()))
l = arr[0]
r = arr[1]
k = arr[2]
if l == 1:
start = 1
elif l % k == 0:
start = k**(l//k)
else:
start = k**((l // k ) + 1)
if start < l or start > r:
print('-1')
else:
while(start <= r):
print(start,'',end='')
start = start*k
```
| 0
|
|
915
|
A
|
Garden
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Luba thinks about watering her garden. The garden can be represented as a segment of length *k*. Luba has got *n* buckets, the *i*-th bucket allows her to water some continuous subsegment of garden of length exactly *a**i* each hour. Luba can't water any parts of the garden that were already watered, also she can't water the ground outside the garden.
Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length *a**i* if she chooses the *i*-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden.
See the examples for better understanding.
|
The first line of input contains two integer numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of buckets and the length of the garden, respectively.
The second line of input contains *n* integer numbers *a**i* (1<=≤<=*a**i*<=≤<=100) — the length of the segment that can be watered by the *i*-th bucket in one hour.
It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket.
|
Print one integer number — the minimum number of hours required to water the garden.
|
[
"3 6\n2 3 5\n",
"6 7\n1 2 3 4 5 6\n"
] |
[
"2\n",
"7\n"
] |
In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can't choose the bucket that allows to water the segment of length 5 because then we can't water the whole garden.
In the second test we can choose only the bucket that allows us to water the segment of length 1.
| 0
|
[
{
"input": "3 6\n2 3 5",
"output": "2"
},
{
"input": "6 7\n1 2 3 4 5 6",
"output": "7"
},
{
"input": "5 97\n1 10 50 97 2",
"output": "1"
},
{
"input": "5 97\n1 10 50 100 2",
"output": "97"
},
{
"input": "100 100\n2 46 24 18 86 90 31 38 84 49 58 28 15 80 14 24 87 56 62 87 41 87 55 71 87 32 41 56 91 32 24 75 43 42 35 30 72 53 31 26 54 61 87 85 36 75 44 31 7 38 77 57 61 54 70 77 45 96 39 57 11 8 91 42 52 15 42 30 92 41 27 26 34 27 3 80 32 86 26 97 63 91 30 75 14 7 19 23 45 11 8 43 44 73 11 56 3 55 63 16",
"output": "50"
},
{
"input": "100 91\n13 13 62 96 74 47 81 46 78 21 20 42 4 73 25 30 76 74 58 28 25 52 42 48 74 40 82 9 25 29 17 22 46 64 57 95 81 39 47 86 40 95 97 35 31 98 45 98 47 78 52 63 58 14 89 97 17 95 28 22 20 36 68 38 95 16 2 26 54 47 42 31 31 81 21 21 65 40 82 53 60 71 75 33 96 98 6 22 95 12 5 48 18 27 58 62 5 96 36 75",
"output": "7"
},
{
"input": "8 8\n8 7 6 5 4 3 2 1",
"output": "1"
},
{
"input": "3 8\n4 3 2",
"output": "2"
},
{
"input": "3 8\n2 4 2",
"output": "2"
},
{
"input": "3 6\n1 3 2",
"output": "2"
},
{
"input": "3 6\n3 2 5",
"output": "2"
},
{
"input": "3 8\n4 2 1",
"output": "2"
},
{
"input": "5 6\n2 3 5 1 2",
"output": "2"
},
{
"input": "2 6\n5 3",
"output": "2"
},
{
"input": "4 12\n6 4 3 1",
"output": "2"
},
{
"input": "3 18\n1 9 6",
"output": "2"
},
{
"input": "3 9\n3 2 1",
"output": "3"
},
{
"input": "3 6\n5 3 2",
"output": "2"
},
{
"input": "2 10\n5 2",
"output": "2"
},
{
"input": "2 18\n6 3",
"output": "3"
},
{
"input": "4 12\n1 2 12 3",
"output": "1"
},
{
"input": "3 7\n3 2 1",
"output": "7"
},
{
"input": "3 6\n3 2 1",
"output": "2"
},
{
"input": "5 10\n5 4 3 2 1",
"output": "2"
},
{
"input": "5 16\n8 4 2 1 7",
"output": "2"
},
{
"input": "6 7\n6 5 4 3 7 1",
"output": "1"
},
{
"input": "2 6\n3 2",
"output": "2"
},
{
"input": "2 4\n4 1",
"output": "1"
},
{
"input": "6 8\n2 4 1 3 5 7",
"output": "2"
},
{
"input": "6 8\n6 5 4 3 2 1",
"output": "2"
},
{
"input": "6 15\n5 2 3 6 4 3",
"output": "3"
},
{
"input": "4 8\n2 4 8 1",
"output": "1"
},
{
"input": "2 5\n5 1",
"output": "1"
},
{
"input": "4 18\n3 1 1 2",
"output": "6"
},
{
"input": "2 1\n2 1",
"output": "1"
},
{
"input": "3 10\n2 10 5",
"output": "1"
},
{
"input": "5 12\n12 4 4 4 3",
"output": "1"
},
{
"input": "3 6\n6 3 2",
"output": "1"
},
{
"input": "2 2\n2 1",
"output": "1"
},
{
"input": "3 18\n1 9 3",
"output": "2"
},
{
"input": "3 8\n7 2 4",
"output": "2"
},
{
"input": "2 100\n99 1",
"output": "100"
},
{
"input": "4 12\n1 3 4 2",
"output": "3"
},
{
"input": "3 6\n2 3 1",
"output": "2"
},
{
"input": "4 6\n3 2 5 12",
"output": "2"
},
{
"input": "4 97\n97 1 50 10",
"output": "1"
},
{
"input": "3 12\n1 12 2",
"output": "1"
},
{
"input": "4 12\n1 4 3 2",
"output": "3"
},
{
"input": "1 1\n1",
"output": "1"
},
{
"input": "3 19\n7 1 1",
"output": "19"
},
{
"input": "5 12\n12 4 3 4 4",
"output": "1"
},
{
"input": "3 8\n8 4 2",
"output": "1"
},
{
"input": "3 3\n3 2 1",
"output": "1"
},
{
"input": "5 6\n3 2 4 2 2",
"output": "2"
},
{
"input": "2 16\n8 4",
"output": "2"
},
{
"input": "3 6\n10 2 3",
"output": "2"
},
{
"input": "5 3\n2 4 5 3 6",
"output": "1"
},
{
"input": "11 99\n1 2 3 6 5 4 7 8 99 33 66",
"output": "1"
},
{
"input": "3 12\n3 12 2",
"output": "1"
},
{
"input": "5 25\n24 5 15 25 23",
"output": "1"
},
{
"input": "2 4\n8 1",
"output": "4"
},
{
"input": "4 100\n2 50 4 1",
"output": "2"
},
{
"input": "3 28\n7 14 1",
"output": "2"
},
{
"input": "4 8\n2 8 4 1",
"output": "1"
},
{
"input": "4 6\n6 1 2 3",
"output": "1"
},
{
"input": "2 12\n4 3",
"output": "3"
},
{
"input": "4 12\n1 2 4 3",
"output": "3"
},
{
"input": "5 12\n2 3 12 6 4",
"output": "1"
},
{
"input": "4 4\n1 2 2 4",
"output": "1"
},
{
"input": "3 6\n2 3 2",
"output": "2"
},
{
"input": "4 21\n21 20 21 2",
"output": "1"
},
{
"input": "3 8\n3 4 2",
"output": "2"
},
{
"input": "1 25\n25",
"output": "1"
},
{
"input": "99 12\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",
"output": "1"
},
{
"input": "98 12\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",
"output": "1"
},
{
"input": "79 12\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",
"output": "1"
},
{
"input": "4 32\n1 1 1 1",
"output": "32"
},
{
"input": "1 100\n1",
"output": "100"
},
{
"input": "2 100\n7 1",
"output": "100"
},
{
"input": "7 24\n1 3 6 4 5 2 7",
"output": "4"
},
{
"input": "6 87\n1 2 8 4 5 7",
"output": "87"
},
{
"input": "1 88\n1",
"output": "88"
},
{
"input": "1 89\n1",
"output": "89"
}
] | 1,593,244,044
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 93
| 6,656,000
|
def sol():
n,k=map(int,input().split())
A=[int(i)for i in input().split()]
p=0
for i in range(2,k//2+1):
if(k%i==0 and i in A):
p=i
print(k//p if p>0 else k)
if(__name__=='__main__'):
sol()
|
Title: Garden
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luba thinks about watering her garden. The garden can be represented as a segment of length *k*. Luba has got *n* buckets, the *i*-th bucket allows her to water some continuous subsegment of garden of length exactly *a**i* each hour. Luba can't water any parts of the garden that were already watered, also she can't water the ground outside the garden.
Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length *a**i* if she chooses the *i*-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden.
See the examples for better understanding.
Input Specification:
The first line of input contains two integer numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=100) — the number of buckets and the length of the garden, respectively.
The second line of input contains *n* integer numbers *a**i* (1<=≤<=*a**i*<=≤<=100) — the length of the segment that can be watered by the *i*-th bucket in one hour.
It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket.
Output Specification:
Print one integer number — the minimum number of hours required to water the garden.
Demo Input:
['3 6\n2 3 5\n', '6 7\n1 2 3 4 5 6\n']
Demo Output:
['2\n', '7\n']
Note:
In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can't choose the bucket that allows to water the segment of length 5 because then we can't water the whole garden.
In the second test we can choose only the bucket that allows us to water the segment of length 1.
|
```python
def sol():
n,k=map(int,input().split())
A=[int(i)for i in input().split()]
p=0
for i in range(2,k//2+1):
if(k%i==0 and i in A):
p=i
print(k//p if p>0 else k)
if(__name__=='__main__'):
sol()
```
| 0
|
|
834
|
B
|
The Festive Evening
|
PROGRAMMING
| 1,100
|
[
"data structures",
"implementation"
] | null | null |
It's the end of July – the time when a festive evening is held at Jelly Castle! Guests from all over the kingdom gather here to discuss new trends in the world of confectionery. Yet some of the things discussed here are not supposed to be disclosed to the general public: the information can cause discord in the kingdom of Sweetland in case it turns out to reach the wrong hands. So it's a necessity to not let any uninvited guests in.
There are 26 entrances in Jelly Castle, enumerated with uppercase English letters from A to Z. Because of security measures, each guest is known to be assigned an entrance he should enter the castle through. The door of each entrance is opened right before the first guest's arrival and closed right after the arrival of the last guest that should enter the castle through this entrance. No two guests can enter the castle simultaneously.
For an entrance to be protected from possible intrusion, a candy guard should be assigned to it. There are *k* such guards in the castle, so if there are more than *k* opened doors, one of them is going to be left unguarded! Notice that a guard can't leave his post until the door he is assigned to is closed.
Slastyona had a suspicion that there could be uninvited guests at the evening. She knows the order in which the invited guests entered the castle, and wants you to help her check whether there was a moment when more than *k* doors were opened.
|
Two integers are given in the first string: the number of guests *n* and the number of guards *k* (1<=≤<=*n*<=≤<=106, 1<=≤<=*k*<=≤<=26).
In the second string, *n* uppercase English letters *s*1*s*2... *s**n* are given, where *s**i* is the entrance used by the *i*-th guest.
|
Output «YES» if at least one door was unguarded during some time, and «NO» otherwise.
You can output each letter in arbitrary case (upper or lower).
|
[
"5 1\nAABBB\n",
"5 1\nABABB\n"
] |
[
"NO\n",
"YES\n"
] |
In the first sample case, the door A is opened right before the first guest's arrival and closed when the second guest enters the castle. The door B is opened right before the arrival of the third guest, and closed after the fifth one arrives. One guard can handle both doors, as the first one is closed before the second one is opened.
In the second sample case, the door B is opened before the second guest's arrival, but the only guard can't leave the door A unattended, as there is still one more guest that should enter the castle through this door.
| 1,000
|
[
{
"input": "5 1\nAABBB",
"output": "NO"
},
{
"input": "5 1\nABABB",
"output": "YES"
},
{
"input": "26 1\nABCDEFGHIJKLMNOPQRSTUVWXYZ",
"output": "NO"
},
{
"input": "27 1\nABCDEFGHIJKLMNOPQRSTUVWXYZA",
"output": "YES"
},
{
"input": "5 2\nABACA",
"output": "NO"
},
{
"input": "6 2\nABCABC",
"output": "YES"
},
{
"input": "8 3\nABCBCDCA",
"output": "NO"
},
{
"input": "73 2\nDEBECECBBADAADEAABEAEEEAEBEAEBCDDBABBAEBACCBEEBBAEADEECACEDEEDABACDCDBBBD",
"output": "YES"
},
{
"input": "44 15\nHGJIFCGGCDGIJDHBIBGAEABCIABIGBDEADBBBAGDFDHA",
"output": "NO"
},
{
"input": "41 19\nTMEYYIIELFDCMBDKWWKYNRNDUPRONYROXQCLVQALP",
"output": "NO"
},
{
"input": "377 3\nEADADBBBBDEAABBAEBABACDBDBBCACAADBEAEACDEAABACADEEDEACACDADABBBBDDEECBDABACACBAECBADAEBDEEBDBCDAEADBCDDACACDCCEEDBCCBBCEDBECBABCDDBBDEADEDAEACDECECBEBACBCCDCDBDAECDECADBCBEDBBDAAEBCAAECCDCCDBDDEBADEEBDCAEABBDEDBBDDEAECCBDDCDEACDAECCBDDABABEAEDCDEDBAECBDEACEBCECEACDCBABCBAAEAADACADBBBBABEADBCADEBCBECCABBDDDEEBCDEBADEBDAAABBEABADEDEAEABCEEBEEDEAEBEABCEDDBACBCCADEBAAAAAEABABBCE",
"output": "YES"
},
{
"input": "433 3\nFZDDHMJGBZCHFUXBBPIEBBEFDWOMXXEPOMDGSMPIUZOMRZQNSJAVNATGIWPDFISKFQXJNVFXPHOZDAEZFDAHDXXQKZMGNSGKQNWGNGJGJZVVITKNFLVCPMZSDMCHBTVAWYVZLIXXIADXNYILEYNIQHKMOGMVOCWGHCWIYMPEPADSJAAKEGTUSEDWAHMNYJDIHBKHVUHLYGNGZDBULRXLSAJHPCMNWCEAAPYMHDTYWPADOTJTXTXUKLCHWKUSZRHEKQEFPVJEJJHRWCKYOIWALRTIBUMNOCRXLSIKQCJVQXEPGOHRUDJDKMUUUDORURWXJNVRVMNOUNRFKSVMTMZGOIJLXEPAMVGESOADYIGZXRBJDIWKNOWTCSROAQTBECHTOZVSQUOOJRZIBAUHMKAXDCIMDZJFMABGRNTGPUJAUNFPFWCJG",
"output": "YES"
},
{
"input": "5 2\nABCAB",
"output": "YES"
},
{
"input": "5 1\nAZAZA",
"output": "YES"
},
{
"input": "7 2\nABCDBCD",
"output": "YES"
},
{
"input": "3 26\nAAB",
"output": "NO"
}
] | 1,598,053,004
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 5
| 1,000
| 9,318,400
|
from sys import stdin
def unguarded_door(n, k, entrances):
first_last = {}
for i in range(n):
entrance = entrances[i]
if entrance in first_last:
first_last[entrance][1] = i
else:
first_last[entrance] = [i, i]
for i in range(n):
guarded_doors = filter(
lambda indices: indices[0] <= i <= indices[1], first_last.values()
)
nb_guarded_doors = len(list(guarded_doors))
if nb_guarded_doors > k:
return "YES"
return "NO"
if __name__ == "__main__":
n, k = [int(c) for c in stdin.readline().split(" ")]
entrances = stdin.readline()
print(unguarded_door(n, k, entrances))
|
Title: The Festive Evening
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's the end of July – the time when a festive evening is held at Jelly Castle! Guests from all over the kingdom gather here to discuss new trends in the world of confectionery. Yet some of the things discussed here are not supposed to be disclosed to the general public: the information can cause discord in the kingdom of Sweetland in case it turns out to reach the wrong hands. So it's a necessity to not let any uninvited guests in.
There are 26 entrances in Jelly Castle, enumerated with uppercase English letters from A to Z. Because of security measures, each guest is known to be assigned an entrance he should enter the castle through. The door of each entrance is opened right before the first guest's arrival and closed right after the arrival of the last guest that should enter the castle through this entrance. No two guests can enter the castle simultaneously.
For an entrance to be protected from possible intrusion, a candy guard should be assigned to it. There are *k* such guards in the castle, so if there are more than *k* opened doors, one of them is going to be left unguarded! Notice that a guard can't leave his post until the door he is assigned to is closed.
Slastyona had a suspicion that there could be uninvited guests at the evening. She knows the order in which the invited guests entered the castle, and wants you to help her check whether there was a moment when more than *k* doors were opened.
Input Specification:
Two integers are given in the first string: the number of guests *n* and the number of guards *k* (1<=≤<=*n*<=≤<=106, 1<=≤<=*k*<=≤<=26).
In the second string, *n* uppercase English letters *s*1*s*2... *s**n* are given, where *s**i* is the entrance used by the *i*-th guest.
Output Specification:
Output «YES» if at least one door was unguarded during some time, and «NO» otherwise.
You can output each letter in arbitrary case (upper or lower).
Demo Input:
['5 1\nAABBB\n', '5 1\nABABB\n']
Demo Output:
['NO\n', 'YES\n']
Note:
In the first sample case, the door A is opened right before the first guest's arrival and closed when the second guest enters the castle. The door B is opened right before the arrival of the third guest, and closed after the fifth one arrives. One guard can handle both doors, as the first one is closed before the second one is opened.
In the second sample case, the door B is opened before the second guest's arrival, but the only guard can't leave the door A unattended, as there is still one more guest that should enter the castle through this door.
|
```python
from sys import stdin
def unguarded_door(n, k, entrances):
first_last = {}
for i in range(n):
entrance = entrances[i]
if entrance in first_last:
first_last[entrance][1] = i
else:
first_last[entrance] = [i, i]
for i in range(n):
guarded_doors = filter(
lambda indices: indices[0] <= i <= indices[1], first_last.values()
)
nb_guarded_doors = len(list(guarded_doors))
if nb_guarded_doors > k:
return "YES"
return "NO"
if __name__ == "__main__":
n, k = [int(c) for c in stdin.readline().split(" ")]
entrances = stdin.readline()
print(unguarded_door(n, k, entrances))
```
| 0
|
|
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"
}
] | 1,645,896,272
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
ilk=list(input())
ikinci=(input())
ilk.reverse()
kıyas=""
for n in ilk:
kıyas=kıyas+n
if kıyas==ikinci:
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
ilk=list(input())
ikinci=(input())
ilk.reverse()
kıyas=""
for n in ilk:
kıyas=kıyas+n
if kıyas==ikinci:
print("YES")
else:
print("NO")
```
| -1
|
652
|
B
|
z-sort
|
PROGRAMMING
| 1,000
|
[
"sortings"
] | null | null |
A student of *z*-school found a kind of sorting called *z*-sort. The array *a* with *n* elements are *z*-sorted if two conditions hold:
1. *a**i*<=≥<=*a**i*<=-<=1 for all even *i*, 1. *a**i*<=≤<=*a**i*<=-<=1 for all odd *i*<=><=1.
For example the arrays [1,2,1,2] and [1,1,1,1] are *z*-sorted while the array [1,2,3,4] isn’t *z*-sorted.
Can you make the array *z*-sorted?
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the array *a*.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*.
|
If it's possible to make the array *a* *z*-sorted print *n* space separated integers *a**i* — the elements after *z*-sort. Otherwise print the only word "Impossible".
|
[
"4\n1 2 2 1\n",
"5\n1 3 2 2 5\n"
] |
[
"1 2 1 2\n",
"1 5 2 3 2\n"
] |
none
| 0
|
[
{
"input": "4\n1 2 2 1",
"output": "1 2 1 2"
},
{
"input": "5\n1 3 2 2 5",
"output": "1 5 2 3 2"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "1 1 1 1 1 1 1 1 1 1"
},
{
"input": "10\n1 9 7 6 2 4 7 8 1 3",
"output": "1 9 1 8 2 7 3 7 4 6"
},
{
"input": "100\n82 51 81 14 37 17 78 92 64 15 8 86 89 8 87 77 66 10 15 12 100 25 92 47 21 78 20 63 13 49 41 36 41 79 16 87 87 69 3 76 80 60 100 49 70 59 72 8 38 71 45 97 71 14 76 54 81 4 59 46 39 29 92 3 49 22 53 99 59 52 74 31 92 43 42 23 44 9 82 47 7 40 12 9 3 55 37 85 46 22 84 52 98 41 21 77 63 17 62 91",
"output": "3 100 3 100 3 99 4 98 7 97 8 92 8 92 8 92 9 92 9 91 10 89 12 87 12 87 13 87 14 86 14 85 15 84 15 82 16 82 17 81 17 81 20 80 21 79 21 78 22 78 22 77 23 77 25 76 29 76 31 74 36 72 37 71 37 71 38 70 39 69 40 66 41 64 41 63 41 63 42 62 43 60 44 59 45 59 46 59 46 55 47 54 47 53 49 52 49 52 49 51"
},
{
"input": "3\n1 2 6",
"output": "1 6 2"
},
{
"input": "136\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 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 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 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"
},
{
"input": "3\n1 2 3",
"output": "1 3 2"
},
{
"input": "7\n999999998 999999999 999999999 999999999 999999999 999999999 1000000000",
"output": "999999998 1000000000 999999999 999999999 999999999 999999999 999999999"
},
{
"input": "3\n100 1 2",
"output": "1 100 2"
}
] | 1,692,457,008
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 15
| 0
|
def insert(arr):
if len(arr) == 0:
return arr
cnt_1 = 1
for j in range(len(arr) - 1):
for i in range(cnt_1):
if arr[i] > arr[cnt_1]:
cnt_1 - 1 > i
number = arr[i]
arr[i] = arr[cnt_1]
for j in range(cnt_1 - i):
arr[cnt_1 - j] = arr[cnt_1 - 1 - j]
arr[i + 1] = number
cnt_1 += 1
if j == len(arr) - 2:
return arr
n = int(input())
arr = list(map(int, input().split()))
ans = insert(arr)
print(*ans)
|
Title: z-sort
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A student of *z*-school found a kind of sorting called *z*-sort. The array *a* with *n* elements are *z*-sorted if two conditions hold:
1. *a**i*<=≥<=*a**i*<=-<=1 for all even *i*, 1. *a**i*<=≤<=*a**i*<=-<=1 for all odd *i*<=><=1.
For example the arrays [1,2,1,2] and [1,1,1,1] are *z*-sorted while the array [1,2,3,4] isn’t *z*-sorted.
Can you make the array *z*-sorted?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of elements in the array *a*.
The second line contains *n* integers *a**i* (1<=≤<=*a**i*<=≤<=109) — the elements of the array *a*.
Output Specification:
If it's possible to make the array *a* *z*-sorted print *n* space separated integers *a**i* — the elements after *z*-sort. Otherwise print the only word "Impossible".
Demo Input:
['4\n1 2 2 1\n', '5\n1 3 2 2 5\n']
Demo Output:
['1 2 1 2\n', '1 5 2 3 2\n']
Note:
none
|
```python
def insert(arr):
if len(arr) == 0:
return arr
cnt_1 = 1
for j in range(len(arr) - 1):
for i in range(cnt_1):
if arr[i] > arr[cnt_1]:
cnt_1 - 1 > i
number = arr[i]
arr[i] = arr[cnt_1]
for j in range(cnt_1 - i):
arr[cnt_1 - j] = arr[cnt_1 - 1 - j]
arr[i + 1] = number
cnt_1 += 1
if j == len(arr) - 2:
return arr
n = int(input())
arr = list(map(int, input().split()))
ans = insert(arr)
print(*ans)
```
| -1
|
|
394
|
D
|
Physical Education and Buns
|
PROGRAMMING
| 0
|
[
"brute force",
"implementation",
"math"
] | null | null |
The Physical education teacher at SESC is a sort of mathematician too. His most favorite topic in mathematics is progressions. That is why the teacher wants the students lined up in non-decreasing height form an arithmetic progression.
To achieve the goal, the gym teacher ordered a lot of magical buns from the dining room. The magic buns come in two types: when a student eats one magic bun of the first type, his height increases by one, when the student eats one magical bun of the second type, his height decreases by one. The physical education teacher, as expected, cares about the health of his students, so he does not want them to eat a lot of buns. More precisely, he wants the maximum number of buns eaten by some student to be minimum.
Help the teacher, get the maximum number of buns that some pupils will have to eat to achieve the goal of the teacher. Also, get one of the possible ways for achieving the objective, namely, the height of the lowest student in the end and the step of the resulting progression.
|
The single line contains integer *n* (2<=≤<=*n*<=≤<=103) — the number of students. The second line contains *n* space-separated integers — the heights of all students. The height of one student is an integer which absolute value doesn't exceed 104.
|
In the first line print the maximum number of buns eaten by some student to achieve the teacher's aim. In the second line, print two space-separated integers — the height of the lowest student in the end and the step of the progression. Please, pay attention that the step should be non-negative.
If there are multiple possible answers, you can print any of them.
|
[
"5\n-3 -4 -2 -3 3\n",
"5\n2 -3 -1 -4 3\n"
] |
[
"2\n-3 1\n",
"1\n-4 2\n"
] |
Lets look at the first sample. We can proceed in the following manner:
- don't feed the 1-st student, his height will stay equal to -3; - give two buns of the first type to the 2-nd student, his height become equal to -2; - give two buns of the first type to the 3-rd student, his height become equal to 0; - give two buns of the first type to the 4-th student, his height become equal to -1; - give two buns of the second type to the 5-th student, his height become equal to 1.
To sum it up, when the students line up in non-decreasing height it will be an arithmetic progression: -3, -2, -1, 0, 1. The height of the lowest student is equal to -3, the step of the progression is equal to 1. The maximum number of buns eaten by one student is equal to 2.
| 2,000
|
[
{
"input": "5\n-3 -4 -2 -3 3",
"output": "2\n-3 1"
},
{
"input": "5\n2 -3 -1 -4 3",
"output": "1\n-4 2"
},
{
"input": "6\n94 65 -33 -43 60 -24",
"output": "25\n-67 34"
},
{
"input": "3\n-10000 10000 -10000",
"output": "5000\n-15000 10000"
},
{
"input": "2\n0 0",
"output": "0\n0 0"
},
{
"input": "7\n-1 -2 -4 -10 6 6 5",
"output": "3\n-9 3"
},
{
"input": "10\n-10 3 -16 -15 14 -16 13 -6 -8 18",
"output": "6\n-20 4"
},
{
"input": "50\n-67 -84 -89 80 40 42 -38 30 74 -12 -66 27 1 11 -45 -44 2 -70 -59 -70 -59 -59 62 100 -5 1 91 79 47 -64 -51 -88 -5 37 82 87 79 46 76 47 60 57 59 -24 47 -49 -63 24 -84 -54",
"output": "15\n-97 4"
},
{
"input": "100\n246 485 -940 -186 -841 -98 711 429 -154 164 -244 -111 886 -447 22 480 224 -132 927 812 -243 -152 -843 403 -320 -346 -407 827 645 -903 -172 540 -359 498 270 284 374 -52 -983 -164 -707 -242 -159 -825 -889 661 -629 212 849 -891 -622 810 957 897 -96 -293 -257 822 690 369 -914 212 -338 -928 -862 525 -537 782 727 665 964 -559 -675 -835 -800 254 -522 -504 239 909 638 -589 -700 907 127 -77 -748 999 152 -253 -505 889 -967 -481 -312 161 28 258 118 -870",
"output": "83\n-1065 21"
},
{
"input": "2\n-9116 9298",
"output": "0\n-9116 18414"
},
{
"input": "2\n-10000 10000",
"output": "0\n-10000 20000"
}
] | 1,520,024,103
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 23
| 139
| 5,632,000
|
q = 10001
n, a = int(input()), list(map(int, input().split()))
a.sort()
for i in range(40000 // (n - 1) + 1):
b = [a[j] - j * i for j in range(n)]
u, v = max(b), min(b)
p = (u - v + 1) // 2
if p < q: q, s, d = p, v + p, i
print(q)
print(s, d)
|
Title: Physical Education and Buns
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Physical education teacher at SESC is a sort of mathematician too. His most favorite topic in mathematics is progressions. That is why the teacher wants the students lined up in non-decreasing height form an arithmetic progression.
To achieve the goal, the gym teacher ordered a lot of magical buns from the dining room. The magic buns come in two types: when a student eats one magic bun of the first type, his height increases by one, when the student eats one magical bun of the second type, his height decreases by one. The physical education teacher, as expected, cares about the health of his students, so he does not want them to eat a lot of buns. More precisely, he wants the maximum number of buns eaten by some student to be minimum.
Help the teacher, get the maximum number of buns that some pupils will have to eat to achieve the goal of the teacher. Also, get one of the possible ways for achieving the objective, namely, the height of the lowest student in the end and the step of the resulting progression.
Input Specification:
The single line contains integer *n* (2<=≤<=*n*<=≤<=103) — the number of students. The second line contains *n* space-separated integers — the heights of all students. The height of one student is an integer which absolute value doesn't exceed 104.
Output Specification:
In the first line print the maximum number of buns eaten by some student to achieve the teacher's aim. In the second line, print two space-separated integers — the height of the lowest student in the end and the step of the progression. Please, pay attention that the step should be non-negative.
If there are multiple possible answers, you can print any of them.
Demo Input:
['5\n-3 -4 -2 -3 3\n', '5\n2 -3 -1 -4 3\n']
Demo Output:
['2\n-3 1\n', '1\n-4 2\n']
Note:
Lets look at the first sample. We can proceed in the following manner:
- don't feed the 1-st student, his height will stay equal to -3; - give two buns of the first type to the 2-nd student, his height become equal to -2; - give two buns of the first type to the 3-rd student, his height become equal to 0; - give two buns of the first type to the 4-th student, his height become equal to -1; - give two buns of the second type to the 5-th student, his height become equal to 1.
To sum it up, when the students line up in non-decreasing height it will be an arithmetic progression: -3, -2, -1, 0, 1. The height of the lowest student is equal to -3, the step of the progression is equal to 1. The maximum number of buns eaten by one student is equal to 2.
|
```python
q = 10001
n, a = int(input()), list(map(int, input().split()))
a.sort()
for i in range(40000 // (n - 1) + 1):
b = [a[j] - j * i for j in range(n)]
u, v = max(b), min(b)
p = (u - v + 1) // 2
if p < q: q, s, d = p, v + p, i
print(q)
print(s, d)
```
| 3
|
|
451
|
A
|
Game With Sticks
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
After winning gold and silver in IOI 2014, Akshat and Malvika want to have some fun. Now they are playing a game on a grid made of *n* horizontal and *m* vertical sticks.
An intersection point is any point on the grid which is formed by the intersection of one horizontal stick and one vertical stick.
In the grid shown below, *n*<==<=3 and *m*<==<=3. There are *n*<=+<=*m*<==<=6 sticks in total (horizontal sticks are shown in red and vertical sticks are shown in green). There are *n*·*m*<==<=9 intersection points, numbered from 1 to 9.
The rules of the game are very simple. The players move in turns. Akshat won gold, so he makes the first move. During his/her move, a player must choose any remaining intersection point and remove from the grid all sticks which pass through this point. A player will lose the game if he/she cannot make a move (i.e. there are no intersection points remaining on the grid at his/her move).
Assume that both players play optimally. Who will win the game?
|
The first line of input contains two space-separated integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
|
Print a single line containing "Akshat" or "Malvika" (without the quotes), depending on the winner of the game.
|
[
"2 2\n",
"2 3\n",
"3 3\n"
] |
[
"Malvika\n",
"Malvika\n",
"Akshat\n"
] |
Explanation of the first sample:
The grid has four intersection points, numbered from 1 to 4.
If Akshat chooses intersection point 1, then he will remove two sticks (1 - 2 and 1 - 3). The resulting grid will look like this.
Now there is only one remaining intersection point (i.e. 4). Malvika must choose it and remove both remaining sticks. After her move the grid will be empty.
In the empty grid, Akshat cannot make any move, hence he will lose.
Since all 4 intersection points of the grid are equivalent, Akshat will lose no matter which one he picks.
| 500
|
[
{
"input": "2 2",
"output": "Malvika"
},
{
"input": "2 3",
"output": "Malvika"
},
{
"input": "3 3",
"output": "Akshat"
},
{
"input": "20 68",
"output": "Malvika"
},
{
"input": "1 1",
"output": "Akshat"
},
{
"input": "1 2",
"output": "Akshat"
},
{
"input": "1 3",
"output": "Akshat"
},
{
"input": "2 1",
"output": "Akshat"
},
{
"input": "2 2",
"output": "Malvika"
},
{
"input": "3 1",
"output": "Akshat"
},
{
"input": "3 2",
"output": "Malvika"
},
{
"input": "68 42",
"output": "Malvika"
},
{
"input": "1 35",
"output": "Akshat"
},
{
"input": "25 70",
"output": "Akshat"
},
{
"input": "59 79",
"output": "Akshat"
},
{
"input": "65 63",
"output": "Akshat"
},
{
"input": "46 6",
"output": "Malvika"
},
{
"input": "28 82",
"output": "Malvika"
},
{
"input": "98 98",
"output": "Malvika"
},
{
"input": "98 99",
"output": "Malvika"
},
{
"input": "98 100",
"output": "Malvika"
},
{
"input": "99 98",
"output": "Malvika"
},
{
"input": "99 99",
"output": "Akshat"
},
{
"input": "99 100",
"output": "Akshat"
},
{
"input": "100 98",
"output": "Malvika"
},
{
"input": "100 99",
"output": "Akshat"
},
{
"input": "100 100",
"output": "Malvika"
},
{
"input": "3 4",
"output": "Akshat"
}
] | 1,679,778,644
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 31
| 0
|
n=int(input())
m=int(input())
if n < m:
if n % 2 ==0:
print ("Malvika")
else:
print ("Akshat")
elif m < n:
if m % 2 ==0:
print ("Malvika")
else:
print ("Akshat")
else:
if m % 2 == 0:
print ("Malvika")
else:
print ("Akshat")
|
Title: Game With Sticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After winning gold and silver in IOI 2014, Akshat and Malvika want to have some fun. Now they are playing a game on a grid made of *n* horizontal and *m* vertical sticks.
An intersection point is any point on the grid which is formed by the intersection of one horizontal stick and one vertical stick.
In the grid shown below, *n*<==<=3 and *m*<==<=3. There are *n*<=+<=*m*<==<=6 sticks in total (horizontal sticks are shown in red and vertical sticks are shown in green). There are *n*·*m*<==<=9 intersection points, numbered from 1 to 9.
The rules of the game are very simple. The players move in turns. Akshat won gold, so he makes the first move. During his/her move, a player must choose any remaining intersection point and remove from the grid all sticks which pass through this point. A player will lose the game if he/she cannot make a move (i.e. there are no intersection points remaining on the grid at his/her move).
Assume that both players play optimally. Who will win the game?
Input Specification:
The first line of input contains two space-separated integers, *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=100).
Output Specification:
Print a single line containing "Akshat" or "Malvika" (without the quotes), depending on the winner of the game.
Demo Input:
['2 2\n', '2 3\n', '3 3\n']
Demo Output:
['Malvika\n', 'Malvika\n', 'Akshat\n']
Note:
Explanation of the first sample:
The grid has four intersection points, numbered from 1 to 4.
If Akshat chooses intersection point 1, then he will remove two sticks (1 - 2 and 1 - 3). The resulting grid will look like this.
Now there is only one remaining intersection point (i.e. 4). Malvika must choose it and remove both remaining sticks. After her move the grid will be empty.
In the empty grid, Akshat cannot make any move, hence he will lose.
Since all 4 intersection points of the grid are equivalent, Akshat will lose no matter which one he picks.
|
```python
n=int(input())
m=int(input())
if n < m:
if n % 2 ==0:
print ("Malvika")
else:
print ("Akshat")
elif m < n:
if m % 2 ==0:
print ("Malvika")
else:
print ("Akshat")
else:
if m % 2 == 0:
print ("Malvika")
else:
print ("Akshat")
```
| -1
|
|
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"
}
] | 1,699,068,250
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 2
| 1,000
| 1,228,800
|
n = int(input())
somapar = 0
somaimpar = 0
tamanho = n//2 + 1
for i in range(1,tamanho):
somapar += 2*i
somaimpar += 2*i-1
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())
somapar = 0
somaimpar = 0
tamanho = n//2 + 1
for i in range(1,tamanho):
somapar += 2*i
somaimpar += 2*i-1
soma = somapar - somaimpar
if n%2!=0:
soma -= n
print(soma)
```
| 0
|
|
485
|
A
|
Factory
|
PROGRAMMING
| 1,400
|
[
"implementation",
"math",
"matrices"
] | null | null |
One industrial factory is reforming working plan. The director suggested to set a mythical detail production norm. If at the beginning of the day there were *x* details in the factory storage, then by the end of the day the factory has to produce (remainder after dividing *x* by *m*) more details. Unfortunately, no customer has ever bought any mythical detail, so all the details produced stay on the factory.
The board of directors are worried that the production by the given plan may eventually stop (that means that there will be а moment when the current number of details on the factory is divisible by *m*).
Given the number of details *a* on the first day and number *m* check if the production stops at some moment.
|
The first line contains two integers *a* and *m* (1<=≤<=*a*,<=*m*<=≤<=105).
|
Print "Yes" (without quotes) if the production will eventually stop, otherwise print "No".
|
[
"1 5\n",
"3 6\n"
] |
[
"No\n",
"Yes\n"
] |
none
| 1,000
|
[
{
"input": "1 5",
"output": "No"
},
{
"input": "3 6",
"output": "Yes"
},
{
"input": "1 8",
"output": "Yes"
},
{
"input": "2 3",
"output": "No"
},
{
"input": "3 24",
"output": "Yes"
},
{
"input": "1 1",
"output": "Yes"
},
{
"input": "100000 100000",
"output": "Yes"
},
{
"input": "1 99989",
"output": "No"
},
{
"input": "512 2",
"output": "Yes"
},
{
"input": "100 24",
"output": "No"
},
{
"input": "1 100000",
"output": "No"
},
{
"input": "100000 1",
"output": "Yes"
},
{
"input": "3 99929",
"output": "No"
},
{
"input": "99961 99971",
"output": "No"
},
{
"input": "1 65536",
"output": "Yes"
},
{
"input": "4 65536",
"output": "Yes"
},
{
"input": "3 65536",
"output": "Yes"
},
{
"input": "32768 65536",
"output": "Yes"
},
{
"input": "65535 65536",
"output": "Yes"
},
{
"input": "1 65535",
"output": "No"
},
{
"input": "98812 100000",
"output": "No"
},
{
"input": "10 5",
"output": "Yes"
},
{
"input": "6 8",
"output": "Yes"
}
] | 1,429,448,889
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 7
| 1,000
| 5,017,600
|
a,m=[int(v) for v in input().split()]
x=a%m
y=[-1]*m
i=0
while True:
if x==0:
print('Yes')
raise SystemExit
if x in y :
print('No')
raise SystemExit
y[i]=x
x=(2*x)%m
i=i+1
|
Title: Factory
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One industrial factory is reforming working plan. The director suggested to set a mythical detail production norm. If at the beginning of the day there were *x* details in the factory storage, then by the end of the day the factory has to produce (remainder after dividing *x* by *m*) more details. Unfortunately, no customer has ever bought any mythical detail, so all the details produced stay on the factory.
The board of directors are worried that the production by the given plan may eventually stop (that means that there will be а moment when the current number of details on the factory is divisible by *m*).
Given the number of details *a* on the first day and number *m* check if the production stops at some moment.
Input Specification:
The first line contains two integers *a* and *m* (1<=≤<=*a*,<=*m*<=≤<=105).
Output Specification:
Print "Yes" (without quotes) if the production will eventually stop, otherwise print "No".
Demo Input:
['1 5\n', '3 6\n']
Demo Output:
['No\n', 'Yes\n']
Note:
none
|
```python
a,m=[int(v) for v in input().split()]
x=a%m
y=[-1]*m
i=0
while True:
if x==0:
print('Yes')
raise SystemExit
if x in y :
print('No')
raise SystemExit
y[i]=x
x=(2*x)%m
i=i+1
```
| 0
|
|
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"
}
] | 1,598,474,204
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 154
| 204,800
|
# https://codeforces.com/contest/26/problem/A
"""
Find the number of almost primes between 1 and n.
A number is almost prime if it has exactly two prime divisors.
Find the number of primes between 1 and n. Then find the number of combinations with product less than 1.
(These are all distinct because of uniqueness of prime factorisation)
"""
from math import floor
from itertools import combinations
def sieve(n):
"""Returns all primes in the range [0, n]"""
is_prime = [True] * (n + 1)
is_prime[0] = is_prime[1] = False
# Can have no prime numbers greater than sqrt n
for i in range(2, floor(n ** 0.5) + 1):
if is_prime[i]:
# Eliminate proper multiples of i
for j in range(2, n // i + 1):
is_prime[i * j] = False
primes = [i for i in range(n + 1) if is_prime[i]]
return primes
def main():
n = int(input())
count = 0
for a, b in combinations(sieve(n), 2):
if a*b <= n:
count += 1
return count // 2
if __name__ == '__main__':
print(main())
|
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
# https://codeforces.com/contest/26/problem/A
"""
Find the number of almost primes between 1 and n.
A number is almost prime if it has exactly two prime divisors.
Find the number of primes between 1 and n. Then find the number of combinations with product less than 1.
(These are all distinct because of uniqueness of prime factorisation)
"""
from math import floor
from itertools import combinations
def sieve(n):
"""Returns all primes in the range [0, n]"""
is_prime = [True] * (n + 1)
is_prime[0] = is_prime[1] = False
# Can have no prime numbers greater than sqrt n
for i in range(2, floor(n ** 0.5) + 1):
if is_prime[i]:
# Eliminate proper multiples of i
for j in range(2, n // i + 1):
is_prime[i * j] = False
primes = [i for i in range(n + 1) if is_prime[i]]
return primes
def main():
n = int(input())
count = 0
for a, b in combinations(sieve(n), 2):
if a*b <= n:
count += 1
return count // 2
if __name__ == '__main__':
print(main())
```
| 0
|
59
|
A
|
Word
|
PROGRAMMING
| 800
|
[
"implementation",
"strings"
] |
A. Word
|
2
|
256
|
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
|
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
|
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
|
[
"HoUse\n",
"ViP\n",
"maTRIx\n"
] |
[
"house\n",
"VIP\n",
"matrix\n"
] |
none
| 500
|
[
{
"input": "HoUse",
"output": "house"
},
{
"input": "ViP",
"output": "VIP"
},
{
"input": "maTRIx",
"output": "matrix"
},
{
"input": "BNHWpnpawg",
"output": "bnhwpnpawg"
},
{
"input": "VTYGP",
"output": "VTYGP"
},
{
"input": "CHNenu",
"output": "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"
}
] | 1,678,133,221
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 30
| 92
| 0
|
word = input()
lcount = 0
ucount = 0
for i in word:
if i.islower():
lcount += 1
else:
ucount += 1
if lcount >= ucount:
print(word.lower())
else:
print(word.upper())
|
Title: Word
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Vasya is very upset that many people on the Net mix uppercase and lowercase letters in one word. That's why he decided to invent an extension for his favorite browser that would change the letters' register in every word so that it either only consisted of lowercase letters or, vice versa, only of uppercase ones. At that as little as possible letters should be changed in the word. For example, the word HoUse must be replaced with house, and the word ViP — with VIP. If a word contains an equal number of uppercase and lowercase letters, you should replace all the letters with lowercase ones. For example, maTRIx should be replaced by matrix. Your task is to use the given method on one given word.
Input Specification:
The first line contains a word *s* — it consists of uppercase and lowercase Latin letters and possesses the length from 1 to 100.
Output Specification:
Print the corrected word *s*. If the given word *s* has strictly more uppercase letters, make the word written in the uppercase register, otherwise - in the lowercase one.
Demo Input:
['HoUse\n', 'ViP\n', 'maTRIx\n']
Demo Output:
['house\n', 'VIP\n', 'matrix\n']
Note:
none
|
```python
word = input()
lcount = 0
ucount = 0
for i in word:
if i.islower():
lcount += 1
else:
ucount += 1
if lcount >= ucount:
print(word.lower())
else:
print(word.upper())
```
| 3.977
|
967
|
B
|
Watering System
|
PROGRAMMING
| 1,000
|
[
"math",
"sortings"
] | null | null |
Arkady wants to water his only flower. Unfortunately, he has a very poor watering system that was designed for $n$ flowers and so it looks like a pipe with $n$ holes. Arkady can only use the water that flows from the first hole.
Arkady can block some of the holes, and then pour $A$ liters of water into the pipe. After that, the water will flow out from the non-blocked holes proportionally to their sizes $s_1, s_2, \ldots, s_n$. In other words, if the sum of sizes of non-blocked holes is $S$, and the $i$-th hole is not blocked, $\frac{s_i \cdot A}{S}$ liters of water will flow out of it.
What is the minimum number of holes Arkady should block to make at least $B$ liters of water flow out of the first hole?
|
The first line contains three integers $n$, $A$, $B$ ($1 \le n \le 100\,000$, $1 \le B \le A \le 10^4$) — the number of holes, the volume of water Arkady will pour into the system, and the volume he wants to get out of the first hole.
The second line contains $n$ integers $s_1, s_2, \ldots, s_n$ ($1 \le s_i \le 10^4$) — the sizes of the holes.
|
Print a single integer — the number of holes Arkady should block.
|
[
"4 10 3\n2 2 2 2\n",
"4 80 20\n3 2 1 4\n",
"5 10 10\n1000 1 1 1 1\n"
] |
[
"1\n",
"0\n",
"4\n"
] |
In the first example Arkady should block at least one hole. After that, $\frac{10 \cdot 2}{6} \approx 3.333$ liters of water will flow out of the first hole, and that suits Arkady.
In the second example even without blocking any hole, $\frac{80 \cdot 3}{10} = 24$ liters will flow out of the first hole, that is not less than $20$.
