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3.99
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
456
|
A
|
Laptops
|
PROGRAMMING
| 1,100
|
[
"sortings"
] | null | null |
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops.
Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality).
All *a**i* are distinct. All *b**i* are distinct.
|
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
|
[
"2\n1 2\n2 1\n"
] |
[
"Happy Alex\n"
] |
none
| 500
|
[
{
"input": "2\n1 2\n2 1",
"output": "Happy Alex"
},
{
"input": "2\n1 1\n2 2",
"output": "Poor Alex"
},
{
"input": "3\n2 2\n3 3\n1 1",
"output": "Poor Alex"
},
{
"input": "3\n3 3\n1 2\n2 1",
"output": "Happy Alex"
},
{
"input": "1\n1 1",
"output": "Poor Alex"
},
{
"input": "3\n2 3\n1 1\n3 2",
"output": "Happy Alex"
},
{
"input": "4\n4 1\n3 2\n2 3\n1 4",
"output": "Happy Alex"
},
{
"input": "2\n2 1\n1 2",
"output": "Happy Alex"
},
{
"input": "3\n3 2\n1 1\n2 3",
"output": "Happy Alex"
},
{
"input": "3\n1 2\n2 3\n3 1",
"output": "Happy Alex"
},
{
"input": "3\n1 2\n2 1\n3 3",
"output": "Happy Alex"
},
{
"input": "3\n1 2\n2 3\n3 1",
"output": "Happy Alex"
}
] | 1,678,635,413
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 46
| 670
| 9,318,400
|
t = int(input())
c = 0
for el in range(t):
a , b = list(map(int , input().split()))
if(a != b):
c += 1
if(c > 0):
print('Happy Alex')
else:
print('Poor Alex')
|
Title: Laptops
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops.
Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality).
All *a**i* are distinct. All *b**i* are distinct.
Output Specification:
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
Demo Input:
['2\n1 2\n2 1\n']
Demo Output:
['Happy Alex\n']
Note:
none
|
```python
t = int(input())
c = 0
for el in range(t):
a , b = list(map(int , input().split()))
if(a != b):
c += 1
if(c > 0):
print('Happy Alex')
else:
print('Poor Alex')
```
| 3
|
|
611
|
A
|
New Year and Days
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Today is Wednesday, the third day of the week. What's more interesting is that tomorrow is the last day of the year 2015.
Limak is a little polar bear. He enjoyed this year a lot. Now, he is so eager to the coming year 2016.
Limak wants to prove how responsible a bear he is. He is going to regularly save candies for the entire year 2016! He considers various saving plans. He can save one candy either on some fixed day of the week or on some fixed day of the month.
Limak chose one particular plan. He isn't sure how many candies he will save in the 2016 with his plan. Please, calculate it and tell him.
|
The only line of the input is in one of the following two formats:
- "*x* of week" where *x* (1<=≤<=*x*<=≤<=7) denotes the day of the week. The 1-st day is Monday and the 7-th one is Sunday. - "*x* of month" where *x* (1<=≤<=*x*<=≤<=31) denotes the day of the month.
|
Print one integer — the number of candies Limak will save in the year 2016.
|
[
"4 of week\n",
"30 of month\n"
] |
[
"52\n",
"11\n"
] |
Polar bears use the Gregorian calendar. It is the most common calendar and you likely use it too. You can read about it on Wikipedia if you want to – [https://en.wikipedia.org/wiki/Gregorian_calendar](https://en.wikipedia.org/wiki/Gregorian_calendar). The week starts with Monday.
In the first sample Limak wants to save one candy on each Thursday (the 4-th day of the week). There are 52 Thursdays in the 2016. Thus, he will save 52 candies in total.
In the second sample Limak wants to save one candy on the 30-th day of each month. There is the 30-th day in exactly 11 months in the 2016 — all months but February. It means that Limak will save 11 candies in total.
| 500
|
[
{
"input": "4 of week",
"output": "52"
},
{
"input": "30 of month",
"output": "11"
},
{
"input": "17 of month",
"output": "12"
},
{
"input": "31 of month",
"output": "7"
},
{
"input": "6 of week",
"output": "53"
},
{
"input": "1 of week",
"output": "52"
},
{
"input": "2 of week",
"output": "52"
},
{
"input": "3 of week",
"output": "52"
},
{
"input": "5 of week",
"output": "53"
},
{
"input": "7 of week",
"output": "52"
},
{
"input": "1 of month",
"output": "12"
},
{
"input": "2 of month",
"output": "12"
},
{
"input": "3 of month",
"output": "12"
},
{
"input": "4 of month",
"output": "12"
},
{
"input": "5 of month",
"output": "12"
},
{
"input": "6 of month",
"output": "12"
},
{
"input": "7 of month",
"output": "12"
},
{
"input": "8 of month",
"output": "12"
},
{
"input": "9 of month",
"output": "12"
},
{
"input": "10 of month",
"output": "12"
},
{
"input": "11 of month",
"output": "12"
},
{
"input": "12 of month",
"output": "12"
},
{
"input": "13 of month",
"output": "12"
},
{
"input": "14 of month",
"output": "12"
},
{
"input": "15 of month",
"output": "12"
},
{
"input": "16 of month",
"output": "12"
},
{
"input": "18 of month",
"output": "12"
},
{
"input": "19 of month",
"output": "12"
},
{
"input": "20 of month",
"output": "12"
},
{
"input": "21 of month",
"output": "12"
},
{
"input": "22 of month",
"output": "12"
},
{
"input": "23 of month",
"output": "12"
},
{
"input": "24 of month",
"output": "12"
},
{
"input": "25 of month",
"output": "12"
},
{
"input": "26 of month",
"output": "12"
},
{
"input": "27 of month",
"output": "12"
},
{
"input": "28 of month",
"output": "12"
},
{
"input": "29 of month",
"output": "12"
}
] | 1,656,151,628
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 38
| 77
| 614,400
|
'''
╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬
╬╬ ▓▓ ▓▓ ╬╬
╬╬ ▓▓ ▓▓ ╬╬
╬╬ ▓▓█████▓▓ ╬╬
╬╬ ▓▓ ▓▓ ╬╬
╬╬ ▓▓ ▓▓ ╬╬
╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬
###########################
// •︿• \\
/\\ //\
/\\ //\
/\\//\
###########################
'''
import sys
import math as mth
from math import ceil as cl
from math import log2 as l2
from math import factorial as fct
from collections import Counter as CNT
mod = 10**9 + 7
def ii():
return int(input())
def fi():
return float(input())
def lii():
return list(map(int, input().split()))
def ss():
return input()
def lss():
return input().split()
def yes():
print("YES")
def no():
print("NO")
############################################
input = lambda : sys.stdin.readline().strip()
'''
╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬
'''
s = ss()
if "of month" in s:
if "31" in s:
print(7)
elif "30" in s:
print("11")
else:
print("12")
else:
if int(s[0]) in [1,2,3,4,7]:
print("52")
else:
print(53)
|
Title: New Year and Days
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today is Wednesday, the third day of the week. What's more interesting is that tomorrow is the last day of the year 2015.
Limak is a little polar bear. He enjoyed this year a lot. Now, he is so eager to the coming year 2016.
Limak wants to prove how responsible a bear he is. He is going to regularly save candies for the entire year 2016! He considers various saving plans. He can save one candy either on some fixed day of the week or on some fixed day of the month.
Limak chose one particular plan. He isn't sure how many candies he will save in the 2016 with his plan. Please, calculate it and tell him.
Input Specification:
The only line of the input is in one of the following two formats:
- "*x* of week" where *x* (1<=≤<=*x*<=≤<=7) denotes the day of the week. The 1-st day is Monday and the 7-th one is Sunday. - "*x* of month" where *x* (1<=≤<=*x*<=≤<=31) denotes the day of the month.
Output Specification:
Print one integer — the number of candies Limak will save in the year 2016.
Demo Input:
['4 of week\n', '30 of month\n']
Demo Output:
['52\n', '11\n']
Note:
Polar bears use the Gregorian calendar. It is the most common calendar and you likely use it too. You can read about it on Wikipedia if you want to – [https://en.wikipedia.org/wiki/Gregorian_calendar](https://en.wikipedia.org/wiki/Gregorian_calendar). The week starts with Monday.
In the first sample Limak wants to save one candy on each Thursday (the 4-th day of the week). There are 52 Thursdays in the 2016. Thus, he will save 52 candies in total.
In the second sample Limak wants to save one candy on the 30-th day of each month. There is the 30-th day in exactly 11 months in the 2016 — all months but February. It means that Limak will save 11 candies in total.
|
```python
'''
╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬
╬╬ ▓▓ ▓▓ ╬╬
╬╬ ▓▓ ▓▓ ╬╬
╬╬ ▓▓█████▓▓ ╬╬
╬╬ ▓▓ ▓▓ ╬╬
╬╬ ▓▓ ▓▓ ╬╬
╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬
###########################
// •︿• \\
/\\ //\
/\\ //\
/\\//\
###########################
'''
import sys
import math as mth
from math import ceil as cl
from math import log2 as l2
from math import factorial as fct
from collections import Counter as CNT
mod = 10**9 + 7
def ii():
return int(input())
def fi():
return float(input())
def lii():
return list(map(int, input().split()))
def ss():
return input()
def lss():
return input().split()
def yes():
print("YES")
def no():
print("NO")
############################################
input = lambda : sys.stdin.readline().strip()
'''
╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬╬
'''
s = ss()
if "of month" in s:
if "31" in s:
print(7)
elif "30" in s:
print("11")
else:
print("12")
else:
if int(s[0]) in [1,2,3,4,7]:
print("52")
else:
print(53)
```
| 3
|
|
576
|
A
|
Vasya and Petya's Game
|
PROGRAMMING
| 1,500
|
[
"math",
"number theory"
] | null | null |
Vasya and Petya are playing a simple game. Vasya thought of number *x* between 1 and *n*, and Petya tries to guess the number.
Petya can ask questions like: "Is the unknown number divisible by number *y*?".
The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of.
Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers *y**i*, he should ask the questions about.
|
A single line contains number *n* (1<=≤<=*n*<=≤<=103).
|
Print the length of the sequence of questions *k* (0<=≤<=*k*<=≤<=*n*), followed by *k* numbers — the questions *y**i* (1<=≤<=*y**i*<=≤<=*n*).
If there are several correct sequences of questions of the minimum length, you are allowed to print any of them.
|
[
"4\n",
"6\n"
] |
[
"3\n2 4 3 \n",
"4\n2 4 3 5 \n"
] |
The sequence from the answer to the first sample test is actually correct.
If the unknown number is not divisible by one of the sequence numbers, it is equal to 1.
If the unknown number is divisible by 4, it is 4.
If the unknown number is divisible by 3, then the unknown number is 3.
Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter.
| 500
|
[
{
"input": "4",
"output": "3\n2 4 3 "
},
{
"input": "6",
"output": "4\n2 4 3 5 "
},
{
"input": "1",
"output": "0"
},
{
"input": "15",
"output": "9\n2 4 8 3 9 5 7 11 13 "
},
{
"input": "19",
"output": "12\n2 4 8 16 3 9 5 7 11 13 17 19 "
},
{
"input": "20",
"output": "12\n2 4 8 16 3 9 5 7 11 13 17 19 "
},
{
"input": "37",
"output": "19\n2 4 8 16 32 3 9 27 5 25 7 11 13 17 19 23 29 31 37 "
},
{
"input": "211",
"output": "61\n2 4 8 16 32 64 128 3 9 27 81 5 25 125 7 49 11 121 13 169 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 "
},
{
"input": "557",
"output": "123\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 5 25 125 7 49 343 11 121 13 169 17 289 19 361 23 529 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 "
},
{
"input": "907",
"output": "179\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 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 ..."
},
{
"input": "953",
"output": "186\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 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 ..."
},
{
"input": "289",
"output": "78\n2 4 8 16 32 64 128 256 3 9 27 81 243 5 25 125 7 49 11 121 13 169 17 289 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 "
},
{
"input": "400",
"output": "97\n2 4 8 16 32 64 128 256 3 9 27 81 243 5 25 125 7 49 343 11 121 13 169 17 289 19 361 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 "
},
{
"input": "900",
"output": "178\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 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 ..."
},
{
"input": "625",
"output": "136\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 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 "
},
{
"input": "729",
"output": "152\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 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 ..."
},
{
"input": "784",
"output": "160\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 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 ..."
},
{
"input": "31",
"output": "17\n2 4 8 16 3 9 27 5 25 7 11 13 17 19 23 29 31 "
},
{
"input": "44",
"output": "21\n2 4 8 16 32 3 9 27 5 25 7 11 13 17 19 23 29 31 37 41 43 "
},
{
"input": "160",
"output": "50\n2 4 8 16 32 64 128 3 9 27 81 5 25 125 7 49 11 121 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 "
},
{
"input": "322",
"output": "83\n2 4 8 16 32 64 128 256 3 9 27 81 243 5 25 125 7 49 11 121 13 169 17 289 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 "
},
{
"input": "894",
"output": "178\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 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 ..."
},
{
"input": "998",
"output": "193\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 961 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 ..."
},
{
"input": "1000",
"output": "193\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 961 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 ..."
},
{
"input": "1000",
"output": "193\n2 4 8 16 32 64 128 256 512 3 9 27 81 243 729 5 25 125 625 7 49 343 11 121 13 169 17 289 19 361 23 529 29 841 31 961 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 ..."
},
{
"input": "30",
"output": "16\n2 4 8 16 3 9 27 5 25 7 11 13 17 19 23 29 "
}
] | 1,652,054,208
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 31
| 0
|
def isPrime(n):
if (n <= 1):
return False
if (n <= 3):
return True
for i in range(n, 3):
if n // i == 0:
return False
else:
return True
import sys
n = int(sys.stdin.readline())
ans = []
for i in range(1, n + 1):
if isPrime(i):
k = 1
while (n // i >= k):
k *= i
ans.append(k)
print(len(ans))
print(*ans)
|
Title: Vasya and Petya's Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya and Petya are playing a simple game. Vasya thought of number *x* between 1 and *n*, and Petya tries to guess the number.
Petya can ask questions like: "Is the unknown number divisible by number *y*?".
The game is played by the following rules: first Petya asks all the questions that interest him (also, he can ask no questions), and then Vasya responds to each question with a 'yes' or a 'no'. After receiving all the answers Petya should determine the number that Vasya thought of.
Unfortunately, Petya is not familiar with the number theory. Help him find the minimum number of questions he should ask to make a guaranteed guess of Vasya's number, and the numbers *y**i*, he should ask the questions about.
Input Specification:
A single line contains number *n* (1<=≤<=*n*<=≤<=103).
Output Specification:
Print the length of the sequence of questions *k* (0<=≤<=*k*<=≤<=*n*), followed by *k* numbers — the questions *y**i* (1<=≤<=*y**i*<=≤<=*n*).
If there are several correct sequences of questions of the minimum length, you are allowed to print any of them.
Demo Input:
['4\n', '6\n']
Demo Output:
['3\n2 4 3 \n', '4\n2 4 3 5 \n']
Note:
The sequence from the answer to the first sample test is actually correct.
If the unknown number is not divisible by one of the sequence numbers, it is equal to 1.
If the unknown number is divisible by 4, it is 4.
If the unknown number is divisible by 3, then the unknown number is 3.
Otherwise, it is equal to 2. Therefore, the sequence of questions allows you to guess the unknown number. It can be shown that there is no correct sequence of questions of length 2 or shorter.
|
```python
def isPrime(n):
if (n <= 1):
return False
if (n <= 3):
return True
for i in range(n, 3):
if n // i == 0:
return False
else:
return True
import sys
n = int(sys.stdin.readline())
ans = []
for i in range(1, n + 1):
if isPrime(i):
k = 1
while (n // i >= k):
k *= i
ans.append(k)
print(len(ans))
print(*ans)
```
| 0
|
|
567
|
B
|
Berland National Library
|
PROGRAMMING
| 1,300
|
[
"implementation"
] | null | null |
Berland National Library has recently been built in the capital of Berland. In addition, in the library you can take any of the collected works of Berland leaders, the library has a reading room.
Today was the pilot launch of an automated reading room visitors' accounting system! The scanner of the system is installed at the entrance to the reading room. It records the events of the form "reader entered room", "reader left room". Every reader is assigned a registration number during the registration procedure at the library — it's a unique integer from 1 to 106. Thus, the system logs events of two forms:
- "+ *r**i*" — the reader with registration number *r**i* entered the room; - "- *r**i*" — the reader with registration number *r**i* left the room.
The first launch of the system was a success, it functioned for some period of time, and, at the time of its launch and at the time of its shutdown, the reading room may already have visitors.
Significant funds of the budget of Berland have been spent on the design and installation of the system. Therefore, some of the citizens of the capital now demand to explain the need for this system and the benefits that its implementation will bring. Now, the developers of the system need to urgently come up with reasons for its existence.
Help the system developers to find the minimum possible capacity of the reading room (in visitors) using the log of the system available to you.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of records in the system log. Next follow *n* events from the system journal in the order in which the were made. Each event was written on a single line and looks as "+ *r**i*" or "- *r**i*", where *r**i* is an integer from 1 to 106, the registration number of the visitor (that is, distinct visitors always have distinct registration numbers).
It is guaranteed that the log is not contradictory, that is, for every visitor the types of any of his two consecutive events are distinct. Before starting the system, and after stopping the room may possibly contain visitors.
|
Print a single integer — the minimum possible capacity of the reading room.
|
[
"6\n+ 12001\n- 12001\n- 1\n- 1200\n+ 1\n+ 7\n",
"2\n- 1\n- 2\n",
"2\n+ 1\n- 1\n"
] |
[
"3",
"2",
"1"
] |
In the first sample test, the system log will ensure that at some point in the reading room were visitors with registration numbers 1, 1200 and 12001. More people were not in the room at the same time based on the log. Therefore, the answer to the test is 3.
| 1,000
|
[
{
"input": "6\n+ 12001\n- 12001\n- 1\n- 1200\n+ 1\n+ 7",
"output": "3"
},
{
"input": "2\n- 1\n- 2",
"output": "2"
},
{
"input": "2\n+ 1\n- 1",
"output": "1"
},
{
"input": "5\n+ 1\n- 1\n+ 2\n+ 3\n- 4",
"output": "3"
},
{
"input": "3\n- 1\n- 2\n- 3",
"output": "3"
},
{
"input": "4\n+ 1\n+ 2\n- 1\n+ 3",
"output": "2"
},
{
"input": "6\n+ 1\n+ 2\n- 1\n+ 3\n- 2\n+ 4",
"output": "2"
},
{
"input": "3\n+ 1\n+ 2\n- 3",
"output": "3"
},
{
"input": "3\n- 1\n+ 2\n- 2",
"output": "1"
},
{
"input": "4\n- 1\n- 2\n+ 3\n+ 4",
"output": "2"
},
{
"input": "1\n+ 1",
"output": "1"
},
{
"input": "1\n- 1",
"output": "1"
},
{
"input": "3\n- 1\n+ 1\n- 1",
"output": "1"
},
{
"input": "10\n+ 1\n+ 2\n+ 3\n+ 4\n+ 5\n+ 6\n+ 7\n+ 8\n+ 9\n+ 10",
"output": "10"
},
{
"input": "5\n+ 5\n+ 4\n- 4\n- 5\n+ 5",
"output": "2"
},
{
"input": "50\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100\n+ 100\n- 100",
"output": "1"
},
{
"input": "10\n- 8\n- 4\n+ 8\n+ 10\n+ 6\n- 8\n+ 9\n- 2\n- 7\n+ 4",
"output": "5"
},
{
"input": "20\n+ 3\n- 3\n- 2\n+ 2\n+ 3\n- 5\n- 1\n+ 1\n- 3\n+ 4\n- 1\n+ 1\n+ 3\n- 3\n+ 5\n- 2\n- 1\n+ 2\n+ 1\n- 5",
"output": "4"
},
{
"input": "50\n+ 4\n+ 5\n+ 3\n+ 2\n- 2\n- 3\n- 4\n+ 3\n+ 2\n- 3\n+ 4\n- 2\n- 4\n+ 2\n+ 3\n- 3\n- 5\n- 1\n+ 4\n+ 5\n- 5\n+ 3\n- 4\n- 3\n- 2\n+ 4\n+ 3\n+ 2\n- 2\n- 4\n+ 5\n+ 1\n+ 4\n+ 2\n- 2\n+ 2\n- 3\n- 5\n- 4\n- 1\n+ 5\n- 2\n- 5\n+ 5\n+ 3\n- 3\n+ 1\n+ 3\n+ 2\n- 1",
"output": "5"
},
{
"input": "10\n- 2\n+ 1\n- 1\n+ 2\n- 2\n+ 2\n+ 1\n- 1\n- 2\n+ 1",
"output": "2"
},
{
"input": "50\n+ 1\n+ 2\n+ 3\n+ 4\n+ 5\n+ 6\n+ 7\n+ 8\n+ 9\n+ 10\n+ 11\n+ 12\n+ 13\n+ 14\n+ 15\n+ 16\n+ 17\n+ 18\n+ 19\n+ 20\n+ 21\n+ 22\n+ 23\n+ 24\n+ 25\n+ 26\n+ 27\n+ 28\n+ 29\n+ 30\n+ 31\n+ 32\n+ 33\n+ 34\n+ 35\n+ 36\n+ 37\n+ 38\n+ 39\n+ 40\n+ 41\n+ 42\n+ 43\n+ 44\n+ 45\n+ 46\n+ 47\n+ 48\n+ 49\n+ 50",
"output": "50"
},
{
"input": "50\n- 1\n- 2\n- 3\n- 4\n- 5\n- 6\n- 7\n- 8\n- 9\n- 10\n- 11\n- 12\n- 13\n- 14\n- 15\n- 16\n- 17\n- 18\n- 19\n- 20\n- 21\n- 22\n- 23\n- 24\n- 25\n- 26\n- 27\n- 28\n- 29\n- 30\n- 31\n- 32\n- 33\n- 34\n- 35\n- 36\n- 37\n- 38\n- 39\n- 40\n- 41\n- 42\n- 43\n- 44\n- 45\n- 46\n- 47\n- 48\n- 49\n- 50",
"output": "50"
}
] | 1,601,542,221
| 1,821
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 78
| 0
|
n, visitors = int(input()), []
mn = 0
for i in range(n):
inp = input().split()
if inp[0] == '+':
visitors.append(inp[2:])
if len(visitors) >= mn:
mn += 1
elif inp[0] == '-':
try:
visitors.remove(inp[2:])
except:
mn += 1
print(mn)
|
Title: Berland National Library
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Berland National Library has recently been built in the capital of Berland. In addition, in the library you can take any of the collected works of Berland leaders, the library has a reading room.
Today was the pilot launch of an automated reading room visitors' accounting system! The scanner of the system is installed at the entrance to the reading room. It records the events of the form "reader entered room", "reader left room". Every reader is assigned a registration number during the registration procedure at the library — it's a unique integer from 1 to 106. Thus, the system logs events of two forms:
- "+ *r**i*" — the reader with registration number *r**i* entered the room; - "- *r**i*" — the reader with registration number *r**i* left the room.
The first launch of the system was a success, it functioned for some period of time, and, at the time of its launch and at the time of its shutdown, the reading room may already have visitors.
Significant funds of the budget of Berland have been spent on the design and installation of the system. Therefore, some of the citizens of the capital now demand to explain the need for this system and the benefits that its implementation will bring. Now, the developers of the system need to urgently come up with reasons for its existence.
Help the system developers to find the minimum possible capacity of the reading room (in visitors) using the log of the system available to you.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of records in the system log. Next follow *n* events from the system journal in the order in which the were made. Each event was written on a single line and looks as "+ *r**i*" or "- *r**i*", where *r**i* is an integer from 1 to 106, the registration number of the visitor (that is, distinct visitors always have distinct registration numbers).
It is guaranteed that the log is not contradictory, that is, for every visitor the types of any of his two consecutive events are distinct. Before starting the system, and after stopping the room may possibly contain visitors.
Output Specification:
Print a single integer — the minimum possible capacity of the reading room.
Demo Input:
['6\n+ 12001\n- 12001\n- 1\n- 1200\n+ 1\n+ 7\n', '2\n- 1\n- 2\n', '2\n+ 1\n- 1\n']
Demo Output:
['3', '2', '1']
Note:
In the first sample test, the system log will ensure that at some point in the reading room were visitors with registration numbers 1, 1200 and 12001. More people were not in the room at the same time based on the log. Therefore, the answer to the test is 3.
|
```python
n, visitors = int(input()), []
mn = 0
for i in range(n):
inp = input().split()
if inp[0] == '+':
visitors.append(inp[2:])
if len(visitors) >= mn:
mn += 1
elif inp[0] == '-':
try:
visitors.remove(inp[2:])
except:
mn += 1
print(mn)
```
| 0
|
|
25
|
A
|
IQ test
|
PROGRAMMING
| 1,300
|
[
"brute force"
] |
A. IQ test
|
2
|
256
|
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
|
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
|
[
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] |
[
"3\n",
"2\n"
] |
none
| 0
|
[
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "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"
}
] | 1,596,624,907
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 248
| 6,963,200
|
length = int(input())
numbers = input().split()
new_arr = []
even_arr = []
odd_arr = []
for num in numbers:
new_arr.append(int(num))
for item in new_arr:
if item % 2 == 0:
even_arr.append(item)
else:
odd_arr.append(item)
if len(even_arr) == 1:
print(new_arr.index(even_arr[0]) + 1)
else:
print(new_arr.index(odd_arr[0]) + 1)
|
Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none
|
```python
length = int(input())
numbers = input().split()
new_arr = []
even_arr = []
odd_arr = []
for num in numbers:
new_arr.append(int(num))
for item in new_arr:
if item % 2 == 0:
even_arr.append(item)
else:
odd_arr.append(item)
if len(even_arr) == 1:
print(new_arr.index(even_arr[0]) + 1)
else:
print(new_arr.index(odd_arr[0]) + 1)
```
| 3.92503
|
158
|
A
|
Next Round
|
PROGRAMMING
| 800
|
[
"*special",
"implementation"
] | null | null |
"Contestant who earns a score equal to or greater than the *k*-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules.
A total of *n* participants took part in the contest (*n*<=≥<=*k*), and you already know their scores. Calculate how many participants will advance to the next round.
|
The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50) separated by a single space.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the score earned by the participant who got the *i*-th place. The given sequence is non-increasing (that is, for all *i* from 1 to *n*<=-<=1 the following condition is fulfilled: *a**i*<=≥<=*a**i*<=+<=1).
|
Output the number of participants who advance to the next round.
|
[
"8 5\n10 9 8 7 7 7 5 5\n",
"4 2\n0 0 0 0\n"
] |
[
"6\n",
"0\n"
] |
In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers.
In the second example nobody got a positive score.
| 500
|
[
{
"input": "8 5\n10 9 8 7 7 7 5 5",
"output": "6"
},
{
"input": "4 2\n0 0 0 0",
"output": "0"
},
{
"input": "5 1\n1 1 1 1 1",
"output": "5"
},
{
"input": "5 5\n1 1 1 1 1",
"output": "5"
},
{
"input": "1 1\n10",
"output": "1"
},
{
"input": "17 14\n16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0",
"output": "14"
},
{
"input": "5 5\n3 2 1 0 0",
"output": "3"
},
{
"input": "8 6\n10 9 8 7 7 7 5 5",
"output": "6"
},
{
"input": "8 7\n10 9 8 7 7 7 5 5",
"output": "8"
},
{
"input": "8 4\n10 9 8 7 7 7 5 5",
"output": "6"
},
{
"input": "8 3\n10 9 8 7 7 7 5 5",
"output": "3"
},
{
"input": "8 1\n10 9 8 7 7 7 5 5",
"output": "1"
},
{
"input": "8 2\n10 9 8 7 7 7 5 5",
"output": "2"
},
{
"input": "1 1\n100",
"output": "1"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "50 25\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": "50"
},
{
"input": "50 25\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 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": "25"
},
{
"input": "50 25\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 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": "26"
},
{
"input": "50 25\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 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 1 1 1",
"output": "50"
},
{
"input": "11 5\n100 99 98 97 96 95 94 93 92 91 90",
"output": "5"
},
{
"input": "10 4\n100 81 70 69 64 43 34 29 15 3",
"output": "4"
},
{
"input": "11 6\n87 71 62 52 46 46 43 35 32 25 12",
"output": "6"
},
{
"input": "17 12\n99 88 86 82 75 75 74 65 58 52 45 30 21 16 7 2 2",
"output": "12"
},
{
"input": "20 3\n98 98 96 89 87 82 82 80 76 74 74 68 61 60 43 32 30 22 4 2",
"output": "3"
},
{
"input": "36 12\n90 87 86 85 83 80 79 78 76 70 69 69 61 61 59 58 56 48 45 44 42 41 33 31 27 25 23 21 20 19 15 14 12 7 5 5",
"output": "12"
},
{
"input": "49 8\n99 98 98 96 92 92 90 89 89 86 86 85 83 80 79 76 74 69 67 67 58 56 55 51 49 47 47 46 45 41 41 40 39 34 34 33 25 23 18 15 13 13 11 9 5 4 3 3 1",
"output": "9"
},
{
"input": "49 29\n100 98 98 96 96 96 95 87 85 84 81 76 74 70 63 63 63 62 57 57 56 54 53 52 50 47 45 41 41 39 38 31 30 28 27 26 23 22 20 15 15 11 7 6 6 4 2 1 0",
"output": "29"
},
{
"input": "49 34\n99 98 96 96 93 92 90 89 88 86 85 85 82 76 73 69 66 64 63 63 60 59 57 57 56 55 54 54 51 48 47 44 42 42 40 39 38 36 33 26 24 23 19 17 17 14 12 7 4",
"output": "34"
},
{
"input": "50 44\n100 100 99 97 95 91 91 84 83 83 79 71 70 69 69 62 61 60 59 59 58 58 58 55 55 54 52 48 47 45 44 44 38 36 32 31 28 28 25 25 24 24 24 22 17 15 14 13 12 4",
"output": "44"
},
{
"input": "50 13\n99 95 94 94 88 87 81 79 78 76 74 72 72 69 68 67 67 67 66 63 62 61 58 57 55 55 54 51 50 50 48 48 42 41 38 35 34 32 31 30 26 24 13 13 12 6 5 4 3 3",
"output": "13"
},
{
"input": "50 30\n100 98 96 94 91 89 88 81 81 81 81 81 76 73 72 71 70 69 66 64 61 59 59 56 52 50 49 48 43 39 36 35 34 34 31 29 27 26 24 22 16 16 15 14 14 14 9 7 4 3",
"output": "30"
},
{
"input": "2 1\n10 10",
"output": "2"
},
{
"input": "2 2\n10 10",
"output": "2"
},
{
"input": "2 2\n10 0",
"output": "1"
},
{
"input": "2 2\n10 1",
"output": "2"
},
{
"input": "2 1\n10 0",
"output": "1"
},
{
"input": "2 1\n10 2",
"output": "1"
},
{
"input": "50 13\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",
"output": "0"
},
{
"input": "50 1\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",
"output": "0"
},
{
"input": "50 50\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",
"output": "0"
},
{
"input": "10 1\n5 5 5 3 3 3 0 0 0 0",
"output": "3"
},
{
"input": "10 2\n5 5 5 3 3 3 0 0 0 0",
"output": "3"
},
{
"input": "10 3\n5 5 5 3 3 3 0 0 0 0",
"output": "3"
},
{
"input": "10 4\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
},
{
"input": "10 5\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
},
{
"input": "10 6\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
},
{
"input": "10 7\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
},
{
"input": "10 8\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
},
{
"input": "10 9\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
},
{
"input": "10 10\n5 5 5 3 3 3 0 0 0 0",
"output": "6"
}
] | 1,699,314,152
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
def scores():
list=[]
participants, place= input().split()
score= input().split()
counter=0
for personal_score in score:
if int(personal_score) > int(place) and int(personal_score)>0:
list.append(personal_score)
counter =counter + 1
return counter
scores()
|
Title: Next Round
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"Contestant who earns a score equal to or greater than the *k*-th place finisher's score will advance to the next round, as long as the contestant earns a positive score..." — an excerpt from contest rules.
A total of *n* participants took part in the contest (*n*<=≥<=*k*), and you already know their scores. Calculate how many participants will advance to the next round.
Input Specification:
The first line of the input contains two integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=50) separated by a single space.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=100), where *a**i* is the score earned by the participant who got the *i*-th place. The given sequence is non-increasing (that is, for all *i* from 1 to *n*<=-<=1 the following condition is fulfilled: *a**i*<=≥<=*a**i*<=+<=1).
Output Specification:
Output the number of participants who advance to the next round.
Demo Input:
['8 5\n10 9 8 7 7 7 5 5\n', '4 2\n0 0 0 0\n']
Demo Output:
['6\n', '0\n']
Note:
In the first example the participant on the 5th place earned 7 points. As the participant on the 6th place also earned 7 points, there are 6 advancers.
In the second example nobody got a positive score.
|
```python
def scores():
list=[]
participants, place= input().split()
score= input().split()
counter=0
for personal_score in score:
if int(personal_score) > int(place) and int(personal_score)>0:
list.append(personal_score)
counter =counter + 1
return counter
scores()
```
| 0
|
|
999
|
E
|
Reachability from the Capital
|
PROGRAMMING
| 2,000
|
[
"dfs and similar",
"graphs",
"greedy"
] | null | null |
There are $n$ cities and $m$ roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
|
The first line of input consists of three integers $n$, $m$ and $s$ ($1 \le n \le 5000, 0 \le m \le 5000, 1 \le s \le n$) — the number of cities, the number of roads and the index of the capital. Cities are indexed from $1$ to $n$.
The following $m$ lines contain roads: road $i$ is given as a pair of cities $u_i$, $v_i$ ($1 \le u_i, v_i \le n$, $u_i \ne v_i$). For each pair of cities $(u, v)$, there can be at most one road from $u$ to $v$. Roads in opposite directions between a pair of cities are allowed (i.e. from $u$ to $v$ and from $v$ to $u$).
|
Print one integer — the minimum number of extra roads needed to make all the cities reachable from city $s$. If all the cities are already reachable from $s$, print 0.
|
[
"9 9 1\n1 2\n1 3\n2 3\n1 5\n5 6\n6 1\n1 8\n9 8\n7 1\n",
"5 4 5\n1 2\n2 3\n3 4\n4 1\n"
] |
[
"3\n",
"1\n"
] |
The first example is illustrated by the following:
For example, you can add roads ($6, 4$), ($7, 9$), ($1, 7$) to make all the cities reachable from $s = 1$.
The second example is illustrated by the following:
In this example, you can add any one of the roads ($5, 1$), ($5, 2$), ($5, 3$), ($5, 4$) to make all the cities reachable from $s = 5$.
| 0
|
[
{
"input": "9 9 1\n1 2\n1 3\n2 3\n1 5\n5 6\n6 1\n1 8\n9 8\n7 1",
"output": "3"
},
{
"input": "5 4 5\n1 2\n2 3\n3 4\n4 1",
"output": "1"
},
{
"input": "5000 0 2956",
"output": "4999"
},
{
"input": "2 0 2",
"output": "1"
},
{
"input": "2 1 1\n1 2",
"output": "0"
},
{
"input": "2 1 2\n1 2",
"output": "1"
},
{
"input": "2 2 2\n1 2\n2 1",
"output": "0"
},
{
"input": "5000 2 238\n3212 238\n238 3212",
"output": "4998"
},
{
"input": "5000 2 3810\n3225 1137\n1137 3225",
"output": "4998"
},
{
"input": "100 1 30\n69 81",
"output": "98"
},
{
"input": "500 1 209\n183 107",
"output": "498"
},
{
"input": "1000 1 712\n542 916",
"output": "998"
},
{
"input": "39 40 38\n4 8\n24 28\n16 17\n7 25\n4 29\n34 35\n16 24\n21 10\n23 36\n36 14\n28 16\n34 19\n15 21\n22 38\n22 37\n37 27\n28 33\n3 29\n32 22\n12 30\n9 15\n5 19\n23 27\n19 17\n25 17\n24 11\n39 10\n6 20\n16 6\n3 18\n34 21\n15 38\n11 19\n11 3\n32 4\n15 13\n16 11\n11 7\n33 7\n3 33",
"output": "12"
},
{
"input": "8 8 1\n3 2\n3 4\n4 5\n5 3\n6 4\n6 7\n7 8\n8 6",
"output": "1"
},
{
"input": "7 7 1\n2 3\n4 2\n7 4\n5 7\n4 5\n4 6\n6 4",
"output": "1"
},
{
"input": "2333 1 1\n1 2",
"output": "2331"
},
{
"input": "7 4 2\n1 2\n3 4\n5 6\n7 1",
"output": "3"
},
{
"input": "1 0 1",
"output": "0"
},
{
"input": "9 9 1\n2 3\n3 4\n4 5\n6 2\n7 6\n8 7\n6 8\n5 9\n9 2",
"output": "1"
},
{
"input": "30 30 29\n29 9\n1 3\n15 5\n14 24\n7 18\n17 12\n5 9\n22 7\n26 20\n7 29\n8 2\n10 12\n17 18\n17 24\n8 12\n23 28\n10 15\n18 2\n29 13\n25 29\n9 2\n7 13\n11 5\n19 26\n10 5\n25 5\n20 10\n5 26\n24 23\n22 8",
"output": "13"
},
{
"input": "30 30 28\n16 28\n5 30\n7 17\n6 1\n12 28\n15 28\n14 30\n25 11\n25 10\n25 8\n12 30\n27 5\n30 10\n22 14\n30 13\n20 27\n15 25\n24 20\n20 15\n1 30\n3 10\n3 4\n3 18\n15 14\n18 22\n20 26\n10 3\n15 27\n23 29\n10 24",
"output": "9"
}
] | 1,694,329,348
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 3
| 140
| 9,318,400
|
def dfs(v):
global used, g
for to in g[v]:
if used[to] == -1:
used[to] = 1
dfs(to)
def dfs2(v):
global used, g, lst
count = 1
used[v] = 2
for to in g[v]:
if used[to] == -1:
count += dfs2(to)
return count
n, m, s = list(map(int, input().split()))
g = [[] for i in range(n + 1)]
used = [-1 for i in range(n + 1)]
used[s] = 0
for i in range(m):
S, to = list(map(int, input().split()))
g[S].append(to)
dfs(s)
ans = 0
bad = []
for i in range(1, n + 1):
if used[i] == -1:
bad.append([i, dfs2(i)])
bad.sort(key=lambda x: x[1])
for S, count in bad:
if used[S] == 2:
ans += 1
dfs(S)
print(ans)
|
Title: Reachability from the Capital
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are $n$ cities and $m$ roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.
What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?
New roads will also be one-way.
Input Specification:
The first line of input consists of three integers $n$, $m$ and $s$ ($1 \le n \le 5000, 0 \le m \le 5000, 1 \le s \le n$) — the number of cities, the number of roads and the index of the capital. Cities are indexed from $1$ to $n$.
The following $m$ lines contain roads: road $i$ is given as a pair of cities $u_i$, $v_i$ ($1 \le u_i, v_i \le n$, $u_i \ne v_i$). For each pair of cities $(u, v)$, there can be at most one road from $u$ to $v$. Roads in opposite directions between a pair of cities are allowed (i.e. from $u$ to $v$ and from $v$ to $u$).
Output Specification:
Print one integer — the minimum number of extra roads needed to make all the cities reachable from city $s$. If all the cities are already reachable from $s$, print 0.
Demo Input:
['9 9 1\n1 2\n1 3\n2 3\n1 5\n5 6\n6 1\n1 8\n9 8\n7 1\n', '5 4 5\n1 2\n2 3\n3 4\n4 1\n']
Demo Output:
['3\n', '1\n']
Note:
The first example is illustrated by the following:
For example, you can add roads ($6, 4$), ($7, 9$), ($1, 7$) to make all the cities reachable from $s = 1$.
The second example is illustrated by the following:
In this example, you can add any one of the roads ($5, 1$), ($5, 2$), ($5, 3$), ($5, 4$) to make all the cities reachable from $s = 5$.
|
```python
def dfs(v):
global used, g
for to in g[v]:
if used[to] == -1:
used[to] = 1
dfs(to)
def dfs2(v):
global used, g, lst
count = 1
used[v] = 2
for to in g[v]:
if used[to] == -1:
count += dfs2(to)
return count
n, m, s = list(map(int, input().split()))
g = [[] for i in range(n + 1)]
used = [-1 for i in range(n + 1)]
used[s] = 0
for i in range(m):
S, to = list(map(int, input().split()))
g[S].append(to)
dfs(s)
ans = 0
bad = []
for i in range(1, n + 1):
if used[i] == -1:
bad.append([i, dfs2(i)])
bad.sort(key=lambda x: x[1])
for S, count in bad:
if used[S] == 2:
ans += 1
dfs(S)
print(ans)
```
| 0
|
|
577
|
A
|
Multiplication Table
|
PROGRAMMING
| 1,000
|
[
"implementation",
"number theory"
] | null | null |
Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=×<=*j*. The rows and columns are numbered starting from 1.
You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*.
|
The single line contains numbers *n* and *x* (1<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=109) — the size of the table and the number that we are looking for in the table.
|
Print a single number: the number of times *x* occurs in the table.
|
[
"10 5\n",
"6 12\n",
"5 13\n"
] |
[
"2\n",
"4\n",
"0\n"
] |
A table for the second sample test is given below. The occurrences of number 12 are marked bold.
| 500
|
[
{
"input": "10 5",
"output": "2"
},
{
"input": "6 12",
"output": "4"
},
{
"input": "5 13",
"output": "0"
},
{
"input": "1 1",
"output": "1"
},
{
"input": "2 1",
"output": "1"
},
{
"input": "100000 1",
"output": "1"
},
{
"input": "1 1000000000",
"output": "0"
},
{
"input": "100000 1000000000",
"output": "16"
},
{
"input": "100000 362880",
"output": "154"
},
{
"input": "1 4",
"output": "0"
},
{
"input": "9 12",
"output": "4"
},
{
"input": "10 123",
"output": "0"
},
{
"input": "9551 975275379",
"output": "0"
},
{
"input": "17286 948615687",
"output": "0"
},
{
"input": "58942 936593001",
"output": "0"
},
{
"input": "50000 989460910",
"output": "4"
},
{
"input": "22741 989460910",
"output": "0"
},
{
"input": "22740 989460910",
"output": "0"
},
{
"input": "100000 989460910",
"output": "4"
},
{
"input": "100000 98280",
"output": "128"
},
{
"input": "100000 997920",
"output": "222"
},
{
"input": "100000 720720",
"output": "226"
},
{
"input": "100000 2162160",
"output": "282"
},
{
"input": "100000 4324320",
"output": "320"
},
{
"input": "100000 8648640",
"output": "348"
},
{
"input": "100000 183783600",
"output": "438"
},
{
"input": "100000 551350800",
"output": "392"
},
{
"input": "40000 551350800",
"output": "150"
},
{
"input": "20000 400000000",
"output": "1"
},
{
"input": "19999 400000000",
"output": "0"
},
{
"input": "19999 399960001",
"output": "1"
},
{
"input": "31621 999887641",
"output": "1"
},
{
"input": "31622 999887641",
"output": "1"
},
{
"input": "31620 999887641",
"output": "0"
},
{
"input": "100000 999887641",
"output": "3"
},
{
"input": "100000 25",
"output": "3"
},
{
"input": "100000 1",
"output": "1"
},
{
"input": "100000 3628800",
"output": "220"
},
{
"input": "100000 39916800",
"output": "328"
},
{
"input": "100000 479001600",
"output": "254"
},
{
"input": "4 9",
"output": "1"
},
{
"input": "2 6",
"output": "0"
},
{
"input": "20 100",
"output": "3"
},
{
"input": "10 3",
"output": "2"
},
{
"input": "4 4",
"output": "3"
},
{
"input": "2 4",
"output": "1"
},
{
"input": "89874 1",
"output": "1"
}
] | 1,643,549,396
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 47
| 109
| 2,252,800
|
N, X = map(int, input().split())
c = 0
for m in range(1, N+1) :
if X%m==0 and X//m<=N :
c += 1
print(c)
|
Title: Multiplication Table
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's consider a table consisting of *n* rows and *n* columns. The cell located at the intersection of *i*-th row and *j*-th column contains number *i*<=×<=*j*. The rows and columns are numbered starting from 1.
You are given a positive integer *x*. Your task is to count the number of cells in a table that contain number *x*.
Input Specification:
The single line contains numbers *n* and *x* (1<=≤<=*n*<=≤<=105, 1<=≤<=*x*<=≤<=109) — the size of the table and the number that we are looking for in the table.
Output Specification:
Print a single number: the number of times *x* occurs in the table.
Demo Input:
['10 5\n', '6 12\n', '5 13\n']
Demo Output:
['2\n', '4\n', '0\n']
Note:
A table for the second sample test is given below. The occurrences of number 12 are marked bold.
|
```python
N, X = map(int, input().split())
c = 0
for m in range(1, N+1) :
if X%m==0 and X//m<=N :
c += 1
print(c)
```
| 3
|
|
519
|
B
|
A and B and Compilation Errors
|
PROGRAMMING
| 1,100
|
[
"data structures",
"implementation",
"sortings"
] | null | null |
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
|
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
|
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
|
[
"5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n",
"6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n"
] |
[
"8\n123\n",
"1\n3\n"
] |
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
| 1,000
|
[
{
"input": "5\n1 5 8 123 7\n123 7 5 1\n5 1 7",
"output": "8\n123"
},
{
"input": "6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5",
"output": "1\n3"
},
{
"input": "3\n1 2 3\n3 2\n2",
"output": "1\n3"
},
{
"input": "10\n460626451 802090732 277246428 661369649 388684428 784303821 376287098 656422756 9301599 25720377\n277246428 388684428 661369649 460626451 656422756 802090732 9301599 784303821 376287098\n376287098 802090732 388684428 9301599 656422756 784303821 460626451 277246428",
"output": "25720377\n661369649"
},
{
"input": "3\n796067435 964699482 819602309\n964699482 796067435\n964699482",
"output": "819602309\n796067435"
},
{
"input": "3\n374054998 726316780 902899520\n902899520 726316780\n726316780",
"output": "374054998\n902899520"
},
{
"input": "3\n168638990 939116221 323703261\n168638990 323703261\n168638990",
"output": "939116221\n323703261"
},
{
"input": "3\n77 77 77\n77 77\n77",
"output": "77\n77"
},
{
"input": "3\n84 30 9\n9 84\n9",
"output": "30\n84"
},
{
"input": "6\n5 4 3 3 5 5\n3 5 5 4 3\n3 5 4 3",
"output": "5\n5"
},
{
"input": "4\n1 5 7 8\n1 5 7\n1 5",
"output": "8\n7"
}
] | 1,698,593,934
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 16
| 2,000
| 23,859,200
|
a = int(input())
arr = list(map(int, input().split()))
arr1 = list(map(int, input().split()))
arr2 = list(map(int, input().split()))
for i in arr:
if arr.count(i) != arr1.count(i):
print(i)
break
for i in arr1:
if arr1.count(i) != arr2.count(i):
print(i)
break
|
Title: A and B and Compilation Errors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output Specification:
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Demo Input:
['5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n', '6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n']
Demo Output:
['8\n123\n', '1\n3\n']
Note:
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
|
```python
a = int(input())
arr = list(map(int, input().split()))
arr1 = list(map(int, input().split()))
arr2 = list(map(int, input().split()))
for i in arr:
if arr.count(i) != arr1.count(i):
print(i)
break
for i in arr1:
if arr1.count(i) != arr2.count(i):
print(i)
break
```
| 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"
}
] | 1,628,452,239
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 28
| 93
| 20,172,800
|
original = input()
s = original
res = 1
while original != s[-1] + s[:-1]:
res += 1
s = s[-1] + s[:-1]
print(res)
|
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
original = input()
s = original
res = 1
while original != s[-1] + s[:-1]:
res += 1
s = s[-1] + s[:-1]
print(res)
```
| 3
|
|
96
|
A
|
Football
|
PROGRAMMING
| 900
|
[
"implementation",
"strings"
] |
A. Football
|
2
|
256
|
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
|
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
|
Print "YES" if the situation is dangerous. Otherwise, print "NO".
|
[
"001001\n",
"1000000001\n"
] |
[
"NO\n",
"YES\n"
] |
none
| 500
|
[
{
"input": "001001",
"output": "NO"
},
{
"input": "1000000001",
"output": "YES"
},
{
"input": "00100110111111101",
"output": "YES"
},
{
"input": "11110111111111111",
"output": "YES"
},
{
"input": "01",
"output": "NO"
},
{
"input": "10100101",
"output": "NO"
},
{
"input": "1010010100000000010",
"output": "YES"
},
{
"input": "101010101",
"output": "NO"
},
{
"input": "000000000100000000000110101100000",
"output": "YES"
},
{
"input": "100001000000110101100000",
"output": "NO"
},
{
"input": "100001000011010110000",
"output": "NO"
},
{
"input": "010",
"output": "NO"
},
{
"input": "10101011111111111111111111111100",
"output": "YES"
},
{
"input": "1001101100",
"output": "NO"
},
{
"input": "1001101010",
"output": "NO"
},
{
"input": "1111100111",
"output": "NO"
},
{
"input": "00110110001110001111",
"output": "NO"
},
{
"input": "11110001001111110001",
"output": "NO"
},
{
"input": "10001111001011111101",
"output": "NO"
},
{
"input": "10000010100000001000110001010100001001001010011",
"output": "YES"
},
{
"input": "01111011111010111100101100001011001010111110000010",
"output": "NO"
},
{
"input": "00100000100100101110011001011011101110110110010100",
"output": "NO"
},
{
"input": "10110100110001001011110101110010100010000000000100101010111110111110100011",
"output": "YES"
},
{
"input": "00011101010101111001011011001101101011111101000010100000111000011100101011",
"output": "NO"
},
{
"input": "01110000110100110101110100111000101101011101011110110100100111100001110111",
"output": "NO"
},
{
"input": "11110110011000100111100111101101011111110100010101011011111101110110110111",
"output": "YES"
},
{
"input": "100100010101110010001011001110100011100010011110100101100011010001001010001001101111001100",
"output": "NO"
},
{
"input": "111110010001011010010011111100110110001111000010100011011100111101111101110010101111011110000001010",
"output": "NO"
},
{
"input": "111110111100010100000100001010111011101011000111011011011010110010100010000101011111000011010011110",
"output": "NO"
},
{
"input": "1011110110111010110111111010010010100011111011110000011000110010011110111010110100011010100010111000",
"output": "NO"
},
{
"input": "0010100111100010110110000011100111110100111110001010000100111111111010111100101101010101001011010110",
"output": "YES"
},
{
"input": "1111010100010100101011101100101101110011000010100010000001111100010011100101010001101111000001011000",
"output": "NO"
},
{
"input": "0101100011001110001110100111100011010101011000000000110110010010111100101111010111100011101100100101",
"output": "YES"
},
{
"input": "0001101110011101110000000010011111101001101111100001001010110000110001100000010001111011011110001101",
"output": "YES"
},
{
"input": "1000010000100000100010000100001000010000100001000010000100001000010000100001000010000100001000010000",
"output": "NO"
},
{
"input": "1000000000100000000010000000001000000000100000000010000000001000000000100000000010000000001000000000",
"output": "YES"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001",
"output": "YES"
},
{
"input": "1111111111111111111111111111111111111111011111111111111111111111111111111111111111111111111111111111",
"output": "YES"
},
{
"input": "10100101000",
"output": "NO"
},
{
"input": "11110111011101",
"output": "NO"
},
{
"input": "10000000",
"output": "YES"
},
{
"input": "00000001",
"output": "YES"
},
{
"input": "01111111",
"output": "YES"
},
{
"input": "11111110",
"output": "YES"
}
] | 1,691,335,390
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 62
| 0
|
s=input()
l=[]
c=0
d=0
for i in s:
l.append(i)
l.sort()
print(l)
for i in l:
if i=='0':
c+=1
else:
d+=1
print(c)
print(d)
if c>=7 or d>=7:
print("YES")
else:
print("NO")
|
Title: Football
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Petya loves football very much. One day, as he was watching a football match, he was writing the players' current positions on a piece of paper. To simplify the situation he depicted it as a string consisting of zeroes and ones. A zero corresponds to players of one team; a one corresponds to players of another team. If there are at least 7 players of some team standing one after another, then the situation is considered dangerous. For example, the situation 00100110111111101 is dangerous and 11110111011101 is not. You are given the current situation. Determine whether it is dangerous or not.
Input Specification:
The first input line contains a non-empty string consisting of characters "0" and "1", which represents players. The length of the string does not exceed 100 characters. There's at least one player from each team present on the field.
Output Specification:
Print "YES" if the situation is dangerous. Otherwise, print "NO".
Demo Input:
['001001\n', '1000000001\n']
Demo Output:
['NO\n', 'YES\n']
Note:
none
|
```python
s=input()
l=[]
c=0
d=0
for i in s:
l.append(i)
l.sort()
print(l)
for i in l:
if i=='0':
c+=1
else:
d+=1
print(c)
print(d)
if c>=7 or d>=7:
print("YES")
else:
print("NO")
```
| 0
|
817
|
C
|
Really Big Numbers
|
PROGRAMMING
| 1,600
|
[
"binary search",
"brute force",
"dp",
"math"
] | null | null |
Ivan likes to learn different things about numbers, but he is especially interested in really big numbers. Ivan thinks that a positive integer number *x* is really big if the difference between *x* and the sum of its digits (in decimal representation) is not less than *s*. To prove that these numbers may have different special properties, he wants to know how rare (or not rare) they are — in fact, he needs to calculate the quantity of really big numbers that are not greater than *n*.
Ivan tried to do the calculations himself, but soon realized that it's too difficult for him. So he asked you to help him in calculations.
|
The first (and the only) line contains two integers *n* and *s* (1<=≤<=*n*,<=*s*<=≤<=1018).
|
Print one integer — the quantity of really big numbers that are not greater than *n*.
|
[
"12 1\n",
"25 20\n",
"10 9\n"
] |
[
"3\n",
"0\n",
"1\n"
] |
In the first example numbers 10, 11 and 12 are really big.
In the second example there are no really big numbers that are not greater than 25 (in fact, the first really big number is 30: 30 - 3 ≥ 20).
In the third example 10 is the only really big number (10 - 1 ≥ 9).
| 0
|
[
{
"input": "12 1",
"output": "3"
},
{
"input": "25 20",
"output": "0"
},
{
"input": "10 9",
"output": "1"
},
{
"input": "300 1000",
"output": "0"
},
{
"input": "500 1000",
"output": "0"
},
{
"input": "1000 2000",
"output": "0"
},
{
"input": "10000 1000",
"output": "8991"
},
{
"input": "1000000000000000000 1000000000000000000",
"output": "0"
},
{
"input": "1000000000000000000 100000000000000000",
"output": "899999999999999991"
},
{
"input": "1000000000000000000 10000000000000000",
"output": "989999999999999991"
},
{
"input": "1000000000000000000 1000000000000000",
"output": "998999999999999991"
},
{
"input": "1000000000000000000 100000000000000",
"output": "999899999999999991"
},
{
"input": "1000000000000000000 200000000000000000",
"output": "799999999999999991"
},
{
"input": "10 5",
"output": "1"
},
{
"input": "20 5",
"output": "11"
},
{
"input": "20 9",
"output": "11"
},
{
"input": "100 9",
"output": "91"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "130 118",
"output": "1"
},
{
"input": "190 181",
"output": "0"
},
{
"input": "1999 1971",
"output": "10"
},
{
"input": "100 99",
"output": "1"
},
{
"input": "6909094398 719694282",
"output": "6189400069"
},
{
"input": "260 258",
"output": "0"
},
{
"input": "35 19",
"output": "6"
},
{
"input": "100 87",
"output": "1"
},
{
"input": "91 89",
"output": "0"
},
{
"input": "109 89",
"output": "10"
},
{
"input": "109 91",
"output": "10"
},
{
"input": "20331 11580",
"output": "8732"
},
{
"input": "405487470 255750281",
"output": "149737161"
},
{
"input": "17382 12863",
"output": "4493"
},
{
"input": "19725 14457",
"output": "5246"
},
{
"input": "24848 15384",
"output": "9449"
},
{
"input": "25727 15982",
"output": "9728"
},
{
"input": "109 90",
"output": "10"
},
{
"input": "1000000000000000000 999999999999999999",
"output": "1"
},
{
"input": "1000000000000000000 999999999999999998",
"output": "1"
},
{
"input": "1009 980",
"output": "10"
},
{
"input": "999999999999999999 999999999999999838",
"output": "0"
},
{
"input": "1000000000000000000 99999999999999800",
"output": "900000000000000061"
},
{
"input": "8785369357 3377262261",
"output": "5408107058"
},
{
"input": "110 109",
"output": "0"
},
{
"input": "999 777",
"output": "200"
},
{
"input": "327170000015578 77230000029054",
"output": "249939999986479"
},
{
"input": "12515000022229 1791000022317",
"output": "10723999999880"
},
{
"input": "9999999999999 9999999999882",
"output": "10"
},
{
"input": "213 196",
"output": "14"
},
{
"input": "92 82",
"output": "0"
},
{
"input": "148 136",
"output": "0"
},
{
"input": "8 9",
"output": "0"
},
{
"input": "309 299",
"output": "0"
},
{
"input": "9999 9963",
"output": "10"
},
{
"input": "82 81",
"output": "0"
},
{
"input": "9999999 9999936",
"output": "10"
},
{
"input": "171 155",
"output": "2"
},
{
"input": "999 972",
"output": "10"
},
{
"input": "999999999999 999999999891",
"output": "10"
},
{
"input": "9 9",
"output": "0"
},
{
"input": "6900 6885",
"output": "1"
},
{
"input": "96 57",
"output": "27"
},
{
"input": "5 4",
"output": "0"
},
{
"input": "17386 5814",
"output": "11557"
},
{
"input": "493679757404593 316259583979965",
"output": "177420173424564"
},
{
"input": "18474 9478",
"output": "8975"
},
{
"input": "270091571496186 250931112649966",
"output": "19160458846177"
},
{
"input": "565751690089037 381448507916936",
"output": "184303182172038"
},
{
"input": "19 10",
"output": "0"
}
] | 1,515,085,312
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 3
| 61
| 5,529,600
|
n,s=map(int,input().split())
def nod(n):
res=0
while n>0:
res+=1
n//=10
return res
def sod(n):
res=0
while n>0:
res+=n%10
n//=10
return res
def maxsod(n):
if n+1==10**nod(n): return sod(n)
else: return 9*(nod(n)-1)
num=n-s
for i in range(s+1,min(s+maxsod(n),n)+1):
if i-sod(i)<s:
num-=1
print(num)
|
Title: Really Big Numbers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Ivan likes to learn different things about numbers, but he is especially interested in really big numbers. Ivan thinks that a positive integer number *x* is really big if the difference between *x* and the sum of its digits (in decimal representation) is not less than *s*. To prove that these numbers may have different special properties, he wants to know how rare (or not rare) they are — in fact, he needs to calculate the quantity of really big numbers that are not greater than *n*.
Ivan tried to do the calculations himself, but soon realized that it's too difficult for him. So he asked you to help him in calculations.
Input Specification:
The first (and the only) line contains two integers *n* and *s* (1<=≤<=*n*,<=*s*<=≤<=1018).
Output Specification:
Print one integer — the quantity of really big numbers that are not greater than *n*.
Demo Input:
['12 1\n', '25 20\n', '10 9\n']
Demo Output:
['3\n', '0\n', '1\n']
Note:
In the first example numbers 10, 11 and 12 are really big.
In the second example there are no really big numbers that are not greater than 25 (in fact, the first really big number is 30: 30 - 3 ≥ 20).
In the third example 10 is the only really big number (10 - 1 ≥ 9).
|
```python
n,s=map(int,input().split())
def nod(n):
res=0
while n>0:
res+=1
n//=10
return res
def sod(n):
res=0
while n>0:
res+=n%10
n//=10
return res
def maxsod(n):
if n+1==10**nod(n): return sod(n)
else: return 9*(nod(n)-1)
num=n-s
for i in range(s+1,min(s+maxsod(n),n)+1):
if i-sod(i)<s:
num-=1
print(num)
```
| 0
|
|
616
|
C
|
The Labyrinth
|
PROGRAMMING
| 1,600
|
[
"dfs and similar"
] | null | null |
You are given a rectangular field of *n*<=×<=*m* cells. Each cell is either empty or impassable (contains an obstacle). Empty cells are marked with '.', impassable cells are marked with '*'. Let's call two empty cells adjacent if they share a side.
Let's call a connected component any non-extendible set of cells such that any two of them are connected by the path of adjacent cells. It is a typical well-known definition of a connected component.
For each impassable cell (*x*,<=*y*) imagine that it is an empty cell (all other cells remain unchanged) and find the size (the number of cells) of the connected component which contains (*x*,<=*y*). You should do it for each impassable cell independently.
The answer should be printed as a matrix with *n* rows and *m* columns. The *j*-th symbol of the *i*-th row should be "." if the cell is empty at the start. Otherwise the *j*-th symbol of the *i*-th row should contain the only digit —- the answer modulo 10. The matrix should be printed without any spaces.
To make your output faster it is recommended to build the output as an array of *n* strings having length *m* and print it as a sequence of lines. It will be much faster than writing character-by-character.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
|
The first line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and columns in the field.
Each of the next *n* lines contains *m* symbols: "." for empty cells, "*" for impassable cells.
|
Print the answer as a matrix as described above. See the examples to precise the format of the output.
|
[
"3 3\n*.*\n.*.\n*.*\n",
"4 5\n**..*\n..***\n.*.*.\n*.*.*\n"
] |
[
"3.3\n.5.\n3.3\n",
"46..3\n..732\n.6.4.\n5.4.3\n"
] |
In first example, if we imagine that the central cell is empty then it will be included to component of size 5 (cross). If any of the corner cell will be empty then it will be included to component of size 3 (corner).
| 0
|
[
{
"input": "3 3\n*.*\n.*.\n*.*",
"output": "3.3\n.5.\n3.3"
},
{
"input": "4 5\n**..*\n..***\n.*.*.\n*.*.*",
"output": "46..3\n..732\n.6.4.\n5.4.3"
},
{
"input": "1 1\n*",
"output": "1"
},
{
"input": "1 1\n.",
"output": "."
},
{
"input": "1 10\n**********",
"output": "1111111111"
},
{
"input": "1 10\n*.***.**.*",
"output": "2.212.22.2"
},
{
"input": "10 1\n*\n*\n*\n*\n*\n.\n*\n.\n*\n*",
"output": "1\n1\n1\n1\n2\n.\n3\n.\n2\n1"
},
{
"input": "10 1\n.\n.\n.\n.\n.\n.\n.\n.\n.\n.",
"output": ".\n.\n.\n.\n.\n.\n.\n.\n.\n."
}
] | 1,642,674,659
| 2,147,483,647
|
PyPy 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 13
| 1,000
| 66,764,800
|
from collections import Counter
import sys
input = sys.stdin.readline
n, m = map(int, input().split())
grid = [[j for j in input()] for i in range(n)]
d = Counter()
def dfs(x, y, g):
grid[x][y] = g
stk, cnt = [(x, y)], 0
while stk:
x, y = stk.pop()
cnt += 1
for dx, dy in (x - 1, y), (x, y + 1), (x + 1, y), (x, y - 1):
if dx < 0 or dx >= n or dy < 0 or dy >= m or grid[dx][dy] == '*':
continue
elif grid[dx][dy] == '.':
grid[dx][dy] = g
stk.append((dx, dy))
d[g] = cnt
g = 0
for x in range(m):
for y in range(n):
if grid[y][x] == ".":
dfs(y, x, g)
g += 1
for y in range(n):
line = []
for x in range(m):
if grid[y][x] == "*":
neibs = set()
for dx, dy in (x - 1, y), (x, y + 1), (x + 1, y), (x, y - 1):
if 0 <= dx < m and 0 <= dy < n:
neibs.add(grid[dy][dx])
line += [str((sum(d[i] for i in neibs) + 1) % 10)]
else:
line += ["."]
print("".join(line))
|
Title: The Labyrinth
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a rectangular field of *n*<=×<=*m* cells. Each cell is either empty or impassable (contains an obstacle). Empty cells are marked with '.', impassable cells are marked with '*'. Let's call two empty cells adjacent if they share a side.
Let's call a connected component any non-extendible set of cells such that any two of them are connected by the path of adjacent cells. It is a typical well-known definition of a connected component.
For each impassable cell (*x*,<=*y*) imagine that it is an empty cell (all other cells remain unchanged) and find the size (the number of cells) of the connected component which contains (*x*,<=*y*). You should do it for each impassable cell independently.
The answer should be printed as a matrix with *n* rows and *m* columns. The *j*-th symbol of the *i*-th row should be "." if the cell is empty at the start. Otherwise the *j*-th symbol of the *i*-th row should contain the only digit —- the answer modulo 10. The matrix should be printed without any spaces.
To make your output faster it is recommended to build the output as an array of *n* strings having length *m* and print it as a sequence of lines. It will be much faster than writing character-by-character.
As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.
Input Specification:
The first line contains two integers *n*,<=*m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and columns in the field.
Each of the next *n* lines contains *m* symbols: "." for empty cells, "*" for impassable cells.
Output Specification:
Print the answer as a matrix as described above. See the examples to precise the format of the output.
Demo Input:
['3 3\n*.*\n.*.\n*.*\n', '4 5\n**..*\n..***\n.*.*.\n*.*.*\n']
Demo Output:
['3.3\n.5.\n3.3\n', '46..3\n..732\n.6.4.\n5.4.3\n']
Note:
In first example, if we imagine that the central cell is empty then it will be included to component of size 5 (cross). If any of the corner cell will be empty then it will be included to component of size 3 (corner).
|
```python
from collections import Counter
import sys
input = sys.stdin.readline
n, m = map(int, input().split())
grid = [[j for j in input()] for i in range(n)]
d = Counter()
def dfs(x, y, g):
grid[x][y] = g
stk, cnt = [(x, y)], 0
while stk:
x, y = stk.pop()
cnt += 1
for dx, dy in (x - 1, y), (x, y + 1), (x + 1, y), (x, y - 1):
if dx < 0 or dx >= n or dy < 0 or dy >= m or grid[dx][dy] == '*':
continue
elif grid[dx][dy] == '.':
grid[dx][dy] = g
stk.append((dx, dy))
d[g] = cnt
g = 0
for x in range(m):
for y in range(n):
if grid[y][x] == ".":
dfs(y, x, g)
g += 1
for y in range(n):
line = []
for x in range(m):
if grid[y][x] == "*":
neibs = set()
for dx, dy in (x - 1, y), (x, y + 1), (x + 1, y), (x, y - 1):
if 0 <= dx < m and 0 <= dy < n:
neibs.add(grid[dy][dx])
line += [str((sum(d[i] for i in neibs) + 1) % 10)]
else:
line += ["."]
print("".join(line))
```
| 0
|
|
43
|
A
|
Football
|
PROGRAMMING
| 1,000
|
[
"strings"
] |
A. Football
|
2
|
256
|
One day Vasya decided to have a look at the results of Berland 1910 Football Championship’s finals. Unfortunately he didn't find the overall score of the match; however, he got hold of a profound description of the match's process. On the whole there are *n* lines in that description each of which described one goal. Every goal was marked with the name of the team that had scored it. Help Vasya, learn the name of the team that won the finals. It is guaranteed that the match did not end in a tie.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of lines in the description. Then follow *n* lines — for each goal the names of the teams that scored it. The names are non-empty lines consisting of uppercase Latin letters whose lengths do not exceed 10 symbols. It is guaranteed that the match did not end in a tie and the description contains no more than two different teams.
|
Print the name of the winning team. We remind you that in football the team that scores more goals is considered the winner.
|
[
"1\nABC\n",
"5\nA\nABA\nABA\nA\nA\n"
] |
[
"ABC\n",
"A\n"
] |
none
| 500
|
[
{
"input": "1\nABC",
"output": "ABC"
},
{
"input": "5\nA\nABA\nABA\nA\nA",
"output": "A"
},
{
"input": "2\nXTSJEP\nXTSJEP",
"output": "XTSJEP"
},
{
"input": "3\nXZYDJAEDZ\nXZYDJAEDZ\nXZYDJAEDZ",
"output": "XZYDJAEDZ"
},
{
"input": "3\nQCCYXL\nQCCYXL\nAXGLFQDD",
"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"
}
] | 1,611,666,342
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 154
| 409,600
|
from collections import Counter, OrderedDict
import operator
lines = int(input())
goals = []
for k in range(lines):
goals.append(input())
dict1 = (Counter(goals))
sorted_tuples = sorted(dict1.items(), key=operator.itemgetter(1), reverse=True)
sorted_dict = OrderedDict()
for k, v in sorted_tuples:
sorted_dict[k] = v
print(list(sorted_dict.keys())[0]) # {1: 1, 3: 4, 2: 9}
|
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
from collections import Counter, OrderedDict
import operator
lines = int(input())
goals = []
for k in range(lines):
goals.append(input())
dict1 = (Counter(goals))
sorted_tuples = sorted(dict1.items(), key=operator.itemgetter(1), reverse=True)
sorted_dict = OrderedDict()
for k, v in sorted_tuples:
sorted_dict[k] = v
print(list(sorted_dict.keys())[0]) # {1: 1, 3: 4, 2: 9}
```
| 3.960737
|
81
|
A
|
Plug-in
|
PROGRAMMING
| 1,400
|
[
"implementation"
] |
A. Plug-in
|
1
|
256
|
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
|
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
|
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
|
[
"hhoowaaaareyyoouu\n",
"reallazy\n",
"abacabaabacabaa\n"
] |
[
"wre",
"rezy",
"a"
] |
none
| 500
|
[
{
"input": "hhoowaaaareyyoouu",
"output": "wre"
},
{
"input": "reallazy",
"output": "rezy"
},
{
"input": "abacabaabacabaa",
"output": "a"
},
{
"input": "xraccabccbry",
"output": "xy"
},
{
"input": "a",
"output": "a"
},
{
"input": "b",
"output": "b"
},
{
"input": "il",
"output": "il"
},
{
"input": "gfj",
"output": "gfj"
},
{
"input": "babbbbbababa",
"output": "babababa"
},
{
"input": "babbbbabbabbbababbabbbbbbabaabaababaaabbbbbabbbbaaaaabbaaabbaabaabbbbabbbababbabaaabbababaaababbbaaa",
"output": "babababababababababababa"
},
{
"input": "aab",
"output": "b"
},
{
"input": "abb",
"output": "a"
},
{
"input": "aba",
"output": "aba"
},
{
"input": "ab",
"output": "ab"
}
] | 1,659,399,056
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 32
| 186
| 15,667,200
|
import sys
input = sys.stdin.readline
s = input()[:-1]
d = []
for i in s:
if d and i == d[-1]:
d.pop()
else:
d.append(i)
print(''.join(d))
|
Title: Plug-in
Time Limit: 1 seconds
Memory Limit: 256 megabytes
Problem Description:
Polycarp thinks about the meaning of life very often. He does this constantly, even when typing in the editor. Every time he starts brooding he can no longer fully concentrate and repeatedly presses the keys that need to be pressed only once. For example, instead of the phrase "how are you" he can type "hhoow aaaare yyoouu".
Polycarp decided to automate the process of correcting such errors. He decided to write a plug-in to the text editor that will remove pairs of identical consecutive letters (if there are any in the text). Of course, this is not exactly what Polycarp needs, but he's got to start from something!
Help Polycarp and write the main plug-in module. Your program should remove from a string all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, the program should remove them as well. Technically, its work should be equivalent to the following: while the string contains a pair of consecutive identical letters, the pair should be deleted. Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.
Input Specification:
The input data consists of a single line to be processed. The length of the line is from 1 to 2·105 characters inclusive. The string contains only lowercase Latin letters.
Output Specification:
Print the given string after it is processed. It is guaranteed that the result will contain at least one character.
Demo Input:
['hhoowaaaareyyoouu\n', 'reallazy\n', 'abacabaabacabaa\n']
Demo Output:
['wre', 'rezy', 'a']
Note:
none
|
```python
import sys
input = sys.stdin.readline
s = input()[:-1]
d = []
for i in s:
if d and i == d[-1]:
d.pop()
else:
d.append(i)
print(''.join(d))
```
| 3.877818
|
510
|
A
|
Fox And Snake
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern.
|
The only line contains two integers: *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=50).