In the third example Arkady has to block all holes except the first to make all water flow out of the first hole.
| 1,000
|
[
{
"input": "4 10 3\n2 2 2 2",
"output": "1"
},
{
"input": "4 80 20\n3 2 1 4",
"output": "0"
},
{
"input": "5 10 10\n1000 1 1 1 1",
"output": "4"
},
{
"input": "10 300 100\n20 1 3 10 8 5 3 6 4 3",
"output": "1"
},
{
"input": "10 300 100\n20 25 68 40 60 37 44 85 23 96",
"output": "8"
},
{
"input": "1 1 1\n1",
"output": "0"
},
{
"input": "1 2 1\n1",
"output": "0"
},
{
"input": "2 2 2\n1 10000",
"output": "1"
},
{
"input": "2 10000 1\n1 9999",
"output": "0"
}
] | 1,525,010,019
| 2,319
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 3
| 77
| 7,065,600
|
n,a,b = map(int,input().split())
s = list(map(int,input().split()))
s[1:].sort(reverse=True)
c = 0
i = 1
while(1):
x = (s[0]*a)/sum(s)
if(x>=b):
print(c)
break
else:
s[i] = 0
c+=1
i+=1
|
Title: Watering System
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Arkady wants to water his only flower. Unfortunately, he has a very poor watering system that was designed for $n$ flowers and so it looks like a pipe with $n$ holes. Arkady can only use the water that flows from the first hole.
Arkady can block some of the holes, and then pour $A$ liters of water into the pipe. After that, the water will flow out from the non-blocked holes proportionally to their sizes $s_1, s_2, \ldots, s_n$. In other words, if the sum of sizes of non-blocked holes is $S$, and the $i$-th hole is not blocked, $\frac{s_i \cdot A}{S}$ liters of water will flow out of it.
What is the minimum number of holes Arkady should block to make at least $B$ liters of water flow out of the first hole?
Input Specification:
The first line contains three integers $n$, $A$, $B$ ($1 \le n \le 100\,000$, $1 \le B \le A \le 10^4$) — the number of holes, the volume of water Arkady will pour into the system, and the volume he wants to get out of the first hole.
The second line contains $n$ integers $s_1, s_2, \ldots, s_n$ ($1 \le s_i \le 10^4$) — the sizes of the holes.
Output Specification:
Print a single integer — the number of holes Arkady should block.
Demo Input:
['4 10 3\n2 2 2 2\n', '4 80 20\n3 2 1 4\n', '5 10 10\n1000 1 1 1 1\n']
Demo Output:
['1\n', '0\n', '4\n']
Note:
In the first example Arkady should block at least one hole. After that, $\frac{10 \cdot 2}{6} \approx 3.333$ liters of water will flow out of the first hole, and that suits Arkady.
In the second example even without blocking any hole, $\frac{80 \cdot 3}{10} = 24$ liters will flow out of the first hole, that is not less than $20$.
In the third example Arkady has to block all holes except the first to make all water flow out of the first hole.
|
```python
n,a,b = map(int,input().split())
s = list(map(int,input().split()))
s[1:].sort(reverse=True)
c = 0
i = 1
while(1):
x = (s[0]*a)/sum(s)
if(x>=b):
print(c)
break
else:
s[i] = 0
c+=1
i+=1
```
| 0
|
|
614
|
A
|
Link/Cut Tree
|
PROGRAMMING
| 1,500
|
[
"brute force",
"implementation"
] | null | null |
Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, which is based on Splay trees. Specifically, he is now studying the *expose* procedure.
Unfortunately, Rostislav is unable to understand the definition of this procedure, so he decided to ask programmer Serezha to help him. Serezha agreed to help if Rostislav solves a simple task (and if he doesn't, then why would he need Splay trees anyway?)
Given integers *l*, *r* and *k*, you need to print all powers of number *k* within range from *l* to *r* inclusive. However, Rostislav doesn't want to spent time doing this, as he got interested in playing a network game called Agar with Gleb. Help him!
|
The first line of the input contains three space-separated integers *l*, *r* and *k* (1<=≤<=*l*<=≤<=*r*<=≤<=1018, 2<=≤<=*k*<=≤<=109).
|
Print all powers of number *k*, that lie within range from *l* to *r* in the increasing order. If there are no such numbers, print "-1" (without the quotes).
|
[
"1 10 2\n",
"2 4 5\n"
] |
[
"1 2 4 8 ",
"-1"
] |
Note to the first sample: numbers 2<sup class="upper-index">0</sup> = 1, 2<sup class="upper-index">1</sup> = 2, 2<sup class="upper-index">2</sup> = 4, 2<sup class="upper-index">3</sup> = 8 lie within the specified range. The number 2<sup class="upper-index">4</sup> = 16 is greater then 10, thus it shouldn't be printed.
| 500
|
[
{
"input": "1 10 2",
"output": "1 2 4 8 "
},
{
"input": "2 4 5",
"output": "-1"
},
{
"input": "18102 43332383920 28554",
"output": "28554 815330916 "
},
{
"input": "19562 31702689720 17701",
"output": "313325401 "
},
{
"input": "11729 55221128400 313",
"output": "97969 30664297 9597924961 "
},
{
"input": "5482 100347128000 342",
"output": "116964 40001688 13680577296 "
},
{
"input": "3680 37745933600 10",
"output": "10000 100000 1000000 10000000 100000000 1000000000 10000000000 "
},
{
"input": "17098 191120104800 43",
"output": "79507 3418801 147008443 6321363049 "
},
{
"input": "10462 418807699200 2",
"output": "16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368 68719476736 137438953472 274877906944 "
},
{
"input": "30061 641846400000 3",
"output": "59049 177147 531441 1594323 4782969 14348907 43046721 129140163 387420489 1162261467 3486784401 10460353203 31381059609 94143178827 282429536481 "
},
{
"input": "1 1000000000000000000 2",
"output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368 68719476736 137438953472 274877906944 549755813888 1099511627776 2199023255552 4398046511104 8796093022208 17592186044416 35184372088832 70368744177664 140737488355328 281474976710656 562949953421312 1125899906842624 2251799813685248 4503599627370496 900719925474099..."
},
{
"input": "32 2498039712000 4",
"output": "64 256 1024 4096 16384 65536 262144 1048576 4194304 16777216 67108864 268435456 1073741824 4294967296 17179869184 68719476736 274877906944 1099511627776 "
},
{
"input": "1 2576683920000 2",
"output": "1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 2097152 4194304 8388608 16777216 33554432 67108864 134217728 268435456 536870912 1073741824 2147483648 4294967296 8589934592 17179869184 34359738368 68719476736 137438953472 274877906944 549755813888 1099511627776 2199023255552 "
},
{
"input": "5 25 5",
"output": "5 25 "
},
{
"input": "1 90 90",
"output": "1 90 "
},
{
"input": "95 2200128528000 68",
"output": "4624 314432 21381376 1453933568 98867482624 "
},
{
"input": "64 426314644000 53",
"output": "2809 148877 7890481 418195493 22164361129 "
},
{
"input": "198765 198765 198765",
"output": "198765 "
},
{
"input": "42 2845016496000 12",
"output": "144 1728 20736 248832 2985984 35831808 429981696 5159780352 61917364224 743008370688 "
},
{
"input": "6 6 3",
"output": "-1"
},
{
"input": "1 10 11",
"output": "1 "
},
{
"input": "2 10 11",
"output": "-1"
},
{
"input": "87 160 41",
"output": "-1"
},
{
"input": "237171123124584251 923523399718980912 7150",
"output": "-1"
},
{
"input": "101021572000739548 453766043506276015 8898",
"output": "-1"
},
{
"input": "366070689449360724 928290634811046396 8230",
"output": "-1"
},
{
"input": "438133886369772308 942612870269666780 7193",
"output": "-1"
},
{
"input": "10 10 10",
"output": "10 "
},
{
"input": "16 16 256",
"output": "-1"
},
{
"input": "1 1000000000000000000 1000000000",
"output": "1 1000000000 1000000000000000000 "
},
{
"input": "1000000000000000000 1000000000000000000 1000000000",
"output": "1000000000000000000 "
},
{
"input": "1000000000 1000000000000000000 1000000000",
"output": "1000000000 1000000000000000000 "
},
{
"input": "1 1 4",
"output": "1 "
},
{
"input": "1 999999999999999999 1000000000",
"output": "1 1000000000 "
},
{
"input": "1 1000000000000000000 999999990",
"output": "1 999999990 999999980000000100 "
},
{
"input": "1 1000000000000000000 999999984",
"output": "1 999999984 999999968000000256 "
},
{
"input": "1 1000000000000000000 324325",
"output": "1 324325 105186705625 34114678301828125 "
},
{
"input": "1 1000000000000000000 999999523",
"output": "1 999999523 999999046000227529 "
},
{
"input": "1 243 3",
"output": "1 3 9 27 81 243 "
},
{
"input": "62769392426654367 567152589733560993 688813",
"output": "326816522793383797 "
},
{
"input": "1 1000000000000000000 690852001",
"output": "1 690852001 477276487285704001 "
},
{
"input": "1 1000000000000000000 918745157",
"output": "1 918745157 844092663510954649 "
},
{
"input": "1 1000000000000000000 131299843",
"output": "1 131299843 17239648771824649 "
},
{
"input": "2861381721051425 2861381721051425 1234",
"output": "-1"
}
] | 1,660,106,342
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 2
| 2,000
| 9,318,400
|
L = []
l,r,k = map(int,input().split())
for i in range(l-1,r):
if k**i<r:
L.append(k**i)
if len(L)==0:
print(-1)
else:print(*L)
|
Title: Link/Cut Tree
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Programmer Rostislav got seriously interested in the Link/Cut Tree data structure, which is based on Splay trees. Specifically, he is now studying the *expose* procedure.
Unfortunately, Rostislav is unable to understand the definition of this procedure, so he decided to ask programmer Serezha to help him. Serezha agreed to help if Rostislav solves a simple task (and if he doesn't, then why would he need Splay trees anyway?)
Given integers *l*, *r* and *k*, you need to print all powers of number *k* within range from *l* to *r* inclusive. However, Rostislav doesn't want to spent time doing this, as he got interested in playing a network game called Agar with Gleb. Help him!
Input Specification:
The first line of the input contains three space-separated integers *l*, *r* and *k* (1<=≤<=*l*<=≤<=*r*<=≤<=1018, 2<=≤<=*k*<=≤<=109).
Output Specification:
Print all powers of number *k*, that lie within range from *l* to *r* in the increasing order. If there are no such numbers, print "-1" (without the quotes).
Demo Input:
['1 10 2\n', '2 4 5\n']
Demo Output:
['1 2 4 8 ', '-1']
Note:
Note to the first sample: numbers 2<sup class="upper-index">0</sup> = 1, 2<sup class="upper-index">1</sup> = 2, 2<sup class="upper-index">2</sup> = 4, 2<sup class="upper-index">3</sup> = 8 lie within the specified range. The number 2<sup class="upper-index">4</sup> = 16 is greater then 10, thus it shouldn't be printed.
|
```python
L = []
l,r,k = map(int,input().split())
for i in range(l-1,r):
if k**i<r:
L.append(k**i)
if len(L)==0:
print(-1)
else:print(*L)
```
| 0
|
|
151
|
A
|
Soft Drinking
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
This winter is so cold in Nvodsk! A group of *n* friends decided to buy *k* bottles of a soft drink called "Take-It-Light" to warm up a bit. Each bottle has *l* milliliters of the drink. Also they bought *c* limes and cut each of them into *d* slices. After that they found *p* grams of salt.
To make a toast, each friend needs *nl* milliliters of the drink, a slice of lime and *np* grams of salt. The friends want to make as many toasts as they can, provided they all drink the same amount. How many toasts can each friend make?
|
The first and only line contains positive integers *n*, *k*, *l*, *c*, *d*, *p*, *nl*, *np*, not exceeding 1000 and no less than 1. The numbers are separated by exactly one space.
|
Print a single integer — the number of toasts each friend can make.
|
[
"3 4 5 10 8 100 3 1\n",
"5 100 10 1 19 90 4 3\n",
"10 1000 1000 25 23 1 50 1\n"
] |
[
"2\n",
"3\n",
"0\n"
] |
A comment to the first sample:
Overall the friends have 4 * 5 = 20 milliliters of the drink, it is enough to make 20 / 3 = 6 toasts. The limes are enough for 10 * 8 = 80 toasts and the salt is enough for 100 / 1 = 100 toasts. However, there are 3 friends in the group, so the answer is *min*(6, 80, 100) / 3 = 2.
| 500
|
[
{
"input": "3 4 5 10 8 100 3 1",
"output": "2"
},
{
"input": "5 100 10 1 19 90 4 3",
"output": "3"
},
{
"input": "10 1000 1000 25 23 1 50 1",
"output": "0"
},
{
"input": "1 7 4 5 5 8 3 2",
"output": "4"
},
{
"input": "2 3 3 5 5 10 1 3",
"output": "1"
},
{
"input": "2 6 4 5 6 5 1 3",
"output": "0"
},
{
"input": "1 7 3 5 3 6 2 1",
"output": "6"
},
{
"input": "2 4 5 4 5 7 3 2",
"output": "1"
},
{
"input": "2 3 6 5 7 8 2 1",
"output": "4"
},
{
"input": "1 4 5 5 3 10 3 1",
"output": "6"
},
{
"input": "1 4 6 7 3 5 1 3",
"output": "1"
},
{
"input": "1 6 5 5 5 8 3 1",
"output": "8"
},
{
"input": "1 7 5 3 3 9 2 1",
"output": "9"
},
{
"input": "3 5 3 7 6 10 3 1",
"output": "1"
},
{
"input": "3 6 3 5 3 6 3 1",
"output": "2"
},
{
"input": "1 7 5 5 5 5 2 2",
"output": "2"
},
{
"input": "2 5 3 5 6 9 2 1",
"output": "3"
},
{
"input": "3 4 3 5 3 6 2 1",
"output": "2"
},
{
"input": "1 5 5 4 7 6 3 1",
"output": "6"
},
{
"input": "2 3 7 6 5 9 3 1",
"output": "3"
},
{
"input": "2 6 5 3 3 8 1 1",
"output": "4"
},
{
"input": "2 4 7 3 4 10 2 1",
"output": "5"
},
{
"input": "1 1000 1000 1000 1000 1000 1 1",
"output": "1000"
},
{
"input": "17 1000 1000 1000 1000 1000 3 7",
"output": "8"
},
{
"input": "115 1000 1000 1000 1000 1000 17 15",
"output": "0"
},
{
"input": "1 587 981 1 2 1 1 1",
"output": "1"
},
{
"input": "1 1 2 1 2 2 1 1",
"output": "2"
}
] | 1,691,935,897
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 156
| 2,867,200
|
n=int(input())
k=int(input())
l=int(input())
c=int(input())
d=int(input())
p=int(input())
nl=int(input())
np=int(input())
v=k*l
o=v//n
e=c*d
g=p/np
le=min(o,e,g)//n
print(le)
|
Title: Soft Drinking
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
This winter is so cold in Nvodsk! A group of *n* friends decided to buy *k* bottles of a soft drink called "Take-It-Light" to warm up a bit. Each bottle has *l* milliliters of the drink. Also they bought *c* limes and cut each of them into *d* slices. After that they found *p* grams of salt.
To make a toast, each friend needs *nl* milliliters of the drink, a slice of lime and *np* grams of salt. The friends want to make as many toasts as they can, provided they all drink the same amount. How many toasts can each friend make?
Input Specification:
The first and only line contains positive integers *n*, *k*, *l*, *c*, *d*, *p*, *nl*, *np*, not exceeding 1000 and no less than 1. The numbers are separated by exactly one space.
Output Specification:
Print a single integer — the number of toasts each friend can make.
Demo Input:
['3 4 5 10 8 100 3 1\n', '5 100 10 1 19 90 4 3\n', '10 1000 1000 25 23 1 50 1\n']
Demo Output:
['2\n', '3\n', '0\n']
Note:
A comment to the first sample:
Overall the friends have 4 * 5 = 20 milliliters of the drink, it is enough to make 20 / 3 = 6 toasts. The limes are enough for 10 * 8 = 80 toasts and the salt is enough for 100 / 1 = 100 toasts. However, there are 3 friends in the group, so the answer is *min*(6, 80, 100) / 3 = 2.
|
```python
n=int(input())
k=int(input())
l=int(input())
c=int(input())
d=int(input())
p=int(input())
nl=int(input())
np=int(input())
v=k*l
o=v//n
e=c*d
g=p/np
le=min(o,e,g)//n
print(le)
```
| -1
|
|
535
|
B
|
Tavas and SaDDas
|
PROGRAMMING
| 1,100
|
[
"bitmasks",
"brute force",
"combinatorics",
"implementation"
] | null | null |
Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you."
The problem is:
You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
If we sort all lucky numbers in increasing order, what's the 1-based index of *n*?
Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back.
|
The first and only line of input contains a lucky number *n* (1<=≤<=*n*<=≤<=109).
|
Print the index of *n* among all lucky numbers.
|
[
"4\n",
"7\n",
"77\n"
] |
[
"1\n",
"2\n",
"6\n"
] |
none
| 1,000
|
[
{
"input": "4",
"output": "1"
},
{
"input": "7",
"output": "2"
},
{
"input": "77",
"output": "6"
},
{
"input": "4",
"output": "1"
},
{
"input": "474744",
"output": "83"
},
{
"input": "777774",
"output": "125"
},
{
"input": "447",
"output": "8"
},
{
"input": "774",
"output": "13"
},
{
"input": "4",
"output": "1"
},
{
"input": "4447747",
"output": "140"
},
{
"input": "7747474",
"output": "233"
},
{
"input": "4444",
"output": "15"
},
{
"input": "4447",
"output": "16"
},
{
"input": "7",
"output": "2"
},
{
"input": "4",
"output": "1"
},
{
"input": "4447744",
"output": "139"
},
{
"input": "77474",
"output": "57"
},
{
"input": "7747",
"output": "28"
},
{
"input": "444",
"output": "7"
},
{
"input": "7",
"output": "2"
},
{
"input": "7774477",
"output": "242"
},
{
"input": "4477774",
"output": "157"
},
{
"input": "7444",
"output": "23"
},
{
"input": "7474747",
"output": "212"
},
{
"input": "77",
"output": "6"
},
{
"input": "774477",
"output": "114"
},
{
"input": "7",
"output": "2"
},
{
"input": "47",
"output": "4"
},
{
"input": "747777",
"output": "110"
},
{
"input": "444444444",
"output": "511"
},
{
"input": "777777777",
"output": "1022"
},
{
"input": "477477447",
"output": "728"
},
{
"input": "777744747",
"output": "996"
}
] | 1,599,523,165
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 33
| 140
| 0
|
from itertools import product
n = input()
count = 1
ans = 0
while count != len(n):
ans += 2 ** count
count += 1
plus = 0
for i in product(['4', '7'], repeat=len(n)):
plus += 1
if ''.join(i) == n:
break
print(ans + plus)
|
Title: Tavas and SaDDas
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Once again Tavas started eating coffee mix without water! Keione told him that it smells awful, but he didn't stop doing that. That's why Keione told his smart friend, SaDDas to punish him! SaDDas took Tavas' headphones and told him: "If you solve the following problem, I'll return it to you."
The problem is:
You are given a lucky number *n*. Lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
If we sort all lucky numbers in increasing order, what's the 1-based index of *n*?
Tavas is not as smart as SaDDas, so he asked you to do him a favor and solve this problem so he can have his headphones back.
Input Specification:
The first and only line of input contains a lucky number *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
Print the index of *n* among all lucky numbers.
Demo Input:
['4\n', '7\n', '77\n']
Demo Output:
['1\n', '2\n', '6\n']
Note:
none
|
```python
from itertools import product
n = input()
count = 1
ans = 0
while count != len(n):
ans += 2 ** count
count += 1
plus = 0
for i in product(['4', '7'], repeat=len(n)):
plus += 1
if ''.join(i) == n:
break
print(ans + plus)
```
| 3
|
|
300
|
A
|
Array
|
PROGRAMMING
| 1,100
|
[
"brute force",
"constructive algorithms",
"implementation"
] | null | null |
Vitaly has an array of *n* distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold:
1. The product of all numbers in the first set is less than zero (<=<<=0). 1. The product of all numbers in the second set is greater than zero (<=><=0). 1. The product of all numbers in the third set is equal to zero. 1. Each number from the initial array must occur in exactly one set.
Help Vitaly. Divide the given array.
|
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=100). The second line contains *n* space-separated distinct integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=103) — the array elements.
|
In the first line print integer *n*1 (*n*1<=><=0) — the number of elements in the first set. Then print *n*1 numbers — the elements that got to the first set.
In the next line print integer *n*2 (*n*2<=><=0) — the number of elements in the second set. Then print *n*2 numbers — the elements that got to the second set.
In the next line print integer *n*3 (*n*3<=><=0) — the number of elements in the third set. Then print *n*3 numbers — the elements that got to the third set.
The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them.
|
[
"3\n-1 2 0\n",
"4\n-1 -2 -3 0\n"
] |
[
"1 -1\n1 2\n1 0\n",
"1 -1\n2 -3 -2\n1 0\n"
] |
none
| 500
|
[
{
"input": "3\n-1 2 0",
"output": "1 -1\n1 2\n1 0"
},
{
"input": "4\n-1 -2 -3 0",
"output": "1 -1\n2 -3 -2\n1 0"
},
{
"input": "5\n-1 -2 1 2 0",
"output": "1 -1\n2 1 2\n2 0 -2"
},
{
"input": "100\n-64 -51 -75 -98 74 -26 -1 -8 -99 -76 -53 -80 -43 -22 -100 -62 -34 -5 -65 -81 -18 -91 -92 -16 -23 -95 -9 -19 -44 -46 -79 52 -35 4 -87 -7 -90 -20 -71 -61 -67 -50 -66 -68 -49 -27 -32 -57 -85 -59 -30 -36 -3 -77 86 -25 -94 -56 60 -24 -37 -72 -41 -31 11 -48 28 -38 -42 -39 -33 -70 -84 0 -93 -73 -14 -69 -40 -97 -6 -55 -45 -54 -10 -29 -96 -12 -83 -15 -21 -47 17 -2 -63 -89 88 13 -58 -82",
"output": "89 -64 -51 -75 -98 -26 -1 -8 -99 -76 -53 -80 -43 -22 -100 -62 -34 -5 -65 -81 -18 -91 -92 -16 -23 -95 -9 -19 -44 -46 -79 -35 -87 -7 -90 -20 -71 -61 -67 -50 -66 -68 -49 -27 -32 -57 -85 -59 -30 -36 -3 -77 -25 -94 -56 -24 -37 -72 -41 -31 -48 -38 -42 -39 -33 -70 -84 -93 -73 -14 -69 -40 -97 -6 -55 -45 -54 -10 -29 -96 -12 -83 -15 -21 -47 -2 -63 -89 -58 -82\n10 74 52 4 86 60 11 28 17 88 13\n1 0"
},
{
"input": "100\n3 -66 -17 54 24 -29 76 89 32 -37 93 -16 99 -25 51 78 23 68 -95 59 18 34 -45 77 9 39 -10 19 8 73 -5 60 12 31 0 2 26 40 48 30 52 49 27 4 87 57 85 58 -61 50 83 80 69 67 91 97 -96 11 100 56 82 53 13 -92 -72 70 1 -94 -63 47 21 14 74 7 6 33 55 65 64 -41 81 42 36 28 38 20 43 71 90 -88 22 84 -86 15 75 62 44 35 98 46",
"output": "19 -66 -17 -29 -37 -16 -25 -95 -45 -10 -5 -61 -96 -92 -72 -94 -63 -41 -88 -86\n80 3 54 24 76 89 32 93 99 51 78 23 68 59 18 34 77 9 39 19 8 73 60 12 31 2 26 40 48 30 52 49 27 4 87 57 85 58 50 83 80 69 67 91 97 11 100 56 82 53 13 70 1 47 21 14 74 7 6 33 55 65 64 81 42 36 28 38 20 43 71 90 22 84 15 75 62 44 35 98 46\n1 0"
},
{
"input": "100\n-17 16 -70 32 -60 75 -100 -9 -68 -30 -42 86 -88 -98 -47 -5 58 -14 -94 -73 -80 -51 -66 -85 -53 49 -25 -3 -45 -69 -11 -64 83 74 -65 67 13 -91 81 6 -90 -54 -12 -39 0 -24 -71 -41 -44 57 -93 -20 -92 18 -43 -52 -55 -84 -89 -19 40 -4 -99 -26 -87 -36 -56 -61 -62 37 -95 -28 63 23 35 -82 1 -2 -78 -96 -21 -77 -76 -27 -10 -97 -8 46 -15 -48 -34 -59 -7 -29 50 -33 -72 -79 22 38",
"output": "75 -17 -70 -60 -100 -9 -68 -30 -42 -88 -98 -47 -5 -14 -94 -73 -80 -51 -66 -85 -53 -25 -3 -45 -69 -11 -64 -65 -91 -90 -54 -12 -39 -24 -71 -41 -44 -93 -20 -92 -43 -52 -55 -84 -89 -19 -4 -99 -26 -87 -36 -56 -61 -62 -95 -28 -82 -2 -78 -96 -21 -77 -76 -27 -10 -97 -8 -15 -48 -34 -59 -7 -29 -33 -72 -79\n24 16 32 75 86 58 49 83 74 67 13 81 6 57 18 40 37 63 23 35 1 46 50 22 38\n1 0"
},
{
"input": "100\n-97 -90 61 78 87 -52 -3 65 83 38 30 -60 35 -50 -73 -77 44 -32 -81 17 -67 58 -6 -34 47 -28 71 -45 69 -80 -4 -7 -57 -79 43 -27 -31 29 16 -89 -21 -93 95 -82 74 -5 -70 -20 -18 36 -64 -66 72 53 62 -68 26 15 76 -40 -99 8 59 88 49 -23 9 10 56 -48 -98 0 100 -54 25 94 13 -63 42 39 -1 55 24 -12 75 51 41 84 -96 -85 -2 -92 14 -46 -91 -19 -11 -86 22 -37",
"output": "51 -97 -90 -52 -3 -60 -50 -73 -77 -32 -81 -67 -6 -34 -28 -45 -80 -4 -7 -57 -79 -27 -31 -89 -21 -93 -82 -5 -70 -20 -18 -64 -66 -68 -40 -99 -23 -48 -98 -54 -63 -1 -12 -96 -85 -2 -92 -46 -91 -19 -11 -86\n47 61 78 87 65 83 38 30 35 44 17 58 47 71 69 43 29 16 95 74 36 72 53 62 26 15 76 8 59 88 49 9 10 56 100 25 94 13 42 39 55 24 75 51 41 84 14 22\n2 0 -37"
},
{
"input": "100\n-75 -60 -18 -92 -71 -9 -37 -34 -82 28 -54 93 -83 -76 -58 -88 -17 -97 64 -39 -96 -81 -10 -98 -47 -100 -22 27 14 -33 -19 -99 87 -66 57 -21 -90 -70 -32 -26 24 -77 -74 13 -44 16 -5 -55 -2 -6 -7 -73 -1 -68 -30 -95 -42 69 0 -20 -79 59 -48 -4 -72 -67 -46 62 51 -52 -86 -40 56 -53 85 -35 -8 49 50 65 29 11 -43 -15 -41 -12 -3 -80 -31 -38 -91 -45 -25 78 94 -23 -63 84 89 -61",
"output": "73 -75 -60 -18 -92 -71 -9 -37 -34 -82 -54 -83 -76 -58 -88 -17 -97 -39 -96 -81 -10 -98 -47 -100 -22 -33 -19 -99 -66 -21 -90 -70 -32 -26 -77 -74 -44 -5 -55 -2 -6 -7 -73 -1 -68 -30 -95 -42 -20 -79 -48 -4 -72 -67 -46 -52 -86 -40 -53 -35 -8 -43 -15 -41 -12 -3 -80 -31 -38 -91 -45 -25 -23 -63\n25 28 93 64 27 14 87 57 24 13 16 69 59 62 51 56 85 49 50 65 29 11 78 94 84 89\n2 0 -61"
},
{
"input": "100\n-87 -48 -76 -1 -10 -17 -22 -19 -27 -99 -43 49 38 -20 -45 -64 44 -96 -35 -74 -65 -41 -21 -75 37 -12 -67 0 -3 5 -80 -93 -81 -97 -47 -63 53 -100 95 -79 -83 -90 -32 88 -77 -16 -23 -54 -28 -4 -73 -98 -25 -39 60 -56 -34 -2 -11 -55 -52 -69 -68 -29 -82 -62 -36 -13 -6 -89 8 -72 18 -15 -50 -71 -70 -92 -42 -78 -61 -9 -30 -85 -91 -94 84 -86 -7 -57 -14 40 -33 51 -26 46 59 -31 -58 -66",
"output": "83 -87 -48 -76 -1 -10 -17 -22 -19 -27 -99 -43 -20 -45 -64 -96 -35 -74 -65 -41 -21 -75 -12 -67 -3 -80 -93 -81 -97 -47 -63 -100 -79 -83 -90 -32 -77 -16 -23 -54 -28 -4 -73 -98 -25 -39 -56 -34 -2 -11 -55 -52 -69 -68 -29 -82 -62 -36 -13 -6 -89 -72 -15 -50 -71 -70 -92 -42 -78 -61 -9 -30 -85 -91 -94 -86 -7 -57 -14 -33 -26 -31 -58 -66\n16 49 38 44 37 5 53 95 88 60 8 18 84 40 51 46 59\n1 0"
},
{
"input": "100\n-95 -28 -43 -72 -11 -24 -37 -35 -44 -66 -45 -62 -96 -51 -55 -23 -31 -26 -59 -17 77 -69 -10 -12 -78 -14 -52 -57 -40 -75 4 -98 -6 7 -53 -3 -90 -63 -8 -20 88 -91 -32 -76 -80 -97 -34 -27 -19 0 70 -38 -9 -49 -67 73 -36 2 81 -39 -65 -83 -64 -18 -94 -79 -58 -16 87 -22 -74 -25 -13 -46 -89 -47 5 -15 -54 -99 56 -30 -60 -21 -86 33 -1 -50 -68 -100 -85 -29 92 -48 -61 42 -84 -93 -41 -82",
"output": "85 -95 -28 -43 -72 -11 -24 -37 -35 -44 -66 -45 -62 -96 -51 -55 -23 -31 -26 -59 -17 -69 -10 -12 -78 -14 -52 -57 -40 -75 -98 -6 -53 -3 -90 -63 -8 -20 -91 -32 -76 -80 -97 -34 -27 -19 -38 -9 -49 -67 -36 -39 -65 -83 -64 -18 -94 -79 -58 -16 -22 -74 -25 -13 -46 -89 -47 -15 -54 -99 -30 -60 -21 -86 -1 -50 -68 -100 -85 -29 -48 -61 -84 -93 -41 -82\n14 77 4 7 88 70 73 2 81 87 5 56 33 92 42\n1 0"
},
{
"input": "100\n-12 -41 57 13 83 -36 53 69 -6 86 -75 87 11 -5 -4 -14 -37 -84 70 2 -73 16 31 34 -45 94 -9 26 27 52 -42 46 96 21 32 7 -18 61 66 -51 95 -48 -76 90 80 -40 89 77 78 54 -30 8 88 33 -24 82 -15 19 1 59 44 64 -97 -60 43 56 35 47 39 50 29 28 -17 -67 74 23 85 -68 79 0 65 55 -3 92 -99 72 93 -71 38 -10 -100 -98 81 62 91 -63 -58 49 -20 22",
"output": "35 -12 -41 -36 -6 -75 -5 -4 -14 -37 -84 -73 -45 -9 -42 -18 -51 -48 -76 -40 -30 -24 -15 -97 -60 -17 -67 -68 -3 -99 -71 -10 -100 -98 -63 -58\n63 57 13 83 53 69 86 87 11 70 2 16 31 34 94 26 27 52 46 96 21 32 7 61 66 95 90 80 89 77 78 54 8 88 33 82 19 1 59 44 64 43 56 35 47 39 50 29 28 74 23 85 79 65 55 92 72 93 38 81 62 91 49 22\n2 0 -20"
},
{
"input": "100\n-34 81 85 -96 50 20 54 86 22 10 -19 52 65 44 30 53 63 71 17 98 -92 4 5 -99 89 -23 48 9 7 33 75 2 47 -56 42 70 -68 57 51 83 82 94 91 45 46 25 95 11 -12 62 -31 -87 58 38 67 97 -60 66 73 -28 13 93 29 59 -49 77 37 -43 -27 0 -16 72 15 79 61 78 35 21 3 8 84 1 -32 36 74 -88 26 100 6 14 40 76 18 90 24 69 80 64 55 41",
"output": "19 -34 -96 -19 -92 -99 -23 -56 -68 -12 -31 -87 -60 -28 -49 -43 -27 -16 -32 -88\n80 81 85 50 20 54 86 22 10 52 65 44 30 53 63 71 17 98 4 5 89 48 9 7 33 75 2 47 42 70 57 51 83 82 94 91 45 46 25 95 11 62 58 38 67 97 66 73 13 93 29 59 77 37 72 15 79 61 78 35 21 3 8 84 1 36 74 26 100 6 14 40 76 18 90 24 69 80 64 55 41\n1 0"
},
{
"input": "100\n-1000 -986 -979 -955 -966 -963 -973 -959 -972 -906 -924 -927 -929 -918 -977 -967 -921 -989 -911 -995 -945 -919 -971 -913 -912 -933 -969 -975 -920 -988 -997 -994 -953 -962 -940 -905 -978 -948 -957 -996 0 -976 -949 -931 -903 -985 -923 -993 -944 -909 -938 -946 -934 -992 -904 -980 -954 -943 -917 -968 -991 -956 -902 -942 -999 -998 -908 -928 -930 -914 -922 -936 -960 -937 -939 -926 -965 -925 -951 -910 -907 -970 -990 -984 -964 -987 -916 -947 -982 -950 -974 -915 -932 -958 -981 -941 -961 -983 -952 -935",
"output": "97 -1000 -986 -979 -955 -966 -963 -973 -959 -972 -906 -924 -927 -929 -918 -977 -967 -921 -989 -911 -995 -945 -919 -971 -913 -912 -933 -969 -975 -920 -988 -997 -994 -953 -962 -940 -905 -978 -948 -957 -996 -976 -949 -931 -903 -985 -923 -993 -944 -909 -938 -946 -934 -992 -904 -980 -954 -943 -917 -968 -991 -956 -902 -942 -999 -998 -908 -928 -930 -914 -922 -936 -960 -937 -939 -926 -965 -925 -951 -910 -907 -970 -990 -984 -964 -987 -916 -947 -982 -950 -974 -915 -932 -958 -981 -941 -961 -983\n2 -935 -952\n1 0"
},
{
"input": "99\n-1000 -986 -979 -955 -966 -963 -973 -959 -972 -906 -924 -927 -929 -918 -977 -967 -921 -989 -911 -995 -945 -919 -971 -913 -912 -933 -969 -975 -920 -988 -997 -994 -953 -962 -940 -905 -978 -948 -957 -996 0 -976 -949 -931 -903 -985 -923 -993 -944 -909 -938 -946 -934 -992 -904 -980 -954 -943 -917 -968 -991 -956 -902 -942 -999 -998 -908 -928 -930 -914 -922 -936 -960 -937 -939 -926 -965 -925 -951 -910 -907 -970 -990 -984 -964 -987 -916 -947 -982 -950 -974 -915 -932 -958 -981 -941 -961 -983 -952",
"output": "95 -1000 -986 -979 -955 -966 -963 -973 -959 -972 -906 -924 -927 -929 -918 -977 -967 -921 -989 -911 -995 -945 -919 -971 -913 -912 -933 -969 -975 -920 -988 -997 -994 -953 -962 -940 -905 -978 -948 -957 -996 -976 -949 -931 -903 -985 -923 -993 -944 -909 -938 -946 -934 -992 -904 -980 -954 -943 -917 -968 -991 -956 -902 -942 -999 -998 -908 -928 -930 -914 -922 -936 -960 -937 -939 -926 -965 -925 -951 -910 -907 -970 -990 -984 -964 -987 -916 -947 -982 -950 -974 -915 -932 -958 -981 -941\n2 -952 -983\n2 0 -961"
},
{
"input": "59\n-990 -876 -641 -726 718 -53 803 -954 894 -265 -587 -665 904 349 754 -978 441 794 -768 -428 -569 -476 188 -620 -290 -333 45 705 -201 109 165 446 13 122 714 -562 -15 -86 -960 43 329 578 287 -776 -14 -71 915 886 -259 337 -495 913 -498 -669 -673 818 225 647 0",
"output": "29 -990 -876 -641 -726 -53 -954 -265 -587 -665 -978 -768 -428 -569 -476 -620 -290 -333 -201 -562 -15 -86 -960 -776 -14 -71 -259 -495 -498 -669\n28 718 803 894 904 349 754 441 794 188 45 705 109 165 446 13 122 714 43 329 578 287 915 886 337 913 818 225 647\n2 0 -673"
},
{
"input": "64\n502 885 -631 -906 735 687 642 -29 -696 -165 -524 15 -129 -663 -846 -501 -651 895 -341 -833 -142 33 -847 688 945 -192 -587 -930 603 849 736 676 788 256 863 -509 319 -49 -807 -158 218 -886 -143 -639 118 -156 -291 325 892 -916 -622 -960 -959 -731 -943 436 -535 861 745 589 -159 376 -182 0",
"output": "35 -631 -906 -29 -696 -165 -524 -129 -663 -846 -501 -651 -341 -833 -142 -847 -192 -587 -930 -509 -49 -807 -158 -886 -143 -639 -156 -291 -916 -622 -960 -959 -731 -943 -535 -159\n27 502 885 735 687 642 15 895 33 688 945 603 849 736 676 788 256 863 319 218 118 325 892 436 861 745 589 376\n2 0 -182"
},
{
"input": "5\n-1 -2 -3 -4 0",
"output": "1 -1\n2 -4 -3\n2 0 -2"
},
{
"input": "3\n-101 101 0",
"output": "1 -101\n1 101\n1 0"
},
{
"input": "21\n-100 -200 -300 -400 -500 -600 -700 -800 -900 -1000 0 100 200 300 400 500 600 700 800 900 1000",
"output": "9 -100 -200 -300 -400 -500 -600 -700 -800 -900\n10 100 200 300 400 500 600 700 800 900 1000\n2 0 -1000"
},
{
"input": "4\n0 -1 -2 -3",
"output": "1 -1\n2 -3 -2\n1 0"
}
] | 1,646,132,833
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 2
| 92
| 0
|
n=int(input())
a=list(map(int,input().split()))
p=[]
n=[]
for i in a:
if i>0:
p.append(i)
elif i<0:
n.append(i)
print(1,n[0])
n.remove(n[0])
if len(n)%2==1:
n.remove(n[0])
print(len(n)+len(p),*p,*n)
print(1,0)
|
Title: Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vitaly has an array of *n* distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold:
1. The product of all numbers in the first set is less than zero (<=<<=0). 1. The product of all numbers in the second set is greater than zero (<=><=0). 1. The product of all numbers in the third set is equal to zero. 1. Each number from the initial array must occur in exactly one set.
Help Vitaly. Divide the given array.
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=100). The second line contains *n* space-separated distinct integers *a*1,<=*a*2,<=...,<=*a**n* (|*a**i*|<=≤<=103) — the array elements.
Output Specification:
In the first line print integer *n*1 (*n*1<=><=0) — the number of elements in the first set. Then print *n*1 numbers — the elements that got to the first set.
In the next line print integer *n*2 (*n*2<=><=0) — the number of elements in the second set. Then print *n*2 numbers — the elements that got to the second set.
In the next line print integer *n*3 (*n*3<=><=0) — the number of elements in the third set. Then print *n*3 numbers — the elements that got to the third set.
The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them.
Demo Input:
['3\n-1 2 0\n', '4\n-1 -2 -3 0\n']
Demo Output:
['1 -1\n1 2\n1 0\n', '1 -1\n2 -3 -2\n1 0\n']
Note:
none
|
```python
n=int(input())
a=list(map(int,input().split()))
p=[]
n=[]
for i in a:
if i>0:
p.append(i)
elif i<0:
n.append(i)
print(1,n[0])
n.remove(n[0])
if len(n)%2==1:
n.remove(n[0])
print(len(n)+len(p),*p,*n)
print(1,0)
```
| 0
|
|
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"
}
] | 1,667,885,768
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 9
| 92
| 0
|
k = int(input())
s = input()
map_alpha = {}
for x in s:
if x in map_alpha:
map_alpha[x] += 1
else:
map_alpha[x] = 1
flag = 0
for key in map_alpha:
if map_alpha[key] % k != 0:
flag = 1
break
if flag:
print("-1")
else:
temp = ""
while k > 0:
for key in map_alpha:
temp += key
k -= 1
print(temp)
|
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()
map_alpha = {}
for x in s:
if x in map_alpha:
map_alpha[x] += 1
else:
map_alpha[x] = 1
flag = 0
for key in map_alpha:
if map_alpha[key] % k != 0:
flag = 1
break
if flag:
print("-1")
else:
temp = ""
while k > 0:
for key in map_alpha:
temp += key
k -= 1
print(temp)
```
| 0
|
|
849
|
B
|
Tell Your World
|
PROGRAMMING
| 1,600
|
[
"brute force",
"geometry"
] | null | null |
Connect the countless points with lines, till we reach the faraway yonder.
There are *n* points on a coordinate plane, the *i*-th of which being (*i*,<=*y**i*).
Determine whether it's possible to draw two parallel and non-overlapping lines, such that every point in the set lies on exactly one of them, and each of them passes through at least one point in the set.
|
The first line of input contains a positive integer *n* (3<=≤<=*n*<=≤<=1<=000) — the number of points.
The second line contains *n* space-separated integers *y*1,<=*y*2,<=...,<=*y**n* (<=-<=109<=≤<=*y**i*<=≤<=109) — the vertical coordinates of each point.
|
Output "Yes" (without quotes) if it's possible to fulfill the requirements, and "No" otherwise.
You can print each letter in any case (upper or lower).
|
[
"5\n7 5 8 6 9\n",
"5\n-1 -2 0 0 -5\n",
"5\n5 4 3 2 1\n",
"5\n1000000000 0 0 0 0\n"
] |
[
"Yes\n",
"No\n",
"No\n",
"Yes\n"
] |
In the first example, there are five points: (1, 7), (2, 5), (3, 8), (4, 6) and (5, 9). It's possible to draw a line that passes through points 1, 3, 5, and another one that passes through points 2, 4 and is parallel to the first one.
In the second example, while it's possible to draw two lines that cover all points, they cannot be made parallel.
In the third example, it's impossible to satisfy both requirements at the same time.
| 1,000
|
[
{
"input": "5\n7 5 8 6 9",
"output": "Yes"
},
{
"input": "5\n-1 -2 0 0 -5",
"output": "No"
},
{
"input": "5\n5 4 3 2 1",
"output": "No"
},
{
"input": "5\n1000000000 0 0 0 0",
"output": "Yes"
},
{
"input": "5\n1000000000 1 0 -999999999 -1000000000",
"output": "Yes"
},
{
"input": "3\n998 244 353",
"output": "Yes"
},
{
"input": "3\n-1000000000 0 1000000000",
"output": "No"
},
{
"input": "5\n-1 -1 -1 -1 1",
"output": "Yes"
},
{
"input": "4\n-9763 530 3595 6660",
"output": "Yes"
},
{
"input": "4\n-253090305 36298498 374072642 711846786",
"output": "Yes"
},
{
"input": "5\n-186772848 -235864239 -191561068 -193955178 -243046569",
"output": "Yes"
},
{
"input": "5\n-954618456 -522919664 -248330428 -130850748 300848044",
"output": "Yes"
},
{
"input": "10\n4846 6705 2530 5757 5283 -944 -2102 -3260 -4418 2913",
"output": "No"
},
{
"input": "10\n-6568 -5920 -5272 -4624 -2435 -635 -2680 -2032 -1384 6565",
"output": "No"
},
{
"input": "20\n319410377 286827025 254243673 221660321 189076969 156493617 123910265 91326913 58743561 26160209 -6423143 -39006495 -71589847 -104173199 -136756551 -169339903 -201923255 -234506607 -267089959 -299673311",
"output": "No"
},
{
"input": "20\n-975467170 758268840 -975467171 758268839 -975467172 758268838 -975467173 758268837 -975467174 758268836 -975467175 758268835 -975467176 758268834 -975467177 758268833 -975467178 758268832 -975467179 758268831",
"output": "Yes"
},
{
"input": "4\n1 0 3 0",
"output": "No"
},
{
"input": "4\n100 2 3 4",
"output": "Yes"
},
{
"input": "5\n7 5 8 6 3",
"output": "No"
},
{
"input": "3\n1000000000 1000000000 -1000000000",
"output": "Yes"
},
{
"input": "4\n1 0 1 4",
"output": "Yes"
},
{
"input": "7\n1 2 -1 0 1 6 7",
"output": "Yes"
},
{
"input": "4\n0 0 4 0",
"output": "Yes"
},
{
"input": "7\n0 0 2 3 4 5 5",
"output": "Yes"
},
{
"input": "5\n7 5 8 6 8",
"output": "No"
},
{
"input": "5\n1 2 9 4 5",
"output": "Yes"
},
{
"input": "8\n1 12 3 14 5 16 7 8",
"output": "Yes"
},
{
"input": "5\n1 6 7 4 9",
"output": "Yes"
},
{
"input": "5\n2 1 0 1 2",
"output": "No"
},
{
"input": "4\n0 0 1 3",
"output": "Yes"
},
{
"input": "4\n100 50 50 10000000",
"output": "No"
},
{
"input": "5\n1 2 3 3 3",
"output": "No"
},
{
"input": "5\n1 2 6 10 17",
"output": "Yes"
},
{
"input": "4\n1 3 4 4",
"output": "Yes"
},
{
"input": "4\n100 50 50 1000000",
"output": "No"
},
{
"input": "6\n1 2 4 5 7 9",
"output": "No"
},
{
"input": "6\n0 0 1 2 3 4",
"output": "Yes"
},
{
"input": "5\n7 5 9 10 8",
"output": "Yes"
},
{
"input": "7\n1 2 2 1 2 2 1",
"output": "Yes"
},
{
"input": "4\n2 2 4 5",
"output": "Yes"
},
{
"input": "6\n1 2 1 3 4 5",
"output": "No"
},
{
"input": "4\n1 3 3 6",
"output": "No"
},
{
"input": "5\n1 2 -3 4 -1",
"output": "Yes"
}
] | 1,504,276,709
| 3,809
|
PyPy 3
|
WRONG_ANSWER
|
PRETESTS
| 6
| 93
| 0
|
from math import *
import itertools as it
from collections import *
EPS = 1e-9
def get_int() : return int(input().strip())
def get_string(): return input().strip()
def get_array() : return list(map(int, input().strip().split(' ')))
def print_array(a, glue = ' '):
print(glue.join(map(str, a)))
def print_grid(grid, glue = ' '):
for row in grid:
print_array(row, glue)
#====================================#
def same(a, i, j, k):
q = (a[i] - a[j]) / (i - j)
return (a[i] - a[k]) / (i - k) == q
def solve():
n = get_int()
a = get_array()
if n == 1: return "No"
if n < 4: return "Yes"
if all(same(a, 0, 1, i) for i in range(2, n)): return "No"
if all(same(a, 1, 2, i) for i in range(3, n)): return "Yes"
for i in range(1, n):
c = [False] * n
c[0] = True
c[i] = True
for j in range(1, n):
if same(a, 0, i, j): c[j] = True
r = [ i for i in range(n) if not c[i] ]
if len(r) == 0: return "No"
if len(r) == 1: return "Yes"
if all(same(a, r[0], r[1], i) for i in r[2:]):
q = (a[i] - a[0]) / i
if (a[r[1]] - a[r[0]]) / (r[1] - r[0]) == q: return "Yes"
return "No"
ntest = 1
# ntest = get_int()
for _ in range(ntest):
print(solve())
|
Title: Tell Your World
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Connect the countless points with lines, till we reach the faraway yonder.
There are *n* points on a coordinate plane, the *i*-th of which being (*i*,<=*y**i*).
Determine whether it's possible to draw two parallel and non-overlapping lines, such that every point in the set lies on exactly one of them, and each of them passes through at least one point in the set.
Input Specification:
The first line of input contains a positive integer *n* (3<=≤<=*n*<=≤<=1<=000) — the number of points.
The second line contains *n* space-separated integers *y*1,<=*y*2,<=...,<=*y**n* (<=-<=109<=≤<=*y**i*<=≤<=109) — the vertical coordinates of each point.
Output Specification:
Output "Yes" (without quotes) if it's possible to fulfill the requirements, and "No" otherwise.
You can print each letter in any case (upper or lower).
Demo Input:
['5\n7 5 8 6 9\n', '5\n-1 -2 0 0 -5\n', '5\n5 4 3 2 1\n', '5\n1000000000 0 0 0 0\n']
Demo Output:
['Yes\n', 'No\n', 'No\n', 'Yes\n']
Note:
In the first example, there are five points: (1, 7), (2, 5), (3, 8), (4, 6) and (5, 9). It's possible to draw a line that passes through points 1, 3, 5, and another one that passes through points 2, 4 and is parallel to the first one.
In the second example, while it's possible to draw two lines that cover all points, they cannot be made parallel.
In the third example, it's impossible to satisfy both requirements at the same time.
|
```python
from math import *
import itertools as it
from collections import *
EPS = 1e-9
def get_int() : return int(input().strip())
def get_string(): return input().strip()
def get_array() : return list(map(int, input().strip().split(' ')))
def print_array(a, glue = ' '):
print(glue.join(map(str, a)))
def print_grid(grid, glue = ' '):
for row in grid:
print_array(row, glue)
#====================================#
def same(a, i, j, k):
q = (a[i] - a[j]) / (i - j)
return (a[i] - a[k]) / (i - k) == q
def solve():
n = get_int()
a = get_array()
if n == 1: return "No"
if n < 4: return "Yes"
if all(same(a, 0, 1, i) for i in range(2, n)): return "No"
if all(same(a, 1, 2, i) for i in range(3, n)): return "Yes"
for i in range(1, n):
c = [False] * n
c[0] = True
c[i] = True
for j in range(1, n):
if same(a, 0, i, j): c[j] = True
r = [ i for i in range(n) if not c[i] ]
if len(r) == 0: return "No"
if len(r) == 1: return "Yes"
if all(same(a, r[0], r[1], i) for i in r[2:]):
q = (a[i] - a[0]) / i
if (a[r[1]] - a[r[0]]) / (r[1] - r[0]) == q: return "Yes"
return "No"
ntest = 1
# ntest = get_int()
for _ in range(ntest):
print(solve())
```
| 0
|
|
745
|
A
|
Hongcow Learns the Cyclic Shift
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] | null | null |
Hongcow is learning to spell! One day, his teacher gives him a word that he needs to learn to spell. Being a dutiful student, he immediately learns how to spell the word.
Hongcow has decided to try to make new words from this one. He starts by taking the word he just learned how to spell, and moves the last character of the word to the beginning of the word. He calls this a cyclic shift. He can apply cyclic shift many times. For example, consecutively applying cyclic shift operation to the word "abracadabra" Hongcow will get words "aabracadabr", "raabracadab" and so on.
Hongcow is now wondering how many distinct words he can generate by doing the cyclic shift arbitrarily many times. The initial string is also counted.
|
The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=50), the word Hongcow initially learns how to spell. The string *s* consists only of lowercase English letters ('a'–'z').
|
Output a single integer equal to the number of distinct strings that Hongcow can obtain by applying the cyclic shift arbitrarily many times to the given string.
|
[
"abcd\n",
"bbb\n",
"yzyz\n"
] |
[
"4\n",
"1\n",
"2\n"
] |
For the first sample, the strings Hongcow can generate are "abcd", "dabc", "cdab", and "bcda".
For the second sample, no matter how many times Hongcow does the cyclic shift, Hongcow can only generate "bbb".
For the third sample, the two strings Hongcow can generate are "yzyz" and "zyzy".
| 500
|
[
{
"input": "abcd",
"output": "4"
},
{
"input": "bbb",
"output": "1"
},
{
"input": "yzyz",
"output": "2"
},
{
"input": "abcdefghijklmnopqrstuvwxyabcdefghijklmnopqrstuvwxy",
"output": "25"
},
{
"input": "zclkjadoprqronzclkjadoprqronzclkjadoprqron",
"output": "14"
},
{
"input": "zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "1"
},
{
"input": "xyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxyxy",
"output": "2"
},
{
"input": "y",
"output": "1"
},
{
"input": "ervbfotfedpozygoumbmxeaqegouaqqzqerlykhmvxvvlcaos",
"output": "49"
},
{
"input": "zyzzzyyzyyyzyyzyzyzyzyzzzyyyzzyzyyzzzzzyyyzzzzyzyy",
"output": "50"
},
{
"input": "zzfyftdezzfyftdezzfyftdezzfyftdezzfyftdezzfyftde",
"output": "8"
},
{
"input": "yehcqdlllqpuxdsaicyjjxiylahgxbygmsopjbxhtimzkashs",
"output": "49"
},
{
"input": "yyyyzzzyzzzyzyzyzyyyyyzzyzyzyyyyyzyzyyyzyzzyyzzzz",
"output": "49"
},
{
"input": "zkqcrhzlzsnwzkqcrhzlzsnwzkqcrhzlzsnwzkqcrhzlzsnw",
"output": "12"
},
{
"input": "xxyxxyxxyxxyxxyxxyxxyxxyxxyxxyxxyxxyxxyxxyxxyxxy",
"output": "3"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaabaaaaaaaaaaaaaaaaaaaaaaaab",
"output": "25"
},
{
"input": "aabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaabaaaba",
"output": "4"
},
{
"input": "pqqpqqpqqpqqpqqpqqpqqpqqpqqpqqpqqppqppqppqppqppq",
"output": "48"
},
{
"input": "zxkljaqzxkljaqzxkljaqzxkljaqzxrljaqzxkljaqzxkljaq",
"output": "49"
},
{
"input": "abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwx",
"output": "50"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaz",
"output": "50"
},
{
"input": "abcddcba",
"output": "8"
},
{
"input": "aabaabaabaacaabaabaabaacaabaabaabaacaabaabaabaac",
"output": "12"
},
{
"input": "aabaabcaabaabcdaabaabcaabaabcd",
"output": "15"
},
{
"input": "ababaababaaababaababaaaababaababaaababaababaaaa",
"output": "47"
},
{
"input": "ababaababaaababaababaaaababaababaaababaababaaa",
"output": "23"
},
{
"input": "aaababaab",
"output": "9"
},
{
"input": "aba",
"output": "3"
}
] | 1,555,835,013
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 28
| 109
| 0
|
s=input()
tool=len(s)-1
tmp={s:1}
ans=1
for i in range(tool,0,-1):
s=s[tool]+s[:tool]
if tmp.get(s,-1)==-1:
tmp[s]=1
ans+=1
print(ans)
'''
abcd
dabc
cdab
bcda
abcd
'''
|
Title: Hongcow Learns the Cyclic Shift
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Hongcow is learning to spell! One day, his teacher gives him a word that he needs to learn to spell. Being a dutiful student, he immediately learns how to spell the word.
Hongcow has decided to try to make new words from this one. He starts by taking the word he just learned how to spell, and moves the last character of the word to the beginning of the word. He calls this a cyclic shift. He can apply cyclic shift many times. For example, consecutively applying cyclic shift operation to the word "abracadabra" Hongcow will get words "aabracadabr", "raabracadab" and so on.
Hongcow is now wondering how many distinct words he can generate by doing the cyclic shift arbitrarily many times. The initial string is also counted.
Input Specification:
The first line of input will be a single string *s* (1<=≤<=|*s*|<=≤<=50), the word Hongcow initially learns how to spell. The string *s* consists only of lowercase English letters ('a'–'z').
Output Specification:
Output a single integer equal to the number of distinct strings that Hongcow can obtain by applying the cyclic shift arbitrarily many times to the given string.
Demo Input:
['abcd\n', 'bbb\n', 'yzyz\n']
Demo Output:
['4\n', '1\n', '2\n']
Note:
For the first sample, the strings Hongcow can generate are "abcd", "dabc", "cdab", and "bcda".
For the second sample, no matter how many times Hongcow does the cyclic shift, Hongcow can only generate "bbb".
For the third sample, the two strings Hongcow can generate are "yzyz" and "zyzy".
|
```python
s=input()
tool=len(s)-1
tmp={s:1}
ans=1
for i in range(tool,0,-1):
s=s[tool]+s[:tool]
if tmp.get(s,-1)==-1:
tmp[s]=1
ans+=1
print(ans)
'''
abcd
dabc
cdab
bcda
abcd
'''
```
| 3
|
|
810
|
A
|
Straight <<A>>
|
PROGRAMMING
| 900
|
[
"implementation",
"math"
] | null | null |
Noora is a student of one famous high school. It's her final year in school — she is going to study in university next year. However, she has to get an «A» graduation certificate in order to apply to a prestigious one.
In school, where Noora is studying, teachers are putting down marks to the online class register, which are integers from 1 to *k*. The worst mark is 1, the best is *k*. Mark that is going to the certificate, is calculated as an average of all the marks, rounded to the closest integer. If several answers are possible, rounding up is produced. For example, 7.3 is rounded to 7, but 7.5 and 7.8784 — to 8.
For instance, if Noora has marks [8,<=9], then the mark to the certificate is 9, because the average is equal to 8.5 and rounded to 9, but if the marks are [8,<=8,<=9], Noora will have graduation certificate with 8.
To graduate with «A» certificate, Noora has to have mark *k*.
Noora got *n* marks in register this year. However, she is afraid that her marks are not enough to get final mark *k*. Noora decided to ask for help in the internet, where hacker Leha immediately responded to her request. He is ready to hack class register for Noora and to add Noora any number of additional marks from 1 to *k*. At the same time, Leha want his hack be unseen to everyone, so he decided to add as less as possible additional marks. Please help Leha to calculate the minimal number of marks he has to add, so that final Noora's mark will become equal to *k*.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*k*<=≤<=100) denoting the number of marks, received by Noora and the value of highest possible mark.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*k*) denoting marks received by Noora before Leha's hack.
|
Print a single integer — minimal number of additional marks, that Leha has to add in order to change Noora's final mark to *k*.
|
[
"2 10\n8 9\n",
"3 5\n4 4 4\n"
] |
[
"4",
"3"
] |
Consider the first example testcase.
Maximal mark is 10, Noora received two marks — 8 and 9, so current final mark is 9. To fix it, Leha can add marks [10, 10, 10, 10] (4 marks in total) to the registry, achieving Noora having average mark equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/1b961585522f76271546da990a6228e7c666277f.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Consequently, new final mark is 10. Less number of marks won't fix the situation.
In the second example Leha can add [5, 5, 5] to the registry, so that making average mark equal to 4.5, which is enough to have 5 in the certificate.
| 500
|
[
{
"input": "2 10\n8 9",
"output": "4"
},
{
"input": "3 5\n4 4 4",
"output": "3"
},
{
"input": "3 10\n10 8 9",
"output": "3"
},
{
"input": "2 23\n21 23",
"output": "2"
},
{
"input": "5 10\n5 10 10 9 10",
"output": "7"
},
{
"input": "12 50\n18 10 26 22 22 23 14 21 27 18 25 12",
"output": "712"
},
{
"input": "38 12\n2 7 10 8 5 3 5 6 3 6 5 1 9 7 7 8 3 4 4 4 5 2 3 6 6 1 6 7 4 4 8 7 4 5 3 6 6 6",
"output": "482"
},
{
"input": "63 86\n32 31 36 29 36 26 28 38 39 32 29 26 33 38 36 38 36 28 43 48 28 33 25 39 39 27 34 25 37 28 40 26 30 31 42 32 36 44 29 36 30 35 48 40 26 34 30 33 33 46 42 24 36 38 33 51 33 41 38 29 29 32 28",
"output": "6469"
},
{
"input": "100 38\n30 24 38 31 31 33 32 32 29 34 29 22 27 23 34 25 32 30 30 26 16 27 38 33 38 38 37 34 32 27 33 23 33 32 24 24 30 36 29 30 33 30 29 30 36 33 33 35 28 24 30 32 38 29 30 36 31 30 27 38 31 36 15 37 32 27 29 24 38 33 28 29 34 21 37 35 32 31 27 25 27 28 31 31 36 38 35 35 36 29 35 22 38 31 38 28 31 27 34 31",
"output": "1340"
},
{
"input": "33 69\n60 69 68 69 69 60 64 60 62 59 54 47 60 62 69 69 69 58 67 69 62 69 68 53 69 69 66 66 57 58 65 69 61",
"output": "329"
},
{
"input": "39 92\n19 17 16 19 15 30 21 25 14 17 19 19 23 16 14 15 17 19 29 15 11 25 19 14 18 20 10 16 11 15 18 20 20 17 18 16 12 17 16",
"output": "5753"
},
{
"input": "68 29\n29 29 29 29 29 28 29 29 29 27 29 29 29 29 29 29 29 23 29 29 26 29 29 29 29 29 29 29 29 29 29 29 29 29 29 29 26 29 29 29 29 29 29 29 29 29 29 29 29 22 29 29 29 29 29 29 29 29 29 29 29 29 29 28 29 29 29 29",
"output": "0"
},
{
"input": "75 30\n22 18 21 26 23 18 28 30 24 24 19 25 28 30 23 29 18 23 23 30 26 30 17 30 18 19 25 26 26 15 27 23 30 21 19 26 25 30 25 28 20 22 22 21 26 17 23 23 24 15 25 19 18 22 30 30 29 21 30 28 28 30 27 25 24 15 22 19 30 21 20 30 18 20 25",
"output": "851"
},
{
"input": "78 43\n2 7 6 5 5 6 4 5 3 4 6 8 4 5 5 4 3 1 2 4 4 6 5 6 4 4 6 4 8 4 6 5 6 1 4 5 6 3 2 5 2 5 3 4 8 8 3 3 4 4 6 6 5 4 5 5 7 9 3 9 6 4 7 3 6 9 6 5 1 7 2 5 6 3 6 2 5 4",
"output": "5884"
},
{
"input": "82 88\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 2 1 1 2 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1",
"output": "14170"
},
{
"input": "84 77\n28 26 36 38 37 44 48 34 40 22 42 35 40 37 30 31 33 35 36 55 47 36 33 47 40 38 27 38 36 33 35 31 47 33 30 38 38 47 49 24 38 37 28 43 39 36 34 33 29 38 36 43 48 38 36 34 33 34 35 31 26 33 39 37 37 37 35 52 47 30 24 46 38 26 43 46 41 50 33 40 36 41 37 30",
"output": "6650"
},
{
"input": "94 80\n21 19 15 16 27 16 20 18 19 19 15 15 20 19 19 21 20 19 13 17 15 9 17 15 23 15 12 18 12 13 15 12 14 13 14 17 20 20 14 21 15 6 10 23 24 8 18 18 13 23 17 22 17 19 19 18 17 24 8 16 18 20 24 19 10 19 15 10 13 14 19 15 16 19 20 15 14 21 16 16 14 14 22 19 12 11 14 13 19 32 16 16 13 20",
"output": "11786"
},
{
"input": "96 41\n13 32 27 34 28 34 30 26 21 24 29 20 25 34 25 16 27 15 22 22 34 22 25 19 23 17 17 22 26 24 23 20 21 27 19 33 13 24 22 18 30 30 27 14 26 24 20 20 22 11 19 31 19 29 18 28 30 22 17 15 28 32 17 24 17 24 24 19 26 23 22 29 18 22 23 29 19 32 26 23 22 22 24 23 27 30 24 25 21 21 33 19 35 27 34 28",
"output": "3182"
},
{
"input": "1 26\n26",
"output": "0"
},
{
"input": "99 39\n25 28 30 28 32 34 31 28 29 28 29 30 33 19 33 31 27 33 29 24 27 30 25 38 28 34 35 31 34 37 30 22 21 24 34 27 34 33 34 33 26 26 36 19 30 22 35 30 21 28 23 35 33 29 21 22 36 31 34 32 34 32 30 32 27 33 38 25 35 26 39 27 29 29 19 33 28 29 34 38 26 30 36 26 29 30 26 34 22 32 29 38 25 27 24 17 25 28 26",
"output": "1807"
},
{
"input": "100 12\n7 6 6 3 5 5 9 8 7 7 4 7 12 6 9 5 6 3 4 7 9 10 7 7 5 3 9 6 9 9 6 7 4 10 4 8 8 6 9 8 6 5 7 4 10 7 5 6 8 9 3 4 8 5 4 8 6 10 5 8 7 5 9 8 5 8 5 6 9 11 4 9 5 5 11 4 6 6 7 3 8 9 6 7 10 4 7 6 9 4 8 11 5 4 10 8 5 10 11 4",
"output": "946"
},
{
"input": "100 18\n1 2 2 2 2 2 1 1 1 2 3 1 3 1 1 4 2 4 1 2 1 2 1 3 2 1 2 1 1 1 2 1 2 2 1 1 4 3 1 1 2 1 3 3 2 1 2 2 1 1 1 1 3 1 1 2 2 1 1 1 5 1 2 1 3 2 2 1 4 2 2 1 1 1 1 1 1 1 1 2 2 1 2 1 1 1 2 1 2 2 2 1 1 3 1 1 2 1 1 2",
"output": "3164"
},
{
"input": "100 27\n16 20 21 10 16 17 18 25 19 18 20 12 11 21 21 23 20 26 20 21 27 16 25 18 25 21 27 12 20 27 18 17 27 13 21 26 12 22 15 21 25 21 18 27 24 15 16 18 23 21 24 27 19 17 24 14 21 16 24 26 13 14 25 18 27 26 22 16 27 27 17 25 17 12 22 10 19 27 19 20 23 22 25 23 17 25 14 20 22 10 22 27 21 20 15 26 24 27 12 16",
"output": "1262"
},
{
"input": "100 29\n20 18 23 24 14 14 16 23 22 17 18 22 21 21 19 19 14 11 18 19 16 22 25 20 14 13 21 24 18 16 18 29 17 25 12 10 18 28 11 16 17 14 15 20 17 20 18 22 10 16 16 20 18 19 29 18 25 27 17 19 24 15 24 25 16 23 19 16 16 20 19 15 12 21 20 13 21 15 15 23 16 23 17 13 17 21 13 18 17 18 18 20 16 12 19 15 27 14 11 18",
"output": "2024"
},
{
"input": "100 30\n16 10 20 11 14 27 15 17 22 26 24 17 15 18 19 22 22 15 21 22 14 21 22 22 21 22 15 17 17 22 18 19 26 18 22 20 22 25 18 18 17 23 18 18 20 13 19 30 17 24 22 19 29 20 20 21 17 18 26 25 22 19 15 18 18 20 19 19 18 18 24 16 19 17 12 21 20 16 23 21 16 17 26 23 25 28 22 20 9 21 17 24 15 19 17 21 29 13 18 15",
"output": "1984"
},
{
"input": "100 59\n56 58 53 59 59 48 59 54 46 59 59 58 48 59 55 59 59 50 59 56 59 59 59 59 59 59 59 57 59 53 45 53 50 59 50 55 58 54 59 56 54 59 59 59 59 48 56 59 59 57 59 59 48 43 55 57 39 59 46 55 55 52 58 57 51 59 59 59 59 53 59 43 51 54 46 59 57 43 50 59 47 58 59 59 59 55 46 56 55 59 56 47 56 56 46 51 47 48 59 55",
"output": "740"
},
{
"input": "100 81\n6 7 6 6 7 6 6 6 3 9 4 5 4 3 4 6 6 6 1 3 9 5 2 3 8 5 6 9 6 6 6 5 4 4 7 7 3 6 11 7 6 4 8 7 12 6 4 10 2 4 9 11 7 4 7 7 8 8 6 7 9 8 4 5 8 13 6 6 6 8 6 2 5 6 7 5 4 4 4 4 2 6 4 8 3 4 7 7 6 7 7 10 5 10 6 7 4 11 8 4",
"output": "14888"
},
{
"input": "100 100\n30 35 23 43 28 49 31 32 30 44 32 37 33 34 38 28 43 32 33 32 50 32 41 38 33 20 40 36 29 21 42 25 23 34 43 32 37 31 30 27 36 32 45 37 33 29 38 34 35 33 28 19 37 33 28 41 31 29 41 27 32 39 30 34 37 40 33 38 35 32 32 34 35 34 28 39 28 34 40 45 31 25 42 28 29 31 33 21 36 33 34 37 40 42 39 30 36 34 34 40",
"output": "13118"
},
{
"input": "100 100\n71 87 100 85 89 98 90 90 71 65 76 75 85 100 81 100 91 80 73 89 86 78 82 89 77 92 78 90 100 81 85 89 73 100 66 60 72 88 91 73 93 76 88 81 86 78 83 77 74 93 97 94 85 78 82 78 91 91 100 78 89 76 78 82 81 78 83 88 87 83 78 98 85 97 98 89 88 75 76 86 74 81 70 76 86 84 99 100 89 94 72 84 82 88 83 89 78 99 87 76",
"output": "3030"
},
{
"input": "100 100\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": "19700"
},
{
"input": "100 100\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": "0"
},
{
"input": "100 100\n1 1 2 1 1 1 1 1 1 1 1 1 1 2 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": "19696"
},
{
"input": "100 100\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": "0"
},
{
"input": "100 100\n100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 98 100 100 100 100 98 100 100 100 100 100 100 99 98 100 100 93 100 100 98 100 100 100 100 93 100 96 100 100 100 94 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 95 88 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100",
"output": "0"
},
{
"input": "100 100\n95 100 100 100 100 100 100 100 100 100 100 100 100 100 87 100 100 100 94 100 100 100 100 100 100 100 100 100 100 100 100 99 100 100 100 100 100 100 100 100 100 100 90 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 97 100 100 100 96 100 98 100 100 100 100 100 96 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 97 100 100 100 100",
"output": "2"
},
{
"input": "100 1\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": "0"
},
{
"input": "100 2\n2 1 1 2 1 1 1 1 2 2 2 2 1 1 1 2 1 1 1 2 2 2 2 1 1 1 1 2 2 2 1 2 2 2 2 1 2 2 1 1 1 1 1 1 2 2 1 2 1 1 1 2 1 2 2 2 2 1 1 1 2 2 1 2 1 1 1 2 1 2 2 1 1 1 2 2 1 1 2 1 1 2 1 1 1 2 1 1 1 1 2 1 1 1 1 2 1 2 1 1",
"output": "16"
},
{
"input": "3 5\n5 5 5",
"output": "0"
},
{
"input": "7 7\n1 1 1 1 1 1 1",
"output": "77"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "100 100\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": "19700"
},
{
"input": "4 10\n10 10 10 10",
"output": "0"
},
{
"input": "1 10\n10",
"output": "0"
},
{
"input": "10 1\n1 1 1 1 1 1 1 1 1 1",
"output": "0"
},
{
"input": "3 10\n10 10 10",
"output": "0"
},
{
"input": "2 4\n3 4",
"output": "0"
},
{
"input": "1 2\n2",
"output": "0"
},
{
"input": "3 4\n4 4 4",
"output": "0"
},
{
"input": "3 2\n2 2 1",
"output": "0"
},
{
"input": "5 5\n5 5 5 5 5",
"output": "0"
},
{
"input": "3 3\n3 3 3",
"output": "0"
},
{
"input": "2 9\n8 9",
"output": "0"
},
{
"input": "3 10\n9 10 10",
"output": "0"
},
{
"input": "1 3\n3",
"output": "0"
},
{
"input": "2 2\n1 2",
"output": "0"
},
{
"input": "2 10\n10 10",
"output": "0"
},
{
"input": "23 14\n7 11 13 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14",
"output": "0"
},
{
"input": "2 10\n9 10",
"output": "0"
},
{
"input": "2 2\n2 2",
"output": "0"
},
{
"input": "10 5\n5 5 5 5 5 5 5 5 5 4",
"output": "0"
},
{
"input": "3 5\n4 5 5",
"output": "0"
},
{
"input": "5 4\n4 4 4 4 4",
"output": "0"
},
{
"input": "2 10\n10 9",
"output": "0"
},
{
"input": "4 5\n3 5 5 5",
"output": "0"
},
{
"input": "10 5\n5 5 5 5 5 5 5 5 5 5",
"output": "0"
},
{
"input": "3 10\n10 10 9",
"output": "0"
},
{
"input": "5 1\n1 1 1 1 1",
"output": "0"
},
{
"input": "2 1\n1 1",
"output": "0"
},
{
"input": "4 10\n9 10 10 10",
"output": "0"
},
{
"input": "5 2\n2 2 2 2 2",
"output": "0"
},
{
"input": "2 5\n4 5",
"output": "0"
},
{
"input": "5 10\n10 10 10 10 10",
"output": "0"
},
{
"input": "2 6\n6 6",
"output": "0"
},
{
"input": "2 9\n9 9",
"output": "0"
},
{
"input": "3 10\n10 9 10",
"output": "0"
},
{
"input": "4 40\n39 40 40 40",
"output": "0"
},
{
"input": "3 4\n3 4 4",
"output": "0"
},
{
"input": "9 9\n9 9 9 9 9 9 9 9 9",
"output": "0"
},
{
"input": "1 4\n4",
"output": "0"
},
{
"input": "4 7\n1 1 1 1",
"output": "44"
},
{
"input": "1 5\n5",
"output": "0"
},
{
"input": "3 1\n1 1 1",
"output": "0"
},
{
"input": "1 100\n100",
"output": "0"
},
{
"input": "2 7\n3 5",
"output": "10"
},
{
"input": "3 6\n6 6 6",
"output": "0"
},
{
"input": "4 2\n1 2 2 2",
"output": "0"
},
{
"input": "4 5\n4 5 5 5",
"output": "0"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "35"
},
{
"input": "66 2\n1 2 2 2 2 1 1 2 1 2 2 2 2 2 2 1 2 1 2 1 2 1 2 1 2 1 1 1 1 2 2 1 2 2 1 1 2 1 2 2 1 1 1 2 1 2 1 2 1 2 1 2 2 2 2 1 2 2 1 2 1 1 1 2 2 1",
"output": "0"
},
{
"input": "2 2\n2 1",
"output": "0"
},
{
"input": "5 5\n5 5 5 4 5",
"output": "0"
},
{
"input": "3 7\n1 1 1",
"output": "33"
},
{
"input": "2 5\n5 5",
"output": "0"
},
{
"input": "1 7\n1",
"output": "11"
},
{
"input": "6 7\n1 1 1 1 1 1",
"output": "66"
},
{
"input": "99 97\n15 80 78 69 12 84 36 51 89 77 88 10 1 19 67 85 6 36 8 70 14 45 88 97 22 13 75 57 83 27 13 97 9 90 68 51 76 37 5 2 16 92 11 48 13 77 35 19 15 74 22 29 21 12 28 42 56 5 32 41 62 75 71 71 68 72 24 77 11 28 78 27 53 88 74 66 1 42 18 16 18 39 75 38 81 5 13 39 40 75 13 36 53 83 9 54 57 63 64",
"output": "10077"
},
{
"input": "8 7\n1 1 1 1 1 1 1 1",
"output": "88"
},
{
"input": "3 2\n2 2 2",
"output": "0"
},
{
"input": "6 5\n5 5 5 5 5 5",
"output": "0"
},
{
"input": "10 5\n5 5 5 5 5 5 5 4 1 1",
"output": "8"
},
{
"input": "1 5\n1",
"output": "7"
},
{
"input": "10 10\n10 10 10 10 10 10 10 10 10 10",
"output": "0"
},
{
"input": "2 3\n2 3",
"output": "0"
},
{
"input": "1 9\n9",
"output": "0"
},
{
"input": "74 2\n2 2 2 2 1 2 2 1 1 1 2 2 1 2 2 2 2 1 2 1 1 1 2 1 1 2 2 1 2 1 1 2 1 1 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 1 1 2 1 1 1 1 1 1 2 2 2 1 1 1 1 1 2 2 2 2 2 2 1 2",
"output": "0"
},
{
"input": "5 5\n5 5 5 5 4",
"output": "0"
}
] | 1,497,468,903
| 902
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 61
| 0
|
n,m=map(int,input().split())
l=list(map(int,input().split()))
k=0
while 1 :
if round((sum(l)+k*m)/(n+k))<m :
k=k+1
else :
break
print(k)
|
Title: Straight <<A>>
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Noora is a student of one famous high school. It's her final year in school — she is going to study in university next year. However, she has to get an «A» graduation certificate in order to apply to a prestigious one.
In school, where Noora is studying, teachers are putting down marks to the online class register, which are integers from 1 to *k*. The worst mark is 1, the best is *k*. Mark that is going to the certificate, is calculated as an average of all the marks, rounded to the closest integer. If several answers are possible, rounding up is produced. For example, 7.3 is rounded to 7, but 7.5 and 7.8784 — to 8.
For instance, if Noora has marks [8,<=9], then the mark to the certificate is 9, because the average is equal to 8.5 and rounded to 9, but if the marks are [8,<=8,<=9], Noora will have graduation certificate with 8.
To graduate with «A» certificate, Noora has to have mark *k*.
Noora got *n* marks in register this year. However, she is afraid that her marks are not enough to get final mark *k*. Noora decided to ask for help in the internet, where hacker Leha immediately responded to her request. He is ready to hack class register for Noora and to add Noora any number of additional marks from 1 to *k*. At the same time, Leha want his hack be unseen to everyone, so he decided to add as less as possible additional marks. Please help Leha to calculate the minimal number of marks he has to add, so that final Noora's mark will become equal to *k*.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*<=≤<=100,<=1<=≤<=*k*<=≤<=100) denoting the number of marks, received by Noora and the value of highest possible mark.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*k*) denoting marks received by Noora before Leha's hack.
Output Specification:
Print a single integer — minimal number of additional marks, that Leha has to add in order to change Noora's final mark to *k*.
Demo Input:
['2 10\n8 9\n', '3 5\n4 4 4\n']
Demo Output:
['4', '3']
Note:
Consider the first example testcase.
Maximal mark is 10, Noora received two marks — 8 and 9, so current final mark is 9. To fix it, Leha can add marks [10, 10, 10, 10] (4 marks in total) to the registry, achieving Noora having average mark equal to <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/1b961585522f76271546da990a6228e7c666277f.png" style="max-width: 100.0%;max-height: 100.0%;"/>. Consequently, new final mark is 10. Less number of marks won't fix the situation.
In the second example Leha can add [5, 5, 5] to the registry, so that making average mark equal to 4.5, which is enough to have 5 in the certificate.
|
```python
n,m=map(int,input().split())
l=list(map(int,input().split()))
k=0
while 1 :
if round((sum(l)+k*m)/(n+k))<m :
k=k+1
else :
break
print(k)
```
| 0
|
|
289
|
B
|
Polo the Penguin and Matrix
|
PROGRAMMING
| 1,400
|
[
"brute force",
"dp",
"implementation",
"sortings",
"ternary search"
] | null | null |
Little penguin Polo has an *n*<=×<=*m* matrix, consisting of integers. Let's index the matrix rows from 1 to *n* from top to bottom and let's index the columns from 1 to *m* from left to right. Let's represent the matrix element on the intersection of row *i* and column *j* as *a**ij*.
In one move the penguin can add or subtract number *d* from some matrix element. Find the minimum number of moves needed to make all matrix elements equal. If the described plan is impossible to carry out, say so.
|
The first line contains three integers *n*, *m* and *d* (1<=≤<=*n*,<=*m*<=≤<=100,<=1<=≤<=*d*<=≤<=104) — the matrix sizes and the *d* parameter. Next *n* lines contain the matrix: the *j*-th integer in the *i*-th row is the matrix element *a**ij* (1<=≤<=*a**ij*<=≤<=104).
|
In a single line print a single integer — the minimum number of moves the penguin needs to make all matrix elements equal. If that is impossible, print "-1" (without the quotes).
|
[
"2 2 2\n2 4\n6 8\n",
"1 2 7\n6 7\n"
] |
[
"4\n",
"-1\n"
] |
none
| 1,000
|
[
{
"input": "2 2 2\n2 4\n6 8",
"output": "4"
},
{
"input": "1 2 7\n6 7",
"output": "-1"
},
{
"input": "3 2 1\n5 7\n1 2\n5 100",
"output": "104"
},
{
"input": "3 3 3\n5 8 5\n11 11 17\n14 5 2",
"output": "12"
},
{
"input": "3 3 3\n5 8 5\n11 11 17\n14 5 3",
"output": "-1"
},
{
"input": "2 2 4\n5 5\n5 5",
"output": "0"
},
{
"input": "7 4 5\n7 7 7 12\n7 12 12 7\n7 7 7 7\n7 7 12 7\n7 7 12 12\n12 12 7 12\n7 7 7 7",
"output": "9"
},
{
"input": "7 7 47\n91 91 91 91 91 91 91\n91 91 91 91 91 91 91\n91 91 91 91 91 91 91\n91 91 91 91 91 91 91\n91 91 91 91 91 91 91\n91 91 91 91 91 91 91\n91 91 91 91 91 91 91",
"output": "0"
},
{
"input": "7 7 47\n47 47 47 47 47 47 47\n47 47 47 47 77 47 47\n47 47 47 47 47 47 47\n47 47 47 47 47 47 47\n47 47 47 47 47 47 47\n47 47 47 47 47 47 47\n47 47 47 47 47 127 47",
"output": "-1"
},
{
"input": "7 5 47\n9583 1734 4601 5353 2110\n3802 5165 7985 6293 324\n7045 653 9160 7891 4930\n1781 3520 4178 2298 3943\n1405 2956 5447 5494 6528\n3097 1640 7750 4883 8032\n4225 8455 1875 4789 4366",
"output": "1508"
},
{
"input": "1 1 1\n1",
"output": "0"
},
{
"input": "1 1 8\n12",
"output": "0"
}
] | 1,699,794,222
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 31
| 92
| 716,800
|
N = 10**4 + 1
n, m, d = map(int,input().split())
a = []
for i in range(n):
a = a + (list(map(int,input().split())))
a.sort()
mid = n*m//2
r = a[0]%d
flag = False
res = 0
for i in range(n*m):
if(a[i]%d!=r):
flag = True
break
res += abs(a[i]-a[mid])//d
if(flag):
print(-1)
else:
print(res)
|
Title: Polo the Penguin and Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little penguin Polo has an *n*<=×<=*m* matrix, consisting of integers. Let's index the matrix rows from 1 to *n* from top to bottom and let's index the columns from 1 to *m* from left to right. Let's represent the matrix element on the intersection of row *i* and column *j* as *a**ij*.
In one move the penguin can add or subtract number *d* from some matrix element. Find the minimum number of moves needed to make all matrix elements equal. If the described plan is impossible to carry out, say so.
Input Specification:
The first line contains three integers *n*, *m* and *d* (1<=≤<=*n*,<=*m*<=≤<=100,<=1<=≤<=*d*<=≤<=104) — the matrix sizes and the *d* parameter. Next *n* lines contain the matrix: the *j*-th integer in the *i*-th row is the matrix element *a**ij* (1<=≤<=*a**ij*<=≤<=104).
Output Specification:
In a single line print a single integer — the minimum number of moves the penguin needs to make all matrix elements equal. If that is impossible, print "-1" (without the quotes).
Demo Input:
['2 2 2\n2 4\n6 8\n', '1 2 7\n6 7\n']
Demo Output:
['4\n', '-1\n']
Note:
none
|
```python
N = 10**4 + 1
n, m, d = map(int,input().split())
a = []
for i in range(n):
a = a + (list(map(int,input().split())))
a.sort()
mid = n*m//2
r = a[0]%d
flag = False
res = 0
for i in range(n*m):
if(a[i]%d!=r):
flag = True
break
res += abs(a[i]-a[mid])//d
if(flag):
print(-1)
else:
print(res)
```
| 3
|
|
908
|
A
|
New Year and Counting Cards
|
PROGRAMMING
| 800
|
[
"brute force",
"implementation"
] | null | null |
Your friend has *n* cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
|
The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit.
|
Print a single integer, the minimum number of cards you must turn over to verify your claim.
|
[
"ee\n",
"z\n",
"0ay1\n"
] |
[
"2\n",
"0\n",
"2\n"
] |
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards.
| 500
|
[
{
"input": "ee",
"output": "2"
},
{
"input": "z",
"output": "0"
},
{
"input": "0ay1",
"output": "2"
},
{
"input": "0abcdefghijklmnopqrstuvwxyz1234567896",
"output": "10"
},
{
"input": "0a0a9e9e2i2i9o9o6u6u9z9z4x4x9b9b",
"output": "18"
},
{
"input": "01234567890123456789012345678901234567890123456789",
"output": "25"
},
{
"input": "qwertyuioplkjhgfdsazxcvbnmqwertyuioplkjhgfdsazxcvb",
"output": "10"
},
{
"input": "cjw2dwmr10pku4yxohe0wglktd",
"output": "4"
},
{
"input": "6z2tx805jie8cfybwtfqvmlveec3iak5z5u3lu62vbxyqht6",
"output": "13"
},
{
"input": "kaq7jyialrfp4ilkni90eq8v3amcbygon7py0hb8z26fbl8ss1",
"output": "13"
},
{
"input": "hpwn50zgbmct80k9rizjqg40nycgs0acwikjqt11nr6m61krfs",
"output": "8"
},
{
"input": "l3rw91a4m25l8iytxyeuixsegzcbm4h41ornf3pixkrmwznrzc",
"output": "14"
},
{
"input": "2222",
"output": "0"
},
{
"input": "13579",
"output": "5"
},
{
"input": "1",
"output": "1"
},
{
"input": "0",
"output": "0"
},
{
"input": "a",
"output": "1"
},
{
"input": "y",
"output": "0"
},
{
"input": "w",
"output": "0"
},
{
"input": "oo",
"output": "2"
},
{
"input": "oy",
"output": "1"
},
{
"input": "yo",
"output": "1"
},
{
"input": "yy",
"output": "0"
},
{
"input": "a0",
"output": "1"
},
{
"input": "a9",
"output": "2"
},
{
"input": "y0",
"output": "0"
},
{
"input": "y7",
"output": "1"
},
{
"input": "0a",
"output": "1"
},
{
"input": "3a",
"output": "2"
},
{
"input": "06",
"output": "0"
},
{
"input": "07",
"output": "1"
},
{
"input": "70",
"output": "1"
},
{
"input": "77",
"output": "2"
},
{
"input": "13570",
"output": "4"
},
{
"input": "0000000000011111",
"output": "5"
},
{
"input": "1357",
"output": "4"
},
{
"input": "uuuuuuuuuuuuuuuuuu",
"output": "18"
},
{
"input": "gabieurat",
"output": "5"
}
] | 1,515,086,064
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 38
| 78
| 5,529,600
|
x=input()
y="aeiou"
c=0
for i in x:
try:
i=int(i)
if i%2==1:
c+=1
except:
for e in y:
if i==e:
c+=1
print(c)
|
Title: New Year and Counting Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Your friend has *n* cards.
You know that each card has a lowercase English letter on one side and a digit on the other.
Currently, your friend has laid out the cards on a table so only one side of each card is visible.
You would like to know if the following statement is true for cards that your friend owns: "If a card has a vowel on one side, then it has an even digit on the other side." More specifically, a vowel is one of 'a', 'e', 'i', 'o' or 'u', and even digit is one of '0', '2', '4', '6' or '8'.
For example, if a card has 'a' on one side, and '6' on the other side, then this statement is true for it. Also, the statement is true, for example, for a card with 'b' and '4', and for a card with 'b' and '3' (since the letter is not a vowel). The statement is false, for example, for card with 'e' and '5'. You are interested if the statement is true for all cards. In particular, if no card has a vowel, the statement is true.
To determine this, you can flip over some cards to reveal the other side. You would like to know what is the minimum number of cards you need to flip in the worst case in order to verify that the statement is true.
Input Specification:
The first and only line of input will contain a string *s* (1<=≤<=|*s*|<=≤<=50), denoting the sides of the cards that you can see on the table currently. Each character of *s* is either a lowercase English letter or a digit.
Output Specification:
Print a single integer, the minimum number of cards you must turn over to verify your claim.
Demo Input:
['ee\n', 'z\n', '0ay1\n']
Demo Output:
['2\n', '0\n', '2\n']
Note:
In the first sample, we must turn over both cards. Note that even though both cards have the same letter, they could possibly have different numbers on the other side.
In the second sample, we don't need to turn over any cards. The statement is vacuously true, since you know your friend has no cards with a vowel on them.
In the third sample, we need to flip the second and fourth cards.
|
```python
x=input()
y="aeiou"
c=0
for i in x:
try:
i=int(i)
if i%2==1:
c+=1
except:
for e in y:
if i==e:
c+=1
print(c)
```
| 3
|
|
154
|
B
|
Colliders
|
PROGRAMMING
| 1,600
|
[
"math",
"number theory"
] | null | null |
By 2312 there were *n* Large Hadron Colliders in the inhabited part of the universe. Each of them corresponded to a single natural number from 1 to *n*. However, scientists did not know what activating several colliders simultaneously could cause, so the colliders were deactivated.
In 2312 there was a startling discovery: a collider's activity is safe if and only if all numbers of activated colliders are pairwise relatively prime to each other (two numbers are relatively prime if their greatest common divisor equals 1)! If two colliders with relatively nonprime numbers are activated, it will cause a global collapse.
Upon learning this, physicists rushed to turn the colliders on and off and carry out all sorts of experiments. To make sure than the scientists' quickness doesn't end with big trouble, the Large Hadron Colliders' Large Remote Control was created. You are commissioned to write the software for the remote (well, you do not expect anybody to operate it manually, do you?).
Initially, all colliders are deactivated. Your program receives multiple requests of the form "activate/deactivate the *i*-th collider". The program should handle requests in the order of receiving them. The program should print the processed results in the format described below.
To the request of "+ i" (that is, to activate the *i*-th collider), the program should print exactly one of the following responses:
- "Success" if the activation was successful. - "Already on", if the *i*-th collider was already activated before the request. - "Conflict with j", if there is a conflict with the *j*-th collider (that is, the *j*-th collider is on, and numbers *i* and *j* are not relatively prime). In this case, the *i*-th collider shouldn't be activated. If a conflict occurs with several colliders simultaneously, you should print the number of any of them.
The request of "- i" (that is, to deactivate the *i*-th collider), should receive one of the following responses from the program:
- "Success", if the deactivation was successful. - "Already off", if the *i*-th collider was already deactivated before the request.
You don't need to print quotes in the output of the responses to the requests.
|
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of colliders and the number of requests, correspondingly.
Next *m* lines contain numbers of requests, one per line, in the form of either "+ i" (without the quotes) — activate the *i*-th collider, or "- i" (without the quotes) — deactivate the *i*-th collider (1<=≤<=*i*<=≤<=*n*).
|
Print *m* lines — the results of executing requests in the above given format. The requests should be processed in the order, in which they are given in the input. Don't forget that the responses to the requests should be printed without quotes.
|
[
"10 10\n+ 6\n+ 10\n+ 5\n- 10\n- 5\n- 6\n+ 10\n+ 3\n+ 6\n+ 3\n"
] |
[
"Success\nConflict with 6\nSuccess\nAlready off\nSuccess\nSuccess\nSuccess\nSuccess\nConflict with 10\nAlready on\n"
] |
Note that in the sample the colliders don't turn on after the second and ninth requests. The ninth request could also receive response "Conflict with 3".
| 1,000
|
[
{
"input": "10 10\n+ 6\n+ 10\n+ 5\n- 10\n- 5\n- 6\n+ 10\n+ 3\n+ 6\n+ 3",
"output": "Success\nConflict with 6\nSuccess\nAlready off\nSuccess\nSuccess\nSuccess\nSuccess\nConflict with 10\nAlready on"
},
{
"input": "7 5\n+ 7\n+ 6\n+ 4\n+ 3\n- 7",
"output": "Success\nSuccess\nConflict with 6\nConflict with 6\nSuccess"
},
{
"input": "10 5\n+ 2\n- 8\n- 4\n- 10\n+ 1",
"output": "Success\nAlready off\nAlready off\nAlready off\nSuccess"
},
{
"input": "10 10\n+ 1\n+ 10\n- 1\n- 10\n+ 1\n- 1\n+ 7\n+ 8\n+ 6\n- 7",
"output": "Success\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nConflict with 8\nSuccess"
},
{
"input": "15 15\n+ 12\n+ 6\n+ 13\n- 13\n+ 7\n+ 14\n+ 8\n+ 13\n- 13\n+ 15\n+ 4\n+ 10\n+ 11\n+ 2\n- 14",
"output": "Success\nConflict with 12\nSuccess\nSuccess\nSuccess\nConflict with 12\nConflict with 12\nSuccess\nSuccess\nConflict with 12\nConflict with 12\nConflict with 12\nSuccess\nConflict with 12\nAlready off"
},
{
"input": "2 20\n+ 1\n+ 2\n- 2\n+ 2\n- 1\n- 2\n+ 2\n- 2\n+ 2\n+ 1\n- 1\n+ 1\n- 1\n- 2\n+ 1\n- 1\n+ 1\n- 1\n+ 2\n+ 1",
"output": "Success\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess"
},
{
"input": "2 20\n- 1\n- 2\n- 1\n- 2\n+ 2\n+ 1\n- 1\n+ 1\n+ 1\n+ 2\n- 2\n+ 1\n- 2\n+ 2\n+ 1\n+ 1\n+ 1\n- 1\n- 1\n- 2",
"output": "Already off\nAlready off\nAlready off\nAlready off\nSuccess\nSuccess\nSuccess\nSuccess\nAlready on\nAlready on\nSuccess\nAlready on\nAlready off\nSuccess\nAlready on\nAlready on\nAlready on\nSuccess\nAlready off\nSuccess"
},
{
"input": "25 20\n+ 7\n+ 14\n- 7\n+ 11\n+ 15\n+ 10\n+ 20\n- 15\n+ 13\n- 14\n+ 4\n- 11\n- 20\n+ 15\n+ 16\n+ 3\n+ 11\n+ 22\n- 16\n- 22",
"output": "Success\nConflict with 7\nSuccess\nSuccess\nSuccess\nConflict with 15\nConflict with 15\nSuccess\nSuccess\nAlready off\nSuccess\nSuccess\nAlready off\nSuccess\nConflict with 4\nConflict with 15\nSuccess\nConflict with 4\nAlready off\nAlready off"
},
{
"input": "50 30\n- 39\n- 2\n+ 37\n- 10\n+ 27\n- 25\n+ 41\n+ 23\n- 36\n+ 49\n+ 5\n- 28\n+ 22\n+ 45\n+ 1\n+ 23\n+ 36\n+ 35\n- 4\n- 28\n- 10\n- 36\n- 38\n- 2\n- 38\n- 38\n- 37\n+ 8\n- 27\n- 28",
"output": "Already off\nAlready off\nSuccess\nAlready off\nSuccess\nAlready off\nSuccess\nSuccess\nAlready off\nSuccess\nSuccess\nAlready off\nSuccess\nConflict with 27\nSuccess\nAlready on\nConflict with 22\nConflict with 5\nAlready off\nAlready off\nAlready off\nAlready off\nAlready off\nAlready off\nAlready off\nAlready off\nSuccess\nConflict with 22\nSuccess\nAlready off"
},
{
"input": "50 50\n+ 14\n+ 4\n+ 20\n+ 37\n+ 50\n+ 46\n+ 19\n- 20\n+ 25\n+ 47\n+ 10\n+ 6\n+ 34\n+ 12\n+ 41\n- 47\n+ 9\n+ 22\n+ 28\n- 41\n- 34\n+ 47\n+ 40\n- 12\n+ 42\n- 9\n- 4\n+ 15\n- 15\n+ 27\n+ 8\n+ 38\n+ 9\n+ 4\n+ 17\n- 8\n+ 13\n- 47\n+ 7\n- 9\n- 38\n+ 30\n+ 48\n- 50\n- 7\n+ 41\n+ 34\n+ 23\n+ 11\n+ 16",
"output": "Success\nConflict with 14\nConflict with 14\nSuccess\nConflict with 14\nConflict with 14\nSuccess\nAlready off\nSuccess\nSuccess\nConflict with 14\nConflict with 14\nConflict with 14\nConflict with 14\nSuccess\nSuccess\nSuccess\nConflict with 14\nConflict with 14\nSuccess\nAlready off\nSuccess\nConflict with 14\nAlready off\nConflict with 14\nSuccess\nAlready off\nConflict with 25\nAlready off\nSuccess\nConflict with 14\nConflict with 14\nConflict with 27\nConflict with 14\nSuccess\nAlready off\nSuccess\nS..."