*n* is an odd number.
|
Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces.
|
[
"3 3\n",
"3 4\n",
"5 3\n",
"9 9\n"
] |
[
"###\n..#\n###\n",
"####\n...#\n####\n",
"###\n..#\n###\n#..\n###\n",
"#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n"
] |
none
| 500
|
[
{
"input": "3 3",
"output": "###\n..#\n###"
},
{
"input": "3 4",
"output": "####\n...#\n####"
},
{
"input": "5 3",
"output": "###\n..#\n###\n#..\n###"
},
{
"input": "9 9",
"output": "#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########"
},
{
"input": "3 5",
"output": "#####\n....#\n#####"
},
{
"input": "3 6",
"output": "######\n.....#\n######"
},
{
"input": "7 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###"
},
{
"input": "7 4",
"output": "####\n...#\n####\n#...\n####\n...#\n####"
},
{
"input": "49 50",
"output": "##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.............................................."
},
{
"input": "43 50",
"output": "##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.............................................."
},
{
"input": "43 27",
"output": "###########################\n..........................#\n###########################\n#..........................\n###########################\n..........................#\n###########################\n#..........................\n###########################\n..........................#\n###########################\n#..........................\n###########################\n..........................#\n###########################\n#..........................\n###########################\n....................."
},
{
"input": "11 15",
"output": "###############\n..............#\n###############\n#..............\n###############\n..............#\n###############\n#..............\n###############\n..............#\n###############"
},
{
"input": "11 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###"
},
{
"input": "19 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###"
},
{
"input": "23 50",
"output": "##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.................................................#\n##################################################\n#.................................................\n##################################################\n.............................................."
},
{
"input": "49 49",
"output": "#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#..."
},
{
"input": "33 43",
"output": "###########################################\n..........................................#\n###########################################\n#..........................................\n###########################################\n..........................................#\n###########################################\n#..........................................\n###########################################\n..........................................#\n###########################################\n#.................."
},
{
"input": "33 44",
"output": "############################################\n...........................................#\n############################################\n#...........................................\n############################################\n...........................................#\n############################################\n#...........................................\n############################################\n...........................................#\n############################################\n#......."
},
{
"input": "45 45",
"output": "#############################################\n............................................#\n#############################################\n#............................................\n#############################################\n............................................#\n#############################################\n#............................................\n#############################################\n............................................#\n#########################################..."
},
{
"input": "45 49",
"output": "#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#################################################\n#................................................\n#################################################\n................................................#\n#..."
},
{
"input": "45 3",
"output": "###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###\n..#\n###\n#..\n###"
},
{
"input": "45 4",
"output": "####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####\n...#\n####\n#...\n####"
},
{
"input": "5 4",
"output": "####\n...#\n####\n#...\n####"
},
{
"input": "5 5",
"output": "#####\n....#\n#####\n#....\n#####"
},
{
"input": "5 6",
"output": "######\n.....#\n######\n#.....\n######"
},
{
"input": "7 5",
"output": "#####\n....#\n#####\n#....\n#####\n....#\n#####"
},
{
"input": "7 6",
"output": "######\n.....#\n######\n#.....\n######\n.....#\n######"
},
{
"input": "7 7",
"output": "#######\n......#\n#######\n#......\n#######\n......#\n#######"
}
] | 1,688,670,661
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 28
| 78
| 4,096,000
|
a,b=map(int,input().split())
c=0
for i in range(a):
for j in range(b):
if i%2==0:
print("#",end="")
else:
if i % 4 == 1 and j == b - 1:
print("#", end="")
elif i % 4 == 3 and j == 0:
print("#", end="")
else:
print(".", end="")
print()
|
Title: Fox And Snake
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel starts to learn programming. The first task is drawing a fox! However, that turns out to be too hard for a beginner, so she decides to draw a snake instead.
A snake is a pattern on a *n* by *m* table. Denote *c*-th cell of *r*-th row as (*r*,<=*c*). The tail of the snake is located at (1,<=1), then it's body extends to (1,<=*m*), then goes down 2 rows to (3,<=*m*), then goes left to (3,<=1) and so on.
Your task is to draw this snake for Fox Ciel: the empty cells should be represented as dot characters ('.') and the snake cells should be filled with number signs ('#').
Consider sample tests in order to understand the snake pattern.
Input Specification:
The only line contains two integers: *n* and *m* (3<=≤<=*n*,<=*m*<=≤<=50).
*n* is an odd number.
Output Specification:
Output *n* lines. Each line should contain a string consisting of *m* characters. Do not output spaces.
Demo Input:
['3 3\n', '3 4\n', '5 3\n', '9 9\n']
Demo Output:
['###\n..#\n###\n', '####\n...#\n####\n', '###\n..#\n###\n#..\n###\n', '#########\n........#\n#########\n#........\n#########\n........#\n#########\n#........\n#########\n']
Note:
none
|
```python
a,b=map(int,input().split())
c=0
for i in range(a):
for j in range(b):
if i%2==0:
print("#",end="")
else:
if i % 4 == 1 and j == b - 1:
print("#", end="")
elif i % 4 == 3 and j == 0:
print("#", end="")
else:
print(".", end="")
print()
```
| 3
|
|
639
|
A
|
Bear and Displayed Friends
|
PROGRAMMING
| 1,200
|
[
"implementation"
] | null | null |
Limak is a little polar bear. He loves connecting with other bears via social networks. He has *n* friends and his relation with the *i*-th of them is described by a unique integer *t**i*. The bigger this value is, the better the friendship is. No two friends have the same value *t**i*.
Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.
The system displays friends who are online. On the screen there is space to display at most *k* friends. If there are more than *k* friends online then the system displays only *k* best of them — those with biggest *t**i*.
Your task is to handle queries of two types:
- "1 id" — Friend *id* becomes online. It's guaranteed that he wasn't online before. - "2 id" — Check whether friend *id* is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?
|
The first line contains three integers *n*, *k* and *q* (1<=≤<=*n*,<=*q*<=≤<=150<=000,<=1<=≤<=*k*<=≤<=*min*(6,<=*n*)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=109) where *t**i* describes how good is Limak's relation with the *i*-th friend.
The *i*-th of the following *q* lines contains two integers *type**i* and *id**i* (1<=≤<=*type**i*<=≤<=2,<=1<=≤<=*id**i*<=≤<=*n*) — the *i*-th query. If *type**i*<==<=1 then a friend *id**i* becomes online. If *type**i*<==<=2 then you should check whether a friend *id**i* is displayed.
It's guaranteed that no two queries of the first type will have the same *id**i* becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (*type**i*<==<=2) so the output won't be empty.
|
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
|
[
"4 2 8\n300 950 500 200\n1 3\n2 4\n2 3\n1 1\n1 2\n2 1\n2 2\n2 3\n",
"6 3 9\n50 20 51 17 99 24\n1 3\n1 4\n1 5\n1 2\n2 4\n2 2\n1 1\n2 4\n2 3\n"
] |
[
"NO\nYES\nNO\nYES\nYES\n",
"NO\nYES\nNO\nYES\n"
] |
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries:
1. "1 3" — Friend 3 becomes online. 1. "2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO". 1. "2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES". 1. "1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3. 1. "1 2" — Friend 2 becomes online. There are 3 friends online now but we were given *k* = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (*t*<sub class="lower-index">1</sub> < *t*<sub class="lower-index">2</sub>, *t*<sub class="lower-index">3</sub>) so friend 1 won't be displayed 1. "2 1" — Print "NO". 1. "2 2" — Print "YES". 1. "2 3" — Print "YES".
| 500
|
[
{
"input": "4 2 8\n300 950 500 200\n1 3\n2 4\n2 3\n1 1\n1 2\n2 1\n2 2\n2 3",
"output": "NO\nYES\nNO\nYES\nYES"
},
{
"input": "6 3 9\n50 20 51 17 99 24\n1 3\n1 4\n1 5\n1 2\n2 4\n2 2\n1 1\n2 4\n2 3",
"output": "NO\nYES\nNO\nYES"
},
{
"input": "6 3 10\n62417580 78150524 410053501 582708235 630200761 760672946\n2 2\n1 5\n1 2\n1 4\n2 4\n2 1\n2 1\n1 6\n2 5\n2 6",
"output": "NO\nYES\nNO\nNO\nYES\nYES"
},
{
"input": "20 2 15\n12698951 55128070 116962690 156763505 188535242 194018601 269939893 428710623 442819431 483000923 516768937 552903993 633087286 656092270 671535141 714291344 717660646 846508634 879748146 937368929\n2 7\n1 2\n2 4\n1 19\n1 12\n1 5\n2 18\n2 11\n1 16\n2 1\n2 3\n2 19\n1 17\n2 9\n2 6",
"output": "NO\nNO\nNO\nNO\nNO\nNO\nYES\nNO\nNO"
},
{
"input": "1 1 1\n1000000000\n2 1",
"output": "NO"
}
] | 1,459,183,765
| 865
|
PyPy 3
|
RUNTIME_ERROR
|
PRETESTS
| 0
| 93
| 23,040,000
|
n, k, q = [int(x) for x in input().split()]
T = [int(x) for x in input().split()]
S = set()
for i in range(q):
t, _id = [int(x) for x in input().split()]
if t == 1:
m = min(S, default = 0)
if len(S) == k:
if m < T[_id-1]:
S.remove(m)
S.add(T[_id-1])
else:
S.add(T[_id-1])
else:
print('YES' if T[_id-1] in S else 'NO')
|
Title: Bear and Displayed Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Limak is a little polar bear. He loves connecting with other bears via social networks. He has *n* friends and his relation with the *i*-th of them is described by a unique integer *t**i*. The bigger this value is, the better the friendship is. No two friends have the same value *t**i*.
Spring is starting and the Winter sleep is over for bears. Limak has just woken up and logged in. All his friends still sleep and thus none of them is online. Some (maybe all) of them will appear online in the next hours, one at a time.
The system displays friends who are online. On the screen there is space to display at most *k* friends. If there are more than *k* friends online then the system displays only *k* best of them — those with biggest *t**i*.
Your task is to handle queries of two types:
- "1 id" — Friend *id* becomes online. It's guaranteed that he wasn't online before. - "2 id" — Check whether friend *id* is displayed by the system. Print "YES" or "NO" in a separate line.
Are you able to help Limak and answer all queries of the second type?
Input Specification:
The first line contains three integers *n*, *k* and *q* (1<=≤<=*n*,<=*q*<=≤<=150<=000,<=1<=≤<=*k*<=≤<=*min*(6,<=*n*)) — the number of friends, the maximum number of displayed online friends and the number of queries, respectively.
The second line contains *n* integers *t*1,<=*t*2,<=...,<=*t**n* (1<=≤<=*t**i*<=≤<=109) where *t**i* describes how good is Limak's relation with the *i*-th friend.
The *i*-th of the following *q* lines contains two integers *type**i* and *id**i* (1<=≤<=*type**i*<=≤<=2,<=1<=≤<=*id**i*<=≤<=*n*) — the *i*-th query. If *type**i*<==<=1 then a friend *id**i* becomes online. If *type**i*<==<=2 then you should check whether a friend *id**i* is displayed.
It's guaranteed that no two queries of the first type will have the same *id**i* becuase one friend can't become online twice. Also, it's guaranteed that at least one query will be of the second type (*type**i*<==<=2) so the output won't be empty.
Output Specification:
For each query of the second type print one line with the answer — "YES" (without quotes) if the given friend is displayed and "NO" (without quotes) otherwise.
Demo Input:
['4 2 8\n300 950 500 200\n1 3\n2 4\n2 3\n1 1\n1 2\n2 1\n2 2\n2 3\n', '6 3 9\n50 20 51 17 99 24\n1 3\n1 4\n1 5\n1 2\n2 4\n2 2\n1 1\n2 4\n2 3\n']
Demo Output:
['NO\nYES\nNO\nYES\nYES\n', 'NO\nYES\nNO\nYES\n']
Note:
In the first sample, Limak has 4 friends who all sleep initially. At first, the system displays nobody because nobody is online. There are the following 8 queries:
1. "1 3" — Friend 3 becomes online. 1. "2 4" — We should check if friend 4 is displayed. He isn't even online and thus we print "NO". 1. "2 3" — We should check if friend 3 is displayed. Right now he is the only friend online and the system displays him. We should print "YES". 1. "1 1" — Friend 1 becomes online. The system now displays both friend 1 and friend 3. 1. "1 2" — Friend 2 becomes online. There are 3 friends online now but we were given *k* = 2 so only two friends can be displayed. Limak has worse relation with friend 1 than with other two online friends (*t*<sub class="lower-index">1</sub> < *t*<sub class="lower-index">2</sub>, *t*<sub class="lower-index">3</sub>) so friend 1 won't be displayed 1. "2 1" — Print "NO". 1. "2 2" — Print "YES". 1. "2 3" — Print "YES".
|
```python
n, k, q = [int(x) for x in input().split()]
T = [int(x) for x in input().split()]
S = set()
for i in range(q):
t, _id = [int(x) for x in input().split()]
if t == 1:
m = min(S, default = 0)
if len(S) == k:
if m < T[_id-1]:
S.remove(m)
S.add(T[_id-1])
else:
S.add(T[_id-1])
else:
print('YES' if T[_id-1] in S else 'NO')
```
| -1
|
|
381
|
A
|
Sereja and Dima
|
PROGRAMMING
| 800
|
[
"greedy",
"implementation",
"two pointers"
] | null | null |
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
|
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
|
[
"4\n4 1 2 10\n",
"7\n1 2 3 4 5 6 7\n"
] |
[
"12 5\n",
"16 12\n"
] |
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.
| 500
|
[
{
"input": "4\n4 1 2 10",
"output": "12 5"
},
{
"input": "7\n1 2 3 4 5 6 7",
"output": "16 12"
},
{
"input": "42\n15 29 37 22 16 5 26 31 6 32 19 3 45 36 33 14 25 20 48 7 42 11 24 28 9 18 8 21 47 17 38 40 44 4 35 1 43 39 41 27 12 13",
"output": "613 418"
},
{
"input": "43\n32 1 15 48 38 26 25 14 20 44 11 30 3 42 49 19 18 46 5 45 10 23 34 9 29 41 2 52 6 17 35 4 50 22 33 51 7 28 47 13 39 37 24",
"output": "644 500"
},
{
"input": "1\n3",
"output": "3 0"
},
{
"input": "45\n553 40 94 225 415 471 126 190 647 394 515 303 189 159 308 6 139 132 326 78 455 75 85 295 135 613 360 614 351 228 578 259 258 591 444 29 33 463 561 174 368 183 140 168 646",
"output": "6848 6568"
},
{
"input": "44\n849 373 112 307 479 608 856 769 526 82 168 143 573 762 115 501 688 36 214 450 396 496 236 309 287 786 397 43 811 141 745 846 350 270 276 677 420 459 403 722 267 54 394 727",
"output": "9562 9561"
},
{
"input": "35\n10 15 18 1 28 16 2 33 6 22 23 4 9 25 35 8 7 26 3 20 30 14 31 19 27 32 11 5 29 24 21 34 13 17 12",
"output": "315 315"
},
{
"input": "17\n580 376 191 496 73 44 520 357 483 149 81 178 514 300 216 598 304",
"output": "3238 2222"
},
{
"input": "30\n334 443 223 424 168 549 189 303 429 559 516 220 459 134 344 346 316 446 209 148 487 526 69 286 102 366 518 280 392 325",
"output": "5246 4864"
},
{
"input": "95\n122 29 188 265 292 287 183 225 222 187 155 256 64 148 173 278 218 136 290 17 31 130 2 87 57 283 255 280 68 166 174 142 102 39 116 206 288 154 26 78 296 172 184 232 77 91 277 8 249 186 94 93 207 251 257 195 101 299 193 124 293 65 58 35 24 302 220 189 252 125 27 284 247 182 141 103 198 97 234 83 281 216 85 180 267 236 109 143 149 239 79 300 191 244 71",
"output": "8147 7807"
},
{
"input": "1\n1",
"output": "1 0"
}
] | 1,694,013,192
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 34
| 46
| 0
|
n = int(input()) # Number of cards
cards = list(map(int, input().split())) # List of card values
# Initialize Sereja's and Dima's scores
sereja_score = 0
dima_score = 0
# Initialize pointers for the leftmost and rightmost cards
left = 0
right = n - 1
# Initialize a variable to keep track of the current player (Sereja or Dima)
current_player = 1 # 1 for Sereja, 2 for Dima
while left <= right:
if cards[left] >= cards[right]:
# Sereja's turn
if current_player == 1:
sereja_score += cards[left]
else:
dima_score += cards[left]
left += 1
else:
# Sereja's turn
if current_player == 1:
sereja_score += cards[right]
else:
dima_score += cards[right]
right -= 1
# Switch to the other player's turn
current_player = 3 - current_player
# Print the final scores
print(sereja_score, dima_score)
|
Title: Sereja and Dima
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Sereja and Dima play a game. The rules of the game are very simple. The players have *n* cards in a row. Each card contains a number, all numbers on the cards are distinct. The players take turns, Sereja moves first. During his turn a player can take one card: either the leftmost card in a row, or the rightmost one. The game ends when there is no more cards. The player who has the maximum sum of numbers on his cards by the end of the game, wins.
Sereja and Dima are being greedy. Each of them chooses the card with the larger number during his move.
Inna is a friend of Sereja and Dima. She knows which strategy the guys are using, so she wants to determine the final score, given the initial state of the game. Help her.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=1000) — the number of cards on the table. The second line contains space-separated numbers on the cards from left to right. The numbers on the cards are distinct integers from 1 to 1000.
Output Specification:
On a single line, print two integers. The first number is the number of Sereja's points at the end of the game, the second number is the number of Dima's points at the end of the game.
Demo Input:
['4\n4 1 2 10\n', '7\n1 2 3 4 5 6 7\n']
Demo Output:
['12 5\n', '16 12\n']
Note:
In the first sample Sereja will take cards with numbers 10 and 2, so Sereja's sum is 12. Dima will take cards with numbers 4 and 1, so Dima's sum is 5.
|
```python
n = int(input()) # Number of cards
cards = list(map(int, input().split())) # List of card values
# Initialize Sereja's and Dima's scores
sereja_score = 0
dima_score = 0
# Initialize pointers for the leftmost and rightmost cards
left = 0
right = n - 1
# Initialize a variable to keep track of the current player (Sereja or Dima)
current_player = 1 # 1 for Sereja, 2 for Dima
while left <= right:
if cards[left] >= cards[right]:
# Sereja's turn
if current_player == 1:
sereja_score += cards[left]
else:
dima_score += cards[left]
left += 1
else:
# Sereja's turn
if current_player == 1:
sereja_score += cards[right]
else:
dima_score += cards[right]
right -= 1
# Switch to the other player's turn
current_player = 3 - current_player
# Print the final scores
print(sereja_score, dima_score)
```
| 3
|
|
71
|
A
|
Way Too Long Words
|
PROGRAMMING
| 800
|
[
"strings"
] |
A. Way Too Long Words
|
1
|
256
|
Sometimes some words like "localization" or "internationalization" are so long that writing them many times in one text is quite tiresome.
Let's consider a word too long, if its length is strictly more than 10 characters. All too long words should be replaced with a special abbreviation.
This abbreviation is made like this: we write down the first and the last letter of a word and between them we write the number of letters between the first and the last letters. That number is in decimal system and doesn't contain any leading zeroes.
Thus, "localization" will be spelt as "l10n", and "internationalization» will be spelt as "i18n".
You are suggested to automatize the process of changing the words with abbreviations. At that all too long words should be replaced by the abbreviation and the words that are not too long should not undergo any changes.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100). Each of the following *n* lines contains one word. All the words consist of lowercase Latin letters and possess the lengths of from 1 to 100 characters.
|
Print *n* lines. The *i*-th line should contain the result of replacing of the *i*-th word from the input data.
|
[
"4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis\n"
] |
[
"word\nl10n\ni18n\np43s\n"
] |
none
| 500
|
[
{
"input": "4\nword\nlocalization\ninternationalization\npneumonoultramicroscopicsilicovolcanoconiosis",
"output": "word\nl10n\ni18n\np43s"
},
{
"input": "5\nabcdefgh\nabcdefghi\nabcdefghij\nabcdefghijk\nabcdefghijklm",
"output": "abcdefgh\nabcdefghi\nabcdefghij\na9k\na11m"
},
{
"input": "3\nnjfngnrurunrgunrunvurn\njfvnjfdnvjdbfvsbdubruvbubvkdb\nksdnvidnviudbvibd",
"output": "n20n\nj27b\nk15d"
},
{
"input": "1\ntcyctkktcctrcyvbyiuhihhhgyvyvyvyvjvytchjckt",
"output": "t41t"
},
{
"input": "24\nyou\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nunofficially\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings",
"output": "you\nare\nregistered\nfor\npractice\nyou\ncan\nsolve\nproblems\nu10y\nresults\ncan\nbe\nfound\nin\nthe\ncontest\nstatus\nand\nin\nthe\nbottom\nof\nstandings"
},
{
"input": "1\na",
"output": "a"
},
{
"input": "26\na\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz",
"output": "a\nb\nc\nd\ne\nf\ng\nh\ni\nj\nk\nl\nm\nn\no\np\nq\nr\ns\nt\nu\nv\nw\nx\ny\nz"
},
{
"input": "1\nabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghijabcdefghij",
"output": "a98j"
},
{
"input": "10\ngyartjdxxlcl\nfzsck\nuidwu\nxbymclornemdmtj\nilppyoapitawgje\ncibzc\ndrgbeu\nhezplmsdekhhbo\nfeuzlrimbqbytdu\nkgdco",
"output": "g10l\nfzsck\nuidwu\nx13j\ni13e\ncibzc\ndrgbeu\nh12o\nf13u\nkgdco"
},
{
"input": "20\nlkpmx\nkovxmxorlgwaomlswjxlpnbvltfv\nhykasjxqyjrmybejnmeumzha\ntuevlumpqbbhbww\nqgqsphvrmupxxc\ntrissbaf\nqfgrlinkzvzqdryckaizutd\nzzqtoaxkvwoscyx\noswytrlnhpjvvnwookx\nlpuzqgec\ngyzqfwxggtvpjhzmzmdw\nrlxjgmvdftvrmvbdwudra\nvsntnjpepnvdaxiporggmglhagv\nxlvcqkqgcrbgtgglj\nlyxwxbiszyhlsrgzeedzprbmcpduvq\nyrmqqvrkqskqukzqrwukpsifgtdc\nxpuohcsjhhuhvr\nvvlfrlxpvqejngwrbfbpmqeirxlw\nsvmasocxdvadmaxtrpakysmeaympy\nyuflqboqfdt",
"output": "lkpmx\nk26v\nh22a\nt13w\nq12c\ntrissbaf\nq21d\nz13x\no17x\nlpuzqgec\ng18w\nr19a\nv25v\nx15j\nl28q\ny26c\nx12r\nv26w\ns27y\ny9t"
},
{
"input": "100\nm\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr",
"output": "m\nz\ns\nv\nd\nr\nv\ny\ny\ne\np\nt\nc\na\nn\nm\np\ng\ni\nj\nc\na\nb\nq\ne\nn\nv\no\nk\nx\nf\ni\nl\na\nq\nr\nu\nb\ns\nl\nc\nl\ne\nv\nj\nm\nx\nb\na\nq\nb\na\nf\nj\nv\nm\nq\nc\nt\nt\nn\nx\no\ny\nr\nu\nh\nm\nj\np\nj\nq\nz\ns\nj\no\ng\nc\nm\nn\no\nm\nr\no\ns\nt\nh\nr\np\nk\nb\nz\ng\no\nc\nc\nz\nz\ng\nr"
}
] | 1,681,166,725
| 2,147,483,647
|
PyPy 3-64
|
RUNTIME_ERROR
|
TESTS
| 0
| 77
| 2,867,200
|
n = int(input())
w = str(input()).split()
dummy = [print(w[i][0], len(w[i])-2, w[i][-1]) if len(w[i])>10 else print(w[i]) for i in range(n)]
|
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
n = int(input())
w = str(input()).split()
dummy = [print(w[i][0], len(w[i])-2, w[i][-1]) if len(w[i])>10 else print(w[i]) for i in range(n)]
```
| -1
|
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"
}
] | 1,697,225,345
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
#fff
n=int(input())
for i in range(n):
word=input()
l=len(word)
l1=len(word[1:l-1])
if(l>10):
word=word[0]+str(l1)+word[0]
print(word)
else:
print(word)
|
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
#fff
n=int(input())
for i in range(n):
word=input()
l=len(word)
l1=len(word[1:l-1])
if(l>10):
word=word[0]+str(l1)+word[0]
print(word)
else:
print(word)
```
| 0
|
592
|
C
|
The Big Race
|
PROGRAMMING
| 1,800
|
[
"math"
] | null | null |
Vector Willman and Array Bolt are the two most famous athletes of Byteforces. They are going to compete in a race with a distance of *L* meters today.
Willman and Bolt have exactly the same speed, so when they compete the result is always a tie. That is a problem for the organizers because they want a winner.
While watching previous races the organizers have noticed that Willman can perform only steps of length equal to *w* meters, and Bolt can perform only steps of length equal to *b* meters. Organizers decided to slightly change the rules of the race. Now, at the end of the racetrack there will be an abyss, and the winner will be declared the athlete, who manages to run farther from the starting point of the the racetrack (which is not the subject to change by any of the athletes).
Note that none of the athletes can run infinitely far, as they both will at some moment of time face the point, such that only one step further will cause them to fall in the abyss. In other words, the athlete will not fall into the abyss if the total length of all his steps will be less or equal to the chosen distance *L*.
Since the organizers are very fair, the are going to set the length of the racetrack as an integer chosen randomly and uniformly in range from 1 to *t* (both are included). What is the probability that Willman and Bolt tie again today?
|
The first line of the input contains three integers *t*, *w* and *b* (1<=≤<=*t*,<=*w*,<=*b*<=≤<=5·1018) — the maximum possible length of the racetrack, the length of Willman's steps and the length of Bolt's steps respectively.
|
Print the answer to the problem as an irreducible fraction . Follow the format of the samples output.
The fraction (*p* and *q* are integers, and both *p*<=≥<=0 and *q*<=><=0 holds) is called irreducible, if there is no such integer *d*<=><=1, that both *p* and *q* are divisible by *d*.
|
[
"10 3 2\n",
"7 1 2\n"
] |
[
"3/10\n",
"3/7\n"
] |
In the first sample Willman and Bolt will tie in case 1, 6 or 7 are chosen as the length of the racetrack.
| 1,500
|
[
{
"input": "10 3 2",
"output": "3/10"
},
{
"input": "7 1 2",
"output": "3/7"
},
{
"input": "1 1 1",
"output": "1/1"
},
{
"input": "5814 31 7",
"output": "94/2907"
},
{
"input": "94268 813 766",
"output": "765/94268"
},
{
"input": "262610 5583 4717",
"output": "2358/131305"
},
{
"input": "3898439 96326 71937",
"output": "71936/3898439"
},
{
"input": "257593781689876390 32561717 4411677",
"output": "7914548537/257593781689876390"
},
{
"input": "111319886766128339 7862842484895022 3003994959686829",
"output": "3003994959686828/111319886766128339"
},
{
"input": "413850294331656955 570110918058849723 409853735661743839",
"output": "409853735661743838/413850294331656955"
},
{
"input": "3000000000000000000 2999999999999999873 2999999999999999977",
"output": "23437499999999999/23437500000000000"
},
{
"input": "9 6 1",
"output": "1/9"
},
{
"input": "32 9 2",
"output": "3/32"
},
{
"input": "976 5 6",
"output": "41/244"
},
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}
] | 1,685,815,802
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 77
| 2,048,000
|
def solve():
n, a, b = map(int, input().split())
if a == b:
print(f'1/{n}')
return
div = (a // math.gcd(a, b)) * b
res = min(a - 1, b - 1) # can't use these numbers as both players would be stuck in the beginning itself
res += (n // div) * min(a, b) # min steps you again can't move
# if last one lies at a place such that div + min(a,b) crosses n then it needs to be substracted.
if ((n // div) * div + min(a, b)) > n:
res -= n - ((n // div) * div + min(a, b))
g = math.gcd(n, res)
res //= g
n //= g
print(f'{res}/{n}')
solve()
|
Title: The Big Race
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vector Willman and Array Bolt are the two most famous athletes of Byteforces. They are going to compete in a race with a distance of *L* meters today.
Willman and Bolt have exactly the same speed, so when they compete the result is always a tie. That is a problem for the organizers because they want a winner.
While watching previous races the organizers have noticed that Willman can perform only steps of length equal to *w* meters, and Bolt can perform only steps of length equal to *b* meters. Organizers decided to slightly change the rules of the race. Now, at the end of the racetrack there will be an abyss, and the winner will be declared the athlete, who manages to run farther from the starting point of the the racetrack (which is not the subject to change by any of the athletes).
Note that none of the athletes can run infinitely far, as they both will at some moment of time face the point, such that only one step further will cause them to fall in the abyss. In other words, the athlete will not fall into the abyss if the total length of all his steps will be less or equal to the chosen distance *L*.
Since the organizers are very fair, the are going to set the length of the racetrack as an integer chosen randomly and uniformly in range from 1 to *t* (both are included). What is the probability that Willman and Bolt tie again today?
Input Specification:
The first line of the input contains three integers *t*, *w* and *b* (1<=≤<=*t*,<=*w*,<=*b*<=≤<=5·1018) — the maximum possible length of the racetrack, the length of Willman's steps and the length of Bolt's steps respectively.
Output Specification:
Print the answer to the problem as an irreducible fraction . Follow the format of the samples output.
The fraction (*p* and *q* are integers, and both *p*<=≥<=0 and *q*<=><=0 holds) is called irreducible, if there is no such integer *d*<=><=1, that both *p* and *q* are divisible by *d*.
Demo Input:
['10 3 2\n', '7 1 2\n']
Demo Output:
['3/10\n', '3/7\n']
Note:
In the first sample Willman and Bolt will tie in case 1, 6 or 7 are chosen as the length of the racetrack.
|
```python
def solve():
n, a, b = map(int, input().split())
if a == b:
print(f'1/{n}')
return
div = (a // math.gcd(a, b)) * b
res = min(a - 1, b - 1) # can't use these numbers as both players would be stuck in the beginning itself
res += (n // div) * min(a, b) # min steps you again can't move
# if last one lies at a place such that div + min(a,b) crosses n then it needs to be substracted.
if ((n // div) * div + min(a, b)) > n:
res -= n - ((n // div) * div + min(a, b))
g = math.gcd(n, res)
res //= g
n //= g
print(f'{res}/{n}')
solve()
```
| -1
|
|
131
|
A
|
cAPS lOCK
|
PROGRAMMING
| 1,000
|
[
"implementation",
"strings"
] | null | null |
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
- either it only contains uppercase letters; - or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
|
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
|
Print the result of the given word's processing.
|
[
"cAPS\n",
"Lock\n"
] |
[
"Caps",
"Lock\n"
] |
none
| 500
|
[
{
"input": "cAPS",
"output": "Caps"
},
{
"input": "Lock",
"output": "Lock"
},
{
"input": "cAPSlOCK",
"output": "cAPSlOCK"
},
{
"input": "CAPs",
"output": "CAPs"
},
{
"input": "LoCK",
"output": "LoCK"
},
{
"input": "OOPS",
"output": "oops"
},
{
"input": "oops",
"output": "oops"
},
{
"input": "a",
"output": "A"
},
{
"input": "A",
"output": "a"
},
{
"input": "aA",
"output": "Aa"
},
{
"input": "Zz",
"output": "Zz"
},
{
"input": "Az",
"output": "Az"
},
{
"input": "zA",
"output": "Za"
},
{
"input": "AAA",
"output": "aaa"
},
{
"input": "AAa",
"output": "AAa"
},
{
"input": "AaR",
"output": "AaR"
},
{
"input": "Tdr",
"output": "Tdr"
},
{
"input": "aTF",
"output": "Atf"
},
{
"input": "fYd",
"output": "fYd"
},
{
"input": "dsA",
"output": "dsA"
},
{
"input": "fru",
"output": "fru"
},
{
"input": "hYBKF",
"output": "Hybkf"
},
{
"input": "XweAR",
"output": "XweAR"
},
{
"input": "mogqx",
"output": "mogqx"
},
{
"input": "eOhEi",
"output": "eOhEi"
},
{
"input": "nkdku",
"output": "nkdku"
},
{
"input": "zcnko",
"output": "zcnko"
},
{
"input": "lcccd",
"output": "lcccd"
},
{
"input": "vwmvg",
"output": "vwmvg"
},
{
"input": "lvchf",
"output": "lvchf"
},
{
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"output": "iunvzcchewenchqqxqypujcrdzluxcljhxphbxeuugnxooopbmobribhhmirilyjgyygfmtmfsvurgyhuwdrlqvibrlpevamjqyo"
},
{
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},
{
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},
{
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},
{
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},
{
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},
{
"input": "upvtbsxswbohxshdrbjxcungzquhuomgxwlryvshshsfvqbrxvcikbglietlpqkiwbhiqpmdwuqosbtdvyxekkaqiwrbsibpoect",
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},
{
"input": "cTKDZNWVYRTFPQLDAUUNSPKTDJTUPPFPRXRSINTVFVNNQNKXWUZUDHZBUSOKTABUEDQKUIVRTTVUREEOBJTSDKJKVEGFXVHXEYPE",
"output": "Ctkdznwvyrtfpqldauunspktdjtuppfprxrsintvfvnnqnkxwuzudhzbusoktabuedqkuivrttvureeobjtsdkjkvegfxvhxeype"
},
{
"input": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu",
"output": "JEGDOFJOFGJHFROGHOFGMHOFGHORGFHMOu"
},
{
"input": "aBACABa",
"output": "aBACABa"
},
{
"input": "AAAAAAAAAAAAAAAAAAAAAAAAaa",
"output": "AAAAAAAAAAAAAAAAAAAAAAAAaa"
},
{
"input": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "aaAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA"
},
{
"input": "dDDDDDDDDDDDDD",
"output": "Dddddddddddddd"
},
{
"input": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "z",
"output": "Z"
},
{
"input": "AZ",
"output": "az"
},
{
"input": "Z",
"output": "z"
},
{
"input": "aAAAA",
"output": "Aaaaa"
},
{
"input": "F",
"output": "f"
}
] | 1,691,706,077
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
word = input()
print(word)
new = ''
for i in range(len(word)) :
if ord(word[i]) >= 97 :
new += chr(ord(word[i]) - 32)
else :
new += chr(ord(word[i]) + 32)
print(new)
|
Title: cAPS lOCK
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
wHAT DO WE NEED cAPS LOCK FOR?
Caps lock is a computer keyboard key. Pressing it sets an input mode in which typed letters are capital by default. If it is pressed by accident, it leads to accidents like the one we had in the first passage.
Let's consider that a word has been typed with the Caps lock key accidentally switched on, if:
- either it only contains uppercase letters; - or all letters except for the first one are uppercase.
In this case we should automatically change the case of all letters. For example, the case of the letters that form words "hELLO", "HTTP", "z" should be changed.
Write a program that applies the rule mentioned above. If the rule cannot be applied, the program should leave the word unchanged.
Input Specification:
The first line of the input data contains a word consisting of uppercase and lowercase Latin letters. The word's length is from 1 to 100 characters, inclusive.
Output Specification:
Print the result of the given word's processing.
Demo Input:
['cAPS\n', 'Lock\n']
Demo Output:
['Caps', 'Lock\n']
Note:
none
|
```python
word = input()
print(word)
new = ''
for i in range(len(word)) :
if ord(word[i]) >= 97 :
new += chr(ord(word[i]) - 32)
else :
new += chr(ord(word[i]) + 32)
print(new)
```
| 0
|
|
20
|
A
|
BerOS file system
|
PROGRAMMING
| 1,700
|
[
"implementation"
] |
A. BerOS file system
|
2
|
64
|
The new operating system BerOS has a nice feature. It is possible to use any number of characters '/' as a delimiter in path instead of one traditional '/'. For example, strings //usr///local//nginx/sbin// and /usr/local/nginx///sbin are equivalent. The character '/' (or some sequence of such characters) at the end of the path is required only in case of the path to the root directory, which can be represented as single character '/'.
A path called normalized if it contains the smallest possible number of characters '/'.
Your task is to transform a given path to the normalized form.
|
The first line of the input contains only lowercase Latin letters and character '/' — the path to some directory. All paths start with at least one character '/'. The length of the given line is no more than 100 characters, it is not empty.
|
The path in normalized form.
|
[
"//usr///local//nginx/sbin\n"
] |
[
"/usr/local/nginx/sbin\n"
] |
none
| 500
|
[
{
"input": "//usr///local//nginx/sbin",
"output": "/usr/local/nginx/sbin"
},
{
"input": "////a//b/////g",
"output": "/a/b/g"
},
{
"input": "/a/b/c",
"output": "/a/b/c"
},
{
"input": "/",
"output": "/"
},
{
"input": "////",
"output": "/"
},
{
"input": "/a//aa/a//",
"output": "/a/aa/a"
},
{
"input": "/aa//b/aa",
"output": "/aa/b/aa"
},
{
"input": "////////////////////////////////////////////////////////////////////////////////////////////////////",
"output": "/"
},
{
"input": "/opt///pokerband///srvb/opt///pokerband///srvb////pokerband///srvb/",
"output": "/opt/pokerband/srvb/opt/pokerband/srvb/pokerband/srvb"
},
{
"input": "/root/pokerband/trunk/server/game/code/src/main/java/com/uosipa/pokerband/gameserver/game/",
"output": "/root/pokerband/trunk/server/game/code/src/main/java/com/uosipa/pokerband/gameserver/game"
},
{
"input": "/root//pokerband/trunk/server/game/code/src///main/java/com/uosipa/pokerband//gameserver/game//",
"output": "/root/pokerband/trunk/server/game/code/src/main/java/com/uosipa/pokerband/gameserver/game"
},
{
"input": "/root//pokerband/trunk/server/game/code/src///main/java/com/uosipa/pokerband//gameserver/game//test/",
"output": "/root/pokerband/trunk/server/game/code/src/main/java/com/uosipa/pokerband/gameserver/game/test"
},
{
"input": "/root//pokerband/trunk/server/game/code/src///main/java/com/uosipa/pokerband//gameserver/game//testt",
"output": "/root/pokerband/trunk/server/game/code/src/main/java/com/uosipa/pokerband/gameserver/game/testt"
},
{
"input": "///a//a////a/a//a//a//a/////",
"output": "/a/a/a/a/a/a/a"
},
{
"input": "/zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz",
"output": "/zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz"
},
{
"input": "/zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz/",
"output": "/zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz"
},
{
"input": "/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a",
"output": "/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a/a"
},
{
"input": "///////////////////////////////////////////////////////////////////////////////////////////////////z",
"output": "/z"
},
{
"input": "/z//////////////////////////////////////////////////////////////////////////////////////////////////",
"output": "/z"
}
] | 1,689,245,612
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 5
| 62
| 0
|
x = input()
for _ in range(100):
x = x.replace("//", "/")
print(x)
|
Title: BerOS file system
Time Limit: 2 seconds
Memory Limit: 64 megabytes
Problem Description:
The new operating system BerOS has a nice feature. It is possible to use any number of characters '/' as a delimiter in path instead of one traditional '/'. For example, strings //usr///local//nginx/sbin// and /usr/local/nginx///sbin are equivalent. The character '/' (or some sequence of such characters) at the end of the path is required only in case of the path to the root directory, which can be represented as single character '/'.
A path called normalized if it contains the smallest possible number of characters '/'.
Your task is to transform a given path to the normalized form.
Input Specification:
The first line of the input contains only lowercase Latin letters and character '/' — the path to some directory. All paths start with at least one character '/'. The length of the given line is no more than 100 characters, it is not empty.
Output Specification:
The path in normalized form.
Demo Input:
['//usr///local//nginx/sbin\n']
Demo Output:
['/usr/local/nginx/sbin\n']
Note:
none
|
```python
x = input()
for _ in range(100):
x = x.replace("//", "/")
print(x)
```
| 0
|
622
|
A
|
Infinite Sequence
|
PROGRAMMING
| 1,000
|
[
"implementation",
"math"
] | null | null |
Consider the infinite sequence of integers: 1,<=1,<=2,<=1,<=2,<=3,<=1,<=2,<=3,<=4,<=1,<=2,<=3,<=4,<=5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).
Find the number on the *n*-th position of the sequence.
|
The only line contains integer *n* (1<=≤<=*n*<=≤<=1014) — the position of the number to find.
Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
|
Print the element in the *n*-th position of the sequence (the elements are numerated from one).
|
[
"3\n",
"5\n",
"10\n",
"55\n",
"56\n"
] |
[
"2\n",
"2\n",
"4\n",
"10\n",
"1\n"
] |
none
| 0
|
[
{
"input": "3",
"output": "2"
},
{
"input": "5",
"output": "2"
},
{
"input": "10",
"output": "4"
},
{
"input": "55",
"output": "10"
},
{
"input": "56",
"output": "1"
},
{
"input": "1000000000000",
"output": "88209"
},
{
"input": "847194127849",
"output": "255708"
},
{
"input": "294719472984",
"output": "593358"
},
{
"input": "999999911791",
"output": "1414213"
},
{
"input": "999999911792",
"output": "1"
},
{
"input": "100000000000000",
"output": "1749820"
},
{
"input": "1",
"output": "1"
},
{
"input": "99993",
"output": "312"
},
{
"input": "99994",
"output": "313"
},
{
"input": "99995",
"output": "314"
},
{
"input": "99990",
"output": "309"
},
{
"input": "2",
"output": "1"
},
{
"input": "99991",
"output": "310"
},
{
"input": "99992",
"output": "311"
},
{
"input": "99996",
"output": "315"
},
{
"input": "99997",
"output": "316"
},
{
"input": "99998",
"output": "317"
},
{
"input": "99999",
"output": "318"
},
{
"input": "1021",
"output": "31"
},
{
"input": "4",
"output": "1"
},
{
"input": "23",
"output": "2"
},
{
"input": "9994",
"output": "124"
},
{
"input": "99939",
"output": "258"
},
{
"input": "99999998250180",
"output": "14142135"
},
{
"input": "6",
"output": "3"
},
{
"input": "8",
"output": "2"
},
{
"input": "35",
"output": "7"
},
{
"input": "100",
"output": "9"
},
{
"input": "10101010",
"output": "745"
},
{
"input": "103",
"output": "12"
},
{
"input": "102",
"output": "11"
},
{
"input": "101",
"output": "10"
}
] | 1,571,070,656
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 10
| 1,000
| 0
|
n = int(input())
j = 1
while n > j:
n -= j
j += 1
print(n)
|
Title: Infinite Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Consider the infinite sequence of integers: 1,<=1,<=2,<=1,<=2,<=3,<=1,<=2,<=3,<=4,<=1,<=2,<=3,<=4,<=5.... The sequence is built in the following way: at first the number 1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number 10 first appears in the sequence in position 55 (the elements are numerated from one).
Find the number on the *n*-th position of the sequence.
Input Specification:
The only line contains integer *n* (1<=≤<=*n*<=≤<=1014) — the position of the number to find.
Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.
Output Specification:
Print the element in the *n*-th position of the sequence (the elements are numerated from one).
Demo Input:
['3\n', '5\n', '10\n', '55\n', '56\n']
Demo Output:
['2\n', '2\n', '4\n', '10\n', '1\n']
Note:
none
|
```python
n = int(input())
j = 1
while n > j:
n -= j
j += 1
print(n)
```
| 0
|
|
552
|
C
|
Vanya and Scales
|
PROGRAMMING
| 1,900
|
[
"brute force",
"dp",
"greedy",
"math",
"meet-in-the-middle",
"number theory"
] | null | null |
Vanya has a scales for weighing loads and weights of masses *w*0,<=*w*1,<=*w*2,<=...,<=*w*100 grams where *w* is some integer not less than 2 (exactly one weight of each nominal value). Vanya wonders whether he can weight an item with mass *m* using the given weights, if the weights can be put on both pans of the scales. Formally speaking, your task is to determine whether it is possible to place an item of mass *m* and some weights on the left pan of the scales, and some weights on the right pan of the scales so that the pans of the scales were in balance.
|
The first line contains two integers *w*,<=*m* (2<=≤<=*w*<=≤<=109, 1<=≤<=*m*<=≤<=109) — the number defining the masses of the weights and the mass of the item.
|
Print word 'YES' if the item can be weighted and 'NO' if it cannot.
|
[
"3 7\n",
"100 99\n",
"100 50\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
Note to the first sample test. One pan can have an item of mass 7 and a weight of mass 3, and the second pan can have two weights of masses 9 and 1, correspondingly. Then 7 + 3 = 9 + 1.
Note to the second sample test. One pan of the scales can have an item of mass 99 and the weight of mass 1, and the second pan can have the weight of mass 100.
Note to the third sample test. It is impossible to measure the weight of the item in the manner described in the input.
| 1,500
|
[
{
"input": "3 7",
"output": "YES"
},
{
"input": "100 99",
"output": "YES"
},
{
"input": "100 50",
"output": "NO"
},
{
"input": "1000000000 1",
"output": "YES"
},
{
"input": "100 10002",
"output": "NO"
},
{
"input": "4 7",
"output": "NO"
},
{
"input": "4 11",
"output": "YES"
},
{
"input": "5 781",
"output": "YES"
},
{
"input": "7 9",
"output": "NO"
},
{
"input": "5077 5988",
"output": "NO"
},
{
"input": "2 9596",
"output": "YES"
},
{
"input": "4 1069",
"output": "YES"
},
{
"input": "4 7134",
"output": "NO"
},
{
"input": "4 9083",
"output": "NO"
},
{
"input": "4 7927",
"output": "NO"
},
{
"input": "4 6772",
"output": "NO"
},
{
"input": "5 782",
"output": "NO"
},
{
"input": "4 1000000000",
"output": "NO"
},
{
"input": "4 357913941",
"output": "YES"
},
{
"input": "4 357918037",
"output": "NO"
},
{
"input": "5 12207031",
"output": "YES"
},
{
"input": "5 41503906",
"output": "YES"
},
{
"input": "5 90332031",
"output": "NO"
},
{
"input": "11 1786324",
"output": "YES"
},
{
"input": "10 999",
"output": "YES"
},
{
"input": "8 28087",
"output": "YES"
},
{
"input": "8 28598",
"output": "NO"
},
{
"input": "32 33586176",
"output": "YES"
},
{
"input": "87 56631258",
"output": "YES"
},
{
"input": "19 20",
"output": "YES"
},
{
"input": "58 11316496",
"output": "YES"
},
{
"input": "89 89",
"output": "YES"
},
{
"input": "21 85756882",
"output": "YES"
},
{
"input": "56 540897225",
"output": "YES"
},
{
"input": "91 8189",
"output": "YES"
},
{
"input": "27 14329927",
"output": "YES"
},
{
"input": "58 198535",
"output": "YES"
},
{
"input": "939 938",
"output": "YES"
},
{
"input": "27463 754243832",
"output": "YES"
},
{
"input": "21427 459137757",
"output": "YES"
},
{
"input": "26045 26045",
"output": "YES"
},
{
"input": "25336 25336",
"output": "YES"
},
{
"input": "24627 24626",
"output": "YES"
},
{
"input": "29245 855299270",
"output": "YES"
},
{
"input": "28536 814274759",
"output": "YES"
},
{
"input": "33154 33155",
"output": "YES"
},
{
"input": "27118 27119",
"output": "YES"
},
{
"input": "70 338171",
"output": "YES"
},
{
"input": "24 346226",
"output": "NO"
},
{
"input": "41 2966964",
"output": "NO"
},
{
"input": "31 29792",
"output": "YES"
},
{
"input": "48 2402",
"output": "NO"
},
{
"input": "65 4159",
"output": "YES"
},
{
"input": "20 67376840",
"output": "NO"
},
{
"input": "72 5111",
"output": "YES"
},
{
"input": "27 14349609",
"output": "YES"
},
{
"input": "44 89146",
"output": "NO"
},
{
"input": "22787 519292944",
"output": "NO"
},
{
"input": "24525 601475624",
"output": "YES"
},
{
"input": "3716 13816089",
"output": "NO"
},
{
"input": "4020 4020",
"output": "YES"
},
{
"input": "13766 13767",
"output": "YES"
},
{
"input": "23512 23511",
"output": "YES"
},
{
"input": "23816 567225671",
"output": "YES"
},
{
"input": "33562 33564",
"output": "NO"
},
{
"input": "33866 33866",
"output": "YES"
},
{
"input": "13057 13059",
"output": "NO"
},
{
"input": "441890232 441890232",
"output": "YES"
},
{
"input": "401739553 401739553",
"output": "YES"
},
{
"input": "285681920 285681919",
"output": "YES"
},
{
"input": "464591587 464591588",
"output": "YES"
},
{
"input": "703722884 703722884",
"output": "YES"
},
{
"input": "982276216 982276216",
"output": "YES"
},
{
"input": "867871061 867871062",
"output": "YES"
},
{
"input": "48433217 48433216",
"output": "YES"
},
{
"input": "8 324818663",
"output": "NO"
},
{
"input": "7 898367507",
"output": "NO"
},
{
"input": "6 471916351",
"output": "NO"
},
{
"input": "5 45465196",
"output": "NO"
},
{
"input": "9 768757144",
"output": "NO"
},
{
"input": "8 342305988",
"output": "NO"
},
{
"input": "6 114457122",
"output": "NO"
},
{
"input": "6 688005966",
"output": "NO"
},
{
"input": "4 556522107",
"output": "NO"
},
{
"input": "3 130070951",
"output": "YES"
},
{
"input": "6 558395604",
"output": "NO"
},
{
"input": "5 131944448",
"output": "NO"
},
{
"input": "2 1000000",
"output": "YES"
},
{
"input": "2 22222222",
"output": "YES"
},
{
"input": "3 100000000",
"output": "YES"
},
{
"input": "3 100000001",
"output": "YES"
},
{
"input": "3 100000002",
"output": "YES"
},
{
"input": "3 100000003",
"output": "YES"
},
{
"input": "3 100000004",
"output": "YES"
},
{
"input": "2 1",
"output": "YES"
},
{
"input": "2 1000000000",
"output": "YES"
},
{
"input": "3 1000000000",
"output": "YES"
},
{
"input": "99999 1000000000",
"output": "NO"
},
{
"input": "10 1000000000",
"output": "YES"
},
{
"input": "1000 1000000000",
"output": "YES"
},
{
"input": "10 999999999",
"output": "YES"
},
{
"input": "100 99999999",
"output": "YES"
},
{
"input": "1000 999999999",
"output": "YES"
},
{
"input": "1000 999999998",
"output": "NO"
},
{
"input": "2 536870912",
"output": "YES"
},
{
"input": "10 99",
"output": "YES"
},
{
"input": "10 8",
"output": "NO"
},
{
"input": "3 5",
"output": "YES"
},
{
"input": "3 26",
"output": "YES"
},
{
"input": "10 8888",
"output": "NO"
},
{
"input": "3 8",
"output": "YES"
},
{
"input": "3 984742145",
"output": "YES"
},
{
"input": "4 43",
"output": "YES"
},
{
"input": "1000000000 1000000000",
"output": "YES"
},
{
"input": "4194304 4194305",
"output": "YES"
},
{
"input": "10 899",
"output": "YES"
},
{
"input": "4 47",
"output": "YES"
},
{
"input": "4 822083581",
"output": "YES"
},
{
"input": "3 999987989",
"output": "YES"
},
{
"input": "4 31",
"output": "NO"
},
{
"input": "4 15",
"output": "YES"
},
{
"input": "100000000 100000001",
"output": "YES"
}
] | 1,542,199,710
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 108
| 0
|
import sys
w,m = input().split()
w = int(w)
m = int(m)
a = []
b = []
k2 = 10000
k = 10000
if w == 4:
print('NO')
sys.exit()
if w<= 3:
print('YES')
sys.exit()
while m!= 0:
if (m % w != (w - 1)) and (m % w != 0) and (m % w != 1):
m = (m+1)/w
print('NO')
sys.exit()
else:
print('YES')
sys.exit()
|
Title: Vanya and Scales
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya has a scales for weighing loads and weights of masses *w*0,<=*w*1,<=*w*2,<=...,<=*w*100 grams where *w* is some integer not less than 2 (exactly one weight of each nominal value). Vanya wonders whether he can weight an item with mass *m* using the given weights, if the weights can be put on both pans of the scales. Formally speaking, your task is to determine whether it is possible to place an item of mass *m* and some weights on the left pan of the scales, and some weights on the right pan of the scales so that the pans of the scales were in balance.
Input Specification:
The first line contains two integers *w*,<=*m* (2<=≤<=*w*<=≤<=109, 1<=≤<=*m*<=≤<=109) — the number defining the masses of the weights and the mass of the item.
Output Specification:
Print word 'YES' if the item can be weighted and 'NO' if it cannot.
Demo Input:
['3 7\n', '100 99\n', '100 50\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
Note to the first sample test. One pan can have an item of mass 7 and a weight of mass 3, and the second pan can have two weights of masses 9 and 1, correspondingly. Then 7 + 3 = 9 + 1.
Note to the second sample test. One pan of the scales can have an item of mass 99 and the weight of mass 1, and the second pan can have the weight of mass 100.
Note to the third sample test. It is impossible to measure the weight of the item in the manner described in the input.
|
```python
import sys
w,m = input().split()
w = int(w)
m = int(m)
a = []
b = []
k2 = 10000
k = 10000
if w == 4:
print('NO')
sys.exit()
if w<= 3:
print('YES')
sys.exit()
while m!= 0:
if (m % w != (w - 1)) and (m % w != 0) and (m % w != 1):
m = (m+1)/w
print('NO')
sys.exit()
else:
print('YES')
sys.exit()
```
| 0
|
|
979
|
A
|
Pizza, Pizza, Pizza!!!
|
PROGRAMMING
| 1,000
|
[
"math"
] | null | null |
Katie, Kuro and Shiro are best friends. They have known each other since kindergarten. That's why they often share everything with each other and work together on some very hard problems.
Today is Shiro's birthday. She really loves pizza so she wants to invite her friends to the pizza restaurant near her house to celebrate her birthday, including her best friends Katie and Kuro.
She has ordered a very big round pizza, in order to serve her many friends. Exactly $n$ of Shiro's friends are here. That's why she has to divide the pizza into $n + 1$ slices (Shiro also needs to eat). She wants the slices to be exactly the same size and shape. If not, some of her friends will get mad and go home early, and the party will be over.
Shiro is now hungry. She wants to cut the pizza with minimum of straight cuts. A cut is a straight segment, it might have ends inside or outside the pizza. But she is too lazy to pick up the calculator.
As usual, she will ask Katie and Kuro for help. But they haven't come yet. Could you help Shiro with this problem?
|
A single line contains one non-negative integer $n$ ($0 \le n \leq 10^{18}$) — the number of Shiro's friends. The circular pizza has to be sliced into $n + 1$ pieces.
|
A single integer — the number of straight cuts Shiro needs.
|
[
"3\n",
"4\n"
] |
[
"2",
"5"
] |
To cut the round pizza into quarters one has to make two cuts through the center with angle $90^{\circ}$ between them.
To cut the round pizza into five equal parts one has to make five cuts.
| 500
|
[
{
"input": "3",
"output": "2"
},
{
"input": "4",
"output": "5"
},
{
"input": "10",
"output": "11"
},
{
"input": "10000000000",
"output": "10000000001"
},
{
"input": "1234567891",
"output": "617283946"
},
{
"input": "7509213957",
"output": "3754606979"
},
{
"input": "99999999999999999",
"output": "50000000000000000"
},
{
"input": "21",
"output": "11"
},
{
"input": "712394453192",
"output": "712394453193"
},
{
"input": "172212168",
"output": "172212169"
},
{
"input": "822981260158260519",
"output": "411490630079130260"
},
{
"input": "28316250877914571",
"output": "14158125438957286"
},
{
"input": "779547116602436424",
"output": "779547116602436425"
},
{
"input": "578223540024979436",
"output": "578223540024979437"
},
{
"input": "335408917861648766",
"output": "335408917861648767"
},
{
"input": "74859962623690078",
"output": "74859962623690079"
},
{
"input": "252509054433933439",
"output": "126254527216966720"
},
{
"input": "760713016476190622",
"output": "760713016476190623"
},
{
"input": "919845426262703496",
"output": "919845426262703497"
},
{
"input": "585335723211047194",
"output": "585335723211047195"
},
{
"input": "522842184971407769",
"output": "261421092485703885"
},
{
"input": "148049062628894320",
"output": "148049062628894321"
},
{
"input": "84324828731963974",
"output": "84324828731963975"
},
{
"input": "354979173822804781",
"output": "177489586911402391"
},
{
"input": "1312150450968413",
"output": "656075225484207"
},
{
"input": "269587449430302150",
"output": "269587449430302151"
},
{
"input": "645762258982631926",
"output": "645762258982631927"
},
{
"input": "615812229161735895",
"output": "307906114580867948"
},
{
"input": "0",
"output": "0"
},
{
"input": "349993004923078531",
"output": "174996502461539266"
},
{
"input": "891351282707723851",
"output": "445675641353861926"
},
{
"input": "563324731189330734",
"output": "563324731189330735"
},
{
"input": "520974001910286909",
"output": "260487000955143455"
},
{
"input": "666729339802329204",
"output": "666729339802329205"
},
{
"input": "856674611404539671",
"output": "428337305702269836"
},
{
"input": "791809296303238499",
"output": "395904648151619250"
},
{
"input": "711066337317063338",
"output": "711066337317063339"
},
{
"input": "931356503492686566",
"output": "931356503492686567"
},
{
"input": "234122432773361866",
"output": "234122432773361867"
},
{
"input": "1000000000000000000",
"output": "1000000000000000001"
},
{
"input": "1",
"output": "1"
},
{
"input": "2",
"output": "3"
},
{
"input": "7",
"output": "4"
},
{
"input": "63",
"output": "32"
},
{
"input": "24",
"output": "25"
},
{
"input": "8",
"output": "9"
},
{
"input": "15",
"output": "8"
}
] | 1,576,819,638
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 47
| 156
| 0
|
x=int(input())
print( 0 if x==0 else (x+1)//2 if (x+1)%2==0 else x+1 )
# My code says who am i
# red is love
# love is not in logic
|
Title: Pizza, Pizza, Pizza!!!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Katie, Kuro and Shiro are best friends. They have known each other since kindergarten. That's why they often share everything with each other and work together on some very hard problems.
Today is Shiro's birthday. She really loves pizza so she wants to invite her friends to the pizza restaurant near her house to celebrate her birthday, including her best friends Katie and Kuro.
She has ordered a very big round pizza, in order to serve her many friends. Exactly $n$ of Shiro's friends are here. That's why she has to divide the pizza into $n + 1$ slices (Shiro also needs to eat). She wants the slices to be exactly the same size and shape. If not, some of her friends will get mad and go home early, and the party will be over.
Shiro is now hungry. She wants to cut the pizza with minimum of straight cuts. A cut is a straight segment, it might have ends inside or outside the pizza. But she is too lazy to pick up the calculator.
As usual, she will ask Katie and Kuro for help. But they haven't come yet. Could you help Shiro with this problem?
Input Specification:
A single line contains one non-negative integer $n$ ($0 \le n \leq 10^{18}$) — the number of Shiro's friends. The circular pizza has to be sliced into $n + 1$ pieces.
Output Specification:
A single integer — the number of straight cuts Shiro needs.
Demo Input:
['3\n', '4\n']
Demo Output:
['2', '5']
Note:
To cut the round pizza into quarters one has to make two cuts through the center with angle $90^{\circ}$ between them.
To cut the round pizza into five equal parts one has to make five cuts.
|
```python
x=int(input())
print( 0 if x==0 else (x+1)//2 if (x+1)%2==0 else x+1 )
# My code says who am i
# red is love
# love is not in logic
```
| 3
|
|
767
|
A
|
Snacktower
|
PROGRAMMING
| 1,100
|
[
"data structures",
"implementation"
] | null | null |
According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents.
|
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*.
|
Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty.
|
[
"3\n3 1 2\n",
"5\n4 5 1 2 3\n"
] |
[
"3\n \n2 1",
"5 4\n \n \n3 2 1\n"
] |
In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before.
| 500
|
[
{
"input": "3\n3 1 2",
"output": "3 \n\n2 1 "
},
{
"input": "5\n4 5 1 2 3",
"output": "5 4 \n\n\n3 2 1 "
},
{
"input": "1\n1",
"output": "1 "
},
{
"input": "2\n1 2",
"output": "2 1 "
},
{
"input": "10\n5 1 6 2 8 3 4 10 9 7",
"output": "10 \n9 8 \n7 6 5 4 3 2 1 "
},
{
"input": "30\n16 10 4 29 5 28 12 21 11 30 18 6 14 3 17 22 20 15 9 1 27 19 24 26 13 25 2 23 8 7",
"output": "30 29 28 \n\n\n\n\n\n\n\n\n\n\n27 \n\n\n26 \n\n25 24 \n\n23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 \n8 \n7 6 5 4 3 2 1 "
},
{
"input": "100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 4 80 72 39",
"output": "100 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n99 98 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 \n\n\n\n\n\n\n\n\n80 79 78 77 76 75 74 73 \n72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 \n39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "2\n2 1",
"output": "2 \n1 "
}
] | 1,671,833,216
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 46
| 0
|
n = int(input())
a = list(map(int, input().split()))
temp = []
ans = []
s = n
p = ''
for i in range(n):
if a[i] >= s:
ans.append(str(a[i])+' ' + p)
s = max(temp)-1 if len(temp) != 0 else s-1
p = ''
if len(temp) >0:
temp.clear()
else:
ans.append(' ')
temp.append(a[i])
p = str(a[i]) + ' ' + p
for i in ans:
print(i)
|
Title: Snacktower
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*.
Output Specification:
Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty.
Demo Input:
['3\n3 1 2\n', '5\n4 5 1 2 3\n']
Demo Output:
['3\n\xa0\n2 1', '5 4\n\xa0\n\xa0\n3 2 1\n']
Note:
In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before.
|
```python
n = int(input())
a = list(map(int, input().split()))
temp = []
ans = []
s = n
p = ''
for i in range(n):
if a[i] >= s:
ans.append(str(a[i])+' ' + p)
s = max(temp)-1 if len(temp) != 0 else s-1
p = ''
if len(temp) >0:
temp.clear()
else:
ans.append(' ')
temp.append(a[i])
p = str(a[i]) + ' ' + p
for i in ans:
print(i)
```
| 0
|
|
681
|
C
|
Heap Operations
|
PROGRAMMING
| 1,600
|
[
"constructive algorithms",
"data structures",
"greedy"
] | null | null |
Petya has recently learned data structure named "Binary heap".
The heap he is now operating with allows the following operations:
- put the given number into the heap; - get the value of the minimum element in the heap; - extract the minimum element from the heap;
Thus, at any moment of time the heap contains several integers (possibly none), some of them might be equal.
In order to better learn this data structure Petya took an empty heap and applied some operations above to it. Also, he carefully wrote down all the operations and their results to his event log, following the format:
- insert *x* — put the element with value *x* in the heap; - getMin *x* — the value of the minimum element contained in the heap was equal to *x*; - removeMin — the minimum element was extracted from the heap (only one instance, if there were many).
All the operations were correct, i.e. there was at least one element in the heap each time getMin or removeMin operations were applied.
While Petya was away for a lunch, his little brother Vova came to the room, took away some of the pages from Petya's log and used them to make paper boats.
Now Vova is worried, if he made Petya's sequence of operations inconsistent. For example, if one apply operations one-by-one in the order they are written in the event log, results of getMin operations might differ from the results recorded by Petya, and some of getMin or removeMin operations may be incorrect, as the heap is empty at the moment they are applied.
Now Vova wants to add some new operation records to the event log in order to make the resulting sequence of operations correct. That is, the result of each getMin operation is equal to the result in the record, and the heap is non-empty when getMin ad removeMin are applied. Vova wants to complete this as fast as possible, as the Petya may get back at any moment. He asks you to add the least possible number of operation records to the current log. Note that arbitrary number of operations may be added at the beginning, between any two other operations, or at the end of the log.
|
The first line of the input contains the only integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of the records left in Petya's journal.
Each of the following *n* lines describe the records in the current log in the order they are applied. Format described in the statement is used. All numbers in the input are integers not exceeding 109 by their absolute value.
|
The first line of the output should contain a single integer *m* — the minimum possible number of records in the modified sequence of operations.
Next *m* lines should contain the corrected sequence of records following the format of the input (described in the statement), one per line and in the order they are applied. All the numbers in the output should be integers not exceeding 109 by their absolute value.
Note that the input sequence of operations must be the subsequence of the output sequence.
It's guaranteed that there exists the correct answer consisting of no more than 1<=000<=000 operations.
|
[
"2\ninsert 3\ngetMin 4\n",
"4\ninsert 1\ninsert 1\nremoveMin\ngetMin 2\n"
] |
[
"4\ninsert 3\nremoveMin\ninsert 4\ngetMin 4\n",
"6\ninsert 1\ninsert 1\nremoveMin\nremoveMin\ninsert 2\ngetMin 2\n"
] |
In the first sample, after number 3 is inserted into the heap, the minimum number is 3. To make the result of the first getMin equal to 4 one should firstly remove number 3 from the heap and then add number 4 into the heap.
In the second sample case number 1 is inserted two times, so should be similarly removed twice.
| 1,500
|
[
{
"input": "2\ninsert 3\ngetMin 4",
"output": "4\ninsert 3\nremoveMin\ninsert 4\ngetMin 4"
},
{
"input": "4\ninsert 1\ninsert 1\nremoveMin\ngetMin 2",
"output": "6\ninsert 1\ninsert 1\nremoveMin\nremoveMin\ninsert 2\ngetMin 2"
},
{
"input": "1\ninsert 1",
"output": "1\ninsert 1"
},
{
"input": "1\ngetMin 31",
"output": "2\ninsert 31\ngetMin 31"
},
{
"input": "1\nremoveMin",
"output": "2\ninsert 0\nremoveMin"
},
{
"input": "2\ninsert 2\ngetMin 2",
"output": "2\ninsert 2\ngetMin 2"
},
{
"input": "2\ninsert 31\nremoveMin",
"output": "2\ninsert 31\nremoveMin"
},
{
"input": "2\ngetMin 31\nremoveMin",
"output": "3\ninsert 31\ngetMin 31\nremoveMin"
},
{
"input": "2\nremoveMin\ngetMin 31",
"output": "4\ninsert 0\nremoveMin\ninsert 31\ngetMin 31"
},
{
"input": "8\ninsert 219147240\nremoveMin\ngetMin 923854124\nremoveMin\ngetMin -876779400\nremoveMin\ninsert 387686853\ngetMin 749998368",
"output": "12\ninsert 219147240\nremoveMin\ninsert 923854124\ngetMin 923854124\nremoveMin\ninsert -876779400\ngetMin -876779400\nremoveMin\ninsert 387686853\nremoveMin\ninsert 749998368\ngetMin 749998368"
},
{
"input": "2\nremoveMin\ninsert 450653162",
"output": "3\ninsert 0\nremoveMin\ninsert 450653162"
},
{
"input": "6\ninsert -799688192\ngetMin 491561656\nremoveMin\ninsert -805250162\ninsert -945439443\nremoveMin",
"output": "8\ninsert -799688192\nremoveMin\ninsert 491561656\ngetMin 491561656\nremoveMin\ninsert -805250162\ninsert -945439443\nremoveMin"
},
{
"input": "30\ninsert 62350949\ngetMin -928976719\nremoveMin\ngetMin 766590157\ngetMin -276914351\ninsert 858958907\ngetMin -794653029\ngetMin 505812710\ngetMin -181182543\ninsert -805198995\nremoveMin\ninsert -200361579\nremoveMin\ninsert 988531216\ninsert -474257426\ninsert 579296921\nremoveMin\ninsert -410043658\ngetMin 716684155\nremoveMin\ngetMin -850837161\ngetMin 368670814\ninsert 579000842\nremoveMin\ngetMin -169833018\ninsert 313148949\nremoveMin\nremoveMin\ngetMin 228901059\ngetMin 599172503",
"output": "52\ninsert 62350949\ninsert -928976719\ngetMin -928976719\nremoveMin\nremoveMin\ninsert 766590157\ngetMin 766590157\ninsert -276914351\ngetMin -276914351\ninsert 858958907\ninsert -794653029\ngetMin -794653029\nremoveMin\nremoveMin\ninsert 505812710\ngetMin 505812710\ninsert -181182543\ngetMin -181182543\ninsert -805198995\nremoveMin\ninsert -200361579\nremoveMin\ninsert 988531216\ninsert -474257426\ninsert 579296921\nremoveMin\ninsert -410043658\nremoveMin\nremoveMin\nremoveMin\nremoveMin\ninsert 71668415..."
},
{
"input": "9\ninsert 3\ninsert 4\ninsert 5\nremoveMin\ngetMin 3\nremoveMin\ngetMin 4\nremoveMin\ngetMin 5",
"output": "10\ninsert 3\ninsert 4\ninsert 5\nremoveMin\ninsert 3\ngetMin 3\nremoveMin\ngetMin 4\nremoveMin\ngetMin 5"
},
{
"input": "9\ninsert 3\ninsert 4\ninsert 5\nremoveMin\ngetMin 5\nremoveMin\ngetMin 4\nremoveMin\ngetMin 3",
"output": "12\ninsert 3\ninsert 4\ninsert 5\nremoveMin\nremoveMin\ngetMin 5\nremoveMin\ninsert 4\ngetMin 4\nremoveMin\ninsert 3\ngetMin 3"
},
{
"input": "1\ngetMin 0",
"output": "2\ninsert 0\ngetMin 0"
},
{
"input": "3\ninsert 3\ninsert 4\ngetMin 4",
"output": "4\ninsert 3\ninsert 4\nremoveMin\ngetMin 4"
},
{
"input": "3\ninsert 1\ninsert 2\ngetMin 2",
"output": "4\ninsert 1\ninsert 2\nremoveMin\ngetMin 2"
},
{
"input": "3\ninsert -1\ninsert 0\ngetMin 0",
"output": "4\ninsert -1\ninsert 0\nremoveMin\ngetMin 0"
},
{
"input": "2\ninsert 0\ngetMin 0",
"output": "2\ninsert 0\ngetMin 0"
},
{
"input": "1\ninsert -1",
"output": "1\ninsert -1"
},
{
"input": "3\ninsert 1\ninsert 0\ngetMin 1",
"output": "4\ninsert 1\ninsert 0\nremoveMin\ngetMin 1"
}
] | 1,623,416,735
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 7
| 109
| 409,600
|
###### ### ####### ####### ## # ##### ### #####
# # # # # # # # # # # # # ###
# # # # # # # # # # # # # ###
###### ######### # # # # # # ######### #
###### ######### # # # # # # ######### #
# # # # # # # # # # #### # # #
# # # # # # # ## # # # # #
###### # # ####### ####### # # ##### # # # #
from __future__ import print_function # for PyPy2
# from itertools import permutations
# from functools import cmp_to_key # for adding custom comparator
# from fractions import Fraction
from collections import *
from sys import stdin
# from bisect import *
from heapq import *
from math import *
g = lambda : stdin.readline().strip()
gl = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
rr = lambda x : reversed(range(x))
mod = int(1e9)+7
inf = float("inf")
ans = []
h = []
for _ in range(gil()[0]):
res = gl()
if len(res) == 1: #removeMin
if h == []:
ans.append('insert 1')
ans.append(res[0])
if h:heappop(h)
elif res[0] == 'insert': # insert
ans.append(' '.join(res))
heappush(h, int(res[1]))
else: # getMin
val = int(res[1])
while h and h[0] < val:
ans.append('removeMin')
heappop(h)
if len(h) == 0:ans.append('insert '+res[1])
ans.append(' '.join(res))
print(len(ans))
print('\n'.join(ans))
|
Title: Heap Operations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya has recently learned data structure named "Binary heap".
The heap he is now operating with allows the following operations:
- put the given number into the heap; - get the value of the minimum element in the heap; - extract the minimum element from the heap;
Thus, at any moment of time the heap contains several integers (possibly none), some of them might be equal.
In order to better learn this data structure Petya took an empty heap and applied some operations above to it. Also, he carefully wrote down all the operations and their results to his event log, following the format:
- insert *x* — put the element with value *x* in the heap; - getMin *x* — the value of the minimum element contained in the heap was equal to *x*; - removeMin — the minimum element was extracted from the heap (only one instance, if there were many).
All the operations were correct, i.e. there was at least one element in the heap each time getMin or removeMin operations were applied.
While Petya was away for a lunch, his little brother Vova came to the room, took away some of the pages from Petya's log and used them to make paper boats.
Now Vova is worried, if he made Petya's sequence of operations inconsistent. For example, if one apply operations one-by-one in the order they are written in the event log, results of getMin operations might differ from the results recorded by Petya, and some of getMin or removeMin operations may be incorrect, as the heap is empty at the moment they are applied.
Now Vova wants to add some new operation records to the event log in order to make the resulting sequence of operations correct. That is, the result of each getMin operation is equal to the result in the record, and the heap is non-empty when getMin ad removeMin are applied. Vova wants to complete this as fast as possible, as the Petya may get back at any moment. He asks you to add the least possible number of operation records to the current log. Note that arbitrary number of operations may be added at the beginning, between any two other operations, or at the end of the log.
Input Specification:
The first line of the input contains the only integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of the records left in Petya's journal.
Each of the following *n* lines describe the records in the current log in the order they are applied. Format described in the statement is used. All numbers in the input are integers not exceeding 109 by their absolute value.
Output Specification:
The first line of the output should contain a single integer *m* — the minimum possible number of records in the modified sequence of operations.
Next *m* lines should contain the corrected sequence of records following the format of the input (described in the statement), one per line and in the order they are applied. All the numbers in the output should be integers not exceeding 109 by their absolute value.
Note that the input sequence of operations must be the subsequence of the output sequence.
It's guaranteed that there exists the correct answer consisting of no more than 1<=000<=000 operations.
Demo Input:
['2\ninsert 3\ngetMin 4\n', '4\ninsert 1\ninsert 1\nremoveMin\ngetMin 2\n']
Demo Output:
['4\ninsert 3\nremoveMin\ninsert 4\ngetMin 4\n', '6\ninsert 1\ninsert 1\nremoveMin\nremoveMin\ninsert 2\ngetMin 2\n']
Note:
In the first sample, after number 3 is inserted into the heap, the minimum number is 3. To make the result of the first getMin equal to 4 one should firstly remove number 3 from the heap and then add number 4 into the heap.
In the second sample case number 1 is inserted two times, so should be similarly removed twice.
|
```python
###### ### ####### ####### ## # ##### ### #####
# # # # # # # # # # # # # ###
# # # # # # # # # # # # # ###
###### ######### # # # # # # ######### #
###### ######### # # # # # # ######### #
# # # # # # # # # # #### # # #
# # # # # # # ## # # # # #
###### # # ####### ####### # # ##### # # # #
from __future__ import print_function # for PyPy2
# from itertools import permutations
# from functools import cmp_to_key # for adding custom comparator
# from fractions import Fraction
from collections import *
from sys import stdin
# from bisect import *
from heapq import *
from math import *
g = lambda : stdin.readline().strip()
gl = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
rr = lambda x : reversed(range(x))
mod = int(1e9)+7
inf = float("inf")
ans = []
h = []
for _ in range(gil()[0]):
res = gl()
if len(res) == 1: #removeMin
if h == []:
ans.append('insert 1')
ans.append(res[0])
if h:heappop(h)
elif res[0] == 'insert': # insert
ans.append(' '.join(res))
heappush(h, int(res[1]))
else: # getMin
val = int(res[1])
while h and h[0] < val:
ans.append('removeMin')
heappop(h)
if len(h) == 0:ans.append('insert '+res[1])
ans.append(' '.join(res))
print(len(ans))
print('\n'.join(ans))
```
| 0
|
|
545
|
C
|
Woodcutters
|
PROGRAMMING
| 1,500
|
[
"dp",
"greedy"
] | null | null |
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are *n* trees located along the road at points with coordinates *x*1,<=*x*2,<=...,<=*x**n*. Each tree has its height *h**i*. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [*x**i*<=-<=*h**i*,<=*x**i*] or [*x**i*;*x**i*<=+<=*h**i*]. The tree that is not cut down occupies a single point with coordinate *x**i*. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of trees.
Next *n* lines contain pairs of integers *x**i*,<=*h**i* (1<=≤<=*x**i*,<=*h**i*<=≤<=109) — the coordinate and the height of the *і*-th tree.
The pairs are given in the order of ascending *x**i*. No two trees are located at the point with the same coordinate.
|
Print a single number — the maximum number of trees that you can cut down by the given rules.
|
[
"5\n1 2\n2 1\n5 10\n10 9\n19 1\n",
"5\n1 2\n2 1\n5 10\n10 9\n20 1\n"
] |
[
"3\n",
"4\n"
] |
In the first sample you can fell the trees like that:
- fell the 1-st tree to the left — now it occupies segment [ - 1;1] - fell the 2-nd tree to the right — now it occupies segment [2;3] - leave the 3-rd tree — it occupies point 5 - leave the 4-th tree — it occupies point 10 - fell the 5-th tree to the right — now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
| 1,750
|
[
{
"input": "5\n1 2\n2 1\n5 10\n10 9\n19 1",
"output": "3"
},
{
"input": "5\n1 2\n2 1\n5 10\n10 9\n20 1",
"output": "4"
},
{
"input": "4\n10 4\n15 1\n19 3\n20 1",
"output": "4"
},
{
"input": "35\n1 7\n3 11\n6 12\n7 6\n8 5\n9 11\n15 3\n16 10\n22 2\n23 3\n25 7\n27 3\n34 5\n35 10\n37 3\n39 4\n40 5\n41 1\n44 1\n47 7\n48 11\n50 6\n52 5\n57 2\n58 7\n60 4\n62 1\n67 3\n68 12\n69 8\n70 1\n71 5\n72 5\n73 6\n74 4",
"output": "10"
},
{
"input": "40\n1 1\n2 1\n3 1\n4 1\n5 1\n6 1\n7 1\n8 1\n9 1\n10 1\n11 1\n12 1\n13 1\n14 1\n15 1\n16 1\n17 1\n18 1\n19 1\n20 1\n21 1\n22 1\n23 1\n24 1\n25 1\n26 1\n27 1\n28 1\n29 1\n30 1\n31 1\n32 1\n33 1\n34 1\n35 1\n36 1\n37 1\n38 1\n39 1\n40 1",
"output": "2"
},
{
"input": "67\n1 1\n3 8\n4 10\n7 8\n9 2\n10 1\n11 5\n12 8\n13 4\n16 6\n18 3\n19 3\n22 5\n24 6\n27 5\n28 3\n29 3\n30 5\n32 5\n33 10\n34 7\n35 8\n36 5\n41 3\n42 2\n43 5\n46 4\n48 4\n49 9\n52 4\n53 9\n55 1\n56 4\n59 7\n68 7\n69 4\n71 9\n72 10\n74 5\n76 4\n77 9\n80 7\n81 9\n82 5\n83 5\n84 9\n85 7\n86 9\n87 4\n88 7\n89 10\n90 3\n91 5\n92 10\n93 5\n94 8\n95 4\n96 2\n97 10\n98 1\n99 3\n100 1\n101 5\n102 4\n103 8\n104 8\n105 8",
"output": "5"
},
{
"input": "1\n1000000000 1000000000",
"output": "1"
},
{
"input": "10\n7 12\n10 2\n12 2\n15 1\n19 2\n20 1\n53 25\n63 10\n75 12\n87 1",
"output": "9"
},
{
"input": "3\n1 1\n1000 1000\n1000000000 1000000000",
"output": "3"
},
{
"input": "2\n1 999999999\n1000000000 1000000000",
"output": "2"
},
{
"input": "10\n999999900 1000000000\n999999901 1000000000\n999999902 1000000000\n999999903 1000000000\n999999904 1000000000\n999999905 1000000000\n999999906 1000000000\n999999907 1000000000\n999999908 1000000000\n999999909 1000000000",
"output": "2"
},
{
"input": "2\n100000000 1000000000\n1000000000 1000000000",
"output": "2"
}
] | 1,667,106,041
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 67
| 358
| 10,240,000
|
n=int(input())
if n==1:
a=map(int,input().split())
print(1)
if n>1:
s=[]
b=0
for i in range(n):
a=list(map(int,input().split()))
s.append(a)
for i in range(1,len(s)-1):
if s[i][0]-s[i-1][0]>s[i][1]:
b+=1
elif s[i+1][0]-s[i][0]>s[i][1]:
b+=1
s[i][0]+=s[i][1]
print(b+2)
|
Title: Woodcutters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are *n* trees located along the road at points with coordinates *x*1,<=*x*2,<=...,<=*x**n*. Each tree has its height *h**i*. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [*x**i*<=-<=*h**i*,<=*x**i*] or [*x**i*;*x**i*<=+<=*h**i*]. The tree that is not cut down occupies a single point with coordinate *x**i*. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of trees.
Next *n* lines contain pairs of integers *x**i*,<=*h**i* (1<=≤<=*x**i*,<=*h**i*<=≤<=109) — the coordinate and the height of the *і*-th tree.
The pairs are given in the order of ascending *x**i*. No two trees are located at the point with the same coordinate.
Output Specification:
Print a single number — the maximum number of trees that you can cut down by the given rules.
Demo Input:
['5\n1 2\n2 1\n5 10\n10 9\n19 1\n', '5\n1 2\n2 1\n5 10\n10 9\n20 1\n']
Demo Output:
['3\n', '4\n']
Note:
In the first sample you can fell the trees like that:
- fell the 1-st tree to the left — now it occupies segment [ - 1;1] - fell the 2-nd tree to the right — now it occupies segment [2;3] - leave the 3-rd tree — it occupies point 5 - leave the 4-th tree — it occupies point 10 - fell the 5-th tree to the right — now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
|
```python
n=int(input())
if n==1:
a=map(int,input().split())
print(1)
if n>1:
s=[]
b=0
for i in range(n):
a=list(map(int,input().split()))
s.append(a)
for i in range(1,len(s)-1):
if s[i][0]-s[i-1][0]>s[i][1]:
b+=1
elif s[i+1][0]-s[i][0]>s[i][1]:
b+=1
s[i][0]+=s[i][1]
print(b+2)
```
| 3
|
|
284
|
B
|
Cows and Poker Game
|
PROGRAMMING
| 1,000
|
[
"brute force",
"implementation"
] | null | null |
There are *n* cows playing poker at a table. For the current betting phase, each player's status is either "ALLIN", "IN", or "FOLDED", and does not change throughout the phase. To increase the suspense, a player whose current status is not "FOLDED" may show his/her hand to the table. However, so as not to affect any betting decisions, he/she may only do so if all other players have a status of either "ALLIN" or "FOLDED". The player's own status may be either "ALLIN" or "IN".
Find the number of cows that can currently show their hands without affecting any betting decisions.
|
The first line contains a single integer, *n* (2<=≤<=*n*<=≤<=2·105). The second line contains *n* characters, each either "A", "I", or "F". The *i*-th character is "A" if the *i*-th player's status is "ALLIN", "I" if the *i*-th player's status is "IN", or "F" if the *i*-th player's status is "FOLDED".
|
The first line should contain a single integer denoting the number of players that can currently show their hands.
|
[
"6\nAFFAAA\n",
"3\nAFI\n"
] |
[
"4\n",
"1\n"
] |
In the first sample, cows 1, 4, 5, and 6 can show their hands. In the second sample, only cow 3 can show her hand.
| 1,000
|
[
{
"input": "6\nAFFAAA",
"output": "4"
},
{
"input": "3\nAFI",
"output": "1"
},
{
"input": "3\nFFF",
"output": "0"
},
{
"input": "3\nFIF",
"output": "1"
},
{
"input": "3\nAAA",
"output": "3"
},
{
"input": "3\nIII",
"output": "0"
},
{
"input": "3\nIIA",
"output": "0"
},
{
"input": "3\nAFF",
"output": "1"
},
{
"input": "5\nFAFFF",
"output": "1"
},
{
"input": "3\nIAA",
"output": "1"
},
{
"input": "3\nIIF",
"output": "0"
},
{
"input": "2\nFA",
"output": "1"
},
{
"input": "2\nFF",
"output": "0"
},
{
"input": "2\nIF",
"output": "1"
},
{
"input": "5\nAAAAI",
"output": "1"
},
{
"input": "5\nIIIIF",
"output": "0"
},
{
"input": "10\nAAAAAAAAAA",
"output": "10"
},
{
"input": "100\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA",
"output": "100"
},
{
"input": "5\nFAIAF",
"output": "1"
},
{
"input": "5\nAIAIF",
"output": "0"
},
{
"input": "5\nFAAII",
"output": "0"
},
{
"input": "5\nAIFFF",
"output": "1"
},
{
"input": "5\nAFAFA",
"output": "3"
},
{
"input": "2\nFA",
"output": "1"
},
{
"input": "8\nAFFFFIAF",
"output": "1"
},
{
"input": "8\nIAAIFFFI",
"output": "0"
},
{
"input": "5\nIIIII",
"output": "0"
}
] | 1,517,748,921
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 66
| 218
| 6,144,000
|
a = int(input())
s = input()
b = {'A':0, 'I':0, 'F':0}
for i in s:
b[i] += 1
if b['I'] > 1:
print(0)
elif b['I'] == 1:
print(1)
else:
print(b['A'])
|
Title: Cows and Poker Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are *n* cows playing poker at a table. For the current betting phase, each player's status is either "ALLIN", "IN", or "FOLDED", and does not change throughout the phase. To increase the suspense, a player whose current status is not "FOLDED" may show his/her hand to the table. However, so as not to affect any betting decisions, he/she may only do so if all other players have a status of either "ALLIN" or "FOLDED". The player's own status may be either "ALLIN" or "IN".
Find the number of cows that can currently show their hands without affecting any betting decisions.
Input Specification:
The first line contains a single integer, *n* (2<=≤<=*n*<=≤<=2·105). The second line contains *n* characters, each either "A", "I", or "F". The *i*-th character is "A" if the *i*-th player's status is "ALLIN", "I" if the *i*-th player's status is "IN", or "F" if the *i*-th player's status is "FOLDED".
Output Specification:
The first line should contain a single integer denoting the number of players that can currently show their hands.
Demo Input:
['6\nAFFAAA\n', '3\nAFI\n']
Demo Output:
['4\n', '1\n']
Note:
In the first sample, cows 1, 4, 5, and 6 can show their hands. In the second sample, only cow 3 can show her hand.
|
```python
a = int(input())
s = input()
b = {'A':0, 'I':0, 'F':0}
for i in s:
b[i] += 1
if b['I'] > 1:
print(0)
elif b['I'] == 1:
print(1)
else:
print(b['A'])
```
| 3
|
|
767
|
A
|
Snacktower
|
PROGRAMMING
| 1,100
|
[
"data structures",
"implementation"
] | null | null |
According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents.
|
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*.
|
Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty.
|
[
"3\n3 1 2\n",
"5\n4 5 1 2 3\n"
] |
[
"3\n \n2 1",
"5 4\n \n \n3 2 1\n"
] |
In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before.
| 500
|
[
{
"input": "3\n3 1 2",
"output": "3 \n\n2 1 "
},
{
"input": "5\n4 5 1 2 3",
"output": "5 4 \n\n\n3 2 1 "
},
{
"input": "1\n1",
"output": "1 "
},
{
"input": "2\n1 2",
"output": "2 1 "
},
{
"input": "10\n5 1 6 2 8 3 4 10 9 7",
"output": "10 \n9 8 \n7 6 5 4 3 2 1 "
},
{
"input": "30\n16 10 4 29 5 28 12 21 11 30 18 6 14 3 17 22 20 15 9 1 27 19 24 26 13 25 2 23 8 7",
"output": "30 29 28 \n\n\n\n\n\n\n\n\n\n\n27 \n\n\n26 \n\n25 24 \n\n23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 \n8 \n7 6 5 4 3 2 1 "
},
{
"input": "100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 4 80 72 39",
"output": "100 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n99 98 \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n97 96 95 94 93 92 91 90 89 88 87 86 85 84 83 82 81 \n\n\n\n\n\n\n\n\n80 79 78 77 76 75 74 73 \n72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 \n39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "2\n2 1",
"output": "2 \n1 "
}
] | 1,592,964,461
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 4
| 140
| 20,172,800
|
n = int(input())
ara = list(int(i) for i in input().split())
a = None
var = []
for i in range(n):
if i == 0:
a = ara[i]
var.append(ara[i])
else:
if a < ara[i]:
print()
var.append(ara[i])
a = ara[i]
else:
for j in range(len(var)-1, -1, -1):
print(var[j], end=" ")
print()
var = []
a = ara[i]
var.append(ara[i])
for j in range(len(var)-1, -1, -1):
print(var[j], end=" ")
print()
|
Title: Snacktower
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time *n* snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.
Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower.
However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.
Write a program that models the behavior of Ankh-Morpork residents.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the total number of snacks.
The second line contains *n* integers, the *i*-th of them equals the size of the snack which fell on the *i*-th day. Sizes are distinct integers from 1 to *n*.
Output Specification:
Print *n* lines. On the *i*-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the *i*-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty.
Demo Input:
['3\n3 1 2\n', '5\n4 5 1 2 3\n']
Demo Output:
['3\n\xa0\n2 1', '5 4\n\xa0\n\xa0\n3 2 1\n']
Note:
In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before.
|
```python
n = int(input())
ara = list(int(i) for i in input().split())
a = None
var = []
for i in range(n):
if i == 0:
a = ara[i]
var.append(ara[i])
else:
if a < ara[i]:
print()
var.append(ara[i])
a = ara[i]
else:
for j in range(len(var)-1, -1, -1):
print(var[j], end=" ")
print()
var = []
a = ara[i]
var.append(ara[i])
for j in range(len(var)-1, -1, -1):
print(var[j], end=" ")
print()
```
| 0
|
|
581
|
A
|
Vasya the Hipster
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
|
The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got.
|
Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
|
[
"3 1\n",
"2 3\n",
"7 3\n"
] |
[
"1 1\n",
"2 0\n",
"3 2\n"
] |
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.
| 500
|
[
{
"input": "3 1",
"output": "1 1"
},
{
"input": "2 3",
"output": "2 0"
},
{
"input": "7 3",
"output": "3 2"
},
{
"input": "100 100",
"output": "100 0"
},
{
"input": "4 10",
"output": "4 3"
},
{
"input": "6 10",
"output": "6 2"
},
{
"input": "6 11",
"output": "6 2"
},
{
"input": "10 40",
"output": "10 15"
},
{
"input": "11 56",
"output": "11 22"
},
{
"input": "34 30",
"output": "30 2"
},
{
"input": "33 33",
"output": "33 0"
},
{
"input": "100 45",
"output": "45 27"
},
{
"input": "100 23",
"output": "23 38"
},
{
"input": "45 12",
"output": "12 16"
},
{
"input": "1 1",
"output": "1 0"
},
{
"input": "1 100",
"output": "1 49"
},
{
"input": "100 1",
"output": "1 49"
},
{
"input": "68 59",
"output": "59 4"
},
{
"input": "45 99",
"output": "45 27"
},
{
"input": "99 100",
"output": "99 0"
},
{
"input": "100 98",
"output": "98 1"
},
{
"input": "59 12",
"output": "12 23"
},
{
"input": "86 4",
"output": "4 41"
},
{
"input": "68 21",
"output": "21 23"
},
{
"input": "100 11",
"output": "11 44"
},
{
"input": "100 10",
"output": "10 45"
},
{
"input": "15 45",
"output": "15 15"
},
{
"input": "11 32",
"output": "11 10"
},
{
"input": "34 96",
"output": "34 31"
},
{
"input": "89 89",
"output": "89 0"
}
] | 1,662,528,539
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 46
| 0
|
li=list(map(int,input().split()))
print(min(li),(abs(li[0]-li[1]))//2)
|
Title: Vasya the Hipster
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
Input Specification:
The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got.
Output Specification:
Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
Demo Input:
['3 1\n', '2 3\n', '7 3\n']
Demo Output:
['1 1\n', '2 0\n', '3 2\n']
Note:
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.
|
```python
li=list(map(int,input().split()))
print(min(li),(abs(li[0]-li[1]))//2)
```
| 3
|
|
61
|
A
|
Ultra-Fast Mathematician
|
PROGRAMMING
| 800
|
[
"implementation"
] |
A. Ultra-Fast Mathematician
|
2
|
256
|
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
|
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
|
Write one line — the corresponding answer. Do not omit the leading 0s.
|
[
"1010100\n0100101\n",
"000\n111\n",
"1110\n1010\n",
"01110\n01100\n"
] |
[
"1110001\n",
"111\n",
"0100\n",
"00010\n"
] |
none
| 500
|
[
{
"input": "1010100\n0100101",
"output": "1110001"
},
{
"input": "000\n111",
"output": "111"
},
{
"input": "1110\n1010",
"output": "0100"
},
{
"input": "01110\n01100",
"output": "00010"
},
{
"input": "011101\n000001",
"output": "011100"
},
{
"input": "10\n01",
"output": "11"
},
{
"input": "00111111\n11011101",
"output": "11100010"
},
{
"input": "011001100\n101001010",
"output": "110000110"
},
{
"input": "1100100001\n0110101100",
"output": "1010001101"
},
{
"input": "00011101010\n10010100101",
"output": "10001001111"
},
{
"input": "100000101101\n111010100011",
"output": "011010001110"
},
{
"input": "1000001111010\n1101100110001",
"output": "0101101001011"
},
{
"input": "01011111010111\n10001110111010",
"output": "11010001101101"
},
{
"input": "110010000111100\n001100101011010",
"output": "111110101100110"
},
{
"input": "0010010111110000\n0000000011010110",
"output": "0010010100100110"
},
{
"input": "00111110111110000\n01111100001100000",
"output": "01000010110010000"
},
{
"input": "101010101111010001\n001001111101111101",
"output": "100011010010101100"
},
{
"input": "0110010101111100000\n0011000101000000110",
"output": "0101010000111100110"
},
{
"input": "11110100011101010111\n00001000011011000000",
"output": "11111100000110010111"
},
{
"input": "101010101111101101001\n111010010010000011111",
"output": "010000111101101110110"
},
{
"input": "0000111111100011000010\n1110110110110000001010",
"output": "1110001001010011001000"
},
{
"input": "10010010101000110111000\n00101110100110111000111",
"output": "10111100001110001111111"
},
{
"input": "010010010010111100000111\n100100111111100011001110",
"output": "110110101101011111001001"
},
{
"input": "0101110100100111011010010\n0101100011010111001010001",
"output": "0000010111110000010000011"
},
{
"input": "10010010100011110111111011\n10000110101100000001000100",
"output": "00010100001111110110111111"
},
{
"input": "000001111000000100001000000\n011100111101111001110110001",
"output": "011101000101111101111110001"
},
{
"input": "0011110010001001011001011100\n0000101101000011101011001010",
"output": "0011011111001010110010010110"
},
{
"input": "11111000000000010011001101111\n11101110011001010100010000000",
"output": "00010110011001000111011101111"
},
{
"input": "011001110000110100001100101100\n001010000011110000001000101001",
"output": "010011110011000100000100000101"
},
{
"input": "1011111010001100011010110101111\n1011001110010000000101100010101",
"output": "0000110100011100011111010111010"
},
{
"input": "10111000100001000001010110000001\n10111000001100101011011001011000",
"output": "00000000101101101010001111011001"
},
{
"input": "000001010000100001000000011011100\n111111111001010100100001100000111",
"output": "111110101001110101100001111011011"
},
{
"input": "1101000000000010011011101100000110\n1110000001100010011010000011011110",
"output": "0011000001100000000001101111011000"
},
{
"input": "01011011000010100001100100011110001\n01011010111000001010010100001110000",
"output": "00000001111010101011110000010000001"
},
{
"input": "000011111000011001000110111100000100\n011011000110000111101011100111000111",
"output": "011000111110011110101101011011000011"
},
{
"input": "1001000010101110001000000011111110010\n0010001011010111000011101001010110000",
"output": "1011001001111001001011101010101000010"
},
{
"input": "00011101011001100101111111000000010101\n10010011011011001011111000000011101011",
"output": "10001110000010101110000111000011111110"
},
{
"input": "111011100110001001101111110010111001010\n111111101101111001110010000101101000100",
"output": "000100001011110000011101110111010001110"
},
{
"input": "1111001001101000001000000010010101001010\n0010111100111110001011000010111110111001",
"output": "1101110101010110000011000000101011110011"
},
{
"input": "00100101111000000101011111110010100011010\n11101110001010010101001000111110101010100",
"output": "11001011110010010000010111001100001001110"
},
{
"input": "101011001110110100101001000111010101101111\n100111100110101011010100111100111111010110",
"output": "001100101000011111111101111011101010111001"
},
{
"input": "1111100001100101000111101001001010011100001\n1000110011000011110010001011001110001000001",
"output": "0111010010100110110101100010000100010100000"
},
{
"input": "01100111011111010101000001101110000001110101\n10011001011111110000000101011001001101101100",
"output": "11111110000000100101000100110111001100011001"
},
{
"input": "110010100111000100100101100000011100000011001\n011001111011100110000110111001110110100111011",
"output": "101011011100100010100011011001101010100100010"
},
{
"input": "0001100111111011010110100100111000000111000110\n1100101011000000000001010010010111001100110001",
"output": "1101001100111011010111110110101111001011110111"
},
{
"input": "00000101110110110001110010100001110100000100000\n10010000110011110001101000111111101010011010001",
"output": "10010101000101000000011010011110011110011110001"
},
{
"input": "110000100101011100100011001111110011111110010001\n101011111001011100110110111101110011010110101100",
"output": "011011011100000000010101110010000000101000111101"
},
{
"input": "0101111101011111010101011101000011101100000000111\n0000101010110110001110101011011110111001010100100",
"output": "0101010111101001011011110110011101010101010100011"
},
{
"input": "11000100010101110011101000011111001010110111111100\n00001111000111001011111110000010101110111001000011",
"output": "11001011010010111000010110011101100100001110111111"
},
{
"input": "101000001101111101101111111000001110110010101101010\n010011100111100001100000010001100101000000111011011",
"output": "111011101010011100001111101001101011110010010110001"
},
{
"input": "0011111110010001010100010110111000110011001101010100\n0111000000100010101010000100101000000100101000111001",
"output": "0100111110110011111110010010010000110111100101101101"
},
{
"input": "11101010000110000011011010000001111101000111011111100\n10110011110001010100010110010010101001010111100100100",
"output": "01011001110111010111001100010011010100010000111011000"
},
{
"input": "011000100001000001101000010110100110011110100111111011\n111011001000001001110011001111011110111110110011011111",
"output": "100011101001001000011011011001111000100000010100100100"
},
{
"input": "0111010110010100000110111011010110100000000111110110000\n1011100100010001101100000100111111101001110010000100110",
"output": "1100110010000101101010111111101001001001110101110010110"
},
{
"input": "10101000100111000111010001011011011011110100110101100011\n11101111000000001100100011111000100100000110011001101110",
"output": "01000111100111001011110010100011111111110010101100001101"
},
{
"input": "000000111001010001000000110001001011100010011101010011011\n110001101000010010000101000100001111101001100100001010010",
"output": "110001010001000011000101110101000100001011111001011001001"
},
{
"input": "0101011100111010000111110010101101111111000000111100011100\n1011111110000010101110111001000011100000100111111111000111",
"output": "1110100010111000101001001011101110011111100111000011011011"
},
{
"input": "11001000001100100111100111100100101011000101001111001001101\n10111110100010000011010100110100100011101001100000001110110",
"output": "01110110101110100100110011010000001000101100101111000111011"
},
{
"input": "010111011011101000000110000110100110001110100001110110111011\n101011110011101011101101011111010100100001100111100100111011",
"output": "111100101000000011101011011001110010101111000110010010000000"
},
{
"input": "1001011110110110000100011001010110000100011010010111010101110\n1101111100001000010111110011010101111010010100000001000010111",
"output": "0100100010111110010011101010000011111110001110010110010111001"
},
{
"input": "10000010101111100111110101111000010100110111101101111111111010\n10110110101100101010011001011010100110111011101100011001100111",
"output": "00110100000011001101101100100010110010001100000001100110011101"
},
{
"input": "011111010011111000001010101001101001000010100010111110010100001\n011111001011000011111001000001111001010110001010111101000010011",
"output": "000000011000111011110011101000010000010100101000000011010110010"
},
{
"input": "1111000000110001011101000100100100001111011100001111001100011111\n1101100110000101100001100000001001011011111011010101000101001010",
"output": "0010100110110100111100100100101101010100100111011010001001010101"
},
{
"input": "01100000101010010011001110100110110010000110010011011001100100011\n10110110010110111100100111000111000110010000000101101110000010111",
"output": "11010110111100101111101001100001110100010110010110110111100110100"
},
{
"input": "001111111010000100001100001010011001111110011110010111110001100111\n110000101001011000100010101100100110000111100000001101001110010111",
"output": "111111010011011100101110100110111111111001111110011010111111110000"
},
{
"input": "1011101011101101011110101101011101011000010011100101010101000100110\n0001000001001111010111100100111101100000000001110001000110000000110",
"output": "1010101010100010001001001001100000111000010010010100010011000100000"
},
{
"input": "01000001011001010011011100010000100100110101111011011011110000001110\n01011110000110011011000000000011000111100001010000000011111001110000",
"output": "00011111011111001000011100010011100011010100101011011000001001111110"
},
{
"input": "110101010100110101000001111110110100010010000100111110010100110011100\n111010010111111011100110101011001011001110110111110100000110110100111",
"output": "001111000011001110100111010101111111011100110011001010010010000111011"
},
{
"input": "1001101011000001011111100110010010000011010001001111011100010100110001\n1111100111110101001111010001010000011001001001010110001111000000100101",
"output": "0110001100110100010000110111000010011010011000011001010011010100010100"
},
{
"input": "00000111110010110001110110001010010101000111011001111111100110011110010\n00010111110100000100110101000010010001100001100011100000001100010100010",
"output": "00010000000110110101000011001000000100100110111010011111101010001010000"
},
{
"input": "100101011100101101000011010001011001101110101110001100010001010111001110\n100001111100101011011111110000001111000111001011111110000010101110111001",
"output": "000100100000000110011100100001010110101001100101110010010011111001110111"
},
{
"input": "1101100001000111001101001011101000111000011110000001001101101001111011010\n0101011101010100011011010110101000010010110010011110101100000110110001000",
"output": "1000111100010011010110011101000000101010101100011111100001101111001010010"
},
{
"input": "01101101010011110101100001110101111011100010000010001101111000011110111111\n00101111001101001100111010000101110000100101101111100111101110010100011011",
"output": "01000010011110111001011011110000001011000111101101101010010110001010100100"
},
{
"input": "101100101100011001101111110110110010100110110010100001110010110011001101011\n000001011010101011110011111101001110000111000010001101000010010000010001101",
"output": "101101110110110010011100001011111100100001110000101100110000100011011100110"
},
{
"input": "0010001011001010001100000010010011110110011000100000000100110000101111001110\n1100110100111000110100001110111001011101001100001010100001010011100110110001",
"output": "1110111111110010111000001100101010101011010100101010100101100011001001111111"
},
{
"input": "00101101010000000101011001101011001100010001100000101011101110000001111001000\n10010110010111000000101101000011101011001010000011011101101011010000000011111",
"output": "10111011000111000101110100101000100111011011100011110110000101010001111010111"
},
{
"input": "111100000100100000101001100001001111001010001000001000000111010000010101101011\n001000100010100101111011111011010110101100001111011000010011011011100010010110",
"output": "110100100110000101010010011010011001100110000111010000010100001011110111111101"
},
{
"input": "0110001101100100001111110101101000100101010010101010011001101001001101110000000\n0111011000000010010111011110010000000001000110001000011001101000000001110100111",
"output": "0001010101100110011000101011111000100100010100100010000000000001001100000100111"
},
{
"input": "10001111111001000101001011110101111010100001011010101100111001010001010010001000\n10000111010010011110111000111010101100000011110001101111001000111010100000000001",
"output": "00001000101011011011110011001111010110100010101011000011110001101011110010001001"
},
{
"input": "100110001110110000100101001110000011110110000110000000100011110100110110011001101\n110001110101110000000100101001101011111100100100001001000110000001111100011110110",
"output": "010111111011000000100001100111101000001010100010001001100101110101001010000111011"
},
{
"input": "0000010100100000010110111100011111111010011101000000100000011001001101101100111010\n0100111110011101010110101011110110010111001111000110101100101110111100101000111111",
"output": "0100101010111101000000010111101001101101010010000110001100110111110001000100000101"
},
{
"input": "11000111001010100001110000001001011010010010110000001110100101000001010101100110111\n11001100100100100001101010110100000111100011101110011010110100001001000011011011010",
"output": "00001011101110000000011010111101011101110001011110010100010001001000010110111101101"
},
{
"input": "010110100010001000100010101001101010011010111110100001000100101000111011100010100001\n110000011111101101010011111000101010111010100001001100001001100101000000111000000000",
"output": "100110111101100101110001010001000000100000011111101101001101001101111011011010100001"
},
{
"input": "0000011110101110010101110110110101100001011001101010101001000010000010000000101001101\n1100111111011100000110000111101110011111100111110001011001000010011111100001001100011",
"output": "1100100001110010010011110001011011111110111110011011110000000000011101100001100101110"
},
{
"input": "10100000101101110001100010010010100101100011010010101000110011100000101010110010000000\n10001110011011010010111011011101101111000111110000111000011010010101001100000001010011",
"output": "00101110110110100011011001001111001010100100100010010000101001110101100110110011010011"
},
{
"input": "001110000011111101101010011111000101010111010100001001100001001100101000000111000000000\n111010000000000000101001110011001000111011001100101010011001000011101001001011110000011",
"output": "110100000011111101000011101100001101101100011000100011111000001111000001001100110000011"
},
{
"input": "1110111100111011010101011011001110001010010010110011110010011111000010011111010101100001\n1001010101011001001010100010101100000110111101011000100010101111111010111100001110010010",
"output": "0111101001100010011111111001100010001100101111101011010000110000111000100011011011110011"
},
{
"input": "11100010001100010011001100001100010011010001101110011110100101110010101101011101000111111\n01110000000110111010110100001010000101011110100101010011000110101110101101110111011110001",
"output": "10010010001010101001111000000110010110001111001011001101100011011100000000101010011001110"
},
{
"input": "001101011001100101101100110000111000101011001001100100000100101000100000110100010111111101\n101001111110000010111101111110001001111001111101111010000110111000100100110010010001011111",
"output": "100100100111100111010001001110110001010010110100011110000010010000000100000110000110100010"
},
{
"input": "1010110110010101000110010010110101011101010100011001101011000110000000100011100100011000000\n0011011111100010001111101101000111001011101110100000110111100100101111010110101111011100011",
"output": "1001101001110111001001111111110010010110111010111001011100100010101111110101001011000100011"
},
{
"input": "10010010000111010111011111110010100101100000001100011100111011100010000010010001011100001100\n00111010100010110010000100010111010001111110100100100011101000101111111111001101101100100100",
"output": "10101000100101100101011011100101110100011110101000111111010011001101111101011100110000101000"
},
{
"input": "010101110001010101100000010111010000000111110011001101100011001000000011001111110000000010100\n010010111011100101010101111110110000000111000100001101101001001000001100101110001010000100001",
"output": "000111001010110000110101101001100000000000110111000000001010000000001111100001111010000110101"
},
{
"input": "1100111110011001000111101001001011000110011010111111100010111111001100111111011101100111101011\n1100000011001000110100110111000001011001010111101000010010100011000001100100111101101000010110",
"output": "0000111101010001110011011110001010011111001101010111110000011100001101011011100000001111111101"
},
{
"input": "00011000100100110111100101100100000000010011110111110010101110110011100001010111010011110100101\n00011011111011111011100101100111100101001110010111000010000111000100100100000001110101111011011",
"output": "00000011011111001100000000000011100101011101100000110000101001110111000101010110100110001111110"
},
{
"input": "000101011001001100000111100010110101111011110101111101000110001101011010111110110011100100000001\n011000101010011111011000111000100000000011011000000001111110001000001111101010110000011100001111",
"output": "011101110011010011011111011010010101111000101101111100111000000101010101010100000011111000001110"
},
{
"input": "1000101001011010000100100100010010011101011001110101111011101111111110010101001101010001010101001\n0110110010011100011111011111110111000000010001110100001010111110101011010011111011111110000110000",
"output": "1110011011000110011011111011100101011101001000000001110001010001010101000110110110101111010011001"
},
{
"input": "01111010010101100111110111111011011010100001011101010000111100101101101110111011001100101011100111\n00001100110110100001111011000010001001001100000010111101000001111011100000010111010010000011000010",
"output": "01110110100011000110001100111001010011101101011111101101111101010110001110101100011110101000100101"
},
{
"input": "111101011101111000001011001101111010110111001101110100100011111011011101001101010101011010111000110\n101111100101101101001100110011000001111010011101110111110110101110011011110011111100001001110101101",
"output": "010010111000010101000111111110111011001101010000000011010101010101000110111110101001010011001101011"
},
{
"input": "1010000011010110011101001101110001110010000111011101110010110110111111001001110100101100010101010001\n0111101000111100101100000101111010100100001000011101010100110011100011010011010101000100101011100011",
"output": "1101101011101010110001001000001011010110001111000000100110000101011100011010100001101000111110110010"
},
{
"input": "0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001\n1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111010",
"output": "1111111010111111101011111110101111111010111111101011111110101111111010111111101011111110101111111011"
},
{
"input": "0\n0",
"output": "0"
},
{
"input": "0\n1",
"output": "1"
}
] | 1,618,514,760
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 102
| 77
| 0
|
n = list(input())
m = list(input())
N = []
M = []
for i in n:
N.append(int(i))
for i in m:
M.append(int(i))
#print(M,N)
res = []
for i in range(len(n)):
if N[i]==0 and M[i] == 0 :
res.append(0)
elif N[i]==1 and M[i] == 1 :
res.append(0)
else:
res.append(1)
final_res = []
for i in res:
final_res.append(str(i))
print(''.join(final_res))
|
Title: Ultra-Fast Mathematician
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Shapur was an extremely gifted student. He was great at everything including Combinatorics, Algebra, Number Theory, Geometry, Calculus, etc. He was not only smart but extraordinarily fast! He could manage to sum 1018 numbers in a single second.
One day in 230 AD Shapur was trying to find out if any one can possibly do calculations faster than him. As a result he made a very great contest and asked every one to come and take part.
In his contest he gave the contestants many different pairs of numbers. Each number is made from digits 0 or 1. The contestants should write a new number corresponding to the given pair of numbers. The rule is simple: The *i*-th digit of the answer is 1 if and only if the *i*-th digit of the two given numbers differ. In the other case the *i*-th digit of the answer is 0.
Shapur made many numbers and first tried his own speed. He saw that he can perform these operations on numbers of length ∞ (length of a number is number of digits in it) in a glance! He always gives correct answers so he expects the contestants to give correct answers, too. He is a good fellow so he won't give anyone very big numbers and he always gives one person numbers of same length.
Now you are going to take part in Shapur's contest. See if you are faster and more accurate.
Input Specification:
There are two lines in each input. Each of them contains a single number. It is guaranteed that the numbers are made from 0 and 1 only and that their length is same. The numbers may start with 0. The length of each number doesn't exceed 100.
Output Specification:
Write one line — the corresponding answer. Do not omit the leading 0s.
Demo Input:
['1010100\n0100101\n', '000\n111\n', '1110\n1010\n', '01110\n01100\n']
Demo Output:
['1110001\n', '111\n', '0100\n', '00010\n']
Note:
none
|
```python
n = list(input())
m = list(input())
N = []
M = []
for i in n:
N.append(int(i))
for i in m:
M.append(int(i))
#print(M,N)
res = []
for i in range(len(n)):
if N[i]==0 and M[i] == 0 :
res.append(0)
elif N[i]==1 and M[i] == 1 :
res.append(0)
else:
res.append(1)
final_res = []
for i in res:
final_res.append(str(i))
print(''.join(final_res))
```
| 3.98075
|
844
|
B
|
Rectangles
|
PROGRAMMING
| 1,300
|
[
"combinatorics",
"math"
] | null | null |
You are given *n*<=×<=*m* table. Each cell of the table is colored white or black. Find the number of non-empty sets of cells such that:
1. All cells in a set have the same color. 1. Every two cells in a set share row or column.
|
The first line of input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) — the number of rows and the number of columns correspondingly.
The next *n* lines of input contain descriptions of rows. There are *m* integers, separated by spaces, in each line. The number equals 0 if the corresponding cell is colored white and equals 1 if the corresponding cell is colored black.
|
Output single integer — the number of non-empty sets from the problem description.
|
[
"1 1\n0\n",
"2 3\n1 0 1\n0 1 0\n"
] |
[
"1\n",
"8\n"
] |
In the second example, there are six one-element sets. Additionally, there are two two-element sets, the first one consists of the first and the third cells of the first row, the second one consists of the first and the third cells of the second row. To sum up, there are 8 sets.
| 1,000
|
[
{
"input": "1 1\n0",
"output": "1"
},
{
"input": "2 3\n1 0 1\n0 1 0",
"output": "8"
},
{
"input": "2 2\n1 1\n1 1",
"output": "8"
},
{
"input": "1 10\n0 0 0 0 0 0 0 0 0 0",
"output": "1023"
},
{
"input": "11 1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1",
"output": "2047"
},
{
"input": "10 11\n1 1 0 1 1 0 0 0 1 0 0\n1 0 0 1 1 1 0 0 1 1 0\n0 0 1 0 1 1 0 1 0 1 1\n0 1 1 1 0 1 0 1 0 0 0\n1 1 1 1 1 1 1 0 1 0 0\n1 1 0 1 1 1 1 0 0 1 1\n1 0 1 0 1 0 0 1 1 1 0\n1 1 0 0 0 0 0 1 0 1 1\n1 1 0 1 1 1 0 0 1 1 0\n1 0 1 1 0 0 1 0 0 1 1",
"output": "2444"
},
{
"input": "50 1\n0\n1\n0\n1\n0\n1\n0\n1\n1\n1\n0\n0\n1\n0\n0\n1\n1\n1\n1\n0\n1\n1\n0\n1\n1\n1\n0\n1\n0\n0\n0\n1\n1\n0\n1\n1\n0\n1\n0\n1\n0\n0\n1\n0\n0\n0\n1\n1\n0\n1",
"output": "142606334"
},
{
"input": "1 50\n0 1 0 1 0 1 0 1 1 1 0 0 1 0 0 1 1 1 1 0 1 1 0 1 1 1 0 1 0 0 0 1 1 0 1 1 0 1 0 1 0 0 1 0 0 0 1 1 0 1",
"output": "142606334"
},
{
"input": "2 20\n0 1 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0",
"output": "589853"
},
{
"input": "5 5\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "285"
},
{
"input": "6 6\n1 1 1 1 1 1\n1 1 1 1 1 1\n1 1 1 1 1 1\n1 1 1 1 1 1\n1 1 1 1 1 1\n1 1 1 1 1 1",
"output": "720"
},
{
"input": "21 2\n0 1\n1 1\n0 1\n0 0\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "1310745"
},
{
"input": "3 15\n1 0 0 0 0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 0 1 0 1 0 0 0 0 0 1 0\n1 0 0 1 0 0 0 0 0 0 0 0 1 0 1",
"output": "22587"
},
{
"input": "10 11\n0 1 0 0 0 0 0 0 0 0 0\n0 1 0 1 0 0 1 0 0 0 0\n0 0 0 0 0 0 1 1 1 0 0\n0 0 0 0 0 0 0 0 0 0 0\n0 0 0 0 1 0 0 0 0 1 0\n0 0 0 0 0 0 1 0 0 0 0\n0 0 0 0 0 0 0 0 0 1 0\n0 0 1 0 0 0 1 1 0 0 0\n0 0 0 0 0 0 0 0 1 0 0\n0 0 1 0 1 0 0 0 0 1 1",
"output": "12047"
},
{
"input": "14 15\n0 1 0 0 0 0 0 0 1 0 0 0 1 0 1\n0 0 0 1 1 1 1 0 1 0 0 1 1 0 0\n1 0 0 0 0 1 1 0 0 0 0 0 0 0 0\n0 1 0 0 0 1 0 1 1 0 0 1 0 0 0\n0 0 1 1 0 1 0 1 0 1 1 0 1 0 0\n0 0 0 1 1 0 0 0 0 0 1 1 0 1 0\n0 0 1 0 0 0 0 0 0 1 0 0 1 1 0\n1 1 0 0 0 1 0 0 0 0 0 0 1 1 0\n0 0 0 0 1 0 1 1 1 0 0 0 1 0 1\n1 0 1 1 0 1 0 0 1 0 0 1 1 1 0\n1 0 0 0 0 1 0 0 0 0 0 1 0 0 0\n0 0 0 1 0 1 0 0 0 0 1 0 0 0 1\n0 0 1 0 1 0 0 0 1 1 1 1 0 0 0\n0 0 0 0 0 0 0 0 0 0 0 1 0 0 0",
"output": "53166"
},
{
"input": "1 50\n0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 0 0 1 0 0",
"output": "1099511628798"
},
{
"input": "50 1\n0\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n0\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n0\n1\n1\n1\n1\n1\n1\n1\n0\n1\n1\n1\n1\n0\n1\n1\n1\n1\n1",
"output": "35184372088862"
},
{
"input": "1 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": "1125899906842623"
},
{
"input": "5 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\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\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\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\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": "5629499534214415"
},
{
"input": "32 2\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "8589934622"
},
{
"input": "1 50\n0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "562949953421312"
},
{
"input": "50 1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1\n1",
"output": "1125899906842623"
},
{
"input": "1 49\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",
"output": "562949953421311"
},
{
"input": "2 50\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\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",
"output": "2251799813685296"
},
{
"input": "3 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\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\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": "3377699720528069"
},
{
"input": "1 50\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",
"output": "1125899906842623"
},
{
"input": "1 40\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",
"output": "1099511627775"
},
{
"input": "1 33\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",
"output": "8589934591"
},
{
"input": "2 40\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\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",
"output": "2199023255590"
},
{
"input": "1 35\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",
"output": "34359738367"
},
{
"input": "50 1\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0\n0",
"output": "1125899906842623"
}
] | 1,645,695,504
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 0
|
n,m=map(int,input().split())
a = [list(map(int,input().split())) for x in range(n)]
s = 0
for row in a:
o,z=0,0
for i in row:
if i==0:
z+=1
else:
o+=1
s+=2**z-1
s+=2**o-1
print(s)
|
Title: Rectangles
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given *n*<=×<=*m* table. Each cell of the table is colored white or black. Find the number of non-empty sets of cells such that:
1. All cells in a set have the same color. 1. Every two cells in a set share row or column.
Input Specification:
The first line of input contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=50) — the number of rows and the number of columns correspondingly.
The next *n* lines of input contain descriptions of rows. There are *m* integers, separated by spaces, in each line. The number equals 0 if the corresponding cell is colored white and equals 1 if the corresponding cell is colored black.
Output Specification:
Output single integer — the number of non-empty sets from the problem description.
Demo Input:
['1 1\n0\n', '2 3\n1 0 1\n0 1 0\n']
Demo Output:
['1\n', '8\n']
Note:
In the second example, there are six one-element sets. Additionally, there are two two-element sets, the first one consists of the first and the third cells of the first row, the second one consists of the first and the third cells of the second row. To sum up, there are 8 sets.
|
```python
n,m=map(int,input().split())
a = [list(map(int,input().split())) for x in range(n)]
s = 0
for row in a:
o,z=0,0
for i in row:
if i==0:
z+=1
else:
o+=1
s+=2**z-1
s+=2**o-1
print(s)
```
| 0
|
|
22
|
A
|
Second Order Statistics
|
PROGRAMMING
| 800
|
[
"brute force"
] |
A. Second Order Statistics
|
2
|
256
|
Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem.
|
The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value.
|
If the given sequence has the second order statistics, output this order statistics, otherwise output NO.
|
[
"4\n1 2 2 -4\n",
"5\n1 2 3 1 1\n"
] |
[
"1\n",
"2\n"
] |
none
| 0
|
[
{
"input": "4\n1 2 2 -4",
"output": "1"
},
{
"input": "5\n1 2 3 1 1",
"output": "2"
},
{
"input": "1\n28",
"output": "NO"
},
{
"input": "2\n-28 12",
"output": "12"
},
{
"input": "3\n-83 40 -80",
"output": "-80"
},
{
"input": "8\n93 77 -92 26 21 -48 53 91",
"output": "-48"
},
{
"input": "20\n-72 -9 -86 80 7 -10 40 -27 -94 92 96 56 28 -19 79 36 -3 -73 -63 -49",
"output": "-86"
},
{
"input": "49\n-74 -100 -80 23 -8 -83 -41 -20 48 17 46 -73 -55 67 85 4 40 -60 -69 -75 56 -74 -42 93 74 -95 64 -46 97 -47 55 0 -78 -34 -31 40 -63 -49 -76 48 21 -1 -49 -29 -98 -11 76 26 94",
"output": "-98"
},
{
"input": "88\n63 48 1 -53 -89 -49 64 -70 -49 71 -17 -16 76 81 -26 -50 67 -59 -56 97 2 100 14 18 -91 -80 42 92 -25 -88 59 8 -56 38 48 -71 -78 24 -14 48 -1 69 73 -76 54 16 -92 44 47 33 -34 -17 -81 21 -59 -61 53 26 10 -76 67 35 -29 70 65 -13 -29 81 80 32 74 -6 34 46 57 1 -45 -55 69 79 -58 11 -2 22 -18 -16 -89 -46",
"output": "-91"
},
{
"input": "100\n34 32 88 20 76 53 -71 -39 -98 -10 57 37 63 -3 -54 -64 -78 -82 73 20 -30 -4 22 75 51 -64 -91 29 -52 -48 83 19 18 -47 46 57 -44 95 89 89 -30 84 -83 67 58 -99 -90 -53 92 -60 -5 -56 -61 27 68 -48 52 -95 64 -48 -30 -67 66 89 14 -33 -31 -91 39 7 -94 -54 92 -96 -99 -83 -16 91 -28 -66 81 44 14 -85 -21 18 40 16 -13 -82 -33 47 -10 -40 -19 10 25 60 -34 -89",
"output": "-98"
},
{
"input": "2\n-1 -1",
"output": "NO"
},
{
"input": "3\n-2 -2 -2",
"output": "NO"
},
{
"input": "100\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0",
"output": "NO"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 -100 100 100 100 100 -100 100 100 100 100 100 100 100 100 100 100 100",
"output": "100"
},
{
"input": "10\n40 71 -85 -85 40 -85 -85 64 -85 47",
"output": "40"
},
{
"input": "23\n-90 -90 -41 -64 -64 -90 -15 10 -43 -90 -64 -64 89 -64 36 47 38 -90 -64 -90 -90 68 -90",
"output": "-64"
},
{
"input": "39\n-97 -93 -42 -93 -97 -93 56 -97 -97 -97 76 -33 -60 91 7 82 17 47 -97 -97 -93 73 -97 12 -97 -97 -97 -97 56 -92 -83 -93 -93 49 -93 -97 -97 -17 -93",
"output": "-93"
},
{
"input": "51\n-21 6 -35 -98 -86 -98 -86 -43 -65 32 -98 -40 96 -98 -98 -98 -98 -86 -86 -98 56 -86 -98 -98 -30 -98 -86 -31 -98 -86 -86 -86 -86 -30 96 -86 -86 -86 -60 25 88 -86 -86 58 31 -47 57 -86 37 44 -83",
"output": "-86"
},
{
"input": "66\n-14 -95 65 -95 -95 -97 -90 -71 -97 -97 70 -95 -95 -97 -95 -27 35 -87 -95 -5 -97 -97 87 34 -49 -95 -97 -95 -97 -95 -30 -95 -97 47 -95 -17 -97 -95 -97 -69 51 -97 -97 -95 -75 87 59 21 63 56 76 -91 98 -97 6 -97 -95 -95 -97 -73 11 -97 -35 -95 -95 -43",
"output": "-95"
},
{
"input": "77\n-67 -93 -93 -92 97 29 93 -93 -93 -5 -93 -7 60 -92 -93 44 -84 68 -92 -93 69 -92 -37 56 43 -93 35 -92 -93 19 -79 18 -92 -93 -93 -37 -93 -47 -93 -92 -92 74 67 19 40 -92 -92 -92 -92 -93 -93 -41 -93 -92 -93 -93 -92 -93 51 -80 6 -42 -92 -92 -66 -12 -92 -92 -3 93 -92 -49 -93 40 62 -92 -92",
"output": "-92"
},
{
"input": "89\n-98 40 16 -87 -98 63 -100 55 -96 -98 -21 -100 -93 26 -98 -98 -100 -89 -98 -5 -65 -28 -100 -6 -66 67 -100 -98 -98 10 -98 -98 -70 7 -98 2 -100 -100 -98 25 -100 -100 -98 23 -68 -100 -98 3 98 -100 -98 -98 -98 -98 -24 -100 -100 -9 -98 35 -100 99 -5 -98 -100 -100 37 -100 -84 57 -98 40 -47 -100 -1 -92 -76 -98 -98 -100 -100 -100 -63 30 21 -100 -100 -100 -12",
"output": "-98"
},
{
"input": "99\n10 -84 -100 -100 73 -64 -100 -94 33 -100 -100 -100 -100 71 64 24 7 -100 -32 -100 -100 77 -100 62 -12 55 45 -100 -100 -80 -100 -100 -100 -100 -100 -100 -100 -100 -100 -39 -48 -100 -34 47 -100 -100 -100 -100 -100 -77 -100 -100 -100 -100 -100 -100 -52 40 -55 -100 -44 -100 72 33 70 -100 -100 -78 -100 -3 100 -77 22 -100 95 -30 -100 10 -69 -100 -100 -100 -100 52 -39 -100 -100 -100 7 -100 -98 -66 95 -17 -100 52 -100 68 -100",
"output": "-98"
},
{
"input": "100\n-99 -98 -64 89 53 57 -99 29 -78 18 -3 -54 76 -98 -99 -98 37 -98 19 -47 89 73 -98 -91 -99 -99 -98 -48 -99 22 -99 -98 -99 -99 -98 -60 84 67 -99 -98 20 -98 88 -98 46 -98 -99 -98 -99 -71 -99 -98 -98 -39 83 95 -98 63 -98 -99 32 -98 -99 -64 57 -30 -53 -83 -4 -99 58 20 -98 -10 -99 -44 -99 -99 -99 -99 75 34 -98 -52 -98 -30 -98 -99 -98 -98 51 -99 -99 -99 -98 -99 -99 -82 -90 92",
"output": "-98"
},
{
"input": "3\n1 2 3",
"output": "2"
},
{
"input": "3\n1 3 2",
"output": "2"
},
{
"input": "3\n2 1 3",
"output": "2"
},
{
"input": "3\n2 3 1",
"output": "2"
},
{
"input": "3\n3 1 2",
"output": "2"
},
{
"input": "3\n3 2 1",
"output": "2"
},
{
"input": "9\n99 99 99 99 100 100 100 100 100",
"output": "100"
},
{
"input": "5\n-100 -100 -100 -100 -100",
"output": "NO"
}
] | 1,551,568,106
| 2,147,483,647
|
PyPy 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 374
| 1,024,000
|
ens=set()
n=int(ch)
tab=list(map(int,input().split()))
ens=set(tab)
liste=list(ens)
if (len(ens)==1) :
printf("NO")
else:
liste.sort()
print(liste[1])
|
Title: Second Order Statistics
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Once Bob needed to find the second order statistics of a sequence of integer numbers. Lets choose each number from the sequence exactly once and sort them. The value on the second position is the second order statistics of the given sequence. In other words it is the smallest element strictly greater than the minimum. Help Bob solve this problem.
Input Specification:
The first input line contains integer *n* (1<=≤<=*n*<=≤<=100) — amount of numbers in the sequence. The second line contains *n* space-separated integer numbers — elements of the sequence. These numbers don't exceed 100 in absolute value.
Output Specification:
If the given sequence has the second order statistics, output this order statistics, otherwise output NO.
Demo Input:
['4\n1 2 2 -4\n', '5\n1 2 3 1 1\n']
Demo Output:
['1\n', '2\n']
Note:
none
|
```python
ens=set()
n=int(ch)
tab=list(map(int,input().split()))
ens=set(tab)
liste=list(ens)
if (len(ens)==1) :
printf("NO")
else:
liste.sort()
print(liste[1])
```
| -1
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of *n* lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation *n*<=-<=1 times:
- Remove any two elements *s* and *t* from the set, and add their concatenation *s*<=+<=*t* to the set.
The cost of such operation is defined to be , where *f*(*s*,<=*c*) denotes the number of times character *c* appears in string *s*.
Given a non-negative integer *k*, construct any valid non-empty set of no more than 100<=000 letters, such that the minimum accumulative cost of the whole process is exactly *k*. It can be shown that a solution always exists.
|
The first and only line of input contains a non-negative integer *k* (0<=≤<=*k*<=≤<=100<=000) — the required minimum cost.
|
Output a non-empty string of no more than 100<=000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
|
[
"12\n",
"3\n"
] |
[
"abababab\n",
"codeforces\n"
] |
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
- {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0; - {"aba", "b", "a", "b", "a", "b"}, with a cost of 1; - {"abab", "a", "b", "a", "b"}, with a cost of 1; - {"abab", "ab", "a", "b"}, with a cost of 0; - {"abab", "aba", "b"}, with a cost of 1; - {"abab", "abab"}, with a cost of 1; - {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process.
| 0
|
[
{
"input": "12",
"output": "abababab"
},
{
"input": "3",
"output": "codeforces"
},
{
"input": "0",
"output": "o"
},
{
"input": "2",
"output": "aabb"
},
{
"input": "5",
"output": "aaabbcc"
},
{
"input": "10",
"output": "aaaaa"
},
{
"input": "233",
"output": "ooououououououououooohhhhhhaaiiiiiibbjjjjjjcckkkkkkddlllllleemmmmmmffnnnnnnggzzzzzz"
},
{
"input": "418",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbcccc"
},
{
"input": "100000",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbcccccccccccc"
},
{
"input": "1",
"output": "aa"
},
{
"input": "4",
"output": "aaabb"
},
{
"input": "6",
"output": "aaaa"
},
{
"input": "7",
"output": "aaaabb"
},
{
"input": "8",
"output": "aaaabbcc"
},
{
"input": "9",
"output": "aaaabbb"
},
{
"input": "11",
"output": "aaaaabb"
},
{
"input": "1317",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbcccc"
},
{
"input": "1926",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbccc"
},
{
"input": "14514",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb"
},
{
"input": "25252",
"output": "niconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconiconicooooquququququququq"
},
{
"input": "99681",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"
},
{
"input": "99998",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbccc..."
},
{
"input": "82944",
"output": "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaabbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb..."
},
{
"input": "831",
"output": "happybirthdayhappybirthdayhappybirthdayhappybirthdayhappybirthdayhappybirthdayhappybirthdayhappybirthdayhappybirthdayqqquqqquqqqu"
},
{
"input": "39393",
"output": "mikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumikumik..."
}
] | 1,504,279,691
| 6,791
|
Python 3
|
WRONG_ANSWER
|
PRETESTS
| 2
| 46
| 0
|
from sys import stdin
from math import sqrt
n = int(stdin.readline().rstrip())
N = n * 2
charset = 'abcdefghijklmnopqrstuvwxyz'
sqr = int(sqrt(N)) // 2 + 1
for i in range(1, sqr + 1):
for j in range(1, i + 1):
for k in range(1, 26 + 1):
N_ = N - k * i * (i + 1)
if N_ % (j * (j + 1)) == 0 and N_ >= 0:
l = N_ // (j * (j + 1))
if l > 26 - k:
continue
s1 = charset[0:k] * (i + 1)
s2 = charset[k:k + l] * (j + 1)
print(s1 + s2)
exit()
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
From beginning till end, this message has been waiting to be conveyed.
For a given unordered multiset of *n* lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation *n*<=-<=1 times:
- Remove any two elements *s* and *t* from the set, and add their concatenation *s*<=+<=*t* to the set.
The cost of such operation is defined to be , where *f*(*s*,<=*c*) denotes the number of times character *c* appears in string *s*.
Given a non-negative integer *k*, construct any valid non-empty set of no more than 100<=000 letters, such that the minimum accumulative cost of the whole process is exactly *k*. It can be shown that a solution always exists.
Input Specification:
The first and only line of input contains a non-negative integer *k* (0<=≤<=*k*<=≤<=100<=000) — the required minimum cost.
Output Specification:
Output a non-empty string of no more than 100<=000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.
Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.
Demo Input:
['12\n', '3\n']
Demo Output:
['abababab\n', 'codeforces\n']
Note:
For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:
- {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0; - {"aba", "b", "a", "b", "a", "b"}, with a cost of 1; - {"abab", "a", "b", "a", "b"}, with a cost of 1; - {"abab", "ab", "a", "b"}, with a cost of 0; - {"abab", "aba", "b"}, with a cost of 1; - {"abab", "abab"}, with a cost of 1; - {"abababab"}, with a cost of 8.
The total cost is 12, and it can be proved to be the minimum cost of the process.
|
```python
from sys import stdin
from math import sqrt
n = int(stdin.readline().rstrip())
N = n * 2
charset = 'abcdefghijklmnopqrstuvwxyz'
sqr = int(sqrt(N)) // 2 + 1
for i in range(1, sqr + 1):
for j in range(1, i + 1):
for k in range(1, 26 + 1):
N_ = N - k * i * (i + 1)
if N_ % (j * (j + 1)) == 0 and N_ >= 0:
l = N_ // (j * (j + 1))
if l > 26 - k:
continue
s1 = charset[0:k] * (i + 1)
s2 = charset[k:k + l] * (j + 1)
print(s1 + s2)
exit()
```
| 0
|
|
349
|
A
|
Cinema Line
|
PROGRAMMING
| 1,100
|
[
"greedy",
"implementation"
] | null | null |
The new "Die Hard" movie has just been released! There are *n* people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 ruble bill. A "Die Hard" ticket costs 25 rubles. Can the booking clerk sell a ticket to each person and give the change if he initially has no money and sells the tickets strictly in the order people follow in the line?
|
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of people in the line. The next line contains *n* integers, each of them equals 25, 50 or 100 — the values of the bills the people have. The numbers are given in the order from the beginning of the line (at the box office) to the end of the line.
|
Print "YES" (without the quotes) if the booking clerk can sell a ticket to each person and give the change. Otherwise print "NO".
|
[
"4\n25 25 50 50\n",
"2\n25 100\n",
"4\n50 50 25 25\n"
] |
[
"YES\n",
"NO\n",
"NO\n"
] |
none
| 500
|
[
{
"input": "4\n25 25 50 50",
"output": "YES"
},
{
"input": "2\n25 100",
"output": "NO"
},
{
"input": "4\n50 50 25 25",
"output": "NO"
},
{
"input": "3\n25 50 100",
"output": "NO"
},
{
"input": "10\n25 25 25 25 25 25 25 25 25 25",
"output": "YES"
},
{
"input": "10\n50 50 50 50 50 50 50 50 50 50",
"output": "NO"
},
{
"input": "10\n100 100 100 100 100 100 100 100 100 100",
"output": "NO"
},
{
"input": "7\n25 25 25 25 50 100 50",
"output": "YES"
},
{
"input": "2\n25 25",
"output": "YES"
},
{
"input": "3\n50 100 100",
"output": "NO"
},
{
"input": "3\n25 50 50",
"output": "NO"
},
{
"input": "2\n50 25",
"output": "NO"
},
{
"input": "14\n25 25 25 25 25 25 25 50 50 50 100 100 100 100",
"output": "NO"
},
{
"input": "5\n25 25 25 100 100",
"output": "NO"
},
{
"input": "7\n25 50 25 50 25 50 100",
"output": "NO"
},
{
"input": "5\n25 50 25 50 100",
"output": "NO"
},
{
"input": "5\n25 25 50 50 50",
"output": "NO"
},
{
"input": "5\n25 25 50 100 50",
"output": "NO"
},
{
"input": "5\n25 25 50 50 100",
"output": "NO"
},
{
"input": "1\n50",
"output": "NO"
}
] | 1,691,403,369
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 124
| 9,830,400
|
import sys
input = sys.stdin.readline
t = 1
# t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
if (a[0] != 25):
print("NO")
else:
flag = 0
d = {25 : 0, 50 : 0}
for i in a:
if (i == 25):
d[25] += 1
elif (i == 50):
if (d[25] > 0):
d[25] -= 1
d[50] += 1
else:
print("NO")
flag = 1
break
else:
if (d[25] == 0):
print("NO")
flag = 1
break
elif (d[50] == 0):
if (d[25] > 2):
d[25] -= 3
else:
print("NO")
flag = 1
break
else:
d[25] -= 1
d[50] -= 1
if (not flag):
print("YES")
|
Title: Cinema Line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The new "Die Hard" movie has just been released! There are *n* people at the cinema box office standing in a huge line. Each of them has a single 100, 50 or 25 ruble bill. A "Die Hard" ticket costs 25 rubles. Can the booking clerk sell a ticket to each person and give the change if he initially has no money and sells the tickets strictly in the order people follow in the line?
Input Specification:
The first line contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of people in the line. The next line contains *n* integers, each of them equals 25, 50 or 100 — the values of the bills the people have. The numbers are given in the order from the beginning of the line (at the box office) to the end of the line.
Output Specification:
Print "YES" (without the quotes) if the booking clerk can sell a ticket to each person and give the change. Otherwise print "NO".
Demo Input:
['4\n25 25 50 50\n', '2\n25 100\n', '4\n50 50 25 25\n']
Demo Output:
['YES\n', 'NO\n', 'NO\n']
Note:
none
|
```python
import sys
input = sys.stdin.readline
t = 1
# t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
if (a[0] != 25):
print("NO")
else:
flag = 0
d = {25 : 0, 50 : 0}
for i in a:
if (i == 25):
d[25] += 1
elif (i == 50):
if (d[25] > 0):
d[25] -= 1
d[50] += 1
else:
print("NO")
flag = 1
break
else:
if (d[25] == 0):
print("NO")
flag = 1
break
elif (d[50] == 0):
if (d[25] > 2):
d[25] -= 3
else:
print("NO")
flag = 1
break
else:
d[25] -= 1
d[50] -= 1
if (not flag):
print("YES")
```
| 3
|
|
122
|
A
|
Lucky Division
|
PROGRAMMING
| 1,000
|
[
"brute force",
"number theory"
] | null | null |
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky.
|
The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked.
|
In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes).
|
[
"47\n",
"16\n",
"78\n"
] |
[
"YES\n",
"YES\n",
"NO\n"
] |
Note that all lucky numbers are almost lucky as any number is evenly divisible by itself.
In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4.
| 500
|
[
{
"input": "47",
"output": "YES"
},
{
"input": "16",
"output": "YES"
},
{
"input": "78",
"output": "NO"
},
{
"input": "48",
"output": "YES"
},
{
"input": "100",
"output": "YES"
},
{
"input": "107",
"output": "NO"
},
{
"input": "77",
"output": "YES"
},
{
"input": "477",
"output": "YES"
},
{
"input": "480",
"output": "YES"
},
{
"input": "1",
"output": "NO"
},
{
"input": "3",
"output": "NO"
},
{
"input": "4",
"output": "YES"
},
{
"input": "49",
"output": "YES"
},
{
"input": "56",
"output": "YES"
},
{
"input": "124",
"output": "YES"
},
{
"input": "1000",
"output": "YES"
},
{
"input": "999",
"output": "NO"
},
{
"input": "298",
"output": "NO"
},
{
"input": "274",
"output": "NO"
},
{
"input": "998",
"output": "NO"
},
{
"input": "42",
"output": "YES"
},
{
"input": "788",
"output": "YES"
},
{
"input": "70",
"output": "YES"
},
{
"input": "444",
"output": "YES"
},
{
"input": "777",
"output": "YES"
},
{
"input": "799",
"output": "YES"
},
{
"input": "25",
"output": "NO"
},
{
"input": "882",
"output": "YES"
},
{
"input": "88",
"output": "YES"
},
{
"input": "11",
"output": "NO"
},
{
"input": "2",
"output": "NO"
},
{
"input": "7",
"output": "YES"
},
{
"input": "8",
"output": "YES"
},
{
"input": "94",
"output": "YES"
},
{
"input": "477",
"output": "YES"
},
{
"input": "141",
"output": "YES"
}
] | 1,690,404,294
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 25
| 92
| 0
|
n = int(input())
def isLuckyNumber(number):
while number > 0:
rest = number % 10
if rest != 4 and rest != 7: return False
number //= 10
return True
if isLuckyNumber(n): print('YES')
else:
if n % 4 == 0 or n % 7 == 0: print('YES')
else: print('NO')
|
Title: Lucky Division
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Petya calls a number almost lucky if it could be evenly divided by some lucky number. Help him find out if the given number *n* is almost lucky.
Input Specification:
The single line contains an integer *n* (1<=≤<=*n*<=≤<=1000) — the number that needs to be checked.
Output Specification:
In the only line print "YES" (without the quotes), if number *n* is almost lucky. Otherwise, print "NO" (without the quotes).
Demo Input:
['47\n', '16\n', '78\n']
Demo Output:
['YES\n', 'YES\n', 'NO\n']
Note:
Note that all lucky numbers are almost lucky as any number is evenly divisible by itself.
In the first sample 47 is a lucky number. In the second sample 16 is divisible by 4.
|
```python
n = int(input())
def isLuckyNumber(number):
while number > 0:
rest = number % 10
if rest != 4 and rest != 7: return False
number //= 10
return True
if isLuckyNumber(n): print('YES')
else:
if n % 4 == 0 or n % 7 == 0: print('YES')
else: print('NO')
```
| 0
|
|
44
|
A
|
Indian Summer
|
PROGRAMMING
| 900
|
[
"implementation"
] |
A. Indian Summer
|
2
|
256
|
Indian summer is such a beautiful time of the year! A girl named Alyona is walking in the forest and picking a bouquet from fallen leaves. Alyona is very choosy — she doesn't take a leaf if it matches the color and the species of the tree of one of the leaves she already has. Find out how many leaves Alyona has picked.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of leaves Alyona has found. The next *n* lines contain the leaves' descriptions. Each leaf is characterized by the species of the tree it has fallen from and by the color. The species of the trees and colors are given in names, consisting of no more than 10 lowercase Latin letters. A name can not be an empty string. The species of a tree and the color are given in each line separated by a space.
|
Output the single number — the number of Alyona's leaves.
|
[
"5\nbirch yellow\nmaple red\nbirch yellow\nmaple yellow\nmaple green\n",
"3\noak yellow\noak yellow\noak yellow\n"
] |
[
"4\n",
"1\n"
] |
none
| 0
|
[
{
"input": "5\nbirch yellow\nmaple red\nbirch yellow\nmaple yellow\nmaple green",
"output": "4"
},
{
"input": "3\noak yellow\noak yellow\noak yellow",
"output": "1"
},
{
"input": "5\nxbnbkzn hp\nkaqkl vrgzbvqstu\nj aqidx\nhos gyul\nwefxmh tygpluae",
"output": "5"
},
{
"input": "1\nqvwli hz",
"output": "1"
},
{
"input": "4\nsrhk x\nsrhk x\nqfoe vnrjuab\nqfoe vnrjuab",
"output": "2"
},
{
"input": "4\nsddqllmmpk syded\nfprsq fnenjnaz\nn hdej\nsddqllmmpk syded",
"output": "3"
},
{
"input": "17\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw\njtbctslqq tosqzw",
"output": "1"
},
{
"input": "18\nb d\nrfdb esp\nrfdb esp\nb d\nrfdb esp\nrfdb esp\nb d\nb d\nrfdb esp\nrfdb esp\nb d\nrfdb esp\nb d\nrfdb esp\nb d\nb d\nrfdb esp\nrfdb esp",
"output": "2"
},
{
"input": "13\nsvpzwtwn rykzfdce\nqweiyeck jkreouy\nhk nnli\ntwxrnbbdt vtuv\nxokqjtylly sz\nesdt dbfidjslq\ng ybqgomvw\nxcpfjmf zcqvz\nifajadhj invzueip\nvdemdnxifb hckwebmi\nsdpnhipam wvowzavh\nuqdlfskhgo vunbpghae\ne dtigwnb",
"output": "13"
},
{
"input": "20\nm vkfh\nvptikamead mvx\nitu mf\nklruxckw aqega\nekqkd enjllwol\ncc uybfdh\nimrfdngvo u\ne uh\ntwt jsslcfuogk\nbljwqsag tuqbdn\nqcv q\nasx gzhvwwmajj\nqcv q\nekqkd enjllwol\nasx gzhvwwmajj\nks vv\nkzyfi cn\ncc uybfdh\nitu mf\ncjbjhtbyvk vatwfmux",
"output": "15"
},
{
"input": "2\nab ab\na bab",
"output": "2"
}
] | 1,651,937,540
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 29
| 92
| 0
|
t=int(input())
h={}
count=0
for _ in range(t):
a=input()
if(a not in h):
h[a]=1
count=count+1
else:
pass
print(count)
|
Title: Indian Summer
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Indian summer is such a beautiful time of the year! A girl named Alyona is walking in the forest and picking a bouquet from fallen leaves. Alyona is very choosy — she doesn't take a leaf if it matches the color and the species of the tree of one of the leaves she already has. Find out how many leaves Alyona has picked.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=100) — the number of leaves Alyona has found. The next *n* lines contain the leaves' descriptions. Each leaf is characterized by the species of the tree it has fallen from and by the color. The species of the trees and colors are given in names, consisting of no more than 10 lowercase Latin letters. A name can not be an empty string. The species of a tree and the color are given in each line separated by a space.