},
{
"input": "100 1\n+ 51",
"output": "Success"
},
{
"input": "1 100\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1\n+ 1\n- 1",
"output": "Success\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess..."
},
{
"input": "100 50\n+ 2\n+ 3\n+ 5\n+ 7\n+ 11\n+ 13\n+ 17\n+ 19\n+ 23\n+ 29\n+ 31\n+ 37\n+ 41\n+ 43\n+ 47\n+ 53\n+ 59\n+ 61\n+ 67\n+ 71\n+ 73\n+ 79\n+ 83\n+ 89\n+ 97\n+ 52\n+ 96\n+ 54\n+ 56\n+ 88\n+ 69\n+ 65\n+ 84\n+ 10\n+ 85\n- 37\n+ 80\n- 53\n+ 25\n- 5\n+ 45\n+ 90\n+ 95\n+ 33\n+ 81\n+ 6\n+ 20\n- 10\n+ 94\n- 61",
"output": "Success\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nSuccess\nConflict with 2\nConflict with 2\nConflict with 2\nConflict with 2\nConflict with 2\nConflict with 3\nConflict with 5\nConflict with 2\nConflict with 2\nConflict with 5\nSuccess\nConflict with 2\nSuccess\nConflict with 5\nSuccess\nConflict with 3\nConflict with 2\nConflict with 19\nCon..."
},
{
"input": "100000 1\n+ 12345",
"output": "Success"
},
{
"input": "4 2\n+ 2\n+ 4",
"output": "Success\nConflict with 2"
},
{
"input": "100000 2\n+ 57314\n+ 85971",
"output": "Success\nConflict with 57314"
},
{
"input": "100000 4\n+ 81799\n+ 81799\n- 81799\n+ 81799",
"output": "Success\nAlready on\nSuccess\nSuccess"
}
] | 1,671,642,664
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 65
| 1,216
| 11,264,000
|
n, m = map(int, input().split())
n += 1
s = [[] for i in range(n)]
for j in range(2, n, 2):
s[j] = [2]
for i in range(3, n, 2):
if s[i]: continue
for j in range(i, n, i):
s[j].append(i)
present, alreadyOn, status = {}, set(), ['']*m
for c in range(m):
inp = list(map(str, input().split()))
collider = int(inp[1])
if (inp[0] == '+'):
if (collider in alreadyOn):
status[c] = 'Already on'
continue
for i in s[collider]:
if (i in present):
status[c] = 'Conflict with ' + str(present[i])
break
else:
status[c] = 'Success'
alreadyOn.add(collider)
for i in s[collider]:
present[i] = collider
else:
if (collider in alreadyOn):
status[c] = 'Success'
for i in s[collider]:
present.pop(i)
alreadyOn.remove(collider)
else:
status[c] = 'Already off'
for i in status:
print(i)
|
Title: Colliders
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
By 2312 there were *n* Large Hadron Colliders in the inhabited part of the universe. Each of them corresponded to a single natural number from 1 to *n*. However, scientists did not know what activating several colliders simultaneously could cause, so the colliders were deactivated.
In 2312 there was a startling discovery: a collider's activity is safe if and only if all numbers of activated colliders are pairwise relatively prime to each other (two numbers are relatively prime if their greatest common divisor equals 1)! If two colliders with relatively nonprime numbers are activated, it will cause a global collapse.
Upon learning this, physicists rushed to turn the colliders on and off and carry out all sorts of experiments. To make sure than the scientists' quickness doesn't end with big trouble, the Large Hadron Colliders' Large Remote Control was created. You are commissioned to write the software for the remote (well, you do not expect anybody to operate it manually, do you?).
Initially, all colliders are deactivated. Your program receives multiple requests of the form "activate/deactivate the *i*-th collider". The program should handle requests in the order of receiving them. The program should print the processed results in the format described below.
To the request of "+ i" (that is, to activate the *i*-th collider), the program should print exactly one of the following responses:
- "Success" if the activation was successful. - "Already on", if the *i*-th collider was already activated before the request. - "Conflict with j", if there is a conflict with the *j*-th collider (that is, the *j*-th collider is on, and numbers *i* and *j* are not relatively prime). In this case, the *i*-th collider shouldn't be activated. If a conflict occurs with several colliders simultaneously, you should print the number of any of them.
The request of "- i" (that is, to deactivate the *i*-th collider), should receive one of the following responses from the program:
- "Success", if the deactivation was successful. - "Already off", if the *i*-th collider was already deactivated before the request.
You don't need to print quotes in the output of the responses to the requests.
Input Specification:
The first line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of colliders and the number of requests, correspondingly.
Next *m* lines contain numbers of requests, one per line, in the form of either "+ i" (without the quotes) — activate the *i*-th collider, or "- i" (without the quotes) — deactivate the *i*-th collider (1<=≤<=*i*<=≤<=*n*).
Output Specification:
Print *m* lines — the results of executing requests in the above given format. The requests should be processed in the order, in which they are given in the input. Don't forget that the responses to the requests should be printed without quotes.
Demo Input:
['10 10\n+ 6\n+ 10\n+ 5\n- 10\n- 5\n- 6\n+ 10\n+ 3\n+ 6\n+ 3\n']
Demo Output:
['Success\nConflict with 6\nSuccess\nAlready off\nSuccess\nSuccess\nSuccess\nSuccess\nConflict with 10\nAlready on\n']
Note:
Note that in the sample the colliders don't turn on after the second and ninth requests. The ninth request could also receive response "Conflict with 3".
|
```python
n, m = map(int, input().split())
n += 1
s = [[] for i in range(n)]
for j in range(2, n, 2):
s[j] = [2]
for i in range(3, n, 2):
if s[i]: continue
for j in range(i, n, i):
s[j].append(i)
present, alreadyOn, status = {}, set(), ['']*m
for c in range(m):
inp = list(map(str, input().split()))
collider = int(inp[1])
if (inp[0] == '+'):
if (collider in alreadyOn):
status[c] = 'Already on'
continue
for i in s[collider]:
if (i in present):
status[c] = 'Conflict with ' + str(present[i])
break
else:
status[c] = 'Success'
alreadyOn.add(collider)
for i in s[collider]:
present[i] = collider
else:
if (collider in alreadyOn):
status[c] = 'Success'
for i in s[collider]:
present.pop(i)
alreadyOn.remove(collider)
else:
status[c] = 'Already off'
for i in status:
print(i)
```
| 3
|
|
699
|
A
|
Launch of Collider
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
There will be a launch of a new, powerful and unusual collider very soon, which located along a straight line. *n* particles will be launched inside it. All of them are located in a straight line and there can not be two or more particles located in the same point. The coordinates of the particles coincide with the distance in meters from the center of the collider, *x**i* is the coordinate of the *i*-th particle and its position in the collider at the same time. All coordinates of particle positions are even integers.
You know the direction of each particle movement — it will move to the right or to the left after the collider's launch start. All particles begin to move simultaneously at the time of the collider's launch start. Each particle will move straight to the left or straight to the right with the constant speed of 1 meter per microsecond. The collider is big enough so particles can not leave it in the foreseeable time.
Write the program which finds the moment of the first collision of any two particles of the collider. In other words, find the number of microseconds before the first moment when any two particles are at the same point.
|
The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of particles.
The second line contains *n* symbols "L" and "R". If the *i*-th symbol equals "L", then the *i*-th particle will move to the left, otherwise the *i*-th symbol equals "R" and the *i*-th particle will move to the right.
The third line contains the sequence of pairwise distinct even integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=109) — the coordinates of particles in the order from the left to the right. It is guaranteed that the coordinates of particles are given in the increasing order.
|
In the first line print the only integer — the first moment (in microseconds) when two particles are at the same point and there will be an explosion.
Print the only integer -1, if the collision of particles doesn't happen.
|
[
"4\nRLRL\n2 4 6 10\n",
"3\nLLR\n40 50 60\n"
] |
[
"1\n",
"-1\n"
] |
In the first sample case the first explosion will happen in 1 microsecond because the particles number 1 and 2 will simultaneously be at the same point with the coordinate 3.
In the second sample case there will be no explosion because there are no particles which will simultaneously be at the same point.
| 500
|
[
{
"input": "4\nRLRL\n2 4 6 10",
"output": "1"
},
{
"input": "3\nLLR\n40 50 60",
"output": "-1"
},
{
"input": "4\nRLLR\n46 230 264 470",
"output": "92"
},
{
"input": "6\nLLRLLL\n446 492 650 844 930 970",
"output": "97"
},
{
"input": "8\nRRLLLLLL\n338 478 512 574 594 622 834 922",
"output": "17"
},
{
"input": "10\nLRLRLLRRLR\n82 268 430 598 604 658 670 788 838 1000",
"output": "3"
},
{
"input": "2\nRL\n0 1000000000",
"output": "500000000"
},
{
"input": "12\nLRLLRRRRLRLL\n254 1260 1476 1768 2924 4126 4150 4602 5578 7142 8134 9082",
"output": "108"
},
{
"input": "14\nRLLRRLRLLRLLLR\n698 2900 3476 3724 3772 3948 4320 4798 5680 6578 7754 8034 8300 8418",
"output": "88"
},
{
"input": "16\nRRLLLRLRLLLLRLLR\n222 306 968 1060 1636 1782 2314 2710 3728 4608 5088 6790 6910 7156 7418 7668",
"output": "123"
},
{
"input": "18\nRLRLLRRRLLLRLRRLRL\n1692 2028 2966 3008 3632 4890 5124 5838 6596 6598 6890 8294 8314 8752 8868 9396 9616 9808",
"output": "10"
},
{
"input": "20\nRLLLLLLLRRRRLRRLRRLR\n380 902 1400 1834 2180 2366 2562 2596 2702 2816 3222 3238 3742 5434 6480 7220 7410 8752 9708 9970",
"output": "252"
},
{
"input": "22\nLRRRRRRRRRRRLLRRRRRLRL\n1790 2150 2178 2456 2736 3282 3622 4114 4490 4772 5204 5240 5720 5840 5910 5912 6586 7920 8584 9404 9734 9830",
"output": "48"
},
{
"input": "24\nLLRLRRLLRLRRRRLLRRLRLRRL\n100 360 864 1078 1360 1384 1438 2320 2618 3074 3874 3916 3964 5178 5578 6278 6630 6992 8648 8738 8922 8930 9276 9720",
"output": "27"
},
{
"input": "26\nRLLLLLLLRLRRLRLRLRLRLLLRRR\n908 1826 2472 2474 2728 3654 3716 3718 3810 3928 4058 4418 4700 5024 5768 6006 6128 6386 6968 7040 7452 7774 7822 8726 9338 9402",
"output": "59"
},
{
"input": "28\nRRLRLRRRRRRLLLRRLRRLLLRRLLLR\n156 172 1120 1362 2512 3326 3718 4804 4990 5810 6242 6756 6812 6890 6974 7014 7088 7724 8136 8596 8770 8840 9244 9250 9270 9372 9400 9626",
"output": "10"
},
{
"input": "30\nRLLRLRLLRRRLRRRLLLLLLRRRLRRLRL\n128 610 1680 2436 2896 2994 3008 3358 3392 4020 4298 4582 4712 4728 5136 5900 6088 6232 6282 6858 6934 7186 7224 7256 7614 8802 8872 9170 9384 9794",
"output": "7"
},
{
"input": "10\nLLLLRRRRRR\n0 2 4 6 8 10 12 14 16 18",
"output": "-1"
},
{
"input": "5\nLLLLL\n0 10 20 30 40",
"output": "-1"
},
{
"input": "6\nRRRRRR\n40 50 60 70 80 100",
"output": "-1"
},
{
"input": "1\nR\n0",
"output": "-1"
},
{
"input": "2\nRL\n2 1000000000",
"output": "499999999"
},
{
"input": "2\nRL\n0 400000",
"output": "200000"
},
{
"input": "2\nRL\n0 200002",
"output": "100001"
},
{
"input": "2\nRL\n2 20000000",
"output": "9999999"
},
{
"input": "4\nLLRL\n2 4 10 100",
"output": "45"
},
{
"input": "4\nRLRL\n2 10 12 14",
"output": "1"
},
{
"input": "2\nRL\n0 100000000",
"output": "50000000"
},
{
"input": "2\nRL\n2 600002",
"output": "300000"
},
{
"input": "1\nL\n0",
"output": "-1"
},
{
"input": "2\nRL\n0 600000",
"output": "300000"
},
{
"input": "5\nRRRRR\n0 2 4 6 8",
"output": "-1"
},
{
"input": "2\nRL\n2 200000000",
"output": "99999999"
},
{
"input": "2\nRL\n0 267382766",
"output": "133691383"
},
{
"input": "3\nRRL\n4 8 999999998",
"output": "499999995"
},
{
"input": "2\nRL\n0 2",
"output": "1"
},
{
"input": "2\nRL\n2 400002",
"output": "200000"
},
{
"input": "2\nLL\n2 4",
"output": "-1"
},
{
"input": "2\nLL\n0 2",
"output": "-1"
},
{
"input": "2\nRL\n0 100000",
"output": "50000"
},
{
"input": "2\nRL\n2 200000020",
"output": "100000009"
},
{
"input": "2\nRL\n2000000 4000000",
"output": "1000000"
},
{
"input": "2\nRL\n0 199998",
"output": "99999"
},
{
"input": "3\nLRR\n40 50 60",
"output": "-1"
},
{
"input": "2\nRL\n200 400400",
"output": "200100"
},
{
"input": "2\nRL\n2 400004",
"output": "200001"
},
{
"input": "2\nRL\n0 200000000",
"output": "100000000"
}
] | 1,620,964,613
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 62
| 6,656,000
|
n = int(input())
d = input()
p = list(map(int,input().split()))
for i in range(n-1):
if(d[i]=='R' and d[i+1]=='L'):
print((p[i]+p[i+1])//2 - p[i])
exit(0)
print(-1)
|
Title: Launch of Collider
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There will be a launch of a new, powerful and unusual collider very soon, which located along a straight line. *n* particles will be launched inside it. All of them are located in a straight line and there can not be two or more particles located in the same point. The coordinates of the particles coincide with the distance in meters from the center of the collider, *x**i* is the coordinate of the *i*-th particle and its position in the collider at the same time. All coordinates of particle positions are even integers.
You know the direction of each particle movement — it will move to the right or to the left after the collider's launch start. All particles begin to move simultaneously at the time of the collider's launch start. Each particle will move straight to the left or straight to the right with the constant speed of 1 meter per microsecond. The collider is big enough so particles can not leave it in the foreseeable time.
Write the program which finds the moment of the first collision of any two particles of the collider. In other words, find the number of microseconds before the first moment when any two particles are at the same point.
Input Specification:
The first line contains the positive integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of particles.
The second line contains *n* symbols "L" and "R". If the *i*-th symbol equals "L", then the *i*-th particle will move to the left, otherwise the *i*-th symbol equals "R" and the *i*-th particle will move to the right.
The third line contains the sequence of pairwise distinct even integers *x*1,<=*x*2,<=...,<=*x**n* (0<=≤<=*x**i*<=≤<=109) — the coordinates of particles in the order from the left to the right. It is guaranteed that the coordinates of particles are given in the increasing order.
Output Specification:
In the first line print the only integer — the first moment (in microseconds) when two particles are at the same point and there will be an explosion.
Print the only integer -1, if the collision of particles doesn't happen.
Demo Input:
['4\nRLRL\n2 4 6 10\n', '3\nLLR\n40 50 60\n']
Demo Output:
['1\n', '-1\n']
Note:
In the first sample case the first explosion will happen in 1 microsecond because the particles number 1 and 2 will simultaneously be at the same point with the coordinate 3.
In the second sample case there will be no explosion because there are no particles which will simultaneously be at the same point.
|
```python
n = int(input())
d = input()
p = list(map(int,input().split()))
for i in range(n-1):
if(d[i]=='R' and d[i+1]=='L'):
print((p[i]+p[i+1])//2 - p[i])
exit(0)
print(-1)
```
| 0
|
|
175
|
C
|
Geometry Horse
|
PROGRAMMING
| 1,600
|
[
"greedy",
"implementation",
"sortings",
"two pointers"
] | null | null |
Vasya plays the Geometry Horse.
The game goal is to destroy geometric figures of the game world. A certain number of points is given for destroying each figure depending on the figure type and the current factor value.
There are *n* types of geometric figures. The number of figures of type *k**i* and figure cost *c**i* is known for each figure type. A player gets *c**i*·*f* points for destroying one figure of type *i*, where *f* is the current factor. The factor value can be an integer number from 1 to *t*<=+<=1, inclusive. At the beginning of the game the factor value is equal to 1. The factor is set to *i*<=+<=1 after destruction of *p**i* (1<=≤<=*i*<=≤<=*t*) figures, so the (*p**i*<=+<=1)-th figure to be destroyed is considered with factor equal to *i*<=+<=1.
Your task is to determine the maximum number of points Vasya can get after he destroys all figures. Take into account that Vasya is so tough that he can destroy figures in any order chosen by him.
|
The first line contains the only integer number *n* (1<=≤<=*n*<=≤<=100) — the number of figure types.
Each of the following *n* lines contains two integer numbers *k**i* and *c**i* (1<=≤<=*k**i*<=≤<=109,<=0<=≤<=*c**i*<=≤<=1000), separated with space — the number of figures of the *i*-th type and the cost of one *i*-type figure, correspondingly.
The next line contains the only integer number *t* (1<=≤<=*t*<=≤<=100) — the number that describe the factor's changes.
The next line contains *t* integer numbers *p**i* (1<=≤<=*p*1<=<<=*p*2<=<<=...<=<<=*p**t*<=≤<=1012), separated with spaces.
Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.
|
Print the only number — the maximum number of points Vasya can get.
|
[
"1\n5 10\n2\n3 6\n",
"2\n3 8\n5 10\n1\n20\n"
] |
[
"70",
"74"
] |
In the first example Vasya destroys three figures first and gets 3·1·10 = 30 points. Then the factor will become equal to 2 and after destroying the last two figures Vasya will get 2·2·10 = 40 points. As a result Vasya will get 70 points.
In the second example all 8 figures will be destroyed with factor 1, so Vasya will get (3·8 + 5·10)·1 = 74 points.
| 1,000
|
[
{
"input": "1\n5 10\n2\n3 6",
"output": "70"
},
{
"input": "2\n3 8\n5 10\n1\n20",
"output": "74"
},
{
"input": "3\n10 3\n20 2\n30 1\n3\n30 50 60",
"output": "200"
},
{
"input": "1\n100 1000\n1\n1",
"output": "199000"
},
{
"input": "1\n1 1000\n1\n1",
"output": "1000"
},
{
"input": "1\n1 1000\n1\n2",
"output": "1000"
},
{
"input": "2\n1000000000 1000\n1 1\n1\n10",
"output": "1999999991001"
},
{
"input": "6\n5 9\n63 3\n30 4\n25 6\n48 2\n29 9\n8\n105 137 172 192 632 722 972 981",
"output": "2251"
},
{
"input": "7\n9902 9\n5809 6\n2358 0\n6868 7\n9630 2\n8302 10\n9422 3\n4\n2148 4563 8488 9575",
"output": "1481866"
},
{
"input": "9\n60129 6\n44235 10\n13131 8\n2012 2\n27536 4\n38950 6\n39080 2\n13892 3\n48709 0\n1\n23853",
"output": "2751752"
},
{
"input": "10\n3466127 4\n3477072 1\n9690039 9\n9885165 6\n2559197 4\n3448456 3\n9169542 1\n6915866 2\n1702896 10\n8934261 5\n6\n3041416 5811699 5920083 8250213 8694306 8899250",
"output": "1843409345"
},
{
"input": "4\n4059578 5\n20774712 1\n64867825 7\n5606945 8\n1\n337246111",
"output": "540002937"
},
{
"input": "16\n196661091 17\n765544213 322\n134522506 115\n914609421 163\n219016066 227\n835576807 856\n682158845 914\n11248128 145\n876496017 854\n141052597 530\n163180278 315\n407245991 60\n294673989 270\n2976249 26\n674132026 519\n347829904 23\n16\n6280951514 53396669509 79113951711 87247958777 121933859963 219062570855 250484361488 292915737777 357877371567 638447479028 646055798354 733144914116 746148995326 752707219571 888597178968 929325038582",
"output": "3493909415554"
},
{
"input": "12\n559720489 0\n961035680 0\n953017025 0\n333351645 0\n840947432 0\n265712969 0\n484023361 0\n215786741 0\n880533785 0\n678800187 0\n817395626 0\n591321601 0\n13\n2165448470 32644841954 456375244913 510187375384 524722185932 628130306204 701569710739 731515209935 745407119699 772031092452 783514111802 933457816308 991905864630",
"output": "0"
},
{
"input": "22\n2 103\n10 84\n7 834\n9 527\n3 415\n10 943\n1 633\n9 444\n7 639\n2 146\n9 208\n5 637\n4 1000\n4 606\n6 43\n2 437\n4 855\n1 70\n4 780\n8 214\n2 196\n1 261\n61\n2 3 6 9 11 12 13 16 17 19 21 24 26 29 31 33 35 36 38 39 40 42 44 46 49 52 53 54 56 59 61 64 66 67 70 72 75 77 78 80 83 84 87 90 92 93 95 98 100 102 105 107 109 112 114 115 116 117 120 122 125",
"output": "2004140"
},
{
"input": "14\n3 689\n4 4\n6 40\n10 309\n2 216\n5 575\n1 203\n5 216\n10 544\n7 979\n1 19\n1 876\n8 505\n4 217\n51\n1 4 5 8 12 18 20 22 26 30 33 39 43 46 50 52 57 59 63 68 73 74 78 79 82 85 90 96 100 104 109 113 118 119 122 126 130 136 138 140 144 147 149 150 156 157 163 164 169 174 178",
"output": "412722"
},
{
"input": "2\n3 116\n3 869\n80\n3 5 11 17 23 31 33 41 42 49 51 53 58 60 65 70 79 84 87 88 89 93 98 102 109 110 111 114 123 129 134 142 143 152 160 162 166 167 174 179 186 191 199 205 214 219 224 227 236 241 249 252 260 268 272 275 282 288 292 293 297 302 310 314 317 319 321 330 336 340 349 358 366 374 378 383 387 390 397 405",
"output": "6431"
},
{
"input": "20\n1 529\n15 864\n1 26\n7 582\n7 914\n4 535\n5 371\n15 500\n13 912\n1 354\n7 327\n7 470\n4 277\n20 656\n8 501\n8 419\n16 569\n2 587\n13 294\n11 37\n77\n1 3 6 8 9 11 12 13 14 16 18 20 22 24 27 30 33 34 35 36 38 40 43 44 46 49 52 54 56 57 60 63 64 66 68 70 73 74 75 77 78 79 80 81 84 86 89 92 93 95 96 97 99 101 103 106 109 111 112 114 115 118 119 120 121 122 123 124 125 128 130 133 134 137 139 140 142",
"output": "4860712"
},
{
"input": "20\n1 676\n10 2\n10 467\n7 826\n7 138\n8 76\n8 148\n2 121\n7 527\n3 571\n10 410\n7 174\n2 318\n6 97\n3 919\n8 684\n3 586\n4 570\n10 494\n8 582\n74\n1 6 10 15 20 22 25 26 27 29 32 33 34 37 39 44 49 52 53 55 56 61 65 66 70 72 74 77 79 80 83 85 88 91 95 98 103 106 107 112 114 119 124 129 133 137 138 140 144 146 147 149 153 155 157 160 165 168 172 173 177 180 181 184 188 193 198 201 206 208 209 213 216 218",
"output": "1497278"
},
{
"input": "1\n555 100\n10\n1 2 3 4 5 6 7 8 9 10",
"output": "605000"
},
{
"input": "1\n1 1\n1\n100000000000",
"output": "1"
},
{
"input": "12\n1000000000 1\n1000000000 2\n1000000000 3\n1000000000 4\n1000000000 5\n1000000000 6\n1000000000 7\n1000000000 8\n1000000000 9\n1000000000 10\n1000000000 11\n1000000000 12\n1\n10000000000",
"output": "101000000000"
},
{
"input": "11\n1000000000 1\n1000000000 2\n1000000000 3\n1000000000 4\n1000000000 5\n1000000000 6\n1000000000 7\n1000000000 8\n1000000000 9\n1000000000 10\n1000000000 11\n1\n10000000000",
"output": "77000000000"
},
{
"input": "1\n10 10\n3\n1 2 3",
"output": "340"
},
{
"input": "1\n1000000000 1000\n2\n3 6",
"output": "2999999991000"
},
{
"input": "1\n100 100\n3\n3 6 9",
"output": "38200"
},
{
"input": "1\n10 1\n10\n1 2 3 4 5 6 7 8 9 10",
"output": "55"
},
{
"input": "1\n10 10\n5\n1 2 3 4 5",
"output": "450"
},
{
"input": "10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n10 10\n1\n1",
"output": "1990"
},
{
"input": "1\n10 10\n2\n3 6",
"output": "210"
},
{
"input": "10\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1\n1000000",
"output": "10000000"
}
] | 1,569,603,501
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 280
| 0
|
if __name__ == '__main__':
n = int(input())
pieces = [input() for _ in range(n)]
pieces = [_.split() for _ in pieces]
pieces = [tuple(_) for _ in pieces]
pieces = [(int(k), int(c)) for k, c in pieces]
t = int(input())
p = input().split()
p = [int(_) for _ in p]
sorted(pieces, key=lambda x: x[1])
destroy_pieces = 0
score = 0
f = 0
for k, c in pieces:
if f >= t or destroy_pieces + k <= p[f]:
score += k * c * (f + 1)
else:
diff = p[f] - destroy_pieces
score += diff * c * (f + 1)
f += 1
score += (k - diff) * c * (f + 1)
destroy_pieces += k
print(score)
|
Title: Geometry Horse
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya plays the Geometry Horse.
The game goal is to destroy geometric figures of the game world. A certain number of points is given for destroying each figure depending on the figure type and the current factor value.
There are *n* types of geometric figures. The number of figures of type *k**i* and figure cost *c**i* is known for each figure type. A player gets *c**i*·*f* points for destroying one figure of type *i*, where *f* is the current factor. The factor value can be an integer number from 1 to *t*<=+<=1, inclusive. At the beginning of the game the factor value is equal to 1. The factor is set to *i*<=+<=1 after destruction of *p**i* (1<=≤<=*i*<=≤<=*t*) figures, so the (*p**i*<=+<=1)-th figure to be destroyed is considered with factor equal to *i*<=+<=1.
Your task is to determine the maximum number of points Vasya can get after he destroys all figures. Take into account that Vasya is so tough that he can destroy figures in any order chosen by him.
Input Specification:
The first line contains the only integer number *n* (1<=≤<=*n*<=≤<=100) — the number of figure types.
Each of the following *n* lines contains two integer numbers *k**i* and *c**i* (1<=≤<=*k**i*<=≤<=109,<=0<=≤<=*c**i*<=≤<=1000), separated with space — the number of figures of the *i*-th type and the cost of one *i*-type figure, correspondingly.
The next line contains the only integer number *t* (1<=≤<=*t*<=≤<=100) — the number that describe the factor's changes.
The next line contains *t* integer numbers *p**i* (1<=≤<=*p*1<=<<=*p*2<=<<=...<=<<=*p**t*<=≤<=1012), separated with spaces.
Please, do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.
Output Specification:
Print the only number — the maximum number of points Vasya can get.
Demo Input:
['1\n5 10\n2\n3 6\n', '2\n3 8\n5 10\n1\n20\n']
Demo Output:
['70', '74']
Note:
In the first example Vasya destroys three figures first and gets 3·1·10 = 30 points. Then the factor will become equal to 2 and after destroying the last two figures Vasya will get 2·2·10 = 40 points. As a result Vasya will get 70 points.
In the second example all 8 figures will be destroyed with factor 1, so Vasya will get (3·8 + 5·10)·1 = 74 points.
|
```python
if __name__ == '__main__':
n = int(input())
pieces = [input() for _ in range(n)]
pieces = [_.split() for _ in pieces]
pieces = [tuple(_) for _ in pieces]
pieces = [(int(k), int(c)) for k, c in pieces]
t = int(input())
p = input().split()
p = [int(_) for _ in p]
sorted(pieces, key=lambda x: x[1])
destroy_pieces = 0
score = 0
f = 0
for k, c in pieces:
if f >= t or destroy_pieces + k <= p[f]:
score += k * c * (f + 1)
else:
diff = p[f] - destroy_pieces
score += diff * c * (f + 1)
f += 1
score += (k - diff) * c * (f + 1)
destroy_pieces += k
print(score)
```
| 0
|
|
967
|
B
|
Watering System
|
PROGRAMMING
| 1,000
|
[
"math",
"sortings"
] | null | null |
Arkady wants to water his only flower. Unfortunately, he has a very poor watering system that was designed for $n$ flowers and so it looks like a pipe with $n$ holes. Arkady can only use the water that flows from the first hole.
Arkady can block some of the holes, and then pour $A$ liters of water into the pipe. After that, the water will flow out from the non-blocked holes proportionally to their sizes $s_1, s_2, \ldots, s_n$. In other words, if the sum of sizes of non-blocked holes is $S$, and the $i$-th hole is not blocked, $\frac{s_i \cdot A}{S}$ liters of water will flow out of it.
What is the minimum number of holes Arkady should block to make at least $B$ liters of water flow out of the first hole?
|
The first line contains three integers $n$, $A$, $B$ ($1 \le n \le 100\,000$, $1 \le B \le A \le 10^4$) — the number of holes, the volume of water Arkady will pour into the system, and the volume he wants to get out of the first hole.
The second line contains $n$ integers $s_1, s_2, \ldots, s_n$ ($1 \le s_i \le 10^4$) — the sizes of the holes.
|
Print a single integer — the number of holes Arkady should block.
|
[
"4 10 3\n2 2 2 2\n",
"4 80 20\n3 2 1 4\n",
"5 10 10\n1000 1 1 1 1\n"
] |
[
"1\n",
"0\n",
"4\n"
] |
In the first example Arkady should block at least one hole. After that, $\frac{10 \cdot 2}{6} \approx 3.333$ liters of water will flow out of the first hole, and that suits Arkady.
In the second example even without blocking any hole, $\frac{80 \cdot 3}{10} = 24$ liters will flow out of the first hole, that is not less than $20$.
In the third example Arkady has to block all holes except the first to make all water flow out of the first hole.
| 1,000
|
[
{
"input": "4 10 3\n2 2 2 2",
"output": "1"
},
{
"input": "4 80 20\n3 2 1 4",
"output": "0"
},
{
"input": "5 10 10\n1000 1 1 1 1",
"output": "4"
},
{
"input": "10 300 100\n20 1 3 10 8 5 3 6 4 3",
"output": "1"
},
{
"input": "10 300 100\n20 25 68 40 60 37 44 85 23 96",
"output": "8"
},
{
"input": "1 1 1\n1",
"output": "0"
},
{
"input": "1 2 1\n1",
"output": "0"
},
{
"input": "2 2 2\n1 10000",
"output": "1"
},
{
"input": "2 10000 1\n1 9999",
"output": "0"
}
] | 1,644,394,082
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 30
| 0
|
n,A,B = map(int, input().split())
s = [int(i) for i in input().split()]
p = 0
for i in range(n):
S = sum(s[i:n])
if s[i]*A/S < B:
p += 1
else:
break
print(p)
|
Title: Watering System
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Arkady wants to water his only flower. Unfortunately, he has a very poor watering system that was designed for $n$ flowers and so it looks like a pipe with $n$ holes. Arkady can only use the water that flows from the first hole.
Arkady can block some of the holes, and then pour $A$ liters of water into the pipe. After that, the water will flow out from the non-blocked holes proportionally to their sizes $s_1, s_2, \ldots, s_n$. In other words, if the sum of sizes of non-blocked holes is $S$, and the $i$-th hole is not blocked, $\frac{s_i \cdot A}{S}$ liters of water will flow out of it.
What is the minimum number of holes Arkady should block to make at least $B$ liters of water flow out of the first hole?
Input Specification:
The first line contains three integers $n$, $A$, $B$ ($1 \le n \le 100\,000$, $1 \le B \le A \le 10^4$) — the number of holes, the volume of water Arkady will pour into the system, and the volume he wants to get out of the first hole.
The second line contains $n$ integers $s_1, s_2, \ldots, s_n$ ($1 \le s_i \le 10^4$) — the sizes of the holes.
Output Specification:
Print a single integer — the number of holes Arkady should block.
Demo Input:
['4 10 3\n2 2 2 2\n', '4 80 20\n3 2 1 4\n', '5 10 10\n1000 1 1 1 1\n']
Demo Output:
['1\n', '0\n', '4\n']
Note:
In the first example Arkady should block at least one hole. After that, $\frac{10 \cdot 2}{6} \approx 3.333$ liters of water will flow out of the first hole, and that suits Arkady.
In the second example even without blocking any hole, $\frac{80 \cdot 3}{10} = 24$ liters will flow out of the first hole, that is not less than $20$.
In the third example Arkady has to block all holes except the first to make all water flow out of the first hole.
|
```python
n,A,B = map(int, input().split())
s = [int(i) for i in input().split()]
p = 0
for i in range(n):
S = sum(s[i:n])
if s[i]*A/S < B:
p += 1
else:
break
print(p)
```
| 0
|
|
186
|
A
|
Comparing Strings
|
PROGRAMMING
| 1,100
|
[
"implementation",
"strings"
] | null | null |
Some dwarves that are finishing the StUDY (State University for Dwarven Youngsters) Bachelor courses, have been told "no genome, no degree". That means that all dwarves should write a thesis on genome. Dwarven genome is far from simple. It is represented by a string that consists of lowercase Latin letters.
Dwarf Misha has already chosen the subject for his thesis: determining by two dwarven genomes, whether they belong to the same race. Two dwarves belong to the same race if we can swap two characters in the first dwarf's genome and get the second dwarf's genome as a result. Help Dwarf Misha and find out whether two gnomes belong to the same race or not.
|
The first line contains the first dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The second line contains the second dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The number of letters in each genome doesn't exceed 105. It is guaranteed that the strings that correspond to the genomes are different. The given genomes may have different length.
|
Print "YES", if the dwarves belong to the same race. Otherwise, print "NO".
|
[
"ab\nba\n",
"aa\nab\n"
] |
[
"YES\n",
"NO\n"
] |
- First example: you can simply swap two letters in string "ab". So we get "ba". - Second example: we can't change string "aa" into string "ab", because "aa" does not contain letter "b".
| 500
|
[
{
"input": "ab\nba",
"output": "YES"
},
{
"input": "aa\nab",
"output": "NO"
},
{
"input": "a\nza",
"output": "NO"
},
{
"input": "vvea\nvvae",
"output": "YES"
},
{
"input": "rtfabanpc\natfabrnpc",
"output": "YES"
},
{
"input": "mt\ntm",
"output": "YES"
},
{
"input": "qxolmbkkt\naovlajmlf",
"output": "NO"
},
{
"input": "b\ng",
"output": "NO"
},
{
"input": "ab\naba",
"output": "NO"
},
{
"input": "ba\na",
"output": "NO"
},
{
"input": "a\nab",
"output": "NO"
},
{
"input": "a\naa",
"output": "NO"
},
{
"input": "a\nz",
"output": "NO"
},
{
"input": "aabb\nbbaa",
"output": "NO"
},
{
"input": "ab\nbd",
"output": "NO"
},
{
"input": "bac\ndae",
"output": "NO"
},
{
"input": "abc\nakl",
"output": "NO"
},
{
"input": "cb\naa",
"output": "NO"
},
{
"input": "abaab\naabba",
"output": "NO"
},
{
"input": "aab\naaa",
"output": "NO"
},
{
"input": "abcde\nedcba",
"output": "NO"
},
{
"input": "abab\nbaba",
"output": "NO"
},
{
"input": "ab\nbac",
"output": "NO"
},
{
"input": "abcd\naqcb",
"output": "NO"
},
{
"input": "abc\nbad",
"output": "NO"
},
{
"input": "ab\nca",
"output": "NO"
},
{
"input": "abc\nab",
"output": "NO"
},
{
"input": "ab\nbae",
"output": "NO"
},
{
"input": "aaaabcccca\naaaadccccb",
"output": "NO"
},
{
"input": "acaa\nabca",
"output": "NO"
},
{
"input": "aba\naab",
"output": "YES"
},
{
"input": "ah\nla",
"output": "NO"
},
{
"input": "aab\naa",
"output": "NO"
},
{
"input": "abab\nabcd",
"output": "NO"
},
{
"input": "ed\nab",
"output": "NO"
}
] | 1,625,795,362
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 53
| 186
| 1,638,400
|
a = list(input())
b = list(input())
t = 0
if(sorted(a) != sorted(b)):
print("NO")
exit()
for x in range(len(a)):
if a[x] != b[x]:
t += 1
if t == 2:
print('YES')
else:
print('NO')
|
Title: Comparing Strings
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Some dwarves that are finishing the StUDY (State University for Dwarven Youngsters) Bachelor courses, have been told "no genome, no degree". That means that all dwarves should write a thesis on genome. Dwarven genome is far from simple. It is represented by a string that consists of lowercase Latin letters.
Dwarf Misha has already chosen the subject for his thesis: determining by two dwarven genomes, whether they belong to the same race. Two dwarves belong to the same race if we can swap two characters in the first dwarf's genome and get the second dwarf's genome as a result. Help Dwarf Misha and find out whether two gnomes belong to the same race or not.
Input Specification:
The first line contains the first dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The second line contains the second dwarf's genome: a non-empty string, consisting of lowercase Latin letters.
The number of letters in each genome doesn't exceed 105. It is guaranteed that the strings that correspond to the genomes are different. The given genomes may have different length.
Output Specification:
Print "YES", if the dwarves belong to the same race. Otherwise, print "NO".
Demo Input:
['ab\nba\n', 'aa\nab\n']
Demo Output:
['YES\n', 'NO\n']
Note:
- First example: you can simply swap two letters in string "ab". So we get "ba". - Second example: we can't change string "aa" into string "ab", because "aa" does not contain letter "b".
|
```python
a = list(input())
b = list(input())
t = 0
if(sorted(a) != sorted(b)):
print("NO")
exit()
for x in range(len(a)):
if a[x] != b[x]:
t += 1
if t == 2:
print('YES')
else:
print('NO')
```
| 3
|
|
180
|
C
|
Letter
|
PROGRAMMING
| 1,400
|
[
"dp"
] | null | null |
Patrick has just finished writing a message to his sweetheart Stacey when he noticed that the message didn't look fancy. Patrick was nervous while writing the message, so some of the letters there were lowercase and some of them were uppercase.
Patrick believes that a message is fancy if any uppercase letter stands to the left of any lowercase one. In other words, this rule describes the strings where first go zero or more uppercase letters, and then — zero or more lowercase letters.
To make the message fancy, Patrick can erase some letter and add the same letter in the same place in the opposite case (that is, he can replace an uppercase letter with the lowercase one and vice versa). Patrick got interested in the following question: what minimum number of actions do we need to make a message fancy? Changing a letter's case in the message counts as one action. Patrick cannot perform any other actions.
|
The only line of the input contains a non-empty string consisting of uppercase and lowercase letters. The string's length does not exceed 105.
|
Print a single number — the least number of actions needed to make the message fancy.
|
[
"PRuvetSTAaYA\n",
"OYPROSTIYAOPECHATALSYAPRIVETSTASYA\n",
"helloworld\n"
] |
[
"5\n",
"0\n",
"0\n"
] |
none
| 0
|
[
{
"input": "PRuvetSTAaYA",
"output": "5"
},
{
"input": "OYPROSTIYAOPECHATALSYAPRIVETSTASYA",
"output": "0"
},
{
"input": "helloworld",
"output": "0"
},
{
"input": "P",
"output": "0"
},
{
"input": "t",
"output": "0"
},
{
"input": "XdJ",
"output": "1"
},
{
"input": "FSFlNEelYY",
"output": "3"
},
{
"input": "lgtyasficu",
"output": "0"
},
{
"input": "WYKUDTDDBT",
"output": "0"
},
{
"input": "yysxwlyqboatikfnpxczmpijziiojbvadlfozjqldssffcxdegyxfrvohoxvgsrvlzjlkcuffoeisrpvagxtbkapkpzcafadzzjd",
"output": "0"
},
{
"input": "mnAkOBuKxaiJwXhKnlcCvjxYXGXDoIqfUYkiLrdSYWhMemgWFzsgpoKOtHqooxbLYFuABWQSXuHdbyPVWyrkeEfqOsnEBikiqhfu",
"output": "43"
},
{
"input": "MMVESdOCALHJCTBTUWWQRGUUVTTTABKKAAdIINAdKLRLLVLODHDXDPMcQfUhPNHFBJSDRGsHZNORSCPNvKOOIuZnZAmTPUCoPNlR",
"output": "13"
},
{
"input": "MMbJIBhgFXPVpdQHLkWJkAHFIfJSpITTCRzRCzvRPRYECCheOknfINZWuKATDBOrEVKTcWXiYPjtzQMGUSGPNTMCUrvYCSWQHqAi",
"output": "27"
},
{
"input": "ZnqXEBOABXVbHRFFqDLWpWBBLqZIagmbRAYMDKJAYyFQXnPYwvDyRXKHLQnMLHGEOnRKmXtGqyQkoBCSmgKPcEIEugwJyxDOtTVn",
"output": "31"
}
] | 1,597,932,026
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 466
| 2,662,400
|
import string
a = input()
if len(a) == 0:
print(0)
dp = [0] * 100000
s = 1 if a[0] not in string.ascii_uppercase else 0
for i in range(1, len(a)):
dp[i] = (dp[i - 1] if a[i] in string.ascii_lowercase else min(1 + dp[i - 1], s))
s += 1 if a[i] not in string.ascii_uppercase else 0
print(dp[len(a) - 1])
|
Title: Letter
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Patrick has just finished writing a message to his sweetheart Stacey when he noticed that the message didn't look fancy. Patrick was nervous while writing the message, so some of the letters there were lowercase and some of them were uppercase.
Patrick believes that a message is fancy if any uppercase letter stands to the left of any lowercase one. In other words, this rule describes the strings where first go zero or more uppercase letters, and then — zero or more lowercase letters.
To make the message fancy, Patrick can erase some letter and add the same letter in the same place in the opposite case (that is, he can replace an uppercase letter with the lowercase one and vice versa). Patrick got interested in the following question: what minimum number of actions do we need to make a message fancy? Changing a letter's case in the message counts as one action. Patrick cannot perform any other actions.
Input Specification:
The only line of the input contains a non-empty string consisting of uppercase and lowercase letters. The string's length does not exceed 105.
Output Specification:
Print a single number — the least number of actions needed to make the message fancy.
Demo Input:
['PRuvetSTAaYA\n', 'OYPROSTIYAOPECHATALSYAPRIVETSTASYA\n', 'helloworld\n']
Demo Output:
['5\n', '0\n', '0\n']
Note:
none
|
```python
import string
a = input()
if len(a) == 0:
print(0)
dp = [0] * 100000
s = 1 if a[0] not in string.ascii_uppercase else 0
for i in range(1, len(a)):
dp[i] = (dp[i - 1] if a[i] in string.ascii_lowercase else min(1 + dp[i - 1], s))
s += 1 if a[i] not in string.ascii_uppercase else 0
print(dp[len(a) - 1])
```
| 3
|
|
926
|
C
|
Is This a Zebra?
|
PROGRAMMING
| 1,700
|
[
"implementation"
] | null | null |
A camera you have accidentally left in a desert has taken an interesting photo. The photo has a resolution of *n* pixels width, and each column of this photo is all white or all black. Thus, we can represent the photo as a sequence of *n* zeros and ones, where 0 means that the corresponding column is all white, and 1 means that the corresponding column is black.
You think that this photo can contain a zebra. In this case the whole photo should consist of several (possibly, only one) alternating black and white stripes of equal width. For example, the photo [0,<=0,<=0,<=1,<=1,<=1,<=0,<=0,<=0] can be a photo of zebra, while the photo [0,<=0,<=0,<=1,<=1,<=1,<=1] can not, because the width of the black stripe is 3, while the width of the white stripe is 4. Can the given photo be a photo of zebra or not?