Output Specification:
Output the single number — the number of Alyona's leaves.
Demo Input:
['5\nbirch yellow\nmaple red\nbirch yellow\nmaple yellow\nmaple green\n', '3\noak yellow\noak yellow\noak yellow\n']
Demo Output:
['4\n', '1\n']
Note:
none
|
```python
t=int(input())
h={}
count=0
for _ in range(t):
a=input()
if(a not in h):
h[a]=1
count=count+1
else:
pass
print(count)
```
| 3.977
|
12
|
B
|
Correct Solution?
|
PROGRAMMING
| 1,100
|
[
"implementation",
"sortings"
] |
B. Correct Solution?
|
2
|
256
|
One cold winter evening Alice and her older brother Bob was sitting at home near the fireplace and giving each other interesting problems to solve. When it was Alice's turn, she told the number *n* to Bob and said:
—Shuffle the digits in this number in order to obtain the smallest possible number without leading zeroes.
—No problem! — said Bob and immediately gave her an answer.
Alice said a random number, so she doesn't know whether Bob's answer is correct. Help her to find this out, because impatient brother is waiting for the verdict.
|
The first line contains one integer *n* (0<=≤<=*n*<=≤<=109) without leading zeroes. The second lines contains one integer *m* (0<=≤<=*m*<=≤<=109) — Bob's answer, possibly with leading zeroes.
|
Print OK if Bob's answer is correct and WRONG_ANSWER otherwise.
|
[
"3310\n1033\n",
"4\n5\n"
] |
[
"OK\n",
"WRONG_ANSWER\n"
] |
none
| 0
|
[
{
"input": "3310\n1033",
"output": "OK"
},
{
"input": "4\n5",
"output": "WRONG_ANSWER"
},
{
"input": "40\n04",
"output": "WRONG_ANSWER"
},
{
"input": "12\n12",
"output": "OK"
},
{
"input": "432\n234",
"output": "OK"
},
{
"input": "17109\n01179",
"output": "WRONG_ANSWER"
},
{
"input": "888\n888",
"output": "OK"
},
{
"input": "912\n9123",
"output": "WRONG_ANSWER"
},
{
"input": "0\n00",
"output": "WRONG_ANSWER"
},
{
"input": "11110\n1111",
"output": "WRONG_ANSWER"
},
{
"input": "7391\n1397",
"output": "WRONG_ANSWER"
},
{
"input": "201\n102",
"output": "OK"
},
{
"input": "111111111\n111111111",
"output": "OK"
},
{
"input": "32352320\n22203335",
"output": "WRONG_ANSWER"
},
{
"input": "1000000000\n1",
"output": "WRONG_ANSWER"
},
{
"input": "994321\n123499",
"output": "OK"
},
{
"input": "10101\n10101",
"output": "WRONG_ANSWER"
},
{
"input": "666\n0666",
"output": "WRONG_ANSWER"
},
{
"input": "2\n02",
"output": "WRONG_ANSWER"
},
{
"input": "313\n133",
"output": "OK"
},
{
"input": "987235645\n234556789",
"output": "OK"
},
{
"input": "90812\n010289",
"output": "WRONG_ANSWER"
},
{
"input": "123\n321",
"output": "WRONG_ANSWER"
},
{
"input": "707\n770",
"output": "WRONG_ANSWER"
},
{
"input": "77774444\n47474747",
"output": "WRONG_ANSWER"
},
{
"input": "1270\n1027",
"output": "OK"
},
{
"input": "320\n23",
"output": "WRONG_ANSWER"
},
{
"input": "123456789\n123456789",
"output": "OK"
},
{
"input": "918273645\n546372819",
"output": "WRONG_ANSWER"
},
{
"input": "300000003\n30000003",
"output": "WRONG_ANSWER"
},
{
"input": "0\n0",
"output": "OK"
},
{
"input": "0\n7",
"output": "WRONG_ANSWER"
}
] | 1,658,812,265
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 2
| 92
| 0
|
def cout(n):
return print(n)
def cin():
return int(input())
def lcin():
return list(map(int,input().split()))
def scin():
return input()
n = scin()
m = cin()
n1 = (n.count('1'))*'1';n2 = (n.count('2'))*'2';n3 = (n.count('3'))*'3'
n4 = (n.count('4'))*'4';n5 = (n.count('5'))*'5';n6 = (n.count('6'))*'6'
n7 = (n.count('7'))*'7';n8 = (n.count('8'))*'8';n9 = (n.count('9'))*'9'
n0 = (n.count('0'))*'0'
s = n1+n0+n2+n3+n4+n5+n6+n7+n8+n9
if int(s) == m:
print('OK')
else:
print('WRONG_ANSWER')
|
Title: Correct Solution?
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One cold winter evening Alice and her older brother Bob was sitting at home near the fireplace and giving each other interesting problems to solve. When it was Alice's turn, she told the number *n* to Bob and said:
—Shuffle the digits in this number in order to obtain the smallest possible number without leading zeroes.
—No problem! — said Bob and immediately gave her an answer.
Alice said a random number, so she doesn't know whether Bob's answer is correct. Help her to find this out, because impatient brother is waiting for the verdict.
Input Specification:
The first line contains one integer *n* (0<=≤<=*n*<=≤<=109) without leading zeroes. The second lines contains one integer *m* (0<=≤<=*m*<=≤<=109) — Bob's answer, possibly with leading zeroes.
Output Specification:
Print OK if Bob's answer is correct and WRONG_ANSWER otherwise.
Demo Input:
['3310\n1033\n', '4\n5\n']
Demo Output:
['OK\n', 'WRONG_ANSWER\n']
Note:
none
|
```python
def cout(n):
return print(n)
def cin():
return int(input())
def lcin():
return list(map(int,input().split()))
def scin():
return input()
n = scin()
m = cin()
n1 = (n.count('1'))*'1';n2 = (n.count('2'))*'2';n3 = (n.count('3'))*'3'
n4 = (n.count('4'))*'4';n5 = (n.count('5'))*'5';n6 = (n.count('6'))*'6'
n7 = (n.count('7'))*'7';n8 = (n.count('8'))*'8';n9 = (n.count('9'))*'9'
n0 = (n.count('0'))*'0'
s = n1+n0+n2+n3+n4+n5+n6+n7+n8+n9
if int(s) == m:
print('OK')
else:
print('WRONG_ANSWER')
```
| 0
|
205
|
A
|
Little Elephant and Rozdil
|
PROGRAMMING
| 900
|
[
"brute force",
"implementation"
] | null | null |
The Little Elephant loves Ukraine very much. Most of all he loves town Rozdol (ukr. "Rozdil").
However, Rozdil is dangerous to settle, so the Little Elephant wants to go to some other town. The Little Elephant doesn't like to spend much time on travelling, so for his journey he will choose a town that needs minimum time to travel to. If there are multiple such cities, then the Little Elephant won't go anywhere.
For each town except for Rozdil you know the time needed to travel to this town. Find the town the Little Elephant will go to or print "Still Rozdil", if he stays in Rozdil.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of cities. The next line contains *n* integers, separated by single spaces: the *i*-th integer represents the time needed to go from town Rozdil to the *i*-th town. The time values are positive integers, not exceeding 109.
You can consider the cities numbered from 1 to *n*, inclusive. Rozdil is not among the numbered cities.
|
Print the answer on a single line — the number of the town the Little Elephant will go to. If there are multiple cities with minimum travel time, print "Still Rozdil" (without the quotes).
|
[
"2\n7 4\n",
"7\n7 4 47 100 4 9 12\n"
] |
[
"2\n",
"Still Rozdil\n"
] |
In the first sample there are only two cities where the Little Elephant can go. The travel time for the first town equals 7, to the second one — 4. The town which is closest to Rodzil (the only one) is the second one, so the answer is 2.
In the second sample the closest cities are cities two and five, the travelling time to both of them equals 4, so the answer is "Still Rozdil".
| 500
|
[
{
"input": "2\n7 4",
"output": "2"
},
{
"input": "7\n7 4 47 100 4 9 12",
"output": "Still Rozdil"
},
{
"input": "1\n47",
"output": "1"
},
{
"input": "2\n1000000000 1000000000",
"output": "Still Rozdil"
},
{
"input": "7\n7 6 5 4 3 2 1",
"output": "7"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "Still Rozdil"
},
{
"input": "4\n1000000000 100000000 1000000 1000000",
"output": "Still Rozdil"
},
{
"input": "20\n7 1 1 2 1 1 8 7 7 8 4 3 7 10 5 3 10 5 10 6",
"output": "Still Rozdil"
},
{
"input": "20\n3 3 6 9 8 2 4 1 7 3 2 9 7 7 9 7 2 6 2 7",
"output": "8"
},
{
"input": "47\n35 79 84 56 67 95 80 34 77 68 14 55 95 32 40 89 58 79 96 66 50 79 35 86 31 74 91 35 22 72 84 38 11 59 73 51 65 11 11 62 30 12 32 71 69 15 11",
"output": "Still Rozdil"
},
{
"input": "47\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3",
"output": "32"
},
{
"input": "25\n73 56 65 70 57 57 49 50 65 68 54 48 66 55 55 66 60 70 47 62 63 70 62 70 63",
"output": "19"
},
{
"input": "100\n922 938 103 689 526 118 314 496 373 47 224 544 495 656 804 754 269 356 354 536 817 597 787 178 516 659 790 993 786 169 866 779 997 350 674 437 219 546 564 893 402 449 751 628 238 686 392 951 716 439 332 563 215 146 746 311 866 773 548 675 731 183 890 802 488 944 473 25 662 559 732 534 777 144 502 593 839 692 84 337 679 718 1005 663 709 940 86 469 310 960 902 233 606 761 116 231 662 903 68 622",
"output": "68"
},
{
"input": "5\n5 5 2 3 1",
"output": "5"
},
{
"input": "1\n1000000000",
"output": "1"
},
{
"input": "3\n7 7 4",
"output": "3"
},
{
"input": "3\n2 2 1",
"output": "3"
},
{
"input": "3\n4 4 3",
"output": "3"
},
{
"input": "2\n10000001 10000002",
"output": "1"
},
{
"input": "4\n3 3 2 1",
"output": "4"
},
{
"input": "3\n1 2 3",
"output": "1"
},
{
"input": "2\n1000000000 100000000",
"output": "2"
},
{
"input": "1\n99999999",
"output": "1"
},
{
"input": "3\n5 5 3",
"output": "3"
},
{
"input": "3\n7 7 5",
"output": "3"
},
{
"input": "3\n3 3 1",
"output": "3"
},
{
"input": "4\n10 10 10 1",
"output": "4"
}
] | 1,694,723,258
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 45
| 186
| 14,438,400
|
import math
from collections import Counter
import io, os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# for _ in range(int(input())):
n = input()
n = int(n)
arr = list(map(int, input().split()))
m = arr.index(min(arr))
arr.sort()
if n == 1:
arr.append(10000000)
if arr[0] == arr[1]:
print("Still Rozdil")
else:
print(m+1)
# for _ in range(int(input())):
# grid = [list(map(int, input().split())) for _ in range(3)]
# grid = [list(input()) for _ in range(n)]
# result = [[1] * 3 for _ in range(3)]
# n, s, r = map(int, input().split())
# arr = list(map(int, input().split()))
# n = input()
# n = int(n)
|
Title: Little Elephant and Rozdil
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Little Elephant loves Ukraine very much. Most of all he loves town Rozdol (ukr. "Rozdil").
However, Rozdil is dangerous to settle, so the Little Elephant wants to go to some other town. The Little Elephant doesn't like to spend much time on travelling, so for his journey he will choose a town that needs minimum time to travel to. If there are multiple such cities, then the Little Elephant won't go anywhere.
For each town except for Rozdil you know the time needed to travel to this town. Find the town the Little Elephant will go to or print "Still Rozdil", if he stays in Rozdil.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105) — the number of cities. The next line contains *n* integers, separated by single spaces: the *i*-th integer represents the time needed to go from town Rozdil to the *i*-th town. The time values are positive integers, not exceeding 109.
You can consider the cities numbered from 1 to *n*, inclusive. Rozdil is not among the numbered cities.
Output Specification:
Print the answer on a single line — the number of the town the Little Elephant will go to. If there are multiple cities with minimum travel time, print "Still Rozdil" (without the quotes).
Demo Input:
['2\n7 4\n', '7\n7 4 47 100 4 9 12\n']
Demo Output:
['2\n', 'Still Rozdil\n']
Note:
In the first sample there are only two cities where the Little Elephant can go. The travel time for the first town equals 7, to the second one — 4. The town which is closest to Rodzil (the only one) is the second one, so the answer is 2.
In the second sample the closest cities are cities two and five, the travelling time to both of them equals 4, so the answer is "Still Rozdil".
|
```python
import math
from collections import Counter
import io, os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
# for _ in range(int(input())):
n = input()
n = int(n)
arr = list(map(int, input().split()))
m = arr.index(min(arr))
arr.sort()
if n == 1:
arr.append(10000000)
if arr[0] == arr[1]:
print("Still Rozdil")
else:
print(m+1)
# for _ in range(int(input())):
# grid = [list(map(int, input().split())) for _ in range(3)]
# grid = [list(input()) for _ in range(n)]
# result = [[1] * 3 for _ in range(3)]
# n, s, r = map(int, input().split())
# arr = list(map(int, input().split()))
# n = input()
# n = int(n)
```
| 3
|
|
416
|
A
|
Guess a number!
|
PROGRAMMING
| 1,400
|
[
"greedy",
"implementation",
"two pointers"
] | null | null |
A TV show called "Guess a number!" is gathering popularity. The whole Berland, the old and the young, are watching the show.
The rules are simple. The host thinks of an integer *y* and the participants guess it by asking questions to the host. There are four types of acceptable questions:
- Is it true that *y* is strictly larger than number *x*? - Is it true that *y* is strictly smaller than number *x*? - Is it true that *y* is larger than or equal to number *x*? - Is it true that *y* is smaller than or equal to number *x*?
On each question the host answers truthfully, "yes" or "no".
Given the sequence of questions and answers, find any integer value of *y* that meets the criteria of all answers. If there isn't such value, print "Impossible".
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10000) — the number of questions (and answers). Next *n* lines each contain one question and one answer to it. The format of each line is like that: "sign x answer", where the sign is:
- ">" (for the first type queries), - "<" (for the second type queries), - ">=" (for the third type queries), - "<=" (for the fourth type queries).
All values of *x* are integer and meet the inequation <=-<=109<=≤<=*x*<=≤<=109. The answer is an English letter "Y" (for "yes") or "N" (for "no").
Consequtive elements in lines are separated by a single space.
|
Print any of such integers *y*, that the answers to all the queries are correct. The printed number *y* must meet the inequation <=-<=2·109<=≤<=*y*<=≤<=2·109. If there are many answers, print any of them. If such value doesn't exist, print word "Impossible" (without the quotes).
|
[
"4\n>= 1 Y\n< 3 N\n<= -3 N\n> 55 N\n",
"2\n> 100 Y\n< -100 Y\n"
] |
[
"17\n",
"Impossible\n"
] |
none
| 500
|
[
{
"input": "4\n>= 1 Y\n< 3 N\n<= -3 N\n> 55 N",
"output": "17"
},
{
"input": "2\n> 100 Y\n< -100 Y",
"output": "Impossible"
},
{
"input": "4\n< 1 N\n> 1 N\n> 1 N\n> 1 N",
"output": "1"
},
{
"input": "4\n<= 1 Y\n>= 1 Y\n>= 1 Y\n<= 1 Y",
"output": "1"
},
{
"input": "4\n< 10 Y\n> -6 Y\n< 10 Y\n< -10 N",
"output": "-5"
},
{
"input": "1\n< 1 N",
"output": "1361956"
},
{
"input": "1\n<= 1 Y",
"output": "-1998638045"
},
{
"input": "1\n> 1 N",
"output": "-1998638045"
},
{
"input": "1\n>= 1 Y",
"output": "1361956"
},
{
"input": "4\n< 1 N\n< 1 N\n< 1 N\n<= 1 Y",
"output": "1"
},
{
"input": "4\n< 1 N\n>= 1 Y\n< 1 N\n< 1 N",
"output": "1361956"
},
{
"input": "4\n> 1 N\n<= 1 Y\n<= 1 Y\n> 1 N",
"output": "-1998638045"
},
{
"input": "4\n>= 1 Y\n> 1 N\n>= 1 Y\n>= 1 Y",
"output": "1"
},
{
"input": "4\n<= 9 Y\n< 3 Y\n< 2 Y\n< 2 Y",
"output": "-1998638045"
},
{
"input": "4\n< 0 N\n< -7 N\n>= 8 N\n>= -5 Y",
"output": "3"
},
{
"input": "4\n<= -6 N\n<= -8 N\n<= 3 Y\n<= 7 Y",
"output": "-2"
},
{
"input": "4\n>= 7 N\n<= -1 N\n>= 5 N\n<= -10 N",
"output": "0"
},
{
"input": "4\n> 5 N\n>= -5 Y\n> -9 Y\n> -9 Y",
"output": "-4"
},
{
"input": "10\n<= -60 N\n>= -59 Y\n> 22 Y\n> 95 N\n<= 91 Y\n> 77 Y\n>= -59 Y\n> -25 Y\n> -22 Y\n>= 52 Y",
"output": "85"
},
{
"input": "10\n>= -18 Y\n>= -35 Y\n> -94 Y\n< -23 N\n< -69 N\n< -68 N\n< 82 Y\n> 92 N\n< 29 Y\n>= -25 Y",
"output": "18"
},
{
"input": "10\n>= 18 Y\n<= -32 N\n>= 85 N\n<= 98 Y\n<= -43 N\n<= -79 N\n>= 97 N\n< -38 N\n< -55 N\n<= -93 N",
"output": "64"
},
{
"input": "10\n<= 2 Y\n< -33 Y\n> 6 N\n> -6 N\n< -28 Y\n> -62 Y\n< 57 Y\n<= 24 Y\n> 23 N\n> -25 N",
"output": "-54"
},
{
"input": "10\n<= -31 N\n>= 66 N\n<= 0 Y\n> -95 Y\n< 27 Y\n< -42 N\n> 3 N\n< 6 Y\n>= -42 Y\n> -70 Y",
"output": "-29"
},
{
"input": "10\n>= 54 N\n<= -52 N\n>= 64 N\n> 65 N\n< 37 Y\n> -84 Y\n>= -94 Y\n>= -95 Y\n> -72 Y\n<= 18 N",
"output": "22"
},
{
"input": "10\n> -24 N\n<= -5 Y\n<= -33 Y\n> 45 N\n> -59 Y\n> -21 N\n<= -48 N\n> 40 N\n< 12 Y\n>= 14 N",
"output": "-47"
},
{
"input": "10\n>= 91 Y\n>= -68 Y\n< 92 N\n>= -15 Y\n> 51 Y\n<= 14 N\n> 17 Y\n< 94 Y\n>= 49 Y\n> -36 Y",
"output": "93"
},
{
"input": "1\n< -1000000000 Y",
"output": "-1998638045"
},
{
"input": "1\n< 1 Y",
"output": "-1998638045"
},
{
"input": "1\n>= -999999999 Y",
"output": "-998638044"
},
{
"input": "1\n> 100000 Y",
"output": "1461956"
},
{
"input": "1\n<= 999999999 Y",
"output": "-1998638045"
},
{
"input": "1\n<= 1000000000 N",
"output": "1001361956"
},
{
"input": "4\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y",
"output": "-1998638045"
},
{
"input": "1\n>= 1000000000 Y",
"output": "1001361955"
},
{
"input": "1\n<= 999999999 N",
"output": "1001361955"
},
{
"input": "1\n<= 100 Y",
"output": "-1998638045"
},
{
"input": "1\n> 1000000000 Y",
"output": "1001361956"
},
{
"input": "1\n<= 1 Y",
"output": "-1998638045"
},
{
"input": "1\n<= 1000000000 Y",
"output": "-1998638045"
},
{
"input": "1\n<= -1000000000 Y",
"output": "-1998638045"
},
{
"input": "1\n<= -999999999 Y",
"output": "-1998638045"
},
{
"input": "1\n> 100 Y",
"output": "1362056"
},
{
"input": "2\n< -1000000000 Y\n< 3 Y",
"output": "-1998638045"
},
{
"input": "1\n<= -1000000 Y",
"output": "-1998638045"
},
{
"input": "8\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y\n< -1000000000 Y",
"output": "-1998638045"
},
{
"input": "1\n<= 15 Y",
"output": "-1998638045"
},
{
"input": "1\n< 1000000000 Y",
"output": "-1998638045"
},
{
"input": "1\n> 10000000 Y",
"output": "11361956"
},
{
"input": "1\n< 0 Y",
"output": "-1998638045"
},
{
"input": "1\n< 100 Y",
"output": "-1998638045"
},
{
"input": "1\n<= 5 Y",
"output": "-1998638045"
},
{
"input": "3\n> 5 Y\n> 0 Y\n< 4 Y",
"output": "Impossible"
},
{
"input": "1\n>= -1000000000 N",
"output": "-1998638045"
},
{
"input": "3\n>= 1 Y\n<= 1 Y\n> 10 Y",
"output": "Impossible"
},
{
"input": "2\n> 1 Y\n< 2 Y",
"output": "Impossible"
},
{
"input": "3\n>= 5 Y\n<= 5 Y\n< 5 Y",
"output": "Impossible"
},
{
"input": "2\n>= 5 N\n> 5 Y",
"output": "Impossible"
},
{
"input": "3\n>= 4 Y\n> 4 Y\n<= 4 Y",
"output": "Impossible"
},
{
"input": "2\n>= 4 Y\n> 4 Y",
"output": "1361960"
}
] | 1,553,709,305
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 109
| 0
|
n = int(input())
simbolo = []
num = []
yn = []
for j in range(n):
s,x,y = input().split()
simbolo.append(s)
num.append(int(x))
yn.append(y)
maximo = float('inf')
minimo = -float('inf')
for i in range(n):
ismax = False
if simbolo[i] == '<' or simbolo[i] == '<=': # se x < y => y é max
ismax = True
if yn[i] == 'N':
ismax = not ismax
if len(simbolo[i]) == 1:
if ismax:
num[i] -= 1
else:
num[i] += 1
if ismax == True:
if num[i]< maximo:
maximo = num[i]
else:
if num[i]>minimo:
minimo = num[i]
if maximo >= minimo:
print (minimo)
else:
print("Impossible")
|
Title: Guess a number!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A TV show called "Guess a number!" is gathering popularity. The whole Berland, the old and the young, are watching the show.
The rules are simple. The host thinks of an integer *y* and the participants guess it by asking questions to the host. There are four types of acceptable questions:
- Is it true that *y* is strictly larger than number *x*? - Is it true that *y* is strictly smaller than number *x*? - Is it true that *y* is larger than or equal to number *x*? - Is it true that *y* is smaller than or equal to number *x*?
On each question the host answers truthfully, "yes" or "no".
Given the sequence of questions and answers, find any integer value of *y* that meets the criteria of all answers. If there isn't such value, print "Impossible".
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=10000) — the number of questions (and answers). Next *n* lines each contain one question and one answer to it. The format of each line is like that: "sign x answer", where the sign is:
- ">" (for the first type queries), - "<" (for the second type queries), - ">=" (for the third type queries), - "<=" (for the fourth type queries).
All values of *x* are integer and meet the inequation <=-<=109<=≤<=*x*<=≤<=109. The answer is an English letter "Y" (for "yes") or "N" (for "no").
Consequtive elements in lines are separated by a single space.
Output Specification:
Print any of such integers *y*, that the answers to all the queries are correct. The printed number *y* must meet the inequation <=-<=2·109<=≤<=*y*<=≤<=2·109. If there are many answers, print any of them. If such value doesn't exist, print word "Impossible" (without the quotes).
Demo Input:
['4\n>= 1 Y\n< 3 N\n<= -3 N\n> 55 N\n', '2\n> 100 Y\n< -100 Y\n']
Demo Output:
['17\n', 'Impossible\n']
Note:
none
|
```python
n = int(input())
simbolo = []
num = []
yn = []
for j in range(n):
s,x,y = input().split()
simbolo.append(s)
num.append(int(x))
yn.append(y)
maximo = float('inf')
minimo = -float('inf')
for i in range(n):
ismax = False
if simbolo[i] == '<' or simbolo[i] == '<=': # se x < y => y é max
ismax = True
if yn[i] == 'N':
ismax = not ismax
if len(simbolo[i]) == 1:
if ismax:
num[i] -= 1
else:
num[i] += 1
if ismax == True:
if num[i]< maximo:
maximo = num[i]
else:
if num[i]>minimo:
minimo = num[i]
if maximo >= minimo:
print (minimo)
else:
print("Impossible")
```
| 0
|
|
703
|
A
|
Mishka and Game
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game.
Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner.
In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw.
Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her!
|
The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds.
The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively.
|
If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line.
If Chris is the winner of the game, print "Chris" (without quotes) in the only line.
If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line.
|
[
"3\n3 5\n2 1\n4 2\n",
"2\n6 1\n1 6\n",
"3\n1 5\n3 3\n2 2\n"
] |
[
"Mishka",
"Friendship is magic!^^",
"Chris"
] |
In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game.
In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1.
In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris.
| 500
|
[
{
"input": "3\n3 5\n2 1\n4 2",
"output": "Mishka"
},
{
"input": "2\n6 1\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "3\n1 5\n3 3\n2 2",
"output": "Chris"
},
{
"input": "6\n4 1\n4 2\n5 3\n5 1\n5 3\n4 1",
"output": "Mishka"
},
{
"input": "8\n2 4\n1 4\n1 5\n2 6\n2 5\n2 5\n2 4\n2 5",
"output": "Chris"
},
{
"input": "8\n4 1\n2 6\n4 2\n2 5\n5 2\n3 5\n5 2\n1 5",
"output": "Friendship is magic!^^"
},
{
"input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n1 3",
"output": "Mishka"
},
{
"input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "9\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "9\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n1 4",
"output": "Mishka"
},
{
"input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "10\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "10\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "100\n2 4\n6 6\n3 2\n1 5\n5 2\n1 5\n1 5\n3 1\n6 5\n4 3\n1 1\n5 1\n3 3\n2 4\n1 5\n3 4\n5 1\n5 5\n2 5\n2 1\n4 3\n6 5\n1 1\n2 1\n1 3\n1 1\n6 4\n4 6\n6 4\n2 1\n2 5\n6 2\n3 4\n5 5\n1 4\n4 6\n3 4\n1 6\n5 1\n4 3\n3 4\n2 2\n1 2\n2 3\n1 3\n4 4\n5 5\n4 5\n4 4\n3 1\n4 5\n2 3\n2 6\n6 5\n6 1\n6 6\n2 3\n6 4\n3 3\n2 5\n4 4\n3 1\n2 4\n6 1\n3 2\n1 3\n5 4\n6 6\n2 5\n5 1\n1 1\n2 5\n6 5\n3 6\n5 6\n4 3\n3 4\n3 4\n6 5\n5 2\n4 2\n1 1\n3 1\n2 6\n1 6\n1 2\n6 1\n3 4\n1 6\n3 1\n5 3\n1 3\n5 6\n2 1\n6 4\n3 1\n1 6\n6 3\n3 3\n4 3",
"output": "Chris"
},
{
"input": "100\n4 1\n3 4\n4 6\n4 5\n6 5\n5 3\n6 2\n6 3\n5 2\n4 5\n1 5\n5 4\n1 4\n4 5\n4 6\n1 6\n4 4\n5 1\n6 4\n6 4\n4 6\n2 3\n6 2\n4 6\n1 4\n2 3\n4 3\n1 3\n6 2\n3 1\n3 4\n2 6\n4 5\n5 4\n2 2\n2 5\n4 1\n2 2\n3 3\n1 4\n5 6\n6 4\n4 2\n6 1\n5 5\n4 1\n2 1\n6 4\n4 4\n4 3\n5 3\n4 5\n5 3\n3 5\n6 3\n1 1\n3 4\n6 3\n6 1\n5 1\n2 4\n4 3\n2 2\n5 5\n1 5\n5 3\n4 6\n1 4\n6 3\n4 3\n2 4\n3 2\n2 4\n3 4\n6 2\n5 6\n1 2\n1 5\n5 5\n2 6\n5 1\n1 6\n5 3\n3 5\n2 6\n4 6\n6 2\n3 1\n5 5\n6 1\n3 6\n4 4\n1 1\n4 6\n5 3\n4 2\n5 1\n3 3\n2 1\n1 4",
"output": "Mishka"
},
{
"input": "100\n6 3\n4 5\n4 3\n5 4\n5 1\n6 3\n4 2\n4 6\n3 1\n2 4\n2 2\n4 6\n5 3\n5 5\n4 2\n6 2\n2 3\n4 4\n6 4\n3 5\n2 4\n2 2\n5 2\n3 5\n2 4\n4 4\n3 5\n6 5\n1 3\n1 6\n2 2\n2 4\n3 2\n5 4\n1 6\n3 4\n4 1\n1 5\n1 4\n5 3\n2 2\n4 5\n6 3\n4 4\n1 1\n4 1\n2 4\n4 1\n4 5\n5 3\n1 1\n1 6\n5 6\n6 6\n4 2\n4 3\n3 4\n3 6\n3 4\n6 5\n3 4\n5 4\n5 1\n5 3\n5 1\n1 2\n2 6\n3 4\n6 5\n4 3\n1 1\n5 5\n5 1\n3 3\n5 2\n1 3\n6 6\n5 6\n1 4\n4 4\n1 4\n3 6\n6 5\n3 3\n3 6\n1 5\n1 2\n3 6\n3 6\n4 1\n5 2\n1 2\n5 2\n3 3\n4 4\n4 2\n6 2\n5 4\n6 1\n6 3",
"output": "Mishka"
},
{
"input": "8\n4 1\n6 2\n4 1\n5 3\n4 1\n5 3\n6 2\n5 3",
"output": "Mishka"
},
{
"input": "5\n3 6\n3 5\n3 5\n1 6\n3 5",
"output": "Chris"
},
{
"input": "4\n4 1\n2 4\n5 3\n3 6",
"output": "Friendship is magic!^^"
},
{
"input": "6\n6 3\n5 1\n6 3\n4 3\n4 3\n5 2",
"output": "Mishka"
},
{
"input": "7\n3 4\n1 4\n2 5\n1 6\n1 6\n1 5\n3 4",
"output": "Chris"
},
{
"input": "6\n6 2\n2 5\n5 2\n3 6\n4 3\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "8\n6 1\n5 3\n4 3\n4 1\n5 1\n4 2\n4 2\n4 1",
"output": "Mishka"
},
{
"input": "9\n2 5\n2 5\n1 4\n2 6\n2 4\n2 5\n2 6\n1 5\n2 5",
"output": "Chris"
},
{
"input": "4\n6 2\n2 4\n4 2\n3 6",
"output": "Friendship is magic!^^"
},
{
"input": "9\n5 2\n4 1\n4 1\n5 1\n6 2\n6 1\n5 3\n6 1\n6 2",
"output": "Mishka"
},
{
"input": "8\n2 4\n3 6\n1 6\n1 6\n2 4\n3 4\n3 6\n3 4",
"output": "Chris"
},
{
"input": "6\n5 3\n3 6\n6 2\n1 6\n5 1\n3 5",
"output": "Friendship is magic!^^"
},
{
"input": "6\n5 2\n5 1\n6 1\n5 2\n4 2\n5 1",
"output": "Mishka"
},
{
"input": "5\n1 4\n2 5\n3 4\n2 6\n3 4",
"output": "Chris"
},
{
"input": "4\n6 2\n3 4\n5 1\n1 6",
"output": "Friendship is magic!^^"
},
{
"input": "93\n4 3\n4 1\n4 2\n5 2\n5 3\n6 3\n4 3\n6 2\n6 3\n5 1\n4 2\n4 2\n5 1\n6 2\n6 3\n6 1\n4 1\n6 2\n5 3\n4 3\n4 1\n4 2\n5 2\n6 3\n5 2\n5 2\n6 3\n5 1\n6 2\n5 2\n4 1\n5 2\n5 1\n4 1\n6 1\n5 2\n4 3\n5 3\n5 3\n5 1\n4 3\n4 3\n4 2\n4 1\n6 2\n6 1\n4 1\n5 2\n5 2\n6 2\n5 3\n5 1\n6 2\n5 1\n6 3\n5 2\n6 2\n6 2\n4 2\n5 2\n6 1\n6 3\n6 3\n5 1\n5 1\n4 1\n5 1\n4 3\n5 3\n6 3\n4 1\n4 3\n6 1\n6 1\n4 2\n6 2\n4 2\n5 2\n4 1\n5 2\n4 1\n5 1\n5 2\n5 1\n4 1\n6 3\n6 2\n4 3\n4 1\n5 2\n4 3\n5 2\n5 1",
"output": "Mishka"
},
{
"input": "11\n1 6\n1 6\n2 4\n2 5\n3 4\n1 5\n1 6\n1 5\n1 6\n2 6\n3 4",
"output": "Chris"
},
{
"input": "70\n6 1\n3 6\n4 3\n2 5\n5 2\n1 4\n6 2\n1 6\n4 3\n1 4\n5 3\n2 4\n5 3\n1 6\n5 1\n3 5\n4 2\n2 4\n5 1\n3 5\n6 2\n1 5\n4 2\n2 5\n5 3\n1 5\n4 2\n1 4\n5 2\n2 6\n4 3\n1 5\n6 2\n3 4\n4 2\n3 5\n6 3\n3 4\n5 1\n1 4\n4 2\n1 4\n6 3\n2 6\n5 2\n1 6\n6 1\n2 6\n5 3\n1 5\n5 1\n1 6\n4 1\n1 5\n4 2\n2 4\n5 1\n2 5\n6 3\n1 4\n6 3\n3 6\n5 1\n1 4\n5 3\n3 5\n4 2\n3 4\n6 2\n1 4",
"output": "Friendship is magic!^^"
},
{
"input": "59\n4 1\n5 3\n6 1\n4 2\n5 1\n4 3\n6 1\n5 1\n4 3\n4 3\n5 2\n5 3\n4 1\n6 2\n5 1\n6 3\n6 3\n5 2\n5 2\n6 1\n4 1\n6 1\n4 3\n5 3\n5 3\n4 3\n4 2\n4 2\n6 3\n6 3\n6 1\n4 3\n5 1\n6 2\n6 1\n4 1\n6 1\n5 3\n4 2\n5 1\n6 2\n6 2\n4 3\n5 3\n4 3\n6 3\n5 2\n5 2\n4 3\n5 1\n5 3\n6 1\n6 3\n6 3\n4 3\n5 2\n5 2\n5 2\n4 3",
"output": "Mishka"
},
{
"input": "42\n1 5\n1 6\n1 6\n1 4\n2 5\n3 6\n1 6\n3 4\n2 5\n2 5\n2 4\n1 4\n3 4\n2 4\n2 6\n1 5\n3 6\n2 6\n2 6\n3 5\n1 4\n1 5\n2 6\n3 6\n1 4\n3 4\n2 4\n1 6\n3 4\n2 4\n2 6\n1 6\n1 4\n1 6\n1 6\n2 4\n1 5\n1 6\n2 5\n3 6\n3 5\n3 4",
"output": "Chris"
},
{
"input": "78\n4 3\n3 5\n4 3\n1 5\n5 1\n1 5\n4 3\n1 4\n6 3\n1 5\n4 1\n2 4\n4 3\n2 4\n5 1\n3 6\n4 2\n3 6\n6 3\n3 4\n4 3\n3 6\n5 3\n1 5\n4 1\n2 6\n4 2\n2 4\n4 1\n3 5\n5 2\n3 6\n4 3\n2 4\n6 3\n1 6\n4 3\n3 5\n6 3\n2 6\n4 1\n2 4\n6 2\n1 6\n4 2\n1 4\n4 3\n1 4\n4 3\n2 4\n6 2\n3 5\n6 1\n3 6\n5 3\n1 6\n6 1\n2 6\n4 2\n1 5\n6 2\n2 6\n6 3\n2 4\n4 2\n3 5\n6 1\n2 5\n5 3\n2 6\n5 1\n3 6\n4 3\n3 6\n6 3\n2 5\n6 1\n2 6",
"output": "Friendship is magic!^^"
},
{
"input": "76\n4 1\n5 2\n4 3\n5 2\n5 3\n5 2\n6 1\n4 2\n6 2\n5 3\n4 2\n6 2\n4 1\n4 2\n5 1\n5 1\n6 2\n5 2\n5 3\n6 3\n5 2\n4 3\n6 3\n6 1\n4 3\n6 2\n6 1\n4 1\n6 1\n5 3\n4 1\n5 3\n4 2\n5 2\n4 3\n6 1\n6 2\n5 2\n6 1\n5 3\n4 3\n5 1\n5 3\n4 3\n5 1\n5 1\n4 1\n4 1\n4 1\n4 3\n5 3\n6 3\n6 3\n5 2\n6 2\n6 3\n5 1\n6 3\n5 3\n6 1\n5 3\n4 1\n5 3\n6 1\n4 2\n6 2\n4 3\n4 1\n6 2\n4 3\n5 3\n5 2\n5 3\n5 1\n6 3\n5 2",
"output": "Mishka"
},
{
"input": "84\n3 6\n3 4\n2 5\n2 4\n1 6\n3 4\n1 5\n1 6\n3 5\n1 6\n2 4\n2 6\n2 6\n2 4\n3 5\n1 5\n3 6\n3 6\n3 4\n3 4\n2 6\n1 6\n1 6\n3 5\n3 4\n1 6\n3 4\n3 5\n2 4\n2 5\n2 5\n3 5\n1 6\n3 4\n2 6\n2 6\n3 4\n3 4\n2 5\n2 5\n2 4\n3 4\n2 5\n3 4\n3 4\n2 6\n2 6\n1 6\n2 4\n1 5\n3 4\n2 5\n2 5\n3 4\n2 4\n2 6\n2 6\n1 4\n3 5\n3 5\n2 4\n2 5\n3 4\n1 5\n1 5\n2 6\n1 5\n3 5\n2 4\n2 5\n3 4\n2 6\n1 6\n2 5\n3 5\n3 5\n3 4\n2 5\n2 6\n3 4\n1 6\n2 5\n2 6\n1 4",
"output": "Chris"
},
{
"input": "44\n6 1\n1 6\n5 2\n1 4\n6 2\n2 5\n5 3\n3 6\n5 2\n1 6\n4 1\n2 4\n6 1\n3 4\n6 3\n3 6\n4 3\n2 4\n6 1\n3 4\n6 1\n1 6\n4 1\n3 5\n6 1\n3 6\n4 1\n1 4\n4 2\n2 6\n6 1\n2 4\n6 2\n1 4\n6 2\n2 4\n5 2\n3 6\n6 3\n2 6\n5 3\n3 4\n5 3\n2 4",
"output": "Friendship is magic!^^"
},
{
"input": "42\n5 3\n5 1\n5 2\n4 1\n6 3\n6 1\n6 2\n4 1\n4 3\n4 1\n5 1\n5 3\n5 1\n4 1\n4 2\n6 1\n6 3\n5 1\n4 1\n4 1\n6 3\n4 3\n6 3\n5 2\n6 1\n4 1\n5 3\n4 3\n5 2\n6 3\n6 1\n5 1\n4 2\n4 3\n5 2\n5 3\n6 3\n5 2\n5 1\n5 3\n6 2\n6 1",
"output": "Mishka"
},
{
"input": "50\n3 6\n2 6\n1 4\n1 4\n1 4\n2 5\n3 4\n3 5\n2 6\n1 6\n3 5\n1 5\n2 6\n2 4\n2 4\n3 5\n1 6\n1 5\n1 5\n1 4\n3 5\n1 6\n3 5\n1 4\n1 5\n1 4\n3 6\n1 6\n1 4\n1 4\n1 4\n1 5\n3 6\n1 6\n1 6\n2 4\n1 5\n2 6\n2 5\n3 5\n3 6\n3 4\n2 4\n2 6\n3 4\n2 5\n3 6\n3 5\n2 4\n2 4",
"output": "Chris"
},
{
"input": "86\n6 3\n2 4\n6 3\n3 5\n6 3\n1 5\n5 2\n2 4\n4 3\n2 6\n4 1\n2 6\n5 2\n1 4\n5 1\n2 4\n4 1\n1 4\n6 2\n3 5\n4 2\n2 4\n6 2\n1 5\n5 3\n2 5\n5 1\n1 6\n6 1\n1 4\n4 3\n3 4\n5 2\n2 4\n5 3\n2 5\n4 3\n3 4\n4 1\n1 5\n6 3\n3 4\n4 3\n3 4\n4 1\n3 4\n5 1\n1 6\n4 2\n1 6\n5 1\n2 4\n5 1\n3 6\n4 1\n1 5\n5 2\n1 4\n4 3\n2 5\n5 1\n1 5\n6 2\n2 6\n4 2\n2 4\n4 1\n2 5\n5 3\n3 4\n5 1\n3 4\n6 3\n3 4\n4 3\n2 6\n6 2\n2 5\n5 2\n3 5\n4 2\n3 6\n6 2\n3 4\n4 2\n2 4",
"output": "Friendship is magic!^^"
},
{
"input": "84\n6 1\n6 3\n6 3\n4 1\n4 3\n4 2\n6 3\n5 3\n6 1\n6 3\n4 3\n5 2\n5 3\n5 1\n6 2\n6 2\n6 1\n4 1\n6 3\n5 2\n4 1\n5 3\n6 3\n4 2\n6 2\n6 3\n4 3\n4 1\n4 3\n5 1\n5 1\n5 1\n4 1\n6 1\n4 3\n6 2\n5 1\n5 1\n6 2\n5 2\n4 1\n6 1\n6 1\n6 3\n6 2\n4 3\n6 3\n6 2\n5 2\n5 1\n4 3\n6 2\n4 1\n6 2\n6 1\n5 2\n5 1\n6 2\n6 1\n5 3\n5 2\n6 1\n6 3\n5 2\n6 1\n6 3\n4 3\n5 1\n6 3\n6 1\n5 3\n4 3\n5 2\n5 1\n6 2\n5 3\n6 1\n5 1\n4 1\n5 1\n5 1\n5 2\n5 2\n5 1",
"output": "Mishka"
},
{
"input": "92\n1 5\n2 4\n3 5\n1 6\n2 5\n1 6\n3 6\n1 6\n2 4\n3 4\n3 4\n3 6\n1 5\n2 5\n1 5\n1 5\n2 6\n2 4\n3 6\n1 4\n1 6\n2 6\n3 4\n2 6\n2 6\n1 4\n3 5\n2 5\n2 6\n1 5\n1 4\n1 5\n3 6\n3 5\n2 5\n1 5\n3 5\n3 6\n2 6\n2 6\n1 5\n3 4\n2 4\n3 6\n2 5\n1 5\n2 4\n1 4\n2 6\n2 6\n2 6\n1 5\n3 6\n3 6\n2 5\n1 4\n2 4\n3 4\n1 5\n2 5\n2 4\n2 5\n3 5\n3 4\n3 6\n2 6\n3 5\n1 4\n3 4\n1 6\n3 6\n2 6\n1 4\n3 6\n3 6\n2 5\n2 6\n1 6\n2 6\n3 5\n2 5\n3 6\n2 5\n2 6\n1 5\n2 4\n1 4\n2 4\n1 5\n2 5\n2 5\n2 6",
"output": "Chris"
},
{
"input": "20\n5 1\n1 4\n4 3\n1 5\n4 2\n3 6\n6 2\n1 6\n4 1\n1 4\n5 2\n3 4\n5 1\n1 6\n5 1\n2 6\n6 3\n2 5\n6 2\n2 4",
"output": "Friendship is magic!^^"
},
{
"input": "100\n4 3\n4 3\n4 2\n4 3\n4 1\n4 3\n5 2\n5 2\n6 2\n4 2\n5 1\n4 2\n5 2\n6 1\n4 1\n6 3\n5 3\n5 1\n5 1\n5 1\n5 3\n6 1\n6 1\n4 1\n5 2\n5 2\n6 1\n6 3\n4 2\n4 1\n5 3\n4 1\n5 3\n5 1\n6 3\n6 3\n6 1\n5 2\n5 3\n5 3\n6 1\n4 1\n6 2\n6 1\n6 2\n6 3\n4 3\n4 3\n6 3\n4 2\n4 2\n5 3\n5 2\n5 2\n4 3\n5 3\n5 2\n4 2\n5 1\n4 2\n5 1\n5 3\n6 3\n5 3\n5 3\n4 2\n4 1\n4 2\n4 3\n6 3\n4 3\n6 2\n6 1\n5 3\n5 2\n4 1\n6 1\n5 2\n6 2\n4 2\n6 3\n4 3\n5 1\n6 3\n5 2\n4 3\n5 3\n5 3\n4 3\n6 3\n4 3\n4 1\n5 1\n6 2\n6 3\n5 3\n6 1\n6 3\n5 3\n6 1",
"output": "Mishka"
},
{
"input": "100\n1 5\n1 4\n1 5\n2 4\n2 6\n3 6\n3 5\n1 5\n2 5\n3 6\n3 5\n1 6\n1 4\n1 5\n1 6\n2 6\n1 5\n3 5\n3 4\n2 6\n2 6\n2 5\n3 4\n1 6\n1 4\n2 4\n1 5\n1 6\n3 5\n1 6\n2 6\n3 5\n1 6\n3 4\n3 5\n1 6\n3 6\n2 4\n2 4\n3 5\n2 6\n1 5\n3 5\n3 6\n2 4\n2 4\n2 6\n3 4\n3 4\n1 5\n1 4\n2 5\n3 4\n1 4\n2 6\n2 5\n2 4\n2 4\n2 5\n1 5\n1 6\n1 5\n1 5\n1 5\n1 6\n3 4\n2 4\n3 5\n3 5\n1 6\n3 5\n1 5\n1 6\n3 6\n3 4\n1 5\n3 5\n3 6\n1 4\n3 6\n1 5\n3 5\n3 6\n3 5\n1 4\n3 4\n2 4\n2 4\n2 5\n3 6\n3 5\n1 5\n2 4\n1 4\n3 4\n1 5\n3 4\n3 6\n3 5\n3 4",
"output": "Chris"
},
{
"input": "100\n4 3\n3 4\n5 1\n2 5\n5 3\n1 5\n6 3\n2 4\n5 2\n2 6\n5 2\n1 5\n6 3\n1 5\n6 3\n3 4\n5 2\n1 5\n6 1\n1 5\n4 2\n3 5\n6 3\n2 6\n6 3\n1 4\n6 2\n3 4\n4 1\n3 6\n5 1\n2 4\n5 1\n3 4\n6 2\n3 5\n4 1\n2 6\n4 3\n2 6\n5 2\n3 6\n6 2\n3 5\n4 3\n1 5\n5 3\n3 6\n4 2\n3 4\n6 1\n3 4\n5 2\n2 6\n5 2\n2 4\n6 2\n3 6\n4 3\n2 4\n4 3\n2 6\n4 2\n3 4\n6 3\n2 4\n6 3\n3 5\n5 2\n1 5\n6 3\n3 6\n4 3\n1 4\n5 2\n1 6\n4 1\n2 5\n4 1\n2 4\n4 2\n2 5\n6 1\n2 4\n6 3\n1 5\n4 3\n2 6\n6 3\n2 6\n5 3\n1 5\n4 1\n1 5\n6 2\n2 5\n5 1\n3 6\n4 3\n3 4",
"output": "Friendship is magic!^^"
},
{
"input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n1 3",
"output": "Mishka"
},
{
"input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "99\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "99\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n2 1\n2 1\n2 1\n1 4",
"output": "Mishka"
},
{
"input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "100\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n1 2\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1\n6 1",
"output": "Chris"
},
{
"input": "100\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n2 1\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6\n1 6",
"output": "Mishka"
},
{
"input": "84\n6 2\n1 5\n6 2\n2 3\n5 5\n1 2\n3 4\n3 4\n6 5\n6 4\n2 5\n4 1\n1 2\n1 1\n1 4\n2 5\n5 6\n6 3\n2 4\n5 5\n2 6\n3 4\n5 1\n3 3\n5 5\n4 6\n4 6\n2 4\n4 1\n5 2\n2 2\n3 6\n3 3\n4 6\n1 1\n2 4\n6 5\n5 2\n6 5\n5 5\n2 5\n6 4\n1 1\n6 2\n3 6\n6 5\n4 4\n1 5\n5 6\n4 4\n3 5\n6 1\n3 4\n1 5\n4 6\n4 6\n4 1\n3 6\n6 2\n1 1\n4 5\n5 4\n5 3\n3 4\n6 4\n1 1\n5 2\n6 5\n6 1\n2 2\n2 4\n3 3\n4 6\n1 3\n6 6\n5 2\n1 6\n6 2\n6 6\n4 1\n3 6\n6 4\n2 3\n3 4",
"output": "Chris"
},
{
"input": "70\n3 4\n2 3\n2 3\n6 5\n6 6\n4 3\n2 3\n3 1\n3 5\n5 6\n1 6\n2 5\n5 3\n2 5\n4 6\n5 1\n6 1\n3 1\n3 3\n5 3\n2 1\n3 3\n6 4\n6 3\n4 3\n4 5\n3 5\n5 5\n5 2\n1 6\n3 4\n5 2\n2 4\n1 6\n4 3\n4 3\n6 2\n1 3\n1 5\n6 1\n3 1\n1 1\n1 3\n2 2\n3 2\n6 4\n1 1\n4 4\n3 1\n4 5\n4 2\n6 3\n4 4\n3 2\n1 2\n2 6\n3 3\n1 5\n1 1\n6 5\n2 2\n3 1\n5 4\n5 2\n6 4\n6 3\n6 6\n6 3\n3 3\n5 4",
"output": "Mishka"
},
{
"input": "56\n6 4\n3 4\n6 1\n3 3\n1 4\n2 3\n1 5\n2 5\n1 5\n5 5\n2 3\n1 1\n3 2\n3 5\n4 6\n4 4\n5 2\n4 3\n3 1\n3 6\n2 3\n3 4\n5 6\n5 2\n5 6\n1 5\n1 5\n4 1\n6 3\n2 2\n2 1\n5 5\n2 1\n4 1\n5 4\n2 5\n4 1\n6 2\n3 4\n4 2\n6 4\n5 4\n4 2\n4 3\n6 2\n6 2\n3 1\n1 4\n3 6\n5 1\n5 5\n3 6\n6 4\n2 3\n6 5\n3 3",
"output": "Mishka"
},
{
"input": "94\n2 4\n6 4\n1 6\n1 4\n5 1\n3 3\n4 3\n6 1\n6 5\n3 2\n2 3\n5 1\n5 3\n1 2\n4 3\n3 2\n2 3\n4 6\n1 3\n6 3\n1 1\n3 2\n4 3\n1 5\n4 6\n3 2\n6 3\n1 6\n1 1\n1 2\n3 5\n1 3\n3 5\n4 4\n4 2\n1 4\n4 5\n1 3\n1 2\n1 1\n5 4\n5 5\n6 1\n2 1\n2 6\n6 6\n4 2\n3 6\n1 6\n6 6\n1 5\n3 2\n1 2\n4 4\n6 4\n4 1\n1 5\n3 3\n1 3\n3 4\n4 4\n1 1\n2 5\n4 5\n3 1\n3 1\n3 6\n3 2\n1 4\n1 6\n6 3\n2 4\n1 1\n2 2\n2 2\n2 1\n5 4\n1 2\n6 6\n2 2\n3 3\n6 3\n6 3\n1 6\n2 3\n2 4\n2 3\n6 6\n2 6\n6 3\n3 5\n1 4\n1 1\n3 5",
"output": "Chris"
},
{
"input": "81\n4 2\n1 2\n2 3\n4 5\n6 2\n1 6\n3 6\n3 4\n4 6\n4 4\n3 5\n4 6\n3 6\n3 5\n3 1\n1 3\n5 3\n3 4\n1 1\n4 1\n1 2\n6 1\n1 3\n6 5\n4 5\n4 2\n4 5\n6 2\n1 2\n2 6\n5 2\n1 5\n2 4\n4 3\n5 4\n1 2\n5 3\n2 6\n6 4\n1 1\n1 3\n3 1\n3 1\n6 5\n5 5\n6 1\n6 6\n5 2\n1 3\n1 4\n2 3\n5 5\n3 1\n3 1\n4 4\n1 6\n6 4\n2 2\n4 6\n4 4\n2 6\n2 4\n2 4\n4 1\n1 6\n1 4\n1 3\n6 5\n5 1\n1 3\n5 1\n1 4\n3 5\n2 6\n1 3\n5 6\n3 5\n4 4\n5 5\n5 6\n4 3",
"output": "Chris"
},
{
"input": "67\n6 5\n3 6\n1 6\n5 3\n5 4\n5 1\n1 6\n1 1\n3 2\n4 4\n3 1\n4 1\n1 5\n5 3\n3 3\n6 4\n2 4\n2 2\n4 3\n1 4\n1 4\n6 1\n1 2\n2 2\n5 1\n6 2\n3 5\n5 5\n2 2\n6 5\n6 2\n4 4\n3 1\n4 2\n6 6\n6 4\n5 1\n2 2\n4 5\n5 5\n4 6\n1 5\n6 3\n4 4\n1 5\n6 4\n3 6\n3 4\n1 6\n2 4\n2 1\n2 5\n6 5\n6 4\n4 1\n3 2\n1 2\n5 1\n5 6\n1 5\n3 5\n3 1\n5 3\n3 2\n5 1\n4 6\n6 6",
"output": "Mishka"
},
{
"input": "55\n6 6\n6 5\n2 2\n2 2\n6 4\n5 5\n6 5\n5 3\n1 3\n2 2\n5 6\n3 3\n3 3\n6 5\n3 5\n5 5\n1 2\n1 1\n4 6\n1 2\n5 5\n6 2\n6 3\n1 2\n5 1\n1 3\n3 3\n4 4\n2 5\n1 1\n5 3\n4 3\n2 2\n4 5\n5 6\n4 5\n6 3\n1 6\n6 4\n3 6\n1 6\n5 2\n6 3\n2 3\n5 5\n4 3\n3 1\n4 2\n1 1\n2 5\n5 3\n2 2\n6 3\n4 5\n2 2",
"output": "Mishka"
},
{
"input": "92\n2 3\n1 3\n2 6\n5 1\n5 5\n3 2\n5 6\n2 5\n3 1\n3 6\n4 5\n2 5\n1 2\n2 3\n6 5\n3 6\n4 4\n6 2\n4 5\n4 4\n5 1\n6 1\n3 4\n3 5\n6 6\n3 2\n6 4\n2 2\n3 5\n6 4\n6 3\n6 6\n3 4\n3 3\n6 1\n5 4\n6 2\n2 6\n5 6\n1 4\n4 6\n6 3\n3 1\n4 1\n6 6\n3 5\n6 3\n6 1\n1 6\n3 2\n6 6\n4 3\n3 4\n1 3\n3 5\n5 3\n6 5\n4 3\n5 5\n4 1\n1 5\n6 4\n2 3\n2 3\n1 5\n1 2\n5 2\n4 3\n3 6\n5 5\n5 4\n1 4\n3 3\n1 6\n5 6\n5 4\n5 3\n1 1\n6 2\n5 5\n2 5\n4 3\n6 6\n5 1\n1 1\n4 6\n4 6\n3 1\n6 4\n2 4\n2 2\n2 1",
"output": "Chris"
},
{
"input": "79\n5 3\n4 6\n3 6\n2 1\n5 2\n2 3\n4 4\n6 2\n2 5\n1 6\n6 6\n2 6\n3 3\n4 5\n6 2\n2 1\n1 5\n5 1\n2 1\n2 6\n5 3\n6 2\n2 6\n2 3\n1 5\n4 4\n6 3\n5 2\n3 2\n1 3\n1 3\n6 3\n2 6\n3 6\n5 3\n4 5\n6 1\n3 5\n3 5\n6 5\n1 5\n4 2\n6 2\n2 3\n4 6\n3 6\n2 5\n4 4\n1 1\n4 6\n2 6\n6 4\n3 2\n4 1\n1 2\n6 4\n5 6\n1 4\n2 2\n5 4\n3 2\n1 2\n2 4\n2 5\n2 1\n3 6\n3 3\n1 1\n2 2\n4 4\n4 5\n3 3\n5 3\n6 2\n4 5\n6 5\n2 5\n5 6\n2 2",
"output": "Chris"
},
{
"input": "65\n1 1\n5 1\n2 2\n5 4\n4 5\n2 5\n3 2\n5 6\n6 3\n1 1\n6 1\n1 5\n1 1\n5 2\n6 4\n1 6\n1 1\n4 3\n2 3\n5 6\n4 4\n6 2\n1 3\n4 3\n1 3\n6 3\n3 5\n4 2\n4 1\n6 1\n3 2\n2 6\n3 2\n3 5\n6 3\n4 3\n1 5\n2 6\n1 3\n4 1\n4 1\n2 5\n2 5\n6 2\n5 3\n3 1\n3 3\n5 1\n2 4\n5 3\n3 3\n1 1\n6 3\n3 3\n5 1\n1 6\n4 5\n6 6\n5 5\n2 5\n4 1\n2 2\n1 4\n1 6\n6 5",
"output": "Mishka"
},
{
"input": "1\n1 1",
"output": "Friendship is magic!^^"
}
] | 1,673,644,278
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 69
| 46
| 0
|
n = int(input())
mishka = 0
chris = 0
for i in range(n):
mish, chri = map(int,input().split())
if mish > chri:
mishka = mishka + 1
elif chri > mish:
chris = chris + 1
if chris == mishka:
print("Friendship is magic!^^")
else:
if chris > mishka:
print("Chris")
elif mishka > chris:
print("Mishka")
|
Title: Mishka and Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mishka is a little polar bear. As known, little bears loves spending their free time playing dice for chocolates. Once in a wonderful sunny morning, walking around blocks of ice, Mishka met her friend Chris, and they started playing the game.
Rules of the game are very simple: at first number of rounds *n* is defined. In every round each of the players throws a cubical dice with distinct numbers from 1 to 6 written on its faces. Player, whose value after throwing the dice is greater, wins the round. In case if player dice values are equal, no one of them is a winner.
In average, player, who won most of the rounds, is the winner of the game. In case if two players won the same number of rounds, the result of the game is draw.
Mishka is still very little and can't count wins and losses, so she asked you to watch their game and determine its result. Please help her!
Input Specification:
The first line of the input contains single integer *n* *n* (1<=≤<=*n*<=≤<=100) — the number of game rounds.
The next *n* lines contains rounds description. *i*-th of them contains pair of integers *m**i* and *c**i* (1<=≤<=*m**i*,<=<=*c**i*<=≤<=6) — values on dice upper face after Mishka's and Chris' throws in *i*-th round respectively.
Output Specification:
If Mishka is the winner of the game, print "Mishka" (without quotes) in the only line.
If Chris is the winner of the game, print "Chris" (without quotes) in the only line.
If the result of the game is draw, print "Friendship is magic!^^" (without quotes) in the only line.
Demo Input:
['3\n3 5\n2 1\n4 2\n', '2\n6 1\n1 6\n', '3\n1 5\n3 3\n2 2\n']
Demo Output:
['Mishka', 'Friendship is magic!^^', 'Chris']
Note:
In the first sample case Mishka loses the first round, but wins second and third rounds and thus she is the winner of the game.
In the second sample case Mishka wins the first round, Chris wins the second round, and the game ends with draw with score 1:1.
In the third sample case Chris wins the first round, but there is no winner of the next two rounds. The winner of the game is Chris.
|
```python
n = int(input())
mishka = 0
chris = 0
for i in range(n):
mish, chri = map(int,input().split())
if mish > chri:
mishka = mishka + 1
elif chri > mish:
chris = chris + 1
if chris == mishka:
print("Friendship is magic!^^")
else:
if chris > mishka:
print("Chris")
elif mishka > chris:
print("Mishka")
```
| 3
|
|
237
|
A
|
Free Cash
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately.
Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe.
Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors.
|
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), that is the number of cafe visitors.
Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=≤<=*h**i*<=≤<=23; 0<=≤<=*m**i*<=≤<=59), representing the time when the *i*-th person comes into the cafe.
Note that the time is given in the chronological order. All time is given within one 24-hour period.
|
Print a single integer — the minimum number of cashes, needed to serve all clients next day.
|
[
"4\n8 0\n8 10\n8 10\n8 45\n",
"3\n0 12\n10 11\n22 22\n"
] |
[
"2\n",
"1\n"
] |
In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away.
In the second sample all visitors will come in different times, so it will be enough one cash.
| 500
|
[
{
"input": "4\n8 0\n8 10\n8 10\n8 45",
"output": "2"
},
{
"input": "3\n0 12\n10 11\n22 22",
"output": "1"
},
{
"input": "5\n12 8\n15 27\n15 27\n16 2\n19 52",
"output": "2"
},
{
"input": "7\n5 6\n7 34\n7 34\n7 34\n12 29\n15 19\n20 23",
"output": "3"
},
{
"input": "8\n0 36\n4 7\n4 7\n4 7\n11 46\n12 4\n15 39\n18 6",
"output": "3"
},
{
"input": "20\n4 12\n4 21\n4 27\n4 56\n5 55\n7 56\n11 28\n11 36\n14 58\n15 59\n16 8\n17 12\n17 23\n17 23\n17 23\n17 23\n17 23\n17 23\n20 50\n22 32",
"output": "6"
},
{
"input": "10\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30\n1 30",
"output": "10"
},
{
"input": "50\n0 23\n1 21\n2 8\n2 45\n3 1\n4 19\n4 37\n7 7\n7 40\n8 43\n9 51\n10 13\n11 2\n11 19\n11 30\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 37\n12 54\n13 32\n13 42\n14 29\n14 34\n14 48\n15 0\n15 27\n16 22\n16 31\n17 25\n17 26\n17 33\n18 14\n18 16\n18 20\n19 0\n19 5\n19 56\n20 22\n21 26\n22 0\n22 10\n22 11\n22 36\n23 17\n23 20",
"output": "8"
},
{
"input": "10\n0 39\n1 35\n1 49\n1 51\n5 24\n7 40\n7 56\n16 42\n23 33\n23 49",
"output": "1"
},
{
"input": "15\n0 16\n6 15\n8 2\n8 6\n8 7\n10 1\n10 1\n10 3\n10 12\n13 5\n14 16\n14 16\n14 16\n14 16\n14 16",
"output": "5"
},
{
"input": "2\n0 24\n1 0",
"output": "1"
},
{
"input": "1\n0 0",
"output": "1"
},
{
"input": "1\n1 5",
"output": "1"
},
{
"input": "1\n1 1",
"output": "1"
},
{
"input": "3\n0 0\n0 0\n0 0",
"output": "3"
},
{
"input": "1\n5 0",
"output": "1"
},
{
"input": "5\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "5"
},
{
"input": "1\n10 10",
"output": "1"
},
{
"input": "1\n8 0",
"output": "1"
},
{
"input": "10\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0\n0 0",
"output": "10"
},
{
"input": "2\n0 0\n0 1",
"output": "1"
},
{
"input": "2\n8 5\n8 5",
"output": "2"
}
] | 1,620,998,577
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 8
| 2,000
| 7,065,600
|
n = int(input())
l = []
for i in range(n):
x, y = map(int, input().split())
l.append(x*60+y)
l.sort()
mx = 1
for i in l:
mx = max(mx, l.count(i))
print(mx)
|
Title: Free Cash
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Valera runs a 24/7 fast food cafe. He magically learned that next day *n* people will visit his cafe. For each person we know the arrival time: the *i*-th person comes exactly at *h**i* hours *m**i* minutes. The cafe spends less than a minute to serve each client, but if a client comes in and sees that there is no free cash, than he doesn't want to wait and leaves the cafe immediately.
Valera is very greedy, so he wants to serve all *n* customers next day (and get more profit). However, for that he needs to ensure that at each moment of time the number of working cashes is no less than the number of clients in the cafe.
Help Valera count the minimum number of cashes to work at his cafe next day, so that they can serve all visitors.
Input Specification:
The first line contains a single integer *n* (1<=≤<=*n*<=≤<=105), that is the number of cafe visitors.
Each of the following *n* lines has two space-separated integers *h**i* and *m**i* (0<=≤<=*h**i*<=≤<=23; 0<=≤<=*m**i*<=≤<=59), representing the time when the *i*-th person comes into the cafe.
Note that the time is given in the chronological order. All time is given within one 24-hour period.
Output Specification:
Print a single integer — the minimum number of cashes, needed to serve all clients next day.
Demo Input:
['4\n8 0\n8 10\n8 10\n8 45\n', '3\n0 12\n10 11\n22 22\n']
Demo Output:
['2\n', '1\n']
Note:
In the first sample it is not enough one cash to serve all clients, because two visitors will come into cafe in 8:10. Therefore, if there will be one cash in cafe, then one customer will be served by it, and another one will not wait and will go away.
In the second sample all visitors will come in different times, so it will be enough one cash.
|
```python
n = int(input())
l = []
for i in range(n):
x, y = map(int, input().split())
l.append(x*60+y)
l.sort()
mx = 1
for i in l:
mx = max(mx, l.count(i))
print(mx)
```
| 0
|
|
507
|
B
|
Amr and Pins
|
PROGRAMMING
| 1,400
|
[
"geometry",
"math"
] | null | null |
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius *r* and center in point (*x*,<=*y*). He wants the circle center to be in new position (*x*',<=*y*').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
|
Input consists of 5 space-separated integers *r*, *x*, *y*, *x*' *y*' (1<=≤<=*r*<=≤<=105, <=-<=105<=≤<=*x*,<=*y*,<=*x*',<=*y*'<=≤<=105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
|
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
|
[
"2 0 0 0 4\n",
"1 1 1 4 4\n",
"4 5 6 5 6\n"
] |
[
"1\n",
"3\n",
"0\n"
] |
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<img class="tex-graphics" src="https://espresso.codeforces.com/4e40fd4cc24a2050a0488aa131e6244369328039.png" style="max-width: 100.0%;max-height: 100.0%;"/>
| 1,000
|
[
{
"input": "2 0 0 0 4",
"output": "1"
},
{
"input": "1 1 1 4 4",
"output": "3"
},
{
"input": "4 5 6 5 6",
"output": "0"
},
{
"input": "10 20 0 40 0",
"output": "1"
},
{
"input": "9 20 0 40 0",
"output": "2"
},
{
"input": "5 -1 -6 -5 1",
"output": "1"
},
{
"input": "99125 26876 -21414 14176 17443",
"output": "1"
},
{
"input": "8066 7339 19155 -90534 -60666",
"output": "8"
},
{
"input": "100000 -100000 -100000 100000 100000",
"output": "2"
},
{
"input": "10 20 0 41 0",
"output": "2"
},
{
"input": "25 -64 -6 -56 64",
"output": "2"
},
{
"input": "125 455 450 439 721",
"output": "2"
},
{
"input": "5 6 3 7 2",
"output": "1"
},
{
"input": "24 130 14786 3147 2140",
"output": "271"
},
{
"input": "125 -363 176 93 330",
"output": "2"
},
{
"input": "1 14 30 30 14",
"output": "12"
},
{
"input": "25 96 13 7 2",
"output": "2"
},
{
"input": "4 100000 -100000 100000 -100000",
"output": "0"
},
{
"input": "1 3 4 2 5",
"output": "1"
},
{
"input": "1 -3 3 2 6",
"output": "3"
},
{
"input": "2 7 20 13 -5",
"output": "7"
},
{
"input": "1 1 1 1 4",
"output": "2"
},
{
"input": "249 -54242 -30537 -45023 -89682",
"output": "121"
},
{
"input": "4 100000 -100000 100000 -99999",
"output": "1"
},
{
"input": "97741 23818 78751 97583 26933",
"output": "1"
},
{
"input": "56767 -29030 51625 79823 -56297",
"output": "2"
},
{
"input": "98260 13729 74998 23701 9253",
"output": "1"
},
{
"input": "67377 -80131 -90254 -57320 14102",
"output": "1"
},
{
"input": "1 100000 100000 100000 -100000",
"output": "100000"
},
{
"input": "19312 19470 82059 58064 62231",
"output": "2"
},
{
"input": "67398 -68747 -79056 -34193 29400",
"output": "1"
},
{
"input": "91099 37184 -71137 75650 -3655",
"output": "1"
},
{
"input": "46456 -2621 -23623 -98302 -99305",
"output": "2"
},
{
"input": "100 100000 -100000 100000 -99999",
"output": "1"
},
{
"input": "1 100000 -100000 100000 -100000",
"output": "0"
},
{
"input": "8 0 0 0 32",
"output": "2"
},
{
"input": "100000 100000 1 -100000 0",
"output": "2"
}
] | 1,641,754,383
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 37
| 93
| 0
|
import math
def main():
r, x, y, a, b = [int(x) for x in input().split()]
d = ((a - x) ** 2 + (b - y) ** 2)**0.5
t = r << 1
ans = math.floor(d) // t
d = d - ans * t
if d > 0:
ans += 1
print(ans)
if __name__ == '__main__':
main()
|
Title: Amr and Pins
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius *r* and center in point (*x*,<=*y*). He wants the circle center to be in new position (*x*',<=*y*').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
Input Specification:
Input consists of 5 space-separated integers *r*, *x*, *y*, *x*' *y*' (1<=≤<=*r*<=≤<=105, <=-<=105<=≤<=*x*,<=*y*,<=*x*',<=*y*'<=≤<=105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
Output Specification:
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
Demo Input:
['2 0 0 0 4\n', '1 1 1 4 4\n', '4 5 6 5 6\n']
Demo Output:
['1\n', '3\n', '0\n']
Note:
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<img class="tex-graphics" src="https://espresso.codeforces.com/4e40fd4cc24a2050a0488aa131e6244369328039.png" style="max-width: 100.0%;max-height: 100.0%;"/>
|
```python
import math
def main():
r, x, y, a, b = [int(x) for x in input().split()]
d = ((a - x) ** 2 + (b - y) ** 2)**0.5
t = r << 1
ans = math.floor(d) // t
d = d - ans * t
if d > 0:
ans += 1
print(ans)
if __name__ == '__main__':
main()
```
| 3
|
|
989
|
C
|
A Mist of Florescence
|
PROGRAMMING
| 1,800
|
[
"constructive algorithms",
"graphs"
] | null | null |
"I've been here once," Mino exclaims with delight, "it's breathtakingly amazing."
"What is it like?"
"Look, Kanno, you've got your paintbrush, and I've got my words. Have a try, shall we?"
There are four kinds of flowers in the wood, Amaranths, Begonias, Centaureas and Dianthuses.
The wood can be represented by a rectangular grid of $n$ rows and $m$ columns. In each cell of the grid, there is exactly one type of flowers.
According to Mino, the numbers of connected components formed by each kind of flowers are $a$, $b$, $c$ and $d$ respectively. Two cells are considered in the same connected component if and only if a path exists between them that moves between cells sharing common edges and passes only through cells containing the same flowers.
You are to help Kanno depict such a grid of flowers, with $n$ and $m$ arbitrarily chosen under the constraints given below. It can be shown that at least one solution exists under the constraints of this problem.
Note that you can choose arbitrary $n$ and $m$ under the constraints below, they are not given in the input.
|
The first and only line of input contains four space-separated integers $a$, $b$, $c$ and $d$ ($1 \leq a, b, c, d \leq 100$) — the required number of connected components of Amaranths, Begonias, Centaureas and Dianthuses, respectively.
|
In the first line, output two space-separated integers $n$ and $m$ ($1 \leq n, m \leq 50$) — the number of rows and the number of columns in the grid respectively.
Then output $n$ lines each consisting of $m$ consecutive English letters, representing one row of the grid. Each letter should be among 'A', 'B', 'C' and 'D', representing Amaranths, Begonias, Centaureas and Dianthuses, respectively.
In case there are multiple solutions, print any. You can output each letter in either case (upper or lower).
|
[
"5 3 2 1\n",
"50 50 1 1\n",
"1 6 4 5\n"
] |
[
"4 7\nDDDDDDD\nDABACAD\nDBABACD\nDDDDDDD",
"4 50\nCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC\nABABABABABABABABABABABABABABABABABABABABABABABABAB\nBABABABABABABABABABABABABABABABABABABABABABABABABA\nDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD",
"7 7\nDDDDDDD\nDDDBDBD\nDDCDCDD\nDBDADBD\nDDCDCDD\nDBDBDDD\nDDDDDDD"
] |
In the first example, each cell of Amaranths, Begonias and Centaureas forms a connected component, while all the Dianthuses form one.
| 1,500
|
[
{
"input": "5 3 2 1",
"output": "5 13\nAABABBBBCDDAD\nABAABBBBCDADD\nAAAABBBBCDDAD\nAAAABCBBCDADD\nAAAABBBBCDDDD"
},
{
"input": "50 50 1 1",
"output": "10 50\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nABABABABABABABABABABABABABABABABABABABABABABABABAA\nBABABABABABABABABABABABABABABABABABABABABABABABABA\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC\nDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD\nDADADADADADADADADADADADADADADADADADADADADADADADADD\nADADADADADADADADADADADADADADADADADADADADADADADADAD\nDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD..."
},
{
"input": "1 6 4 5",
"output": "6 13\nAABABBCBCCDCD\nABAABBBBCCCCD\nAABABBCBCCDCD\nABAABCBBCDCCD\nAABABBBBCCDCD\nAAAABBBBCCCCD"
},
{
"input": "1 1 1 1",
"output": "2 4\nABCD\nABCD"
},
{
"input": "4 8 16 32",
"output": "16 32\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nABAAAAAAABAAAAAAAAAAAAAAABABAAAA\nBAAAAAAAAAAABAAAAAAAAAAABAAAAAAA\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBCBBBBBCBCBCBCBCBBBCBCBBBBBBBBBB\nCBCBBBBBBBBBCBBBCBBBCBBBBBCBBBCB\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC\nCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCC\nDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDC\nCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC\nDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD\nDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD\nADDDDDDDDDDDDDDDADDDDDDDDDDD..."
},
{
"input": "1 1 1 50",
"output": "7 50\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC\nCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCC\nDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDC\nCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC\nDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD"
},
{
"input": "19 58 20 18",
"output": "19 50\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nABABABABABAAABABABABABAAABABAAAAABABAAABABAAAAABAA\nAAAABABABAAABABABABABAAABAAABAAABAAAAAAABAAAAAAAAA\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nAABAAABABABAAABABABABAAAAAAABABAAABAAABAAABAAABABA\nABABAAABAAABABAAABAAAAABAAABABAAABAAAAABAAABAAAAAB\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBBBCBBBBBBBBBCBBBCBCBBBBBBBBBCBBBCBBBBBBBBBBBCBCBB\nBBBBCBCBCBCBBBBBCBBBBBCBCBCBBBCBBBBB..."
},
{
"input": "100 100 100 100",
"output": "40 50\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nAAABAAABAAABAAABABABABABABAAABABABABABABAAAAABABAA\nAABAAAAAAAAAAABAAABAAABABABAAABABAAABABABABABABAAA\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nAAAABABABAAAAAAABABABABABAAABABABABABABAAAAAAABABA\nABABABAAABABABAAABABAAABAAABABABABABAAABABABABABAB\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nABABABAAABAAABABAAAAAAABAAAAAAABABAAABAAABAAAAABAA\nBABABABAAABABABABABAAABABABABAAABABAAABABABAAABABA\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA..."
},
{
"input": "1 1 1 2",
"output": "2 7\nABCCDCD\nABCCCCD"
},
{
"input": "1 1 3 1",
"output": "3 7\nABBCBCD\nABCBBCD\nABBBBCD"
},
{
"input": "1 4 1 1",
"output": "4 7\nAABABCD\nABAABCD\nAABABCD\nAAAABCD"
},
{
"input": "5 1 1 1",
"output": "5 7\nABCDDAD\nABCDADD\nABCDDAD\nABCDADD\nABCDDDD"
},
{
"input": "1 4 7 3",
"output": "7 13\nAAAABBCBCCDCD\nABAABCBBCCCCD\nAAAABBCBCCCCD\nABAABCBBCCCCD\nAABABBCBCCCCD\nAAAABCBBCDCCD\nAAAABBBBCCCCD"
},
{
"input": "6 2 5 1",
"output": "6 13\nAAAABBCBCDDAD\nAAAABBBBCDADD\nAAAABBCBCDDAD\nAAAABCBBCDADD\nAABABBCBCDDAD\nAAAABBBBCDDDD"
},
{
"input": "1 5 6 3",
"output": "6 13\nAAAABBCBCCCCD\nABAABCBBCCCCD\nAABABBCBCCCCD\nABAABCBBCDCCD\nAABABBCBCCDCD\nAAAABBBBCCCCD"
},
{
"input": "4 1 4 5",
"output": "5 13\nABBCBCCDCDDAD\nABCBBCDCCDDDD\nABBBBCCDCDDAD\nABCBBCDCCDADD\nABBBBCCCCDDDD"
},
{
"input": "4 5 3 6",
"output": "6 16\nAAAABBCBCCDCDDAD\nABAABBBBCDCCDDDD\nAABABBCBCCDCDDAD\nABAABBBBCDCCDDDD\nAABABBBBCCDCDDAD\nAAAABBBBCCCCDDDD"
},
{
"input": "2 5 1 17",
"output": "13 17\nAAAAAAAAAAAAAAAAA\nABAAAAAAABAAAAAAA\nAABAAAAAAAAABAAAA\nAAAAAAAAAAAAAAAAA\nBBBBBBBBBBBBBBBBB\nCCCCCCCCCCCCCCCCC\nCDCDCDCDCDCDCDCDC\nDCDCDCDCDCDCDCDCC\nCCCCCCCCCCCCCCCCC\nDDDDDDDDDDDDDDDDD\nDDDDDDDDDDDDDDDDD\nDDADDDDDDDDDDDDDD\nDDDDDDDDDDDDDDDDD"
},
{
"input": "11 4 5 14",
"output": "14 16\nAAAABBBBCCDCDDDD\nABAABBBBCDCCDADD\nAAAABBBBCCDCDDAD\nAAAABBBBCDCCDADD\nAAAABBBBCCDCDDAD\nAAAABBBBCDCCDADD\nAAAABBBBCCDCDDAD\nAAAABCBBCDCCDDDD\nAAAABBCBCCDCDDDD\nABAABCBBCDCCDADD\nAAAABBBBCCDCDDAD\nAAAABCBBCDCCDADD\nAABABBBBCCDCDDAD\nAAAABBBBCCCCDDDD"
},
{
"input": "19 19 8 10",
"output": "16 19\nAAAAAAAAAAAAAAAAAAA\nABABABABABABABABABA\nBABABABABABABABABAA\nAAAAAAAAAAAAAAAAAAA\nBBBBBBBBBBBBBBBBBBB\nBBBCBBBBBBBBBCBBBCB\nBBCBBBCBCBBBBBCBBBB\nBBBBBBBBBBBBBBBBBBB\nCCCCCCCCCCCCCCCCCCC\nCCCDCCCCCDCCCDCCCDC\nCCCCDCDCCCDCCCDCDCC\nCCCCCCCCCCCCCCCCCCC\nDDDDDDDDDDDDDDDDDDD\nDADADADADADADADADAD\nADADADADADADADADADD\nDDDDDDDDDDDDDDDDDDD"
},
{
"input": "49 49 49 49",
"output": "16 49\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nABABABABABABABABABABABABABABABABABABABABABABABABA\nBABABABABABABABABABABABABABABABABABABABABABABABAA\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCB\nCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBB\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC\nCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDC..."
},
{
"input": "49 50 50 50",
"output": "16 50\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nABABABABABABABABABABABABABABABABABABABABABABABABAA\nBABABABABABABABABABABABABABABABABABABABABABABABABA\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBB\nCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCB\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC\nCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCDCD..."
},
{
"input": "50 50 51 50",
"output": "19 50\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nABABABABABABABABABABABABABABABABABABABABABABABABAA\nBABABABABABABABABABABABABABABABABABABABABABABABABA\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBBBCBBBBBCBCBBBBBCBBBCBBBCBBBBBBBCBCBBBCBBBBBCBBBB\nCBCBCBCBCBBBBBBBBBBBBBBBBBBBCBCBBBCBCBCBBBCBCBCBBB\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBBCBCBBBBBCBBBCBBBCBCBCBCBCBCBCBCBCBCBBBCBBBBBCBCB\nBBBBBCBBBCBBBBBCBBBBBCBBBCBBBCBBBCBB..."
},
{
"input": "15 63 41 45",
"output": "19 50\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nABABABABABAAABABABABABAAABABAAAAABABAAABABAAAAABAA\nAAAABABABAAABABABABABAAABAAABAAABABABABABAAABAAAAA\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nAABAAABABABAAABABABABAAAAAAABABAAABAAABAAABAAABABA\nABABAAABAAABABAAABAAAAABAAABABAAABABAAABAAABAAAAAB\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBBBCBCBCBCBCBCBCBCBCBCBBBCBBBCBBBCBCBCBCBCBCBCBCBB\nCBBBCBCBCBCBCBCBCBBBCBCBCBCBCBCBBBBB..."
},
{
"input": "45 36 25 13",
"output": "16 45\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nABAAABABABABAAABABABABABAAABABABABABAAAAABAAA\nBABABABABABABAAAAABABABABABABABABABABAAABABAA\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBCBCBCBCBCBBBCBBBBBBBBBBBBBBBCBCBCBBBCBCBBBCB\nCBBBCBCBCBCBBBBBCBBBBBBBBBCBBBCBBBBBCBCBCBCBB\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC\nCCCCCCCCCCCCCCCCCCCDCCCCCCCCCDCCCCCDCCCCCDCCC\nCCCCCCCCCCCCDCCCDCCCCCDCDCCCCCDCCC..."
},
{
"input": "31 41 59 26",
"output": "19 50\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nABABABABABABAAAAABAAABABABABABABABABAAABABABABABAA\nBABABABABABABAAABABAAABAAABAAABABABABAAABABABABABA\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBBBCBBBBBCBCBCBBBCBBBCBBBCBBBBBBBCBCBBBCBBBBBCBBBB\nCBCBCBCBCBCBBBBBBBBBBBBBCBBBCBCBBBCBCBCBBBCBCBCBBB\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBBCBCBBBBBCBBBCBBBCBCBCBCBCBCBCBCBCBCBBBCBBBBBCBCB\nBCBBBCBBBCBBBCBCBBBBBCBBBCBBBCBCBCBC..."
},
{
"input": "18 90 64 16",
"output": "22 50\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nABABABABABABABABABABABABABABABABABABABABABABAAABAA\nAABABABABAAABABABABABABABAAABABABABABABABABABABAAA\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nAABABABABABABABABABABABABABABABAAABABABABABABABABA\nABABABABABABABAAABABAAABABABABABABABABABABABAAABAB\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBBBBBCBCBCBBBCBBBBBCBCBCBBBCBCBCBBBCBBBCBCBCBCBCBB\nCBCBCBBBCBCBBBCBBBCBCBBBCBCBBBCBCBBB..."
},
{
"input": "77 88 99 1",
"output": "22 50\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nABABABABABABABABABABABABABABABABABABABABABAAAAABAA\nAABABABABAAABABABABABABABAAABABABABABABABABABABAAA\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nAABABABABABABABABABABABABAAABABAAABABABABABABABABA\nABABABABABABABAAABABAAABABABABABABABABABABABAAABAB\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBB\nCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCBCB..."
},
{
"input": "99 100 1 100",
"output": "28 50\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nAAABAAABAAABAAABABABABABABAAABABABABABABAAAAABABAA\nAABAAAAAAAAAAABAAABAAABABABAAABABAAABABABABABABAAA\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nAAAABABABAAAAAAABABABABABAAABABABABABABAAAAAAABABA\nABABABAAABABABAAABABAAABAAABABABABABAAABABABABABAB\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nABABABAAABAAABABAAAAAAABAAAAAAABABAAABAAABAAAAABAA\nBABABABAAABABABABABAAABABABABAAABABAAABABABAAABABA\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA..."
},
{
"input": "100 50 100 49",
"output": "28 50\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nABABABABABABABABABABABABABABABABABABABABABABABABAA\nBABABABABABABABABABABABABABABABABABABABABABABABABA\nAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBCBCBBBBBCBCBCBBBCBBBBBCBCBBBCBCBCBBBCBCBCBCBBBBBB\nBBBBCBCBCBCBBBCBCBCBCBCBBBCBCBCBCBBBCBCBCBCBCBCBBB\nBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB\nBBCBCBBBBBCBCBCBCBCBCBCBCBCBCBBBCBCBCBBBBBCBCBCBCB\nBCBBBBBCBCBCBBBCBCBBBCBCBCBBBBBCBCBB..."
}
] | 1,528,812,213
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 108
| 0
|
abcd = input().split(' ')
a = int(abcd[0])
b = int(abcd[1])
c = int(abcd[2])
d = int(abcd[3])
out = []
h = int(50 / 4)
for letter in 'D C B A'.split(' '):
for i in range(h):
out.append([letter for j in range(50)])
y = 0
for counts in [a - 1,b - 1,c - 1,d - 1]:
counter = 0
shift = 1
for q in range(3):
for x in range(49):
if counter == counts:
break
out[y + shift + x % 2][x] = ['A', 'B', 'C', 'D'][int(y / h)]
counter += 1
shift += 3
if counter == counts:
break
if counter < counts:
for i in range(counts - counter):
out[y + shift][2 * x] = ['A', 'B', 'C', 'D'][4 - int(y / h)]
y += h
print(4 * h, 50)
for line in out:
print(''.join(line))
|
Title: A Mist of Florescence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
"I've been here once," Mino exclaims with delight, "it's breathtakingly amazing."
"What is it like?"
"Look, Kanno, you've got your paintbrush, and I've got my words. Have a try, shall we?"
There are four kinds of flowers in the wood, Amaranths, Begonias, Centaureas and Dianthuses.
The wood can be represented by a rectangular grid of $n$ rows and $m$ columns. In each cell of the grid, there is exactly one type of flowers.
According to Mino, the numbers of connected components formed by each kind of flowers are $a$, $b$, $c$ and $d$ respectively. Two cells are considered in the same connected component if and only if a path exists between them that moves between cells sharing common edges and passes only through cells containing the same flowers.
You are to help Kanno depict such a grid of flowers, with $n$ and $m$ arbitrarily chosen under the constraints given below. It can be shown that at least one solution exists under the constraints of this problem.
Note that you can choose arbitrary $n$ and $m$ under the constraints below, they are not given in the input.
Input Specification:
The first and only line of input contains four space-separated integers $a$, $b$, $c$ and $d$ ($1 \leq a, b, c, d \leq 100$) — the required number of connected components of Amaranths, Begonias, Centaureas and Dianthuses, respectively.
Output Specification:
In the first line, output two space-separated integers $n$ and $m$ ($1 \leq n, m \leq 50$) — the number of rows and the number of columns in the grid respectively.
Then output $n$ lines each consisting of $m$ consecutive English letters, representing one row of the grid. Each letter should be among 'A', 'B', 'C' and 'D', representing Amaranths, Begonias, Centaureas and Dianthuses, respectively.
In case there are multiple solutions, print any. You can output each letter in either case (upper or lower).
Demo Input:
['5 3 2 1\n', '50 50 1 1\n', '1 6 4 5\n']
Demo Output:
['4 7\nDDDDDDD\nDABACAD\nDBABACD\nDDDDDDD', '4 50\nCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC\nABABABABABABABABABABABABABABABABABABABABABABABABAB\nBABABABABABABABABABABABABABABABABABABABABABABABABA\nDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD', '7 7\nDDDDDDD\nDDDBDBD\nDDCDCDD\nDBDADBD\nDDCDCDD\nDBDBDDD\nDDDDDDD']
Note:
In the first example, each cell of Amaranths, Begonias and Centaureas forms a connected component, while all the Dianthuses form one.
|
```python
abcd = input().split(' ')
a = int(abcd[0])
b = int(abcd[1])
c = int(abcd[2])
d = int(abcd[3])
out = []
h = int(50 / 4)
for letter in 'D C B A'.split(' '):
for i in range(h):
out.append([letter for j in range(50)])
y = 0
for counts in [a - 1,b - 1,c - 1,d - 1]:
counter = 0
shift = 1
for q in range(3):
for x in range(49):
if counter == counts:
break
out[y + shift + x % 2][x] = ['A', 'B', 'C', 'D'][int(y / h)]
counter += 1
shift += 3
if counter == counts:
break
if counter < counts:
for i in range(counts - counter):
out[y + shift][2 * x] = ['A', 'B', 'C', 'D'][4 - int(y / h)]
y += h
print(4 * h, 50)
for line in out:
print(''.join(line))
```
| 3
|
|
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"
}
] | 1,698,055,745
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 92
| 0
|
s = input();a = s[0];b = s[1:]
print(a.upper(),b,sep='')
|
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];b = s[1:]
print(a.upper(),b,sep='')
```
| 3
|
|
677
|
E
|
Vanya and Balloons
|
PROGRAMMING
| 2,300
|
[
"binary search",
"brute force",
"dp",
"implementation"
] | null | null |
Vanya plays a game of balloons on the field of size *n*<=×<=*n*, where each cell contains a balloon with one of the values 0, 1, 2 or 3. The goal is to destroy a cross, such that the product of all values of balloons in the cross is maximum possible. There are two types of crosses: normal and rotated. For example:
or
Formally, the cross is given by three integers *r*, *c* and *d*, such that *d*<=≤<=*r*,<=*c*<=≤<=*n*<=-<=*d*<=+<=1. The normal cross consists of balloons located in cells (*x*,<=*y*) (where *x* stay for the number of the row and *y* for the number of the column), such that |*x*<=-<=*r*|·|*y*<=-<=*c*|<==<=0 and |*x*<=-<=*r*|<=+<=|*y*<=-<=*c*|<=<<=*d*. Rotated cross consists of balloons located in cells (*x*,<=*y*), such that |*x*<=-<=*r*|<==<=|*y*<=-<=*c*| and |*x*<=-<=*r*|<=<<=*d*.
Vanya wants to know the maximum possible product of the values of balls forming one cross. As this value can be large, output it modulo 109<=+<=7.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of rows and columns in the table with balloons.
The each of the following *n* lines contains *n* characters '0', '1', '2' or '3' — the description of the values in balloons.
|
Print the maximum possible product modulo 109<=+<=7. Note, that you are not asked to maximize the remainder modulo 109<=+<=7, but to find the maximum value and print it this modulo.
|
[
"4\n1233\n0213\n2020\n0303\n",
"5\n00300\n00300\n33333\n00300\n00300\n",
"5\n00003\n02030\n00300\n03020\n30000\n",
"5\n21312\n10003\n10002\n10003\n23231\n",
"5\n12131\n12111\n12112\n21311\n21212\n"
] |
[
"108\n",
"19683\n",
"108\n",
"3\n",
"24\n"
] |
In the first sample, the maximum product is achieved for a rotated cross with a center in the cell (3, 3) and radius 1: 2·2·3·3·3 = 108.
| 2,250
|
[] | 1,692,524,195
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 30
| 0
|
print("_RANDOM_GUESS_1692524194.9703364")# 1692524194.9703534
|
Title: Vanya and Balloons
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vanya plays a game of balloons on the field of size *n*<=×<=*n*, where each cell contains a balloon with one of the values 0, 1, 2 or 3. The goal is to destroy a cross, such that the product of all values of balloons in the cross is maximum possible. There are two types of crosses: normal and rotated. For example:
or
Formally, the cross is given by three integers *r*, *c* and *d*, such that *d*<=≤<=*r*,<=*c*<=≤<=*n*<=-<=*d*<=+<=1. The normal cross consists of balloons located in cells (*x*,<=*y*) (where *x* stay for the number of the row and *y* for the number of the column), such that |*x*<=-<=*r*|·|*y*<=-<=*c*|<==<=0 and |*x*<=-<=*r*|<=+<=|*y*<=-<=*c*|<=<<=*d*. Rotated cross consists of balloons located in cells (*x*,<=*y*), such that |*x*<=-<=*r*|<==<=|*y*<=-<=*c*| and |*x*<=-<=*r*|<=<<=*d*.
Vanya wants to know the maximum possible product of the values of balls forming one cross. As this value can be large, output it modulo 109<=+<=7.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=1000) — the number of rows and columns in the table with balloons.
The each of the following *n* lines contains *n* characters '0', '1', '2' or '3' — the description of the values in balloons.
Output Specification:
Print the maximum possible product modulo 109<=+<=7. Note, that you are not asked to maximize the remainder modulo 109<=+<=7, but to find the maximum value and print it this modulo.
Demo Input:
['4\n1233\n0213\n2020\n0303\n', '5\n00300\n00300\n33333\n00300\n00300\n', '5\n00003\n02030\n00300\n03020\n30000\n', '5\n21312\n10003\n10002\n10003\n23231\n', '5\n12131\n12111\n12112\n21311\n21212\n']
Demo Output:
['108\n', '19683\n', '108\n', '3\n', '24\n']
Note:
In the first sample, the maximum product is achieved for a rotated cross with a center in the cell (3, 3) and radius 1: 2·2·3·3·3 = 108.
|
```python
print("_RANDOM_GUESS_1692524194.9703364")# 1692524194.9703534
```
| 0
|
|
489
|
C
|
Given Length and Sum of Digits...
|
PROGRAMMING
| 1,400
|
[
"dp",
"greedy",
"implementation"
] | null | null |
You have a positive integer *m* and a non-negative integer *s*. Your task is to find the smallest and the largest of the numbers that have length *m* and sum of digits *s*. The required numbers should be non-negative integers written in the decimal base without leading zeroes.
|
The single line of the input contains a pair of integers *m*, *s* (1<=≤<=*m*<=≤<=100,<=0<=≤<=*s*<=≤<=900) — the length and the sum of the digits of the required numbers.
|
In the output print the pair of the required non-negative integer numbers — first the minimum possible number, then — the maximum possible number. If no numbers satisfying conditions required exist, print the pair of numbers "-1 -1" (without the quotes).
|
[
"2 15\n",
"3 0\n"
] |
[
"69 96\n",
"-1 -1\n"
] |
none
| 1,500
|
[
{
"input": "2 15",
"output": "69 96"
},
{
"input": "3 0",
"output": "-1 -1"
},
{
"input": "2 1",
"output": "10 10"
},
{
"input": "3 10",
"output": "109 910"
},
{
"input": "100 100",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000099999999999 9999999999910000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "1 900",
"output": "-1 -1"
},
{
"input": "1 9",
"output": "9 9"
},
{
"input": "1 0",
"output": "0 0"
},
{
"input": "1 1",
"output": "1 1"
},
{
"input": "1 2",
"output": "2 2"
},
{
"input": "1 8",
"output": "8 8"
},
{
"input": "1 10",
"output": "-1 -1"
},
{
"input": "1 11",
"output": "-1 -1"
},
{
"input": "2 0",
"output": "-1 -1"
},
{
"input": "2 1",
"output": "10 10"
},
{
"input": "2 2",
"output": "11 20"
},
{
"input": "2 8",
"output": "17 80"
},
{
"input": "2 10",
"output": "19 91"
},
{
"input": "2 11",
"output": "29 92"
},
{
"input": "2 16",
"output": "79 97"
},
{
"input": "2 17",
"output": "89 98"
},
{
"input": "2 18",
"output": "99 99"
},
{
"input": "2 19",
"output": "-1 -1"
},
{
"input": "2 20",
"output": "-1 -1"
},
{
"input": "2 900",
"output": "-1 -1"
},
{
"input": "3 1",
"output": "100 100"
},
{
"input": "3 2",
"output": "101 200"
},
{
"input": "3 3",
"output": "102 300"
},
{
"input": "3 9",
"output": "108 900"
},
{
"input": "3 10",
"output": "109 910"
},
{
"input": "3 20",
"output": "299 992"
},
{
"input": "3 21",
"output": "399 993"
},
{
"input": "3 26",
"output": "899 998"
},
{
"input": "3 27",
"output": "999 999"
},
{
"input": "3 28",
"output": "-1 -1"
},
{
"input": "3 100",
"output": "-1 -1"
},
{
"input": "100 0",
"output": "-1 -1"
},
{
"input": "100 1",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 2",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001 2000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 9",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000008 9000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 10",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000009 9100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 11",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000019 9200000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 296",
"output": "1000000000000000000000000000000000000000000000000000000000000000000799999999999999999999999999999999 9999999999999999999999999999999980000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 297",
"output": "1000000000000000000000000000000000000000000000000000000000000000000899999999999999999999999999999999 9999999999999999999999999999999990000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 298",
"output": "1000000000000000000000000000000000000000000000000000000000000000000999999999999999999999999999999999 9999999999999999999999999999999991000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 299",
"output": "1000000000000000000000000000000000000000000000000000000000000000001999999999999999999999999999999999 9999999999999999999999999999999992000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 300",
"output": "1000000000000000000000000000000000000000000000000000000000000000002999999999999999999999999999999999 9999999999999999999999999999999993000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 301",
"output": "1000000000000000000000000000000000000000000000000000000000000000003999999999999999999999999999999999 9999999999999999999999999999999994000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 895",
"output": "4999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999994"
},
{
"input": "100 896",
"output": "5999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999995"
},
{
"input": "100 897",
"output": "6999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999996"
},
{
"input": "100 898",
"output": "7999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999997"
},
{
"input": "100 899",
"output": "8999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999998"
},
{
"input": "100 900",
"output": "9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999"
},
{
"input": "99 900",
"output": "-1 -1"
},
{
"input": "99 891",
"output": "999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999"
},
{
"input": "99 892",
"output": "-1 -1"
},
{
"input": "96 364",
"output": "100000000000000000000000000000000000000000000000000000039999999999999999999999999999999999999999 999999999999999999999999999999999999999940000000000000000000000000000000000000000000000000000000"
},
{
"input": "97 206",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000079999999999999999999999 9999999999999999999999800000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "98 250",
"output": "10000000000000000000000000000000000000000000000000000000000000000000006999999999999999999999999999 99999999999999999999999999970000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "99 96",
"output": "100000000000000000000000000000000000000000000000000000000000000000000000000000000000000059999999999 999999999960000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "100 215",
"output": "1000000000000000000000000000000000000000000000000000000000000000000000000000799999999999999999999999 9999999999999999999999980000000000000000000000000000000000000000000000000000000000000000000000000000"
},
{
"input": "5 18",
"output": "10089 99000"
},
{
"input": "2 19",
"output": "-1 -1"
},
{
"input": "3 3",
"output": "102 300"
}
] | 1,677,070,574
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 31
| 0
|
m,s=map(int,input().split())
low=[0]*m
hie=[0]*m
x=s
if s>0:
low[0]=1
x-=1
i=m-1
while x>0 and i>0:
temp=min(9,x)
low[i]+=temp
x-=temp
i-=1
xx=s
i=0
while xx>0 and i<m:
temp=min(9,xx)
hie[i]=temp
xx-=temp
i+=1
if (low == [0]*len(low) and hie == [0]*len(hie) )or x!=0 or xx!=0:
print("-1 -1")
else:
print(*low,sep='',end=' ')
print(*hie,sep='')
|
Title: Given Length and Sum of Digits...
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You have a positive integer *m* and a non-negative integer *s*. Your task is to find the smallest and the largest of the numbers that have length *m* and sum of digits *s*. The required numbers should be non-negative integers written in the decimal base without leading zeroes.
Input Specification:
The single line of the input contains a pair of integers *m*, *s* (1<=≤<=*m*<=≤<=100,<=0<=≤<=*s*<=≤<=900) — the length and the sum of the digits of the required numbers.
Output Specification:
In the output print the pair of the required non-negative integer numbers — first the minimum possible number, then — the maximum possible number. If no numbers satisfying conditions required exist, print the pair of numbers "-1 -1" (without the quotes).
Demo Input:
['2 15\n', '3 0\n']
Demo Output:
['69 96\n', '-1 -1\n']
Note:
none
|
```python
m,s=map(int,input().split())
low=[0]*m
hie=[0]*m
x=s
if s>0:
low[0]=1
x-=1
i=m-1
while x>0 and i>0:
temp=min(9,x)
low[i]+=temp
x-=temp
i-=1
xx=s
i=0
while xx>0 and i<m:
temp=min(9,xx)
hie[i]=temp
xx-=temp
i+=1
if (low == [0]*len(low) and hie == [0]*len(hie) )or x!=0 or xx!=0:
print("-1 -1")
else:
print(*low,sep='',end=' ')
print(*hie,sep='')
```
| 0
|
|
723
|
A
|
The New Year: Meeting Friends
|
PROGRAMMING
| 800
|
[
"implementation",
"math",
"sortings"
] | null | null |
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
|
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
|
Print one integer — the minimum total distance the friends need to travel in order to meet together.
|
[
"7 1 4\n",
"30 20 10\n"
] |
[
"6\n",
"20\n"
] |
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
| 500
|
[
{
"input": "7 1 4",
"output": "6"
},
{
"input": "30 20 10",
"output": "20"
},
{
"input": "1 4 100",
"output": "99"
},
{
"input": "100 1 91",
"output": "99"
},
{
"input": "1 45 100",
"output": "99"
},
{
"input": "1 2 3",
"output": "2"
},
{
"input": "71 85 88",
"output": "17"
},
{
"input": "30 38 99",
"output": "69"
},
{
"input": "23 82 95",
"output": "72"
},
{
"input": "22 41 47",
"output": "25"
},
{
"input": "9 94 77",
"output": "85"
},
{
"input": "1 53 51",
"output": "52"
},
{
"input": "25 97 93",
"output": "72"
},
{
"input": "42 53 51",
"output": "11"
},
{
"input": "81 96 94",
"output": "15"
},
{
"input": "21 5 93",
"output": "88"
},
{
"input": "50 13 75",
"output": "62"
},
{
"input": "41 28 98",
"output": "70"
},
{
"input": "69 46 82",
"output": "36"
},
{
"input": "87 28 89",
"output": "61"
},
{
"input": "44 45 40",
"output": "5"
},
{
"input": "86 97 68",
"output": "29"
},
{
"input": "43 92 30",
"output": "62"
},
{
"input": "16 70 1",
"output": "69"
},
{
"input": "40 46 19",
"output": "27"
},
{
"input": "71 38 56",
"output": "33"
},
{
"input": "82 21 80",
"output": "61"
},
{
"input": "75 8 35",
"output": "67"
},
{
"input": "75 24 28",
"output": "51"
},
{
"input": "78 23 56",
"output": "55"
},
{
"input": "85 31 10",
"output": "75"
},
{
"input": "76 50 9",
"output": "67"
},
{
"input": "95 37 34",
"output": "61"
},
{
"input": "84 61 35",
"output": "49"
},
{
"input": "87 85 37",
"output": "50"
},
{
"input": "1 3 2",
"output": "2"
},
{
"input": "4 2 6",
"output": "4"
},
{
"input": "6 9 3",
"output": "6"
},
{
"input": "12 4 8",
"output": "8"
},
{
"input": "15 10 5",
"output": "10"
},
{
"input": "1 50 17",
"output": "49"
},
{
"input": "10 5 15",
"output": "10"
},
{
"input": "8 1 9",
"output": "8"
},
{
"input": "3 5 4",
"output": "2"
},
{
"input": "2 1 3",
"output": "2"
},
{
"input": "1 8 2",
"output": "7"
},
{
"input": "1 100 2",
"output": "99"
},
{
"input": "1 4 6",
"output": "5"
}
] | 1,693,930,881
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 48
| 46
| 0
|
x,y,z=map(int,input().split(' '))
distance =[abs(x-y),abs(x-z),abs(z-y)]
min_distance=max(distance)
print(min_distance)
|
Title: The New Year: Meeting Friends
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
There are three friend living on the straight line *Ox* in Lineland. The first friend lives at the point *x*1, the second friend lives at the point *x*2, and the third friend lives at the point *x*3. They plan to celebrate the New Year together, so they need to meet at one point. What is the minimum total distance they have to travel in order to meet at some point and celebrate the New Year?
It's guaranteed that the optimal answer is always integer.
Input Specification:
The first line of the input contains three distinct integers *x*1, *x*2 and *x*3 (1<=≤<=*x*1,<=*x*2,<=*x*3<=≤<=100) — the coordinates of the houses of the first, the second and the third friends respectively.
Output Specification:
Print one integer — the minimum total distance the friends need to travel in order to meet together.
Demo Input:
['7 1 4\n', '30 20 10\n']
Demo Output:
['6\n', '20\n']
Note:
In the first sample, friends should meet at the point 4. Thus, the first friend has to travel the distance of 3 (from the point 7 to the point 4), the second friend also has to travel the distance of 3 (from the point 1 to the point 4), while the third friend should not go anywhere because he lives at the point 4.
|
```python
x,y,z=map(int,input().split(' '))
distance =[abs(x-y),abs(x-z),abs(z-y)]
min_distance=max(distance)
print(min_distance)
```
| 3
|
|
443
|
A
|
Anton and Letters
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
|
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
|
Print a single number — the number of distinct letters in Anton's set.
|
[
"{a, b, c}\n",
"{b, a, b, a}\n",
"{}\n"
] |
[
"3\n",
"2\n",
"0\n"
] |
none
| 500
|
[
{
"input": "{a, b, c}",
"output": "3"
},
{
"input": "{b, a, b, a}",
"output": "2"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{a, a, c, b, b, b, c, c, c, c}",
"output": "3"
},
{
"input": "{a, c, b, b}",
"output": "3"
},
{
"input": "{a, b}",
"output": "2"
},
{
"input": "{a}",
"output": "1"
},
{
"input": "{b, a, b, a, b, c, c, b, c, b}",
"output": "3"
},
{
"input": "{e, g, c, e}",
"output": "3"
},
{
"input": "{a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a, a}",
"output": "1"
},
{
"input": "{a, a, b}",
"output": "2"
},
{
"input": "{a, b, b, b, a, b, a, a, a, a, a, a, b, a, b, a, a, a, a, a, b, a, b, a}",
"output": "2"
},
{
"input": "{j, u, a, c, f, w, e, w, x, t, h, p, v, n, i, l, x, n, i, b, u, c, a, a}",
"output": "16"
},
{
"input": "{x, i, w, c, p, e, h, z, k, i}",
"output": "9"
},
{
"input": "{t, k, o, x, r, d, q, j, k, e, z, w, y, r, z, s, s, e, s, b, k, i}",
"output": "15"
},
{
"input": "{y}",
"output": "1"
},
{
"input": "{x}",
"output": "1"
},
{
"input": "{b, z, a, z}",
"output": "3"
},
{
"input": "{z}",
"output": "1"
},
{
"input": "{a, z}",
"output": "2"
},
{
"input": "{a, b, z}",
"output": "3"
},
{
"input": "{s, q, z, r, t, a, b, h, j, i, o, z, r, q}",
"output": "11"
}
] | 1,690,226,558
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 0
| 15
| 0
|
s=list(input())
count=0
for i in range (len(s)):
if s[i]!=',' and s[i]!=' ' and s[i]!='{' and s[i]!='}' and s[i]!=-1:
count+=1
for j in range (i+1,len(s)):
if s[i]==s[j]:
s[j]=-1
print(s)
print(count)
|
Title: Anton and Letters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Recently, Anton has found a set. The set consists of small English letters. Anton carefully wrote out all the letters from the set in one line, separated by a comma. He also added an opening curved bracket at the beginning of the line and a closing curved bracket at the end of the line.
Unfortunately, from time to time Anton would forget writing some letter and write it again. He asks you to count the total number of distinct letters in his set.
Input Specification:
The first and the single line contains the set of letters. The length of the line doesn't exceed 1000. It is guaranteed that the line starts from an opening curved bracket and ends with a closing curved bracket. Between them, small English letters are listed, separated by a comma. Each comma is followed by a space.
Output Specification:
Print a single number — the number of distinct letters in Anton's set.
Demo Input:
['{a, b, c}\n', '{b, a, b, a}\n', '{}\n']
Demo Output:
['3\n', '2\n', '0\n']
Note:
none
|
```python
s=list(input())
count=0
for i in range (len(s)):
if s[i]!=',' and s[i]!=' ' and s[i]!='{' and s[i]!='}' and s[i]!=-1:
count+=1
for j in range (i+1,len(s)):
if s[i]==s[j]:
s[j]=-1
print(s)
print(count)
```
| 0
|
|
421
|
A
|
Pasha and Hamsters
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"implementation"
] | null | null |
Pasha has two hamsters: Arthur and Alexander. Pasha put *n* apples in front of them. Pasha knows which apples Arthur likes. Similarly, Pasha knows which apples Alexander likes. Pasha doesn't want any conflict between the hamsters (as they may like the same apple), so he decided to distribute the apples between the hamsters on his own. He is going to give some apples to Arthur and some apples to Alexander. It doesn't matter how many apples each hamster gets but it is important that each hamster gets only the apples he likes. It is possible that somebody doesn't get any apples.
Help Pasha distribute all the apples between the hamsters. Note that Pasha wants to distribute all the apples, not just some of them.
|
The first line contains integers *n*, *a*, *b* (1<=≤<=*n*<=≤<=100; 1<=≤<=*a*,<=*b*<=≤<=*n*) — the number of apples Pasha has, the number of apples Arthur likes and the number of apples Alexander likes, correspondingly.
The next line contains *a* distinct integers — the numbers of the apples Arthur likes. The next line contains *b* distinct integers — the numbers of the apples Alexander likes.
Assume that the apples are numbered from 1 to *n*. The input is such that the answer exists.
|
Print *n* characters, each of them equals either 1 or 2. If the *i*-h character equals 1, then the *i*-th apple should be given to Arthur, otherwise it should be given to Alexander. If there are multiple correct answers, you are allowed to print any of them.
|
[
"4 2 3\n1 2\n2 3 4\n",
"5 5 2\n3 4 1 2 5\n2 3\n"
] |
[
"1 1 2 2\n",
"1 1 1 1 1\n"
] |
none
| 500
|
[
{
"input": "4 2 3\n1 2\n2 3 4",
"output": "1 1 2 2"
},
{
"input": "5 5 2\n3 4 1 2 5\n2 3",
"output": "1 1 1 1 1"
},
{
"input": "100 69 31\n1 3 4 5 6 7 8 9 10 11 12 14 15 16 17 18 19 20 21 24 26 27 29 31 37 38 39 40 44 46 48 49 50 51 53 55 56 57 58 59 60 61 63 64 65 66 67 68 69 70 71 72 74 76 77 78 79 80 81 82 83 89 92 94 95 97 98 99 100\n2 13 22 23 25 28 30 32 33 34 35 36 41 42 43 45 47 52 54 62 73 75 84 85 86 87 88 90 91 93 96",
"output": "1 2 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 2 2 1 2 1 1 2 1 2 1 2 2 2 2 2 1 1 1 1 2 2 2 1 2 1 2 1 1 1 1 2 1 2 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 2 1 2 1 1 1 1 1 1 1 1 2 2 2 2 2 1 2 2 1 2 1 1 2 1 1 1 1"
},
{
"input": "100 56 44\n1 2 5 8 14 15 17 18 20 21 23 24 25 27 30 33 34 35 36 38 41 42 44 45 46 47 48 49 50 53 56 58 59 60 62 63 64 65 68 69 71 75 76 80 81 84 87 88 90 91 92 94 95 96 98 100\n3 4 6 7 9 10 11 12 13 16 19 22 26 28 29 31 32 37 39 40 43 51 52 54 55 57 61 66 67 70 72 73 74 77 78 79 82 83 85 86 89 93 97 99",
"output": "1 1 2 2 1 2 2 1 2 2 2 2 2 1 1 2 1 1 2 1 1 2 1 1 1 2 1 2 2 1 2 2 1 1 1 1 2 1 2 2 1 1 2 1 1 1 1 1 1 1 2 2 1 2 2 1 2 1 1 1 2 1 1 1 1 2 2 1 1 2 1 2 2 2 1 1 2 2 2 1 1 2 2 1 2 2 1 1 2 1 1 1 2 1 1 1 2 1 2 1"
},
{
"input": "100 82 18\n1 2 3 4 5 6 7 8 9 10 11 13 14 15 16 17 18 19 20 22 23 25 27 29 30 31 32 33 34 35 36 37 38 42 43 44 45 46 47 48 49 50 51 53 54 55 57 58 59 60 61 62 63 64 65 66 67 68 69 71 72 73 74 75 77 78 79 80 82 83 86 88 90 91 92 93 94 96 97 98 99 100\n12 21 24 26 28 39 40 41 52 56 70 76 81 84 85 87 89 95",
"output": "1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 2 1 1 2 1 2 1 2 1 1 1 1 1 1 1 1 1 1 2 2 2 1 1 1 1 1 1 1 1 1 1 2 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 2 1 1 1 1 2 1 1 2 2 1 2 1 2 1 1 1 1 1 2 1 1 1 1 1"
},
{
"input": "99 72 27\n1 2 3 4 5 6 7 8 10 11 12 13 14 15 16 17 20 23 25 26 28 29 30 32 33 34 35 36 39 41 42 43 44 45 46 47 50 51 52 54 55 56 58 59 60 61 62 67 70 71 72 74 75 76 77 80 81 82 84 85 86 88 90 91 92 93 94 95 96 97 98 99\n9 18 19 21 22 24 27 31 37 38 40 48 49 53 57 63 64 65 66 68 69 73 78 79 83 87 89",
"output": "1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 2 2 1 2 2 1 2 1 1 2 1 1 1 2 1 1 1 1 1 2 2 1 2 1 1 1 1 1 1 1 2 2 1 1 1 2 1 1 1 2 1 1 1 1 1 2 2 2 2 1 2 2 1 1 1 2 1 1 1 1 2 2 1 1 1 2 1 1 1 2 1 2 1 1 1 1 1 1 1 1 1 1"
},
{
"input": "99 38 61\n1 3 10 15 16 22 23 28 31 34 35 36 37 38 39 43 44 49 50 53 56 60 63 68 69 70 72 74 75 77 80 81 83 85 96 97 98 99\n2 4 5 6 7 8 9 11 12 13 14 17 18 19 20 21 24 25 26 27 29 30 32 33 40 41 42 45 46 47 48 51 52 54 55 57 58 59 61 62 64 65 66 67 71 73 76 78 79 82 84 86 87 88 89 90 91 92 93 94 95",
"output": "1 2 1 2 2 2 2 2 2 1 2 2 2 2 1 1 2 2 2 2 2 1 1 2 2 2 2 1 2 2 1 2 2 1 1 1 1 1 1 2 2 2 1 1 2 2 2 2 1 1 2 2 1 2 2 1 2 2 2 1 2 2 1 2 2 2 2 1 1 1 2 1 2 1 1 2 1 2 2 1 1 2 1 2 1 2 2 2 2 2 2 2 2 2 2 1 1 1 1"
},
{
"input": "99 84 15\n1 2 3 5 6 7 8 9 10 11 12 13 14 15 16 17 19 20 21 22 23 24 25 26 27 28 29 30 31 32 34 35 36 37 38 39 40 41 42 43 44 47 48 50 51 52 53 55 56 58 59 60 61 62 63 64 65 68 69 70 71 72 73 74 75 77 79 80 81 82 83 84 85 86 87 89 90 91 92 93 94 97 98 99\n4 18 33 45 46 49 54 57 66 67 76 78 88 95 96",
"output": "1 1 1 2 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 2 1 1 1 1 1 1 1 1 1 1 1 2 2 1 1 2 1 1 1 1 2 1 1 2 1 1 1 1 1 1 1 1 2 2 1 1 1 1 1 1 1 1 2 1 2 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 2 2 1 1 1"
},
{
"input": "4 3 1\n1 3 4\n2",
"output": "1 2 1 1"
},
{
"input": "4 3 1\n1 2 4\n3",
"output": "1 1 2 1"
},
{
"input": "4 2 2\n2 3\n1 4",
"output": "2 1 1 2"
},
{
"input": "4 3 1\n2 3 4\n1",
"output": "2 1 1 1"
},
{
"input": "1 1 1\n1\n1",
"output": "1"
},
{
"input": "2 1 1\n2\n1",
"output": "2 1"
},
{
"input": "2 1 1\n1\n2",
"output": "1 2"
},
{
"input": "3 3 1\n1 2 3\n1",
"output": "1 1 1"
},
{
"input": "3 3 1\n1 2 3\n3",
"output": "1 1 1"
},
{
"input": "3 2 1\n1 3\n2",
"output": "1 2 1"
},
{
"input": "100 1 100\n84\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": "2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2"
},
{
"input": "100 100 1\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\n17",
"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"
},
{
"input": "98 51 47\n1 2 3 4 6 7 8 10 13 15 16 18 19 21 22 23 25 26 27 29 31 32 36 37 39 40 41 43 44 48 49 50 51 52 54 56 58 59 65 66 68 79 80 84 86 88 89 90 94 95 97\n5 9 11 12 14 17 20 24 28 30 33 34 35 38 42 45 46 47 53 55 57 60 61 62 63 64 67 69 70 71 72 73 74 75 76 77 78 81 82 83 85 87 91 92 93 96 98",
"output": "1 1 1 1 2 1 1 1 2 1 2 2 1 2 1 1 2 1 1 2 1 1 1 2 1 1 1 2 1 2 1 1 2 2 2 1 1 2 1 1 1 2 1 1 2 2 2 1 1 1 1 1 2 1 2 1 2 1 1 2 2 2 2 2 1 1 2 1 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2 1 2 1 2 1 1 1 2 2 2 1 1 2 1 2"
},
{
"input": "98 28 70\n1 13 15 16 19 27 28 40 42 43 46 53 54 57 61 63 67 68 69 71 75 76 78 80 88 93 97 98\n2 3 4 5 6 7 8 9 10 11 12 14 17 18 20 21 22 23 24 25 26 29 30 31 32 33 34 35 36 37 38 39 41 44 45 47 48 49 50 51 52 55 56 58 59 60 62 64 65 66 70 72 73 74 77 79 81 82 83 84 85 86 87 89 90 91 92 94 95 96",
"output": "1 2 2 2 2 2 2 2 2 2 2 2 1 2 1 1 2 2 1 2 2 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2 2 1 2 1 1 2 2 1 2 2 2 2 2 2 1 1 2 2 1 2 2 2 1 2 1 2 2 2 1 1 1 2 1 2 2 2 1 1 2 1 2 1 2 2 2 2 2 2 2 1 2 2 2 2 1 2 2 2 1 1"
},
{
"input": "97 21 76\n7 10 16 17 26 30 34 39 40 42 44 46 53 54 56 64 67 72 78 79 94\n1 2 3 4 5 6 8 9 11 12 13 14 15 18 19 20 21 22 23 24 25 27 28 29 31 32 33 35 36 37 38 41 43 45 47 48 49 50 51 52 55 57 58 59 60 61 62 63 65 66 68 69 70 71 73 74 75 76 77 80 81 82 83 84 85 86 87 88 89 90 91 92 93 95 96 97",
"output": "2 2 2 2 2 2 1 2 2 1 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 1 2 2 2 2 1 1 2 1 2 1 2 1 2 2 2 2 2 2 1 1 2 1 2 2 2 2 2 2 2 1 2 2 1 2 2 2 2 1 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2"
},
{
"input": "97 21 76\n1 10 12 13 17 18 22 25 31 48 50 54 61 64 67 74 78 81 86 88 94\n2 3 4 5 6 7 8 9 11 14 15 16 19 20 21 23 24 26 27 28 29 30 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 49 51 52 53 55 56 57 58 59 60 62 63 65 66 68 69 70 71 72 73 75 76 77 79 80 82 83 84 85 87 89 90 91 92 93 95 96 97",
"output": "1 2 2 2 2 2 2 2 2 1 2 1 1 2 2 2 1 1 2 2 2 1 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 1 2 2 2 2 2 2 1 2 2 1 2 2 1 2 2 2 2 2 2 1 2 2 2 1 2 2 1 2 2 2 2 1 2 1 2 2 2 2 2 1 2 2 2"
},
{
"input": "96 10 86\n2 5 31 37 68 69 80 82 90 91\n1 3 4 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 32 33 34 35 36 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 70 71 72 73 74 75 76 77 78 79 81 83 84 85 86 87 88 89 92 93 94 95 96",
"output": "2 1 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 2 2 2 2 2 2 2 2 2 1 2 1 2 2 2 2 2 2 2 1 1 2 2 2 2 2"
},
{
"input": "95 4 91\n58 65 70 93\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 59 60 61 62 63 64 66 67 68 69 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 94 95",
"output": "2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 1 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2"
},
{
"input": "98 88 10\n1 2 4 5 6 8 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 38 39 40 41 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 62 63 64 65 66 67 68 69 71 72 73 74 75 76 77 79 80 81 83 84 85 86 87 88 89 90 92 93 94 95 96 97 98\n3 7 32 37 42 61 70 78 82 91",
"output": "1 1 2 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 2 1 1 1 1 2 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 2 1 1 1 2 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1"
},
{
"input": "98 96 2\n1 2 3 4 5 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\n6 7",
"output": "1 1 1 1 1 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 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 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": "97 97 1\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\n94",
"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"
},
{
"input": "97 97 1\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\n20",
"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"
},
{
"input": "96 96 1\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\n48",
"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"
},
{
"input": "95 95 1\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\n55",
"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"
}
] | 1,650,167,496
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 15
| 0
|
n = int(input()[0])
a = set(map(int, input().split()))
b = set(map(int, input().split())) - a
for i in range(1, n+1):
print(1 if i in a else 2, end=' ')
|
Title: Pasha and Hamsters
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha has two hamsters: Arthur and Alexander. Pasha put *n* apples in front of them. Pasha knows which apples Arthur likes. Similarly, Pasha knows which apples Alexander likes. Pasha doesn't want any conflict between the hamsters (as they may like the same apple), so he decided to distribute the apples between the hamsters on his own. He is going to give some apples to Arthur and some apples to Alexander. It doesn't matter how many apples each hamster gets but it is important that each hamster gets only the apples he likes. It is possible that somebody doesn't get any apples.
Help Pasha distribute all the apples between the hamsters. Note that Pasha wants to distribute all the apples, not just some of them.
Input Specification:
The first line contains integers *n*, *a*, *b* (1<=≤<=*n*<=≤<=100; 1<=≤<=*a*,<=*b*<=≤<=*n*) — the number of apples Pasha has, the number of apples Arthur likes and the number of apples Alexander likes, correspondingly.
The next line contains *a* distinct integers — the numbers of the apples Arthur likes. The next line contains *b* distinct integers — the numbers of the apples Alexander likes.
Assume that the apples are numbered from 1 to *n*. The input is such that the answer exists.
Output Specification:
Print *n* characters, each of them equals either 1 or 2. If the *i*-h character equals 1, then the *i*-th apple should be given to Arthur, otherwise it should be given to Alexander. If there are multiple correct answers, you are allowed to print any of them.
Demo Input:
['4 2 3\n1 2\n2 3 4\n', '5 5 2\n3 4 1 2 5\n2 3\n']
Demo Output:
['1 1 2 2\n', '1 1 1 1 1\n']
Note:
none
|
```python
n = int(input()[0])
a = set(map(int, input().split()))
b = set(map(int, input().split())) - a
for i in range(1, n+1):
print(1 if i in a else 2, end=' ')
```
| 0
|
|
390
|
A
|
Inna and Alarm Clock
|
PROGRAMMING
| 0
|
[
"implementation"
] | null | null |
Inna loves sleeping very much, so she needs *n* alarm clocks in total to wake up. Let's suppose that Inna's room is a 100<=×<=100 square with the lower left corner at point (0,<=0) and with the upper right corner at point (100,<=100). Then the alarm clocks are points with integer coordinates in this square.
The morning has come. All *n* alarm clocks in Inna's room are ringing, so Inna wants to turn them off. For that Inna has come up with an amusing game:
- First Inna chooses a type of segments that she will use throughout the game. The segments can be either vertical or horizontal. - Then Inna makes multiple moves. In a single move, Inna can paint a segment of any length on the plane, she chooses its type at the beginning of the game (either vertical or horizontal), then all alarm clocks that are on this segment switch off. The game ends when all the alarm clocks are switched off.
Inna is very sleepy, so she wants to get through the alarm clocks as soon as possible. Help her, find the minimum number of moves in the game that she needs to turn off all the alarm clocks!
|
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of the alarm clocks. The next *n* lines describe the clocks: the *i*-th line contains two integers *x**i*, *y**i* — the coordinates of the *i*-th alarm clock (0<=≤<=*x**i*,<=*y**i*<=≤<=100).
Note that a single point in the room can contain any number of alarm clocks and the alarm clocks can lie on the sides of the square that represents the room.
|
In a single line print a single integer — the minimum number of segments Inna will have to draw if she acts optimally.
|
[
"4\n0 0\n0 1\n0 2\n1 0\n",
"4\n0 0\n0 1\n1 0\n1 1\n",
"4\n1 1\n1 2\n2 3\n3 3\n"
] |
[
"2\n",
"2\n",
"3\n"
] |
In the first sample, Inna first chooses type "vertical segments", and then she makes segments with ends at : (0, 0), (0, 2); and, for example, (1, 0), (1, 1). If she paints horizontal segments, she will need at least 3 segments.
In the third sample it is important to note that Inna doesn't have the right to change the type of the segments during the game. That's why she will need 3 horizontal or 3 vertical segments to end the game.
| 500
|
[
{
"input": "4\n0 0\n0 1\n0 2\n1 0",
"output": "2"
},
{
"input": "4\n0 0\n0 1\n1 0\n1 1",
"output": "2"
},
{
"input": "4\n1 1\n1 2\n2 3\n3 3",
"output": "3"
},
{
"input": "1\n0 0",
"output": "1"
},
{
"input": "42\n28 87\n26 16\n59 90\n47 61\n28 83\n36 30\n67 10\n6 95\n9 49\n86 94\n52 24\n74 9\n86 24\n28 51\n25 99\n40 98\n57 33\n18 96\n43 36\n3 79\n4 86\n38 61\n25 61\n6 100\n58 81\n28 19\n64 4\n3 40\n2 56\n41 49\n97 100\n86 34\n42 36\n44 40\n14 85\n21 60\n76 99\n64 47\n69 13\n49 37\n97 37\n3 70",
"output": "31"
},
{
"input": "21\n54 85\n69 37\n42 87\n53 18\n28 22\n13 3\n62 97\n38 91\n67 19\n100 79\n29 18\n48 40\n68 84\n44 20\n37 34\n73 53\n21 5\n20 73\n24 94\n23 52\n7 55",
"output": "20"
},
{
"input": "19\n1 1\n1 2\n1 3\n1 4\n1 5\n1 6\n1 7\n1 8\n1 9\n1 10\n1 11\n1 12\n1 13\n1 14\n1 15\n1 16\n1 17\n1 18\n1 19",
"output": "1"
},
{
"input": "12\n1 1\n1 3\n1 5\n2 1\n2 2\n2 4\n3 1\n3 3\n3 5\n4 1\n4 2\n4 3",
"output": "4"
}
] | 1,654,092,978
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 19
| 171
| 0
|
h=set()
v=set()
for i in range(int(input())):
k = input().split()
v.add(k[0])
h.add(k[1])
print(min(len(v),len(h)))
|
Title: Inna and Alarm Clock
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Inna loves sleeping very much, so she needs *n* alarm clocks in total to wake up. Let's suppose that Inna's room is a 100<=×<=100 square with the lower left corner at point (0,<=0) and with the upper right corner at point (100,<=100). Then the alarm clocks are points with integer coordinates in this square.
The morning has come. All *n* alarm clocks in Inna's room are ringing, so Inna wants to turn them off. For that Inna has come up with an amusing game:
- First Inna chooses a type of segments that she will use throughout the game. The segments can be either vertical or horizontal. - Then Inna makes multiple moves. In a single move, Inna can paint a segment of any length on the plane, she chooses its type at the beginning of the game (either vertical or horizontal), then all alarm clocks that are on this segment switch off. The game ends when all the alarm clocks are switched off.
Inna is very sleepy, so she wants to get through the alarm clocks as soon as possible. Help her, find the minimum number of moves in the game that she needs to turn off all the alarm clocks!
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=105) — the number of the alarm clocks. The next *n* lines describe the clocks: the *i*-th line contains two integers *x**i*, *y**i* — the coordinates of the *i*-th alarm clock (0<=≤<=*x**i*,<=*y**i*<=≤<=100).
Note that a single point in the room can contain any number of alarm clocks and the alarm clocks can lie on the sides of the square that represents the room.
Output Specification:
In a single line print a single integer — the minimum number of segments Inna will have to draw if she acts optimally.
Demo Input:
['4\n0 0\n0 1\n0 2\n1 0\n', '4\n0 0\n0 1\n1 0\n1 1\n', '4\n1 1\n1 2\n2 3\n3 3\n']
Demo Output:
['2\n', '2\n', '3\n']
Note:
In the first sample, Inna first chooses type "vertical segments", and then she makes segments with ends at : (0, 0), (0, 2); and, for example, (1, 0), (1, 1). If she paints horizontal segments, she will need at least 3 segments.
In the third sample it is important to note that Inna doesn't have the right to change the type of the segments during the game. That's why she will need 3 horizontal or 3 vertical segments to end the game.
|
```python
h=set()
v=set()
for i in range(int(input())):
k = input().split()
v.add(k[0])
h.add(k[1])
print(min(len(v),len(h)))
```
| 3
|
|
322
|
B
|
Ciel and Flowers
|
PROGRAMMING
| 1,600
|
[
"combinatorics",
"math"
] | null | null |
Fox Ciel has some flowers: *r* red flowers, *g* green flowers and *b* blue flowers. She wants to use these flowers to make several bouquets. There are 4 types of bouquets:
- To make a "red bouquet", it needs 3 red flowers. - To make a "green bouquet", it needs 3 green flowers. - To make a "blue bouquet", it needs 3 blue flowers. - To make a "mixing bouquet", it needs 1 red, 1 green and 1 blue flower.
Help Fox Ciel to find the maximal number of bouquets she can make.
|
The first line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=109) — the number of red, green and blue flowers.
|
Print the maximal number of bouquets Fox Ciel can make.
|
[
"3 6 9\n",
"4 4 4\n",
"0 0 0\n"
] |
[
"6\n",
"4\n",
"0\n"
] |
In test case 1, we can make 1 red bouquet, 2 green bouquets and 3 blue bouquets.
In test case 2, we can make 1 red, 1 green, 1 blue and 1 mixing bouquet.
| 1,000
|
[
{
"input": "3 6 9",
"output": "6"
},
{
"input": "4 4 4",
"output": "4"
},
{
"input": "0 0 0",
"output": "0"
},
{
"input": "0 3 6",
"output": "3"
},
{
"input": "7 8 9",
"output": "7"
},
{
"input": "8 8 9",
"output": "8"
},
{
"input": "15 3 999",
"output": "339"
},
{
"input": "32 62 92",
"output": "62"
},
{
"input": "123456789 123456789 123456789",
"output": "123456789"
},
{
"input": "3 5 5",
"output": "4"
},
{
"input": "666806767 385540591 357848286",
"output": "470065214"
},
{
"input": "80010646 727118126 817880463",
"output": "541669744"
},
{
"input": "829651016 732259171 572879931",
"output": "711596705"
},
{
"input": "242854896 442432924 180395753",
"output": "288561190"
},
{
"input": "139978911 5123031 935395222",
"output": "360165721"
},
{
"input": "553182792 10264076 395427398",
"output": "319624755"
},
{
"input": "597790453 720437830 855459575",
"output": "724562619"
},
{
"input": "494914467 356982656 757942689",
"output": "536613270"
},
{
"input": "908118348 67156409 217974865",
"output": "397749873"
},
{
"input": "952726009 629846517 972974334",
"output": "851848953"
},
{
"input": "775140200 616574841 630329230",
"output": "674014756"
},
{
"input": "524780569 326748594 90361407",
"output": "313963523"
},
{
"input": "937984449 184405994 992844522",
"output": "705078321"
},
{
"input": "835108464 525983528 452876698",
"output": "604656229"
},
{
"input": "879716125 531124573 207876166",
"output": "539572288"
},
{
"input": "292920005 241298326 667908343",
"output": "400708891"
},
{
"input": "1000000000 1000000000 1000000000",
"output": "1000000000"
},
{
"input": "1000000000 999999999 999999998",
"output": "999999998"
},
{
"input": "999999998 999999998 999999999",
"output": "999999998"
},
{
"input": "0 1 1",
"output": "0"
},
{
"input": "0 1000000000 0",
"output": "333333333"
},
{
"input": "0 1 0",
"output": "0"
},
{
"input": "1 0 0",
"output": "0"
},
{
"input": "0 2 2",
"output": "0"
},
{
"input": "3 3 5",
"output": "3"
},
{
"input": "2 2 0",
"output": "0"
},
{
"input": "0 5 5",
"output": "2"
},
{
"input": "2 0 11",
"output": "3"
},
{
"input": "9 9 7",
"output": "8"
},
{
"input": "65 30 74",
"output": "56"
}
] | 1,688,069,933
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 5
| 124
| 0
|
import sys
input = lambda: sys.stdin.readline().rstrip()
from collections import deque,defaultdict,Counter
from itertools import permutations,combinations
from bisect import *
from heapq import *
from math import ceil,gcd,lcm,floor,comb
A,B,C = map(int,input().split())
ans = A//3+B//3+C//3
A%=3
B%=3
C%=3
print(ans+min(A,B,C))
|
Title: Ciel and Flowers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Fox Ciel has some flowers: *r* red flowers, *g* green flowers and *b* blue flowers. She wants to use these flowers to make several bouquets. There are 4 types of bouquets:
- To make a "red bouquet", it needs 3 red flowers. - To make a "green bouquet", it needs 3 green flowers. - To make a "blue bouquet", it needs 3 blue flowers. - To make a "mixing bouquet", it needs 1 red, 1 green and 1 blue flower.
Help Fox Ciel to find the maximal number of bouquets she can make.
Input Specification:
The first line contains three integers *r*, *g* and *b* (0<=≤<=*r*,<=*g*,<=*b*<=≤<=109) — the number of red, green and blue flowers.
Output Specification:
Print the maximal number of bouquets Fox Ciel can make.
Demo Input:
['3 6 9\n', '4 4 4\n', '0 0 0\n']
Demo Output:
['6\n', '4\n', '0\n']
Note:
In test case 1, we can make 1 red bouquet, 2 green bouquets and 3 blue bouquets.
In test case 2, we can make 1 red, 1 green, 1 blue and 1 mixing bouquet.
|
```python
import sys
input = lambda: sys.stdin.readline().rstrip()
from collections import deque,defaultdict,Counter
from itertools import permutations,combinations
from bisect import *
from heapq import *
from math import ceil,gcd,lcm,floor,comb
A,B,C = map(int,input().split())
ans = A//3+B//3+C//3
A%=3
B%=3
C%=3
print(ans+min(A,B,C))
```
| 0
|
|
501
|
A
|
Contest
|
PROGRAMMING
| 900
|
[
"implementation"
] | null | null |
Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points.
Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth.
|
The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180).
It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round).
|
Output on a single line:
"Misha" (without the quotes), if Misha got more points than Vasya.
"Vasya" (without the quotes), if Vasya got more points than Misha.
"Tie" (without the quotes), if both of them got the same number of points.
|
[
"500 1000 20 30\n",
"1000 1000 1 1\n",
"1500 1000 176 177\n"
] |
[
"Vasya\n",
"Tie\n",
"Misha\n"
] |
none
| 500
|
[
{
"input": "500 1000 20 30",
"output": "Vasya"
},
{
"input": "1000 1000 1 1",
"output": "Tie"
},
{
"input": "1500 1000 176 177",
"output": "Misha"
},
{
"input": "1500 1000 74 177",
"output": "Misha"
},
{
"input": "750 2500 175 178",
"output": "Vasya"
},
{
"input": "750 1000 54 103",
"output": "Tie"
},
{
"input": "2000 1250 176 130",
"output": "Tie"
},
{
"input": "1250 1750 145 179",
"output": "Tie"
},
{
"input": "2000 2000 176 179",
"output": "Tie"
},
{
"input": "1500 1500 148 148",
"output": "Tie"
},
{
"input": "2750 1750 134 147",
"output": "Misha"
},
{
"input": "3250 250 175 173",
"output": "Misha"
},
{
"input": "500 500 170 176",
"output": "Misha"
},
{
"input": "250 1000 179 178",
"output": "Vasya"
},
{
"input": "3250 1000 160 138",
"output": "Misha"
},
{
"input": "3000 2000 162 118",
"output": "Tie"
},
{
"input": "1500 1250 180 160",
"output": "Tie"
},
{
"input": "1250 2500 100 176",
"output": "Tie"
},
{
"input": "3500 3500 177 178",
"output": "Tie"
},
{
"input": "3000 3250 16 34",
"output": "Tie"
},
{
"input": "1750 3000 137 49",
"output": "Vasya"
},
{
"input": "500 1500 179 71",
"output": "Vasya"
},
{
"input": "1250 2000 101 180",
"output": "Misha"
},
{
"input": "250 750 180 176",
"output": "Vasya"
},
{
"input": "2250 2250 163 145",
"output": "Vasya"
},
{
"input": "3000 3000 176 78",
"output": "Vasya"
},
{
"input": "250 3500 8 178",
"output": "Vasya"
},
{
"input": "1750 1250 179 180",
"output": "Misha"
},
{
"input": "2750 1750 13 164",
"output": "Misha"
},
{
"input": "1750 2250 178 53",
"output": "Vasya"
},
{
"input": "2500 2750 73 179",
"output": "Misha"
},
{
"input": "1000 3500 178 175",
"output": "Vasya"
},
{
"input": "1000 500 7 162",
"output": "Misha"
},
{
"input": "1000 250 175 48",
"output": "Misha"
},
{
"input": "1750 500 166 177",
"output": "Misha"
},
{
"input": "250 250 0 0",
"output": "Tie"
},
{
"input": "250 3500 0 0",
"output": "Vasya"
},
{
"input": "250 3500 0 180",
"output": "Vasya"
},
{
"input": "3500 3500 180 180",
"output": "Tie"
},
{
"input": "3500 250 0 180",
"output": "Misha"
}
] | 1,633,290,004
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 40
| 77
| 6,758,400
|
a, b, c, d = map(int, input().split())
e = max(3*a/10, a-a*c/250)
f = max(3*b/10, b-b*d/250)
if e>f:
print("Misha")
elif e<f :
print("Vasya")
else:
print("Tie")
|
Title: Contest
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Misha and Vasya participated in a Codeforces contest. Unfortunately, each of them solved only one problem, though successfully submitted it at the first attempt. Misha solved the problem that costs *a* points and Vasya solved the problem that costs *b* points. Besides, Misha submitted the problem *c* minutes after the contest started and Vasya submitted the problem *d* minutes after the contest started. As you know, on Codeforces the cost of a problem reduces as a round continues. That is, if you submit a problem that costs *p* points *t* minutes after the contest started, you get points.
Misha and Vasya are having an argument trying to find out who got more points. Help them to find out the truth.
Input Specification:
The first line contains four integers *a*, *b*, *c*, *d* (250<=≤<=*a*,<=*b*<=≤<=3500, 0<=≤<=*c*,<=*d*<=≤<=180).
It is guaranteed that numbers *a* and *b* are divisible by 250 (just like on any real Codeforces round).
Output Specification:
Output on a single line:
"Misha" (without the quotes), if Misha got more points than Vasya.
"Vasya" (without the quotes), if Vasya got more points than Misha.
"Tie" (without the quotes), if both of them got the same number of points.
Demo Input:
['500 1000 20 30\n', '1000 1000 1 1\n', '1500 1000 176 177\n']
Demo Output:
['Vasya\n', 'Tie\n', 'Misha\n']
Note:
none
|
```python
a, b, c, d = map(int, input().split())
e = max(3*a/10, a-a*c/250)
f = max(3*b/10, b-b*d/250)
if e>f:
print("Misha")
elif e<f :
print("Vasya")
else:
print("Tie")
```
| 3
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,635,664,399
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 92
| 4,300,800
|
a = map(int, input().split())
data = list(a)
print (int(data[0] * data[1] / 2)) if (data[0] % 2 == 0 or data[1] % 2 == 0) \
else print (int((data[0] * data[1] - 1) / 2))
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none
|
```python
a = map(int, input().split())
data = list(a)
print (int(data[0] * data[1] / 2)) if (data[0] % 2 == 0 or data[1] % 2 == 0) \
else print (int((data[0] * data[1] - 1) / 2))
```
| 3.968989
|
456
|
A
|
Laptops
|
PROGRAMMING
| 1,100
|
[
"sortings"
] | null | null |
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist.
|
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops.
Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality).
All *a**i* are distinct. All *b**i* are distinct.
|
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
|
[
"2\n1 2\n2 1\n"
] |
[
"Happy Alex\n"
] |
none
| 500
|
[
{
"input": "2\n1 2\n2 1",
"output": "Happy Alex"
},
{
"input": "2\n1 1\n2 2",
"output": "Poor Alex"
},
{
"input": "3\n2 2\n3 3\n1 1",
"output": "Poor Alex"
},
{
"input": "3\n3 3\n1 2\n2 1",
"output": "Happy Alex"
},
{
"input": "1\n1 1",
"output": "Poor Alex"
},
{
"input": "3\n2 3\n1 1\n3 2",
"output": "Happy Alex"
},
{
"input": "4\n4 1\n3 2\n2 3\n1 4",
"output": "Happy Alex"
},
{
"input": "2\n2 1\n1 2",
"output": "Happy Alex"
},
{
"input": "3\n3 2\n1 1\n2 3",
"output": "Happy Alex"
},
{
"input": "3\n1 2\n2 3\n3 1",
"output": "Happy Alex"
},
{
"input": "3\n1 2\n2 1\n3 3",
"output": "Happy Alex"
},
{
"input": "3\n1 2\n2 3\n3 1",
"output": "Happy Alex"
}
] | 1,698,343,429
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 1
| 61
| 0
|
t = int(input())
arr = []
for i in range(t):
p,q = map(int,input().split())
arr.append([p,q])
arr.sort()
arr = [[1, 2], [2, 1]]
found = False
for i in range(len(arr)):
for j in range(len(arr)):
if arr[i][0] < arr[j][0] and arr[i][1] > arr[j][1]:
print("Happy Alex")
found = True
break;
if not found:
print("Poor Alex")
|
Title: Laptops
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of *n* laptops. Determine whether two described above laptops exist.
Input Specification:
The first line contains an integer *n* (1<=≤<=*n*<=≤<=105) — the number of laptops.
Next *n* lines contain two integers each, *a**i* and *b**i* (1<=≤<=*a**i*,<=*b**i*<=≤<=*n*), where *a**i* is the price of the *i*-th laptop, and *b**i* is the number that represents the quality of the *i*-th laptop (the larger the number is, the higher is the quality).
All *a**i* are distinct. All *b**i* are distinct.