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the width of the photo.
The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=1) — the description of the photo. If *a**i* is zero, the *i*-th column is all black. If *a**i* is one, then the *i*-th column is all white.
|
If the photo can be a photo of zebra, print "YES" (without quotes). Otherwise, print "NO".
You can print each letter in any case (upper or lower).
|
[
"9\n0 0 0 1 1 1 0 0 0\n",
"7\n0 0 0 1 1 1 1\n",
"5\n1 1 1 1 1\n",
"8\n1 1 1 0 0 0 1 1\n",
"9\n1 1 0 1 1 0 1 1 0\n"
] |
[
"YES\n",
"NO\n",
"YES\n",
"NO\n",
"NO\n"
] |
The first two examples are described in the statements.
In the third example all pixels are white, so the photo can be a photo of zebra.
In the fourth example the width of the first stripe is equal to three (white color), the width of the second stripe is equal to three (black), and the width of the third stripe is equal to two (white). Thus, not all stripes have equal length, so this photo is not a photo of zebra.
| 0
|
[
{
"input": "9\n0 0 0 1 1 1 0 0 0",
"output": "YES"
},
{
"input": "7\n0 0 0 1 1 1 1",
"output": "NO"
},
{
"input": "5\n1 1 1 1 1",
"output": "YES"
},
{
"input": "8\n1 1 1 0 0 0 1 1",
"output": "NO"
},
{
"input": "9\n1 1 0 1 1 0 1 1 0",
"output": "NO"
},
{
"input": "1\n0",
"output": "YES"
},
{
"input": "1\n1",
"output": "YES"
},
{
"input": "2\n0 0",
"output": "YES"
},
{
"input": "2\n0 1",
"output": "YES"
},
{
"input": "2\n1 0",
"output": "YES"
},
{
"input": "2\n1 1",
"output": "YES"
},
{
"input": "3\n1 1 0",
"output": "NO"
},
{
"input": "7\n0 0 0 1 1 1 0",
"output": "NO"
},
{
"input": "3\n0 1 1",
"output": "NO"
},
{
"input": "3\n0 0 1",
"output": "NO"
},
{
"input": "6\n0 0 1 0 1 0",
"output": "NO"
},
{
"input": "4\n0 1 1 0",
"output": "NO"
},
{
"input": "5\n0 1 1 0 0",
"output": "NO"
},
{
"input": "4\n0 1 0 0",
"output": "NO"
},
{
"input": "5\n1 1 1 0 0",
"output": "NO"
},
{
"input": "10\n0 0 1 1 0 0 0 1 1 1",
"output": "NO"
},
{
"input": "5\n0 0 0 0 1",
"output": "NO"
},
{
"input": "14\n0 0 0 1 1 1 1 0 0 0 0 1 1 1",
"output": "NO"
},
{
"input": "4\n1 0 1 0",
"output": "YES"
},
{
"input": "5\n1 0 0 0 1",
"output": "NO"
},
{
"input": "6\n1 1 1 0 1 1",
"output": "NO"
},
{
"input": "7\n1 1 1 1 1 0 1",
"output": "NO"
},
{
"input": "8\n1 1 0 0 1 1 0 0",
"output": "YES"
},
{
"input": "9\n0 1 1 0 0 0 1 1 1",
"output": "NO"
},
{
"input": "10\n0 0 0 0 0 1 1 1 1 1",
"output": "YES"
},
{
"input": "11\n0 1 0 1 0 1 0 0 0 1 0",
"output": "NO"
},
{
"input": "20\n1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0",
"output": "YES"
},
{
"input": "50\n0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 1 1 1 1",
"output": "NO"
},
{
"input": "100\n1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1",
"output": "YES"
},
{
"input": "3\n0 0 0",
"output": "YES"
},
{
"input": "3\n0 1 0",
"output": "YES"
},
{
"input": "3\n1 0 0",
"output": "NO"
},
{
"input": "3\n1 0 1",
"output": "YES"
},
{
"input": "3\n1 1 1",
"output": "YES"
}
] | 1,521,302,950
| 2,050
|
PyPy 3
|
OK
|
TESTS
| 64
| 140
| 27,545,600
|
n = int(input())
a = input().split()
now = -1
white = 0
black = 0
for i in range(n):
if(int(a[i]) == 0):
if(white > 0):
if(now == -1):
now = white
elif(now != white):
print("NO")
exit(0)
white = 0
black += 1
else:
if(black > 0):
if(now == -1):
now = black
elif(now != black):
print("NO")
exit(0)
black = 0
white += 1
if(white > 0):
if(now == -1):
now = white
elif(now != white):
print("NO")
exit(0)
if(black > 0):
if(now == -1):
now = black
elif(now != black):
print("NO")
exit(0)
print("YES")
|
Title: Is This a Zebra?
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A camera you have accidentally left in a desert has taken an interesting photo. The photo has a resolution of *n* pixels width, and each column of this photo is all white or all black. Thus, we can represent the photo as a sequence of *n* zeros and ones, where 0 means that the corresponding column is all white, and 1 means that the corresponding column is black.
You think that this photo can contain a zebra. In this case the whole photo should consist of several (possibly, only one) alternating black and white stripes of equal width. For example, the photo [0,<=0,<=0,<=1,<=1,<=1,<=0,<=0,<=0] can be a photo of zebra, while the photo [0,<=0,<=0,<=1,<=1,<=1,<=1] can not, because the width of the black stripe is 3, while the width of the white stripe is 4. Can the given photo be a photo of zebra or not?
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the width of the photo.
The second line contains a sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=1) — the description of the photo. If *a**i* is zero, the *i*-th column is all black. If *a**i* is one, then the *i*-th column is all white.
Output Specification:
If the photo can be a photo of zebra, print "YES" (without quotes). Otherwise, print "NO".
You can print each letter in any case (upper or lower).
Demo Input:
['9\n0 0 0 1 1 1 0 0 0\n', '7\n0 0 0 1 1 1 1\n', '5\n1 1 1 1 1\n', '8\n1 1 1 0 0 0 1 1\n', '9\n1 1 0 1 1 0 1 1 0\n']
Demo Output:
['YES\n', 'NO\n', 'YES\n', 'NO\n', 'NO\n']
Note:
The first two examples are described in the statements.
In the third example all pixels are white, so the photo can be a photo of zebra.
In the fourth example the width of the first stripe is equal to three (white color), the width of the second stripe is equal to three (black), and the width of the third stripe is equal to two (white). Thus, not all stripes have equal length, so this photo is not a photo of zebra.
|
```python
n = int(input())
a = input().split()
now = -1
white = 0
black = 0
for i in range(n):
if(int(a[i]) == 0):
if(white > 0):
if(now == -1):
now = white
elif(now != white):
print("NO")
exit(0)
white = 0
black += 1
else:
if(black > 0):
if(now == -1):
now = black
elif(now != black):
print("NO")
exit(0)
black = 0
white += 1
if(white > 0):
if(now == -1):
now = white
elif(now != white):
print("NO")
exit(0)
if(black > 0):
if(now == -1):
now = black
elif(now != black):
print("NO")
exit(0)
print("YES")
```
| 3
|
|
27
|
A
|
Next Test
|
PROGRAMMING
| 1,200
|
[
"implementation",
"sortings"
] |
A. Next Test
|
2
|
256
|
«Polygon» is a system which allows to create programming tasks in a simple and professional way. When you add a test to the problem, the corresponding form asks you for the test index. As in most cases it is clear which index the next test will have, the system suggests the default value of the index. It is calculated as the smallest positive integer which is not used as an index for some previously added test.
You are to implement this feature. Create a program which determines the default index of the next test, given the indexes of the previously added tests.
|
The first line contains one integer *n* (1<=≤<=*n*<=≤<=3000) — the amount of previously added tests. The second line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3000) — indexes of these tests.
|
Output the required default value for the next test index.
|
[
"3\n1 7 2\n"
] |
[
"3\n"
] |
none
| 500
|
[
{
"input": "1\n1",
"output": "2"
},
{
"input": "2\n2 1",
"output": "3"
},
{
"input": "3\n3 4 1",
"output": "2"
},
{
"input": "4\n6 4 3 5",
"output": "1"
},
{
"input": "5\n3 2 1 7 4",
"output": "5"
},
{
"input": "6\n4 1 2 5 3 7",
"output": "6"
},
{
"input": "7\n3 2 1 6 5 7 4",
"output": "8"
},
{
"input": "8\n2 8 3 7 6 9 1 5",
"output": "4"
},
{
"input": "9\n10 5 9 3 8 7 1 2 4",
"output": "6"
},
{
"input": "10\n7 2 3 8 9 6 5 4 1 10",
"output": "11"
},
{
"input": "1\n1",
"output": "2"
},
{
"input": "2\n1 2",
"output": "3"
},
{
"input": "3\n2 4 1",
"output": "3"
},
{
"input": "4\n4 2 3 1",
"output": "5"
},
{
"input": "5\n3 1 4 2 5",
"output": "6"
},
{
"input": "6\n1 3 6 7 2 4",
"output": "5"
},
{
"input": "7\n1 5 4 7 2 3 6",
"output": "8"
},
{
"input": "8\n12 1 6 5 2 8 3 4",
"output": "7"
},
{
"input": "9\n3 2 7 5 6 4 1 9 10",
"output": "8"
},
{
"input": "10\n1 7 13 6 5 10 3 8 2 4",
"output": "9"
},
{
"input": "1\n2",
"output": "1"
},
{
"input": "1\n3",
"output": "1"
},
{
"input": "1\n3000",
"output": "1"
},
{
"input": "2\n2 3",
"output": "1"
},
{
"input": "2\n3000 1",
"output": "2"
}
] | 1,624,604,276
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 124
| 0
|
n=int(input())
a=sorted([int(x) for x in input().split()])
for i in range(n):
if i+1!=a[i]:
print(i+1)
break
|
Title: Next Test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
«Polygon» is a system which allows to create programming tasks in a simple and professional way. When you add a test to the problem, the corresponding form asks you for the test index. As in most cases it is clear which index the next test will have, the system suggests the default value of the index. It is calculated as the smallest positive integer which is not used as an index for some previously added test.
You are to implement this feature. Create a program which determines the default index of the next test, given the indexes of the previously added tests.
Input Specification:
The first line contains one integer *n* (1<=≤<=*n*<=≤<=3000) — the amount of previously added tests. The second line contains *n* distinct integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=3000) — indexes of these tests.
Output Specification:
Output the required default value for the next test index.
Demo Input:
['3\n1 7 2\n']
Demo Output:
['3\n']
Note:
none
|
```python
n=int(input())
a=sorted([int(x) for x in input().split()])
for i in range(n):
if i+1!=a[i]:
print(i+1)
break
```
| 0
|
492
|
B
|
Vanya and Lanterns
|
PROGRAMMING
| 1,200
|
[
"binary search",
"implementation",
"math",
"sortings"
] | null | null |
Vanya walks late at night along a straight street of length *l*, lit by *n* lanterns. Consider the coordinate system with the beginning of the street corresponding to the point 0, and its end corresponding to the point *l*. Then the *i*-th lantern is at the point *a**i*. The lantern lights all points of the street that are at the distance of at most *d* from it, where *d* is some positive number, common for all lanterns.
Vanya wonders: what is the minimum light radius *d* should the lanterns have to light the whole street?
|
The first line contains two integers *n*, *l* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*l*<=≤<=109) — the number of lanterns and the length of the street respectively.
The next line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=*l*). Multiple lanterns can be located at the same point. The lanterns may be located at the ends of the street.
|
Print the minimum light radius *d*, needed to light the whole street. The answer will be considered correct if its absolute or relative error doesn't exceed 10<=-<=9.
|
[
"7 15\n15 5 3 7 9 14 0\n",
"2 5\n2 5\n"
] |
[
"2.5000000000\n",
"2.0000000000\n"
] |
Consider the second sample. At *d* = 2 the first lantern will light the segment [0, 4] of the street, and the second lantern will light segment [3, 5]. Thus, the whole street will be lit.
| 1,000
|
[
{
"input": "7 15\n15 5 3 7 9 14 0",
"output": "2.5000000000"
},
{
"input": "2 5\n2 5",
"output": "2.0000000000"
},
{
"input": "46 615683844\n431749087 271781274 274974690 324606253 480870261 401650581 13285442 478090364 266585394 425024433 588791449 492057200 391293435 563090494 317950 173675329 473068378 356306865 311731938 192959832 321180686 141984626 578985584 512026637 175885185 590844074 47103801 212211134 330150 509886963 565955809 315640375 612907074 500474373 524310737 568681652 315339618 478782781 518873818 271322031 74600969 539099112 85129347 222068995 106014720 77282307",
"output": "22258199.5000000000"
},
{
"input": "2 1000000000\n0 1000000000",
"output": "500000000.0000000000"
},
{
"input": "2 555\n200 300",
"output": "255.0000000000"
},
{
"input": "1 1\n1",
"output": "1.0000000000"
},
{
"input": "1 1\n0",
"output": "1.0000000000"
},
{
"input": "1 1000000000\n0",
"output": "1000000000.0000000000"
},
{
"input": "1 1000000000\n1000000000",
"output": "1000000000.0000000000"
},
{
"input": "1 999999999\n499999999",
"output": "500000000.0000000000"
},
{
"input": "4 700\n0 250 475 700",
"output": "125.0000000000"
},
{
"input": "4 700\n0 225 450 700",
"output": "125.0000000000"
},
{
"input": "5 1000\n0 250 500 750 1000",
"output": "125.0000000000"
},
{
"input": "5 2000\n1001 1001 1001 1001 1001",
"output": "1001.0000000000"
},
{
"input": "1 999\n501",
"output": "501.0000000000"
},
{
"input": "2 1000\n400 600",
"output": "400.0000000000"
},
{
"input": "4 15\n1 3 9 11",
"output": "4.0000000000"
},
{
"input": "1 5\n2",
"output": "3.0000000000"
}
] | 1,692,273,432
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 32
| 61
| 0
|
a,b=map(int,input().split())
c=list(map(int,input().split()))
c.sort()
l=[]
if len(c)==1:
if c[0]==0:
print(b-c[0])
elif c[0]==b:
print(b-0)
else:
print(max((b-c[0]),(c[0]-0)))
else:
if c[0] == 0 and c[len(c) - 1] == b:
for i in range(a - 1):
l.append((c[i + 1] - c[i]) / 2)
elif c[0] == 0 and c[len(c) - 1] != b:
for i in range(a):
l.append((c[i + 1] - c[i]) / 2)
l.append(b - c[len(c) - 1])
elif c[0] != 0 and c[len(c) - 1] == b:
l.append(c[0])
for i in range(a - 1):
l.append((c[i + 1] - c[i]) / 2)
else:
l.append(c[0])
for i in range(a - 1):
l.append((c[i + 1] - c[i]) / 2)
l.append(b - c[len(c) - 1])
n = max(l)
if n != float:
n = ((n / 10) * 10)
v = str(n)
v += "000000000"
print(v)
else:
k = str(n)
k += "000000000"
print(k)
|
Title: Vanya and Lanterns
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya walks late at night along a straight street of length *l*, lit by *n* lanterns. Consider the coordinate system with the beginning of the street corresponding to the point 0, and its end corresponding to the point *l*. Then the *i*-th lantern is at the point *a**i*. The lantern lights all points of the street that are at the distance of at most *d* from it, where *d* is some positive number, common for all lanterns.
Vanya wonders: what is the minimum light radius *d* should the lanterns have to light the whole street?
Input Specification:
The first line contains two integers *n*, *l* (1<=≤<=*n*<=≤<=1000, 1<=≤<=*l*<=≤<=109) — the number of lanterns and the length of the street respectively.
The next line contains *n* integers *a**i* (0<=≤<=*a**i*<=≤<=*l*). Multiple lanterns can be located at the same point. The lanterns may be located at the ends of the street.
Output Specification:
Print the minimum light radius *d*, needed to light the whole street. The answer will be considered correct if its absolute or relative error doesn't exceed 10<=-<=9.
Demo Input:
['7 15\n15 5 3 7 9 14 0\n', '2 5\n2 5\n']
Demo Output:
['2.5000000000\n', '2.0000000000\n']
Note:
Consider the second sample. At *d* = 2 the first lantern will light the segment [0, 4] of the street, and the second lantern will light segment [3, 5]. Thus, the whole street will be lit.
|
```python
a,b=map(int,input().split())
c=list(map(int,input().split()))
c.sort()
l=[]
if len(c)==1:
if c[0]==0:
print(b-c[0])
elif c[0]==b:
print(b-0)
else:
print(max((b-c[0]),(c[0]-0)))
else:
if c[0] == 0 and c[len(c) - 1] == b:
for i in range(a - 1):
l.append((c[i + 1] - c[i]) / 2)
elif c[0] == 0 and c[len(c) - 1] != b:
for i in range(a):
l.append((c[i + 1] - c[i]) / 2)
l.append(b - c[len(c) - 1])
elif c[0] != 0 and c[len(c) - 1] == b:
l.append(c[0])
for i in range(a - 1):
l.append((c[i + 1] - c[i]) / 2)
else:
l.append(c[0])
for i in range(a - 1):
l.append((c[i + 1] - c[i]) / 2)
l.append(b - c[len(c) - 1])
n = max(l)
if n != float:
n = ((n / 10) * 10)
v = str(n)
v += "000000000"
print(v)
else:
k = str(n)
k += "000000000"
print(k)
```
| 3
|
|
270
|
A
|
Fancy Fence
|
PROGRAMMING
| 1,100
|
[
"geometry",
"implementation",
"math"
] | null | null |
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
|
The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees.
|
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
|
[
"3\n30\n60\n90\n"
] |
[
"NO\nYES\nYES\n"
] |
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square.
| 500
|
[
{
"input": "3\n30\n60\n90",
"output": "NO\nYES\nYES"
},
{
"input": "6\n1\n2\n3\n170\n179\n25",
"output": "NO\nNO\nNO\nYES\nYES\nNO"
}
] | 1,689,636,149
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 3
| 92
| 0
|
w = int(input())
for i in range(w):
x = int(input())
if (360 / (180 - x)) % 1 == 0:
print("YES")
else:
print("NO")
|
Title: Fancy Fence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Emuskald needs a fence around his farm, but he is too lazy to build it himself. So he purchased a fence-building robot.
He wants the fence to be a regular polygon. The robot builds the fence along a single path, but it can only make fence corners at a single angle *a*.
Will the robot be able to build the fence Emuskald wants? In other words, is there a regular polygon which angles are equal to *a*?
Input Specification:
The first line of input contains an integer *t* (0<=<<=*t*<=<<=180) — the number of tests. Each of the following *t* lines contains a single integer *a* (0<=<<=*a*<=<<=180) — the angle the robot can make corners at measured in degrees.
Output Specification:
For each test, output on a single line "YES" (without quotes), if the robot can build a fence Emuskald wants, and "NO" (without quotes), if it is impossible.
Demo Input:
['3\n30\n60\n90\n']
Demo Output:
['NO\nYES\nYES\n']
Note:
In the first test case, it is impossible to build the fence, since there is no regular polygon with angle <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/df5f4b07dd5316fde165b43657b2696e2919e791.png" style="max-width: 100.0%;max-height: 100.0%;"/>.
In the second test case, the fence is a regular triangle, and in the last test case — a square.
|
```python
w = int(input())
for i in range(w):
x = int(input())
if (360 / (180 - x)) % 1 == 0:
print("YES")
else:
print("NO")
```
| 3
|
|
382
|
A
|
Ksenia and Pan Scales
|
PROGRAMMING
| 1,100
|
[
"greedy",
"implementation"
] | null | null |
Ksenia has ordinary pan scales and several weights of an equal mass. Ksenia has already put some weights on the scales, while other weights are untouched. Ksenia is now wondering whether it is possible to put all the remaining weights on the scales so that the scales were in equilibrium.
The scales is in equilibrium if the total sum of weights on the left pan is equal to the total sum of weights on the right pan.
|
The first line has a non-empty sequence of characters describing the scales. In this sequence, an uppercase English letter indicates a weight, and the symbol "|" indicates the delimiter (the character occurs in the sequence exactly once). All weights that are recorded in the sequence before the delimiter are initially on the left pan of the scale. All weights that are recorded in the sequence after the delimiter are initially on the right pan of the scale.
The second line contains a non-empty sequence containing uppercase English letters. Each letter indicates a weight which is not used yet.
It is guaranteed that all the English letters in the input data are different. It is guaranteed that the input does not contain any extra characters.
|
If you cannot put all the weights on the scales so that the scales were in equilibrium, print string "Impossible". Otherwise, print the description of the resulting scales, copy the format of the input.
If there are multiple answers, print any of them.
|
[
"AC|T\nL\n",
"|ABC\nXYZ\n",
"W|T\nF\n",
"ABC|\nD\n"
] |
[
"AC|TL\n",
"XYZ|ABC\n",
"Impossible\n",
"Impossible\n"
] |
none
| 500
|
[
{
"input": "AC|T\nL",
"output": "AC|TL"
},
{
"input": "|ABC\nXYZ",
"output": "XYZ|ABC"
},
{
"input": "W|T\nF",
"output": "Impossible"
},
{
"input": "ABC|\nD",
"output": "Impossible"
},
{
"input": "A|BC\nDEF",
"output": "ADF|BCE"
},
{
"input": "|\nABC",
"output": "Impossible"
},
{
"input": "|\nZXCVBANMIO",
"output": "XVAMO|ZCBNI"
},
{
"input": "|C\nA",
"output": "A|C"
},
{
"input": "|\nAB",
"output": "B|A"
},
{
"input": "A|XYZ\nUIOPL",
"output": "Impossible"
},
{
"input": "K|B\nY",
"output": "Impossible"
},
{
"input": "EQJWDOHKZRBISPLXUYVCMNFGT|\nA",
"output": "Impossible"
},
{
"input": "|MACKERIGZPVHNDYXJBUFLWSO\nQT",
"output": "Impossible"
},
{
"input": "ERACGIZOVPT|WXUYMDLJNQS\nKB",
"output": "ERACGIZOVPTB|WXUYMDLJNQSK"
},
{
"input": "CKQHRUZMISGE|FBVWPXDLTJYN\nOA",
"output": "CKQHRUZMISGEA|FBVWPXDLTJYNO"
},
{
"input": "V|CMOEUTAXBFWSK\nDLRZJGIYNQHP",
"output": "VDLRZJGIYNQHP|CMOEUTAXBFWSK"
},
{
"input": "QWHNMALDGKTJ|\nPBRYVXZUESCOIF",
"output": "QWHNMALDGKTJF|PBRYVXZUESCOI"
},
{
"input": "|\nFXCVMUEWZAHNDOSITPRLKQJYBG",
"output": "XVUWANOIPLQYG|FCMEZHDSTRKJB"
},
{
"input": "IB|PCGHZ\nFXWTJQNEKAUM",
"output": "Impossible"
},
{
"input": "EC|IWAXQ\nJUHSRKGZTOMYN",
"output": "ECJUHRGTMN|IWAXQSKZOY"
},
{
"input": "VDINYMA|UQKWBCLRHZJ\nXEGOF",
"output": "Impossible"
},
{
"input": "ZLTPSIQUBAR|XFDEMYC\nHNOJWG",
"output": "ZLTPSIQUBARG|XFDEMYCHNOJW"
},
{
"input": "R|FLZOTJNU\nGIYHKVX",
"output": "RGIYHKVX|FLZOTJNU"
},
{
"input": "W|TL\nQROFSADYPKHEJNMXBZVUCIG",
"output": "WQOSDPHJMBVCG|TLRFAYKENXZUI"
},
{
"input": "NRDFQSEKLAYMOT|ZH\nGUXIBJCVPW",
"output": "Impossible"
},
{
"input": "FGRT|\nAC",
"output": "Impossible"
},
{
"input": "|FGRT\nAC",
"output": "Impossible"
},
{
"input": "A|\nB",
"output": "A|B"
},
{
"input": "|A\nB",
"output": "B|A"
},
{
"input": "|\nA",
"output": "Impossible"
},
{
"input": "|\nQWERTYUIOPASDFGHJKLZXCVBNM",
"output": "WRYIPSFHKZCBM|QETUOADGJLXVN"
},
{
"input": "QWERTYUIOPASDFGHJKLZXCVBN|\nM",
"output": "Impossible"
},
{
"input": "QWERTY|VBN\nUIOPASDFGHJKLZXC",
"output": "Impossible"
},
{
"input": "ABC|D\nKSL",
"output": "Impossible"
},
{
"input": "A|BCDEF\nGH",
"output": "Impossible"
},
{
"input": "|ABC\nD",
"output": "Impossible"
},
{
"input": "A|BC\nDE",
"output": "Impossible"
},
{
"input": "|ASD\nX",
"output": "Impossible"
},
{
"input": "AB|CDEF\nXYZRT",
"output": "Impossible"
}
] | 1,648,300,508
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 4
| 46
| 0
|
lst = input().split('|')
[str1, str2] = sorted(lst, key=len)
inp2 = input()
if len(str1) + len(inp2) < len(str2) or len(str1) == len(str2):
print('Impossible')
exit()
while len(str1) != len(str2):
str1 += inp2[0]
inp2 = inp2[1:]
if str2 == lst[0]:
print(str2+'|'+str1)
else:
print(str1+'|'+str2)
|
Title: Ksenia and Pan Scales
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ksenia has ordinary pan scales and several weights of an equal mass. Ksenia has already put some weights on the scales, while other weights are untouched. Ksenia is now wondering whether it is possible to put all the remaining weights on the scales so that the scales were in equilibrium.
The scales is in equilibrium if the total sum of weights on the left pan is equal to the total sum of weights on the right pan.
Input Specification:
The first line has a non-empty sequence of characters describing the scales. In this sequence, an uppercase English letter indicates a weight, and the symbol "|" indicates the delimiter (the character occurs in the sequence exactly once). All weights that are recorded in the sequence before the delimiter are initially on the left pan of the scale. All weights that are recorded in the sequence after the delimiter are initially on the right pan of the scale.
The second line contains a non-empty sequence containing uppercase English letters. Each letter indicates a weight which is not used yet.
It is guaranteed that all the English letters in the input data are different. It is guaranteed that the input does not contain any extra characters.
Output Specification:
If you cannot put all the weights on the scales so that the scales were in equilibrium, print string "Impossible". Otherwise, print the description of the resulting scales, copy the format of the input.
If there are multiple answers, print any of them.
Demo Input:
['AC|T\nL\n', '|ABC\nXYZ\n', 'W|T\nF\n', 'ABC|\nD\n']
Demo Output:
['AC|TL\n', 'XYZ|ABC\n', 'Impossible\n', 'Impossible\n']
Note:
none
|
```python
lst = input().split('|')
[str1, str2] = sorted(lst, key=len)
inp2 = input()
if len(str1) + len(inp2) < len(str2) or len(str1) == len(str2):
print('Impossible')
exit()
while len(str1) != len(str2):
str1 += inp2[0]
inp2 = inp2[1:]
if str2 == lst[0]:
print(str2+'|'+str1)
else:
print(str1+'|'+str2)
```
| 0
|
|
514
|
A
|
Chewbaсca and Number
|
PROGRAMMING
| 1,200
|
[
"greedy",
"implementation"
] | null | null |
Luke Skywalker gave Chewbacca an integer number *x*. Chewbacca isn't good at numbers but he loves inverting digits in them. Inverting digit *t* means replacing it with digit 9<=-<=*t*.
Help Chewbacca to transform the initial number *x* to the minimum possible positive number by inverting some (possibly, zero) digits. The decimal representation of the final number shouldn't start with a zero.
|
The first line contains a single integer *x* (1<=≤<=*x*<=≤<=1018) — the number that Luke Skywalker gave to Chewbacca.
|
Print the minimum possible positive number that Chewbacca can obtain after inverting some digits. The number shouldn't contain leading zeroes.
|
[
"27\n",
"4545\n"
] |
[
"22\n",
"4444\n"
] |
none
| 500
|
[
{
"input": "27",
"output": "22"
},
{
"input": "4545",
"output": "4444"
},
{
"input": "1",
"output": "1"
},
{
"input": "9",
"output": "9"
},
{
"input": "8772",
"output": "1222"
},
{
"input": "81",
"output": "11"
},
{
"input": "71723447",
"output": "21223442"
},
{
"input": "91730629",
"output": "91230320"
},
{
"input": "420062703497",
"output": "420032203402"
},
{
"input": "332711047202",
"output": "332211042202"
},
{
"input": "3395184971407775",
"output": "3304114021402224"
},
{
"input": "8464062628894325",
"output": "1434032321104324"
},
{
"input": "164324828731963982",
"output": "134324121231033012"
},
{
"input": "384979173822804784",
"output": "314020123122104214"
},
{
"input": "41312150450968417",
"output": "41312140440031412"
},
{
"input": "2156",
"output": "2143"
},
{
"input": "1932",
"output": "1032"
},
{
"input": "5902",
"output": "4002"
},
{
"input": "5728",
"output": "4221"
},
{
"input": "8537",
"output": "1432"
},
{
"input": "55403857",
"output": "44403142"
},
{
"input": "270739",
"output": "220230"
},
{
"input": "28746918",
"output": "21243011"
},
{
"input": "10279211",
"output": "10220211"
},
{
"input": "40289679",
"output": "40210320"
},
{
"input": "545203238506",
"output": "444203231403"
},
{
"input": "461117063340",
"output": "431112033340"
},
{
"input": "658492686568",
"output": "341402313431"
},
{
"input": "857373361868",
"output": "142323331131"
},
{
"input": "429325660016",
"output": "420324330013"
},
{
"input": "9894448650287940",
"output": "9104441340212040"
},
{
"input": "6354510839296263",
"output": "3344410130203233"
},
{
"input": "6873575462224593",
"output": "3123424432224403"
},
{
"input": "4237951492601449",
"output": "4232041402301440"
},
{
"input": "2680352384836991",
"output": "2310342314133001"
},
{
"input": "606187734191890310",
"output": "303112234101100310"
},
{
"input": "351499943576823355",
"output": "341400043423123344"
},
{
"input": "180593481782177068",
"output": "110403411212122031"
},
{
"input": "999999999999999999",
"output": "900000000000000000"
},
{
"input": "1000000000000000000",
"output": "1000000000000000000"
},
{
"input": "9999",
"output": "9000"
},
{
"input": "99",
"output": "90"
},
{
"input": "9991",
"output": "9001"
}
] | 1,666,218,628
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 31
| 0
|
n = input()
ans = ''
for i in range(len(n)):
if int(n[i]) < 5:
ans += n[i]
else:
temp = 9 - int(n[i])
ans += str(temp)
print(int(ans))
|
Title: Chewbaсca and Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Luke Skywalker gave Chewbacca an integer number *x*. Chewbacca isn't good at numbers but he loves inverting digits in them. Inverting digit *t* means replacing it with digit 9<=-<=*t*.
Help Chewbacca to transform the initial number *x* to the minimum possible positive number by inverting some (possibly, zero) digits. The decimal representation of the final number shouldn't start with a zero.
Input Specification:
The first line contains a single integer *x* (1<=≤<=*x*<=≤<=1018) — the number that Luke Skywalker gave to Chewbacca.
Output Specification:
Print the minimum possible positive number that Chewbacca can obtain after inverting some digits. The number shouldn't contain leading zeroes.
Demo Input:
['27\n', '4545\n']
Demo Output:
['22\n', '4444\n']
Note:
none
|
```python
n = input()
ans = ''
for i in range(len(n)):
if int(n[i]) < 5:
ans += n[i]
else:
temp = 9 - int(n[i])
ans += str(temp)
print(int(ans))
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Santa Claus likes palindromes very much. There was his birthday recently. *k* of his friends came to him to congratulate him, and each of them presented to him a string *s**i* having the same length *n*. We denote the beauty of the *i*-th string by *a**i*. It can happen that *a**i* is negative — that means that Santa doesn't find this string beautiful at all.
Santa Claus is crazy about palindromes. He is thinking about the following question: what is the maximum possible total beauty of a palindrome which can be obtained by concatenating some (possibly all) of the strings he has? Each present can be used at most once. Note that all strings have the same length *n*.
Recall that a palindrome is a string that doesn't change after one reverses it.
Since the empty string is a palindrome too, the answer can't be negative. Even if all *a**i*'s are negative, Santa can obtain the empty string.
|
The first line contains two positive integers *k* and *n* divided by space and denoting the number of Santa friends and the length of every string they've presented, respectively (1<=≤<=*k*,<=*n*<=≤<=100<=000; *n*·*k* <=≤<=100<=000).
*k* lines follow. The *i*-th of them contains the string *s**i* and its beauty *a**i* (<=-<=10<=000<=≤<=*a**i*<=≤<=10<=000). The string consists of *n* lowercase English letters, and its beauty is integer. Some of strings may coincide. Also, equal strings can have different beauties.
|
In the only line print the required maximum possible beauty.
|
[
"7 3\nabb 2\naaa -3\nbba -1\nzyz -4\nabb 5\naaa 7\nxyx 4\n",
"3 1\na 1\na 2\na 3\n",
"2 5\nabcde 10000\nabcde 10000\n"
] |
[
"12\n",
"6\n",
"0\n"
] |
In the first example Santa can obtain abbaaaxyxaaabba by concatenating strings 5, 2, 7, 6 and 3 (in this order).
| 0
|
[
{
"input": "7 3\nabb 2\naaa -3\nbba -1\nzyz -4\nabb 5\naaa 7\nxyx 4",
"output": "12"
},
{
"input": "3 1\na 1\na 2\na 3",
"output": "6"
},
{
"input": "2 5\nabcde 10000\nabcde 10000",
"output": "0"
},
{
"input": "10 10\nnjxbzflaka -1\nfelbvvtkja 6\ngxiuztqkcw 5\naomvscmtti 6\njsqmkoyuca -2\nwckqtzuixg 5\najktvvblef -5\nittmcsvmoa -1\nakalfzbxjn 10\nacuyokmqsj 8",
"output": "31"
},
{
"input": "10 20\njvyxocgomfmrtllgmagp 13\ngvtjnyaofrswcnnifzfq 17\nqisxpseggpjfoijmqnel -5\nlenqmjiofjpggespxsiq 14\nqfzfinncwsrfoaynjtvg 14\ncaayidazlylxyisihdhx 14\npgamglltrmfmogcoxyvj 11\nxhdhisiyxlylzadiyaac 2\ntbirihfpjgbbtclpxwhv 19\nvhwxplctbbgjpfhiribt 10",
"output": "109"
},
{
"input": "1 1\ne -1",
"output": "0"
},
{
"input": "2 1\nt 1\nt 2",
"output": "3"
},
{
"input": "1 2\nyy 1",
"output": "1"
},
{
"input": "2 2\nsn 1\nns 2",
"output": "3"
},
{
"input": "3 3\nada -1\nxuv -1\nvux 3",
"output": "2"
},
{
"input": "4 3\ndbd 24\naba 90\ncbc 54\naba 46",
"output": "190"
},
{
"input": "2 3\naaa 5\naaa -2",
"output": "5"
},
{
"input": "4 3\naba 4\naba 3\naba 3\naba -2",
"output": "10"
},
{
"input": "4 3\naba 4\naba 2\naba 2\naba -1",
"output": "8"
},
{
"input": "3 2\naa 5\naa -2\nbb 1",
"output": "5"
},
{
"input": "2 2\naa 500\naa -50",
"output": "500"
},
{
"input": "2 1\na 5\na -1",
"output": "5"
},
{
"input": "2 3\naba 10\naba -3",
"output": "10"
},
{
"input": "2 3\naba 10\naba -9",
"output": "10"
}
] | 1,482,663,236
| 6,536
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 13
| 374
| 8,806,400
|
import sys
from collections import defaultdict
k, n = [int(x) for x in input().split()]
strs = []
scores = []
for i in range(k):
words = input().split()
strs.append(words[0])
scores.append(int(words[1]))
strs_set = set(strs)
palin_scores = defaultdict(list)
str8_scores = defaultdict(list)
rev_scores = defaultdict(list)
for s, score in zip(strs, scores):
srev = s[::-1]
if s == srev:
palin_scores[s].append(score)
elif srev in strs_set:
if s < srev:
str8_scores[s].append(score)
else:
rev_scores[srev].append(score)
for dd in (palin_scores, str8_scores, rev_scores):
for v in dd.values():
v.sort(reverse=True)
# print(palin_scores, file=sys.stderr)
# print(str8_scores, file=sys.stderr)
# print(rev_scores, file=sys.stderr)
non_palin_score = 0
for s in str8_scores:
for score1, score2 in zip(str8_scores[s], rev_scores[s]):
score_sum = score1 + score2
if score_sum > 0:
non_palin_score += score1 + score2
else:
break
palin_score = 0
middle_score = 0
for s, scorelist in palin_scores.items():
if len(scorelist) >= 2:
for i in range(0, len(scorelist) - 1, 2):
score_sum = scorelist[i] + scorelist[i + 1]
if score_sum > 0:
palin_score += score_sum
else:
middle_score = max(middle_score, scorelist[i])
break
else:
i += 2
if i < len(scorelist):
middle_score = max(middle_score, scorelist[i])
else:
middle_score = max(middle_score, scorelist[0])
# print(non_palin_score, palin_score, middle_score, file=sys.stderr)
print(non_palin_score + palin_score + middle_score)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Santa Claus likes palindromes very much. There was his birthday recently. *k* of his friends came to him to congratulate him, and each of them presented to him a string *s**i* having the same length *n*. We denote the beauty of the *i*-th string by *a**i*. It can happen that *a**i* is negative — that means that Santa doesn't find this string beautiful at all.
Santa Claus is crazy about palindromes. He is thinking about the following question: what is the maximum possible total beauty of a palindrome which can be obtained by concatenating some (possibly all) of the strings he has? Each present can be used at most once. Note that all strings have the same length *n*.
Recall that a palindrome is a string that doesn't change after one reverses it.
Since the empty string is a palindrome too, the answer can't be negative. Even if all *a**i*'s are negative, Santa can obtain the empty string.
Input Specification:
The first line contains two positive integers *k* and *n* divided by space and denoting the number of Santa friends and the length of every string they've presented, respectively (1<=≤<=*k*,<=*n*<=≤<=100<=000; *n*·*k* <=≤<=100<=000).
*k* lines follow. The *i*-th of them contains the string *s**i* and its beauty *a**i* (<=-<=10<=000<=≤<=*a**i*<=≤<=10<=000). The string consists of *n* lowercase English letters, and its beauty is integer. Some of strings may coincide. Also, equal strings can have different beauties.
Output Specification:
In the only line print the required maximum possible beauty.
Demo Input:
['7 3\nabb 2\naaa -3\nbba -1\nzyz -4\nabb 5\naaa 7\nxyx 4\n', '3 1\na 1\na 2\na 3\n', '2 5\nabcde 10000\nabcde 10000\n']
Demo Output:
['12\n', '6\n', '0\n']
Note:
In the first example Santa can obtain abbaaaxyxaaabba by concatenating strings 5, 2, 7, 6 and 3 (in this order).
|
```python
import sys
from collections import defaultdict
k, n = [int(x) for x in input().split()]
strs = []
scores = []
for i in range(k):
words = input().split()
strs.append(words[0])
scores.append(int(words[1]))
strs_set = set(strs)
palin_scores = defaultdict(list)
str8_scores = defaultdict(list)
rev_scores = defaultdict(list)
for s, score in zip(strs, scores):
srev = s[::-1]
if s == srev:
palin_scores[s].append(score)
elif srev in strs_set:
if s < srev:
str8_scores[s].append(score)
else:
rev_scores[srev].append(score)
for dd in (palin_scores, str8_scores, rev_scores):
for v in dd.values():
v.sort(reverse=True)
# print(palin_scores, file=sys.stderr)
# print(str8_scores, file=sys.stderr)
# print(rev_scores, file=sys.stderr)
non_palin_score = 0
for s in str8_scores:
for score1, score2 in zip(str8_scores[s], rev_scores[s]):
score_sum = score1 + score2
if score_sum > 0:
non_palin_score += score1 + score2
else:
break
palin_score = 0
middle_score = 0
for s, scorelist in palin_scores.items():
if len(scorelist) >= 2:
for i in range(0, len(scorelist) - 1, 2):
score_sum = scorelist[i] + scorelist[i + 1]
if score_sum > 0:
palin_score += score_sum
else:
middle_score = max(middle_score, scorelist[i])
break
else:
i += 2
if i < len(scorelist):
middle_score = max(middle_score, scorelist[i])
else:
middle_score = max(middle_score, scorelist[0])
# print(non_palin_score, palin_score, middle_score, file=sys.stderr)
print(non_palin_score + palin_score + middle_score)
```
| 0
|
|
376
|
A
|
Lever
|
PROGRAMMING
| 900
|
[
"implementation",
"math"
] | null | null |
You have a description of a lever as string *s*. We'll represent the string length as record |*s*|, then the lever looks as a horizontal bar with weights of length |*s*|<=-<=1 with exactly one pivot. We will assume that the bar is a segment on the *Ox* axis between points 0 and |*s*|<=-<=1.
The decoding of the lever description is given below.
- If the *i*-th character of the string equals "^", that means that at coordinate *i* there is the pivot under the bar. - If the *i*-th character of the string equals "=", that means that at coordinate *i* there is nothing lying on the bar. - If the *i*-th character of the string equals digit *c* (1-9), that means that at coordinate *i* there is a weight of mass *c* on the bar.
Your task is, given the lever description, print if it will be in balance or not. Assume that the bar doesn't weight anything. Assume that the bar initially is in balance then all weights are simultaneously put on it. After that the bar either tilts to the left, or tilts to the right, or is in balance.
|
The first line contains the lever description as a non-empty string *s* (3<=≤<=|*s*|<=≤<=106), consisting of digits (1-9) and characters "^" and "=". It is guaranteed that the line contains exactly one character "^". It is guaranteed that the pivot of the lever isn't located in any end of the lever bar.
To solve the problem you may need 64-bit integer numbers. Please, do not forget to use them in your programs.
|
Print "left" if the given lever tilts to the left, "right" if it tilts to the right and "balance", if it is in balance.
|
[
"=^==\n",
"9===^==1\n",
"2==^7==\n",
"41^52==\n"
] |
[
"balance\n",
"left\n",
"right\n",
"balance\n"
] |
As you solve the problem, you may find the following link useful to better understand how a lever functions: http://en.wikipedia.org/wiki/Lever.
The pictures to the examples:
| 500
|
[
{
"input": "=^==",
"output": "balance"
},
{
"input": "9===^==1",
"output": "left"
},
{
"input": "2==^7==",
"output": "right"
},
{
"input": "41^52==",
"output": "balance"
},
{
"input": "=^2=4=1===1=",
"output": "right"
},
{
"input": "9=6===5==3=9=1=1^7=1==",
"output": "left"
},
{
"input": "85=61=36=^93===4==44==35==94===39===15===",
"output": "right"
},
{
"input": "==88=^95==83=45===8====73===7==7====1=29====29=8=85=",
"output": "right"
},
{
"input": "==41^52==",
"output": "balance"
},
{
"input": "2===================^2",
"output": "left"
},
{
"input": "9^=============1",
"output": "right"
},
{
"input": "4=========^=55",
"output": "left"
},
{
"input": "123^321",
"output": "balance"
},
{
"input": "7^1=2",
"output": "balance"
},
{
"input": "589==^==958",
"output": "right"
}
] | 1,594,911,197
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 60
| 265
| 26,828,800
|
s=input()
n=len(s)
x,y=0,0
z=s.index('^')
for i in range(n):
if(s[i].isdigit()):
x+=(int(s[i])*(i-z))
if(x==0):
print('balance')
elif(x<0):
print('left')
else:
print("right")
|
Title: Lever
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a description of a lever as string *s*. We'll represent the string length as record |*s*|, then the lever looks as a horizontal bar with weights of length |*s*|<=-<=1 with exactly one pivot. We will assume that the bar is a segment on the *Ox* axis between points 0 and |*s*|<=-<=1.
The decoding of the lever description is given below.
- If the *i*-th character of the string equals "^", that means that at coordinate *i* there is the pivot under the bar. - If the *i*-th character of the string equals "=", that means that at coordinate *i* there is nothing lying on the bar. - If the *i*-th character of the string equals digit *c* (1-9), that means that at coordinate *i* there is a weight of mass *c* on the bar.
Your task is, given the lever description, print if it will be in balance or not. Assume that the bar doesn't weight anything. Assume that the bar initially is in balance then all weights are simultaneously put on it. After that the bar either tilts to the left, or tilts to the right, or is in balance.
Input Specification:
The first line contains the lever description as a non-empty string *s* (3<=≤<=|*s*|<=≤<=106), consisting of digits (1-9) and characters "^" and "=". It is guaranteed that the line contains exactly one character "^". It is guaranteed that the pivot of the lever isn't located in any end of the lever bar.
To solve the problem you may need 64-bit integer numbers. Please, do not forget to use them in your programs.
Output Specification:
Print "left" if the given lever tilts to the left, "right" if it tilts to the right and "balance", if it is in balance.
Demo Input:
['=^==\n', '9===^==1\n', '2==^7==\n', '41^52==\n']
Demo Output:
['balance\n', 'left\n', 'right\n', 'balance\n']
Note:
As you solve the problem, you may find the following link useful to better understand how a lever functions: http://en.wikipedia.org/wiki/Lever.
The pictures to the examples:
|
```python
s=input()
n=len(s)
x,y=0,0
z=s.index('^')
for i in range(n):
if(s[i].isdigit()):
x+=(int(s[i])*(i-z))
if(x==0):
print('balance')
elif(x<0):
print('left')
else:
print("right")
```
| 3
|
|
352
|
B
|
Jeff and Periods
|
PROGRAMMING
| 1,300
|
[
"implementation",
"sortings"
] | null | null |
One day Jeff got hold of an integer sequence *a*1, *a*2, ..., *a**n* of length *n*. The boy immediately decided to analyze the sequence. For that, he needs to find all values of *x*, for which these conditions hold:
- *x* occurs in sequence *a*. - Consider all positions of numbers *x* in the sequence *a* (such *i*, that *a**i*<==<=*x*). These numbers, sorted in the increasing order, must form an arithmetic progression.