Output Specification:
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
Demo Input:
['2\n1 2\n2 1\n']
Demo Output:
['Happy Alex\n']
Note:
none
|
```python
t = int(input())
arr = []
for i in range(t):
p,q = map(int,input().split())
arr.append([p,q])
arr.sort()
arr = [[1, 2], [2, 1]]
found = False
for i in range(len(arr)):
for j in range(len(arr)):
if arr[i][0] < arr[j][0] and arr[i][1] > arr[j][1]:
print("Happy Alex")
found = True
break;
if not found:
print("Poor Alex")
```
| 0
|
|
25
|
A
|
IQ test
|
PROGRAMMING
| 1,300
|
[
"brute force"
] |
A. IQ test
|
2
|
256
|
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
|
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
|
[
"5\n2 4 7 8 10\n",
"4\n1 2 1 1\n"
] |
[
"3\n",
"2\n"
] |
none
| 0
|
[
{
"input": "5\n2 4 7 8 10",
"output": "3"
},
{
"input": "4\n1 2 1 1",
"output": "2"
},
{
"input": "3\n1 2 2",
"output": "1"
},
{
"input": "3\n100 99 100",
"output": "2"
},
{
"input": "3\n5 3 2",
"output": "3"
},
{
"input": "4\n43 28 1 91",
"output": "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"
}
] | 1,610,904,008
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 32
| 154
| 0
|
n = int(input())
l = list(map(int,input().split()))
e1 = []
o1 = []
for i in range(n) :
if l[i]%2==0:
e1.append(i)
else:
o1.append(i)
if len(e1)==1:
print(e1[0]+1)
else:
print(o1[0]+1)
|
Title: IQ test
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Bob is preparing to pass IQ test. The most frequent task in this test is to find out which one of the given *n* numbers differs from the others. Bob observed that one number usually differs from the others in evenness. Help Bob — to check his answers, he needs a program that among the given *n* numbers finds one that is different in evenness.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — amount of numbers in the task. The second line contains *n* space-separated natural numbers, not exceeding 100. It is guaranteed, that exactly one of these numbers differs from the others in evenness.
Output Specification:
Output index of number that differs from the others in evenness. Numbers are numbered from 1 in the input order.
Demo Input:
['5\n2 4 7 8 10\n', '4\n1 2 1 1\n']
Demo Output:
['3\n', '2\n']
Note:
none
|
```python
n = int(input())
l = list(map(int,input().split()))
e1 = []
o1 = []
for i in range(n) :
if l[i]%2==0:
e1.append(i)
else:
o1.append(i)
if len(e1)==1:
print(e1[0]+1)
else:
print(o1[0]+1)
```
| 3.9615
|
263
|
A
|
Beautiful Matrix
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
|
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
|
Print a single integer — the minimum number of moves needed to make the matrix beautiful.
|
[
"0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n",
"0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n"
] |
[
"3\n",
"1\n"
] |
none
| 500
|
[
{
"input": "0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "0"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1",
"output": "4"
},
{
"input": "0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0\n0 0 0 0 0",
"output": "1"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0\n0 0 0 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n1 0 0 0 0",
"output": "4"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0",
"output": "3"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 1 0 0",
"output": "2"
},
{
"input": "0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 1 0",
"output": "3"
}
] | 1,696,762,359
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 25
| 92
| 0
|
i = 1
for _ in range(5):
r = input().split()
if "1" in r:
j = r.index("1")+1
print((abs(3-j)+abs(3-i)))
break
i += 1
|
Title: Beautiful Matrix
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You've got a 5<=×<=5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
1. Swap two neighboring matrix rows, that is, rows with indexes *i* and *i*<=+<=1 for some integer *i* (1<=≤<=*i*<=<<=5). 1. Swap two neighboring matrix columns, that is, columns with indexes *j* and *j*<=+<=1 for some integer *j* (1<=≤<=*j*<=<<=5).
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
Input Specification:
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
Output Specification:
Print a single integer — the minimum number of moves needed to make the matrix beautiful.
Demo Input:
['0 0 0 0 0\n0 0 0 0 1\n0 0 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n', '0 0 0 0 0\n0 0 0 0 0\n0 1 0 0 0\n0 0 0 0 0\n0 0 0 0 0\n']
Demo Output:
['3\n', '1\n']
Note:
none
|
```python
i = 1
for _ in range(5):
r = input().split()
if "1" in r:
j = r.index("1")+1
print((abs(3-j)+abs(3-i)))
break
i += 1
```
| 3
|
|
830
|
B
|
Cards Sorting
|
PROGRAMMING
| 1,600
|
[
"data structures",
"implementation",
"sortings"
] | null | null |
Vasily has a deck of cards consisting of *n* cards. There is an integer on each of the cards, this integer is between 1 and 100<=000, inclusive. It is possible that some cards have the same integers on them.
Vasily decided to sort the cards. To do this, he repeatedly takes the top card from the deck, and if the number on it equals the minimum number written on the cards in the deck, then he places the card away. Otherwise, he puts it under the deck and takes the next card from the top, and so on. The process ends as soon as there are no cards in the deck. You can assume that Vasily always knows the minimum number written on some card in the remaining deck, but doesn't know where this card (or these cards) is.
You are to determine the total number of times Vasily takes the top card from the deck.
|
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of cards in the deck.
The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000), where *a**i* is the number written on the *i*-th from top card in the deck.
|
Print the total number of times Vasily takes the top card from the deck.
|
[
"4\n6 3 1 2\n",
"1\n1000\n",
"7\n3 3 3 3 3 3 3\n"
] |
[
"7\n",
"1\n",
"7\n"
] |
In the first example Vasily at first looks at the card with number 6 on it, puts it under the deck, then on the card with number 3, puts it under the deck, and then on the card with number 1. He places away the card with 1, because the number written on it is the minimum among the remaining cards. After that the cards from top to bottom are [2, 6, 3]. Then Vasily looks at the top card with number 2 and puts it away. After that the cards from top to bottom are [6, 3]. Then Vasily looks at card 6, puts it under the deck, then at card 3 and puts it away. Then there is only one card with number 6 on it, and Vasily looks at it and puts it away. Thus, in total Vasily looks at 7 cards.
| 1,000
|
[
{
"input": "4\n6 3 1 2",
"output": "7"
},
{
"input": "1\n1000",
"output": "1"
},
{
"input": "7\n3 3 3 3 3 3 3",
"output": "7"
},
{
"input": "64\n826 142 89 337 897 891 1004 704 281 644 910 852 147 193 289 384 625 695 416 944 162 939 164 1047 359 114 499 99 713 300 268 316 256 404 852 496 373 322 716 202 689 857 936 806 556 153 137 863 1047 678 564 474 282 135 610 176 855 360 814 144 77 112 354 154",
"output": "1042"
},
{
"input": "87\n12 2 2 10 12 1 5 9 15 2 4 7 7 14 8 10 1 6 7 6 13 15 10 6 2 11 13 1 15 14 8 8 4 7 11 12 3 15 9 2 13 1 7 11 2 1 13 11 8 14 2 2 12 7 13 4 13 3 13 3 11 1 7 13 15 8 12 4 12 4 1 4 9 3 13 12 10 15 14 10 7 7 7 2 7 6 10",
"output": "580"
},
{
"input": "10\n4 3 4 3 3 3 4 4 4 3",
"output": "15"
},
{
"input": "20\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "20"
},
{
"input": "30\n6283 14661 69188 39640 41261 48019 86266 70517 4592 69008 20602 33339 29980 96844 76008 96294 27120 22671 5243 742 33692 18068 29056 48033 1223 82728 99765 38350 36425 10671",
"output": "235"
},
{
"input": "100\n9 9 72 55 14 8 55 58 35 67 3 18 73 92 41 49 15 60 18 66 9 26 97 47 43 88 71 97 19 34 48 96 79 53 8 24 69 49 12 23 77 12 21 88 66 9 29 13 61 69 54 77 41 13 4 68 37 74 7 6 29 76 55 72 89 4 78 27 29 82 18 83 12 4 32 69 89 85 66 13 92 54 38 5 26 56 17 55 29 4 17 39 29 94 3 67 85 98 21 14",
"output": "1805"
}
] | 1,669,363,437
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 4
| 61
| 0
|
import sys
input = sys.stdin.readline
n = int(input())
w = list(map(int, input().split()))
s = sorted(enumerate(w), key=lambda x:x[1])
c = 0
q = n
a, b = 0, 0
for i in range(n):
if s[i][0] >= a:
b += 1
else:
c += q
q -= b
b = 1
a = s[i][0]
c += q
print(c)
|
Title: Cards Sorting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily has a deck of cards consisting of *n* cards. There is an integer on each of the cards, this integer is between 1 and 100<=000, inclusive. It is possible that some cards have the same integers on them.
Vasily decided to sort the cards. To do this, he repeatedly takes the top card from the deck, and if the number on it equals the minimum number written on the cards in the deck, then he places the card away. Otherwise, he puts it under the deck and takes the next card from the top, and so on. The process ends as soon as there are no cards in the deck. You can assume that Vasily always knows the minimum number written on some card in the remaining deck, but doesn't know where this card (or these cards) is.
You are to determine the total number of times Vasily takes the top card from the deck.
Input Specification:
The first line contains single integer *n* (1<=≤<=*n*<=≤<=100<=000) — the number of cards in the deck.
The second line contains a sequence of *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100<=000), where *a**i* is the number written on the *i*-th from top card in the deck.
Output Specification:
Print the total number of times Vasily takes the top card from the deck.
Demo Input:
['4\n6 3 1 2\n', '1\n1000\n', '7\n3 3 3 3 3 3 3\n']
Demo Output:
['7\n', '1\n', '7\n']
Note:
In the first example Vasily at first looks at the card with number 6 on it, puts it under the deck, then on the card with number 3, puts it under the deck, and then on the card with number 1. He places away the card with 1, because the number written on it is the minimum among the remaining cards. After that the cards from top to bottom are [2, 6, 3]. Then Vasily looks at the top card with number 2 and puts it away. After that the cards from top to bottom are [6, 3]. Then Vasily looks at card 6, puts it under the deck, then at card 3 and puts it away. Then there is only one card with number 6 on it, and Vasily looks at it and puts it away. Thus, in total Vasily looks at 7 cards.
|
```python
import sys
input = sys.stdin.readline
n = int(input())
w = list(map(int, input().split()))
s = sorted(enumerate(w), key=lambda x:x[1])
c = 0
q = n
a, b = 0, 0
for i in range(n):
if s[i][0] >= a:
b += 1
else:
c += q
q -= b
b = 1
a = s[i][0]
c += q
print(c)
```
| 0
|
|
278
|
A
|
Circle Line
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
The circle line of the Berland subway has *n* stations. We know the distances between all pairs of neighboring stations:
- *d*1 is the distance between the 1-st and the 2-nd station;- *d*2 is the distance between the 2-nd and the 3-rd station;...- *d**n*<=-<=1 is the distance between the *n*<=-<=1-th and the *n*-th station;- *d**n* is the distance between the *n*-th and the 1-st station.
The trains go along the circle line in both directions. Find the shortest distance between stations with numbers *s* and *t*.
|
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — the number of stations on the circle line. The second line contains *n* integers *d*1,<=*d*2,<=...,<=*d**n* (1<=≤<=*d**i*<=≤<=100) — the distances between pairs of neighboring stations. The third line contains two integers *s* and *t* (1<=≤<=*s*,<=*t*<=≤<=*n*) — the numbers of stations, between which you need to find the shortest distance. These numbers can be the same.
The numbers in the lines are separated by single spaces.
|
Print a single number — the length of the shortest path between stations number *s* and *t*.
|
[
"4\n2 3 4 9\n1 3\n",
"4\n5 8 2 100\n4 1\n",
"3\n1 1 1\n3 1\n",
"3\n31 41 59\n1 1\n"
] |
[
"5\n",
"15\n",
"1\n",
"0\n"
] |
In the first sample the length of path 1 → 2 → 3 equals 5, the length of path 1 → 4 → 3 equals 13.
In the second sample the length of path 4 → 1 is 100, the length of path 4 → 3 → 2 → 1 is 15.
In the third sample the length of path 3 → 1 is 1, the length of path 3 → 2 → 1 is 2.
In the fourth sample the numbers of stations are the same, so the shortest distance equals 0.
| 500
|
[
{
"input": "4\n2 3 4 9\n1 3",
"output": "5"
},
{
"input": "4\n5 8 2 100\n4 1",
"output": "15"
},
{
"input": "3\n1 1 1\n3 1",
"output": "1"
},
{
"input": "3\n31 41 59\n1 1",
"output": "0"
},
{
"input": "5\n16 13 10 30 15\n4 2",
"output": "23"
},
{
"input": "6\n89 82 87 32 67 33\n4 4",
"output": "0"
},
{
"input": "7\n2 3 17 10 2 2 2\n4 2",
"output": "18"
},
{
"input": "3\n4 37 33\n3 3",
"output": "0"
},
{
"input": "8\n87 40 96 7 86 86 72 97\n6 8",
"output": "158"
},
{
"input": "10\n91 94 75 99 100 91 79 86 79 92\n2 8",
"output": "348"
},
{
"input": "19\n1 1 1 1 2 1 1 1 1 1 2 1 3 2 2 1 1 1 2\n7 7",
"output": "0"
},
{
"input": "34\n96 65 24 99 74 76 97 93 99 69 94 82 92 91 98 83 95 97 96 81 90 95 86 87 43 78 88 86 82 62 76 99 83 96\n21 16",
"output": "452"
},
{
"input": "50\n75 98 65 75 99 89 84 65 9 53 62 61 61 53 80 7 6 47 86 1 89 27 67 1 31 39 53 92 19 20 76 41 60 15 29 94 76 82 87 89 93 38 42 6 87 36 100 97 93 71\n2 6",
"output": "337"
},
{
"input": "99\n1 15 72 78 23 22 26 98 7 2 75 58 100 98 45 79 92 69 79 72 33 88 62 9 15 87 17 73 68 54 34 89 51 91 28 44 20 11 74 7 85 61 30 46 95 72 36 18 48 22 42 46 29 46 86 53 96 55 98 34 60 37 75 54 1 81 20 68 84 19 18 18 75 84 86 57 73 34 23 43 81 87 47 96 57 41 69 1 52 44 54 7 85 35 5 1 19 26 7\n4 64",
"output": "1740"
},
{
"input": "100\n33 63 21 27 49 82 86 93 43 55 4 72 89 85 5 34 80 7 23 13 21 49 22 73 89 65 81 25 6 92 82 66 58 88 48 96 1 1 16 48 67 96 84 63 87 76 20 100 36 4 31 41 35 62 55 76 74 70 68 41 4 16 39 81 2 41 34 73 66 57 41 89 78 93 68 96 87 47 92 60 40 58 81 12 19 74 56 83 56 61 83 97 26 92 62 52 39 57 89 95\n71 5",
"output": "2127"
},
{
"input": "100\n95 98 99 81 98 96 100 92 96 90 99 91 98 98 91 78 97 100 96 98 87 93 96 99 91 92 96 92 90 97 85 83 99 95 66 91 87 89 100 95 100 88 99 84 96 79 99 100 94 100 99 99 92 89 99 91 100 94 98 97 91 92 90 87 84 99 97 98 93 100 90 85 75 95 86 71 98 93 91 87 92 95 98 94 95 94 100 98 96 100 97 96 95 95 86 86 94 97 98 96\n67 57",
"output": "932"
},
{
"input": "100\n100 100 100 100 100 100 100 100 100 100 97 100 100 100 100 100 99 100 100 99 99 100 99 100 100 100 100 100 100 100 100 100 97 99 98 98 100 98 98 100 99 100 100 100 100 99 100 98 100 99 98 99 98 98 100 100 100 100 100 100 100 100 100 100 99 100 100 100 100 100 98 100 99 99 100 96 100 96 100 99 100 100 99 100 99 100 100 100 99 100 100 100 100 98 98 97 100 100 99 98\n16 6",
"output": "997"
},
{
"input": "100\n3 6 23 4 23 1 2 14 2 3 3 9 17 8 10 5 1 14 8 5 7 4 13 8 5 6 24 3 12 3 4 9 2 8 2 1 2 1 3 2 1 6 14 23 8 6 3 5 7 8 18 9 2 5 22 6 13 16 2 4 31 20 4 3 3 6 6 1 1 18 5 11 1 14 4 16 6 37 11 1 8 3 7 11 21 14 3 3 12 2 5 1 9 16 3 1 3 4 4 2\n98 24",
"output": "195"
},
{
"input": "100\n1 1 3 1 1 2 1 2 1 1 2 2 2 1 1 1 1 1 1 3 1 1 1 3 1 3 3 1 1 2 1 1 1 1 1 2 1 1 1 4 1 1 3 3 2 1 1 1 1 1 2 2 1 3 1 1 1 2 4 1 1 2 5 2 1 1 2 1 1 1 2 3 1 1 1 1 2 1 1 1 1 2 1 1 1 1 1 2 2 3 1 7 3 1 3 1 2 1 2 1\n49 10",
"output": "60"
},
{
"input": "100\n75 62 31 96 62 76 93 96 72 67 88 35 67 34 60 56 95 86 82 48 64 61 74 100 56 98 76 98 78 55 53 10 12 78 58 45 86 90 93 77 69 73 88 66 92 88 33 50 95 69 89 12 93 57 93 89 59 53 71 86 15 13 61 93 24 100 58 76 46 95 76 82 50 20 79 38 5 72 99 81 55 90 90 65 85 44 63 39 6 34 98 72 88 30 59 73 84 61 25 67\n86 25",
"output": "2523"
},
{
"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\n1 51",
"output": "5000"
},
{
"input": "4\n1 1 1 1\n2 4",
"output": "2"
},
{
"input": "4\n1 1 2 1\n2 4",
"output": "2"
}
] | 1,616,292,313
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 23
| 124
| 307,200
|
def swap(num1,num2):
arr = [num2,num1]
return arr
num = int(input())
distance = list(map(int, input() .split()))
cities = list(map(int, input() .split()))
answer = 0
sum1 = 0
sum2 = 0
if cities[0] == cities[1]:
answer = 0
if (cities[0] > cities[1]):
cities = swap(cities[0], cities[1])
i = cities[0] - 1
while(i < cities[1] - 1):
sum1 += distance[i]
i += 1
i = cities[1] - 1
while(i < num):
sum2 += distance[i]
i += 1
i = 0
while(i < cities[0] - 1):
sum2 += distance[i]
i += 1
if(sum1 < sum2):
print(sum1)
else:
print(sum2)
|
Title: Circle Line
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The circle line of the Berland subway has *n* stations. We know the distances between all pairs of neighboring stations:
- *d*1 is the distance between the 1-st and the 2-nd station;- *d*2 is the distance between the 2-nd and the 3-rd station;...- *d**n*<=-<=1 is the distance between the *n*<=-<=1-th and the *n*-th station;- *d**n* is the distance between the *n*-th and the 1-st station.
The trains go along the circle line in both directions. Find the shortest distance between stations with numbers *s* and *t*.
Input Specification:
The first line contains integer *n* (3<=≤<=*n*<=≤<=100) — the number of stations on the circle line. The second line contains *n* integers *d*1,<=*d*2,<=...,<=*d**n* (1<=≤<=*d**i*<=≤<=100) — the distances between pairs of neighboring stations. The third line contains two integers *s* and *t* (1<=≤<=*s*,<=*t*<=≤<=*n*) — the numbers of stations, between which you need to find the shortest distance. These numbers can be the same.
The numbers in the lines are separated by single spaces.
Output Specification:
Print a single number — the length of the shortest path between stations number *s* and *t*.
Demo Input:
['4\n2 3 4 9\n1 3\n', '4\n5 8 2 100\n4 1\n', '3\n1 1 1\n3 1\n', '3\n31 41 59\n1 1\n']
Demo Output:
['5\n', '15\n', '1\n', '0\n']
Note:
In the first sample the length of path 1 → 2 → 3 equals 5, the length of path 1 → 4 → 3 equals 13.
In the second sample the length of path 4 → 1 is 100, the length of path 4 → 3 → 2 → 1 is 15.
In the third sample the length of path 3 → 1 is 1, the length of path 3 → 2 → 1 is 2.
In the fourth sample the numbers of stations are the same, so the shortest distance equals 0.
|
```python
def swap(num1,num2):
arr = [num2,num1]
return arr
num = int(input())
distance = list(map(int, input() .split()))
cities = list(map(int, input() .split()))
answer = 0
sum1 = 0
sum2 = 0
if cities[0] == cities[1]:
answer = 0
if (cities[0] > cities[1]):
cities = swap(cities[0], cities[1])
i = cities[0] - 1
while(i < cities[1] - 1):
sum1 += distance[i]
i += 1
i = cities[1] - 1
while(i < num):
sum2 += distance[i]
i += 1
i = 0
while(i < cities[0] - 1):
sum2 += distance[i]
i += 1
if(sum1 < sum2):
print(sum1)
else:
print(sum2)
```
| 3
|
|
793
|
B
|
Igor and his way to work
|
PROGRAMMING
| 1,600
|
[
"dfs and similar",
"graphs",
"implementation",
"shortest paths"
] | null | null |
Woken up by the alarm clock Igor the financial analyst hurried up to the work. He ate his breakfast and sat in his car. Sadly, when he opened his GPS navigator, he found that some of the roads in Bankopolis, the city where he lives, are closed due to road works. Moreover, Igor has some problems with the steering wheel, so he can make no more than two turns on his way to his office in bank.
Bankopolis looks like a grid of *n* rows and *m* columns. Igor should find a way from his home to the bank that has no more than two turns and doesn't contain cells with road works, or determine that it is impossible and he should work from home. A turn is a change in movement direction. Igor's car can only move to the left, to the right, upwards and downwards. Initially Igor can choose any direction. Igor is still sleepy, so you should help him.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and the number of columns in the grid.
Each of the next *n* lines contains *m* characters denoting the corresponding row of the grid. The following characters can occur:
- "." — an empty cell; - "*" — a cell with road works; - "S" — the cell where Igor's home is located; - "T" — the cell where Igor's office is located.
It is guaranteed that "S" and "T" appear exactly once each.
|
In the only line print "YES" if there is a path between Igor's home and Igor's office with no more than two turns, and "NO" otherwise.
|
[
"5 5\n..S..\n****.\nT....\n****.\n.....\n",
"5 5\nS....\n****.\n.....\n.****\n..T..\n"
] |
[
"YES",
"NO"
] |
The first sample is shown on the following picture:
In the second sample it is impossible to reach Igor's office using less that 4 turns, thus there exists no path using no more than 2 turns. The path using exactly 4 turns is shown on this picture:
| 1,000
|
[
{
"input": "5 5\nS....\n****.\n.....\n.****\n..T..",
"output": "NO"
},
{
"input": "1 2\nST",
"output": "YES"
},
{
"input": "3 1\nS\n*\nT",
"output": "NO"
},
{
"input": "3 3\n*..\n...\nTS.",
"output": "YES"
},
{
"input": "3 3\nT.*\n*.*\n*S*",
"output": "YES"
},
{
"input": "7 7\n.S.****\n...*.*.\n.****..\n.*.**.*\n..T*...\n***..*.\n*******",
"output": "YES"
},
{
"input": "3 3\n**T\n*S*\n***",
"output": "NO"
},
{
"input": "2 2\nST\n*.",
"output": "YES"
},
{
"input": "2 2\nS.\n.T",
"output": "YES"
},
{
"input": "2 2\nTS\n.*",
"output": "YES"
},
{
"input": "2 2\n.T\nS*",
"output": "YES"
}
] | 1,501,906,587
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 62
| 4,915,200
|
turns = {}
score_dict = {}
def has_turn(from_cell, direction):
return not turns[from_cell] & direction
def move(matrix, from_cell, to_cell, direction):
# print('from_cell, to_cell, direction ', from_cell, to_cell, direction)
can_move = False
if not to_cell in turns:
turns[to_cell] = 0
score_dict[to_cell] = 0
turn = has_turn(from_cell, direction) and 1 or 0
# print('turn ', turn)
new_score = score_dict[from_cell] + turn
if new_score < 3:
score_dict[to_cell] = new_score
turns[to_cell] |= direction
can_move = True
return can_move
def should_move(matrix, cell):
return matrix[cell[0]][cell[1]] in ['.', 'T'] \
and cell not in score_dict
def solve(n, m, matrix):
turns.clear()
score_dict.clear()
start = None
stop = None
for i in range(n):
matrix[i] = list(matrix[i])
if not start and 'S' in matrix[i]:
start = (i, matrix[i].index('S'))
if not stop and 'T' in matrix[i]:
stop = (i, matrix[i].index('T'))
turns[start] = 0b1111
score_dict[start] = 0
queue = [start]
move_offset = [(-1, 0), (0, 1), (1, 0), (0, -1)]
directions = [8, 4, 2, 1]
while queue:
# print(queue)
cur = queue.pop(0)
if cur == stop:
return 'YES'
for i in range(len(directions)):
next_cell = (cur[0] + move_offset[i][0], cur[1] + move_offset[i][1])
direction = directions[i]
if next_cell[0] >= 0 and next_cell[0] < n \
and next_cell[1] >= 0 and next_cell[1] < m:
if should_move(matrix, next_cell) \
and move(matrix, cur, next_cell, direction):
queue.append(next_cell)
return 'NO'
n, m = map(int, input().split())
rows = []
for i in range(n):
rows.append(input())
print(solve(n, m, rows))
|
Title: Igor and his way to work
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Woken up by the alarm clock Igor the financial analyst hurried up to the work. He ate his breakfast and sat in his car. Sadly, when he opened his GPS navigator, he found that some of the roads in Bankopolis, the city where he lives, are closed due to road works. Moreover, Igor has some problems with the steering wheel, so he can make no more than two turns on his way to his office in bank.
Bankopolis looks like a grid of *n* rows and *m* columns. Igor should find a way from his home to the bank that has no more than two turns and doesn't contain cells with road works, or determine that it is impossible and he should work from home. A turn is a change in movement direction. Igor's car can only move to the left, to the right, upwards and downwards. Initially Igor can choose any direction. Igor is still sleepy, so you should help him.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=1000) — the number of rows and the number of columns in the grid.
Each of the next *n* lines contains *m* characters denoting the corresponding row of the grid. The following characters can occur:
- "." — an empty cell; - "*" — a cell with road works; - "S" — the cell where Igor's home is located; - "T" — the cell where Igor's office is located.
It is guaranteed that "S" and "T" appear exactly once each.
Output Specification:
In the only line print "YES" if there is a path between Igor's home and Igor's office with no more than two turns, and "NO" otherwise.
Demo Input:
['5 5\n..S..\n****.\nT....\n****.\n.....\n', '5 5\nS....\n****.\n.....\n.****\n..T..\n']
Demo Output:
['YES', 'NO']
Note:
The first sample is shown on the following picture:
In the second sample it is impossible to reach Igor's office using less that 4 turns, thus there exists no path using no more than 2 turns. The path using exactly 4 turns is shown on this picture:
|
```python
turns = {}
score_dict = {}
def has_turn(from_cell, direction):
return not turns[from_cell] & direction
def move(matrix, from_cell, to_cell, direction):
# print('from_cell, to_cell, direction ', from_cell, to_cell, direction)
can_move = False
if not to_cell in turns:
turns[to_cell] = 0
score_dict[to_cell] = 0
turn = has_turn(from_cell, direction) and 1 or 0
# print('turn ', turn)
new_score = score_dict[from_cell] + turn
if new_score < 3:
score_dict[to_cell] = new_score
turns[to_cell] |= direction
can_move = True
return can_move
def should_move(matrix, cell):
return matrix[cell[0]][cell[1]] in ['.', 'T'] \
and cell not in score_dict
def solve(n, m, matrix):
turns.clear()
score_dict.clear()
start = None
stop = None
for i in range(n):
matrix[i] = list(matrix[i])
if not start and 'S' in matrix[i]:
start = (i, matrix[i].index('S'))
if not stop and 'T' in matrix[i]:
stop = (i, matrix[i].index('T'))
turns[start] = 0b1111
score_dict[start] = 0
queue = [start]
move_offset = [(-1, 0), (0, 1), (1, 0), (0, -1)]
directions = [8, 4, 2, 1]
while queue:
# print(queue)
cur = queue.pop(0)
if cur == stop:
return 'YES'
for i in range(len(directions)):
next_cell = (cur[0] + move_offset[i][0], cur[1] + move_offset[i][1])
direction = directions[i]
if next_cell[0] >= 0 and next_cell[0] < n \
and next_cell[1] >= 0 and next_cell[1] < m:
if should_move(matrix, next_cell) \
and move(matrix, cur, next_cell, direction):
queue.append(next_cell)
return 'NO'
n, m = map(int, input().split())
rows = []
for i in range(n):
rows.append(input())
print(solve(n, m, rows))
```
| 0
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,678,601,945
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 35
| 124
| 0
|
m,n = (input().split())
area = int(m)*int(n)
print(int(area/2))
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none
|
```python
m,n = (input().split())
area = int(m)*int(n)
print(int(area/2))
```
| 3.969
|
508
|
A
|
Pasha and Pixels
|
PROGRAMMING
| 1,100
|
[
"brute force"
] | null | null |
Pasha loves his phone and also putting his hair up... But the hair is now irrelevant.
Pasha has installed a new game to his phone. The goal of the game is following. There is a rectangular field consisting of *n* row with *m* pixels in each row. Initially, all the pixels are colored white. In one move, Pasha can choose any pixel and color it black. In particular, he can choose the pixel that is already black, then after the boy's move the pixel does not change, that is, it remains black. Pasha loses the game when a 2<=×<=2 square consisting of black pixels is formed.
Pasha has made a plan of *k* moves, according to which he will paint pixels. Each turn in his plan is represented as a pair of numbers *i* and *j*, denoting respectively the row and the column of the pixel to be colored on the current move.
Determine whether Pasha loses if he acts in accordance with his plan, and if he does, on what move the 2<=×<=2 square consisting of black pixels is formed.
|
The first line of the input contains three integers *n*,<=*m*,<=*k* (1<=≤<=*n*,<=*m*<=≤<=1000, 1<=≤<=*k*<=≤<=105) — the number of rows, the number of columns and the number of moves that Pasha is going to perform.
The next *k* lines contain Pasha's moves in the order he makes them. Each line contains two integers *i* and *j* (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*j*<=≤<=*m*), representing the row number and column number of the pixel that was painted during a move.
|
If Pasha loses, print the number of the move when the 2<=×<=2 square consisting of black pixels is formed.
If Pasha doesn't lose, that is, no 2<=×<=2 square consisting of black pixels is formed during the given *k* moves, print 0.
|
[
"2 2 4\n1 1\n1 2\n2 1\n2 2\n",
"2 3 6\n2 3\n2 2\n1 3\n2 2\n1 2\n1 1\n",
"5 3 7\n2 3\n1 2\n1 1\n4 1\n3 1\n5 3\n3 2\n"
] |
[
"4\n",
"5\n",
"0\n"
] |
none
| 500
|
[
{
"input": "2 2 4\n1 1\n1 2\n2 1\n2 2",
"output": "4"
},
{
"input": "2 3 6\n2 3\n2 2\n1 3\n2 2\n1 2\n1 1",
"output": "5"
},
{
"input": "5 3 7\n2 3\n1 2\n1 1\n4 1\n3 1\n5 3\n3 2",
"output": "0"
},
{
"input": "3 3 11\n2 1\n3 1\n1 1\n1 3\n1 2\n2 3\n3 3\n3 2\n2 2\n1 3\n3 3",
"output": "9"
},
{
"input": "2 2 5\n1 1\n2 1\n2 1\n1 2\n2 2",
"output": "5"
},
{
"input": "518 518 10\n37 97\n47 278\n17 467\n158 66\n483 351\n83 123\n285 219\n513 187\n380 75\n304 352",
"output": "0"
},
{
"input": "1 1 5\n1 1\n1 1\n1 1\n1 1\n1 1",
"output": "0"
},
{
"input": "1 5 5\n1 1\n1 2\n1 3\n1 4\n1 5",
"output": "0"
},
{
"input": "5 1 5\n1 1\n2 1\n3 1\n4 1\n5 1",
"output": "0"
},
{
"input": "1 1 1\n1 1",
"output": "0"
},
{
"input": "10 10 4\n5 9\n6 9\n6 10\n5 10",
"output": "4"
},
{
"input": "1000 1000 4\n999 999\n999 1000\n1000 999\n1000 1000",
"output": "4"
},
{
"input": "2 3 5\n2 3\n1 3\n1 2\n1 1\n2 2",
"output": "5"
},
{
"input": "1000 1000 4\n1000 1000\n999 999\n1000 999\n999 1000",
"output": "4"
}
] | 1,668,086,699
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 15
| 0
|
n,m,k=map(int,input().split())
x=[[0]*(m+2) for i in range(n+2)]
def check(i,j):
if mx[i][j+1] and mx[i+1][j] and mx[i+1][j+1]:
return True
if mx[i][j-1] and mx[i+1][j-1] and mx[i+1][j]:
return True
if mx[i-1][j] and mx[i-1][j+1] and mx[i][j+1]:
return True
if mx[i-1][j-1] and mx[i-1][j] and mx[i][j-1]:
return True
return False
for i in range(k):
a,b=map(int,input().split())
x[a][b]=1
if check(a,b):
print(i+1)
break
else:
print(0)
|
Title: Pasha and Pixels
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pasha loves his phone and also putting his hair up... But the hair is now irrelevant.
Pasha has installed a new game to his phone. The goal of the game is following. There is a rectangular field consisting of *n* row with *m* pixels in each row. Initially, all the pixels are colored white. In one move, Pasha can choose any pixel and color it black. In particular, he can choose the pixel that is already black, then after the boy's move the pixel does not change, that is, it remains black. Pasha loses the game when a 2<=×<=2 square consisting of black pixels is formed.
Pasha has made a plan of *k* moves, according to which he will paint pixels. Each turn in his plan is represented as a pair of numbers *i* and *j*, denoting respectively the row and the column of the pixel to be colored on the current move.
Determine whether Pasha loses if he acts in accordance with his plan, and if he does, on what move the 2<=×<=2 square consisting of black pixels is formed.
Input Specification:
The first line of the input contains three integers *n*,<=*m*,<=*k* (1<=≤<=*n*,<=*m*<=≤<=1000, 1<=≤<=*k*<=≤<=105) — the number of rows, the number of columns and the number of moves that Pasha is going to perform.
The next *k* lines contain Pasha's moves in the order he makes them. Each line contains two integers *i* and *j* (1<=≤<=*i*<=≤<=*n*, 1<=≤<=*j*<=≤<=*m*), representing the row number and column number of the pixel that was painted during a move.
Output Specification:
If Pasha loses, print the number of the move when the 2<=×<=2 square consisting of black pixels is formed.
If Pasha doesn't lose, that is, no 2<=×<=2 square consisting of black pixels is formed during the given *k* moves, print 0.
Demo Input:
['2 2 4\n1 1\n1 2\n2 1\n2 2\n', '2 3 6\n2 3\n2 2\n1 3\n2 2\n1 2\n1 1\n', '5 3 7\n2 3\n1 2\n1 1\n4 1\n3 1\n5 3\n3 2\n']
Demo Output:
['4\n', '5\n', '0\n']
Note:
none
|
```python
n,m,k=map(int,input().split())
x=[[0]*(m+2) for i in range(n+2)]
def check(i,j):
if mx[i][j+1] and mx[i+1][j] and mx[i+1][j+1]:
return True
if mx[i][j-1] and mx[i+1][j-1] and mx[i+1][j]:
return True
if mx[i-1][j] and mx[i-1][j+1] and mx[i][j+1]:
return True
if mx[i-1][j-1] and mx[i-1][j] and mx[i][j-1]:
return True
return False
for i in range(k):
a,b=map(int,input().split())
x[a][b]=1
if check(a,b):
print(i+1)
break
else:
print(0)
```
| -1
|
|
581
|
A
|
Vasya the Hipster
|
PROGRAMMING
| 800
|
[
"implementation",
"math"
] | null | null |
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
|
The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got.
|
Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
|
[
"3 1\n",
"2 3\n",
"7 3\n"
] |
[
"1 1\n",
"2 0\n",
"3 2\n"
] |
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.
| 500
|
[
{
"input": "3 1",
"output": "1 1"
},
{
"input": "2 3",
"output": "2 0"
},
{
"input": "7 3",
"output": "3 2"
},
{
"input": "100 100",
"output": "100 0"
},
{
"input": "4 10",
"output": "4 3"
},
{
"input": "6 10",
"output": "6 2"
},
{
"input": "6 11",
"output": "6 2"
},
{
"input": "10 40",
"output": "10 15"
},
{
"input": "11 56",
"output": "11 22"
},
{
"input": "34 30",
"output": "30 2"
},
{
"input": "33 33",
"output": "33 0"
},
{
"input": "100 45",
"output": "45 27"
},
{
"input": "100 23",
"output": "23 38"
},
{
"input": "45 12",
"output": "12 16"
},
{
"input": "1 1",
"output": "1 0"
},
{
"input": "1 100",
"output": "1 49"
},
{
"input": "100 1",
"output": "1 49"
},
{
"input": "68 59",
"output": "59 4"
},
{
"input": "45 99",
"output": "45 27"
},
{
"input": "99 100",
"output": "99 0"
},
{
"input": "100 98",
"output": "98 1"
},
{
"input": "59 12",
"output": "12 23"
},
{
"input": "86 4",
"output": "4 41"
},
{
"input": "68 21",
"output": "21 23"
},
{
"input": "100 11",
"output": "11 44"
},
{
"input": "100 10",
"output": "10 45"
},
{
"input": "15 45",
"output": "15 15"
},
{
"input": "11 32",
"output": "11 10"
},
{
"input": "34 96",
"output": "34 31"
},
{
"input": "89 89",
"output": "89 0"
}
] | 1,691,085,404
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 30
| 46
| 0
|
a,b=map(int,input().split())
s=0
d=0
if a>b:
d+=b
a-=b
if a %2 == 0:
s+=int(a/2)
else:
s+=int((a-1)/2)
else:
d+=a
b-=a
if b % 2 ==0:
s+=int(b/2)
else:
s+=int((b-1)/2)
print(d,s)
|
Title: Vasya the Hipster
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day Vasya the Hipster decided to count how many socks he had. It turned out that he had *a* red socks and *b* blue socks.
According to the latest fashion, hipsters should wear the socks of different colors: a red one on the left foot, a blue one on the right foot.
Every day Vasya puts on new socks in the morning and throws them away before going to bed as he doesn't want to wash them.
Vasya wonders, what is the maximum number of days when he can dress fashionable and wear different socks, and after that, for how many days he can then wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Can you help him?
Input Specification:
The single line of the input contains two positive integers *a* and *b* (1<=≤<=*a*,<=*b*<=≤<=100) — the number of red and blue socks that Vasya's got.
Output Specification:
Print two space-separated integers — the maximum number of days when Vasya can wear different socks and the number of days when he can wear the same socks until he either runs out of socks or cannot make a single pair from the socks he's got.
Keep in mind that at the end of the day Vasya throws away the socks that he's been wearing on that day.
Demo Input:
['3 1\n', '2 3\n', '7 3\n']
Demo Output:
['1 1\n', '2 0\n', '3 2\n']
Note:
In the first sample Vasya can first put on one pair of different socks, after that he has two red socks left to wear on the second day.
|
```python
a,b=map(int,input().split())
s=0
d=0
if a>b:
d+=b
a-=b
if a %2 == 0:
s+=int(a/2)
else:
s+=int((a-1)/2)
else:
d+=a
b-=a
if b % 2 ==0:
s+=int(b/2)
else:
s+=int((b-1)/2)
print(d,s)
```
| 3
|
|
513
|
A
|
Game
|
PROGRAMMING
| 800
|
[
"constructive algorithms",
"math"
] | null | null |
Two players play a simple game. Each player is provided with a box with balls. First player's box contains exactly *n*1 balls and second player's box contains exactly *n*2 balls. In one move first player can take from 1 to *k*1 balls from his box and throw them away. Similarly, the second player can take from 1 to *k*2 balls from his box in his move. Players alternate turns and the first player starts the game. The one who can't make a move loses. Your task is to determine who wins if both players play optimally.
|
The first line contains four integers *n*1,<=*n*2,<=*k*1,<=*k*2. All numbers in the input are from 1 to 50.
This problem doesn't have subproblems. You will get 3 points for the correct submission.
|
Output "First" if the first player wins and "Second" otherwise.
|
[
"2 2 1 2\n",
"2 1 1 1\n"
] |
[
"Second\n",
"First\n"
] |
Consider the first sample test. Each player has a box with 2 balls. The first player draws a single ball from his box in one move and the second player can either take 1 or 2 balls from his box in one move. No matter how the first player acts, the second player can always win if he plays wisely.
| 3
|
[
{
"input": "2 2 1 2",
"output": "Second"
},
{
"input": "2 1 1 1",
"output": "First"
},
{
"input": "5 7 4 1",
"output": "Second"
},
{
"input": "5 7 1 4",
"output": "Second"
},
{
"input": "5 7 10 10",
"output": "Second"
},
{
"input": "5 7 1 10",
"output": "Second"
},
{
"input": "1 1 1 1",
"output": "Second"
},
{
"input": "50 50 50 50",
"output": "Second"
},
{
"input": "50 49 1 2",
"output": "First"
},
{
"input": "50 48 3 1",
"output": "First"
},
{
"input": "48 50 12 11",
"output": "Second"
},
{
"input": "49 50 11 12",
"output": "Second"
},
{
"input": "49 49 4 1",
"output": "Second"
},
{
"input": "49 49 3 3",
"output": "Second"
},
{
"input": "1 50 1 50",
"output": "Second"
},
{
"input": "1 50 50 50",
"output": "Second"
},
{
"input": "50 1 1 1",
"output": "First"
},
{
"input": "50 1 1 50",
"output": "First"
},
{
"input": "32 31 10 9",
"output": "First"
},
{
"input": "32 4 17 3",
"output": "First"
}
] | 1,423,330,051
| 1,651
|
Python 3
|
OK
|
TESTS
| 20
| 62
| 0
|
n1, n2, k1, k2=map(int, input().split())
if(n1>n2):
print("First")
if(n1<=n2):
print("Second")
|
Title: Game
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Two players play a simple game. Each player is provided with a box with balls. First player's box contains exactly *n*1 balls and second player's box contains exactly *n*2 balls. In one move first player can take from 1 to *k*1 balls from his box and throw them away. Similarly, the second player can take from 1 to *k*2 balls from his box in his move. Players alternate turns and the first player starts the game. The one who can't make a move loses. Your task is to determine who wins if both players play optimally.
Input Specification:
The first line contains four integers *n*1,<=*n*2,<=*k*1,<=*k*2. All numbers in the input are from 1 to 50.
This problem doesn't have subproblems. You will get 3 points for the correct submission.
Output Specification:
Output "First" if the first player wins and "Second" otherwise.
Demo Input:
['2 2 1 2\n', '2 1 1 1\n']
Demo Output:
['Second\n', 'First\n']
Note:
Consider the first sample test. Each player has a box with 2 balls. The first player draws a single ball from his box in one move and the second player can either take 1 or 2 balls from his box in one move. No matter how the first player acts, the second player can always win if he plays wisely.
|
```python
n1, n2, k1, k2=map(int, input().split())
if(n1>n2):
print("First")
if(n1<=n2):
print("Second")
```
| 3
|
|
977
|
C
|
Less or Equal
|
PROGRAMMING
| 1,200
|
[
"sortings"
] | null | null |
You are given a sequence of integers of length $n$ and integer number $k$. You should print any integer number $x$ in the range of $[1; 10^9]$ (i.e. $1 \le x \le 10^9$) such that exactly $k$ elements of given sequence are less than or equal to $x$.
Note that the sequence can contain equal elements.
If there is no such $x$, print "-1" (without quotes).
|
The first line of the input contains integer numbers $n$ and $k$ ($1 \le n \le 2 \cdot 10^5$, $0 \le k \le n$). The second line of the input contains $n$ integer numbers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the sequence itself.
|
Print any integer number $x$ from range $[1; 10^9]$ such that exactly $k$ elements of given sequence is less or equal to $x$.
If there is no such $x$, print "-1" (without quotes).
|
[
"7 4\n3 7 5 1 10 3 20\n",
"7 2\n3 7 5 1 10 3 20\n"
] |
[
"6",
"-1\n"
] |
In the first example $5$ is also a valid answer because the elements with indices $[1, 3, 4, 6]$ is less than or equal to $5$ and obviously less than or equal to $6$.
In the second example you cannot choose any number that only $2$ elements of the given sequence will be less than or equal to this number because $3$ elements of the given sequence will be also less than or equal to this number.
| 0
|
[
{
"input": "7 4\n3 7 5 1 10 3 20",
"output": "5"
},
{
"input": "7 2\n3 7 5 1 10 3 20",
"output": "-1"
},
{
"input": "1 0\n1",
"output": "-1"
},
{
"input": "1 0\n2",
"output": "1"
},
{
"input": "1 1\n1000000000",
"output": "1000000000"
},
{
"input": "3 0\n3 3 3",
"output": "2"
},
{
"input": "3 0\n2 2 3",
"output": "1"
},
{
"input": "5 0\n3 4 5 6 7",
"output": "2"
},
{
"input": "4 0\n2 3 4 5",
"output": "1"
},
{
"input": "2 2\n1000000000 1000000000",
"output": "1000000000"
},
{
"input": "7 2\n2 7 5 1 10 2 20",
"output": "-1"
},
{
"input": "2 1\n1 1",
"output": "-1"
},
{
"input": "5 3\n1 3 3 4 5",
"output": "3"
},
{
"input": "4 4\n1000000000 1000000000 1000000000 1000000000",
"output": "1000000000"
}
] | 1,697,752,615
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 2
| 46
| 0
|
def solve():
n,k = input().split()
nums = list(map(int,input().split()))
nums.sort()
n,k = int(n),int(k)
id = k-1
person = nums[id]
l = 0
r = len(nums)-1
while l < r:
m = l+(r-l)//2
if nums[m] > person:
r = m
else:
l = m+1
if l-1 != id:
return -1
return person
print(solve())
|
Title: Less or Equal
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given a sequence of integers of length $n$ and integer number $k$. You should print any integer number $x$ in the range of $[1; 10^9]$ (i.e. $1 \le x \le 10^9$) such that exactly $k$ elements of given sequence are less than or equal to $x$.
Note that the sequence can contain equal elements.
If there is no such $x$, print "-1" (without quotes).
Input Specification:
The first line of the input contains integer numbers $n$ and $k$ ($1 \le n \le 2 \cdot 10^5$, $0 \le k \le n$). The second line of the input contains $n$ integer numbers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 10^9$) — the sequence itself.
Output Specification:
Print any integer number $x$ from range $[1; 10^9]$ such that exactly $k$ elements of given sequence is less or equal to $x$.
If there is no such $x$, print "-1" (without quotes).
Demo Input:
['7 4\n3 7 5 1 10 3 20\n', '7 2\n3 7 5 1 10 3 20\n']
Demo Output:
['6', '-1\n']
Note:
In the first example $5$ is also a valid answer because the elements with indices $[1, 3, 4, 6]$ is less than or equal to $5$ and obviously less than or equal to $6$.
In the second example you cannot choose any number that only $2$ elements of the given sequence will be less than or equal to this number because $3$ elements of the given sequence will be also less than or equal to this number.
|
```python
def solve():
n,k = input().split()
nums = list(map(int,input().split()))
nums.sort()
n,k = int(n),int(k)
id = k-1
person = nums[id]
l = 0
r = len(nums)-1
while l < r:
m = l+(r-l)//2
if nums[m] > person:
r = m
else:
l = m+1
if l-1 != id:
return -1
return person
print(solve())
```
| 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"
}
] | 1,632,430,269
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 92
| 6,963,200
|
k = int(input())
s = input()
ns = set(s)
if len(ns) > k:
print(-1)
else:
if len(s) % len(ns) == 0:
count = 0
for i in ns:
count += (s.count(i) / k)
if count == (len(s) / k):
temp = " "
for i in ns:
temp += ((s.count(i) // k) * i)
print(temp[1:] * k)
else:
print(-1)
else:
print(-1)
|
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()
ns = set(s)
if len(ns) > k:
print(-1)
else:
if len(s) % len(ns) == 0:
count = 0
for i in ns:
count += (s.count(i) / k)
if count == (len(s) / k):
temp = " "
for i in ns:
temp += ((s.count(i) // k) * i)
print(temp[1:] * k)
else:
print(-1)
else:
print(-1)
```
| 0
|
|
599
|
A
|
Patrick and Shopping
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.
|
The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.
|
Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.
|
[
"10 20 30\n",
"1 1 5\n"
] |
[
"60\n",
"4\n"
] |
The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
| 500
|
[
{
"input": "10 20 30",
"output": "60"
},
{
"input": "1 1 5",
"output": "4"
},
{
"input": "100 33 34",
"output": "134"
},
{
"input": "777 777 777",
"output": "2331"
},
{
"input": "2 2 8",
"output": "8"
},
{
"input": "12 34 56",
"output": "92"
},
{
"input": "789 101112 131415",
"output": "203802"
},
{
"input": "27485716 99999999 35182",
"output": "55041796"
},
{
"input": "1 293548 5",
"output": "12"
},
{
"input": "12059 259855 5874875",
"output": "543828"
},
{
"input": "46981 105809 585858",
"output": "305580"
},
{
"input": "9889 1221 2442",
"output": "7326"
},
{
"input": "100500 200600 300700",
"output": "601800"
},
{
"input": "318476 318476 318476",
"output": "955428"
},
{
"input": "23985 3353 75633",
"output": "54676"
},
{
"input": "120 1298 2222",
"output": "2836"
},
{
"input": "98437 23487 666672",
"output": "243848"
},
{
"input": "100000000 100000000 100000000",
"output": "300000000"
},
{
"input": "2 5 2",
"output": "8"
},
{
"input": "1 1000 1",
"output": "4"
},
{
"input": "1 100000000 1",
"output": "4"
}
] | 1,667,018,085
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 30
| 0
|
d_list = []
d = input().split()
for u in range(len(d)):
d[u] = int(d[u])
d_list.append(d[u])
if d[0] + d[1] > d[2]:
print(d[2]*2)
elif d[0] + d[1] < d[2]:
print((d[0] + d[1])*2)
else:
print(d[2]*2)
|
Title: Patrick and Shopping
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Today Patrick waits for a visit from his friend Spongebob. To prepare for the visit, Patrick needs to buy some goodies in two stores located near his house. There is a *d*1 meter long road between his house and the first shop and a *d*2 meter long road between his house and the second shop. Also, there is a road of length *d*3 directly connecting these two shops to each other. Help Patrick calculate the minimum distance that he needs to walk in order to go to both shops and return to his house.
Patrick always starts at his house. He should visit both shops moving only along the three existing roads and return back to his house. He doesn't mind visiting the same shop or passing the same road multiple times. The only goal is to minimize the total distance traveled.
Input Specification:
The first line of the input contains three integers *d*1, *d*2, *d*3 (1<=≤<=*d*1,<=*d*2,<=*d*3<=≤<=108) — the lengths of the paths.
- *d*1 is the length of the path connecting Patrick's house and the first shop; - *d*2 is the length of the path connecting Patrick's house and the second shop; - *d*3 is the length of the path connecting both shops.
Output Specification:
Print the minimum distance that Patrick will have to walk in order to visit both shops and return to his house.
Demo Input:
['10 20 30\n', '1 1 5\n']
Demo Output:
['60\n', '4\n']
Note:
The first sample is shown on the picture in the problem statement. One of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
In the second sample one of the optimal routes is: house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> first shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> second shop <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/70a0795f45d32287dba0eb83fc4a3f470c6e5537.png" style="max-width: 100.0%;max-height: 100.0%;"/> house.
|
```python
d_list = []
d = input().split()
for u in range(len(d)):
d[u] = int(d[u])
d_list.append(d[u])
if d[0] + d[1] > d[2]:
print(d[2]*2)
elif d[0] + d[1] < d[2]:
print((d[0] + d[1])*2)
else:
print(d[2]*2)
```
| 0
|
|
365
|
A
|
Good Number
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*).
|
The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109).
|
Print a single integer — the number of *k*-good numbers in *a*.
|
[
"10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n",
"2 1\n1\n10\n"
] |
[
"10\n",
"1\n"
] |
none
| 500
|
[
{
"input": "10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560",
"output": "10"
},
{
"input": "2 1\n1\n10",
"output": "1"
},
{
"input": "1 0\n1000000000",
"output": "1"
},
{
"input": "1 1\n1000000000",
"output": "1"
},
{
"input": "6 0\n10\n102\n120\n1032\n1212103\n1999999",
"output": "5"
},
{
"input": "1 3\n1000000000",
"output": "0"
},
{
"input": "1 9\n1000000000",
"output": "0"
},
{
"input": "1 0\n1",
"output": "0"
},
{
"input": "1 1\n1",
"output": "0"
},
{
"input": "1 3\n1",
"output": "0"
},
{
"input": "1 9\n1",
"output": "0"
},
{
"input": "2 8\n123456780\n123",
"output": "1"
},
{
"input": "2 8\n12345678\n1230",
"output": "0"
},
{
"input": "6 1\n10\n102\n120\n1032\n1212103\n1999999",
"output": "5"
},
{
"input": "6 2\n10\n102\n120\n1032\n1212103\n1999999",
"output": "4"
},
{
"input": "6 3\n10\n102\n120\n1032\n1212103\n1999999",
"output": "2"
},
{
"input": "6 9\n10\n102\n120\n1032\n1212103\n1999999",
"output": "0"
},
{
"input": "1 2\n2220",
"output": "0"
}
] | 1,673,427,213
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 3
| 46
| 0
|
a, k = map(int,input().split())
count = 0
k_list = [str(i) for i in range(1,k)]
k_list.append(str(k))
k_list.append(str(0))
l = len(k_list)
n_l ="".join(k_list)
for x in range(a):
list_a = [(x) for x in input()]
str_a = "".join(list_a)
if n_l in str_a:
count += 1
print(count)
|
Title: Good Number
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Let's call a number *k*-good if it contains all digits not exceeding *k* (0,<=...,<=*k*). You've got a number *k* and an array *a* containing *n* numbers. Find out how many *k*-good numbers are in *a* (count each number every time it occurs in array *a*).
Input Specification:
The first line contains integers *n* and *k* (1<=≤<=*n*<=≤<=100, 0<=≤<=*k*<=≤<=9). The *i*-th of the following *n* lines contains integer *a**i* without leading zeroes (1<=≤<=*a**i*<=≤<=109).
Output Specification:
Print a single integer — the number of *k*-good numbers in *a*.
Demo Input:
['10 6\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n1234560\n', '2 1\n1\n10\n']
Demo Output:
['10\n', '1\n']
Note:
none
|
```python
a, k = map(int,input().split())
count = 0
k_list = [str(i) for i in range(1,k)]
k_list.append(str(k))
k_list.append(str(0))
l = len(k_list)
n_l ="".join(k_list)
for x in range(a):
list_a = [(x) for x in input()]
str_a = "".join(list_a)
if n_l in str_a:
count += 1
print(count)
```
| 0
|
|
50
|
A
|
Domino piling
|
PROGRAMMING
| 800
|
[
"greedy",
"math"
] |
A. Domino piling
|
2
|
256
|
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
|
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
|
Output one number — the maximal number of dominoes, which can be placed.
|
[
"2 4\n",
"3 3\n"
] |
[
"4\n",
"4\n"
] |
none
| 500
|
[
{
"input": "2 4",
"output": "4"
},
{
"input": "3 3",
"output": "4"
},
{
"input": "1 5",
"output": "2"
},
{
"input": "1 6",
"output": "3"
},
{
"input": "1 15",
"output": "7"
},
{
"input": "1 16",
"output": "8"
},
{
"input": "2 5",
"output": "5"
},
{
"input": "2 6",
"output": "6"
},
{
"input": "2 7",
"output": "7"
},
{
"input": "2 14",
"output": "14"
},
{
"input": "2 15",
"output": "15"
},
{
"input": "1 4",
"output": "2"
},
{
"input": "2 16",
"output": "16"
},
{
"input": "3 5",
"output": "7"
},
{
"input": "3 6",
"output": "9"
},
{
"input": "3 10",
"output": "15"
},
{
"input": "3 14",
"output": "21"
},
{
"input": "3 15",
"output": "22"
},
{
"input": "3 16",
"output": "24"
},
{
"input": "5 7",
"output": "17"
},
{
"input": "16 16",
"output": "128"
},
{
"input": "15 16",
"output": "120"
},
{
"input": "2 3",
"output": "3"
},
{
"input": "15 15",
"output": "112"
},
{
"input": "14 16",
"output": "112"
},
{
"input": "11 13",
"output": "71"
},
{
"input": "5 16",
"output": "40"
},
{
"input": "8 15",
"output": "60"
},
{
"input": "2 2",
"output": "2"
},
{
"input": "3 4",
"output": "6"
},
{
"input": "4 4",
"output": "8"
},
{
"input": "1 1",
"output": "0"
},
{
"input": "1 2",
"output": "1"
},
{
"input": "1 3",
"output": "1"
},
{
"input": "14 15",
"output": "105"
}
] | 1,693,733,584
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 92
| 0
|
m,n = map(int,input().split())
x = m*n
print(x//2)
|
Title: Domino piling
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
You are given a rectangular board of *M*<=×<=*N* squares. Also you are given an unlimited number of standard domino pieces of 2<=×<=1 squares. You are allowed to rotate the pieces. You are asked to place as many dominoes as possible on the board so as to meet the following conditions:
1. Each domino completely covers two squares.
2. No two dominoes overlap.
3. Each domino lies entirely inside the board. It is allowed to touch the edges of the board.
Find the maximum number of dominoes, which can be placed under these restrictions.
Input Specification:
In a single line you are given two integers *M* and *N* — board sizes in squares (1<=≤<=*M*<=≤<=*N*<=≤<=16).
Output Specification:
Output one number — the maximal number of dominoes, which can be placed.
Demo Input:
['2 4\n', '3 3\n']
Demo Output:
['4\n', '4\n']
Note:
none
|
```python
m,n = map(int,input().split())
x = m*n
print(x//2)
```
| 3.977
|
246
|
A
|
Buggy Sorting
|
PROGRAMMING
| 900
|
[
"constructive algorithms",
"greedy",
"sortings"
] | null | null |
Little boy Valera studies an algorithm of sorting an integer array. After studying the theory, he went on to the practical tasks. As a result, he wrote a program that sorts an array of *n* integers *a*1,<=*a*2,<=...,<=*a**n* in the non-decreasing order. The pseudocode of the program, written by Valera, is given below. The input of the program gets number *n* and array *a*.
But Valera could have made a mistake, because he hasn't yet fully learned the sorting algorithm. If Valera made a mistake in his program, you need to give a counter-example that makes his program work improperly (that is, the example that makes the program sort the array not in the non-decreasing order). If such example for the given value of *n* doesn't exist, print -1.
|
You've got a single integer *n* (1<=≤<=*n*<=≤<=50) — the size of the sorted array.
|
Print *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the counter-example, for which Valera's algorithm won't work correctly. If the counter-example that meets the described conditions is impossible to give, print -1.
If there are several counter-examples, consisting of *n* numbers, you are allowed to print any of them.
|
[
"1\n"
] |
[
"-1\n"
] |
none
| 500
|
[
{
"input": "1",
"output": "-1"
},
{
"input": "2",
"output": "-1"
},
{
"input": "3",
"output": "3 2 1 "
},
{
"input": "4",
"output": "4 3 2 1 "
},
{
"input": "5",
"output": "5 4 3 2 1 "
},
{
"input": "6",
"output": "6 5 4 3 2 1 "
},
{
"input": "7",
"output": "7 6 5 4 3 2 1 "
},
{
"input": "8",
"output": "8 7 6 5 4 3 2 1 "
},
{
"input": "9",
"output": "9 8 7 6 5 4 3 2 1 "
},
{
"input": "50",
"output": "50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "22",
"output": "22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "34",
"output": "34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "50",
"output": "50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "12",
"output": "12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "26",
"output": "26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "38",
"output": "38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "4",
"output": "4 3 2 1 "
},
{
"input": "18",
"output": "18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "30",
"output": "30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "46",
"output": "46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
},
{
"input": "32",
"output": "32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 "
}
] | 1,623,499,939
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 21
| 108
| 0
|
def main_function():
a = int(input())
if a > 2:
return " ".join([str(a - i) for i in range(a)])
return -1
print(main_function())
|
Title: Buggy Sorting
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Little boy Valera studies an algorithm of sorting an integer array. After studying the theory, he went on to the practical tasks. As a result, he wrote a program that sorts an array of *n* integers *a*1,<=*a*2,<=...,<=*a**n* in the non-decreasing order. The pseudocode of the program, written by Valera, is given below. The input of the program gets number *n* and array *a*.
But Valera could have made a mistake, because he hasn't yet fully learned the sorting algorithm. If Valera made a mistake in his program, you need to give a counter-example that makes his program work improperly (that is, the example that makes the program sort the array not in the non-decreasing order). If such example for the given value of *n* doesn't exist, print -1.
Input Specification:
You've got a single integer *n* (1<=≤<=*n*<=≤<=50) — the size of the sorted array.
Output Specification:
Print *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=100) — the counter-example, for which Valera's algorithm won't work correctly. If the counter-example that meets the described conditions is impossible to give, print -1.
If there are several counter-examples, consisting of *n* numbers, you are allowed to print any of them.
Demo Input:
['1\n']
Demo Output:
['-1\n']
Note:
none
|
```python
def main_function():
a = int(input())
if a > 2:
return " ".join([str(a - i) for i in range(a)])
return -1
print(main_function())
```
| 3
|
|
344
|
A
|
Magnets
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.
|
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.
|
On the single line of the output print the number of groups of magnets.
|
[
"6\n10\n10\n10\n01\n10\n10\n",
"4\n01\n01\n10\n10\n"
] |
[
"3\n",
"2\n"
] |
The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets.
| 500
|
[
{
"input": "6\n10\n10\n10\n01\n10\n10",
"output": "3"
},
{
"input": "4\n01\n01\n10\n10",
"output": "2"
},
{
"input": "1\n10",
"output": "1"
},
{
"input": "2\n01\n10",
"output": "2"
},
{
"input": "2\n10\n10",
"output": "1"
},
{
"input": "3\n10\n01\n10",
"output": "3"
},
{
"input": "1\n01",
"output": "1"
},
{
"input": "2\n01\n01",
"output": "1"
},
{
"input": "2\n10\n01",
"output": "2"
},
{
"input": "3\n01\n01\n01",
"output": "1"
},
{
"input": "3\n10\n10\n01",
"output": "2"
},
{
"input": "3\n01\n10\n10",
"output": "2"
},
{
"input": "115\n10\n10\n10\n10\n01\n01\n10\n10\n10\n01\n01\n10\n01\n01\n10\n10\n10\n01\n10\n01\n10\n10\n01\n01\n10\n10\n10\n10\n01\n10\n01\n01\n10\n10\n10\n10\n01\n10\n10\n10\n01\n10\n01\n10\n10\n10\n10\n01\n01\n01\n10\n10\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n10\n10\n01\n10\n01\n01\n01\n01\n01\n10\n01\n10\n10\n01\n01\n01\n10\n01\n01\n10\n10\n01\n01\n01\n01\n01\n10\n01\n10\n01\n10\n01\n01\n01\n10\n01\n10\n10\n01\n10\n10\n01\n01\n01\n10\n10\n10\n10\n10\n10\n10\n10",
"output": "55"
}
] | 1,695,979,397
| 2,147,483,647
|
PyPy 3-64
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 6
| 1,000
| 11,161,600
|
n=int(input())
magnets=[]
for i in range(n):
magnet=input()
magnets.append(magnet)
group=1
for i in range(len(magnets)-1):
if magnets[i]!=magnets[i+1]:
group+=1
print(group)
|
Title: Magnets
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mad scientist Mike entertains himself by arranging rows of dominoes. He doesn't need dominoes, though: he uses rectangular magnets instead. Each magnet has two poles, positive (a "plus") and negative (a "minus"). If two magnets are put together at a close distance, then the like poles will repel each other and the opposite poles will attract each other.
Mike starts by laying one magnet horizontally on the table. During each following step Mike adds one more magnet horizontally to the right end of the row. Depending on how Mike puts the magnet on the table, it is either attracted to the previous one (forming a group of multiple magnets linked together) or repelled by it (then Mike lays this magnet at some distance to the right from the previous one). We assume that a sole magnet not linked to others forms a group of its own.
Mike arranged multiple magnets in a row. Determine the number of groups that the magnets formed.
Input Specification:
The first line of the input contains an integer *n* (1<=≤<=*n*<=≤<=100000) — the number of magnets. Then *n* lines follow. The *i*-th line (1<=≤<=*i*<=≤<=*n*) contains either characters "01", if Mike put the *i*-th magnet in the "plus-minus" position, or characters "10", if Mike put the magnet in the "minus-plus" position.
Output Specification:
On the single line of the output print the number of groups of magnets.
Demo Input:
['6\n10\n10\n10\n01\n10\n10\n', '4\n01\n01\n10\n10\n']
Demo Output:
['3\n', '2\n']
Note:
The first testcase corresponds to the figure. The testcase has three groups consisting of three, one and two magnets.
The second testcase has two groups, each consisting of two magnets.
|
```python
n=int(input())
magnets=[]
for i in range(n):
magnet=input()
magnets.append(magnet)
group=1
for i in range(len(magnets)-1):
if magnets[i]!=magnets[i+1]:
group+=1
print(group)
```
| 0
|
|
0
|
none
|
none
|
none
| 0
|
[
"none"
] | null | null |
Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are *n* watchmen on a plane, the *i*-th watchman is located at point (*x**i*,<=*y**i*).
They need to arrange a plan, but there are some difficulties on their way. As you know, Doctor Manhattan considers the distance between watchmen *i* and *j* to be |*x**i*<=-<=*x**j*|<=+<=|*y**i*<=-<=*y**j*|. Daniel, as an ordinary person, calculates the distance using the formula .
The success of the operation relies on the number of pairs (*i*,<=*j*) (1<=≤<=*i*<=<<=*j*<=≤<=*n*), such that the distance between watchman *i* and watchmen *j* calculated by Doctor Manhattan is equal to the distance between them calculated by Daniel. You were asked to compute the number of such pairs.
|
The first line of the input contains the single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of watchmen.
Each of the following *n* lines contains two integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=≤<=109).
Some positions may coincide.
|
Print the number of pairs of watchmen such that the distance between them calculated by Doctor Manhattan is equal to the distance calculated by Daniel.
|
[
"3\n1 1\n7 5\n1 5\n",
"6\n0 0\n0 1\n0 2\n-1 1\n0 1\n1 1\n"
] |
[
"2\n",
"11\n"
] |
In the first sample, the distance between watchman 1 and watchman 2 is equal to |1 - 7| + |1 - 5| = 10 for Doctor Manhattan and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bcb5b7064b5f02088da0fdcf677e6fda495dd0df.png" style="max-width: 100.0%;max-height: 100.0%;"/> for Daniel. For pairs (1, 1), (1, 5) and (7, 5), (1, 5) Doctor Manhattan and Daniel will calculate the same distances.
| 0
|
[
{
"input": "3\n1 1\n7 5\n1 5",
"output": "2"
},
{
"input": "6\n0 0\n0 1\n0 2\n-1 1\n0 1\n1 1",
"output": "11"
},
{
"input": "10\n46 -55\n46 45\n46 45\n83 -55\n46 45\n83 -55\n46 45\n83 45\n83 45\n46 -55",
"output": "33"
},
{
"input": "1\n-5 -90",
"output": "0"
},
{
"input": "2\n315 845\n-669 -762",
"output": "0"
},
{
"input": "3\n8911 7861\n-6888 7861\n8911 7861",
"output": "3"
},
{
"input": "2\n-1 1000000000\n0 -1",
"output": "0"
},
{
"input": "2\n1000000000 0\n-7 1",
"output": "0"
},
{
"input": "2\n1 4\n2 1",
"output": "0"
},
{
"input": "2\n1 0\n0 2333333",
"output": "0"
},
{
"input": "2\n2 1\n1 2",
"output": "0"
},
{
"input": "2\n1 1000000000\n2 -1000000000",
"output": "0"
},
{
"input": "2\n0 1000000000\n1 -7",
"output": "0"
},
{
"input": "2\n1 0\n0 19990213",
"output": "0"
}
] | 1,587,101,463
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 27
| 3,000
| 41,062,400
|
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
n = int(input())
l = []
for i in range(n):
l.append(list(map(int,input().split())))
xd = {}
yd = {}
dup = {}
for [x,y] in l:
if x not in xd.keys():
xd[x] = 1
else:
xd[x] += 1
if y not in yd.keys():
yd[y] = 1
else:
yd[y] += 1
if (x,y) not in dup.keys():
dup[(x,y)] = 1
else:
dup[(x,y)] += 1
count = 0
for x in xd.keys() :
if xd[x] != 1:
count += nCr(xd[x],2)
for y in yd.keys() :
if yd[y] != 1:
count += nCr(yd[y],2)
for (x,y) in dup.keys() :
if dup[(x,y)] != 1:
count -= nCr(dup[(x,y)],2)
print(count)
|
Title: none
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are *n* watchmen on a plane, the *i*-th watchman is located at point (*x**i*,<=*y**i*).
They need to arrange a plan, but there are some difficulties on their way. As you know, Doctor Manhattan considers the distance between watchmen *i* and *j* to be |*x**i*<=-<=*x**j*|<=+<=|*y**i*<=-<=*y**j*|. Daniel, as an ordinary person, calculates the distance using the formula .
The success of the operation relies on the number of pairs (*i*,<=*j*) (1<=≤<=*i*<=<<=*j*<=≤<=*n*), such that the distance between watchman *i* and watchmen *j* calculated by Doctor Manhattan is equal to the distance between them calculated by Daniel. You were asked to compute the number of such pairs.
Input Specification:
The first line of the input contains the single integer *n* (1<=≤<=*n*<=≤<=200<=000) — the number of watchmen.
Each of the following *n* lines contains two integers *x**i* and *y**i* (|*x**i*|,<=|*y**i*|<=≤<=109).
Some positions may coincide.
Output Specification:
Print the number of pairs of watchmen such that the distance between them calculated by Doctor Manhattan is equal to the distance calculated by Daniel.
Demo Input:
['3\n1 1\n7 5\n1 5\n', '6\n0 0\n0 1\n0 2\n-1 1\n0 1\n1 1\n']
Demo Output:
['2\n', '11\n']
Note:
In the first sample, the distance between watchman 1 and watchman 2 is equal to |1 - 7| + |1 - 5| = 10 for Doctor Manhattan and <img align="middle" class="tex-formula" src="https://espresso.codeforces.com/bcb5b7064b5f02088da0fdcf677e6fda495dd0df.png" style="max-width: 100.0%;max-height: 100.0%;"/> for Daniel. For pairs (1, 1), (1, 5) and (7, 5), (1, 5) Doctor Manhattan and Daniel will calculate the same distances.
|
```python
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
n = int(input())
l = []
for i in range(n):
l.append(list(map(int,input().split())))
xd = {}
yd = {}
dup = {}
for [x,y] in l:
if x not in xd.keys():
xd[x] = 1
else:
xd[x] += 1
if y not in yd.keys():
yd[y] = 1
else:
yd[y] += 1
if (x,y) not in dup.keys():
dup[(x,y)] = 1
else:
dup[(x,y)] += 1
count = 0
for x in xd.keys() :
if xd[x] != 1:
count += nCr(xd[x],2)
for y in yd.keys() :
if yd[y] != 1:
count += nCr(yd[y],2)
for (x,y) in dup.keys() :
if dup[(x,y)] != 1:
count -= nCr(dup[(x,y)],2)
print(count)
```
| 0
|
|
25
|
B
|
Phone numbers
|
PROGRAMMING
| 1,100
|
[
"implementation"
] |
B. Phone numbers
|
2
|
256
|
Phone number in Berland is a sequence of *n* digits. Often, to make it easier to memorize the number, it is divided into groups of two or three digits. For example, the phone number 1198733 is easier to remember as 11-987-33. Your task is to find for a given phone number any of its divisions into groups of two or three digits.
|
The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of digits in the phone number. The second line contains *n* digits — the phone number to divide into groups.
|
Output any of divisions of the given phone number into groups of two or three digits. Separate groups by single character -. If the answer is not unique, output any.
|
[
"6\n549871\n",
"7\n1198733\n"
] |
[
"54-98-71",
"11-987-33\n"
] |
none
| 0
|
[
{
"input": "6\n549871",
"output": "54-98-71"
},
{
"input": "7\n1198733",
"output": "119-87-33"
},
{
"input": "2\n74",
"output": "74"
},
{
"input": "2\n33",
"output": "33"
},
{
"input": "3\n074",
"output": "074"
},
{
"input": "3\n081",
"output": "081"
},
{
"input": "4\n3811",
"output": "38-11"
},
{
"input": "5\n21583",
"output": "215-83"
},
{
"input": "8\n33408349",
"output": "33-40-83-49"
},
{
"input": "9\n988808426",
"output": "988-80-84-26"
},
{
"input": "10\n0180990956",
"output": "01-80-99-09-56"
},
{
"input": "15\n433488906230138",
"output": "433-48-89-06-23-01-38"
},
{
"input": "22\n7135498415686025907059",
"output": "71-35-49-84-15-68-60-25-90-70-59"
},
{
"input": "49\n2429965524999668169991253653390090510755018570235",
"output": "242-99-65-52-49-99-66-81-69-99-12-53-65-33-90-09-05-10-75-50-18-57-02-35"
},
{
"input": "72\n491925337784111770500147619881727525570039735507439360627744863794794290",
"output": "49-19-25-33-77-84-11-17-70-50-01-47-61-98-81-72-75-25-57-00-39-73-55-07-43-93-60-62-77-44-86-37-94-79-42-90"
},
{
"input": "95\n32543414456047900690980198395035321172843693417425457554204776648220562494524275489599199209210",
"output": "325-43-41-44-56-04-79-00-69-09-80-19-83-95-03-53-21-17-28-43-69-34-17-42-54-57-55-42-04-77-66-48-22-05-62-49-45-24-27-54-89-59-91-99-20-92-10"
},
{
"input": "97\n9362344595153688016434451101547661156123505108492010669557671355055642365998461003851354321478898",
"output": "936-23-44-59-51-53-68-80-16-43-44-51-10-15-47-66-11-56-12-35-05-10-84-92-01-06-69-55-76-71-35-50-55-64-23-65-99-84-61-00-38-51-35-43-21-47-88-98"
},
{
"input": "98\n65521815795893886057122984634320900545031770769333931308009346017867969790810907868670369236928568",
"output": "65-52-18-15-79-58-93-88-60-57-12-29-84-63-43-20-90-05-45-03-17-70-76-93-33-93-13-08-00-93-46-01-78-67-96-97-90-81-09-07-86-86-70-36-92-36-92-85-68"
},
{
"input": "99\n455213856470326729480192345541970106407563996625458559297407682539801838244443866898560852503660390",
"output": "455-21-38-56-47-03-26-72-94-80-19-23-45-54-19-70-10-64-07-56-39-96-62-54-58-55-92-97-40-76-82-53-98-01-83-82-44-44-38-66-89-85-60-85-25-03-66-03-90"
},
{
"input": "100\n4004223124942730640235383244438257614581534320356060987241659784249551110165034719443327659510644224",
"output": "40-04-22-31-24-94-27-30-64-02-35-38-32-44-43-82-57-61-45-81-53-43-20-35-60-60-98-72-41-65-97-84-24-95-51-11-01-65-03-47-19-44-33-27-65-95-10-64-42-24"
}
] | 1,671,875,638
| 2,038
|
PyPy 3-64
|
OK
|
TESTS
| 20
| 124
| 0
|
n = int(input())
s = input()
ans=""
if(n%2==0):
for i in range(0,n,2):
if(i<n-2):
ans = ans + s[i:i+2]+"-"
if(i==n-2):
ans = ans + s[i:i+2]
else :
j = n-3
k = 0
while (k<j):
ans = ans + s[k:k+2]+"-"
k+=2
ans = ans+s[j:]
print(ans)
|
Title: Phone numbers
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Phone number in Berland is a sequence of *n* digits. Often, to make it easier to memorize the number, it is divided into groups of two or three digits. For example, the phone number 1198733 is easier to remember as 11-987-33. Your task is to find for a given phone number any of its divisions into groups of two or three digits.
Input Specification:
The first line contains integer *n* (2<=≤<=*n*<=≤<=100) — amount of digits in the phone number. The second line contains *n* digits — the phone number to divide into groups.
Output Specification:
Output any of divisions of the given phone number into groups of two or three digits. Separate groups by single character -. If the answer is not unique, output any.
Demo Input:
['6\n549871\n', '7\n1198733\n']
Demo Output:
['54-98-71', '11-987-33\n']
Note:
none
|
```python
n = int(input())
s = input()
ans=""
if(n%2==0):
for i in range(0,n,2):
if(i<n-2):
ans = ans + s[i:i+2]+"-"
if(i==n-2):
ans = ans + s[i:i+2]
else :
j = n-3
k = 0
while (k<j):
ans = ans + s[k:k+2]+"-"
k+=2
ans = ans+s[j:]
print(ans)
```
| 3.969
|
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"
}
] | 1,690,039,718
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 1
| 31
| 0
|
inp=input()
pos=inp.index("|")
world1=inp[:pos]
world2=inp[pos+1:]
rest=input()
len3=len(rest)
ans=False
if len(world1)<=len(world2) and pos!=0:
ans=True
rec=len(world2)-len(world1)
len3-=rec
if len3<0 or len3%2==1:
print("Impossible")
exit(0)
world1+=rest[:rec+int(len3/2)]
if len3!=0:
world2+=rest[len3:]
elif len(world1)>=len(world2) and pos!=0:
ans=True
rec=len(world1)-len(world2)
len3-=rec
if len3<0 or len3%2==1 :
print("Impossible")
exit(0)
world2+=rest[:rec+int(len3/2)]
if len3!=0:
world1+=rest[len3:]
if ans:
print(world1+"|"+world2)
else:
if len3%2==0:
print(rest[0:int(len3/2)]+"|"+rest[int(len3/2):])
else:
print("Impossible")
|
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
inp=input()
pos=inp.index("|")
world1=inp[:pos]
world2=inp[pos+1:]
rest=input()
len3=len(rest)
ans=False
if len(world1)<=len(world2) and pos!=0:
ans=True
rec=len(world2)-len(world1)
len3-=rec
if len3<0 or len3%2==1:
print("Impossible")
exit(0)
world1+=rest[:rec+int(len3/2)]
if len3!=0:
world2+=rest[len3:]
elif len(world1)>=len(world2) and pos!=0:
ans=True
rec=len(world1)-len(world2)
len3-=rec
if len3<0 or len3%2==1 :
print("Impossible")
exit(0)
world2+=rest[:rec+int(len3/2)]
if len3!=0:
world1+=rest[len3:]
if ans:
print(world1+"|"+world2)
else:
if len3%2==0:
print(rest[0:int(len3/2)]+"|"+rest[int(len3/2):])
else:
print("Impossible")
```
| 0
|
|
596
|
A
|
Wilbur and Swimming Pool
|
PROGRAMMING
| 1,100
|
[
"geometry",
"implementation"
] | null | null |
After making bad dives into swimming pools, Wilbur wants to build a swimming pool in the shape of a rectangle in his backyard. He has set up coordinate axes, and he wants the sides of the rectangle to be parallel to them. Of course, the area of the rectangle must be positive. Wilbur had all four vertices of the planned pool written on a paper, until his friend came along and erased some of the vertices.
Now Wilbur is wondering, if the remaining *n* vertices of the initial rectangle give enough information to restore the area of the planned swimming pool.
|
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=4) — the number of vertices that were not erased by Wilbur's friend.
Each of the following *n* lines contains two integers *x**i* and *y**i* (<=-<=1000<=≤<=*x**i*,<=*y**i*<=≤<=1000) —the coordinates of the *i*-th vertex that remains. Vertices are given in an arbitrary order.
It's guaranteed that these points are distinct vertices of some rectangle, that has positive area and which sides are parallel to the coordinate axes.
|
Print the area of the initial rectangle if it could be uniquely determined by the points remaining. Otherwise, print <=-<=1.
|
[
"2\n0 0\n1 1\n",
"1\n1 1\n"
] |
[
"1\n",
"-1\n"
] |
In the first sample, two opposite corners of the initial rectangle are given, and that gives enough information to say that the rectangle is actually a unit square.
In the second sample there is only one vertex left and this is definitely not enough to uniquely define the area.
| 500
|
[
{
"input": "2\n0 0\n1 1",
"output": "1"
},
{
"input": "1\n1 1",
"output": "-1"
},
{
"input": "1\n-188 17",
"output": "-1"
},
{
"input": "1\n71 -740",
"output": "-1"
},
{
"input": "4\n-56 -858\n-56 -174\n778 -858\n778 -174",
"output": "570456"
},
{
"input": "2\n14 153\n566 -13",
"output": "91632"
},
{
"input": "2\n-559 894\n314 127",
"output": "669591"
},
{
"input": "1\n-227 -825",
"output": "-1"
},
{
"input": "2\n-187 583\n25 13",
"output": "120840"
},
{
"input": "2\n-337 451\n32 -395",
"output": "312174"
},
{
"input": "4\n-64 -509\n-64 960\n634 -509\n634 960",
"output": "1025362"
},
{
"input": "2\n-922 -505\n712 -683",
"output": "290852"
},
{
"input": "2\n-1000 -1000\n-1000 0",
"output": "-1"
},
{
"input": "2\n-1000 -1000\n0 -1000",
"output": "-1"
},
{
"input": "4\n-414 -891\n-414 896\n346 -891\n346 896",
"output": "1358120"
},
{
"input": "2\n56 31\n704 -121",
"output": "98496"
},
{
"input": "4\n-152 198\n-152 366\n458 198\n458 366",
"output": "102480"
},
{
"input": "3\n-890 778\n-418 296\n-890 296",
"output": "227504"
},
{
"input": "4\n852 -184\n852 724\n970 -184\n970 724",
"output": "107144"
},
{
"input": "1\n858 -279",
"output": "-1"
},
{
"input": "2\n-823 358\n446 358",
"output": "-1"
},
{
"input": "2\n-739 -724\n-739 443",
"output": "-1"
},
{
"input": "2\n686 664\n686 -590",
"output": "-1"
},
{
"input": "3\n-679 301\n240 -23\n-679 -23",
"output": "297756"
},
{
"input": "2\n-259 -978\n978 -978",
"output": "-1"
},
{
"input": "1\n627 -250",
"output": "-1"
},
{
"input": "3\n-281 598\n679 -990\n-281 -990",
"output": "1524480"
},
{
"input": "2\n-414 -431\n-377 -688",
"output": "9509"
},
{
"input": "3\n-406 566\n428 426\n-406 426",
"output": "116760"
},
{
"input": "3\n-686 695\n-547 308\n-686 308",
"output": "53793"
},
{
"input": "1\n-164 -730",
"output": "-1"
},
{
"input": "2\n980 -230\n980 592",
"output": "-1"
},
{
"input": "4\n-925 306\n-925 602\n398 306\n398 602",
"output": "391608"
},
{
"input": "3\n576 -659\n917 -739\n576 -739",
"output": "27280"
},
{
"input": "1\n720 -200",
"output": "-1"
},
{
"input": "4\n-796 -330\n-796 758\n171 -330\n171 758",
"output": "1052096"
},
{
"input": "2\n541 611\n-26 611",
"output": "-1"
},
{
"input": "3\n-487 838\n134 691\n-487 691",
"output": "91287"
},
{
"input": "2\n-862 -181\n-525 -181",
"output": "-1"
},
{
"input": "1\n-717 916",
"output": "-1"
},
{
"input": "1\n-841 -121",
"output": "-1"
},
{
"input": "4\n259 153\n259 999\n266 153\n266 999",
"output": "5922"
},
{
"input": "2\n295 710\n295 254",
"output": "-1"
},
{
"input": "4\n137 -184\n137 700\n712 -184\n712 700",
"output": "508300"
},
{
"input": "2\n157 994\n377 136",
"output": "188760"
},
{
"input": "1\n193 304",
"output": "-1"
},
{
"input": "4\n5 -952\n5 292\n553 -952\n553 292",
"output": "681712"
},
{
"input": "2\n-748 697\n671 575",
"output": "173118"
},
{
"input": "2\n-457 82\n260 -662",
"output": "533448"
},
{
"input": "2\n-761 907\n967 907",
"output": "-1"
},
{
"input": "3\n-639 51\n-321 -539\n-639 -539",
"output": "187620"
},
{
"input": "2\n-480 51\n89 -763",
"output": "463166"
},
{
"input": "4\n459 -440\n459 -94\n872 -440\n872 -94",
"output": "142898"
},
{
"input": "2\n380 -849\n68 -849",
"output": "-1"
},
{
"input": "2\n-257 715\n102 715",
"output": "-1"
},
{
"input": "2\n247 -457\n434 -921",
"output": "86768"
},
{
"input": "4\n-474 -894\n-474 -833\n-446 -894\n-446 -833",
"output": "1708"
},
{
"input": "3\n-318 831\n450 31\n-318 31",
"output": "614400"
},
{
"input": "3\n-282 584\n696 488\n-282 488",
"output": "93888"
},
{
"input": "3\n258 937\n395 856\n258 856",
"output": "11097"
},
{
"input": "1\n-271 -499",
"output": "-1"
},
{
"input": "2\n-612 208\n326 -559",
"output": "719446"
},
{
"input": "2\n115 730\n562 -546",
"output": "570372"
},
{
"input": "2\n-386 95\n-386 750",
"output": "-1"
},
{
"input": "3\n0 0\n0 1\n1 0",
"output": "1"
},
{
"input": "3\n0 4\n3 4\n3 1",
"output": "9"
},
{
"input": "3\n1 1\n1 2\n2 1",
"output": "1"
},
{
"input": "3\n1 4\n4 4\n4 1",
"output": "9"
},
{
"input": "3\n1 1\n2 1\n1 2",
"output": "1"
},
{
"input": "3\n0 0\n1 0\n1 1",
"output": "1"
},
{
"input": "3\n0 0\n0 5\n5 0",
"output": "25"
},
{
"input": "3\n0 0\n0 1\n1 1",
"output": "1"
},
{
"input": "4\n0 0\n1 0\n1 1\n0 1",
"output": "1"
},
{
"input": "3\n4 4\n1 4\n4 1",
"output": "9"
},
{
"input": "3\n0 0\n2 0\n2 1",
"output": "2"
},
{
"input": "3\n0 0\n2 0\n0 2",
"output": "4"
},
{
"input": "3\n0 0\n0 1\n5 0",
"output": "5"
},
{
"input": "3\n1 1\n1 3\n3 1",
"output": "4"
},
{
"input": "4\n0 0\n1 0\n0 1\n1 1",
"output": "1"
},
{
"input": "2\n1 0\n2 1",
"output": "1"
},
{
"input": "3\n0 0\n1 0\n0 1",
"output": "1"
},
{
"input": "3\n1 0\n0 0\n0 1",
"output": "1"
},
{
"input": "3\n0 0\n0 5\n5 5",
"output": "25"
},
{
"input": "3\n1 0\n5 0\n5 10",
"output": "40"
},
{
"input": "3\n0 0\n1 0\n1 2",
"output": "2"
},
{
"input": "4\n0 1\n0 0\n1 0\n1 1",
"output": "1"
},
{
"input": "3\n0 0\n2 0\n0 1",
"output": "2"
},
{
"input": "3\n-2 -1\n-1 -1\n-1 -2",
"output": "1"
},
{
"input": "2\n1 0\n0 1",
"output": "1"
},
{
"input": "4\n1 1\n3 3\n3 1\n1 3",
"output": "4"
},
{
"input": "3\n2 1\n1 2\n2 2",
"output": "1"
},
{
"input": "3\n0 0\n0 3\n3 0",
"output": "9"
},
{
"input": "2\n0 3\n3 3",
"output": "-1"
},
{
"input": "4\n2 0\n2 8\n5 8\n5 0",
"output": "24"
},
{
"input": "2\n0 999\n100 250",
"output": "74900"
},
{
"input": "3\n1 1\n1 5\n5 1",
"output": "16"
},
{
"input": "3\n0 1\n0 0\n1 1",
"output": "1"
},
{
"input": "3\n0 0\n10 0\n0 10",
"output": "100"
},
{
"input": "2\n0 0\n-1 -1",
"output": "1"
},
{
"input": "3\n1 5\n2 2\n2 5",
"output": "3"
},
{
"input": "3\n0 0\n0 1\n2 0",
"output": "2"
},
{
"input": "3\n0 1\n1 0\n0 0",
"output": "1"
},
{
"input": "3\n0 0\n0 -1\n1 -1",
"output": "1"
},
{
"input": "3\n0 1\n1 0\n1 1",
"output": "1"
},
{
"input": "3\n3 5\n3 2\n7 2",
"output": "12"
},
{
"input": "3\n1 2\n1 3\n2 2",
"output": "1"
},
{
"input": "3\n5 0\n0 0\n0 5",
"output": "25"
},
{
"input": "3\n1 0\n1 3\n5 0",
"output": "12"
},
{
"input": "3\n0 0\n0 2\n2 0",
"output": "4"
},
{
"input": "3\n1 1\n0 0\n1 0",
"output": "1"
},
{
"input": "3\n1 2\n1 3\n2 3",
"output": "1"
},
{
"input": "4\n0 0\n0 1\n1 1\n1 0",
"output": "1"
},
{
"input": "2\n-3 0\n3 3",
"output": "18"
},
{
"input": "3\n1 1\n0 1\n1 0",
"output": "1"
},
{
"input": "3\n0 0\n5 0\n5 5",
"output": "25"
},
{
"input": "3\n79 79\n79 158\n158 79",
"output": "6241"
},
{
"input": "3\n1 0\n1 -1\n0 0",
"output": "1"
},
{
"input": "3\n1 1\n1 2\n2 2",
"output": "1"
},
{
"input": "3\n0 1\n0 0\n1 0",
"output": "1"
},
{
"input": "3\n2 1\n2 4\n6 1",
"output": "12"
},
{
"input": "3\n5 0\n0 0\n5 5",
"output": "25"
}
] | 1,649,230,194
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 121
| 62
| 0
|
import math
def rl():
return map(int,input().split())
def ri():
return int(input())
n=ri()
x=[]
y=[]
while n:
a,b=rl()
x.append(a)
y.append(b)
n-=1
s=(max(x)-min(x))*(max(y)-min(y))
if s==0:
print(-1)
else:
print(s)
|
Title: Wilbur and Swimming Pool
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After making bad dives into swimming pools, Wilbur wants to build a swimming pool in the shape of a rectangle in his backyard. He has set up coordinate axes, and he wants the sides of the rectangle to be parallel to them. Of course, the area of the rectangle must be positive. Wilbur had all four vertices of the planned pool written on a paper, until his friend came along and erased some of the vertices.
Now Wilbur is wondering, if the remaining *n* vertices of the initial rectangle give enough information to restore the area of the planned swimming pool.
Input Specification:
The first line of the input contains a single integer *n* (1<=≤<=*n*<=≤<=4) — the number of vertices that were not erased by Wilbur's friend.
Each of the following *n* lines contains two integers *x**i* and *y**i* (<=-<=1000<=≤<=*x**i*,<=*y**i*<=≤<=1000) —the coordinates of the *i*-th vertex that remains. Vertices are given in an arbitrary order.
It's guaranteed that these points are distinct vertices of some rectangle, that has positive area and which sides are parallel to the coordinate axes.
Output Specification:
Print the area of the initial rectangle if it could be uniquely determined by the points remaining. Otherwise, print <=-<=1.
Demo Input:
['2\n0 0\n1 1\n', '1\n1 1\n']
Demo Output:
['1\n', '-1\n']
Note:
In the first sample, two opposite corners of the initial rectangle are given, and that gives enough information to say that the rectangle is actually a unit square.
In the second sample there is only one vertex left and this is definitely not enough to uniquely define the area.
|
```python
import math
def rl():
return map(int,input().split())
def ri():
return int(input())
n=ri()
x=[]
y=[]
while n:
a,b=rl()
x.append(a)
y.append(b)
n-=1
s=(max(x)-min(x))*(max(y)-min(y))
if s==0:
print(-1)
else:
print(s)
```
| 3
|
|
440
|
A
|
Forgotten Episode
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Polycarpus adores TV series. Right now he is ready to finish watching a season of a popular sitcom "Graph Theory". In total, the season has *n* episodes, numbered with integers from 1 to *n*.
Polycarpus watches episodes not one by one but in a random order. He has already watched all the episodes except for one. Which episode has Polycaprus forgotten to watch?
|
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100000) — the number of episodes in a season. Assume that the episodes are numbered by integers from 1 to *n*.
The second line contains *n*<=-<=1 integer *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the numbers of episodes that Polycarpus has watched. All values of *a**i* are distinct.
|
Print the number of the episode that Polycarpus hasn't watched.
|
[
"10\n3 8 10 1 7 9 6 5 2\n"
] |
[
"4\n"
] |
none
| 500
|
[
{
"input": "10\n3 8 10 1 7 9 6 5 2",
"output": "4"
},
{
"input": "5\n4 3 2 1",
"output": "5"
},
{
"input": "2\n1",
"output": "2"
},
{
"input": "2\n2",
"output": "1"
},
{
"input": "3\n1 2",
"output": "3"
},
{
"input": "3\n1 3",
"output": "2"
},
{
"input": "3\n2 3",
"output": "1"
},
{
"input": "3\n2 1",
"output": "3"
},
{
"input": "3\n3 1",
"output": "2"
},
{
"input": "3\n3 2",
"output": "1"
},
{
"input": "5\n2 3 4 5",
"output": "1"
},
{
"input": "5\n5 4 3 2",
"output": "1"
},
{
"input": "5\n5 2 4 3",
"output": "1"
},
{
"input": "5\n1 2 3 4",
"output": "5"
},
{
"input": "5\n2 1 3 4",
"output": "5"
},
{
"input": "5\n1 5 3 4",
"output": "2"
},
{
"input": "5\n1 4 5 2",
"output": "3"
},
{
"input": "5\n2 1 5 3",
"output": "4"
},
{
"input": "5\n2 3 4 5",
"output": "1"
}
] | 1,691,772,486
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 35
| 93
| 17,510,400
|
n = int(input())
L = set(map(int, input().split())) # Use a set for O(1) membership check
missing = set(range(1, n+1)) - L # Calculate the missing elements
print(min(missing)) # Print the smallest missing element
|
Title: Forgotten Episode
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarpus adores TV series. Right now he is ready to finish watching a season of a popular sitcom "Graph Theory". In total, the season has *n* episodes, numbered with integers from 1 to *n*.
Polycarpus watches episodes not one by one but in a random order. He has already watched all the episodes except for one. Which episode has Polycaprus forgotten to watch?
Input Specification:
The first line of the input contains integer *n* (2<=≤<=*n*<=≤<=100000) — the number of episodes in a season. Assume that the episodes are numbered by integers from 1 to *n*.
The second line contains *n*<=-<=1 integer *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=*n*) — the numbers of episodes that Polycarpus has watched. All values of *a**i* are distinct.
Output Specification:
Print the number of the episode that Polycarpus hasn't watched.
Demo Input:
['10\n3 8 10 1 7 9 6 5 2\n']
Demo Output:
['4\n']
Note:
none
|
```python
n = int(input())
L = set(map(int, input().split())) # Use a set for O(1) membership check
missing = set(range(1, n+1)) - L # Calculate the missing elements
print(min(missing)) # Print the smallest missing element
```
| 3
|
|
612
|
C
|
Replace To Make Regular Bracket Sequence
|
PROGRAMMING
| 1,400
|
[
"data structures",
"expression parsing",
"math"
] | null | null |
You are given string *s* consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let *s*1 and *s*2 be a RBS then the strings <*s*1>*s*2, {*s*1}*s*2, [*s*1]*s*2, (*s*1)*s*2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string *s* RBS.
|
The only line contains a non empty string *s*, consisting of only opening and closing brackets of four kinds. The length of *s* does not exceed 106.
|
If it's impossible to get RBS from *s* print Impossible.
Otherwise print the least number of replaces needed to get RBS from *s*.
|
[
"[<}){}\n",
"{()}[]\n",
"]]\n"
] |
[
"2",
"0",
"Impossible"
] |
none
| 0
|
[
{
"input": "[<}){}",
"output": "2"
},
{
"input": "{()}[]",
"output": "0"
},
{
"input": "]]",
"output": "Impossible"
},
{
"input": ">",
"output": "Impossible"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{}",
"output": "0"
},
{
"input": "{]",
"output": "1"
},
{
"input": "{]",
"output": "1"
},
{
"input": "{]",
"output": "1"
},
{
"input": "[]{[]({)([",
"output": "Impossible"
},
{
"input": "(([{>}{[{[)]]>>]",
"output": "7"
},
{
"input": "((<>)[]<]><]",
"output": "3"
},
{
"input": "[[([[(>]>)))[<)>",
"output": "6"
},
{
"input": "({)[}<)](}",
"output": "5"
},
{
"input": "(}{)[<][)(]}",
"output": "6"
},
{
"input": ">}({>]{[}<{<{{)[]]{)]>]]]<(][{)<<<{<<)>)()[>{<]]{}<>}}}}(>}<})(][>{((<{<)]}>)))][>[}[])<]){]]][<[)([",
"output": "Impossible"
},
{
"input": "<<[<{{<([({<<[)<>(]]){})>[](])[)))[[}>]<)>[[>{>>>[<]}<>>)[>]<{)<[><(<]][>(>]>][(<][{]}(()<[()[>><<])<]})]<]}{)",
"output": "45"
},
{
"input": "[<<{{((}[}<<)<)>})(][{>}})((>)<[)[>}[})[)>()[()[((}<<(>)<>](<>(}[>})[[[{)<}<<(}{>>}[<([[])<><)]<{>}[>>>{({>)}]})>)",
"output": "42"
},
{
"input": "(<[([(<({>(}{]>[(})])}])()<<}{]{[>]>(>>[(>>}[){(}<[{(()]{{<(<{][[{<><{<{)<>>]}}}{)(}{})}[<))>>}((({>){({}{{]}]>>}})>))",
"output": "43"
},
{
"input": "{(]}<([]<]{>]<{<({{{<>))}[({(}{)[}({>]}}<<)}<]))(<>(>{>{{{)<}({<]<>{)(>[)>{({}<([<[[)]><>{]}}(>]{}<)[})]}]]}]}>}",
"output": "40"
},
{
"input": "<(([)]<[}>{)>][[(<()({{{>>((]<}<{{<>}><<[)}[[)([[)[)}<]<{(<>){([)><{[([[][({])})<][(}]}>>[){)[({>){(}[}})>}>]>)]}))})>}]",
"output": "45"
},
{
"input": ")(",
"output": "Impossible"
},
{
"input": "[[[[[[[[",
"output": "Impossible"
}
] | 1,614,776,144
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 6
| 46
| 307,200
|
try:
s=input()
p=s
p='M'+p
def io(x):
x='M'+x
while '()' in x:
x=x.replace('()','')
if len(x)==1:
return True
else:
return False
while '()' in p or '{}' in p or '<>' in p or '[]' in p:
if '()' in p:
p=p.replace('()','')
if '[]' in p:
p=p.replace('[]','')
if '<>' in p:
p=p.replace('<>','')
if '{}' in p:
p=p.replace('{}','')
if len(p)==1:
print(0)
elif len(p)==len(s)+1:
print('Impossible')
else:
p=p.replace('M','')
if len(p)%2==1:
print("Impossible")
else:
p=list(p)
a=p.count('[')
a1=p.count(']')
b=p.count('(')
b1=p.count(')')
c=p.count('{')
c1=p.count('}')
d=p.count("<")
d1=p.count(">")
if (a+b+c+d)!=a1+b1+c1+d1:
print("Impossible")
else:
p=''.join(p)
t=p
if '[' in p:
p=p.replace('[','(')
if '<' in p:
p=p.replace('<','(')
if '{' in p:
p=p.replace('{','(')
if ']' in p:
p=p.replace(']',')')
if '>' in p:
p=p.replace('>',')')
if '}' in p:
p=p.replace('}',')')
if io(p)==True:
op0=abs(a-a1)
op1=abs(b-b1)
op2=abs(c-c1)
op3=abs(d-d1)
lo=[op0,op1,op2,op3]
lp=max(lo)
lo.remove(lp)
print(sum(lo)-lp)
else:
print("Impossible")
except:
pass
|
Title: Replace To Make Regular Bracket Sequence
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given string *s* consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.
The following definition of a regular bracket sequence is well-known, so you can be familiar with it.
Let's define a regular bracket sequence (RBS). Empty string is RBS. Let *s*1 and *s*2 be a RBS then the strings <*s*1>*s*2, {*s*1}*s*2, [*s*1]*s*2, (*s*1)*s*2 are also RBS.
For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.
Determine the least number of replaces to make the string *s* RBS.
Input Specification:
The only line contains a non empty string *s*, consisting of only opening and closing brackets of four kinds. The length of *s* does not exceed 106.
Output Specification:
If it's impossible to get RBS from *s* print Impossible.
Otherwise print the least number of replaces needed to get RBS from *s*.
Demo Input:
['[<}){}\n', '{()}[]\n', ']]\n']
Demo Output:
['2', '0', 'Impossible']
Note:
none
|
```python
try:
s=input()
p=s
p='M'+p
def io(x):
x='M'+x
while '()' in x:
x=x.replace('()','')
if len(x)==1:
return True
else:
return False
while '()' in p or '{}' in p or '<>' in p or '[]' in p:
if '()' in p:
p=p.replace('()','')
if '[]' in p:
p=p.replace('[]','')
if '<>' in p:
p=p.replace('<>','')
if '{}' in p:
p=p.replace('{}','')
if len(p)==1:
print(0)
elif len(p)==len(s)+1:
print('Impossible')
else:
p=p.replace('M','')
if len(p)%2==1:
print("Impossible")
else:
p=list(p)
a=p.count('[')
a1=p.count(']')
b=p.count('(')
b1=p.count(')')
c=p.count('{')
c1=p.count('}')
d=p.count("<")
d1=p.count(">")
if (a+b+c+d)!=a1+b1+c1+d1:
print("Impossible")
else:
p=''.join(p)
t=p
if '[' in p:
p=p.replace('[','(')
if '<' in p:
p=p.replace('<','(')
if '{' in p:
p=p.replace('{','(')
if ']' in p:
p=p.replace(']',')')
if '>' in p:
p=p.replace('>',')')
if '}' in p:
p=p.replace('}',')')
if io(p)==True:
op0=abs(a-a1)
op1=abs(b-b1)
op2=abs(c-c1)
op3=abs(d-d1)
lo=[op0,op1,op2,op3]
lp=max(lo)
lo.remove(lp)
print(sum(lo)-lp)
else:
print("Impossible")
except:
pass
```
| 0
|
|
825
|
A
|
Binary Protocol
|
PROGRAMMING
| 1,100
|
[
"implementation"
] | null | null |
Polycarp has just invented a new binary protocol for data transmission. He is encoding positive integer decimal number to binary string using following algorithm:
- Each digit is represented with number of '1' characters equal to the value of that digit (for 0 it is zero ones). - Digits are written one by one in order corresponding to number and separated by single '0' character.
Though Polycarp learnt how to encode the numbers, he has no idea how to decode them back. Help him calculate the decoded number.
|
The first line contains one integer number *n* (1<=≤<=*n*<=≤<=89) — length of the string *s*.
The second line contains string *s* — sequence of '0' and '1' characters, number in its encoded format. It is guaranteed that the number corresponding to the string is positive and doesn't exceed 109. The string always starts with '1'.
|
Print the decoded number.
|
[
"3\n111\n",
"9\n110011101\n"
] |
[
"3\n",
"2031\n"
] |
none
| 0
|
[
{
"input": "3\n111",
"output": "3"
},
{
"input": "9\n110011101",
"output": "2031"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "3\n100",
"output": "100"
},
{
"input": "5\n10001",
"output": "1001"
},
{
"input": "14\n11001100011000",
"output": "202002000"
},
{
"input": "31\n1000011111111100011110111111111",
"output": "100090049"
},
{
"input": "53\n10110111011110111110111111011111110111111110111111111",
"output": "123456789"
},
{
"input": "89\n11111111101111111110111111111011111111101111111110111111111011111111101111111110111111111",
"output": "999999999"
},
{
"input": "10\n1000000000",
"output": "1000000000"
},
{
"input": "2\n10",
"output": "10"
},
{
"input": "4\n1110",
"output": "30"
},
{
"input": "8\n10101010",
"output": "11110"
}
] | 1,608,817,462
| 2,147,483,647
|
PyPy 3
|
WRONG_ANSWER
|
TESTS
| 3
| 124
| 0
|
n = int(input())
s = str(input())
res, p = '', 0
while p < n:
if p + 1 < n:
if s[p] == '0' and s[p + 1] == '0':
res += '0'
p += 2
continue
if s[p] == '1':
count = 1
p += 1
while p < n:
if s[p] == '1':
p += 1
count += 1
else:
break
res += str(count)
else:
p += 1
if res == '':
print(0)
else:
print(res)
|
Title: Binary Protocol
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Polycarp has just invented a new binary protocol for data transmission. He is encoding positive integer decimal number to binary string using following algorithm:
- Each digit is represented with number of '1' characters equal to the value of that digit (for 0 it is zero ones). - Digits are written one by one in order corresponding to number and separated by single '0' character.
Though Polycarp learnt how to encode the numbers, he has no idea how to decode them back. Help him calculate the decoded number.
Input Specification:
The first line contains one integer number *n* (1<=≤<=*n*<=≤<=89) — length of the string *s*.
The second line contains string *s* — sequence of '0' and '1' characters, number in its encoded format. It is guaranteed that the number corresponding to the string is positive and doesn't exceed 109. The string always starts with '1'.
Output Specification:
Print the decoded number.
Demo Input:
['3\n111\n', '9\n110011101\n']
Demo Output:
['3\n', '2031\n']
Note:
none
|
```python
n = int(input())
s = str(input())
res, p = '', 0
while p < n:
if p + 1 < n:
if s[p] == '0' and s[p + 1] == '0':
res += '0'
p += 2
continue
if s[p] == '1':
count = 1
p += 1
while p < n:
if s[p] == '1':
p += 1
count += 1
else:
break
res += str(count)
else:
p += 1
if res == '':
print(0)
else:
print(res)
```
| 0
|
|
600
|
F
|
Edge coloring of bipartite graph
|
PROGRAMMING
| 2,800
|
[
"graphs"
] | null | null |
You are given an undirected bipartite graph without multiple edges. You should paint the edges of graph to minimal number of colours, so that no two adjacent edges have the same colour.
|
The first line contains three integers *a*,<=*b*,<=*m* (1<=≤<=*a*,<=*b*<=≤<=1000, 0<=≤<=*m*<=≤<=105), *a* is the size of the first part, *b* is the size of the second part, *m* is the number of edges in the graph.
Each of the next *m* lines contains two integers *x*,<=*y* (1<=≤<=*x*<=≤<=*a*,<=1<=≤<=*y*<=≤<=*b*), where *x* is the number of the vertex in the first part and *y* is the number of the vertex in the second part. It is guaranteed that there are no multiple edges.
|
In the first line print integer *c* — the minimal number of colours. The second line should contain *m* integers from 1 to *c* — the colours of the edges (in the order they appear in the input).
If there are several solutions, you can print any one of them.
|
[
"4 3 5\n1 2\n2 2\n3 2\n4 1\n4 3\n"
] |
[
"3\n1 2 3 1 2\n"
] |
none
| 0
|
[
{
"input": "4 3 5\n1 2\n2 2\n3 2\n4 1\n4 3",
"output": "3\n1 2 3 1 2"
},
{
"input": "4 3 5\n1 2\n2 2\n3 2\n4 1\n4 3",
"output": "3\n1 2 3 1 2"
},
{
"input": "4 3 0",
"output": "0"
},
{
"input": "10 10 67\n1 1\n1 2\n1 3\n1 7\n1 9\n1 10\n2 1\n2 2\n2 3\n2 6\n2 8\n2 10\n3 2\n3 3\n3 6\n3 8\n3 9\n3 10\n4 1\n4 4\n4 5\n4 6\n4 7\n4 8\n5 2\n5 4\n5 7\n5 8\n5 9\n5 10\n6 1\n6 2\n6 3\n6 4\n6 6\n6 8\n6 9\n6 10\n7 2\n7 4\n7 6\n7 9\n7 10\n8 3\n8 4\n8 5\n8 6\n8 7\n8 8\n8 9\n8 10\n9 1\n9 2\n9 3\n9 5\n9 6\n9 7\n9 8\n9 9\n9 10\n10 1\n10 3\n10 4\n10 5\n10 8\n10 9\n10 10",
"output": "9\n3 2 1 4 5 6 2 6 4 3 5 1 1 2 4 3 6 5 1 3 4 5 6 2 3 5 1 6 4 2 4 5 3 2 6 1 7 8 4 8 2 3 7 5 4 6 7 2 8 1 3 5 7 6 1 8 3 4 2 9 6 7 1 2 9 8 4"
},
{
"input": "10 10 27\n1 10\n2 1\n2 3\n2 6\n2 8\n3 2\n3 4\n3 5\n4 1\n4 3\n4 5\n5 2\n5 5\n5 6\n6 1\n6 6\n7 8\n7 9\n8 1\n8 3\n8 6\n8 8\n9 1\n9 10\n10 2\n10 4\n10 5",
"output": "5\n1 1 2 3 4 1 2 3 2 1 4 2 1 4 3 2 1 2 4 3 1 2 5 2 3 1 2"
},
{
"input": "10 10 10\n1 7\n1 10\n2 3\n3 3\n4 5\n4 6\n4 7\n5 5\n8 10\n10 9",
"output": "3\n1 2 1 2 1 2 3 2 1 1"
},
{
"input": "100 100 50\n6 1\n6 89\n12 34\n14 4\n16 12\n20 45\n22 41\n22 87\n25 81\n30 92\n30 98\n31 16\n31 89\n32 84\n33 45\n33 94\n34 97\n36 94\n37 81\n39 23\n40 55\n40 60\n42 82\n44 80\n46 57\n46 86\n50 48\n55 33\n56 59\n56 76\n64 27\n64 60\n65 24\n71 95\n72 28\n74 23\n76 11\n80 34\n80 46\n81 22\n81 46\n85 2\n87 9\n91 97\n92 35\n95 22\n97 87\n98 29\n98 74\n100 7",
"output": "2\n1 2 1 1 1 1 1 2 1 1 2 2 1 1 2 1 1 2 2 1 1 2 1 1 1 2 1 1 1 2 2 1 1 1 1 2 1 2 1 1 2 1 1 2 1 2 1 1 2 1"
},
{
"input": "100 100 50\n3 71\n3 97\n5 65\n7 49\n9 85\n10 92\n12 60\n16 52\n17 13\n18 22\n22 85\n24 16\n27 47\n29 18\n31 83\n36 10\n37 68\n37 75\n38 1\n41 48\n43 99\n45 65\n45 96\n46 33\n50 39\n51 43\n53 55\n59 4\n63 1\n64 58\n64 92\n65 95\n70 49\n74 52\n75 51\n76 29\n76 43\n80 92\n84 51\n85 25\n85 37\n86 24\n86 81\n87 51\n91 7\n93 33\n97 50\n100 39\n100 59\n100 66",
"output": "3\n1 2 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 2 1 1 1 2 1 1 1 1 1 1 2 1 2 1 2 2 1 1 2 3 2 1 2 1 2 3 1 2 1 2 1 3"
},
{
"input": "100 100 50\n4 76\n7 17\n8 6\n8 58\n11 56\n12 79\n14 38\n19 39\n22 50\n24 33\n27 41\n29 5\n29 35\n30 20\n31 37\n31 80\n32 50\n38 39\n42 49\n42 59\n48 1\n48 80\n49 36\n49 70\n50 95\n51 3\n51 33\n57 28\n59 71\n59 94\n59 95\n61 70\n63 5\n63 98\n64 73\n66 65\n74 85\n77 13\n77 59\n78 61\n79 4\n80 39\n82 91\n85 82\n85 92\n86 45\n88 32\n89 7\n93 21\n96 36",
"output": "3\n1 1 1 2 1 1 1 1 1 1 1 1 2 1 1 2 2 2 1 2 2 1 1 2 1 1 2 1 1 2 3 1 2 1 1 1 1 1 3 1 1 3 1 1 2 1 1 1 1 2"
},
{
"input": "15 15 54\n1 1\n1 3\n1 5\n1 10\n1 14\n1 15\n2 3\n2 5\n2 14\n3 4\n3 10\n4 2\n4 13\n4 15\n5 4\n5 8\n5 10\n6 4\n6 6\n6 7\n6 8\n6 15\n7 3\n7 6\n7 7\n7 10\n8 1\n8 4\n8 6\n8 13\n9 2\n9 3\n10 2\n10 7\n10 15\n11 3\n11 6\n11 7\n11 10\n11 11\n12 5\n12 9\n12 10\n13 11\n14 2\n14 8\n14 12\n14 14\n14 15\n15 4\n15 5\n15 6\n15 10\n15 15",
"output": "7\n1 2 3 4 5 6 1 2 3 1 2 1 2 3 2 1 3 3 1 2 4 5 3 2 1 5 2 4 3 1 2 4 3 4 1 5 4 3 1 2 4 2 6 1 4 3 1 6 2 5 1 6 7 4"
},
{
"input": "15 15 49\n1 4\n1 7\n1 9\n1 11\n1 13\n2 1\n2 2\n2 4\n2 6\n2 8\n2 12\n2 13\n3 1\n3 2\n3 5\n3 9\n3 10\n4 2\n4 5\n4 6\n5 1\n5 8\n5 12\n6 1\n6 6\n6 15\n7 14\n8 2\n8 5\n8 6\n8 15\n9 1\n9 6\n9 13\n10 9\n10 11\n11 1\n11 2\n12 3\n12 7\n12 14\n13 5\n13 9\n13 14\n14 2\n14 3\n14 13\n15 10\n15 15",
"output": "7\n1 2 3 4 5 1 2 3 4 5 6 7 2 1 3 4 5 3 1 2 3 1 2 4 1 2 1 4 2 3 1 5 6 1 1 2 6 5 1 3 2 4 2 3 6 2 3 1 3"
},
{
"input": "15 15 49\n1 4\n1 7\n1 9\n1 11\n1 13\n2 1\n2 2\n2 4\n2 6\n2 8\n2 12\n2 13\n3 1\n3 2\n3 5\n3 9\n3 10\n4 2\n4 5\n4 6\n5 1\n5 8\n5 12\n6 1\n6 6\n6 15\n7 14\n8 2\n8 5\n8 6\n8 15\n9 1\n9 6\n9 13\n10 9\n10 11\n11 1\n11 2\n12 3\n12 7\n12 14\n13 5\n13 9\n13 14\n14 2\n14 3\n14 13\n15 10\n15 15",
"output": "7\n1 2 3 4 5 1 2 3 4 5 6 7 2 1 3 4 5 3 1 2 3 1 2 4 1 2 1 4 2 3 1 5 6 1 1 2 6 5 1 3 2 4 2 3 6 2 3 1 3"
},
{
"input": "15 15 53\n1 6\n2 4\n2 10\n3 3\n3 4\n3 11\n3 13\n3 14\n4 2\n4 12\n5 7\n5 8\n5 10\n6 1\n6 9\n7 10\n7 15\n8 7\n8 8\n9 4\n9 5\n9 6\n9 8\n9 11\n9 15\n10 1\n10 11\n10 15\n11 5\n11 7\n11 10\n11 12\n11 13\n12 1\n12 2\n12 4\n12 8\n12 10\n13 6\n13 7\n13 9\n13 13\n13 14\n14 9\n14 10\n14 15\n15 5\n15 8\n15 10\n15 11\n15 12\n15 14\n15 15",
"output": "7\n1 1 2 1 2 3 4 5 1 2 1 2 3 1 2 1 2 2 1 3 1 2 4 5 6 2 1 3 2 3 4 1 5 3 2 4 5 6 3 4 1 2 6 3 5 1 3 6 7 2 4 1 5"
},
{
"input": "139 1000 0",
"output": "0"
},
{
"input": "139 1000 1\n75 791",
"output": "1\n1"
}
] | 1,657,806,830
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 47
| 889
| 52,326,400
|
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from heapq import *
from collections import deque
class HopCroftKarp:
def __init__(self, n, m):
self.n = n
self.m = m
self.G = [[] for _ in range(n)]
self.RG = [[] for _ in range(m)]
self.match_l = [-1] * n
self.match_r = [-1] * m
self.used = [0] * n
self.time_stamp = 0
def add_edges(self, u, v):
self.G[u].append(v)
def _build_argument_path(self):
queue = deque()
self.dist = [-1] * self.n
for i in range(self.n):
if self.match_l[i] == -1:
queue.append(i)
self.dist[i] = 0
while queue:
a = queue.popleft()
for b in self.G[a]:
c = self.match_r[b]
if c >= 0 and self.dist[c] == -1:
self.dist[c] = self.dist[a] + 1
queue.append(c)
def _find_min_dist_argument_path(self, a):
self.used[a] = self.time_stamp
for b in self.G[a]:
c = self.match_r[b]
if c < 0 or (self.used[c] != self.time_stamp and self.dist[c] == self.dist[a] + 1 and self._find_min_dist_argument_path(c)):
self.match_r[b] = a
self.match_l[a] = b
return True
return False
def max_matching(self):
while 1:
self._build_argument_path()
self.time_stamp += 1
flow = 0
for i in range(self.n):
if self.match_l[i] == -1:
flow += self._find_min_dist_argument_path(i)
if flow == 0:
break
ret = []
for i in range(self.n):
if self.match_l[i] >= 0:
ret.append((i, self.match_l[i]))
return ret
class UnionFind:
def __init__(self, n):
self.n = n
self.parents = [-1] * n
self.group = n
def find(self, x):
if self.parents[x] < 0:
return x
else:
self.parents[x] = self.find(self.parents[x])
return self.parents[x]
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if x == y:
return
self.group -= 1
if self.parents[x] > self.parents[y]:
x, y = y, x
self.parents[x] += self.parents[y]
self.parents[y] = x
def size(self, x):
return -self.parents[self.find(x)]
def same(self, x, y):
return self.find(x) == self.find(y)
def members(self, x):
root = self.find(x)
return [i for i in range(self.n) if self.find(i) == root]
def roots(self):
return [i for i, x in enumerate(self.parents) if x < 0]
def group_count(self):
return self.group
def all_group_members(self):
dic = {r:[] for r in self.roots()}
for i in range(self.n):
dic[self.find(i)].append(i)
return dic
def __str__(self):
return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())
def contract(deg, k):
hq = []
for i, d in enumerate(deg):
hq.append([d, i])
heapify(hq)
UF = UnionFind(len(deg))
while len(hq) >= 2:
p = heappop(hq)
q = heappop(hq)
if p[0] + q[0] > k:
continue
p[0] += q[0]
UF.union(p[1], q[1])
heappush(hq, p)
return UF
def build_k_regular_graph(n, m, A, B):
dega = [0] * n
degb = [0] * m
for a in A:
dega[a] += 1
for b in B:
degb[b] += 1
K = max(*dega, *degb)
UFa = contract(dega, K)
ida = [-1] * n
pa = 0
for i in range(n):
if UFa.find(i) == i:
ida[i] = pa
pa += 1
UFb = contract(degb, K)
idb = [-1] * m
pb = 0
for i in range(m):
if UFb.find(i) == i:
idb[i] = pb
pb += 1
p = max(pa, pb)
dega = [0] * p
degb = [0] * p
C = []
D = []
for i in range(len(A)):
u = ida[UFa.find(A[i])]
v = idb[UFb.find(B[i])]
C.append(u)
D.append(v)
dega[u] += 1
degb[v] += 1
j = 0
for i in range(p):
while dega[i] < K:
while degb[j] == K:
j += 1
C.append(i)
D.append(j)
dega[i] += 1
degb[j] += 1
return K, p, C, D
def EdgeColoring(a, b, A, B):
K, n, A, B = build_k_regular_graph(a, b, A, B)
ord = [i for i in range(len(A))]
ans = []
def euler_trail(ord):
V = 2 * n
G = [[] for _ in range(V)]
m = 0
for i in ord:
G[A[i]].append((B[i] + n, m))
G[B[i] + n].append((A[i], m))
m += 1
used_v = [0] * V
used_e = [0] * m
ans = []
for i in range(V):
if used_v[i]:
continue
st = []
ord2 = []
st.append((i, -1))
while st:
id_ = st[-1][0]
used_v[id_] = True
if len(G[id_]) == 0:
ord2.append(st[-1][1])
st.pop()
else:
e = G[id_][-1]
G[id_].pop()
if used_e[e[1]]:
continue
used_e[e[1]] = True
st.append(e)
ord2.pop()
ord2 = ord2[::-1]
ans += ord2
for i, a in enumerate(ans):
ans[i] = ord[a]
return ans
def rec(ord, K):
if K == 0:
return
elif K == 1:
ans.append(ord)
return
elif K & 1:
G = HopCroftKarp(n, n)
for i in ord:
G.add_edges(A[i], B[i])
G.max_matching()
lst = []
ans.append([])
for i in ord:
if G.match_l[A[i]] == B[i]:
G.match_l[A[i]] = -1
ans[-1].append(i)
else:
lst.append(i)
rec(lst, K - 1)
else:
path = euler_trail(ord)
L = []
R = []
for i, p in enumerate(path):
if i & 1:
L.append(p)
else:
R.append(p)
rec(L, K // 2)
rec(R, K // 2)
rec(ord, K)
return K, ans
def solve():
a, b, m = map(int, input().split())
A = [-1] * m
B = [-1] * m
for i in range(m):
A[i], B[i] = map(int, input().split())
A[i] -= 1
B[i] -= 1
K, ans = EdgeColoring(a, b, A, B)
color = [-1] * m
for i in range(len(ans)):
for j in ans[i]:
if j < m:
color[j] = i + 1
print(K)
print(*color)
for _ in range(1):
solve()
|
Title: Edge coloring of bipartite graph
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
You are given an undirected bipartite graph without multiple edges. You should paint the edges of graph to minimal number of colours, so that no two adjacent edges have the same colour.
Input Specification:
The first line contains three integers *a*,<=*b*,<=*m* (1<=≤<=*a*,<=*b*<=≤<=1000, 0<=≤<=*m*<=≤<=105), *a* is the size of the first part, *b* is the size of the second part, *m* is the number of edges in the graph.
Each of the next *m* lines contains two integers *x*,<=*y* (1<=≤<=*x*<=≤<=*a*,<=1<=≤<=*y*<=≤<=*b*), where *x* is the number of the vertex in the first part and *y* is the number of the vertex in the second part. It is guaranteed that there are no multiple edges.
Output Specification:
In the first line print integer *c* — the minimal number of colours. The second line should contain *m* integers from 1 to *c* — the colours of the edges (in the order they appear in the input).
If there are several solutions, you can print any one of them.
Demo Input:
['4 3 5\n1 2\n2 2\n3 2\n4 1\n4 3\n']
Demo Output:
['3\n1 2 3 1 2\n']
Note:
none
|
```python
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from heapq import *
from collections import deque
class HopCroftKarp:
def __init__(self, n, m):
self.n = n
self.m = m
self.G = [[] for _ in range(n)]
self.RG = [[] for _ in range(m)]
self.match_l = [-1] * n
self.match_r = [-1] * m
self.used = [0] * n
self.time_stamp = 0
def add_edges(self, u, v):
self.G[u].append(v)
def _build_argument_path(self):
queue = deque()
self.dist = [-1] * self.n
for i in range(self.n):
if self.match_l[i] == -1:
queue.append(i)
self.dist[i] = 0
while queue:
a = queue.popleft()
for b in self.G[a]:
c = self.match_r[b]
if c >= 0 and self.dist[c] == -1:
self.dist[c] = self.dist[a] + 1
queue.append(c)
def _find_min_dist_argument_path(self, a):
self.used[a] = self.time_stamp
for b in self.G[a]:
c = self.match_r[b]
if c < 0 or (self.used[c] != self.time_stamp and self.dist[c] == self.dist[a] + 1 and self._find_min_dist_argument_path(c)):
self.match_r[b] = a
self.match_l[a] = b
return True
return False
def max_matching(self):
while 1:
self._build_argument_path()
self.time_stamp += 1
flow = 0
for i in range(self.n):
if self.match_l[i] == -1:
flow += self._find_min_dist_argument_path(i)
if flow == 0:
break
ret = []
for i in range(self.n):
if self.match_l[i] >= 0:
ret.append((i, self.match_l[i]))
return ret
class UnionFind:
def __init__(self, n):
self.n = n
self.parents = [-1] * n
self.group = n
def find(self, x):
if self.parents[x] < 0:
return x
else:
self.parents[x] = self.find(self.parents[x])
return self.parents[x]
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if x == y:
return
self.group -= 1
if self.parents[x] > self.parents[y]:
x, y = y, x
self.parents[x] += self.parents[y]
self.parents[y] = x
def size(self, x):
return -self.parents[self.find(x)]
def same(self, x, y):
return self.find(x) == self.find(y)
def members(self, x):
root = self.find(x)
return [i for i in range(self.n) if self.find(i) == root]
def roots(self):
return [i for i, x in enumerate(self.parents) if x < 0]
def group_count(self):
return self.group
def all_group_members(self):
dic = {r:[] for r in self.roots()}
for i in range(self.n):
dic[self.find(i)].append(i)
return dic
def __str__(self):
return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())
def contract(deg, k):
hq = []
for i, d in enumerate(deg):
hq.append([d, i])
heapify(hq)
UF = UnionFind(len(deg))
while len(hq) >= 2:
p = heappop(hq)
q = heappop(hq)
if p[0] + q[0] > k:
continue
p[0] += q[0]
UF.union(p[1], q[1])
heappush(hq, p)
return UF
def build_k_regular_graph(n, m, A, B):
dega = [0] * n
degb = [0] * m
for a in A:
dega[a] += 1
for b in B:
degb[b] += 1
K = max(*dega, *degb)
UFa = contract(dega, K)
ida = [-1] * n
pa = 0
for i in range(n):
if UFa.find(i) == i:
ida[i] = pa
pa += 1
UFb = contract(degb, K)
idb = [-1] * m
pb = 0
for i in range(m):
if UFb.find(i) == i:
idb[i] = pb
pb += 1
p = max(pa, pb)
dega = [0] * p
degb = [0] * p
C = []
D = []
for i in range(len(A)):
u = ida[UFa.find(A[i])]
v = idb[UFb.find(B[i])]
C.append(u)
D.append(v)
dega[u] += 1
degb[v] += 1
j = 0
for i in range(p):
while dega[i] < K:
while degb[j] == K:
j += 1
C.append(i)
D.append(j)
dega[i] += 1
degb[j] += 1
return K, p, C, D
def EdgeColoring(a, b, A, B):
K, n, A, B = build_k_regular_graph(a, b, A, B)
ord = [i for i in range(len(A))]
ans = []
def euler_trail(ord):
V = 2 * n
G = [[] for _ in range(V)]
m = 0
for i in ord:
G[A[i]].append((B[i] + n, m))
G[B[i] + n].append((A[i], m))
m += 1
used_v = [0] * V
used_e = [0] * m
ans = []
for i in range(V):
if used_v[i]:
continue
st = []
ord2 = []
st.append((i, -1))
while st:
id_ = st[-1][0]
used_v[id_] = True
if len(G[id_]) == 0:
ord2.append(st[-1][1])
st.pop()
else:
e = G[id_][-1]
G[id_].pop()
if used_e[e[1]]:
continue
used_e[e[1]] = True
st.append(e)
ord2.pop()
ord2 = ord2[::-1]
ans += ord2
for i, a in enumerate(ans):
ans[i] = ord[a]
return ans
def rec(ord, K):
if K == 0:
return
elif K == 1:
ans.append(ord)
return
elif K & 1:
G = HopCroftKarp(n, n)
for i in ord:
G.add_edges(A[i], B[i])
G.max_matching()
lst = []
ans.append([])
for i in ord:
if G.match_l[A[i]] == B[i]:
G.match_l[A[i]] = -1
ans[-1].append(i)
else:
lst.append(i)
rec(lst, K - 1)
else:
path = euler_trail(ord)
L = []
R = []
for i, p in enumerate(path):
if i & 1:
L.append(p)
else:
R.append(p)
rec(L, K // 2)
rec(R, K // 2)
rec(ord, K)
return K, ans
def solve():
a, b, m = map(int, input().split())
A = [-1] * m
B = [-1] * m
for i in range(m):
A[i], B[i] = map(int, input().split())
A[i] -= 1
B[i] -= 1
K, ans = EdgeColoring(a, b, A, B)
color = [-1] * m
for i in range(len(ans)):
for j in ans[i]:
if j < m:
color[j] = i + 1
print(K)
print(*color)
for _ in range(1):
solve()
```
| 3
|
|
66
|
B
|
Petya and Countryside
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation"
] |
B. Petya and Countryside
|
2
|
256
|
Little Petya often travels to his grandmother in the countryside. The grandmother has a large garden, which can be represented as a rectangle 1<=×<=*n* in size, when viewed from above. This rectangle is divided into *n* equal square sections. The garden is very unusual as each of the square sections possesses its own fixed height and due to the newest irrigation system we can create artificial rain above each section.
Creating artificial rain is an expensive operation. That's why we limit ourselves to creating the artificial rain only above one section. At that, the water from each watered section will flow into its neighbouring sections if their height does not exceed the height of the section. That is, for example, the garden can be represented by a 1<=×<=5 rectangle, where the section heights are equal to 4, 2, 3, 3, 2. Then if we create an artificial rain over any of the sections with the height of 3, the water will flow over all the sections, except the ones with the height of 4. See the illustration of this example at the picture:
As Petya is keen on programming, he decided to find such a section that if we create artificial rain above it, the number of watered sections will be maximal. Help him.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=1000). The second line contains *n* positive integers which are the height of the sections. All the numbers are no less than 1 and not more than 1000.
|
Print a single number, the maximal number of watered sections if we create artificial rain above exactly one section.
|
[
"1\n2\n",
"5\n1 2 1 2 1\n",
"8\n1 2 1 1 1 3 3 4\n"
] |
[
"1\n",
"3\n",
"6\n"
] |
none
| 1,000
|
[
{
"input": "1\n2",
"output": "1"
},
{
"input": "5\n1 2 1 2 1",
"output": "3"
},
{
"input": "8\n1 2 1 1 1 3 3 4",
"output": "6"
},
{
"input": "10\n1 2 3 4 5 6 7 8 9 10",
"output": "10"
},
{
"input": "10\n10 9 8 7 6 5 4 3 2 1",
"output": "10"
},
{
"input": "2\n100 100",
"output": "2"
},
{
"input": "3\n100 100 100",
"output": "3"
},
{
"input": "11\n1 2 3 4 5 6 5 4 3 2 1",
"output": "11"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 100 88 87 86 85 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 1 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "61"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 1 82 83 84 85 86 87 88 89 90 91 92 93 94 100 5 4 3 2 1",
"output": "81"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 1 86 87 88 89 90 91 92 93 100 6 5 4 3 2 1",
"output": "85"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 1 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 92 100 7 6 5 4 3 2 1",
"output": "61"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 100 8 7 6 1 4 3 2 1",
"output": "96"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 100 10 9 8 7 6 5 4 3 2 1",
"output": "100"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 1 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 88 100 11 10 9 8 7 6 5 4 3 2 1",
"output": "55"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 1 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 87 100 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "59"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 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 86 100 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "86"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 1 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 100 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "83"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 100 63 62 61 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 1 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "74"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 100 9 8 7 6 5 4 3 2 1",
"output": "100"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 100 84 83 82 81 80 79 78 77 76 75 74 73 72 71 70 69 68 67 66 65 64 63 62 61 60 59 58 57 56 55 54 53 1 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1",
"output": "52"
},
{
"input": "100\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 100 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 1 2 1",
"output": "98"
},
{
"input": "10\n1 4 4 4 4 4 1 2 4 3",
"output": "7"
}
] | 1,591,128,093
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 80
| 310
| 102,400
|
# Author : debugster
# Email : [email protected]
# Date : 2020-06-03 01:51:41
import sys
import os
def get_int():
return map(int, input().split())
def get_array():
return list(map(int, input().split()))
if os.environ.get("DEBUGSTER_PYTHON"):
sys.stdin = open('in.txt', 'r')
sys.stdout = open('out.txt','w')
n = int(input())
h = get_array()
left = dict()
right = dict()
for i in range(n):
left[i] = 0
right[i] = 0
for i in range(1, n):
if h[i - 1] <= h[i]:
left[i] = left[i - 1] + 1
for i in range(n - 2, -1, -1):
if h[i] >= h[i + 1]:
right[i] = right[i + 1] + 1
ans = -1
for i in range(n):
ans = max(ans, left[i] + right[i] + 1)
print(ans)
|
Title: Petya and Countryside
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
Little Petya often travels to his grandmother in the countryside. The grandmother has a large garden, which can be represented as a rectangle 1<=×<=*n* in size, when viewed from above. This rectangle is divided into *n* equal square sections. The garden is very unusual as each of the square sections possesses its own fixed height and due to the newest irrigation system we can create artificial rain above each section.
Creating artificial rain is an expensive operation. That's why we limit ourselves to creating the artificial rain only above one section. At that, the water from each watered section will flow into its neighbouring sections if their height does not exceed the height of the section. That is, for example, the garden can be represented by a 1<=×<=5 rectangle, where the section heights are equal to 4, 2, 3, 3, 2. Then if we create an artificial rain over any of the sections with the height of 3, the water will flow over all the sections, except the ones with the height of 4. See the illustration of this example at the picture:
As Petya is keen on programming, he decided to find such a section that if we create artificial rain above it, the number of watered sections will be maximal. Help him.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=1000). The second line contains *n* positive integers which are the height of the sections. All the numbers are no less than 1 and not more than 1000.
Output Specification:
Print a single number, the maximal number of watered sections if we create artificial rain above exactly one section.
Demo Input:
['1\n2\n', '5\n1 2 1 2 1\n', '8\n1 2 1 1 1 3 3 4\n']
Demo Output:
['1\n', '3\n', '6\n']
Note:
none
|
```python
# Author : debugster
# Email : [email protected]
# Date : 2020-06-03 01:51:41
import sys
import os
def get_int():
return map(int, input().split())
def get_array():
return list(map(int, input().split()))
if os.environ.get("DEBUGSTER_PYTHON"):
sys.stdin = open('in.txt', 'r')
sys.stdout = open('out.txt','w')
n = int(input())
h = get_array()
left = dict()
right = dict()
for i in range(n):
left[i] = 0
right[i] = 0
for i in range(1, n):
if h[i - 1] <= h[i]:
left[i] = left[i - 1] + 1
for i in range(n - 2, -1, -1):
if h[i] >= h[i + 1]:
right[i] = right[i + 1] + 1
ans = -1
for i in range(n):
ans = max(ans, left[i] + right[i] + 1)
print(ans)
```
| 3.922309
|
727
|
A
|
Transformation: from A to B
|
PROGRAMMING
| 1,000
|
[
"brute force",
"dfs and similar",
"math"
] | null | null |
Vasily has a number *a*, which he wants to turn into a number *b*. For this purpose, he can do two types of operations:
- multiply the current number by 2 (that is, replace the number *x* by 2·*x*); - append the digit 1 to the right of current number (that is, replace the number *x* by 10·*x*<=+<=1).
You need to help Vasily to transform the number *a* into the number *b* using only the operations described above, or find that it is impossible.
Note that in this task you are not required to minimize the number of operations. It suffices to find any way to transform *a* into *b*.
|
The first line contains two positive integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=109) — the number which Vasily has and the number he wants to have.
|
If there is no way to get *b* from *a*, print "NO" (without quotes).
Otherwise print three lines. On the first line print "YES" (without quotes). The second line should contain single integer *k* — the length of the transformation sequence. On the third line print the sequence of transformations *x*1,<=*x*2,<=...,<=*x**k*, where:
- *x*1 should be equal to *a*, - *x**k* should be equal to *b*, - *x**i* should be obtained from *x**i*<=-<=1 using any of two described operations (1<=<<=*i*<=≤<=*k*).
If there are multiple answers, print any of them.
|
[
"2 162\n",
"4 42\n",
"100 40021\n"
] |
[
"YES\n5\n2 4 8 81 162 \n",
"NO\n",
"YES\n5\n100 200 2001 4002 40021 \n"
] |
none
| 1,000
|
[
{
"input": "2 162",
"output": "YES\n5\n2 4 8 81 162 "
},
{
"input": "4 42",
"output": "NO"
},
{
"input": "100 40021",
"output": "YES\n5\n100 200 2001 4002 40021 "
},
{
"input": "1 111111111",
"output": "YES\n9\n1 11 111 1111 11111 111111 1111111 11111111 111111111 "
},
{
"input": "1 1000000000",
"output": "NO"
},
{
"input": "999999999 1000000000",
"output": "NO"
},
{
"input": "1 2",
"output": "YES\n2\n1 2 "
},
{
"input": "1 536870912",
"output": "YES\n30\n1 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 "
},
{
"input": "11111 11111111",
"output": "YES\n4\n11111 111111 1111111 11111111 "
},
{
"input": "59139 946224",
"output": "YES\n5\n59139 118278 236556 473112 946224 "
},
{
"input": "9859 19718",
"output": "YES\n2\n9859 19718 "
},
{
"input": "25987 51974222",
"output": "YES\n5\n25987 259871 2598711 25987111 51974222 "
},
{
"input": "9411 188222222",
"output": "YES\n6\n9411 94111 941111 9411111 94111111 188222222 "
},
{
"input": "25539 510782222",
"output": "YES\n6\n25539 255391 2553911 25539111 255391111 510782222 "
},
{
"input": "76259 610072",
"output": "YES\n4\n76259 152518 305036 610072 "
},
{
"input": "92387 184774",
"output": "YES\n2\n92387 184774 "
},
{
"input": "8515 85151111",
"output": "YES\n5\n8515 85151 851511 8515111 85151111 "
},
{
"input": "91939 9193911",
"output": "YES\n3\n91939 919391 9193911 "
},
{
"input": "30518 610361",
"output": "YES\n3\n30518 61036 610361 "
},
{
"input": "46646 373168844",
"output": "YES\n7\n46646 466461 932922 9329221 93292211 186584422 373168844 "
},
{
"input": "30070 300701",
"output": "YES\n2\n30070 300701 "
},
{
"input": "13494 1079528",
"output": "YES\n5\n13494 134941 269882 539764 1079528 "
},
{
"input": "96918 775344422",
"output": "YES\n7\n96918 193836 1938361 3876722 38767221 387672211 775344422 "
},
{
"input": "13046 260921",
"output": "YES\n3\n13046 26092 260921 "
},
{
"input": "29174 5834811",
"output": "YES\n4\n29174 58348 583481 5834811 "
},
{
"input": "79894 319576421",
"output": "YES\n6\n79894 798941 1597882 15978821 31957642 319576421 "
},
{
"input": "96022 1920442",
"output": "YES\n3\n96022 960221 1920442 "
},
{
"input": "79446 6355681",
"output": "YES\n5\n79446 158892 317784 635568 6355681 "
},
{
"input": "5440 27853056",
"output": "YES\n11\n5440 10880 108801 217602 435204 870408 1740816 3481632 6963264 13926528 27853056 "
},
{
"input": "250000000 705032705",
"output": "NO"
},
{
"input": "17 35",
"output": "NO"
},
{
"input": "1 3",
"output": "NO"
},
{
"input": "2 11",
"output": "NO"
}
] | 1,685,122,319
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 46
| 0
|
a, b = list(map(int, input().split()))
lista = []
lista.append(b)
flag = False
while a <= b:
# print("~~", a, b)
if a == b:
flag = True
break
elif b % 2 == 0:
b = b // 2
lista.append(b)
elif b % 10 == 1:
b = (b - 1) // 10
lista.append(b)
else:
break
if not (flag):
print("NO")
else:
rev = reversed(lista)
print("YES")
print(len(lista))
print(*rev)
|
Title: Transformation: from A to B
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasily has a number *a*, which he wants to turn into a number *b*. For this purpose, he can do two types of operations:
- multiply the current number by 2 (that is, replace the number *x* by 2·*x*); - append the digit 1 to the right of current number (that is, replace the number *x* by 10·*x*<=+<=1).
You need to help Vasily to transform the number *a* into the number *b* using only the operations described above, or find that it is impossible.
Note that in this task you are not required to minimize the number of operations. It suffices to find any way to transform *a* into *b*.
Input Specification:
The first line contains two positive integers *a* and *b* (1<=≤<=*a*<=<<=*b*<=≤<=109) — the number which Vasily has and the number he wants to have.
Output Specification:
If there is no way to get *b* from *a*, print "NO" (without quotes).
Otherwise print three lines. On the first line print "YES" (without quotes). The second line should contain single integer *k* — the length of the transformation sequence. On the third line print the sequence of transformations *x*1,<=*x*2,<=...,<=*x**k*, where:
- *x*1 should be equal to *a*, - *x**k* should be equal to *b*, - *x**i* should be obtained from *x**i*<=-<=1 using any of two described operations (1<=<<=*i*<=≤<=*k*).
If there are multiple answers, print any of them.
Demo Input:
['2 162\n', '4 42\n', '100 40021\n']
Demo Output:
['YES\n5\n2 4 8 81 162 \n', 'NO\n', 'YES\n5\n100 200 2001 4002 40021 \n']
Note:
none
|
```python
a, b = list(map(int, input().split()))
lista = []
lista.append(b)
flag = False
while a <= b:
# print("~~", a, b)
if a == b:
flag = True
break
elif b % 2 == 0:
b = b // 2
lista.append(b)
elif b % 10 == 1:
b = (b - 1) // 10
lista.append(b)
else:
break
if not (flag):
print("NO")
else:
rev = reversed(lista)
print("YES")
print(len(lista))
print(*rev)
```
| 3
|
|
979
|
B
|
Treasure Hunt
|
PROGRAMMING
| 1,800
|
[
"greedy"
] | null | null |
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of $7$ because its subribbon a appears $7$ times, and the ribbon abcdabc has the beauty of $2$ because its subribbon abc appears twice.