Help Jeff, find all *x* that meet the problem conditions.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105). The numbers are separated by spaces.
|
In the first line print integer *t* — the number of valid *x*. On each of the next *t* lines print two integers *x* and *p**x*, where *x* is current suitable value, *p**x* is the common difference between numbers in the progression (if *x* occurs exactly once in the sequence, *p**x* must equal 0). Print the pairs in the order of increasing *x*.
|
[
"1\n2\n",
"8\n1 2 1 3 1 2 1 5\n"
] |
[
"1\n2 0\n",
"4\n1 2\n2 4\n3 0\n5 0\n"
] |
In the first test 2 occurs exactly once in the sequence, ergo *p*<sub class="lower-index">2</sub> = 0.
| 1,000
|
[
{
"input": "1\n2",
"output": "1\n2 0"
},
{
"input": "8\n1 2 1 3 1 2 1 5",
"output": "4\n1 2\n2 4\n3 0\n5 0"
},
{
"input": "3\n1 10 5",
"output": "3\n1 0\n5 0\n10 0"
},
{
"input": "4\n9 9 3 5",
"output": "3\n3 0\n5 0\n9 1"
},
{
"input": "6\n1 2 2 1 1 2",
"output": "0"
},
{
"input": "6\n2 6 3 8 7 2",
"output": "5\n2 5\n3 0\n6 0\n7 0\n8 0"
},
{
"input": "7\n2 1 2 1 2 1 2",
"output": "2\n1 2\n2 2"
},
{
"input": "8\n1 1 1 1 1 1 1 1",
"output": "1\n1 1"
},
{
"input": "9\n2 3 3 3 2 1 2 3 2",
"output": "1\n1 0"
},
{
"input": "10\n3 1 1 1 1 3 1 2 2 1",
"output": "2\n2 1\n3 5"
},
{
"input": "12\n10 9 8 7 7 8 9 10 10 9 8 7",
"output": "0"
}
] | 1,622,816,956
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 2
| 280
| 0
|
n=int(input())
lst = [int(i) for i in input().split()][:n]
s=set(lst)
dt={}
ans=[]
di=[]
def check(l):
flag=0
if(len(l)==1):
di.append(0)
return True
diff=l[1]-l[0]
for i in range(0,len(l)-1):
if((l[i+1]-l[i])!=diff):
flag=1
if(flag==1):
return False
else:
di.append(diff)
return True
for i in s:
l=[]
for j in range(0,n):
if(i==lst[j]):
dt[i]=j
l.append(j)
l.sort()
if(check(l)):
ans.append(i)
print(len(ans))
for i in range(0,len(ans)):
print(ans[i],di[i])
|
Title: Jeff and Periods
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Jeff got hold of an integer sequence *a*1, *a*2, ..., *a**n* of length *n*. The boy immediately decided to analyze the sequence. For that, he needs to find all values of *x*, for which these conditions hold:
- *x* occurs in sequence *a*. - Consider all positions of numbers *x* in the sequence *a* (such *i*, that *a**i*<==<=*x*). These numbers, sorted in the increasing order, must form an arithmetic progression.
Help Jeff, find all *x* that meet the problem conditions.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105). The next line contains integers *a*1, *a*2, ..., *a**n* (1<=≤<=*a**i*<=≤<=105). The numbers are separated by spaces.
Output Specification:
In the first line print integer *t* — the number of valid *x*. On each of the next *t* lines print two integers *x* and *p**x*, where *x* is current suitable value, *p**x* is the common difference between numbers in the progression (if *x* occurs exactly once in the sequence, *p**x* must equal 0). Print the pairs in the order of increasing *x*.
Demo Input:
['1\n2\n', '8\n1 2 1 3 1 2 1 5\n']
Demo Output:
['1\n2 0\n', '4\n1 2\n2 4\n3 0\n5 0\n']
Note:
In the first test 2 occurs exactly once in the sequence, ergo *p*<sub class="lower-index">2</sub> = 0.
|
```python
n=int(input())
lst = [int(i) for i in input().split()][:n]
s=set(lst)
dt={}
ans=[]
di=[]
def check(l):
flag=0
if(len(l)==1):
di.append(0)
return True
diff=l[1]-l[0]
for i in range(0,len(l)-1):
if((l[i+1]-l[i])!=diff):
flag=1
if(flag==1):
return False
else:
di.append(diff)
return True
for i in s:
l=[]
for j in range(0,n):
if(i==lst[j]):
dt[i]=j
l.append(j)
l.sort()
if(check(l)):
ans.append(i)
print(len(ans))
for i in range(0,len(ans)):
print(ans[i],di[i])
```
| 0
|
|
807
|
B
|
T-Shirt Hunt
|
PROGRAMMING
| 1,300
|
[
"brute force",
"implementation"
] | null | null |
Not so long ago the Codecraft-17 contest was held on Codeforces. The top 25 participants, and additionally random 25 participants out of those who got into top 500, will receive a Codeforces T-shirt.
Unfortunately, you didn't manage to get into top 25, but you got into top 500, taking place *p*.
Now the elimination round of 8VC Venture Cup 2017 is being held. It has been announced that the Codecraft-17 T-shirt winners will be chosen as follows. Let *s* be the number of points of the winner of the elimination round of 8VC Venture Cup 2017. Then the following pseudocode will be executed:
Here "div" is the integer division operator, "mod" is the modulo (the remainder of division) operator.
As the result of pseudocode execution, 25 integers between 26 and 500, inclusive, will be printed. These will be the numbers of places of the participants who get the Codecraft-17 T-shirts. It is guaranteed that the 25 printed integers will be pairwise distinct for any value of *s*.
You're in the lead of the elimination round of 8VC Venture Cup 2017, having *x* points. You believe that having at least *y* points in the current round will be enough for victory.
To change your final score, you can make any number of successful and unsuccessful hacks. A successful hack brings you 100 points, an unsuccessful one takes 50 points from you. It's difficult to do successful hacks, though.
You want to win the current round and, at the same time, ensure getting a Codecraft-17 T-shirt. What is the smallest number of successful hacks you have to do to achieve that?
|
The only line contains three integers *p*, *x* and *y* (26<=≤<=*p*<=≤<=500; 1<=≤<=*y*<=≤<=*x*<=≤<=20000) — your place in Codecraft-17, your current score in the elimination round of 8VC Venture Cup 2017, and the smallest number of points you consider sufficient for winning the current round.
|
Output a single integer — the smallest number of successful hacks you have to do in order to both win the elimination round of 8VC Venture Cup 2017 and ensure getting a Codecraft-17 T-shirt.
It's guaranteed that your goal is achievable for any valid input data.
|
[
"239 10880 9889\n",
"26 7258 6123\n",
"493 8000 8000\n",
"101 6800 6500\n",
"329 19913 19900\n"
] |
[
"0\n",
"2\n",
"24\n",
"0\n",
"8\n"
] |
In the first example, there is no need to do any hacks since 10880 points already bring the T-shirt to the 239-th place of Codecraft-17 (that is, you). In this case, according to the pseudocode, the T-shirts will be given to the participants at the following places:
In the second example, you have to do two successful and one unsuccessful hack to make your score equal to 7408.
In the third example, you need to do as many as 24 successful hacks to make your score equal to 10400.
In the fourth example, it's sufficient to do 6 unsuccessful hacks (and no successful ones) to make your score equal to 6500, which is just enough for winning the current round and also getting the T-shirt.
| 1,000
|
[
{
"input": "239 10880 9889",
"output": "0"
},
{
"input": "26 7258 6123",
"output": "2"
},
{
"input": "493 8000 8000",
"output": "24"
},
{
"input": "101 6800 6500",
"output": "0"
},
{
"input": "329 19913 19900",
"output": "8"
},
{
"input": "264 19252 10888",
"output": "0"
},
{
"input": "176 9670 9174",
"output": "6"
},
{
"input": "42 11 6",
"output": "27"
},
{
"input": "412 17647 15917",
"output": "8"
},
{
"input": "91 4883 4302",
"output": "12"
},
{
"input": "200 16031 15842",
"output": "24"
},
{
"input": "186 18666 18329",
"output": "23"
},
{
"input": "486 9748 9598",
"output": "25"
},
{
"input": "180 4213 4207",
"output": "27"
},
{
"input": "329 19989 1",
"output": "0"
},
{
"input": "390 11676 2570",
"output": "0"
},
{
"input": "173 7017 4512",
"output": "0"
},
{
"input": "38 6404 5034",
"output": "0"
},
{
"input": "364 17243 16625",
"output": "0"
},
{
"input": "57 11066 9738",
"output": "1"
},
{
"input": "419 9142 8622",
"output": "1"
},
{
"input": "31 12956 10515",
"output": "2"
},
{
"input": "412 5027 4975",
"output": "2"
},
{
"input": "94 1231 986",
"output": "3"
},
{
"input": "173 7783 7674",
"output": "3"
},
{
"input": "338 8291 8008",
"output": "7"
},
{
"input": "424 10906 10346",
"output": "13"
},
{
"input": "168 2953 2292",
"output": "17"
},
{
"input": "406 16527 16314",
"output": "22"
},
{
"input": "368 1597 1506",
"output": "26"
},
{
"input": "111 14627 14479",
"output": "26"
},
{
"input": "400 15224 15212",
"output": "27"
},
{
"input": "427 19269 19231",
"output": "27"
},
{
"input": "26 10232 10220",
"output": "27"
},
{
"input": "500 7030 7023",
"output": "27"
},
{
"input": "26 13819 13682",
"output": "0"
},
{
"input": "500 18737 18069",
"output": "0"
},
{
"input": "26 20000 20000",
"output": "7"
},
{
"input": "26 1 1",
"output": "6"
},
{
"input": "26 20000 1",
"output": "0"
},
{
"input": "68 51 1",
"output": "0"
},
{
"input": "198 6550 6549",
"output": "5"
},
{
"input": "68 50 49",
"output": "5"
},
{
"input": "239 10927 10880",
"output": "11"
},
{
"input": "239 10830 9889",
"output": "1"
},
{
"input": "329 2150 1900",
"output": "0"
},
{
"input": "164 49 48",
"output": "1"
},
{
"input": "329 2150 2101",
"output": "6"
}
] | 1,495,191,700
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 48
| 62
| 0
|
l=input().split()
p,x,y=int(l[0])-26,int(l[1]),int(l[2])
def ch(n):
i=(n//50)%475
for c in range(25):
i=(i*96+42)%475
if i==p:return 0
return 1
z=y+(x-y)%50
while ch(z):z+=50
if z>x:
n=(z-x)//50
print((n>>1)+(n&1))
else:
print('0')
|
Title: T-Shirt Hunt
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Not so long ago the Codecraft-17 contest was held on Codeforces. The top 25 participants, and additionally random 25 participants out of those who got into top 500, will receive a Codeforces T-shirt.
Unfortunately, you didn't manage to get into top 25, but you got into top 500, taking place *p*.
Now the elimination round of 8VC Venture Cup 2017 is being held. It has been announced that the Codecraft-17 T-shirt winners will be chosen as follows. Let *s* be the number of points of the winner of the elimination round of 8VC Venture Cup 2017. Then the following pseudocode will be executed:
Here "div" is the integer division operator, "mod" is the modulo (the remainder of division) operator.
As the result of pseudocode execution, 25 integers between 26 and 500, inclusive, will be printed. These will be the numbers of places of the participants who get the Codecraft-17 T-shirts. It is guaranteed that the 25 printed integers will be pairwise distinct for any value of *s*.
You're in the lead of the elimination round of 8VC Venture Cup 2017, having *x* points. You believe that having at least *y* points in the current round will be enough for victory.
To change your final score, you can make any number of successful and unsuccessful hacks. A successful hack brings you 100 points, an unsuccessful one takes 50 points from you. It's difficult to do successful hacks, though.
You want to win the current round and, at the same time, ensure getting a Codecraft-17 T-shirt. What is the smallest number of successful hacks you have to do to achieve that?
Input Specification:
The only line contains three integers *p*, *x* and *y* (26<=≤<=*p*<=≤<=500; 1<=≤<=*y*<=≤<=*x*<=≤<=20000) — your place in Codecraft-17, your current score in the elimination round of 8VC Venture Cup 2017, and the smallest number of points you consider sufficient for winning the current round.
Output Specification:
Output a single integer — the smallest number of successful hacks you have to do in order to both win the elimination round of 8VC Venture Cup 2017 and ensure getting a Codecraft-17 T-shirt.
It's guaranteed that your goal is achievable for any valid input data.
Demo Input:
['239 10880 9889\n', '26 7258 6123\n', '493 8000 8000\n', '101 6800 6500\n', '329 19913 19900\n']
Demo Output:
['0\n', '2\n', '24\n', '0\n', '8\n']
Note:
In the first example, there is no need to do any hacks since 10880 points already bring the T-shirt to the 239-th place of Codecraft-17 (that is, you). In this case, according to the pseudocode, the T-shirts will be given to the participants at the following places:
In the second example, you have to do two successful and one unsuccessful hack to make your score equal to 7408.
In the third example, you need to do as many as 24 successful hacks to make your score equal to 10400.
In the fourth example, it's sufficient to do 6 unsuccessful hacks (and no successful ones) to make your score equal to 6500, which is just enough for winning the current round and also getting the T-shirt.
|
```python
l=input().split()
p,x,y=int(l[0])-26,int(l[1]),int(l[2])
def ch(n):
i=(n//50)%475
for c in range(25):
i=(i*96+42)%475
if i==p:return 0
return 1
z=y+(x-y)%50
while ch(z):z+=50
if z>x:
n=(z-x)//50
print((n>>1)+(n&1))
else:
print('0')
```
| 3
|
|
1,004
|
C
|
Sonya and Robots
|
PROGRAMMING
| 1,400
|
[
"constructive algorithms",
"implementation"
] | null | null |
Since Sonya is interested in robotics too, she decided to construct robots that will read and recognize numbers.
Sonya has drawn $n$ numbers in a row, $a_i$ is located in the $i$-th position. She also has put a robot at each end of the row (to the left of the first number and to the right of the last number). Sonya will give a number to each robot (they can be either same or different) and run them. When a robot is running, it is moving toward to another robot, reading numbers in the row. When a robot is reading a number that is equal to the number that was given to that robot, it will turn off and stay in the same position.
Sonya does not want robots to break, so she will give such numbers that robots will stop before they meet. That is, the girl wants them to stop at different positions so that the first robot is to the left of the second one.
For example, if the numbers $[1, 5, 4, 1, 3]$ are written, and Sonya gives the number $1$ to the first robot and the number $4$ to the second one, the first robot will stop in the $1$-st position while the second one in the $3$-rd position. In that case, robots will not meet each other. As a result, robots will not be broken. But if Sonya gives the number $4$ to the first robot and the number $5$ to the second one, they will meet since the first robot will stop in the $3$-rd position while the second one is in the $2$-nd position.
Sonya understands that it does not make sense to give a number that is not written in the row because a robot will not find this number and will meet the other robot.
Sonya is now interested in finding the number of different pairs that she can give to robots so that they will not meet. In other words, she wants to know the number of pairs ($p$, $q$), where she will give $p$ to the first robot and $q$ to the second one. Pairs ($p_i$, $q_i$) and ($p_j$, $q_j$) are different if $p_i\neq p_j$ or $q_i\neq q_j$.
Unfortunately, Sonya is busy fixing robots that broke after a failed launch. That is why she is asking you to find the number of pairs that she can give to robots so that they will not meet.
|
The first line contains a single integer $n$ ($1\leq n\leq 10^5$) — the number of numbers in a row.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1\leq a_i\leq 10^5$) — the numbers in a row.
|
Print one number — the number of possible pairs that Sonya can give to robots so that they will not meet.
|
[
"5\n1 5 4 1 3\n",
"7\n1 2 1 1 1 3 2\n"
] |
[
"9\n",
"7\n"
] |
In the first example, Sonya can give pairs ($1$, $1$), ($1$, $3$), ($1$, $4$), ($1$, $5$), ($4$, $1$), ($4$, $3$), ($5$, $1$), ($5$, $3$), and ($5$, $4$).
In the second example, Sonya can give pairs ($1$, $1$), ($1$, $2$), ($1$, $3$), ($2$, $1$), ($2$, $2$), ($2$, $3$), and ($3$, $2$).
| 1,500
|
[
{
"input": "5\n1 5 4 1 3",
"output": "9"
},
{
"input": "7\n1 2 1 1 1 3 2",
"output": "7"
},
{
"input": "10\n2 2 4 4 3 1 1 2 3 2",
"output": "14"
},
{
"input": "15\n1 2 2 1 2 4 2 1 1 6 6 4 2 5 4",
"output": "20"
},
{
"input": "1\n1",
"output": "0"
}
] | 1,530,810,729
| 2,229
|
PyPy 3
|
OK
|
TESTS
| 42
| 233
| 14,643,200
|
def main():
n = int(input())
a = tuple(map(int, input().split()))
seen = set()
count = 0
distinct = [0]
for x in reversed(a):
if x in seen:
distinct.append(distinct[-1])
continue
seen.add(x)
distinct.append(distinct[-1]+1)
distinct.pop()
seen.clear()
#print(tuple(reversed(distinct)))
for x, distinct_right in zip(a, reversed(distinct)):
#print(x, distinct_right)
if x in seen:
continue
seen.add(x)
count += distinct_right
print(count)
main()
|
Title: Sonya and Robots
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Since Sonya is interested in robotics too, she decided to construct robots that will read and recognize numbers.
Sonya has drawn $n$ numbers in a row, $a_i$ is located in the $i$-th position. She also has put a robot at each end of the row (to the left of the first number and to the right of the last number). Sonya will give a number to each robot (they can be either same or different) and run them. When a robot is running, it is moving toward to another robot, reading numbers in the row. When a robot is reading a number that is equal to the number that was given to that robot, it will turn off and stay in the same position.
Sonya does not want robots to break, so she will give such numbers that robots will stop before they meet. That is, the girl wants them to stop at different positions so that the first robot is to the left of the second one.
For example, if the numbers $[1, 5, 4, 1, 3]$ are written, and Sonya gives the number $1$ to the first robot and the number $4$ to the second one, the first robot will stop in the $1$-st position while the second one in the $3$-rd position. In that case, robots will not meet each other. As a result, robots will not be broken. But if Sonya gives the number $4$ to the first robot and the number $5$ to the second one, they will meet since the first robot will stop in the $3$-rd position while the second one is in the $2$-nd position.
Sonya understands that it does not make sense to give a number that is not written in the row because a robot will not find this number and will meet the other robot.
Sonya is now interested in finding the number of different pairs that she can give to robots so that they will not meet. In other words, she wants to know the number of pairs ($p$, $q$), where she will give $p$ to the first robot and $q$ to the second one. Pairs ($p_i$, $q_i$) and ($p_j$, $q_j$) are different if $p_i\neq p_j$ or $q_i\neq q_j$.
Unfortunately, Sonya is busy fixing robots that broke after a failed launch. That is why she is asking you to find the number of pairs that she can give to robots so that they will not meet.
Input Specification:
The first line contains a single integer $n$ ($1\leq n\leq 10^5$) — the number of numbers in a row.
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1\leq a_i\leq 10^5$) — the numbers in a row.
Output Specification:
Print one number — the number of possible pairs that Sonya can give to robots so that they will not meet.
Demo Input:
['5\n1 5 4 1 3\n', '7\n1 2 1 1 1 3 2\n']
Demo Output:
['9\n', '7\n']
Note:
In the first example, Sonya can give pairs ($1$, $1$), ($1$, $3$), ($1$, $4$), ($1$, $5$), ($4$, $1$), ($4$, $3$), ($5$, $1$), ($5$, $3$), and ($5$, $4$).
In the second example, Sonya can give pairs ($1$, $1$), ($1$, $2$), ($1$, $3$), ($2$, $1$), ($2$, $2$), ($2$, $3$), and ($3$, $2$).
|
```python
def main():
n = int(input())
a = tuple(map(int, input().split()))
seen = set()
count = 0
distinct = [0]
for x in reversed(a):
if x in seen:
distinct.append(distinct[-1])
continue
seen.add(x)
distinct.append(distinct[-1]+1)
distinct.pop()
seen.clear()
#print(tuple(reversed(distinct)))
for x, distinct_right in zip(a, reversed(distinct)):
#print(x, distinct_right)
if x in seen:
continue
seen.add(x)
count += distinct_right
print(count)
main()
```
| 3
|
|
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"
}
] | 1,487,260,175
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 33
| 171
| 23,142,400
|
s = input()
l = len(s)
print(26+25*l)
|
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
s = input()
l = len(s)
print(26+25*l)
```
| 3
|
|
493
|
D
|
Vasya and Chess
|
PROGRAMMING
| 1,700
|
[
"constructive algorithms",
"games",
"math"
] | null | null |
Vasya decided to learn to play chess. Classic chess doesn't seem interesting to him, so he plays his own sort of chess.
The queen is the piece that captures all squares on its vertical, horizontal and diagonal lines. If the cell is located on the same vertical, horizontal or diagonal line with queen, and the cell contains a piece of the enemy color, the queen is able to move to this square. After that the enemy's piece is removed from the board. The queen cannot move to a cell containing an enemy piece if there is some other piece between it and the queen.
There is an *n*<=×<=*n* chessboard. We'll denote a cell on the intersection of the *r*-th row and *c*-th column as (*r*,<=*c*). The square (1,<=1) contains the white queen and the square (1,<=*n*) contains the black queen. All other squares contain green pawns that don't belong to anyone.
The players move in turns. The player that moves first plays for the white queen, his opponent plays for the black queen.
On each move the player has to capture some piece with his queen (that is, move to a square that contains either a green pawn or the enemy queen). The player loses if either he cannot capture any piece during his move or the opponent took his queen during the previous move.
Help Vasya determine who wins if both players play with an optimal strategy on the board *n*<=×<=*n*.
|
The input contains a single number *n* (2<=≤<=*n*<=≤<=109) — the size of the board.
|
On the first line print the answer to problem — string "white" or string "black", depending on who wins if the both players play optimally.
If the answer is "white", then you should also print two integers *r* and *c* representing the cell (*r*,<=*c*), where the first player should make his first move to win. If there are multiple such cells, print the one with the minimum *r*. If there are still multiple squares, print the one with the minimum *c*.
|
[
"2\n",
"3\n"
] |
[
"white\n1 2\n",
"black\n"
] |
In the first sample test the white queen can capture the black queen at the first move, so the white player wins.
In the second test from the statement if the white queen captures the green pawn located on the central vertical line, then it will be captured by the black queen during the next move. So the only move for the white player is to capture the green pawn located at (2, 1).
Similarly, the black queen doesn't have any other options but to capture the green pawn located at (2, 3), otherwise if it goes to the middle vertical line, it will be captured by the white queen.
During the next move the same thing happens — neither the white, nor the black queen has other options rather than to capture green pawns situated above them. Thus, the white queen ends up on square (3, 1), and the black queen ends up on square (3, 3).
In this situation the white queen has to capture any of the green pawns located on the middle vertical line, after that it will be captured by the black queen. Thus, the player who plays for the black queen wins.
| 1,500
|
[
{
"input": "2",
"output": "white\n1 2"
},
{
"input": "3",
"output": "black"
},
{
"input": "4",
"output": "white\n1 2"
},
{
"input": "6",
"output": "white\n1 2"
},
{
"input": "10",
"output": "white\n1 2"
},
{
"input": "16",
"output": "white\n1 2"
},
{
"input": "100",
"output": "white\n1 2"
},
{
"input": "10006",
"output": "white\n1 2"
},
{
"input": "99966246",
"output": "white\n1 2"
},
{
"input": "1000000000",
"output": "white\n1 2"
},
{
"input": "999999999",
"output": "black"
},
{
"input": "999999997",
"output": "black"
},
{
"input": "900001",
"output": "black"
},
{
"input": "775681",
"output": "black"
},
{
"input": "666666",
"output": "white\n1 2"
},
{
"input": "12345",
"output": "black"
},
{
"input": "111111",
"output": "black"
},
{
"input": "346367",
"output": "black"
},
{
"input": "13",
"output": "black"
},
{
"input": "11",
"output": "black"
},
{
"input": "9",
"output": "black"
},
{
"input": "7",
"output": "black"
},
{
"input": "5",
"output": "black"
},
{
"input": "19",
"output": "black"
},
{
"input": "939698497",
"output": "black"
},
{
"input": "999999996",
"output": "white\n1 2"
}
] | 1,593,777,255
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 26
| 109
| 0
|
if int(input())&1 :
print('black')
else :
print('white\n1 2')
|
Title: Vasya and Chess
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya decided to learn to play chess. Classic chess doesn't seem interesting to him, so he plays his own sort of chess.
The queen is the piece that captures all squares on its vertical, horizontal and diagonal lines. If the cell is located on the same vertical, horizontal or diagonal line with queen, and the cell contains a piece of the enemy color, the queen is able to move to this square. After that the enemy's piece is removed from the board. The queen cannot move to a cell containing an enemy piece if there is some other piece between it and the queen.
There is an *n*<=×<=*n* chessboard. We'll denote a cell on the intersection of the *r*-th row and *c*-th column as (*r*,<=*c*). The square (1,<=1) contains the white queen and the square (1,<=*n*) contains the black queen. All other squares contain green pawns that don't belong to anyone.
The players move in turns. The player that moves first plays for the white queen, his opponent plays for the black queen.
On each move the player has to capture some piece with his queen (that is, move to a square that contains either a green pawn or the enemy queen). The player loses if either he cannot capture any piece during his move or the opponent took his queen during the previous move.
Help Vasya determine who wins if both players play with an optimal strategy on the board *n*<=×<=*n*.
Input Specification:
The input contains a single number *n* (2<=≤<=*n*<=≤<=109) — the size of the board.
Output Specification:
On the first line print the answer to problem — string "white" or string "black", depending on who wins if the both players play optimally.
If the answer is "white", then you should also print two integers *r* and *c* representing the cell (*r*,<=*c*), where the first player should make his first move to win. If there are multiple such cells, print the one with the minimum *r*. If there are still multiple squares, print the one with the minimum *c*.
Demo Input:
['2\n', '3\n']
Demo Output:
['white\n1 2\n', 'black\n']
Note:
In the first sample test the white queen can capture the black queen at the first move, so the white player wins.
In the second test from the statement if the white queen captures the green pawn located on the central vertical line, then it will be captured by the black queen during the next move. So the only move for the white player is to capture the green pawn located at (2, 1).
Similarly, the black queen doesn't have any other options but to capture the green pawn located at (2, 3), otherwise if it goes to the middle vertical line, it will be captured by the white queen.
During the next move the same thing happens — neither the white, nor the black queen has other options rather than to capture green pawns situated above them. Thus, the white queen ends up on square (3, 1), and the black queen ends up on square (3, 3).
In this situation the white queen has to capture any of the green pawns located on the middle vertical line, after that it will be captured by the black queen. Thus, the player who plays for the black queen wins.
|
```python
if int(input())&1 :
print('black')
else :
print('white\n1 2')
```
| 3
|
|
742
|
A
|
Arpa’s hard exam and Mehrdad’s naive cheat
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math",
"number theory"
] | null | null |
There exists an island called Arpa’s land, some beautiful girls live there, as ugly ones do.
Mehrdad wants to become minister of Arpa’s land. Arpa has prepared an exam. Exam has only one question, given *n*, print the last digit of 1378*n*.
Mehrdad has become quite confused and wants you to help him. Please help, although it's a naive cheat.
|
The single line of input contains one integer *n* (0<=<=≤<=<=*n*<=<=≤<=<=109).
|
Print single integer — the last digit of 1378*n*.
|
[
"1\n",
"2\n"
] |
[
"8",
"4"
] |
In the first example, last digit of 1378<sup class="upper-index">1</sup> = 1378 is 8.
In the second example, last digit of 1378<sup class="upper-index">2</sup> = 1378·1378 = 1898884 is 4.
| 500
|
[
{
"input": "1",
"output": "8"
},
{
"input": "2",
"output": "4"
},
{
"input": "1000",
"output": "6"
},
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "6"
},
{
"input": "1000000000",
"output": "6"
},
{
"input": "5",
"output": "8"
},
{
"input": "6",
"output": "4"
},
{
"input": "999999999",
"output": "2"
},
{
"input": "1378",
"output": "4"
},
{
"input": "13781378",
"output": "4"
},
{
"input": "51202278",
"output": "4"
},
{
"input": "999999998",
"output": "4"
},
{
"input": "999999997",
"output": "8"
},
{
"input": "12193721",
"output": "8"
},
{
"input": "0",
"output": "1"
},
{
"input": "989898989",
"output": "8"
},
{
"input": "7",
"output": "2"
},
{
"input": "8",
"output": "6"
},
{
"input": "9",
"output": "8"
},
{
"input": "10",
"output": "4"
},
{
"input": "11",
"output": "2"
},
{
"input": "12",
"output": "6"
},
{
"input": "13",
"output": "8"
},
{
"input": "14",
"output": "4"
},
{
"input": "15",
"output": "2"
},
{
"input": "16",
"output": "6"
},
{
"input": "999999996",
"output": "6"
},
{
"input": "999999995",
"output": "2"
},
{
"input": "999999994",
"output": "4"
}
] | 1,645,698,980
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 15
| 46
| 0
|
a=int(input())
if a%4==0:print(6)
elif a%2==0:print(4)
elif a%3==0:print(2)
elif a%1==0:print(8)
|
Title: Arpa’s hard exam and Mehrdad’s naive cheat
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There exists an island called Arpa’s land, some beautiful girls live there, as ugly ones do.
Mehrdad wants to become minister of Arpa’s land. Arpa has prepared an exam. Exam has only one question, given *n*, print the last digit of 1378*n*.
Mehrdad has become quite confused and wants you to help him. Please help, although it's a naive cheat.
Input Specification:
The single line of input contains one integer *n* (0<=<=≤<=<=*n*<=<=≤<=<=109).
Output Specification:
Print single integer — the last digit of 1378*n*.
Demo Input:
['1\n', '2\n']
Demo Output:
['8', '4']
Note:
In the first example, last digit of 1378<sup class="upper-index">1</sup> = 1378 is 8.
In the second example, last digit of 1378<sup class="upper-index">2</sup> = 1378·1378 = 1898884 is 4.
|
```python
a=int(input())
if a%4==0:print(6)
elif a%2==0:print(4)
elif a%3==0:print(2)
elif a%1==0:print(8)
```
| 0
|
|
83
|
A
|
Magical Array
|
PROGRAMMING
| 1,300
|
[
"math"
] |
A. Magical Array
|
2
|
256
|
Valery is very interested in magic. Magic attracts him so much that he sees it everywhere. He explains any strange and weird phenomenon through intervention of supernatural forces. But who would have thought that even in a regular array of numbers Valera manages to see something beautiful and magical.
Valera absolutely accidentally got a piece of ancient parchment on which an array of numbers was written. He immediately thought that the numbers in this array were not random. As a result of extensive research Valera worked out a wonderful property that a magical array should have: an array is defined as magic if its minimum and maximum coincide.
He decided to share this outstanding discovery with you, but he asks you for help in return. Despite the tremendous intelligence and wit, Valera counts very badly and so you will have to complete his work. All you have to do is count the number of magical subarrays of the original array of numbers, written on the parchment. Subarray is defined as non-empty sequence of consecutive elements.
|
The first line of the input data contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains an array of original integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109).
|
Print on the single line the answer to the problem: the amount of subarrays, which are magical.
Please do not use the %lld specificator to read or write 64-bit numbers in C++. It is recommended to use cin, cout streams (you can also use the %I64d specificator).
|
[
"4\n2 1 1 4\n",
"5\n-2 -2 -2 0 1\n"
] |
[
"5\n",
"8\n"
] |
Notes to sample tests:
Magical subarrays are shown with pairs of indices [a;b] of the beginning and the end.
In the first sample: [1;1], [2;2], [3;3], [4;4], [2;3].
In the second sample: [1;1], [2;2], [3;3], [4;4], [5;5], [1;2], [2;3], [1;3].
| 500
|
[
{
"input": "4\n2 1 1 4",
"output": "5"
},
{
"input": "5\n-2 -2 -2 0 1",
"output": "8"
},
{
"input": "1\n10",
"output": "1"
},
{
"input": "2\n5 6",
"output": "2"
},
{
"input": "5\n5 5 4 5 5",
"output": "7"
},
{
"input": "8\n1 2 0 0 0 0 3 3",
"output": "15"
},
{
"input": "12\n-4 3 3 2 3 3 3 -4 2 -4 -4 -4",
"output": "19"
},
{
"input": "10\n7 1 0 10 0 -5 -3 -2 0 0",
"output": "11"
},
{
"input": "20\n6 0 0 -3 1 -3 0 -8 1 3 5 2 -1 -5 -1 9 0 6 -2 4",
"output": "21"
},
{
"input": "100\n0 -18 -9 -15 3 16 -28 0 -28 0 28 -20 -9 9 -11 0 18 -15 -18 -26 0 -27 -25 -22 6 -5 8 14 -17 24 20 3 -6 24 -27 1 -23 0 4 12 -20 0 -10 30 22 -6 13 16 0 15 17 -8 -2 0 -5 13 11 23 -17 -29 10 15 -28 0 -23 4 20 17 -7 -5 -16 -17 16 2 20 19 -8 0 8 -5 12 0 0 -14 -15 -28 -10 20 0 8 -1 10 14 9 0 4 -16 15 13 -10",
"output": "101"
},
{
"input": "50\n2 0 2 0 0 0 0 -1 -2 -2 -2 1 1 2 2 0 2 0 2 -3 0 0 0 0 3 1 -2 0 -1 0 -2 3 -1 2 0 2 0 0 0 0 2 0 1 0 0 3 0 0 -2 0",
"output": "75"
},
{
"input": "2\n-510468670 0",
"output": "2"
},
{
"input": "150\n0 -2 1 -2 0 0 0 0 -2 0 -2 -1 0 0 2 0 1 -2 1 -1 0 0 0 2 -2 2 -1 0 0 0 -2 0 2 0 1 0 -2 0 -2 -1 -1 -2 -2 2 0 0 1 -2 -2 -1 -2 0 2 1 1 -1 1 0 -2 2 0 0 0 1 -1 0 -2 -1 0 -2 2 1 1 0 0 2 0 0 2 -1 0 0 2 0 2 0 -2 -1 1 -2 1 0 0 -2 -1 -1 0 0 2 -1 -1 -1 -1 -2 0 0 2 -1 -1 1 0 0 1 -1 0 0 -1 2 2 0 0 0 -1 -2 0 1 0 -1 0 -1 -1 0 2 0 1 2 0 0 2 0 0 1 2 0 2 -2 2 0 2 2",
"output": "196"
}
] | 1,647,271,246
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 50
| 248
| 9,728,000
|
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n = int(input())
a = list(map(int, input().split()))
c, x = 0, a[0]
ans = 0
for i in a:
if not i ^ x:
c += 1
else:
ans += c * (c + 1) // 2
c, x = 1, i
ans += c * (c + 1) // 2
print(ans)
|
Title: Magical Array
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Valery is very interested in magic. Magic attracts him so much that he sees it everywhere. He explains any strange and weird phenomenon through intervention of supernatural forces. But who would have thought that even in a regular array of numbers Valera manages to see something beautiful and magical.
Valera absolutely accidentally got a piece of ancient parchment on which an array of numbers was written. He immediately thought that the numbers in this array were not random. As a result of extensive research Valera worked out a wonderful property that a magical array should have: an array is defined as magic if its minimum and maximum coincide.
He decided to share this outstanding discovery with you, but he asks you for help in return. Despite the tremendous intelligence and wit, Valera counts very badly and so you will have to complete his work. All you have to do is count the number of magical subarrays of the original array of numbers, written on the parchment. Subarray is defined as non-empty sequence of consecutive elements.
Input Specification:
The first line of the input data contains an integer *n* (1<=≤<=*n*<=≤<=105). The second line contains an array of original integers *a*1,<=*a*2,<=...,<=*a**n* (<=-<=109<=≤<=*a**i*<=≤<=109).
Output Specification:
Print on the single line the answer to the problem: the amount of subarrays, which are magical.
Please do not use the %lld specificator to read or write 64-bit numbers in C++. It is recommended to use cin, cout streams (you can also use the %I64d specificator).
Demo Input:
['4\n2 1 1 4\n', '5\n-2 -2 -2 0 1\n']
Demo Output:
['5\n', '8\n']
Note:
Notes to sample tests:
Magical subarrays are shown with pairs of indices [a;b] of the beginning and the end.
In the first sample: [1;1], [2;2], [3;3], [4;4], [2;3].
In the second sample: [1;1], [2;2], [3;3], [4;4], [5;5], [1;2], [2;3], [1;3].
|
```python
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
n = int(input())
a = list(map(int, input().split()))
c, x = 0, a[0]
ans = 0
for i in a:
if not i ^ x:
c += 1
else:
ans += c * (c + 1) // 2
c, x = 1, i
ans += c * (c + 1) // 2
print(ans)
```
| 3.91988
|
574
|
B
|
Bear and Three Musketeers
|
PROGRAMMING
| 1,500
|
[
"brute force",
"dfs and similar",
"graphs",
"hashing"
] | null | null |
Do you know a story about the three musketeers? Anyway, you will learn about its origins now.
Richelimakieu is a cardinal in the city of Bearis. He is tired of dealing with crime by himself. He needs three brave warriors to help him to fight against bad guys.
There are *n* warriors. Richelimakieu wants to choose three of them to become musketeers but it's not that easy. The most important condition is that musketeers must know each other to cooperate efficiently. And they shouldn't be too well known because they could be betrayed by old friends. For each musketeer his recognition is the number of warriors he knows, excluding other two musketeers.
Help Richelimakieu! Find if it is possible to choose three musketeers knowing each other, and what is minimum possible sum of their recognitions.
|
The first line contains two space-separated integers, *n* and *m* (3<=≤<=*n*<=≤<=4000, 0<=≤<=*m*<=≤<=4000) — respectively number of warriors and number of pairs of warriors knowing each other.
*i*-th of the following *m* lines contains two space-separated integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*). Warriors *a**i* and *b**i* know each other. Each pair of warriors will be listed at most once.
|
If Richelimakieu can choose three musketeers, print the minimum possible sum of their recognitions. Otherwise, print "-1" (without the quotes).
|
[
"5 6\n1 2\n1 3\n2 3\n2 4\n3 4\n4 5\n",
"7 4\n2 1\n3 6\n5 1\n1 7\n"
] |
[
"2\n",
"-1\n"
] |
In the first sample Richelimakieu should choose a triple 1, 2, 3. The first musketeer doesn't know anyone except other two musketeers so his recognition is 0. The second musketeer has recognition 1 because he knows warrior number 4. The third musketeer also has recognition 1 because he knows warrior 4. Sum of recognitions is 0 + 1 + 1 = 2.
The other possible triple is 2, 3, 4 but it has greater sum of recognitions, equal to 1 + 1 + 1 = 3.
In the second sample there is no triple of warriors knowing each other.
| 1,000
|
[
{
"input": "5 6\n1 2\n1 3\n2 3\n2 4\n3 4\n4 5",
"output": "2"
},
{
"input": "7 4\n2 1\n3 6\n5 1\n1 7",
"output": "-1"
},
{
"input": "5 0",
"output": "-1"
},
{
"input": "7 14\n3 6\n2 3\n5 2\n5 6\n7 5\n7 4\n6 2\n3 5\n7 1\n4 1\n6 1\n7 6\n6 4\n5 4",
"output": "5"
},
{
"input": "15 15\n4 15\n12 1\n15 6\n11 6\n15 7\n6 8\n15 10\n6 12\n12 8\n15 8\n15 3\n11 9\n7 3\n6 4\n12 11",
"output": "4"
},
{
"input": "12 66\n9 12\n1 4\n8 4\n5 3\n10 5\n12 2\n3 2\n2 7\n1 7\n3 7\n6 2\n4 2\n6 10\n8 10\n4 6\n8 5\n12 6\n11 9\n7 12\n5 4\n11 7\n9 4\n10 4\n6 3\n1 6\n9 7\n3 8\n6 11\n10 9\n3 11\n11 1\n5 12\n8 2\n2 1\n3 1\n12 4\n3 9\n10 12\n8 11\n7 10\n11 5\n9 5\n8 7\n11 4\n8 1\n2 11\n5 1\n3 4\n8 12\n9 2\n10 11\n9 1\n5 7\n10 3\n11 12\n7 4\n2 10\n12 3\n6 8\n7 6\n2 5\n1 10\n12 1\n9 6\n8 9\n6 5",
"output": "27"
},
{
"input": "3 0",
"output": "-1"
},
{
"input": "3 2\n2 3\n2 1",
"output": "-1"
},
{
"input": "3 3\n3 1\n3 2\n2 1",
"output": "0"
},
{
"input": "4 6\n3 4\n1 3\n4 1\n3 2\n2 1\n4 2",
"output": "3"
},
{
"input": "8 10\n1 5\n4 1\n1 2\n2 8\n2 7\n6 3\n5 8\n3 5\n7 8\n1 6",
"output": "2"
},
{
"input": "15 17\n1 3\n7 10\n7 9\n8 13\n6 15\n8 2\n13 6\n10 5\n15 3\n4 15\n4 6\n5 11\n13 9\n12 2\n11 14\n4 12\n14 1",
"output": "3"
},
{
"input": "25 10\n19 11\n19 13\n13 11\n13 22\n19 23\n19 20\n13 17\n19 14\n13 15\n19 4",
"output": "7"
},
{
"input": "987 50\n221 959\n221 553\n959 695\n553 959\n819 437\n371 295\n695 553\n959 347\n595 699\n652 628\n553 347\n868 589\n695 221\n282 714\n351 703\n104 665\n755 436\n556 511\n695 347\n221 347\n243 874\n695 847\n863 501\n583 145\n786 221\n38 286\n72 397\n808 658\n724 437\n911 548\n405 759\n681 316\n648 328\n327 199\n772 139\n932 609\n859 576\n915 507\n379 316\n381 348\n918 871\n261 450\n443 389\n549 246\n901 515\n930 923\n336 545\n179 225\n213 677\n458 204",
"output": "6"
},
{
"input": "4000 0",
"output": "-1"
}
] | 1,465,302,330
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 967
| 6,041,600
|
from collections import defaultdict,deque
def main():
n,m = map(int, input().split())
d = defaultdict(set)
for i in range(m):
a,b = map(int,input().split())
d[a].add(b)
d[b].add(a)
ans = 1e10
for key,v in d.items():
for j in v:
for k in v:
if k in d[j]:
ans = min(ans,len(v)+len(d[j])+len(d[k])-6)
print(ans if ans != 1e10 else "-1")
if __name__ == "__main__":
main()
|
Title: Bear and Three Musketeers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Do you know a story about the three musketeers? Anyway, you will learn about its origins now.
Richelimakieu is a cardinal in the city of Bearis. He is tired of dealing with crime by himself. He needs three brave warriors to help him to fight against bad guys.
There are *n* warriors. Richelimakieu wants to choose three of them to become musketeers but it's not that easy. The most important condition is that musketeers must know each other to cooperate efficiently. And they shouldn't be too well known because they could be betrayed by old friends. For each musketeer his recognition is the number of warriors he knows, excluding other two musketeers.
Help Richelimakieu! Find if it is possible to choose three musketeers knowing each other, and what is minimum possible sum of their recognitions.
Input Specification:
The first line contains two space-separated integers, *n* and *m* (3<=≤<=*n*<=≤<=4000, 0<=≤<=*m*<=≤<=4000) — respectively number of warriors and number of pairs of warriors knowing each other.
*i*-th of the following *m* lines contains two space-separated integers *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*, *a**i*<=≠<=*b**i*). Warriors *a**i* and *b**i* know each other. Each pair of warriors will be listed at most once.
Output Specification:
If Richelimakieu can choose three musketeers, print the minimum possible sum of their recognitions. Otherwise, print "-1" (without the quotes).
Demo Input:
['5 6\n1 2\n1 3\n2 3\n2 4\n3 4\n4 5\n', '7 4\n2 1\n3 6\n5 1\n1 7\n']
Demo Output:
['2\n', '-1\n']
Note:
In the first sample Richelimakieu should choose a triple 1, 2, 3. The first musketeer doesn't know anyone except other two musketeers so his recognition is 0. The second musketeer has recognition 1 because he knows warrior number 4. The third musketeer also has recognition 1 because he knows warrior 4. Sum of recognitions is 0 + 1 + 1 = 2.
The other possible triple is 2, 3, 4 but it has greater sum of recognitions, equal to 1 + 1 + 1 = 3.
In the second sample there is no triple of warriors knowing each other.
|
```python
from collections import defaultdict,deque
def main():
n,m = map(int, input().split())
d = defaultdict(set)
for i in range(m):
a,b = map(int,input().split())
d[a].add(b)
d[b].add(a)
ans = 1e10
for key,v in d.items():
for j in v:
for k in v:
if k in d[j]:
ans = min(ans,len(v)+len(d[j])+len(d[k])-6)
print(ans if ans != 1e10 else "-1")
if __name__ == "__main__":
main()
```
| 3
|
|
686
|
A
|
Free Ice Cream
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
|
The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
|
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
|
[
"5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n",
"5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n"
] |
[
"22 1\n",
"3 2\n"
] |
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.
| 500
|
[
{
"input": "5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20",
"output": "22 1"
},
{
"input": "5 17\n- 16\n- 2\n- 98\n+ 100\n- 98",
"output": "3 2"
},
{
"input": "6 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000\n+ 1000000000",
"output": "7000000000 0"
},
{
"input": "5 12\n- 12\n+ 7\n- 6\n- 1\n+ 46",
"output": "46 0"
},
{
"input": "11 1000\n- 100\n+ 100\n+ 100\n+ 100\n+ 100\n- 100\n- 100\n- 100\n- 100\n- 100\n- 100",
"output": "700 0"
},
{
"input": "1 0\n- 526403222",
"output": "0 1"
},
{
"input": "1 897986543\n- 371188251",
"output": "526798292 0"
},
{
"input": "1 0\n+ 1",
"output": "1 0"
},
{
"input": "1 0\n- 1",
"output": "0 1"
},
{
"input": "1 10\n+ 10",
"output": "20 0"
},
{
"input": "1 3\n- 5",
"output": "3 1"
},
{
"input": "1 0\n- 5",
"output": "0 1"
},
{
"input": "1 0\n+ 5",
"output": "5 0"
}
] | 1,658,510,813
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 46
| 0
|
n,x=input().split()
n=int(n);x=int(x)
c=0
for i in range(n):
a,b=input().split()
b=int(b)
if a=="+":
x+=b
elif b>x:
c+=1
else :
x-=b
print(x,c)
|
Title: Free Ice Cream
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After their adventure with the magic mirror Kay and Gerda have returned home and sometimes give free ice cream to kids in the summer.
At the start of the day they have *x* ice cream packs. Since the ice cream is free, people start standing in the queue before Kay and Gerda's house even in the night. Each person in the queue wants either to take several ice cream packs for himself and his friends or to give several ice cream packs to Kay and Gerda (carriers that bring ice cream have to stand in the same queue).
If a carrier with *d* ice cream packs comes to the house, then Kay and Gerda take all his packs. If a child who wants to take *d* ice cream packs comes to the house, then Kay and Gerda will give him *d* packs if they have enough ice cream, otherwise the child will get no ice cream at all and will leave in distress.
Kay wants to find the amount of ice cream they will have after all people will leave from the queue, and Gerda wants to find the number of distressed kids.
Input Specification:
The first line contains two space-separated integers *n* and *x* (1<=≤<=*n*<=≤<=1000, 0<=≤<=*x*<=≤<=109).
Each of the next *n* lines contains a character '+' or '-', and an integer *d**i*, separated by a space (1<=≤<=*d**i*<=≤<=109). Record "+ *d**i*" in *i*-th line means that a carrier with *d**i* ice cream packs occupies *i*-th place from the start of the queue, and record "- *d**i*" means that a child who wants to take *d**i* packs stands in *i*-th place.
Output Specification:
Print two space-separated integers — number of ice cream packs left after all operations, and number of kids that left the house in distress.
Demo Input:
['5 7\n+ 5\n- 10\n- 20\n+ 40\n- 20\n', '5 17\n- 16\n- 2\n- 98\n+ 100\n- 98\n']
Demo Output:
['22 1\n', '3 2\n']
Note:
Consider the first sample.
1. Initially Kay and Gerda have 7 packs of ice cream. 1. Carrier brings 5 more, so now they have 12 packs. 1. A kid asks for 10 packs and receives them. There are only 2 packs remaining. 1. Another kid asks for 20 packs. Kay and Gerda do not have them, so the kid goes away distressed. 1. Carrier bring 40 packs, now Kay and Gerda have 42 packs. 1. Kid asks for 20 packs and receives them. There are 22 packs remaining.
|
```python
n,x=input().split()
n=int(n);x=int(x)
c=0
for i in range(n):
a,b=input().split()
b=int(b)
if a=="+":
x+=b
elif b>x:
c+=1
else :
x-=b
print(x,c)
```
| 3
|
|
411
|
A
|
Password Check
|
PROGRAMMING
| 800
|
[
"*special",
"implementation"
] | null | null |
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check.
Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions:
- the password length is at least 5 characters; - the password contains at least one large English letter; - the password contains at least one small English letter; - the password contains at least one digit.
You are given a password. Please implement the automatic check of its complexity for company Q.
|
The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_".
|
If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes).
|
[
"abacaba\n",
"X12345\n",
"CONTEST_is_STARTED!!11\n"
] |
[
"Too weak\n",
"Too weak\n",
"Correct\n"
] |
none
| 0
|
[
{
"input": "abacaba",
"output": "Too weak"
},
{
"input": "X12345",
"output": "Too weak"
},
{
"input": "CONTEST_is_STARTED!!11",
"output": "Correct"
},
{
"input": "1zA__",
"output": "Correct"
},
{
"input": "1zA_",
"output": "Too weak"
},
{
"input": "zA___",
"output": "Too weak"
},
{
"input": "1A___",
"output": "Too weak"
},
{
"input": "z1___",
"output": "Too weak"
},
{
"input": "0",
"output": "Too weak"
},
{
"input": "_",
"output": "Too weak"
},
{
"input": "a",
"output": "Too weak"
},
{
"input": "D",
"output": "Too weak"
},
{
"input": "_",
"output": "Too weak"
},
{
"input": "?",
"output": "Too weak"
},
{
"input": "?",
"output": "Too weak"
},
{
"input": "._,.!.,...?_,!.",
"output": "Too weak"
},
{
"input": "!_?_,?,?.,.,_!!!.!,.__,?!!,_!,?_,!??,?!..._!?_,?_!,?_.,._,,_.,.",
"output": "Too weak"
},
{
"input": "?..!.,,?,__.,...????_???__!,?...?.,,,,___!,.!,_,,_,??!_?_,!!?_!_??.?,.!!?_?_.,!",
"output": "Too weak"
},
{
"input": "XZX",
"output": "Too weak"
},
{
"input": "R",
"output": "Too weak"
},
{
"input": "H.FZ",
"output": "Too weak"
},
{
"input": "KSHMICWPK,LSBM_JVZ!IPDYDG_GOPCHXFJTKJBIFY,FPHMY,CB?PZEAG..,X,.GFHPIDBB,IQ?MZ",
"output": "Too weak"
},
{
"input": "EFHI,,Y?HMMUI,,FJGAY?FYPBJQMYM!DZHLFCTFWT?JOPDW,S_!OR?ATT?RWFBMAAKUHIDMHSD?LCZQY!UD_CGYGBAIRDPICYS",
"output": "Too weak"
},
{
"input": "T,NDMUYCCXH_L_FJHMCCAGX_XSCPGOUZSY?D?CNDSYRITYS,VAT!PJVKNTBMXGGRYKACLYU.RJQ_?UWKXYIDE_AE",
"output": "Too weak"
},
{
"input": "y",
"output": "Too weak"
},
{
"input": "qgw",
"output": "Too weak"
},
{
"input": "g",
"output": "Too weak"
},
{
"input": "loaray",
"output": "Too weak"
},
{
"input": "d_iymyvxolmjayhwpedocopqwmy.oalrdg!_n?.lrxpamhygps?kkzxydsbcaihfs.j?eu!oszjsy.vzu?!vs.bprz_j",
"output": "Too weak"
},
{
"input": "txguglvclyillwnono",
"output": "Too weak"
},
{
"input": "FwX",
"output": "Too weak"
},
{
"input": "Zi",
"output": "Too weak"
},
{
"input": "PodE",
"output": "Too weak"
},
{
"input": "SdoOuJ?nj_wJyf",
"output": "Too weak"
},
{
"input": "MhnfZjsUyXYw?f?ubKA",
"output": "Too weak"
},
{
"input": "CpWxDVzwHfYFfoXNtXMFuAZr",
"output": "Too weak"
},
{
"input": "9.,0",
"output": "Too weak"
},
{
"input": "5,8",
"output": "Too weak"
},
{
"input": "7",
"output": "Too weak"
},
{
"input": "34__39_02!,!,82!129!2!566",
"output": "Too weak"
},
{
"input": "96156027.65935663!_87!,44,..7914_!0_1,.4!!62!.8350!17_282!!9.2584,!!7__51.526.7",
"output": "Too weak"
},
{
"input": "90328_",
"output": "Too weak"
},
{
"input": "B9",
"output": "Too weak"
},
{
"input": "P1H",
"output": "Too weak"
},
{
"input": "J2",
"output": "Too weak"
},
{
"input": "M6BCAKW!85OSYX1D?.53KDXP42F",
"output": "Too weak"
},
{
"input": "C672F429Y8X6XU7S,.K9111UD3232YXT81S4!729ER7DZ.J7U1R_7VG6.FQO,LDH",
"output": "Too weak"
},
{
"input": "W2PI__!.O91H8OFY6AB__R30L9XOU8800?ZUD84L5KT99818NFNE35V.8LJJ5P2MM.B6B",
"output": "Too weak"
},
{
"input": "z1",
"output": "Too weak"
},
{
"input": "p1j",
"output": "Too weak"
},
{
"input": "j9",
"output": "Too weak"
},
{
"input": "v8eycoylzv0qkix5mfs_nhkn6k!?ovrk9!b69zy!4frc?k",
"output": "Too weak"
},
{
"input": "l4!m_44kpw8.jg!?oh,?y5oraw1tg7_x1.osl0!ny?_aihzhtt0e2!mr92tnk0es!1f,9he40_usa6c50l",
"output": "Too weak"
},
{
"input": "d4r!ak.igzhnu!boghwd6jl",
"output": "Too weak"
},
{
"input": "It0",
"output": "Too weak"
},
{
"input": "Yb1x",
"output": "Too weak"
},
{
"input": "Qf7",
"output": "Too weak"
},
{
"input": "Vu7jQU8.!FvHBYTsDp6AphaGfnEmySP9te",
"output": "Correct"
},
{
"input": "Ka4hGE,vkvNQbNolnfwp",
"output": "Correct"
},
{
"input": "Ee9oluD?amNItsjeQVtOjwj4w_ALCRh7F3eaZah",
"output": "Correct"
},
{
"input": "Um3Fj?QLhNuRE_Gx0cjMLOkGCm",
"output": "Correct"
},
{
"input": "Oq2LYmV9HmlaW",
"output": "Correct"
},
{
"input": "Cq7r3Wrb.lDb_0wsf7!ruUUGSf08RkxD?VsBEDdyE?SHK73TFFy0f8gmcATqGafgTv8OOg8or2HyMPIPiQ2Hsx8q5rn3_WZe",
"output": "Correct"
},
{
"input": "Wx4p1fOrEMDlQpTlIx0p.1cnFD7BnX2K8?_dNLh4cQBx_Zqsv83BnL5hGKNcBE9g3QB,!fmSvgBeQ_qiH7",
"output": "Correct"
},
{
"input": "k673,",
"output": "Too weak"
},
{
"input": "LzuYQ",
"output": "Too weak"
},
{
"input": "Pasq!",
"output": "Too weak"
},
{
"input": "x5hve",
"output": "Too weak"
},
{
"input": "b27fk",
"output": "Too weak"
},
{
"input": "h6y1l",
"output": "Too weak"
},
{
"input": "i9nij",
"output": "Too weak"
},
{
"input": "Gf5Q6",
"output": "Correct"
},
{
"input": "Uf24o",
"output": "Correct"
},
{
"input": "Oj9vu",
"output": "Correct"
},
{
"input": "c7jqaudcqmv8o7zvb5x_gp6zcgl6nwr7tz5or!28.tj8s1m2.wxz5a4id03!rq07?662vy.7.p5?vk2f2mc7ag8q3861rgd0rmbr",
"output": "Too weak"
},
{
"input": "i6a.,8jb,n0kv4.1!7h?p.96pnhhgy6cl7dg7e4o6o384ys3z.t71kkq,,w,oqi4?u,,m5!rzu6wym_4hm,ohjy!.vvksl?pt,,1",
"output": "Too weak"
},
{
"input": "M10V_MN_1K8YX2LA!89EYV7!5V9?,.IDHDP6JEC.OGLY.180LMZ6KW3Z5E17IT94ZNHS!79GN09Q6LH0,F3AYNKP?KM,QP_?XRD6",
"output": "Too weak"
},
{
"input": "Hi7zYuVXCPhaho68YgCMzzgLILM6toQTJq8akMqqrnUn6ZCD36iA1yVVpvlsIiMpCu!1QZd4ycIrQ5Kcrhk5k0jTrwdAAEEP_T2f",
"output": "Correct"
},
{
"input": "Bk2Q38vDSW5JqYu.077iYC.9YoiPc!Dh6FJWOVze6?YXiFjPNa4F1RG?154m9mY2jQobBnbxM,cDV8l1UX1?v?p.tTYIyJO!NYmE",
"output": "Correct"
},
{
"input": "Ro1HcZ.piN,JRR88DLh,WtW!pbFM076?wCSbqfK7N2s5zUySFBtzk7HV,BxHXR0zALAr016z5jvvB.WUdEcKgYFav5TygwHQC..C",
"output": "Correct"
},
{
"input": "!?.,_",
"output": "Too weak"
}
] | 1,645,845,257
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 0
|
j = input()
f = False
f1 = False
f2 = False
for i in j:
if i >= 'a' and i <= 'z':
f1 = True
if i >= 'A' and i<= 'Z':
f = True
if i>= '1' and i<= '9':
f2 = True
if f == f1==f2==True:
break
if f == f1==f2==True and len(j)>=5:
print('Correct')
else:
print('Two weak')
|
Title: Password Check
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have probably registered on Internet sites many times. And each time you should enter your invented password. Usually the registration form automatically checks the password's crypt resistance. If the user's password isn't complex enough, a message is displayed. Today your task is to implement such an automatic check.
Web-developers of the company Q assume that a password is complex enough, if it meets all of the following conditions:
- the password length is at least 5 characters; - the password contains at least one large English letter; - the password contains at least one small English letter; - the password contains at least one digit.
You are given a password. Please implement the automatic check of its complexity for company Q.
Input Specification:
The first line contains a non-empty sequence of characters (at most 100 characters). Each character is either a large English letter, or a small English letter, or a digit, or one of characters: "!", "?", ".", ",", "_".
Output Specification:
If the password is complex enough, print message "Correct" (without the quotes), otherwise print message "Too weak" (without the quotes).
Demo Input:
['abacaba\n', 'X12345\n', 'CONTEST_is_STARTED!!11\n']
Demo Output:
['Too weak\n', 'Too weak\n', 'Correct\n']
Note:
none
|
```python
j = input()
f = False
f1 = False
f2 = False
for i in j:
if i >= 'a' and i <= 'z':
f1 = True
if i >= 'A' and i<= 'Z':
f = True
if i>= '1' and i<= '9':
f2 = True
if f == f1==f2==True:
break
if f == f1==f2==True and len(j)>=5:
print('Correct')
else:
print('Two weak')
```
| 0
|
|
825
|
D
|
Suitable Replacement
|
PROGRAMMING
| 1,500
|
[
"binary search",
"greedy",
"implementation"
] | null | null |
You are given two strings *s* and *t* consisting of small Latin letters, string *s* can also contain '?' characters.
Suitability of string *s* is calculated by following metric:
Any two letters can be swapped positions, these operations can be performed arbitrary number of times over any pair of positions. Among all resulting strings *s*, you choose the one with the largest number of non-intersecting occurrences of string *t*. Suitability is this number of occurrences.
You should replace all '?' characters with small Latin letters in such a way that the suitability of string *s* is maximal.
|
The first line contains string *s* (1<=≤<=|*s*|<=≤<=106).
The second line contains string *t* (1<=≤<=|*t*|<=≤<=106).
|
Print string *s* with '?' replaced with small Latin letters in such a way that suitability of that string is maximal.
If there are multiple strings with maximal suitability then print any of them.
|
[
"?aa?\nab\n",
"??b?\nza\n",
"abcd\nabacaba\n"
] |
[
"baab\n",
"azbz\n",
"abcd\n"
] |
In the first example string "baab" can be transformed to "abab" with swaps, this one has suitability of 2. That means that string "baab" also has suitability of 2.
In the second example maximal suitability you can achieve is 1 and there are several dozens of such strings, "azbz" is just one of them.
In the third example there are no '?' characters and the suitability of the string is 0.
| 0
|
[
{
"input": "?aa?\nab",
"output": "baab"
},
{
"input": "??b?\nza",
"output": "azbz"
},
{
"input": "abcd\nabacaba",
"output": "abcd"
},
{
"input": "mqwstphetbfrsyxuzdww\nrutseqtsbh",
"output": "mqwstphetbfrsyxuzdww"
},
{
"input": "????????????????????\nxwkxsxlrre",
"output": "eekkllrrrrsswwxxxxxx"
},
{
"input": "fzjqgrgzzzghrwgwhfjq\nggwfrjzzqh",
"output": "fzjqgrgzzzghrwgwhfjq"
},
{
"input": "ng?\nvu",
"output": "ngz"
},
{
"input": "???\nb",
"output": "bbb"
},
{
"input": "??a?\na",
"output": "aaaa"
},
{
"input": "c?aa???a?cdcc??\nabcadbaccb",
"output": "cbaabbzazcdcczz"
},
{
"input": "cc?cdb?????b?cc?b?\ncabdacd",
"output": "ccacdbaaaddbdcczbz"
},
{
"input": "ccaaa?bc?baccbccccbca?cccab??aacccbaac?bccacc?accccaccbcbbcbc\ncbaaacccc",
"output": "ccaaaabcabaccbccccbcaacccabaaaacccbaaczbccacczaccccaccbcbbcbc"
},
{
"input": "aacaacbabc?bcca?ccca\nbc",
"output": "aacaacbabcbbccabccca"
}
] | 1,674,284,539
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 15
| 0
|
a=input()
b=input()
s=''
for i in a:
if i=='?':
n=random.randint(0,len(b))
s[i]=b[n]
else:
s[i]=a[i]
print(s)
|
Title: Suitable Replacement
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given two strings *s* and *t* consisting of small Latin letters, string *s* can also contain '?' characters.
Suitability of string *s* is calculated by following metric:
Any two letters can be swapped positions, these operations can be performed arbitrary number of times over any pair of positions. Among all resulting strings *s*, you choose the one with the largest number of non-intersecting occurrences of string *t*. Suitability is this number of occurrences.
You should replace all '?' characters with small Latin letters in such a way that the suitability of string *s* is maximal.
Input Specification:
The first line contains string *s* (1<=≤<=|*s*|<=≤<=106).
The second line contains string *t* (1<=≤<=|*t*|<=≤<=106).
Output Specification:
Print string *s* with '?' replaced with small Latin letters in such a way that suitability of that string is maximal.
If there are multiple strings with maximal suitability then print any of them.
Demo Input:
['?aa?\nab\n', '??b?\nza\n', 'abcd\nabacaba\n']
Demo Output:
['baab\n', 'azbz\n', 'abcd\n']
Note:
In the first example string "baab" can be transformed to "abab" with swaps, this one has suitability of 2. That means that string "baab" also has suitability of 2.
In the second example maximal suitability you can achieve is 1 and there are several dozens of such strings, "azbz" is just one of them.
In the third example there are no '?' characters and the suitability of the string is 0.
|
```python
a=input()
b=input()
s=''
for i in a:
if i=='?':
n=random.randint(0,len(b))
s[i]=b[n]
else:
s[i]=a[i]
print(s)
```
| -1
|
|
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 "
}
] | 1,634,195,992
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 63
| 608
| 11,878,400
|
d={}
n,m=map(int,input().split())
for i in range(m):
k=[3,2,1]
l=list(map(int,input().split()))
temp=4
for i in range(len(l)):
if l[i] in d:
temp=i
k.remove(d[l[i]])
break
for i in range(len(l)):
if i!=temp:
d[l[i]]=k[-1]
k.pop()
s=list(d.keys())
s.sort()
for i in s:
print(d[i],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
d={}
n,m=map(int,input().split())
for i in range(m):
k=[3,2,1]
l=list(map(int,input().split()))
temp=4
for i in range(len(l)):
if l[i] in d:
temp=i
k.remove(d[l[i]])
break
for i in range(len(l)):
if i!=temp:
d[l[i]]=k[-1]
k.pop()
s=list(d.keys())
s.sort()
for i in s:
print(d[i],end=" ")
```
| 3
|
|
466
|
D
|
Increase Sequence
|
PROGRAMMING
| 2,100
|
[
"combinatorics",
"dp"
] | null | null |
Peter has a sequence of integers *a*1,<=*a*2,<=...,<=*a**n*. Peter wants all numbers in the sequence to equal *h*. He can perform the operation of "adding one on the segment [*l*,<=*r*]": add one to all elements of the sequence with indices from *l* to *r* (inclusive). At that, Peter never chooses any element as the beginning of the segment twice. Similarly, Peter never chooses any element as the end of the segment twice. In other words, for any two segments [*l*1,<=*r*1] and [*l*2,<=*r*2], where Peter added one, the following inequalities hold: *l*1<=≠<=*l*2 and *r*1<=≠<=*r*2.
How many distinct ways are there to make all numbers in the sequence equal *h*? Print this number of ways modulo 1000000007 (109<=+<=7). Two ways are considered distinct if one of them has a segment that isn't in the other way.
|
The first line contains two integers *n*,<=*h* (1<=≤<=*n*,<=*h*<=≤<=2000). The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=2000).
|
Print a single integer — the answer to the problem modulo 1000000007 (109<=+<=7).
|
[
"3 2\n1 1 1\n",
"5 1\n1 1 1 1 1\n",
"4 3\n3 2 1 1\n"
] |
[
"4\n",
"1\n",
"0\n"
] |
none
| 2,000
|
[
{
"input": "3 2\n1 1 1",
"output": "4"
},
{
"input": "5 1\n1 1 1 1 1",
"output": "1"
},
{
"input": "4 3\n3 2 1 1",
"output": "0"
},
{
"input": "1 2000\n2000",
"output": "1"
},
{
"input": "3 2\n2 1 1",
"output": "2"
},
{
"input": "3 4\n4 3 2",
"output": "0"
},
{
"input": "10 6\n5 4 4 4 5 4 4 4 4 5",
"output": "972"
},
{
"input": "10 6\n5 4 4 4 5 4 4 4 4 5",
"output": "972"
},
{
"input": "7 4\n3 3 4 4 4 3 3",
"output": "4"
},
{
"input": "3 6\n5 6 5",
"output": "1"
},
{
"input": "3 2\n1 0 1",
"output": "2"
},
{
"input": "6 1\n0 1 0 0 0 1",
"output": "4"
},
{
"input": "6 3\n2 1 2 2 1 2",
"output": "8"
},
{
"input": "99 1999\n1999 1998 1997 1996 1995 1994 1993 1992 1991 1990 1989 1988 1987 1986 1985 1984 1983 1982 1981 1980 1979 1978 1977 1976 1975 1974 1973 1972 1971 1970 1969 1968 1967 1966 1965 1964 1963 1962 1961 1960 1959 1958 1957 1956 1955 1954 1953 1952 1951 1950 1951 1952 1953 1954 1955 1956 1957 1958 1959 1960 1961 1962 1963 1964 1965 1966 1967 1968 1969 1970 1971 1972 1973 1974 1975 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999",
"output": "726372166"
},
{
"input": "9 4\n2 3 2 2 2 3 3 3 2",
"output": "0"
},
{
"input": "1 349\n348",
"output": "1"
},
{
"input": "10 20\n20 19 18 17 16 15 16 17 18 19",
"output": "120"
},
{
"input": "19 10\n10 9 8 7 6 5 4 3 2 10 2 3 4 5 6 7 8 9 10",
"output": "0"
},
{
"input": "19 11\n10 9 8 7 6 5 4 3 3 3 3 3 4 5 6 7 8 9 10",
"output": "264539520"
},
{
"input": "7 3\n2 1 1 2 1 1 2",
"output": "36"
}
] | 1,689,253,765
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 46
| 0
|
print("_RANDOM_GUESS_1689253765.7700129")# 1689253765.770045
|
Title: Increase Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Peter has a sequence of integers *a*1,<=*a*2,<=...,<=*a**n*. Peter wants all numbers in the sequence to equal *h*. He can perform the operation of "adding one on the segment [*l*,<=*r*]": add one to all elements of the sequence with indices from *l* to *r* (inclusive). At that, Peter never chooses any element as the beginning of the segment twice. Similarly, Peter never chooses any element as the end of the segment twice. In other words, for any two segments [*l*1,<=*r*1] and [*l*2,<=*r*2], where Peter added one, the following inequalities hold: *l*1<=≠<=*l*2 and *r*1<=≠<=*r*2.
How many distinct ways are there to make all numbers in the sequence equal *h*? Print this number of ways modulo 1000000007 (109<=+<=7). Two ways are considered distinct if one of them has a segment that isn't in the other way.
Input Specification:
The first line contains two integers *n*,<=*h* (1<=≤<=*n*,<=*h*<=≤<=2000). The next line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=2000).
Output Specification:
Print a single integer — the answer to the problem modulo 1000000007 (109<=+<=7).
Demo Input:
['3 2\n1 1 1\n', '5 1\n1 1 1 1 1\n', '4 3\n3 2 1 1\n']
Demo Output:
['4\n', '1\n', '0\n']
Note:
none
|
```python
print("_RANDOM_GUESS_1689253765.7700129")# 1689253765.770045
```
| 0
|
|
499
|
B
|
Lecture
|
PROGRAMMING
| 1,000
|
[
"implementation",
"strings"
] | null | null |
You have a new professor of graph theory and he speaks very quickly. You come up with the following plan to keep up with his lecture and make notes.
You know two languages, and the professor is giving the lecture in the first one. The words in both languages consist of lowercase English characters, each language consists of several words. For each language, all words are distinct, i.e. they are spelled differently. Moreover, the words of these languages have a one-to-one correspondence, that is, for each word in each language, there exists exactly one word in the other language having has the same meaning.
You can write down every word the professor says in either the first language or the second language. Of course, during the lecture you write down each word in the language in which the word is shorter. In case of equal lengths of the corresponding words you prefer the word of the first language.
You are given the text of the lecture the professor is going to read. Find out how the lecture will be recorded in your notes.
|
The first line contains two integers, *n* and *m* (1<=≤<=*n*<=≤<=3000, 1<=≤<=*m*<=≤<=3000) — the number of words in the professor's lecture and the number of words in each of these languages.
The following *m* lines contain the words. The *i*-th line contains two strings *a**i*, *b**i* meaning that the word *a**i* belongs to the first language, the word *b**i* belongs to the second language, and these two words have the same meaning. It is guaranteed that no word occurs in both languages, and each word occurs in its language exactly once.
The next line contains *n* space-separated strings *c*1,<=*c*2,<=...,<=*c**n* — the text of the lecture. It is guaranteed that each of the strings *c**i* belongs to the set of strings {*a*1,<=*a*2,<=... *a**m*}.
All the strings in the input are non-empty, each consisting of no more than 10 lowercase English letters.
|
Output exactly *n* words: how you will record the lecture in your notebook. Output the words of the lecture in the same order as in the input.
|
[
"4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest\n",
"5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll\n"
] |
[
"codeforces round letter round\n",
"hbnyiyc joll joll un joll\n"
] |
none
| 500
|
[
{
"input": "4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest",
"output": "codeforces round letter round"
},
{
"input": "5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll",
"output": "hbnyiyc joll joll un joll"
},
{
"input": "5 5\nqueyqj f\nb vn\ntabzvk qpfzoqx\nytnyonoc hnxsd\njpggvr lchinjmt\nqueyqj jpggvr b ytnyonoc b",
"output": "f jpggvr b hnxsd b"
},
{
"input": "10 22\nazbrll oen\ngh vdyayei\njphveblohx vfglv\nmfyxib jepnvhcuwo\nrpikazqj uam\nl rx\nokjenof qpnyi\nj tixqrno\nod itozmfct\nikkdxmirx ev\nqexftojc p\nkdazb zjs\nmbk ykvqjrxaxu\nhbcwhouzq pwt\nmirpsz zfaegpl\nuhkkvcj rlvwj\nef iqnnwtolrc\npjzfcpmeag ecdayth\nsa qcthz\ncbfhfxi qrnbvdryz\nwqel tj\natx smkbid\nef hbcwhouzq cbfhfxi hbcwhouzq mirpsz cbfhfxi cbfhfxi okjenof pjzfcpmeag kdazb",
"output": "ef pwt cbfhfxi pwt mirpsz cbfhfxi cbfhfxi qpnyi ecdayth zjs"
},
{
"input": "1 1\namit am\namit",
"output": "am"
},
{
"input": "1 1\na c\na",
"output": "a"
}
] | 1,632,234,449
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 37
| 280
| 7,372,800
|
def main():
a,b=input().split()
a,b=int(a),int(b)
prepod=[]
neprepod=[]
for i in range(b):
s1,s2=input().split()
prepod.append(s1)
neprepod.append(s2)
otvet=''
text=input().split()
for i in range(a):
if len(neprepod[prepod.index(text[i])])>=len(text[i]):
otvet+=text[i]+" "
else:
otvet+=neprepod[prepod.index(text[i])]+" "
return otvet
print(main())
|
Title: Lecture
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a new professor of graph theory and he speaks very quickly. You come up with the following plan to keep up with his lecture and make notes.
You know two languages, and the professor is giving the lecture in the first one. The words in both languages consist of lowercase English characters, each language consists of several words. For each language, all words are distinct, i.e. they are spelled differently. Moreover, the words of these languages have a one-to-one correspondence, that is, for each word in each language, there exists exactly one word in the other language having has the same meaning.
You can write down every word the professor says in either the first language or the second language. Of course, during the lecture you write down each word in the language in which the word is shorter. In case of equal lengths of the corresponding words you prefer the word of the first language.
You are given the text of the lecture the professor is going to read. Find out how the lecture will be recorded in your notes.
Input Specification:
The first line contains two integers, *n* and *m* (1<=≤<=*n*<=≤<=3000, 1<=≤<=*m*<=≤<=3000) — the number of words in the professor's lecture and the number of words in each of these languages.
The following *m* lines contain the words. The *i*-th line contains two strings *a**i*, *b**i* meaning that the word *a**i* belongs to the first language, the word *b**i* belongs to the second language, and these two words have the same meaning. It is guaranteed that no word occurs in both languages, and each word occurs in its language exactly once.
The next line contains *n* space-separated strings *c*1,<=*c*2,<=...,<=*c**n* — the text of the lecture. It is guaranteed that each of the strings *c**i* belongs to the set of strings {*a*1,<=*a*2,<=... *a**m*}.
All the strings in the input are non-empty, each consisting of no more than 10 lowercase English letters.
Output Specification:
Output exactly *n* words: how you will record the lecture in your notebook. Output the words of the lecture in the same order as in the input.
Demo Input:
['4 3\ncodeforces codesecrof\ncontest round\nletter message\ncodeforces contest letter contest\n', '5 3\njoll wuqrd\neuzf un\nhbnyiyc rsoqqveh\nhbnyiyc joll joll euzf joll\n']
Demo Output:
['codeforces round letter round\n', 'hbnyiyc joll joll un joll\n']
Note:
none
|
```python
def main():
a,b=input().split()
a,b=int(a),int(b)
prepod=[]
neprepod=[]
for i in range(b):
s1,s2=input().split()
prepod.append(s1)
neprepod.append(s2)
otvet=''
text=input().split()
for i in range(a):
if len(neprepod[prepod.index(text[i])])>=len(text[i]):
otvet+=text[i]+" "
else:
otvet+=neprepod[prepod.index(text[i])]+" "
return otvet
print(main())
```
| 3
|
|
753
|
A
|
Santa Claus and Candies
|
PROGRAMMING
| 1,000
|
[
"dp",
"greedy",
"math"
] | null | null |
Santa Claus has *n* candies, he dreams to give them as gifts to children.
What is the maximal number of children for whose he can give candies if Santa Claus want each kid should get distinct positive integer number of candies. Santa Class wants to give all *n* candies he has.
|
The only line contains positive integer number *n* (1<=≤<=*n*<=≤<=1000) — number of candies Santa Claus has.
|
Print to the first line integer number *k* — maximal number of kids which can get candies.
Print to the second line *k* distinct integer numbers: number of candies for each of *k* kid. The sum of *k* printed numbers should be exactly *n*.
If there are many solutions, print any of them.
|
[
"5\n",
"9\n",
"2\n"
] |
[
"2\n2 3\n",
"3\n3 5 1\n",
"1\n2 \n"
] |
none
| 500
|
[
{
"input": "5",
"output": "2\n1 4 "
},
{
"input": "9",
"output": "3\n1 2 6 "
},
{
"input": "2",
"output": "1\n2 "
},
{
"input": "1",
"output": "1\n1 "
},
{
"input": "3",
"output": "2\n1 2 "
},
{
"input": "1000",
"output": "44\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 54 "
},
{
"input": "4",
"output": "2\n1 3 "
},
{
"input": "6",
"output": "3\n1 2 3 "
},
{
"input": "7",
"output": "3\n1 2 4 "
},
{
"input": "8",
"output": "3\n1 2 5 "
},
{
"input": "10",
"output": "4\n1 2 3 4 "
},
{
"input": "11",
"output": "4\n1 2 3 5 "
},
{
"input": "12",
"output": "4\n1 2 3 6 "
},
{
"input": "13",
"output": "4\n1 2 3 7 "
},
{
"input": "14",
"output": "4\n1 2 3 8 "
},
{
"input": "15",
"output": "5\n1 2 3 4 5 "
},
{
"input": "16",
"output": "5\n1 2 3 4 6 "
},
{
"input": "20",
"output": "5\n1 2 3 4 10 "
},
{
"input": "21",
"output": "6\n1 2 3 4 5 6 "
},
{
"input": "22",
"output": "6\n1 2 3 4 5 7 "
},
{
"input": "27",
"output": "6\n1 2 3 4 5 12 "
},
{
"input": "28",
"output": "7\n1 2 3 4 5 6 7 "
},
{
"input": "29",
"output": "7\n1 2 3 4 5 6 8 "
},
{
"input": "35",
"output": "7\n1 2 3 4 5 6 14 "
},
{
"input": "36",
"output": "8\n1 2 3 4 5 6 7 8 "
},
{
"input": "37",
"output": "8\n1 2 3 4 5 6 7 9 "
},
{
"input": "44",
"output": "8\n1 2 3 4 5 6 7 16 "
},
{
"input": "45",
"output": "9\n1 2 3 4 5 6 7 8 9 "
},
{
"input": "46",
"output": "9\n1 2 3 4 5 6 7 8 10 "
},
{
"input": "230",
"output": "20\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 40 "
},
{
"input": "231",
"output": "21\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 "
},
{
"input": "232",
"output": "21\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 22 "
},
{
"input": "239",
"output": "21\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 29 "
},
{
"input": "629",
"output": "34\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 68 "
},
{
"input": "630",
"output": "35\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 "
},
{
"input": "631",
"output": "35\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 36 "
},
{
"input": "945",
"output": "42\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 84 "
},
{
"input": "946",
"output": "43\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 "
},
{
"input": "947",
"output": "43\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 44 "
},
{
"input": "989",
"output": "43\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 86 "
},
{
"input": "990",
"output": "44\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 "
},
{
"input": "991",
"output": "44\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 45 "
},
{
"input": "956",
"output": "43\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 53 "
},
{
"input": "981",
"output": "43\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 78 "
},
{
"input": "867",
"output": "41\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 47 "
},
{
"input": "906",
"output": "42\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 45 "
},
{
"input": "999",
"output": "44\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 53 "
},
{
"input": "100",
"output": "13\n1 2 3 4 5 6 7 8 9 10 11 12 22 "
},
{
"input": "126",
"output": "15\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 21 "
}
] | 1,614,756,082
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 49
| 62
| 0
|
n = int(input())
tmp = 0
res = []
i = 0
while n > tmp:
i += 1
tmp += i
if tmp > n:
tmp -= i
break
else:
res.append(i)
tmp = tmp - res.pop()
res.append(n-tmp)
print(len(res))
print(*res)
|
Title: Santa Claus and Candies
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Santa Claus has *n* candies, he dreams to give them as gifts to children.
What is the maximal number of children for whose he can give candies if Santa Claus want each kid should get distinct positive integer number of candies. Santa Class wants to give all *n* candies he has.
Input Specification:
The only line contains positive integer number *n* (1<=≤<=*n*<=≤<=1000) — number of candies Santa Claus has.
Output Specification:
Print to the first line integer number *k* — maximal number of kids which can get candies.
Print to the second line *k* distinct integer numbers: number of candies for each of *k* kid. The sum of *k* printed numbers should be exactly *n*.
If there are many solutions, print any of them.
Demo Input:
['5\n', '9\n', '2\n']
Demo Output:
['2\n2 3\n', '3\n3 5 1\n', '1\n2 \n']
Note:
none
|
```python
n = int(input())
tmp = 0
res = []
i = 0
while n > tmp:
i += 1
tmp += i
if tmp > n:
tmp -= i
break
else:
res.append(i)
tmp = tmp - res.pop()
res.append(n-tmp)
print(len(res))
print(*res)
```
| 3
|
|
803
|
C
|
Maximal GCD
|
PROGRAMMING
| 1,900
|
[
"constructive algorithms",
"greedy",
"math"
] | null | null |
You are given positive integer number *n*. You should create such strictly increasing sequence of *k* positive numbers *a*1,<=*a*2,<=...,<=*a**k*, that their sum is equal to *n* and greatest common divisor is maximal.
Greatest common divisor of sequence is maximum of such numbers that every element of sequence is divisible by them.
If there is no possible sequence then output -1.
|
The first line consists of two numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1010).
|
If the answer exists then output *k* numbers — resulting sequence. Otherwise output -1. If there are multiple answers, print any of them.
|
[
"6 3\n",
"8 2\n",
"5 3\n"
] |
[
"1 2 3\n",
"2 6\n",
"-1\n"
] |
none
| 0
|
[
{
"input": "6 3",
"output": "1 2 3"
},
{
"input": "8 2",
"output": "2 6"
},
{
"input": "5 3",
"output": "-1"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 2",
"output": "-1"
},
{
"input": "2 1",
"output": "2"
},
{
"input": "2 10000000000",
"output": "-1"
},
{
"input": "5 1",
"output": "5"
},
{
"input": "6 2",
"output": "2 4"
},
{
"input": "24 2",
"output": "8 16"
},
{
"input": "24 3",
"output": "4 8 12"
},
{
"input": "24 4",
"output": "2 4 6 12"
},
{
"input": "24 5",
"output": "1 2 3 4 14"
},
{
"input": "479001600 2",
"output": "159667200 319334400"
},
{
"input": "479001600 3",
"output": "79833600 159667200 239500800"
},
{
"input": "479001600 4",
"output": "47900160 95800320 143700480 191600640"
},
{
"input": "479001600 5",
"output": "31933440 63866880 95800320 127733760 159667200"
},
{
"input": "479001600 6",
"output": "22809600 45619200 68428800 91238400 114048000 136857600"
},
{
"input": "3000000021 1",
"output": "3000000021"
},
{
"input": "3000000021 2",
"output": "1000000007 2000000014"
},
{
"input": "3000000021 3",
"output": "3 6 3000000012"
},
{
"input": "3000000021 4",
"output": "3 6 9 3000000003"
},
{
"input": "3000000021 50000",
"output": "1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "3000000021 100000",
"output": "-1"
},
{
"input": "10000000000 100",
"output": "1953125 3906250 5859375 7812500 9765625 11718750 13671875 15625000 17578125 19531250 21484375 23437500 25390625 27343750 29296875 31250000 33203125 35156250 37109375 39062500 41015625 42968750 44921875 46875000 48828125 50781250 52734375 54687500 56640625 58593750 60546875 62500000 64453125 66406250 68359375 70312500 72265625 74218750 76171875 78125000 80078125 82031250 83984375 85937500 87890625 89843750 91796875 93750000 95703125 97656250 99609375 101562500 103515625 105468750 107421875 109375000 1113281..."
},
{
"input": "10000000000 2000",
"output": "4000 8000 12000 16000 20000 24000 28000 32000 36000 40000 44000 48000 52000 56000 60000 64000 68000 72000 76000 80000 84000 88000 92000 96000 100000 104000 108000 112000 116000 120000 124000 128000 132000 136000 140000 144000 148000 152000 156000 160000 164000 168000 172000 176000 180000 184000 188000 192000 196000 200000 204000 208000 212000 216000 220000 224000 228000 232000 236000 240000 244000 248000 252000 256000 260000 264000 268000 272000 276000 280000 284000 288000 292000 296000 300000 304000 30800..."
},
{
"input": "10000000000 5000",
"output": "640 1280 1920 2560 3200 3840 4480 5120 5760 6400 7040 7680 8320 8960 9600 10240 10880 11520 12160 12800 13440 14080 14720 15360 16000 16640 17280 17920 18560 19200 19840 20480 21120 21760 22400 23040 23680 24320 24960 25600 26240 26880 27520 28160 28800 29440 30080 30720 31360 32000 32640 33280 33920 34560 35200 35840 36480 37120 37760 38400 39040 39680 40320 40960 41600 42240 42880 43520 44160 44800 45440 46080 46720 47360 48000 48640 49280 49920 50560 51200 51840 52480 53120 53760 54400 55040 55680 56320..."
},
{
"input": "10000000000 100000",
"output": "1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "10000000000 100000000",
"output": "-1"
},
{
"input": "10000000000 10000000000",
"output": "-1"
},
{
"input": "10000000000 100001",
"output": "1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "1 4000000000",
"output": "-1"
},
{
"input": "4294967296 4294967296",
"output": "-1"
},
{
"input": "71227122 9603838834",
"output": "-1"
},
{
"input": "10000000000 9603838835",
"output": "-1"
},
{
"input": "5 5999999999",
"output": "-1"
},
{
"input": "2 9324327498",
"output": "-1"
},
{
"input": "9 2",
"output": "3 6"
},
{
"input": "10000000000 4294967296",
"output": "-1"
},
{
"input": "1 3500000000",
"output": "-1"
},
{
"input": "10000000000 4000000000",
"output": "-1"
},
{
"input": "2000 9324327498",
"output": "-1"
},
{
"input": "10000000000 8589934592",
"output": "-1"
},
{
"input": "5000150001 100001",
"output": "1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "10000000000 3037000500",
"output": "-1"
},
{
"input": "9400000000 9324327498",
"output": "-1"
},
{
"input": "10000000000 3307000500",
"output": "-1"
},
{
"input": "2 4000000000",
"output": "-1"
},
{
"input": "1000 4294967295",
"output": "-1"
},
{
"input": "36 3",
"output": "6 12 18"
},
{
"input": "2147483648 4294967296",
"output": "-1"
},
{
"input": "999 4294967295",
"output": "-1"
},
{
"input": "10000000000 130000",
"output": "1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "10000000000 140000",
"output": "1 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 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155..."
},
{
"input": "10000000000 6074001000",
"output": "-1"
},
{
"input": "12344321 1",
"output": "12344321"
},
{
"input": "2 2",
"output": "-1"
},
{
"input": "28 7",
"output": "1 2 3 4 5 6 7"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "1 2",
"output": "-1"
},
{
"input": "1 3",
"output": "-1"
},
{
"input": "1 4",
"output": "-1"
},
{
"input": "1 5",
"output": "-1"
},
{
"input": "1 6",
"output": "-1"
},
{
"input": "1 7",
"output": "-1"
},
{
"input": "1 8",
"output": "-1"
},
{
"input": "1 9",
"output": "-1"
},
{
"input": "1 10",
"output": "-1"
},
{
"input": "2 1",
"output": "2"
},
{
"input": "2 2",
"output": "-1"
},
{
"input": "2 3",
"output": "-1"
},
{
"input": "2 4",
"output": "-1"
},
{
"input": "2 5",
"output": "-1"
},
{
"input": "2 6",
"output": "-1"
},
{
"input": "2 7",
"output": "-1"
},
{
"input": "2 8",
"output": "-1"
},
{
"input": "2 9",
"output": "-1"
},
{
"input": "2 10",
"output": "-1"
},
{
"input": "3 1",
"output": "3"
},
{
"input": "3 2",
"output": "1 2"
},
{
"input": "3 3",
"output": "-1"
},
{
"input": "3 4",
"output": "-1"
},
{
"input": "3 5",
"output": "-1"
},
{
"input": "3 6",
"output": "-1"
},
{
"input": "3 7",
"output": "-1"
},
{
"input": "3 8",
"output": "-1"
},
{
"input": "3 9",
"output": "-1"
},
{
"input": "3 10",
"output": "-1"
},
{
"input": "4 1",
"output": "4"
},
{
"input": "4 2",
"output": "1 3"
},
{
"input": "4 3",
"output": "-1"
},
{
"input": "4 4",
"output": "-1"
},
{
"input": "4 5",
"output": "-1"
},
{
"input": "4 6",
"output": "-1"
},
{
"input": "4 7",
"output": "-1"
},
{
"input": "4 8",
"output": "-1"
},
{
"input": "4 9",
"output": "-1"
},
{
"input": "4 10",
"output": "-1"
},
{
"input": "5 1",
"output": "5"
},
{
"input": "5 2",
"output": "1 4"
},
{
"input": "5 3",
"output": "-1"
},
{
"input": "5 4",
"output": "-1"
},
{
"input": "5 5",
"output": "-1"
},
{
"input": "5 6",
"output": "-1"
},
{
"input": "5 7",
"output": "-1"
},
{
"input": "5 8",
"output": "-1"
},
{
"input": "5 9",
"output": "-1"
},
{
"input": "5 10",
"output": "-1"
},
{
"input": "6 1",
"output": "6"
},
{
"input": "6 2",
"output": "2 4"
},
{
"input": "6 3",
"output": "1 2 3"
},
{
"input": "6 4",
"output": "-1"
},
{
"input": "6 5",
"output": "-1"
},
{
"input": "6 6",
"output": "-1"
},
{
"input": "6 7",
"output": "-1"
},
{
"input": "6 8",
"output": "-1"
},
{
"input": "6 9",
"output": "-1"
},
{
"input": "6 10",
"output": "-1"
},
{
"input": "7 1",
"output": "7"
},
{
"input": "7 2",
"output": "1 6"
},
{
"input": "7 3",
"output": "1 2 4"
},
{
"input": "7 4",
"output": "-1"
},
{
"input": "7 5",
"output": "-1"
},
{
"input": "7 6",
"output": "-1"
},
{
"input": "7 7",
"output": "-1"
},
{
"input": "7 8",
"output": "-1"
},
{
"input": "7 9",
"output": "-1"
},
{
"input": "7 10",
"output": "-1"
},
{
"input": "8 1",
"output": "8"
},
{
"input": "8 2",
"output": "2 6"
},
{
"input": "8 3",
"output": "1 2 5"
},
{
"input": "8 4",
"output": "-1"
},
{
"input": "8 5",
"output": "-1"
},
{
"input": "8 6",
"output": "-1"
},
{
"input": "8 7",
"output": "-1"
},
{
"input": "8 8",
"output": "-1"
},
{
"input": "8 9",
"output": "-1"
},
{
"input": "8 10",
"output": "-1"
}
] | 1,696,315,319
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 20
| 1,000
| 1,331,200
|
import sys
def input() : return sys.stdin.readline().strip()
def getints() : return map(int,sys.stdin.readline().strip().split())
n,k = getints()
t = (k*(k+1))//2
if n < t: print(-1); exit()
a = n%t
b = n//t
while a%b != 0:
a += t
b -= 1
print(*[x*b for x in range(1,k)],k*b+a)
|
Title: Maximal GCD
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given positive integer number *n*. You should create such strictly increasing sequence of *k* positive numbers *a*1,<=*a*2,<=...,<=*a**k*, that their sum is equal to *n* and greatest common divisor is maximal.
Greatest common divisor of sequence is maximum of such numbers that every element of sequence is divisible by them.
If there is no possible sequence then output -1.
Input Specification:
The first line consists of two numbers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1010).
Output Specification:
If the answer exists then output *k* numbers — resulting sequence. Otherwise output -1. If there are multiple answers, print any of them.
Demo Input:
['6 3\n', '8 2\n', '5 3\n']
Demo Output:
['1 2 3\n', '2 6\n', '-1\n']
Note:
none
|
```python
import sys
def input() : return sys.stdin.readline().strip()
def getints() : return map(int,sys.stdin.readline().strip().split())
n,k = getints()
t = (k*(k+1))//2
if n < t: print(-1); exit()
a = n%t
b = n//t
while a%b != 0:
a += t
b -= 1
print(*[x*b for x in range(1,k)],k*b+a)
```
| 0
|
|
625
|
B
|
War of the Corporations
|
PROGRAMMING
| 1,200
|
[
"constructive algorithms",
"greedy",
"strings"
] | null | null |
A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Gogol continue their fierce competition. Crucial moment is just around the corner: Gogol is ready to release it's new tablet Lastus 3000.
This new device is equipped with specially designed artificial intelligence (AI). Employees of Pineapple did their best to postpone the release of Lastus 3000 as long as possible. Finally, they found out, that the name of the new artificial intelligence is similar to the name of the phone, that Pineapple released 200 years ago. As all rights on its name belong to Pineapple, they stand on changing the name of Gogol's artificial intelligence.
Pineapple insists, that the name of their phone occurs in the name of AI as a substring. Because the name of technology was already printed on all devices, the Gogol's director decided to replace some characters in AI name with "#". As this operation is pretty expensive, you should find the minimum number of characters to replace with "#", such that the name of AI doesn't contain the name of the phone as a substring.
Substring is a continuous subsequence of a string.
|
The first line of the input contains the name of AI designed by Gogol, its length doesn't exceed 100<=000 characters. Second line contains the name of the phone released by Pineapple 200 years ago, its length doesn't exceed 30. Both string are non-empty and consist of only small English letters.
|
Print the minimum number of characters that must be replaced with "#" in order to obtain that the name of the phone doesn't occur in the name of AI as a substring.
|
[
"intellect\ntell\n",
"google\napple\n",
"sirisiri\nsir\n"
] |
[
"1",
"0",
"2"
] |
In the first sample AI's name may be replaced with "int#llect".
In the second sample Gogol can just keep things as they are.
In the third sample one of the new possible names of AI may be "s#ris#ri".
| 750
|
[
{
"input": "intellect\ntell",
"output": "1"
},
{
"input": "google\napple",
"output": "0"
},
{
"input": "sirisiri\nsir",
"output": "2"
},
{
"input": "sirisiri\nsiri",
"output": "2"
},
{
"input": "aaaaaaa\naaaa",
"output": "1"
},
{
"input": "bbbbbb\nbb",
"output": "3"
},
{
"input": "abc\nabcabc",
"output": "0"
},
{
"input": "kek\nkekekek",
"output": "0"
},
{
"input": "aaaaa\naaa",
"output": "1"
},
{
"input": "abcdabcv\nabcd",
"output": "1"
},
{
"input": "abcabcabczabcabcabcz\ncab",
"output": "4"
},
{
"input": "aatopotopotopotaa\ntopot",
"output": "2"
},
{
"input": "abcabcabcabcabcabcabcabcabcabc\nabcabcabcabcabcabcabcabcabcabc",
"output": "1"
},
{
"input": "sosossosos\nsos",
"output": "2"
},
{
"input": "sosossosossosossosossosossosossosossosossosossosossosossosossosossosossosossosossosossosossosossosos\nsos",
"output": "20"
},
{
"input": "tatatx\ntatx",
"output": "1"
},
{
"input": "sxxsxxsxxd\nsxxsxxd",
"output": "1"
}
] | 1,578,459,823
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 55
| 1,000
| 1,433,600
|
#py3
import sys, threading, os.path
import string
import collections, heapq, math, bisect
sys.setrecursionlimit(10 ** 6)
threading.stack_size(2 ** 27)
def main():
if os.path.exists('in.rtf'):
input = open('in.rtf', 'r')
else:
input = sys.stdin
# --------------------------------INPUT---------------------------------
s1 = list(str(input.readline().rstrip('\n')))
s2 = list(str(input.readline().rstrip('\n')))
num, count = 0, 0
for i in range(len(s1)):
match = True
for j in range(len(s2)):
if i+j==len(s1) or s1[i+j]!=s2[j]:
match=False
break
if match:
count+=1
s1[i+len(s2)-1] = '#'
output = count
# -------------------------------OUTPUT----------------------------------
if os.path.exists('out.rtf'):
open('out.rtf', 'w').writelines(str(output))
else:
sys.stdout.write(str(output))
if __name__ == '__main__':
main()
# threading.Thread(target=main).start()
|
Title: War of the Corporations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Gogol continue their fierce competition. Crucial moment is just around the corner: Gogol is ready to release it's new tablet Lastus 3000.
This new device is equipped with specially designed artificial intelligence (AI). Employees of Pineapple did their best to postpone the release of Lastus 3000 as long as possible. Finally, they found out, that the name of the new artificial intelligence is similar to the name of the phone, that Pineapple released 200 years ago. As all rights on its name belong to Pineapple, they stand on changing the name of Gogol's artificial intelligence.
Pineapple insists, that the name of their phone occurs in the name of AI as a substring. Because the name of technology was already printed on all devices, the Gogol's director decided to replace some characters in AI name with "#". As this operation is pretty expensive, you should find the minimum number of characters to replace with "#", such that the name of AI doesn't contain the name of the phone as a substring.
Substring is a continuous subsequence of a string.
Input Specification:
The first line of the input contains the name of AI designed by Gogol, its length doesn't exceed 100<=000 characters. Second line contains the name of the phone released by Pineapple 200 years ago, its length doesn't exceed 30. Both string are non-empty and consist of only small English letters.
Output Specification:
Print the minimum number of characters that must be replaced with "#" in order to obtain that the name of the phone doesn't occur in the name of AI as a substring.
Demo Input:
['intellect\ntell\n', 'google\napple\n', 'sirisiri\nsir\n']
Demo Output:
['1', '0', '2']
Note:
In the first sample AI's name may be replaced with "int#llect".
In the second sample Gogol can just keep things as they are.
In the third sample one of the new possible names of AI may be "s#ris#ri".
|
```python
#py3
import sys, threading, os.path
import string
import collections, heapq, math, bisect
sys.setrecursionlimit(10 ** 6)
threading.stack_size(2 ** 27)
def main():
if os.path.exists('in.rtf'):
input = open('in.rtf', 'r')
else:
input = sys.stdin
# --------------------------------INPUT---------------------------------
s1 = list(str(input.readline().rstrip('\n')))
s2 = list(str(input.readline().rstrip('\n')))
num, count = 0, 0
for i in range(len(s1)):
match = True
for j in range(len(s2)):
if i+j==len(s1) or s1[i+j]!=s2[j]:
match=False
break
if match:
count+=1
s1[i+len(s2)-1] = '#'
output = count
# -------------------------------OUTPUT----------------------------------
if os.path.exists('out.rtf'):
open('out.rtf', 'w').writelines(str(output))
else:
sys.stdout.write(str(output))
if __name__ == '__main__':
main()
# threading.Thread(target=main).start()
```
| 0
|
|
710
|
A
|
King Moves
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
The only king stands on the standard chess board. You are given his position in format "cd", where *c* is the column from 'a' to 'h' and *d* is the row from '1' to '8'. Find the number of moves permitted for the king.
Check the king's moves here [https://en.wikipedia.org/wiki/King_(chess)](https://en.wikipedia.org/wiki/King_(chess)).
|
The only line contains the king's position in the format "cd", where 'c' is the column from 'a' to 'h' and 'd' is the row from '1' to '8'.
|
Print the only integer *x* — the number of moves permitted for the king.
|
[
"e4\n"
] |
[
"8\n"
] |
none
| 0
|
[
{
"input": "e4",
"output": "8"
},
{
"input": "a1",
"output": "3"
},
{
"input": "h8",
"output": "3"
},
{
"input": "a4",
"output": "5"
},
{
"input": "g7",
"output": "8"
},
{
"input": "e1",
"output": "5"
},
{
"input": "b2",
"output": "8"
},
{
"input": "c7",
"output": "8"
},
{
"input": "h6",
"output": "5"
},
{
"input": "c8",
"output": "5"
},
{
"input": "h2",
"output": "5"
},
{
"input": "h5",
"output": "5"
},
{
"input": "a8",
"output": "3"
},
{
"input": "f8",
"output": "5"
},
{
"input": "h1",
"output": "3"
},
{
"input": "f2",
"output": "8"
},
{
"input": "e8",
"output": "5"
},
{
"input": "h3",
"output": "5"
},
{
"input": "b8",
"output": "5"
},
{
"input": "g8",
"output": "5"
},
{
"input": "d8",
"output": "5"
},
{
"input": "h4",
"output": "5"
},
{
"input": "b1",
"output": "5"
},
{
"input": "a2",
"output": "5"
}
] | 1,472,391,157
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 1
| 77
| 22,835,200
|
s = input()
c = s[0]
d = s[1]
f1 = 'a' < c < 'h'
f2 = '1' < d < '8'
if f1 and f2: ans = 8
elif f1 or f2: ans = 6
else: ans = 4
print(ans)
|
Title: King Moves
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The only king stands on the standard chess board. You are given his position in format "cd", where *c* is the column from 'a' to 'h' and *d* is the row from '1' to '8'. Find the number of moves permitted for the king.
Check the king's moves here [https://en.wikipedia.org/wiki/King_(chess)](https://en.wikipedia.org/wiki/King_(chess)).
Input Specification:
The only line contains the king's position in the format "cd", where 'c' is the column from 'a' to 'h' and 'd' is the row from '1' to '8'.
Output Specification:
Print the only integer *x* — the number of moves permitted for the king.
Demo Input:
['e4\n']
Demo Output:
['8\n']
Note:
none
|
```python
s = input()
c = s[0]
d = s[1]
f1 = 'a' < c < 'h'
f2 = '1' < d < '8'
if f1 and f2: ans = 8
elif f1 or f2: ans = 6
else: ans = 4
print(ans)
```
| 0
|
|
313
|
A
|
Ilya and Bank Account
|
PROGRAMMING
| 900
|
[
"implementation",
"number theory"
] | null | null |
Ilya is a very clever lion, he lives in an unusual city ZooVille. In this city all the animals have their rights and obligations. Moreover, they even have their own bank accounts. The state of a bank account is an integer. The state of a bank account can be a negative number. This means that the owner of the account owes the bank money.
Ilya the Lion has recently had a birthday, so he got a lot of gifts. One of them (the gift of the main ZooVille bank) is the opportunity to delete the last digit or the digit before last from the state of his bank account no more than once. For example, if the state of Ilya's bank account is -123, then Ilya can delete the last digit and get his account balance equal to -12, also he can remove its digit before last and get the account balance equal to -13. Of course, Ilya is permitted not to use the opportunity to delete a digit from the balance.
Ilya is not very good at math, and that's why he asks you to help him maximize his bank account. Find the maximum state of the bank account that can be obtained using the bank's gift.
|
The single line contains integer *n* (10<=≤<=|*n*|<=≤<=109) — the state of Ilya's bank account.
|
In a single line print an integer — the maximum state of the bank account that Ilya can get.
|
[
"2230\n",
"-10\n",
"-100003\n"
] |
[
"2230\n",
"0\n",
"-10000\n"
] |
In the first test sample Ilya doesn't profit from using the present.
In the second test sample you can delete digit 1 and get the state of the account equal to 0.
| 500
|
[
{
"input": "2230",
"output": "2230"
},
{
"input": "-10",
"output": "0"
},
{
"input": "-100003",
"output": "-10000"
},
{
"input": "544883178",
"output": "544883178"
},
{
"input": "-847251738",
"output": "-84725173"
},
{
"input": "423654797",
"output": "423654797"
},
{
"input": "-623563697",
"output": "-62356367"
},
{
"input": "645894116",
"output": "645894116"
},
{
"input": "-384381709",
"output": "-38438170"
},
{
"input": "437587210",
"output": "437587210"
},
{
"input": "-297534606",
"output": "-29753460"
},
{
"input": "891773002",
"output": "891773002"
},
{
"input": "-56712976",
"output": "-5671296"
},
{
"input": "963662765",
"output": "963662765"
},
{
"input": "-272656295",
"output": "-27265625"
},
{
"input": "383441522",
"output": "383441522"
},
{
"input": "-477665112",
"output": "-47766511"
},
{
"input": "791725034",
"output": "791725034"
},
{
"input": "-812168727",
"output": "-81216872"
},
{
"input": "528894922",
"output": "528894922"
},
{
"input": "-479977172",
"output": "-47997712"
},
{
"input": "568044564",
"output": "568044564"
},
{
"input": "-392784794",
"output": "-39278474"
},
{
"input": "925596634",
"output": "925596634"
},
{
"input": "-836078769",
"output": "-83607876"
},
{
"input": "71036059",
"output": "71036059"
},
{
"input": "-337396162",
"output": "-33739612"
},
{
"input": "87129297",
"output": "87129297"
},
{
"input": "-648171877",
"output": "-64817187"
},
{
"input": "20218400",
"output": "20218400"
},
{
"input": "10",
"output": "10"
},
{
"input": "1000000000",
"output": "1000000000"
},
{
"input": "-1000000000",
"output": "-100000000"
},
{
"input": "-102",
"output": "-10"
},
{
"input": "-120",
"output": "-10"
},
{
"input": "-20",
"output": "0"
},
{
"input": "-15",
"output": "-1"
},
{
"input": "-5575533",
"output": "-557553"
},
{
"input": "-50",
"output": "0"
},
{
"input": "-31",
"output": "-1"
},
{
"input": "-55",
"output": "-5"
},
{
"input": "-81",
"output": "-1"
},
{
"input": "-99",
"output": "-9"
},
{
"input": "-23",
"output": "-2"
},
{
"input": "-41",
"output": "-1"
},
{
"input": "-24",
"output": "-2"
},
{
"input": "46",
"output": "46"
}
] | 1,694,998,461
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 32
| 92
| 0
|
def Ilya_Bank_Account(n):
if n > 0 :
return n
new_n_1 = n // 10 + 1
new_n_2 = (n // 100) * 10 + n % 10
return max(new_n_1 , new_n_2)
if __name__ == "__main__" :
number_bank_account = int(input())
print(Ilya_Bank_Account(number_bank_account))
|
Title: Ilya and Bank Account
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ilya is a very clever lion, he lives in an unusual city ZooVille. In this city all the animals have their rights and obligations. Moreover, they even have their own bank accounts. The state of a bank account is an integer. The state of a bank account can be a negative number. This means that the owner of the account owes the bank money.
Ilya the Lion has recently had a birthday, so he got a lot of gifts. One of them (the gift of the main ZooVille bank) is the opportunity to delete the last digit or the digit before last from the state of his bank account no more than once. For example, if the state of Ilya's bank account is -123, then Ilya can delete the last digit and get his account balance equal to -12, also he can remove its digit before last and get the account balance equal to -13. Of course, Ilya is permitted not to use the opportunity to delete a digit from the balance.
Ilya is not very good at math, and that's why he asks you to help him maximize his bank account. Find the maximum state of the bank account that can be obtained using the bank's gift.
Input Specification:
The single line contains integer *n* (10<=≤<=|*n*|<=≤<=109) — the state of Ilya's bank account.
Output Specification:
In a single line print an integer — the maximum state of the bank account that Ilya can get.
Demo Input:
['2230\n', '-10\n', '-100003\n']
Demo Output:
['2230\n', '0\n', '-10000\n']
Note:
In the first test sample Ilya doesn't profit from using the present.
In the second test sample you can delete digit 1 and get the state of the account equal to 0.
|
```python
def Ilya_Bank_Account(n):
if n > 0 :
return n
new_n_1 = n // 10 + 1
new_n_2 = (n // 100) * 10 + n % 10
return max(new_n_1 , new_n_2)
if __name__ == "__main__" :
number_bank_account = int(input())
print(Ilya_Bank_Account(number_bank_account))
```
| 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"
}
] | 1,663,550,393
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 81
| 124
| 0
|
n = int(input())
result = list(map(sum, zip(*[[int(i) for i in input().split()] for j in range(n)])))
if result == [0, 0, 0]:
print("YES")
else:
print("NO")
|
Title: Young Physicist
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" — thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100), then follow *n* lines containing three integers each: the *x**i* coordinate, the *y**i* coordinate and the *z**i* coordinate of the force vector, applied to the body (<=-<=100<=≤<=*x**i*,<=*y**i*,<=*z**i*<=≤<=100).
Output Specification:
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Demo Input:
['3\n4 1 7\n-2 4 -1\n1 -5 -3\n', '3\n3 -1 7\n-5 2 -4\n2 -1 -3\n']
Demo Output:
['NO', 'YES']
Note:
none
|
```python
n = int(input())
result = list(map(sum, zip(*[[int(i) for i in input().split()] for j in range(n)])))
if result == [0, 0, 0]:
print("YES")
else:
print("NO")
```
| 3.969
|
389
|
A
|
Fox and Number Game
|
PROGRAMMING
| 1,000
|
[
"greedy",
"math"
] | null | null |
Fox Ciel is playing a game with numbers now.
Ciel has *n* positive integers: *x*1, *x*2, ..., *x**n*. She can do the following operation as many times as needed: select two different indexes *i* and *j* such that *x**i* > *x**j* hold, and then apply assignment *x**i* = *x**i* - *x**j*. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum.
|
The first line contains an integer *n* (2<=≤<=*n*<=≤<=100). Then the second line contains *n* integers: *x*1, *x*2, ..., *x**n* (1<=≤<=*x**i*<=≤<=100).
|
Output a single integer — the required minimal sum.
|
[
"2\n1 2\n",
"3\n2 4 6\n",
"2\n12 18\n",
"5\n45 12 27 30 18\n"
] |
[
"2\n",
"6\n",
"12\n",
"15\n"
] |
In the first example the optimal way is to do the assignment: *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
In the second example the optimal sequence of operations is: *x*<sub class="lower-index">3</sub> = *x*<sub class="lower-index">3</sub> - *x*<sub class="lower-index">2</sub>, *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
| 500
|
[
{
"input": "2\n1 2",
"output": "2"
},
{
"input": "3\n2 4 6",
"output": "6"
},
{
"input": "2\n12 18",
"output": "12"
},
{
"input": "5\n45 12 27 30 18",
"output": "15"
},
{
"input": "2\n1 1",
"output": "2"
},
{
"input": "2\n100 100",
"output": "200"
},
{
"input": "2\n87 58",
"output": "58"
},
{
"input": "39\n52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52 52",
"output": "2028"
},
{
"input": "59\n96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96 96",
"output": "5664"
},
{
"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": "10000"
},
{
"input": "100\n70 70 77 42 98 84 56 91 35 21 7 70 77 77 56 63 14 84 56 14 77 77 63 70 14 7 28 91 63 49 21 84 98 56 77 98 98 84 98 14 7 56 49 28 91 98 7 56 14 91 14 98 49 28 98 14 98 98 14 70 35 28 63 28 49 63 63 56 91 98 35 42 42 35 63 35 42 14 63 21 77 56 42 77 35 91 56 21 28 84 56 70 70 91 98 70 84 63 21 98",
"output": "700"
},
{
"input": "39\n63 21 21 42 21 63 21 84 42 21 84 63 42 63 84 84 84 42 42 84 21 63 42 63 42 42 63 42 42 63 84 42 21 84 21 63 42 21 42",
"output": "819"
},
{
"input": "59\n70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70 70",
"output": "4130"
},
{
"input": "87\n44 88 88 88 88 66 88 22 22 88 88 44 88 22 22 22 88 88 88 88 66 22 88 88 88 88 66 66 44 88 44 44 66 22 88 88 22 44 66 44 88 66 66 22 22 22 22 88 22 22 44 66 88 22 22 88 66 66 88 22 66 88 66 88 66 44 88 44 22 44 44 22 44 88 44 44 44 44 22 88 88 88 66 66 88 44 22",
"output": "1914"
},
{
"input": "15\n63 63 63 63 63 63 63 63 63 63 63 63 63 63 63",
"output": "945"
},
{
"input": "39\n63 77 21 14 14 35 21 21 70 42 21 70 28 77 28 77 7 42 63 7 98 49 98 84 35 70 70 91 14 42 98 7 42 7 98 42 56 35 91",
"output": "273"
},
{
"input": "18\n18 18 18 36 36 36 54 72 54 36 72 54 36 36 36 36 18 36",
"output": "324"
},
{
"input": "46\n71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71 71",
"output": "3266"
},
{
"input": "70\n66 11 66 11 44 11 44 99 55 22 88 11 11 22 55 44 22 77 44 77 77 22 44 55 88 11 99 99 88 22 77 77 66 11 11 66 99 55 55 44 66 44 77 44 44 55 33 55 44 88 77 77 22 66 33 44 11 22 55 44 22 66 77 33 33 44 44 44 22 33",
"output": "770"
},
{
"input": "10\n60 12 96 48 60 24 60 36 60 60",
"output": "120"
},
{
"input": "20\n51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51 51",
"output": "1020"
},
{
"input": "50\n58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58 58",
"output": "2900"
},
{
"input": "98\n70 60 100 30 70 20 30 50 50 30 90 40 30 40 60 80 60 60 80 50 10 80 20 10 20 10 50 70 30 80 30 50 60 90 90 100 60 30 90 20 30 60 90 80 60 60 10 90 10 50 40 40 80 90 100 40 70 40 30 50 60 50 60 30 40 20 90 60 20 20 20 70 60 70 50 100 90 50 20 40 80 60 10 60 50 40 40 10 50 10 40 10 80 100 100 90 10 90",
"output": "980"
},
{
"input": "100\n82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82 82",
"output": "8200"
},
{
"input": "100\n11 87 77 93 3 54 21 93 9 71 37 23 69 85 74 3 48 99 51 31 56 19 21 96 39 6 4 4 29 69 100 42 1 22 81 53 48 49 81 61 10 7 40 61 7 71 51 59 79 44 50 35 95 80 83 8 98 40 18 94 84 49 52 74 66 69 39 37 100 44 38 62 2 80 46 31 35 53 5 60 21 49 63 55 20 53 80 53 66 34 23 92 77 50 86 63 65 24 12 70",
"output": "100"
},
{
"input": "2\n100 1",
"output": "2"
},
{
"input": "2\n18 30",
"output": "12"
},
{
"input": "2\n3 5",
"output": "2"
},
{
"input": "2\n1 10",
"output": "2"
},
{
"input": "2\n8 5",
"output": "2"
},
{
"input": "5\n2 3 5 8 18",
"output": "5"
},
{
"input": "5\n2 4 1 6 8",
"output": "5"
},
{
"input": "3\n12 10 5",
"output": "3"
},
{
"input": "3\n6 10 15",
"output": "3"
}
] | 1,568,131,601
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 10
| 109
| 0
|
n=int(input())
a=list(map(int,input().split()))
while(1):
a.sort()
for i in range(n-1):
if(a[len(a)-1-i]>a[len(a)-2-i]):
f=0
a[len(a)-1-i]=a[len(a)-1-i]-a[len(a)-2-i]
else:
f=1
if(f==1):
break
print(sum(a))
|
Title: Fox and Number Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel is playing a game with numbers now.
Ciel has *n* positive integers: *x*1, *x*2, ..., *x**n*. She can do the following operation as many times as needed: select two different indexes *i* and *j* such that *x**i* > *x**j* hold, and then apply assignment *x**i* = *x**i* - *x**j*. The goal is to make the sum of all numbers as small as possible.
Please help Ciel to find this minimal sum.
Input Specification:
The first line contains an integer *n* (2<=≤<=*n*<=≤<=100). Then the second line contains *n* integers: *x*1, *x*2, ..., *x**n* (1<=≤<=*x**i*<=≤<=100).
Output Specification:
Output a single integer — the required minimal sum.
Demo Input:
['2\n1 2\n', '3\n2 4 6\n', '2\n12 18\n', '5\n45 12 27 30 18\n']
Demo Output:
['2\n', '6\n', '12\n', '15\n']
Note:
In the first example the optimal way is to do the assignment: *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
In the second example the optimal sequence of operations is: *x*<sub class="lower-index">3</sub> = *x*<sub class="lower-index">3</sub> - *x*<sub class="lower-index">2</sub>, *x*<sub class="lower-index">2</sub> = *x*<sub class="lower-index">2</sub> - *x*<sub class="lower-index">1</sub>.
|
```python
n=int(input())
a=list(map(int,input().split()))
while(1):
a.sort()
for i in range(n-1):
if(a[len(a)-1-i]>a[len(a)-2-i]):
f=0
a[len(a)-1-i]=a[len(a)-1-i]-a[len(a)-2-i]
else:
f=1
if(f==1):
break
print(sum(a))
```
| 0
|
|
116
|
A
|
Tram
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
|
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
|
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
|
[
"4\n0 3\n2 5\n4 2\n4 0\n"
] |
[
"6\n"
] |
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.
| 500
|
[
{
"input": "4\n0 3\n2 5\n4 2\n4 0",
"output": "6"
},
{
"input": "5\n0 4\n4 6\n6 5\n5 4\n4 0",
"output": "6"
},
{
"input": "10\n0 5\n1 7\n10 8\n5 3\n0 5\n3 3\n8 8\n0 6\n10 1\n9 0",
"output": "18"
},
{
"input": "3\n0 1\n1 1\n1 0",
"output": "1"
},
{
"input": "4\n0 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "3\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "3\n0 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "5\n0 73\n73 189\n189 766\n766 0\n0 0",
"output": "766"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 1\n1 0",
"output": "1"
},
{
"input": "5\n0 917\n917 923\n904 992\n1000 0\n11 0",
"output": "1011"
},
{
"input": "5\n0 1\n1 2\n2 1\n1 2\n2 0",
"output": "2"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "20\n0 7\n2 1\n2 2\n5 7\n2 6\n6 10\n2 4\n0 4\n7 4\n8 0\n10 6\n2 1\n6 1\n1 7\n0 3\n8 7\n6 3\n6 3\n1 1\n3 0",
"output": "22"
},
{
"input": "5\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "10\n0 592\n258 598\n389 203\n249 836\n196 635\n478 482\n994 987\n1000 0\n769 0\n0 0",
"output": "1776"
},
{
"input": "10\n0 1\n1 0\n0 0\n0 0\n0 0\n0 1\n1 1\n0 1\n1 0\n1 0",
"output": "2"
},
{
"input": "10\n0 926\n926 938\n938 931\n931 964\n937 989\n983 936\n908 949\n997 932\n945 988\n988 0",
"output": "1016"
},
{
"input": "10\n0 1\n1 2\n1 2\n2 2\n2 2\n2 2\n1 1\n1 1\n2 1\n2 0",
"output": "3"
},
{
"input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "10\n0 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 1000\n1000 0",
"output": "1000"
},
{
"input": "50\n0 332\n332 268\n268 56\n56 711\n420 180\n160 834\n149 341\n373 777\n763 93\n994 407\n86 803\n700 132\n471 608\n429 467\n75 5\n638 305\n405 853\n316 478\n643 163\n18 131\n648 241\n241 766\n316 847\n640 380\n923 759\n789 41\n125 421\n421 9\n9 388\n388 829\n408 108\n462 856\n816 411\n518 688\n290 7\n405 912\n397 772\n396 652\n394 146\n27 648\n462 617\n514 433\n780 35\n710 705\n460 390\n194 508\n643 56\n172 469\n1000 0\n194 0",
"output": "2071"
},
{
"input": "50\n0 0\n0 1\n1 1\n0 1\n0 0\n1 0\n0 0\n1 0\n0 0\n0 0\n0 0\n0 0\n0 1\n0 0\n0 0\n0 1\n1 0\n0 1\n0 0\n1 1\n1 0\n0 1\n0 0\n1 1\n0 1\n1 0\n1 1\n1 0\n0 0\n1 1\n1 0\n0 1\n0 0\n0 1\n1 1\n1 1\n1 1\n1 0\n1 1\n1 0\n0 1\n1 0\n0 0\n0 1\n1 1\n1 1\n0 1\n0 0\n1 0\n1 0",
"output": "3"
},
{
"input": "50\n0 926\n926 971\n915 980\n920 965\n954 944\n928 952\n955 980\n916 980\n906 935\n944 913\n905 923\n912 922\n965 934\n912 900\n946 930\n931 983\n979 905\n925 969\n924 926\n910 914\n921 977\n934 979\n962 986\n942 909\n976 903\n982 982\n991 941\n954 929\n902 980\n947 983\n919 924\n917 943\n916 905\n907 913\n964 977\n984 904\n905 999\n950 970\n986 906\n993 970\n960 994\n963 983\n918 986\n980 900\n931 986\n993 997\n941 909\n907 909\n1000 0\n278 0",
"output": "1329"
},
{
"input": "2\n0 863\n863 0",
"output": "863"
},
{
"input": "50\n0 1\n1 2\n2 2\n1 1\n1 1\n1 2\n1 2\n1 1\n1 2\n1 1\n1 1\n1 2\n1 2\n1 1\n2 1\n2 2\n1 2\n2 2\n1 2\n2 1\n2 1\n2 2\n2 1\n1 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n2 2\n2 1\n1 2\n2 2\n1 2\n1 1\n1 1\n2 1\n2 1\n2 2\n2 1\n2 1\n1 2\n1 2\n1 2\n1 2\n2 0\n2 0\n2 0\n0 0",
"output": "8"
},
{
"input": "50\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "100\n0 1\n0 0\n0 0\n1 0\n0 0\n0 1\n0 1\n1 1\n0 0\n0 0\n1 1\n0 0\n1 1\n0 1\n1 1\n0 1\n1 1\n1 0\n1 0\n0 0\n1 0\n0 1\n1 0\n0 0\n0 0\n1 1\n1 1\n0 1\n0 0\n1 0\n1 1\n0 1\n1 0\n1 1\n0 1\n1 1\n1 0\n0 0\n0 0\n0 1\n0 0\n0 1\n1 1\n0 0\n1 1\n1 1\n0 0\n0 1\n1 0\n0 1\n0 0\n0 1\n0 1\n1 1\n1 1\n1 1\n0 0\n0 0\n1 1\n0 1\n0 1\n1 0\n0 0\n0 0\n1 1\n0 1\n0 1\n1 1\n1 1\n0 1\n1 1\n1 1\n0 0\n1 0\n0 1\n0 0\n0 0\n1 1\n1 1\n1 1\n1 1\n0 1\n1 0\n1 0\n1 0\n1 0\n1 0\n0 0\n1 0\n1 0\n0 0\n1 0\n0 0\n0 1\n1 0\n0 1\n1 0\n1 0\n1 0\n1 0",
"output": "11"
},
{
"input": "100\n0 2\n1 2\n2 1\n1 2\n1 2\n2 1\n2 2\n1 1\n1 1\n2 1\n1 2\n2 1\n1 2\n2 2\n2 2\n2 2\n1 2\n2 2\n2 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 2\n1 1\n1 1\n1 1\n1 1\n2 2\n1 2\n2 1\n1 1\n2 2\n1 1\n2 1\n1 1\n2 2\n2 1\n1 2\n1 1\n1 2\n2 1\n2 2\n1 1\n2 1\n1 1\n2 1\n1 1\n1 2\n2 2\n2 2\n1 1\n2 2\n1 2\n2 1\n2 1\n1 1\n1 1\n1 2\n1 2\n1 1\n1 1\n2 1\n1 2\n1 2\n2 1\n2 2\n2 2\n2 2\n2 1\n2 2\n1 1\n1 2\n1 2\n1 1\n2 2\n2 2\n1 1\n2 1\n1 1\n1 2\n1 2\n1 2\n1 1\n1 1\n2 2\n1 2\n2 1\n2 1\n2 1\n1 2\n1 2\n1 1\n2 2\n1 2\n2 0\n2 0\n2 0\n1 0",
"output": "7"
},
{
"input": "100\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 900\n900 0",
"output": "900"
},
{
"input": "2\n0 1\n1 0",
"output": "1"
},
{
"input": "2\n0 0\n0 0",
"output": "0"
},
{
"input": "2\n0 1000\n1000 0",
"output": "1000"
},
{
"input": "3\n0 802\n175 188\n815 0",
"output": "815"
},
{
"input": "3\n0 910\n910 976\n976 0",
"output": "976"
},
{
"input": "3\n0 2\n2 1\n1 0",
"output": "2"
}
] | 1,692,874,248
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 71
| 122
| 0
|
t=int(input())
max1=0
total=0
while(t>0):
x,y=map(int,input().split())
total-=x
total+=y
if(max1<total):
max1=total
t-=1
print(max1)
|
Title: Tram
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Linear Kingdom has exactly one tram line. It has *n* stops, numbered from 1 to *n* in the order of tram's movement. At the *i*-th stop *a**i* passengers exit the tram, while *b**i* passengers enter it. The tram is empty before it arrives at the first stop. Also, when the tram arrives at the last stop, all passengers exit so that it becomes empty.
Your task is to calculate the tram's minimum capacity such that the number of people inside the tram at any time never exceeds this capacity. Note that at each stop all exiting passengers exit before any entering passenger enters the tram.
Input Specification:
The first line contains a single number *n* (2<=≤<=*n*<=≤<=1000) — the number of the tram's stops.
Then *n* lines follow, each contains two integers *a**i* and *b**i* (0<=≤<=*a**i*,<=*b**i*<=≤<=1000) — the number of passengers that exits the tram at the *i*-th stop, and the number of passengers that enter the tram at the *i*-th stop. The stops are given from the first to the last stop in the order of tram's movement.
- The number of people who exit at a given stop does not exceed the total number of people in the tram immediately before it arrives at the stop. More formally, . This particularly means that *a*1<==<=0. - At the last stop, all the passengers exit the tram and it becomes empty. More formally, . - No passenger will enter the train at the last stop. That is, *b**n*<==<=0.
Output Specification:
Print a single integer denoting the minimum possible capacity of the tram (0 is allowed).
Demo Input:
['4\n0 3\n2 5\n4 2\n4 0\n']
Demo Output:
['6\n']
Note:
For the first example, a capacity of 6 is sufficient:
- At the first stop, the number of passengers inside the tram before arriving is 0. Then, 3 passengers enter the tram, and the number of passengers inside the tram becomes 3. - At the second stop, 2 passengers exit the tram (1 passenger remains inside). Then, 5 passengers enter the tram. There are 6 passengers inside the tram now. - At the third stop, 4 passengers exit the tram (2 passengers remain inside). Then, 2 passengers enter the tram. There are 4 passengers inside the tram now. - Finally, all the remaining passengers inside the tram exit the tram at the last stop. There are no passenger inside the tram now, which is in line with the constraints.
Since the number of passengers inside the tram never exceeds 6, a capacity of 6 is sufficient. Furthermore it is not possible for the tram to have a capacity less than 6. Hence, 6 is the correct answer.
|
```python
t=int(input())
max1=0
total=0
while(t>0):
x,y=map(int,input().split())
total-=x
total+=y
if(max1<total):
max1=total
t-=1
print(max1)
```
| 3
|
|
398
|
A
|
Cards
|
PROGRAMMING
| 0
|
[
"constructive algorithms",
"implementation"
] | null | null |
User ainta loves to play with cards. He has *a* cards containing letter "o" and *b* cards containing letter "x". He arranges the cards in a row, and calculates the score of the deck by the formula below.
1. At first, the score is 0. 1. For each block of contiguous "o"s with length *x* the score increases by *x*2. 1. For each block of contiguous "x"s with length *y* the score decreases by *y*2.
For example, if *a*<==<=6,<=*b*<==<=3 and ainta have arranged the cards in the order, that is described by string "ooxoooxxo", the score of the deck equals 22<=-<=12<=+<=32<=-<=22<=+<=12<==<=9. That is because the deck has 5 blocks in total: "oo", "x", "ooo", "xx", "o".
User ainta likes big numbers, so he wants to maximize the score with the given cards. Help ainta make the score as big as possible. Note, that he has to arrange all his cards.
|
The first line contains two space-separated integers *a* and *b* (0<=≤<=*a*,<=*b*<=≤<=105; *a*<=+<=*b*<=≥<=1) — the number of "o" cards and the number of "x" cards.
|
In the first line print a single integer *v* — the maximum score that ainta can obtain.
In the second line print *a*<=+<=*b* characters describing the deck. If the *k*-th card of the deck contains "o", the *k*-th character must be "o". If the *k*-th card of the deck contains "x", the *k*-th character must be "x". The number of "o" characters must be equal to *a*, and the number of "x " characters must be equal to *b*. If there are many ways to maximize *v*, print any.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
|
[
"2 3\n",
"4 0\n",
"0 4\n"
] |
[
"-1\nxoxox\n",
"16\noooo",
"-16\nxxxx"
] |
none
| 500
|
[
{
"input": "2 3",
"output": "-1\nxoxox"
},
{
"input": "4 0",
"output": "16\noooo"
},
{
"input": "0 4",
"output": "-16\nxxxx"
},
{
"input": "8 6",
"output": "46\nxxxooooooooxxx"
},
{
"input": "28691 28312",
"output": "809737773\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "1 1",
"output": "0\nox"
},
{
"input": "46000 39000",
"output": "2092541530\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "1234 5678",
"output": "976892\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "19310 18",
"output": "372875938\nxxxxxxxxxooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "38 5",
"output": "1431\nxxxooooooooooooooooooooooooooooooooooooooxx"
},
{
"input": "2 122",
"output": "-4960\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
},
{
"input": "9966 12376",
"output": "95873950\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "4 2",
"output": "14\nxoooox"
},
{
"input": "0 26501",
"output": "-702303001\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx..."
},
{
"input": "500 500",
"output": "220582\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "98751 29491",
"output": "9725946462\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "1 18468",
"output": "-170533511\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx..."
},
{
"input": "75232 0",
"output": "5659853824\nooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "83093 94343",
"output": "6827912284\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "86224 91008",
"output": "7359384778\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "92608 85844",
"output": "8502762302\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "94989 92701",
"output": "8942524504\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "83195 80484",
"output": "6856118621\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "4 9",
"output": "-13\nxxoxxoxxoxxox"
},
{
"input": "8 10",
"output": "16\nxxxxoooooooxxxoxxx"
},
{
"input": "223 874",
"output": "15479\nxxxxxxxxxxxxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxx..."
},
{
"input": "206 209",
"output": "34847\nxxxxxxxxxxxxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxx"
},
{
"input": "493 442",
"output": "217415\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "18931 31308",
"output": "346300009\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "21944 37439",
"output": "465971835\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "29626 16323",
"output": "869876049\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "78912 100000",
"output": "6148027918\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "5 17",
"output": "-44\nxxxoxxxoxxxoxxxoxxxoxx"
},
{
"input": "2 60570",
"output": "-1222908298\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx..."
},
{
"input": "23 89946",
"output": "-337095103\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx..."
},
{
"input": "7 18030",
"output": "-40635107\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx..."
},
{
"input": "27813 15",
"output": "773562856\nxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "29648 34",
"output": "879003326\nxxxxxxxxxxxxxxxxxooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "25661 14735",
"output": "651917342\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "2596 14758",
"output": "4665910\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "21478 14813",
"output": "455254951\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "1454 26690",
"output": "-444804\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "31161 18112",
"output": "962094403\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "1698 32709",
"output": "-523357\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "749 9800",
"output": "-113239\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoxxxxxxxxxxxxxxxxxxxxxxxxxxxxxo..."
},
{
"input": "79123 95821",
"output": "6184587446\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "79979 92032",
"output": "6323396597\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "99979 12032",
"output": "9985440319\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxooooooooooooooooooooooooooooooooooooooooooooooooooooo..."
},
{
"input": "1 2",
"output": "-1\nxox"
},
{
"input": "2 1",
"output": "3\noox"
},
{
"input": "1 1",
"output": "0\nox"
},
{
"input": "1 0",
"output": "1\no"
},
{
"input": "0 1",
"output": "-1\nx"
},
{
"input": "2 2",
"output": "2\nxoox"
},
{
"input": "4 1",
"output": "15\noooox"
},
{
"input": "4 2",
"output": "14\nxoooox"
},
{
"input": "4 3",
"output": "11\nxxoooox"
},
{
"input": "4 4",
"output": "8\nxxooooxx"
},
{
"input": "4 5",
"output": "3\nxxxooooxx"
},
{
"input": "4 6",
"output": "-2\nxxxooooxxx"
},
{
"input": "4 7",
"output": "-7\nxxxoooxxoxx"
},
{
"input": "99999 99997",
"output": "9910809718\nxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxoooooooooooooooooooooooooooooooooooooooooooooooooooo..."
}
] | 1,554,191,270
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 0
| 108
| 0
|
cards=input().split(' ')
if int(cards[0])<int(cards[1]):
print(pow(int(cards[1])%int(cards[0]),2)+pow(int(cards[1])//int(cards[0]),2)-int(cards[0]))
print('x'*(int(cards[1])%int(cards[0]))+('o'+'x'*(int(cards[1])//int(cards[0])))*int(cards[0]))
else:
print(pow(int(cards[0])-int(cards[1])+1,2)-1)
print('xo'*int(cards[1])+'o'*(int(cards[0])-int(cards[1])))
|
Title: Cards
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
User ainta loves to play with cards. He has *a* cards containing letter "o" and *b* cards containing letter "x". He arranges the cards in a row, and calculates the score of the deck by the formula below.
1. At first, the score is 0. 1. For each block of contiguous "o"s with length *x* the score increases by *x*2. 1. For each block of contiguous "x"s with length *y* the score decreases by *y*2.
For example, if *a*<==<=6,<=*b*<==<=3 and ainta have arranged the cards in the order, that is described by string "ooxoooxxo", the score of the deck equals 22<=-<=12<=+<=32<=-<=22<=+<=12<==<=9. That is because the deck has 5 blocks in total: "oo", "x", "ooo", "xx", "o".
User ainta likes big numbers, so he wants to maximize the score with the given cards. Help ainta make the score as big as possible. Note, that he has to arrange all his cards.
Input Specification:
The first line contains two space-separated integers *a* and *b* (0<=≤<=*a*,<=*b*<=≤<=105; *a*<=+<=*b*<=≥<=1) — the number of "o" cards and the number of "x" cards.
Output Specification:
In the first line print a single integer *v* — the maximum score that ainta can obtain.
In the second line print *a*<=+<=*b* characters describing the deck. If the *k*-th card of the deck contains "o", the *k*-th character must be "o". If the *k*-th card of the deck contains "x", the *k*-th character must be "x". The number of "o" characters must be equal to *a*, and the number of "x " characters must be equal to *b*. If there are many ways to maximize *v*, print any.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Demo Input:
['2 3\n', '4 0\n', '0 4\n']
Demo Output:
['-1\nxoxox\n', '16\noooo', '-16\nxxxx']
Note:
none
|
```python
cards=input().split(' ')
if int(cards[0])<int(cards[1]):
print(pow(int(cards[1])%int(cards[0]),2)+pow(int(cards[1])//int(cards[0]),2)-int(cards[0]))
print('x'*(int(cards[1])%int(cards[0]))+('o'+'x'*(int(cards[1])//int(cards[0])))*int(cards[0]))
else:
print(pow(int(cards[0])-int(cards[1])+1,2)-1)
print('xo'*int(cards[1])+'o'*(int(cards[0])-int(cards[1])))
```
| 0
|
|
9
|
A
|
Die Roll
|
PROGRAMMING
| 800
|
[
"math",
"probabilities"
] |
A. Die Roll
|
1
|
64
|
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
|
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
|
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
|
[
"4 2\n"
] |
[
"1/2\n"
] |
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
| 0
|
[
{
"input": "4 2",
"output": "1/2"
},
{
"input": "1 1",
"output": "1/1"
},
{
"input": "1 2",
"output": "5/6"
},
{
"input": "1 3",
"output": "2/3"
},
{
"input": "1 4",
"output": "1/2"
},
{
"input": "1 5",
"output": "1/3"
},
{
"input": "1 6",
"output": "1/6"
},
{
"input": "2 1",
"output": "5/6"
},
{
"input": "2 2",
"output": "5/6"
},
{
"input": "2 3",
"output": "2/3"
},
{
"input": "2 4",
"output": "1/2"
},
{
"input": "2 5",
"output": "1/3"
},
{
"input": "2 6",
"output": "1/6"
},
{
"input": "3 1",
"output": "2/3"
},
{
"input": "3 2",
"output": "2/3"
},
{
"input": "3 3",
"output": "2/3"
},
{
"input": "3 4",
"output": "1/2"
},
{
"input": "3 5",
"output": "1/3"
},
{
"input": "3 6",
"output": "1/6"
},
{
"input": "4 1",
"output": "1/2"
},
{
"input": "4 3",
"output": "1/2"
},
{
"input": "4 4",
"output": "1/2"
},
{
"input": "4 5",
"output": "1/3"
},
{
"input": "4 6",
"output": "1/6"
},
{
"input": "5 1",
"output": "1/3"
},
{
"input": "5 2",
"output": "1/3"
},
{
"input": "5 3",
"output": "1/3"
},
{
"input": "5 4",
"output": "1/3"
},
{
"input": "5 5",
"output": "1/3"
},
{
"input": "5 6",
"output": "1/6"
},
{
"input": "6 1",
"output": "1/6"
},
{
"input": "6 2",
"output": "1/6"
},
{
"input": "6 3",
"output": "1/6"
},
{
"input": "6 4",
"output": "1/6"
},
{
"input": "6 5",
"output": "1/6"
},
{
"input": "6 6",
"output": "1/6"
}
] | 1,591,174,525
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 216
| 0
|
a,b = map(int,input().split())
m = max(a,b)
p = 6 - m
if(6%p==0):
print('1'+'/'+str(6//p))
else:
print('p'+'/'+'6')
|
Title: Die Roll
Time Limit: 1 seconds
Memory Limit: 64 megabytes
Problem Description:
Yakko, Wakko and Dot, world-famous animaniacs, decided to rest from acting in cartoons, and take a leave to travel a bit. Yakko dreamt to go to Pennsylvania, his Motherland and the Motherland of his ancestors. Wakko thought about Tasmania, its beaches, sun and sea. Dot chose Transylvania as the most mysterious and unpredictable place.
But to their great regret, the leave turned to be very short, so it will be enough to visit one of the three above named places. That's why Yakko, as the cleverest, came up with a truly genius idea: let each of the three roll an ordinary six-sided die, and the one with the highest amount of points will be the winner, and will take the other two to the place of his/her dreams.
Yakko thrown a die and got Y points, Wakko — W points. It was Dot's turn. But she didn't hurry. Dot wanted to know for sure what were her chances to visit Transylvania.
It is known that Yakko and Wakko are true gentlemen, that's why if they have the same amount of points with Dot, they will let Dot win.
Input Specification:
The only line of the input file contains two natural numbers Y and W — the results of Yakko's and Wakko's die rolls.
Output Specification:
Output the required probability in the form of irreducible fraction in format «A/B», where A — the numerator, and B — the denominator. If the required probability equals to zero, output «0/1». If the required probability equals to 1, output «1/1».
Demo Input:
['4 2\n']
Demo Output:
['1/2\n']
Note:
Dot will go to Transylvania, if she is lucky to roll 4, 5 or 6 points.
|
```python
a,b = map(int,input().split())
m = max(a,b)
p = 6 - m
if(6%p==0):
print('1'+'/'+str(6//p))
else:
print('p'+'/'+'6')
```
| 0
|
Subsets and Splits
Successful Python Submissions
Retrieves all records from the train dataset where the verdict is 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Retrieves records of users with a rating of 1600 or higher and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a rating above 2000 and a verdict of 'OK', providing basic filtering but limited analytical value.
SQL Console for MatrixStudio/Codeforces-Python-Submissions
Counts the number of entries with a 'OK' verdict, providing a basic overview of a specific category within the dataset.