The rules are simple. The game will have $n$ turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after $n$ turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
|
The first line contains an integer $n$ ($0 \leq n \leq 10^{9}$) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than $10^{5}$ uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
|
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
|
[
"3\nKuroo\nShiro\nKatie\n",
"7\ntreasurehunt\nthreefriends\nhiCodeforces\n",
"1\nabcabc\ncbabac\nababca\n",
"15\nfoPaErcvJ\nmZaxowpbt\nmkuOlaHRE\n"
] |
[
"Kuro\n",
"Shiro\n",
"Katie\n",
"Draw\n"
] |
In the first example, after $3$ turns, Kuro can change his ribbon into ooooo, which has the beauty of $5$, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most $4$, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is $9$ and the number of turn is $15$, everyone can change their ribbons in some way to reach the maximal beauty of $9$ by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
| 1,000
|
[
{
"input": "3\nKuroo\nShiro\nKatie",
"output": "Kuro"
},
{
"input": "7\ntreasurehunt\nthreefriends\nhiCodeforces",
"output": "Shiro"
},
{
"input": "1\nabcabc\ncbabac\nababca",
"output": "Katie"
},
{
"input": "15\nfoPaErcvJ\nmZaxowpbt\nmkuOlaHRE",
"output": "Draw"
},
{
"input": "1\naaaaaaaaaa\nAAAAAAcAAA\nbbbbbbzzbb",
"output": "Shiro"
},
{
"input": "60\nddcZYXYbZbcXYcZdYbddaddYaZYZdaZdZZdXaaYdaZZZaXZXXaaZbb\ndcdXcYbcaXYaXYcacYabYcbZYdacaYbYdXaccYXZZZdYbbYdcZZZbY\nXaZXbbdcXaadcYdYYcbZdcaXaYZabbXZZYbYbcXbaXabcXbXadbZYZ",
"output": "Draw"
},
{
"input": "9174\nbzbbbzzzbbzzccczzccczzbzbzcbzbbzccbzcccbccczzbbcbbzbzzzcbczbzbzzbbbczbbcbzzzbcbzczbcczb\ndbzzzccdcdczzzzzcdczbbzcdzbcdbzzdczbzddcddbdbzzzczcczzbdcbbzccbzzzdzbzddcbzbdzdcczccbdb\nzdczddzcdddddczdczdczdcdzczddzczdzddczdcdcdzczczzdzccdccczczdzczczdzcdddzddzccddcczczzd",
"output": "Draw"
},
{
"input": "727\nbaabbabbbababbbbaaaabaabbaabababaaababaaababbbbababbbbbbbbbbaaabaabbbbbbbbaaaabaabbaaabaabbabaa\nddcdcccccccdccdcdccdddcddcddcddddcdddcdcdccddcdddddccddcccdcdddcdcccdccccccdcdcdccccccdccccccdc\nfffeefeffeefeeeeffefffeeefffeefffefeefefeeeffefefefefefefffffffeeeeeffffeefeeeeffffeeeeeefeffef",
"output": "Draw"
},
{
"input": "61\nbzqiqprzfwddqwctcrhnkqcsnbmcmfmrgaljwieajfouvuiunmfbrehxchupmsdpwilwu\njyxxujvxkwilikqeegzxlyiugflxqqbwbujzedqnlzucdnuipacatdhcozuvgktwvirhs\ntqiahohijwfcetyyjlkfhfvkhdgllxmhyyhhtlhltcdspusyhwpwqzyagtsbaswaobwub",
"output": "Katie"
},
{
"input": "30\njAjcdwkvcTYSYBBLniJIIIiubKWnqeDtUiaXSIPfhDTOrCWBQetm\nPQPOTgqfBWzQvPNeEaUaPQGdUgldmOZsBtsIqZGGyXozntMpOsyY\nNPfvGxMqIULNWOmUrHJfsqORUHkzKQfecXsTzgFCmUtFmIBudCJr",
"output": "Draw"
},
{
"input": "3\nabcabcabcabcdddabc\nzxytzytxxtytxyzxyt\nfgffghfghffgghghhh",
"output": "Katie"
},
{
"input": "3\naaaaa\naaaaa\naaaab",
"output": "Draw"
},
{
"input": "3\naaaaaaa\naaaabcd\nabcdefg",
"output": "Draw"
},
{
"input": "3\naaaaaaa\naaabcde\nabcdefg",
"output": "Kuro"
},
{
"input": "3\naaaaaaa\naaaabbb\nabcdefg",
"output": "Draw"
},
{
"input": "3\naaa\nbbb\nabc",
"output": "Draw"
},
{
"input": "3\naaaaa\nabcde\nabcde",
"output": "Kuro"
},
{
"input": "3\naaaaa\nqwert\nlkjhg",
"output": "Kuro"
},
{
"input": "3\naaaaa\nbbbbb\naabcd",
"output": "Draw"
},
{
"input": "3\nabcde\nfghij\nkkkkk",
"output": "Katie"
},
{
"input": "4\naaaabcd\naaaabcd\naaaaaaa",
"output": "Draw"
},
{
"input": "3\naaaabb\naabcde\nabcdef",
"output": "Kuro"
},
{
"input": "2\naaab\nabcd\naaaa",
"output": "Draw"
},
{
"input": "3\naaaaaa\naaaaaa\nabcdef",
"output": "Draw"
},
{
"input": "1\nAAAAA\nBBBBB\nABCDE",
"output": "Draw"
},
{
"input": "1\nabcde\naaaaa\naaaaa",
"output": "Draw"
},
{
"input": "4\naaabbb\nabfcde\nabfcde",
"output": "Kuro"
},
{
"input": "0\naaa\naab\nccd",
"output": "Kuro"
},
{
"input": "3\naaaaa\naaaaa\naabbb",
"output": "Draw"
},
{
"input": "3\nxxxxxx\nxxxooo\nabcdef",
"output": "Draw"
},
{
"input": "2\noooo\naaac\nabcd",
"output": "Draw"
},
{
"input": "1\naaaaaaa\naaabcde\nabcdefg",
"output": "Kuro"
},
{
"input": "3\nooooo\naaabb\nabcde",
"output": "Draw"
},
{
"input": "3\naaaaa\nqwert\nqwery",
"output": "Kuro"
},
{
"input": "2\naaaaaa\nbbbbbb\naaaaab",
"output": "Draw"
},
{
"input": "3\naabb\naabb\naabc",
"output": "Draw"
},
{
"input": "2\naaa\naab\naab",
"output": "Draw"
},
{
"input": "3\nbbbbcc\nbbbbbb\nsadfgh",
"output": "Draw"
},
{
"input": "3\naaaaaacc\nxxxxkkkk\nxxxxkkkk",
"output": "Kuro"
},
{
"input": "2\naaaac\nbbbbc\nccccc",
"output": "Draw"
},
{
"input": "3\naaaaaaaaa\naaabbbbbb\nabcdewert",
"output": "Draw"
},
{
"input": "3\naaabc\naaaab\nabcde",
"output": "Draw"
},
{
"input": "3\naaaaaaaa\naaaaaaab\naaaabbbb",
"output": "Draw"
},
{
"input": "2\nabcdefg\nabccccc\nacccccc",
"output": "Draw"
},
{
"input": "3\naaaaa\naabcd\nabcde",
"output": "Draw"
},
{
"input": "4\naaabbb\nabcdef\nabcdef",
"output": "Kuro"
},
{
"input": "4\naaabbb\naabdef\nabcdef",
"output": "Draw"
},
{
"input": "3\nabba\nbbbb\naaaa",
"output": "Draw"
},
{
"input": "3\naaaaa\nbbaaa\nabcde",
"output": "Draw"
},
{
"input": "2\naaa\naaa\nabc",
"output": "Draw"
},
{
"input": "3\naaaaa\nabcda\nabcde",
"output": "Draw"
},
{
"input": "3\naaaaa\nabcde\nbcdef",
"output": "Kuro"
},
{
"input": "3\naaabb\naabbc\nqwert",
"output": "Draw"
},
{
"input": "3\naaaaaa\naabbcc\naabbcc",
"output": "Kuro"
},
{
"input": "3\nAAAAAA\nAAAAAB\nABCDEF",
"output": "Draw"
},
{
"input": "3\nabc\naac\nbbb",
"output": "Draw"
},
{
"input": "2\naaaab\naabbc\naabbc",
"output": "Kuro"
},
{
"input": "2\naaaaaab\naaaaabb\nabcdefg",
"output": "Draw"
},
{
"input": "3\naaaaaaaaaaa\nbbbbbbbbaaa\nqwertyuiasd",
"output": "Draw"
},
{
"input": "3\naaaa\nbbbb\naabb",
"output": "Draw"
},
{
"input": "3\naaaabb\naaabcd\nabcdef",
"output": "Draw"
},
{
"input": "3\naaa\nabc\nbbb",
"output": "Draw"
},
{
"input": "1\naa\nab\nbb",
"output": "Shiro"
},
{
"input": "1\naacb\nabcd\naaaa",
"output": "Draw"
},
{
"input": "3\naaaabb\naaabbb\nabcdef",
"output": "Draw"
},
{
"input": "3\naaaa\naaaa\nabcd",
"output": "Draw"
},
{
"input": "2\nabcd\nabcd\naaad",
"output": "Katie"
},
{
"input": "3\naaa\nbbb\naab",
"output": "Draw"
},
{
"input": "3\naaaaaa\naaaaab\naaaaaa",
"output": "Draw"
},
{
"input": "2\naaab\nabcd\nabcd",
"output": "Kuro"
},
{
"input": "3\nooooo\nShiro\nKatie",
"output": "Kuro"
},
{
"input": "3\naaabb\naabcd\nabcde",
"output": "Draw"
},
{
"input": "4\nabcd\nabcd\naaaa",
"output": "Draw"
},
{
"input": "4\naaa\nbbb\naab",
"output": "Draw"
},
{
"input": "2\nxxxx\nyyyx\nabcd",
"output": "Draw"
},
{
"input": "3\nAAAAA\nAAAAB\nABCDE",
"output": "Draw"
},
{
"input": "3\naaaacdc\naaaaabc\naaaaabc",
"output": "Draw"
},
{
"input": "3\naaaaaa\naabcde\naabcde",
"output": "Kuro"
},
{
"input": "3\naaabb\naaabb\naaaaa",
"output": "Draw"
},
{
"input": "5\nabbbbb\ncbbbbb\nabcdef",
"output": "Draw"
},
{
"input": "3\naaaaaaaaa\naaaaabbbb\naaaaabbbb",
"output": "Kuro"
},
{
"input": "4\naaaaaab\naaabbbb\naaabbbb",
"output": "Draw"
},
{
"input": "3\naaaabb\naaaabb\naaabbb",
"output": "Draw"
},
{
"input": "2\naaaabb\naaaaab\nabcdef",
"output": "Draw"
},
{
"input": "2\naaaaa\naaaae\nabcde",
"output": "Draw"
},
{
"input": "3\naaaaaa\nbbbcde\nabcdef",
"output": "Draw"
},
{
"input": "4\naaaabbb\naabcdef\naabcdef",
"output": "Kuro"
},
{
"input": "2\naaaaa\naaaab\nabcde",
"output": "Draw"
},
{
"input": "3\naabbbbb\naaabbbb\nabcdefg",
"output": "Draw"
},
{
"input": "3\nabcde\naabcd\naaaaa",
"output": "Draw"
},
{
"input": "5\naaabbcc\nabcdefg\nabcdefg",
"output": "Kuro"
},
{
"input": "3\naabbb\nabcde\nabcde",
"output": "Kuro"
},
{
"input": "0\nbbb\nabb\nqer",
"output": "Kuro"
},
{
"input": "5\naabbbbb\naaaaaaa\nabcdefg",
"output": "Draw"
},
{
"input": "2\naaaab\naaaab\naaabb",
"output": "Draw"
},
{
"input": "2\naaaaaab\naaaabbb\naaaaccc",
"output": "Kuro"
},
{
"input": "3\naaaaaaaaaaaa\naaaaaaaaaaab\naaaaaabbbbbb",
"output": "Draw"
},
{
"input": "3\naaabb\nabcde\naaaaa",
"output": "Draw"
},
{
"input": "3\naaaaaac\naaaaebc\naaaaaac",
"output": "Draw"
},
{
"input": "3\naaaaaa\naaabbb\nqwerty",
"output": "Draw"
},
{
"input": "3\ncccca\nabcde\nabcde",
"output": "Kuro"
},
{
"input": "100005\nAA\nBC\nCC",
"output": "Draw"
},
{
"input": "3\naaaa\nbbbb\nccca",
"output": "Draw"
},
{
"input": "3\naaaaa\nbcdef\nbcdef",
"output": "Kuro"
},
{
"input": "2\naaab\naabb\nqwer",
"output": "Draw"
},
{
"input": "3\nabcddd\nabcdef\nbbaaaa",
"output": "Draw"
},
{
"input": "2\naaaa\naaaa\naabc",
"output": "Draw"
},
{
"input": "3\naaaa\naaaa\naaab",
"output": "Draw"
},
{
"input": "3\nabcddd\nabcdef\naaaaaa",
"output": "Draw"
},
{
"input": "1\naaaa\nabcd\naaab",
"output": "Katie"
}
] | 1,530,262,308
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 93
| 0
|
n = int(input())
scores = []
for _ in range(3):
letters = {}
word = input()
if n >= len(word) - 1:
print("Draw")
exit(0)
for c in word:
if c in letters:
letters[c] += 1
else:
letters[c] = 0
scores.append(max(letters.values()))
if scores[_] > len(word):
scores[_] = len(word)
if sorted(scores)[2] == sorted(scores)[1]:
print("Draw")
else:
if scores[0] == max(scores):
print("Kuro")
elif scores[1] == max(scores):
print("Shiro")
else:
print("Katie")
|
Title: Treasure Hunt
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
After the big birthday party, Katie still wanted Shiro to have some more fun. Later, she came up with a game called treasure hunt. Of course, she invited her best friends Kuro and Shiro to play with her.
The three friends are very smart so they passed all the challenges very quickly and finally reached the destination. But the treasure can only belong to one cat so they started to think of something which can determine who is worthy of the treasure. Instantly, Kuro came up with some ribbons.
A random colorful ribbon is given to each of the cats. Each color of the ribbon can be represented as an uppercase or lowercase Latin letter. Let's call a consecutive subsequence of colors that appears in the ribbon a subribbon. The beauty of a ribbon is defined as the maximum number of times one of its subribbon appears in the ribbon. The more the subribbon appears, the more beautiful is the ribbon. For example, the ribbon aaaaaaa has the beauty of $7$ because its subribbon a appears $7$ times, and the ribbon abcdabc has the beauty of $2$ because its subribbon abc appears twice.
The rules are simple. The game will have $n$ turns. Every turn, each of the cats must change strictly one color (at one position) in his/her ribbon to an arbitrary color which is different from the unchanged one. For example, a ribbon aaab can be changed into acab in one turn. The one having the most beautiful ribbon after $n$ turns wins the treasure.
Could you find out who is going to be the winner if they all play optimally?
Input Specification:
The first line contains an integer $n$ ($0 \leq n \leq 10^{9}$) — the number of turns.
Next 3 lines contain 3 ribbons of Kuro, Shiro and Katie one per line, respectively. Each ribbon is a string which contains no more than $10^{5}$ uppercase and lowercase Latin letters and is not empty. It is guaranteed that the length of all ribbons are equal for the purpose of fairness. Note that uppercase and lowercase letters are considered different colors.
Output Specification:
Print the name of the winner ("Kuro", "Shiro" or "Katie"). If there are at least two cats that share the maximum beauty, print "Draw".
Demo Input:
['3\nKuroo\nShiro\nKatie\n', '7\ntreasurehunt\nthreefriends\nhiCodeforces\n', '1\nabcabc\ncbabac\nababca\n', '15\nfoPaErcvJ\nmZaxowpbt\nmkuOlaHRE\n']
Demo Output:
['Kuro\n', 'Shiro\n', 'Katie\n', 'Draw\n']
Note:
In the first example, after $3$ turns, Kuro can change his ribbon into ooooo, which has the beauty of $5$, while reaching such beauty for Shiro and Katie is impossible (both Shiro and Katie can reach the beauty of at most $4$, for example by changing Shiro's ribbon into SSiSS and changing Katie's ribbon into Kaaaa). Therefore, the winner is Kuro.
In the fourth example, since the length of each of the string is $9$ and the number of turn is $15$, everyone can change their ribbons in some way to reach the maximal beauty of $9$ by changing their strings into zzzzzzzzz after 9 turns, and repeatedly change their strings into azzzzzzzz and then into zzzzzzzzz thrice. Therefore, the game ends in a draw.
|
```python
n = int(input())
scores = []
for _ in range(3):
letters = {}
word = input()
if n >= len(word) - 1:
print("Draw")
exit(0)
for c in word:
if c in letters:
letters[c] += 1
else:
letters[c] = 0
scores.append(max(letters.values()))
if scores[_] > len(word):
scores[_] = len(word)
if sorted(scores)[2] == sorted(scores)[1]:
print("Draw")
else:
if scores[0] == max(scores):
print("Kuro")
elif scores[1] == max(scores):
print("Shiro")
else:
print("Katie")
```
| 0
|
|
733
|
A
|
Grasshopper And the String
|
PROGRAMMING
| 1,000
|
[
"implementation"
] | null | null |
One day, the Grasshopper was jumping on the lawn and found a piece of paper with a string. Grasshopper became interested what is the minimum jump ability he should have in order to be able to reach the far end of the string, jumping only on vowels of the English alphabet. Jump ability is the maximum possible length of his jump.
Formally, consider that at the begginning the Grasshopper is located directly in front of the leftmost character of the string. His goal is to reach the position right after the rightmost character of the string. In one jump the Grasshopper could jump to the right any distance from 1 to the value of his jump ability.
The following letters are vowels: 'A', 'E', 'I', 'O', 'U' and 'Y'.
|
The first line contains non-empty string consisting of capital English letters. It is guaranteed that the length of the string does not exceed 100.
|
Print single integer *a* — the minimum jump ability of the Grasshopper (in the number of symbols) that is needed to overcome the given string, jumping only on vowels.
|
[
"ABABBBACFEYUKOTT\n",
"AAA\n"
] |
[
"4",
"1"
] |
none
| 500
|
[
{
"input": "ABABBBACFEYUKOTT",
"output": "4"
},
{
"input": "AAA",
"output": "1"
},
{
"input": "A",
"output": "1"
},
{
"input": "B",
"output": "2"
},
{
"input": "AEYUIOAEIYAEOUIYOEIUYEAOIUEOEAYOEIUYAEOUIYEOIKLMJNHGTRWSDZXCVBNMHGFDSXVWRTPPPLKMNBXIUOIUOIUOIUOOIU",
"output": "39"
},
{
"input": "AEYUIOAEIYAEOUIYOEIUYEAOIUEOEAYOEIUYAEOUIYEOIAEYUIOAEIYAEOUIYOEIUYEAOIUEOEAYOEIUYAEOUIYEOI",
"output": "1"
},
{
"input": "KMLPTGFHNBVCDRFGHNMBVXWSQFDCVBNHTJKLPMNFVCKMLPTGFHNBVCDRFGHNMBVXWSQFDCVBNHTJKLPMNFVC",
"output": "85"
},
{
"input": "QWERTYUIOPASDFGHJKLZXCVBNMQWERTYUIOPASDFGHJKLZXCVBNMQWERTYUIOPASDFGHJKLZXCVBNMQWERTYUIOPASDFGHJKLZ",
"output": "18"
},
{
"input": "PKLKBWTXVJ",
"output": "11"
},
{
"input": "CFHFPTGMOKXVLJJZJDQW",
"output": "12"
},
{
"input": "TXULTFSBUBFLRNQORMMULWNVLPWTYJXZBPBGAWNX",
"output": "9"
},
{
"input": "DAIUSEAUEUYUWEIOOEIOUYVYYOPEEWEBZOOOAOXUOIEUKYYOJOYAUYUUIYUXOUJLGIYEIIYUOCUAACRY",
"output": "4"
},
{
"input": "VRPHBNWNWVWBWMFJJDCTJQJDJBKSJRZLVQRVVFLTZFSGCGDXCWQVWWWMFVCQHPKXXVRKTGWGPSMQTPKNDQJHNSKLXPCXDJDQDZZD",
"output": "101"
},
{
"input": "SGDDFCDRDWGPNNFBBZZJSPXFYMZKPRXTCHVJSJJBWZXXQMDZBNKDHRGSRLGLRKPMWXNSXJPNJLDPXBSRCQMHJKPZNTPNTZXNPCJC",
"output": "76"
},
{
"input": "NVTQVNLGWFDBCBKSDLTBGWBMNQZWZQJWNGVCTCQBGWNTYJRDBPZJHXCXFMIXNRGSTXHQPCHNFQPCMDZWJGLJZWMRRFCVLBKDTDSC",
"output": "45"
},
{
"input": "SREZXQFVPQCLRCQGMKXCBRWKYZKWKRMZGXPMKWNMFZTRDPHJFCSXVPPXWKZMZTBFXGNLPLHZIPLFXNRRQFDTLFPKBGCXKTMCFKKT",
"output": "48"
},
{
"input": "ICKJKMVPDNZPLKDSLTPZNRLSQSGHQJQQPJJSNHNWVDLJRLZEJSXZDPHYXGGWXHLCTVQSKWNWGTLJMOZVJNZPVXGVPJKHFVZTGCCX",
"output": "47"
},
{
"input": "XXFPZDRPXLNHGDVCBDKJMKLGUQZXLLWYLOKFZVGXVNPJWZZZNRMQBRJCZTSDRHSNCVDMHKVXCXPCRBWSJCJWDRDPVZZLCZRTDRYA",
"output": "65"
},
{
"input": "HDDRZDKCHHHEDKHZMXQSNQGSGNNSCCPVJFGXGNCEKJMRKSGKAPQWPCWXXWHLSMRGSJWEHWQCSJJSGLQJXGVTBYALWMLKTTJMFPFS",
"output": "28"
},
{
"input": "PXVKJHXVDPWGLHWFWMJPMCCNHCKSHCPZXGIHHNMYNFQBUCKJJTXXJGKRNVRTQFDFMLLGPQKFOVNNLTNDIEXSARRJKGSCZKGGJCBW",
"output": "35"
},
{
"input": "EXNMTTFPJLDHXDQBJJRDRYBZVFFHUDCHCPNFZWXSMZXNFVJGHZWXVBRQFNUIDVLZOVPXQNVMFNBTJDSCKRLNGXPSADTGCAHCBJKL",
"output": "30"
},
{
"input": "NRNLSQQJGIJBCZFTNKJCXMGPARGWXPSHZXOBNSFOLDQVXTVAGJZNLXULHBRDGMNQKQGWMRRDPYCSNFVPUFTFBUBRXVJGNGSPJKLL",
"output": "19"
},
{
"input": "SRHOKCHQQMVZKTCVQXJJCFGYFXGMBZSZFNAFETXILZHPGHBWZRZQFMGSEYRUDVMCIQTXTBTSGFTHRRNGNTHHWWHCTDFHSVARMCMB",
"output": "30"
},
{
"input": "HBSVZHDKGNIRQUBYKYHUPJCEETGFMVBZJTHYHFQPFBVBSMQACYAVWZXSBGNKWXFNMQJFMSCHJVWBZXZGSNBRUHTHAJKVLEXFBOFB",
"output": "34"
},
{
"input": "NXKMUGOPTUQNSRYTKUKSCWCRQSZKKFPYUMDIBJAHJCEKZJVWZAWOLOEFBFXLQDDPNNZKCQHUPBFVDSXSUCVLMZXQROYQYIKPQPWR",
"output": "17"
},
{
"input": "TEHJDICFNOLQVQOAREVAGUAWODOCXJXIHYXFAEPEXRHPKEIIRCRIVASKNTVYUYDMUQKSTSSBYCDVZKDDHTSDWJWACPCLYYOXGCLT",
"output": "15"
},
{
"input": "LCJJUZZFEIUTMSEXEYNOOAIZMORQDOANAMUCYTFRARDCYHOYOPHGGYUNOGNXUAOYSEMXAZOOOFAVHQUBRNGORSPNQWZJYQQUNPEB",
"output": "9"
},
{
"input": "UUOKAOOJBXUTSMOLOOOOSUYYFTAVBNUXYFVOOGCGZYQEOYISIYOUULUAIJUYVVOENJDOCLHOSOHIHDEJOIGZNIXEMEGZACHUAQFW",
"output": "5"
},
{
"input": "OUUBEHXOOURMOAIAEHXCUOIYHUJEVAWYRCIIAGDRIPUIPAIUYAIWJEVYEYYUYBYOGVYESUJCFOJNUAHIOOKBUUHEJFEWPOEOUHYA",
"output": "4"
},
{
"input": "EMNOYEEUIOUHEWZITIAEZNCJUOUAOQEAUYEIHYUSUYUUUIAEDIOOERAEIRBOJIEVOMECOGAIAIUIYYUWYIHIOWVIJEYUEAFYULSE",
"output": "5"
},
{
"input": "BVOYEAYOIEYOREJUYEUOEOYIISYAEOUYAAOIOEOYOOOIEFUAEAAESUOOIIEUAAGAEISIAPYAHOOEYUJHUECGOYEIDAIRTBHOYOYA",
"output": "5"
},
{
"input": "GOIEOAYIEYYOOEOAIAEOOUWYEIOTNYAANAYOOXEEOEAVIOIAAIEOIAUIAIAAUEUAOIAEUOUUZYIYAIEUEGOOOOUEIYAEOSYAEYIO",
"output": "3"
},
{
"input": "AUEAOAYIAOYYIUIOAULIOEUEYAIEYYIUOEOEIEYRIYAYEYAEIIMMAAEAYAAAAEOUICAUAYOUIAOUIAIUOYEOEEYAEYEYAAEAOYIY",
"output": "3"
},
{
"input": "OAIIYEYYAOOEIUOEEIOUOIAEFIOAYETUYIOAAAEYYOYEYOEAUIIUEYAYYIIAOIEEYGYIEAAOOWYAIEYYYIAOUUOAIAYAYYOEUEOY",
"output": "2"
},
{
"input": "EEEAOEOEEIOUUUEUEAAOEOIUYJEYAIYIEIYYEAUOIIYIUOOEUCYEOOOYYYIUUAYIAOEUEIEAOUOIAACAOOUAUIYYEAAAOOUYIAAE",
"output": "2"
},
{
"input": "AYEYIIEUIYOYAYEUEIIIEUYUUAUEUIYAIAAUYONIEYIUIAEUUOUOYYOUUUIUIAEYEOUIIUOUUEOAIUUYAAEOAAEOYUUIYAYRAIII",
"output": "2"
},
{
"input": "YOOAAUUAAAYEUYIUIUYIUOUAEIEEIAUEOAUIIAAIUYEUUOYUIYEAYAAAYUEEOEEAEOEEYYOUAEUYEEAIIYEUEYJOIIYUIOIUOIEE",
"output": "2"
},
{
"input": "UYOIIIAYOOAIUUOOEEUYIOUAEOOEIOUIAIEYOAEAIOOEOOOIUYYUYIAAUIOUYYOOUAUIEYYUOAAUUEAAIEUIAUEUUIAUUOYOAYIU",
"output": "1"
},
{
"input": "ABBABBB",
"output": "4"
},
{
"input": "ABCD",
"output": "4"
},
{
"input": "XXYC",
"output": "3"
},
{
"input": "YYY",
"output": "1"
},
{
"input": "ABABBBBBBB",
"output": "8"
},
{
"input": "YYYY",
"output": "1"
},
{
"input": "YYYYY",
"output": "1"
},
{
"input": "AXXX",
"output": "4"
},
{
"input": "YYYYYYY",
"output": "1"
},
{
"input": "BYYBBB",
"output": "4"
},
{
"input": "YYYYYYYYY",
"output": "1"
},
{
"input": "CAAAAA",
"output": "2"
},
{
"input": "CCCACCCC",
"output": "5"
},
{
"input": "ABABBBACFEYUKOTTTT",
"output": "5"
},
{
"input": "AABBYYYYYYYY",
"output": "3"
},
{
"input": "BYBACYC",
"output": "2"
},
{
"input": "Y",
"output": "1"
},
{
"input": "ABBBBBB",
"output": "7"
},
{
"input": "BACDYDI",
"output": "3"
},
{
"input": "XEXXXXXXXXXXXXXXX",
"output": "16"
},
{
"input": "TTYTT",
"output": "3"
},
{
"input": "AAYBC",
"output": "3"
},
{
"input": "ABABBBACFEYUKOTTTTT",
"output": "6"
},
{
"input": "YYAYY",
"output": "1"
},
{
"input": "YZZY",
"output": "3"
},
{
"input": "YY",
"output": "1"
},
{
"input": "ZZYZZ",
"output": "3"
},
{
"input": "YBBBY",
"output": "4"
},
{
"input": "BBBACCCCCCC",
"output": "8"
},
{
"input": "YBBBBY",
"output": "5"
},
{
"input": "YYYYYYYYYY",
"output": "1"
},
{
"input": "ABABBBBBBBBBBBB",
"output": "13"
}
] | 1,563,392,089
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 70
| 124
| 0
|
s=input()
m=1
l=0
d={'A','E','I','O','U','Y'}
for i in range(len(s)):
if s[i] in d:
l=max(l,m)
m=1
else:
m+=1
l=max(l,m)
if l==0:
print(len(s)+1)
else:
print(l)
|
Title: Grasshopper And the String
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
One day, the Grasshopper was jumping on the lawn and found a piece of paper with a string. Grasshopper became interested what is the minimum jump ability he should have in order to be able to reach the far end of the string, jumping only on vowels of the English alphabet. Jump ability is the maximum possible length of his jump.
Formally, consider that at the begginning the Grasshopper is located directly in front of the leftmost character of the string. His goal is to reach the position right after the rightmost character of the string. In one jump the Grasshopper could jump to the right any distance from 1 to the value of his jump ability.
The following letters are vowels: 'A', 'E', 'I', 'O', 'U' and 'Y'.
Input Specification:
The first line contains non-empty string consisting of capital English letters. It is guaranteed that the length of the string does not exceed 100.
Output Specification:
Print single integer *a* — the minimum jump ability of the Grasshopper (in the number of symbols) that is needed to overcome the given string, jumping only on vowels.
Demo Input:
['ABABBBACFEYUKOTT\n', 'AAA\n']
Demo Output:
['4', '1']
Note:
none
|
```python
s=input()
m=1
l=0
d={'A','E','I','O','U','Y'}
for i in range(len(s)):
if s[i] in d:
l=max(l,m)
m=1
else:
m+=1
l=max(l,m)
if l==0:
print(len(s)+1)
else:
print(l)
```
| 3
|
|
161
|
A
|
Dress'em in Vests!
|
PROGRAMMING
| 1,300
|
[
"binary search",
"brute force",
"greedy",
"two pointers"
] | null | null |
The Two-dimensional kingdom is going through hard times... This morning the Three-Dimensional kingdom declared war on the Two-dimensional one. This (possibly armed) conflict will determine the ultimate owner of the straight line.
The Two-dimensional kingdom has a regular army of *n* people. Each soldier registered himself and indicated the desired size of the bulletproof vest: the *i*-th soldier indicated size *a**i*. The soldiers are known to be unpretentious, so the command staff assumes that the soldiers are comfortable in any vests with sizes from *a**i*<=-<=*x* to *a**i*<=+<=*y*, inclusive (numbers *x*,<=*y*<=≥<=0 are specified).
The Two-dimensional kingdom has *m* vests at its disposal, the *j*-th vest's size equals *b**j*. Help mobilize the Two-dimensional kingdom's army: equip with vests as many soldiers as possible. Each vest can be used only once. The *i*-th soldier can put on the *j*-th vest, if *a**i*<=-<=*x*<=≤<=*b**j*<=≤<=*a**i*<=+<=*y*.
|
The first input line contains four integers *n*, *m*, *x* and *y* (1<=≤<=*n*,<=*m*<=≤<=105, 0<=≤<=*x*,<=*y*<=≤<=109) — the number of soldiers, the number of vests and two numbers that specify the soldiers' unpretentiousness, correspondingly.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) in non-decreasing order, separated by single spaces — the desired sizes of vests.
The third line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**j*<=≤<=109) in non-decreasing order, separated by single spaces — the sizes of the available vests.
|
In the first line print a single integer *k* — the maximum number of soldiers equipped with bulletproof vests.
In the next *k* lines print *k* pairs, one pair per line, as "*u**i* *v**i*" (without the quotes). Pair (*u**i*, *v**i*) means that soldier number *u**i* must wear vest number *v**i*. Soldiers and vests are numbered starting from one in the order in which they are specified in the input. All numbers of soldiers in the pairs should be pairwise different, all numbers of vests in the pairs also should be pairwise different. You can print the pairs in any order.
If there are multiple optimal answers, you are allowed to print any of them.
|
[
"5 3 0 0\n1 2 3 3 4\n1 3 5\n",
"3 3 2 2\n1 5 9\n3 5 7\n"
] |
[
"2\n1 1\n3 2\n",
"3\n1 1\n2 2\n3 3\n"
] |
In the first sample you need the vests' sizes to match perfectly: the first soldier gets the first vest (size 1), the third soldier gets the second vest (size 3). This sample allows another answer, which gives the second vest to the fourth soldier instead of the third one.
In the second sample the vest size can differ from the desired size by at most 2 sizes, so all soldiers can be equipped.
| 1,000
|
[
{
"input": "5 3 0 0\n1 2 3 3 4\n1 3 5",
"output": "2\n1 1\n3 2"
},
{
"input": "3 3 2 2\n1 5 9\n3 5 7",
"output": "3\n1 1\n2 2\n3 3"
},
{
"input": "1 1 0 0\n1\n1",
"output": "1\n1 1"
},
{
"input": "1 1 0 0\n1\n2",
"output": "0"
},
{
"input": "2 3 1 4\n1 5\n1 2 2",
"output": "1\n1 1"
},
{
"input": "20 30 1 4\n1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 4 4 4 5\n1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 4 4 5 5",
"output": "20\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n10 10\n11 11\n12 12\n13 15\n14 16\n15 17\n16 18\n17 19\n18 20\n19 21\n20 22"
},
{
"input": "33 23 17 2\n1 1 2 2 2 3 3 3 3 3 3 4 4 4 4 4 5 5 5 6 6 7 7 7 8 8 8 8 8 9 9 10 10\n1 1 3 3 4 4 4 5 5 6 6 6 7 8 8 8 8 8 8 9 9 10 10",
"output": "23\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n12 10\n13 11\n14 12\n17 13\n20 14\n21 15\n22 16\n23 17\n24 18\n25 19\n26 20\n27 21\n28 22\n29 23"
},
{
"input": "2 2 1 4\n1 4\n3 6",
"output": "2\n1 1\n2 2"
},
{
"input": "20 20 1 4\n1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 4 4 5 5\n3 3 3 3 3 4 4 4 4 4 4 4 4 5 5 5 6 6 7 7",
"output": "20\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n10 10\n11 11\n12 12\n13 13\n14 14\n15 15\n16 16\n17 17\n18 18\n19 19\n20 20"
},
{
"input": "33 23 17 2\n1 1 1 2 3 3 3 3 3 4 4 4 4 5 6 6 6 6 6 6 7 7 7 7 7 8 8 8 8 8 8 10 10\n10 10 10 11 12 12 12 12 12 13 13 13 13 14 15 15 15 15 15 15 16 16 16",
"output": "5\n26 1\n27 2\n28 3\n32 4\n33 5"
},
{
"input": "1 1 1 2\n783266931\n783266932",
"output": "1\n1 1"
},
{
"input": "2 3 1 4\n1 1\n3 3 4",
"output": "2\n1 1\n2 2"
},
{
"input": "20 30 1 4\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4",
"output": "20\n1 1\n2 2\n3 3\n4 4\n5 5\n6 6\n7 7\n8 8\n9 9\n10 10\n11 11\n12 12\n13 13\n14 14\n15 15\n16 16\n17 17\n18 18\n19 19\n20 20"
},
{
"input": "33 23 17 2\n1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 4 5 5 5 5 5 5\n10 10 10 10 10 10 10 11 11 11 11 11 11 11 12 12 12 12 12 12 13 13 13",
"output": "0"
},
{
"input": "10 1 1 1\n1 1 1 5 5 7 7 8 8 8\n3",
"output": "0"
},
{
"input": "10 1 2 1\n1 2 5 8 9 9 9 10 10 10\n7",
"output": "1\n4 1"
},
{
"input": "1 1 2 1\n7\n8",
"output": "1\n1 1"
},
{
"input": "1 1 0 2\n9\n7",
"output": "0"
},
{
"input": "2 2 0 2\n2 8\n3 5",
"output": "1\n1 1"
},
{
"input": "2 2 1 0\n5 5\n6 7",
"output": "0"
},
{
"input": "2 2 3 0\n8 9\n1 5",
"output": "1\n1 2"
}
] | 1,637,251,075
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 0
| 124
| 512,000
|
from sys import stdin
input = stdin.readline
def answer():
ans = []
j = 0
for i in range(m):
if((a[j] - x) > b[i]):continue
while(j < n and (a[j] + y) < b[i]):
j += 1
if(j == n):break
ans.append([i + 1 , j + 1])
j += 1
print(len(ans))
for v in ans:print(*v)
for T in range(1):
n , m , x , y = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
answer()
|
Title: Dress'em in Vests!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The Two-dimensional kingdom is going through hard times... This morning the Three-Dimensional kingdom declared war on the Two-dimensional one. This (possibly armed) conflict will determine the ultimate owner of the straight line.
The Two-dimensional kingdom has a regular army of *n* people. Each soldier registered himself and indicated the desired size of the bulletproof vest: the *i*-th soldier indicated size *a**i*. The soldiers are known to be unpretentious, so the command staff assumes that the soldiers are comfortable in any vests with sizes from *a**i*<=-<=*x* to *a**i*<=+<=*y*, inclusive (numbers *x*,<=*y*<=≥<=0 are specified).
The Two-dimensional kingdom has *m* vests at its disposal, the *j*-th vest's size equals *b**j*. Help mobilize the Two-dimensional kingdom's army: equip with vests as many soldiers as possible. Each vest can be used only once. The *i*-th soldier can put on the *j*-th vest, if *a**i*<=-<=*x*<=≤<=*b**j*<=≤<=*a**i*<=+<=*y*.
Input Specification:
The first input line contains four integers *n*, *m*, *x* and *y* (1<=≤<=*n*,<=*m*<=≤<=105, 0<=≤<=*x*,<=*y*<=≤<=109) — the number of soldiers, the number of vests and two numbers that specify the soldiers' unpretentiousness, correspondingly.
The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) in non-decreasing order, separated by single spaces — the desired sizes of vests.
The third line contains *m* integers *b*1,<=*b*2,<=...,<=*b**m* (1<=≤<=*b**j*<=≤<=109) in non-decreasing order, separated by single spaces — the sizes of the available vests.
Output Specification:
In the first line print a single integer *k* — the maximum number of soldiers equipped with bulletproof vests.
In the next *k* lines print *k* pairs, one pair per line, as "*u**i* *v**i*" (without the quotes). Pair (*u**i*, *v**i*) means that soldier number *u**i* must wear vest number *v**i*. Soldiers and vests are numbered starting from one in the order in which they are specified in the input. All numbers of soldiers in the pairs should be pairwise different, all numbers of vests in the pairs also should be pairwise different. You can print the pairs in any order.
If there are multiple optimal answers, you are allowed to print any of them.
Demo Input:
['5 3 0 0\n1 2 3 3 4\n1 3 5\n', '3 3 2 2\n1 5 9\n3 5 7\n']
Demo Output:
['2\n1 1\n3 2\n', '3\n1 1\n2 2\n3 3\n']
Note:
In the first sample you need the vests' sizes to match perfectly: the first soldier gets the first vest (size 1), the third soldier gets the second vest (size 3). This sample allows another answer, which gives the second vest to the fourth soldier instead of the third one.
In the second sample the vest size can differ from the desired size by at most 2 sizes, so all soldiers can be equipped.
|
```python
from sys import stdin
input = stdin.readline
def answer():
ans = []
j = 0
for i in range(m):
if((a[j] - x) > b[i]):continue
while(j < n and (a[j] + y) < b[i]):
j += 1
if(j == n):break
ans.append([i + 1 , j + 1])
j += 1
print(len(ans))
for v in ans:print(*v)
for T in range(1):
n , m , x , y = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
answer()
```
| 0
|
|
698
|
A
|
Vacations
|
PROGRAMMING
| 1,400
|
[
"dp"
] | null | null |
Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
|
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out.
|
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days.
|
[
"4\n1 3 2 0\n",
"7\n1 3 3 2 1 2 3\n",
"2\n2 2\n"
] |
[
"2\n",
"0\n",
"1\n"
] |
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
| 500
|
[
{
"input": "4\n1 3 2 0",
"output": "2"
},
{
"input": "7\n1 3 3 2 1 2 3",
"output": "0"
},
{
"input": "2\n2 2",
"output": "1"
},
{
"input": "1\n0",
"output": "1"
},
{
"input": "10\n0 0 1 1 0 0 0 0 1 0",
"output": "8"
},
{
"input": "100\n3 2 3 3 3 2 3 1 3 2 2 3 2 3 3 3 3 3 3 1 2 2 3 1 3 3 2 2 2 3 1 0 3 3 3 2 3 3 1 1 3 1 3 3 3 1 3 1 3 0 1 3 2 3 2 1 1 3 2 3 3 3 2 3 1 3 3 3 3 2 2 2 1 3 1 3 3 3 3 1 3 2 3 3 0 3 3 3 3 3 1 0 2 1 3 3 0 2 3 3",
"output": "16"
},
{
"input": "10\n2 3 0 1 3 1 2 2 1 0",
"output": "3"
},
{
"input": "45\n3 3 2 3 2 3 3 3 0 3 3 3 3 3 3 3 1 3 2 3 2 3 2 2 2 3 2 3 3 3 3 3 1 2 3 3 2 2 2 3 3 3 3 1 3",
"output": "6"
},
{
"input": "1\n1",
"output": "0"
},
{
"input": "1\n2",
"output": "0"
},
{
"input": "1\n3",
"output": "0"
},
{
"input": "2\n1 1",
"output": "1"
},
{
"input": "2\n1 3",
"output": "0"
},
{
"input": "2\n0 1",
"output": "1"
},
{
"input": "2\n0 0",
"output": "2"
},
{
"input": "2\n3 3",
"output": "0"
},
{
"input": "3\n3 3 3",
"output": "0"
},
{
"input": "2\n3 2",
"output": "0"
},
{
"input": "2\n0 2",
"output": "1"
},
{
"input": "10\n2 2 3 3 3 3 2 1 3 2",
"output": "2"
},
{
"input": "15\n0 1 0 0 0 2 0 1 0 0 0 2 0 0 0",
"output": "11"
},
{
"input": "15\n1 3 2 2 2 3 3 3 3 2 3 2 2 1 1",
"output": "4"
},
{
"input": "15\n3 1 3 2 3 2 2 2 3 3 3 3 2 3 2",
"output": "3"
},
{
"input": "20\n0 2 0 1 0 0 0 1 2 0 1 1 1 0 1 1 0 1 1 0",
"output": "12"
},
{
"input": "20\n2 3 2 3 3 3 3 2 0 3 1 1 2 3 0 3 2 3 0 3",
"output": "5"
},
{
"input": "20\n3 3 3 3 2 3 3 2 1 3 3 2 2 2 3 2 2 2 2 2",
"output": "4"
},
{
"input": "25\n0 0 1 0 0 1 0 0 1 0 0 1 0 2 0 0 2 0 0 1 0 2 0 1 1",
"output": "16"
},
{
"input": "25\n1 3 3 2 2 3 3 3 3 3 1 2 2 3 2 0 2 1 0 1 3 2 2 3 3",
"output": "5"
},
{
"input": "25\n2 3 1 3 3 2 1 3 3 3 1 3 3 1 3 2 3 3 1 3 3 3 2 3 3",
"output": "3"
},
{
"input": "30\n0 0 1 0 1 0 1 1 0 0 0 0 0 0 1 0 0 1 1 0 0 2 0 0 1 1 2 0 0 0",
"output": "22"
},
{
"input": "30\n1 1 3 2 2 0 3 2 3 3 1 2 0 1 1 2 3 3 2 3 1 3 2 3 0 2 0 3 3 2",
"output": "9"
},
{
"input": "30\n1 2 3 2 2 3 3 3 3 3 3 3 3 3 3 1 2 2 3 2 3 3 3 2 1 3 3 3 1 3",
"output": "2"
},
{
"input": "35\n0 1 1 0 0 2 0 0 1 0 0 0 1 0 1 0 1 0 0 0 1 2 1 0 2 2 1 0 1 0 1 1 1 0 0",
"output": "21"
},
{
"input": "35\n2 2 0 3 2 2 0 3 3 1 1 3 3 1 2 2 0 2 2 2 2 3 1 0 2 1 3 2 2 3 2 3 3 1 2",
"output": "11"
},
{
"input": "35\n1 2 2 3 3 3 3 3 2 2 3 3 2 3 3 2 3 2 3 3 2 2 2 3 3 2 3 3 3 1 3 3 2 2 2",
"output": "7"
},
{
"input": "40\n2 0 1 1 0 0 0 0 2 0 1 1 1 0 0 1 0 0 0 0 0 2 0 0 0 2 1 1 1 3 0 0 0 0 0 0 0 1 1 0",
"output": "28"
},
{
"input": "40\n2 2 3 2 0 2 3 2 1 2 3 0 2 3 2 1 1 3 1 1 0 2 3 1 3 3 1 1 3 3 2 2 1 3 3 3 2 3 3 1",
"output": "10"
},
{
"input": "40\n1 3 2 3 3 2 3 3 2 2 3 1 2 1 2 2 3 1 2 2 1 2 2 2 1 2 2 3 2 3 2 3 2 3 3 3 1 3 2 3",
"output": "8"
},
{
"input": "45\n2 1 0 0 0 2 1 0 1 0 0 2 2 1 1 0 0 2 0 0 0 0 0 0 1 0 0 2 0 0 1 1 0 0 1 0 0 1 1 2 0 0 2 0 2",
"output": "29"
},
{
"input": "45\n3 3 2 3 3 3 2 2 3 2 3 1 3 2 3 2 2 1 1 3 2 3 2 1 3 1 2 3 2 2 0 3 3 2 3 2 3 2 3 2 0 3 1 1 3",
"output": "8"
},
{
"input": "50\n3 0 0 0 2 0 0 0 0 0 0 0 2 1 0 2 0 1 0 1 3 0 2 1 1 0 0 1 1 0 0 1 2 1 1 2 1 1 0 0 0 0 0 0 0 1 2 2 0 0",
"output": "32"
},
{
"input": "50\n3 3 3 3 1 0 3 3 0 2 3 1 1 1 3 2 3 3 3 3 3 1 0 1 2 2 3 3 2 3 0 0 0 2 1 0 1 2 2 2 2 0 2 2 2 1 2 3 3 2",
"output": "16"
},
{
"input": "50\n3 2 3 1 2 1 2 3 3 2 3 3 2 1 3 3 3 3 3 3 2 3 2 3 2 2 3 3 3 2 3 3 3 3 2 3 1 2 3 3 2 3 3 1 2 2 1 1 3 3",
"output": "7"
},
{
"input": "55\n0 0 1 1 0 1 0 0 1 0 1 0 0 0 2 0 0 1 0 0 0 1 0 0 0 0 3 1 0 0 0 1 0 0 0 0 2 0 0 0 2 0 2 1 0 0 0 0 0 0 0 0 2 0 0",
"output": "40"
},
{
"input": "55\n3 0 3 3 3 2 0 2 3 0 3 2 3 3 0 3 3 1 3 3 1 2 3 2 0 3 3 2 1 2 3 2 3 0 3 2 2 1 2 3 2 2 1 3 2 2 3 1 3 2 2 3 3 2 2",
"output": "13"
},
{
"input": "55\n3 3 1 3 2 3 2 3 2 2 3 3 3 3 3 1 1 3 3 2 3 2 3 2 0 1 3 3 3 3 2 3 2 3 1 1 2 2 2 3 3 3 3 3 2 2 2 3 2 3 3 3 3 1 3",
"output": "7"
},
{
"input": "60\n0 1 0 0 0 0 0 0 0 2 1 1 3 0 0 0 0 0 1 0 1 1 0 0 0 3 0 1 0 1 0 2 0 0 0 0 0 1 0 0 0 0 1 1 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0",
"output": "44"
},
{
"input": "60\n3 2 1 3 2 2 3 3 3 1 1 3 2 2 3 3 1 3 2 2 3 3 2 2 2 2 0 2 2 3 2 3 0 3 3 3 2 3 3 0 1 3 2 1 3 1 1 2 1 3 1 1 2 2 1 3 3 3 2 2",
"output": "15"
},
{
"input": "60\n3 2 2 3 2 3 2 3 3 2 3 2 3 3 2 3 3 3 3 3 3 2 3 3 1 2 3 3 3 2 1 3 3 1 3 1 3 0 3 3 3 2 3 2 3 2 3 3 1 1 2 3 3 3 3 2 1 3 2 3",
"output": "8"
},
{
"input": "65\n1 0 2 1 1 0 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 1 2 0 2 1 0 2 1 0 1 0 1 1 0 1 1 1 2 1 0 1 0 0 0 0 1 2 2 1 0 0 1 2 1 2 0 2 0 0 0 1 1",
"output": "35"
},
{
"input": "65\n2 2 2 3 0 2 1 2 3 3 1 3 1 2 1 3 2 3 2 2 2 1 2 0 3 1 3 1 1 3 1 3 3 3 3 3 1 3 0 3 1 3 1 2 2 3 2 0 3 1 3 2 1 2 2 2 3 3 2 3 3 3 2 2 3",
"output": "13"
},
{
"input": "65\n3 2 3 3 3 2 3 2 3 3 3 3 3 3 3 3 3 2 3 2 3 2 2 3 3 3 3 3 2 2 2 3 3 2 3 3 2 3 3 3 3 2 3 3 3 2 2 3 3 3 3 3 3 2 2 3 3 2 3 3 1 3 3 3 3",
"output": "6"
},
{
"input": "70\n1 0 0 0 1 0 1 0 0 0 1 1 0 1 0 0 1 1 1 0 1 1 0 0 1 1 1 3 1 1 0 1 2 0 2 1 0 0 0 1 1 1 1 1 0 0 1 0 0 0 1 1 1 3 0 0 1 0 0 0 1 0 0 0 0 0 1 0 1 1",
"output": "43"
},
{
"input": "70\n2 3 3 3 1 3 3 1 2 1 1 2 2 3 0 2 3 3 1 3 3 2 2 3 3 3 2 2 2 2 1 3 3 0 2 1 1 3 2 3 3 2 2 3 1 3 1 2 3 2 3 3 2 2 2 3 1 1 2 1 3 3 2 2 3 3 3 1 1 1",
"output": "16"
},
{
"input": "70\n3 3 2 2 1 2 1 2 2 2 2 2 3 3 2 3 3 3 3 2 2 2 2 3 3 3 1 3 3 3 2 3 3 3 3 2 3 3 1 3 1 3 2 3 3 2 3 3 3 2 3 2 3 3 1 2 3 3 2 2 2 3 2 3 3 3 3 3 3 1",
"output": "10"
},
{
"input": "75\n1 0 0 1 1 0 0 1 0 1 2 0 0 2 1 1 0 0 0 0 0 0 2 1 1 0 0 0 0 1 0 1 0 1 1 1 0 1 0 0 1 0 0 0 0 0 0 1 1 0 0 1 2 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 1 1 0 1 0",
"output": "51"
},
{
"input": "75\n1 3 3 3 1 1 3 2 3 3 1 3 3 3 2 1 3 2 2 3 1 1 1 1 1 1 2 3 3 3 3 3 3 2 3 3 3 3 3 2 3 3 2 2 2 1 2 3 3 2 2 3 0 1 1 3 3 0 0 1 1 3 2 3 3 3 3 1 2 2 3 3 3 3 1",
"output": "16"
},
{
"input": "75\n3 3 3 3 2 2 3 2 2 3 2 2 1 2 3 3 2 2 3 3 1 2 2 2 1 3 3 3 1 2 2 3 3 3 2 3 2 2 2 3 3 1 3 2 2 3 3 3 0 3 2 1 3 3 2 3 3 3 3 1 2 3 3 3 2 2 3 3 3 3 2 2 3 3 1",
"output": "11"
},
{
"input": "80\n0 0 0 0 2 0 1 1 1 1 1 0 0 0 0 2 0 0 1 0 0 0 0 1 1 0 2 2 1 1 0 1 0 1 0 1 1 1 0 1 2 1 1 0 0 0 1 1 0 1 1 0 1 0 0 1 0 0 1 0 0 0 0 0 0 0 2 2 0 1 1 0 0 0 0 0 0 0 0 1",
"output": "56"
},
{
"input": "80\n2 2 3 3 2 1 0 1 0 3 2 2 3 2 1 3 1 3 3 2 3 3 3 2 3 3 3 2 1 3 3 1 3 3 3 3 3 3 2 2 2 1 3 2 1 3 2 1 1 0 1 1 2 1 3 0 1 2 3 2 2 3 2 3 1 3 3 2 1 1 0 3 3 3 3 1 2 1 2 0",
"output": "17"
},
{
"input": "80\n2 3 3 2 2 2 3 3 2 3 3 3 3 3 2 3 2 3 2 3 3 3 3 3 3 3 3 3 2 3 1 3 2 3 3 0 3 1 2 3 3 1 2 3 2 3 3 2 3 3 3 3 3 2 2 3 0 3 3 3 3 3 2 2 3 2 3 3 3 3 3 2 3 2 3 3 3 3 2 3",
"output": "9"
},
{
"input": "85\n0 1 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 2 0 1 0 0 2 0 1 1 0 0 0 0 2 2 0 0 0 1 0 0 0 1 2 0 1 0 0 0 2 1 1 2 0 3 1 0 2 2 1 0 0 1 1 0 0 0 0 1 0 2 1 1 2 1 0 0 1 2 1 2 0 0 1 0 1 0",
"output": "54"
},
{
"input": "85\n2 3 1 3 2 3 1 3 3 2 1 2 1 2 2 3 2 2 3 2 0 3 3 2 1 2 2 2 3 3 2 3 3 3 2 1 1 3 1 3 2 2 2 3 3 2 3 2 3 1 1 3 2 3 1 3 3 2 3 3 2 2 3 0 1 1 2 2 2 2 1 2 3 1 3 3 1 3 2 2 3 2 3 3 3",
"output": "19"
},
{
"input": "85\n1 2 1 2 3 2 3 3 3 3 3 3 3 2 1 3 2 3 3 3 3 2 3 3 3 1 3 3 3 3 2 3 3 3 3 3 3 2 2 1 3 3 3 3 2 2 3 1 1 2 3 3 3 2 3 3 3 3 3 2 3 3 3 2 2 3 3 1 1 1 3 3 3 3 1 3 3 3 1 3 3 1 3 2 3",
"output": "9"
},
{
"input": "90\n2 0 1 0 0 0 0 0 0 1 1 2 0 0 0 0 0 0 0 2 2 0 2 0 0 2 1 0 2 0 1 0 1 0 0 1 2 2 0 0 1 0 0 1 0 1 0 2 0 1 1 1 0 1 1 0 1 0 2 0 1 0 1 0 0 0 1 0 0 1 2 0 0 0 1 0 0 2 2 0 0 0 0 0 1 3 1 1 0 1",
"output": "57"
},
{
"input": "90\n2 3 3 3 2 3 2 1 3 0 3 2 3 3 2 1 3 3 2 3 2 3 3 2 1 3 1 3 3 1 2 2 3 3 2 1 2 3 2 3 0 3 3 2 2 3 1 0 3 3 1 3 3 3 3 2 1 2 2 1 3 2 1 3 3 1 2 0 2 2 3 2 2 3 3 3 1 3 2 1 2 3 3 2 3 2 3 3 2 1",
"output": "17"
},
{
"input": "90\n2 3 2 3 2 2 3 3 2 3 2 1 2 3 3 3 2 3 2 3 3 2 3 3 3 1 3 3 1 3 2 3 2 2 1 3 3 3 3 3 3 3 3 3 3 2 3 2 3 2 1 3 3 3 3 2 2 3 3 3 3 3 3 3 3 3 3 3 3 2 2 3 3 3 3 1 3 2 3 3 3 2 2 3 2 3 2 1 3 2",
"output": "9"
},
{
"input": "95\n0 0 3 0 2 0 1 0 0 2 0 0 0 0 0 0 0 1 0 0 0 2 0 0 0 0 0 1 0 0 2 1 0 0 1 0 0 0 1 0 0 0 0 1 0 1 0 0 1 0 1 2 0 1 2 2 0 0 1 0 2 0 0 0 1 0 2 1 2 1 0 1 0 0 0 1 0 0 1 1 2 1 1 1 1 2 0 0 0 0 0 1 1 0 1",
"output": "61"
},
{
"input": "95\n2 3 3 2 1 1 3 3 3 2 3 3 3 2 3 2 3 3 3 2 3 2 2 3 3 2 1 2 3 3 3 1 3 0 3 3 1 3 3 1 0 1 3 3 3 0 2 1 3 3 3 3 0 1 3 2 3 3 2 1 3 1 2 1 1 2 3 0 3 3 2 1 3 2 1 3 3 3 2 2 3 2 3 3 3 2 1 3 3 3 2 3 3 1 2",
"output": "15"
},
{
"input": "95\n2 3 3 2 3 2 2 1 3 1 2 1 2 3 1 2 3 3 1 3 3 3 1 2 3 2 2 2 2 3 3 3 2 2 3 3 3 3 3 1 2 2 3 3 3 3 2 3 2 2 2 3 3 2 3 3 3 3 3 3 3 0 3 2 0 3 3 1 3 3 3 2 3 2 3 2 3 3 3 3 2 2 1 1 3 3 3 3 3 1 3 3 3 3 2",
"output": "14"
},
{
"input": "100\n1 0 2 0 0 0 0 2 0 0 0 1 0 1 0 0 1 0 1 2 0 1 1 0 0 1 0 1 1 0 0 0 2 0 1 0 0 2 0 0 0 0 0 1 1 1 0 0 1 0 2 0 0 0 0 1 0 1 0 1 0 1 0 1 2 2 0 0 2 0 1 0 1 0 1 0 0 0 1 0 0 2 1 1 1 0 0 1 0 0 0 2 0 0 2 1 1 0 0 2",
"output": "63"
},
{
"input": "100\n3 2 1 3 2 3 2 3 2 2 3 1 3 3 3 3 3 2 2 3 2 2 3 2 3 3 3 2 3 1 2 1 3 3 3 3 1 3 3 3 3 3 2 3 2 1 3 3 1 2 2 3 1 3 3 1 2 2 1 3 1 3 2 2 3 3 1 3 2 3 1 2 1 2 3 3 2 2 1 2 3 3 3 3 3 1 3 3 3 3 2 1 3 0 3 3 3 2 3 3",
"output": "15"
},
{
"input": "100\n1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2",
"output": "0"
},
{
"input": "100\n3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3",
"output": "0"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1",
"output": "50"
},
{
"input": "100\n2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2",
"output": "50"
},
{
"input": "99\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",
"output": "49"
},
{
"input": "100\n2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1",
"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": "100"
},
{
"input": "2\n0 3",
"output": "1"
},
{
"input": "2\n1 0",
"output": "1"
},
{
"input": "2\n1 2",
"output": "0"
},
{
"input": "2\n2 0",
"output": "1"
},
{
"input": "2\n2 1",
"output": "0"
},
{
"input": "2\n2 3",
"output": "0"
},
{
"input": "2\n3 0",
"output": "1"
},
{
"input": "2\n3 1",
"output": "0"
},
{
"input": "100\n3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0 3 0",
"output": "50"
}
] | 1,686,714,385
| 2,147,483,647
|
PyPy 3-64
|
WRONG_ANSWER
|
TESTS
| 5
| 46
| 0
|
n=int(input())
x=list(map(int,input().split()))
def solve(i,p=0):
if i==n:
return i+1
if x[i]==1 and p!=1:
return solve(i+1,1)
elif x[i]==2 and p!=2:
return solve(i+1,2)
elif x[i]==3:
if p:
return solve(i+1,[2,1][p-1])
else:
return min(solve(i+1,1),solve(i+1,2))
else:
return i+1
f=-1
e=0
while e<=n:
f+=1
e=solve(e)
print(f)
|
Title: Vacations
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Vasya has *n* days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this *n* days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the *i*-th day there are four options:
1. on this day the gym is closed and the contest is not carried out; 1. on this day the gym is closed and the contest is carried out; 1. on this day the gym is open and the contest is not carried out; 1. on this day the gym is open and the contest is carried out.
On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).
Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.
Input Specification:
The first line contains a positive integer *n* (1<=≤<=*n*<=≤<=100) — the number of days of Vasya's vacations.
The second line contains the sequence of integers *a*1,<=*a*2,<=...,<=*a**n* (0<=≤<=*a**i*<=≤<=3) separated by space, where:
- *a**i* equals 0, if on the *i*-th day of vacations the gym is closed and the contest is not carried out; - *a**i* equals 1, if on the *i*-th day of vacations the gym is closed, but the contest is carried out; - *a**i* equals 2, if on the *i*-th day of vacations the gym is open and the contest is not carried out; - *a**i* equals 3, if on the *i*-th day of vacations the gym is open and the contest is carried out.
Output Specification:
Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:
- to do sport on any two consecutive days, - to write the contest on any two consecutive days.
Demo Input:
['4\n1 3 2 0\n', '7\n1 3 3 2 1 2 3\n', '2\n2 2\n']
Demo Output:
['2\n', '0\n', '1\n']
Note:
In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.
In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.
In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.
|
```python
n=int(input())
x=list(map(int,input().split()))
def solve(i,p=0):
if i==n:
return i+1
if x[i]==1 and p!=1:
return solve(i+1,1)
elif x[i]==2 and p!=2:
return solve(i+1,2)
elif x[i]==3:
if p:
return solve(i+1,[2,1][p-1])
else:
return min(solve(i+1,1),solve(i+1,2))
else:
return i+1
f=-1
e=0
while e<=n:
f+=1
e=solve(e)
print(f)
```
| 0
|
|
302
|
A
|
Eugeny and Array
|
PROGRAMMING
| 800
|
[
"implementation"
] | null | null |
Eugeny has array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n*, consisting of *n* integers. Each integer *a**i* equals to -1, or to 1. Also, he has *m* queries:
- Query number *i* is given as a pair of integers *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). - The response to the query will be integer 1, if the elements of array *a* can be rearranged so as the sum *a**l**i*<=+<=*a**l**i*<=+<=1<=+<=...<=+<=*a**r**i*<==<=0, otherwise the response to the query will be integer 0.
Help Eugeny, answer all his queries.
|
The first line contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=2·105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (*a**i*<==<=-1,<=1). Next *m* lines contain Eugene's queries. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*).
|
Print *m* integers — the responses to Eugene's queries in the order they occur in the input.
|
[
"2 3\n1 -1\n1 1\n1 2\n2 2\n",
"5 5\n-1 1 1 1 -1\n1 1\n2 3\n3 5\n2 5\n1 5\n"
] |
[
"0\n1\n0\n",
"0\n1\n0\n1\n0\n"
] |
none
| 500
|
[
{
"input": "2 3\n1 -1\n1 1\n1 2\n2 2",
"output": "0\n1\n0"
},
{
"input": "5 5\n-1 1 1 1 -1\n1 1\n2 3\n3 5\n2 5\n1 5",
"output": "0\n1\n0\n1\n0"
},
{
"input": "3 3\n1 1 1\n2 2\n1 1\n1 1",
"output": "0\n0\n0"
},
{
"input": "4 4\n-1 -1 -1 -1\n1 3\n1 2\n1 2\n1 1",
"output": "0\n0\n0\n0"
},
{
"input": "5 5\n-1 -1 -1 -1 -1\n1 1\n1 1\n3 4\n1 1\n1 4",
"output": "0\n0\n0\n0\n0"
},
{
"input": "6 6\n-1 -1 1 -1 -1 1\n1 1\n3 4\n1 1\n1 1\n1 3\n1 4",
"output": "0\n1\n0\n0\n0\n1"
},
{
"input": "7 7\n-1 -1 -1 1 -1 -1 -1\n1 1\n2 7\n1 3\n1 5\n4 7\n1 7\n6 7",
"output": "0\n0\n0\n0\n0\n0\n1"
},
{
"input": "8 8\n1 1 1 1 1 1 1 1\n5 8\n2 6\n2 3\n1 7\n7 7\n1 6\n1 8\n1 3",
"output": "0\n0\n0\n0\n0\n0\n0\n0"
},
{
"input": "9 9\n-1 1 1 1 1 1 1 1 1\n1 7\n5 6\n1 4\n1 1\n1 1\n6 8\n1 1\n6 7\n3 5",
"output": "0\n1\n0\n0\n0\n0\n0\n1\n0"
},
{
"input": "10 10\n-1 1 -1 1 -1 -1 -1 -1 -1 -1\n6 7\n2 5\n3 6\n1 3\n3 5\n4 5\n3 4\n1 6\n1 1\n1 1",
"output": "1\n1\n1\n0\n0\n1\n1\n0\n0\n0"
},
{
"input": "1 1\n-1\n1 1",
"output": "0"
},
{
"input": "1 1\n1\n1 1",
"output": "0"
}
] | 1,616,594,111
| 2,147,483,647
|
Python 3
|
TIME_LIMIT_EXCEEDED
|
TESTS
| 10
| 1,000
| 3,788,800
|
n, cases = map(int, input().split())
arr = list(map(int, input().split()))
while cases:
cases -= 1
a, b = map(int, input().split())
minus = 0
plus = 0
for n in arr:
if n > 0:
plus += 1
else:
minus += 1
if (b-a+1) % 2 != 0:
print(0)
else:
if (b-a+1)//2 <= min(minus, plus):
print(1)
else:
print(0)
|
Title: Eugeny and Array
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Eugeny has array *a*<==<=*a*1,<=*a*2,<=...,<=*a**n*, consisting of *n* integers. Each integer *a**i* equals to -1, or to 1. Also, he has *m* queries:
- Query number *i* is given as a pair of integers *l**i*, *r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*). - The response to the query will be integer 1, if the elements of array *a* can be rearranged so as the sum *a**l**i*<=+<=*a**l**i*<=+<=1<=+<=...<=+<=*a**r**i*<==<=0, otherwise the response to the query will be integer 0.
Help Eugeny, answer all his queries.
Input Specification:
The first line contains integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=2·105). The second line contains *n* integers *a*1,<=*a*2,<=...,<=*a**n* (*a**i*<==<=-1,<=1). Next *m* lines contain Eugene's queries. The *i*-th line contains integers *l**i*,<=*r**i* (1<=≤<=*l**i*<=≤<=*r**i*<=≤<=*n*).
Output Specification:
Print *m* integers — the responses to Eugene's queries in the order they occur in the input.
Demo Input:
['2 3\n1 -1\n1 1\n1 2\n2 2\n', '5 5\n-1 1 1 1 -1\n1 1\n2 3\n3 5\n2 5\n1 5\n']
Demo Output:
['0\n1\n0\n', '0\n1\n0\n1\n0\n']
Note:
none
|
```python
n, cases = map(int, input().split())
arr = list(map(int, input().split()))
while cases:
cases -= 1
a, b = map(int, input().split())
minus = 0
plus = 0
for n in arr:
if n > 0:
plus += 1
else:
minus += 1
if (b-a+1) % 2 != 0:
print(0)
else:
if (b-a+1)//2 <= min(minus, plus):
print(1)
else:
print(0)
```
| 0
|
|
386
|
B
|
Fly, freebies, fly!
|
PROGRAMMING
| 1,400
|
[
"binary search",
"brute force",
"implementation"
] | null | null |
Everyone loves a freebie. Especially students.
It is well-known that if in the night before exam a student opens window, opens the student's record-book and shouts loudly three times "Fly, freebie, fly!" — then flown freebie helps him to pass the upcoming exam.
In the night before the exam on mathematical analysis *n* students living in dormitory shouted treasured words. The *i*-th student made a sacrament at the time *t**i*, where *t**i* is the number of seconds elapsed since the beginning of the night.
It is known that the freebie is a capricious and willful lady. That night the freebie was near dormitory only for *T* seconds. Therefore, if for two students their sacrament times differ for more than *T*, then the freebie didn't visit at least one of them.
Since all students are optimists, they really want to know what is the maximal number of students visited by the freebie can be.
|
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=100), where *n* — the number of students shouted "Fly, freebie, fly!" The second line contains *n* positive integers *t**i* (1<=≤<=*t**i*<=≤<=1000).
The last line contains integer *T* (1<=≤<=*T*<=≤<=1000) — the time interval during which the freebie was near the dormitory.
|
Print a single integer — the largest number of people who will pass exam tomorrow because of the freebie visit.
|
[
"6\n4 1 7 8 3 8\n1\n"
] |
[
"3\n"
] |
none
| 1,000
|
[
{
"input": "6\n4 1 7 8 3 8\n1",
"output": "3"
},
{
"input": "4\n4 2 1 5\n2",
"output": "2"
},
{
"input": "10\n4 7 1 3 8 5 2 1 8 4\n3",
"output": "6"
},
{
"input": "8\n39 49 37 28 40 17 50 2\n10",
"output": "3"
},
{
"input": "2\n1 1\n1",
"output": "2"
},
{
"input": "2\n1 1\n2",
"output": "2"
},
{
"input": "2\n1 1\n1000",
"output": "2"
},
{
"input": "2\n1 2\n2",
"output": "2"
},
{
"input": "2\n450 826\n1000",
"output": "2"
},
{
"input": "3\n3 1 1\n1",
"output": "2"
},
{
"input": "3\n3 1 2\n2",
"output": "3"
},
{
"input": "3\n3 4 3\n1",
"output": "3"
},
{
"input": "3\n3 4 3\n1",
"output": "3"
},
{
"input": "100\n63 69 36 40 74 31 86 42 81 95 60 55 98 98 2 16 84 37 61 47 81 91 85 62 85 32 79 74 65 48 39 60 97 90 59 76 98 73 58 5 16 54 59 42 9 27 95 24 9 6 42 49 64 61 22 27 43 60 39 87 99 57 5 62 48 67 81 36 27 87 41 88 5 33 43 81 82 65 46 52 43 68 85 75 81 99 30 56 67 55 92 4 3 3 66 32 30 45 22 88\n5",
"output": "11"
},
{
"input": "100\n97 29 39 42 68 100 44 54 6 70 17 100 52 85 67 1 43 49 1 47 98 35 5 38 37 73 84 20 13 15 78 65 29 92 20 40 38 11 12 100 24 94 29 92 83 47 25 63 23 85 85 93 61 60 35 40 96 50 19 15 28 19 98 59 42 14 54 65 2 53 38 9 15 69 43 63 63 8 55 12 81 57 69 21 57 11 99 45 23 31 59 2 16 61 43 36 12 39 42 13\n50",
"output": "62"
},
{
"input": "100\n31 1 56 82 96 98 25 41 74 73 8 66 95 50 89 77 98 12 69 45 6 10 48 59 1 77 15 77 9 52 66 8 6 71 39 3 58 73 66 45 8 22 67 83 58 6 96 79 46 43 44 90 13 67 56 32 83 96 93 22 49 10 100 79 99 41 13 71 42 96 89 10 84 95 89 7 18 49 16 54 61 35 25 71 26 68 22 40 68 19 30 51 18 20 12 61 11 23 86 72\n1",
"output": "6"
},
{
"input": "100\n30 74 20 6 3 63 48 45 36 26 33 24 60 71 45 5 19 37 74 100 98 82 67 76 37 46 68 48 56 29 33 19 15 84 76 92 50 53 42 19 5 91 23 38 93 50 39 45 89 17 57 14 86 81 31 6 16 5 80 6 86 49 18 75 30 30 85 94 38 33 50 76 72 32 73 96 28 3 18 20 96 84 89 48 71 64 6 59 87 31 94 24 9 64 15 86 66 11 32 40\n90",
"output": "94"
},
{
"input": "100\n398 82 739 637 913 962 680 125 963 931 311 680 20 530 795 126 881 666 226 323 594 416 176 6 820 317 866 723 831 432 139 706 608 218 963 550 592 544 874 927 763 468 121 424 91 956 42 442 883 66 299 654 964 730 160 615 515 255 709 278 224 223 304 292 41 450 445 556 477 327 647 518 90 470 894 837 655 495 612 113 746 610 751 486 116 933 314 348 736 58 219 429 976 773 678 642 696 522 161 422\n1",
"output": "3"
},
{
"input": "100\n760 621 622 793 66 684 411 813 474 404 304 934 319 411 99 965 722 156 681 400 481 462 571 726 696 244 124 350 403 566 564 641 381 494 703 3 348 213 343 390 27 660 46 591 990 931 477 823 890 21 936 267 282 753 599 269 387 443 622 673 473 745 646 224 911 7 155 880 332 932 51 994 144 666 789 691 323 738 192 372 191 246 903 666 929 252 132 614 11 938 298 286 309 596 210 18 143 760 759 584\n10",
"output": "6"
},
{
"input": "100\n923 357 749 109 685 126 961 437 859 91 985 488 644 777 950 144 479 667 1 535 475 38 843 606 672 333 798 42 595 854 410 914 934 586 329 595 861 321 603 924 434 636 475 395 619 449 336 790 279 931 605 898 276 47 537 935 508 576 168 465 115 884 960 593 883 581 468 426 848 289 525 309 589 106 924 238 829 975 897 373 650 41 952 621 817 46 366 488 924 561 960 449 311 32 517 737 20 765 799 3\n100",
"output": "18"
},
{
"input": "100\n98 63 672 100 254 218 623 415 426 986 920 915 736 795 407 541 382 213 935 743 961 59 660 512 134 935 248 378 739 356 543 714 28 667 602 596 759 791 103 564 225 520 159 542 966 332 983 655 517 273 95 242 593 940 286 236 41 318 941 727 384 225 319 627 982 359 232 769 854 172 643 598 215 231 305 30 347 469 929 919 90 294 739 641 368 270 932 452 234 741 309 234 357 392 707 873 808 398 417 483\n1000",
"output": "100"
},
{
"input": "100\n1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1\n1",
"output": "100"
},
{
"input": "100\n2 1 1 1 2 2 2 2 2 2 1 1 1 1 2 2 1 1 1 2 2 1 1 1 1 2 1 2 1 2 1 2 1 2 2 2 1 1 2 1 2 2 1 1 2 2 2 2 2 1 1 2 1 1 1 2 1 2 1 2 1 2 1 1 2 1 2 1 2 1 2 1 2 1 1 2 2 1 2 2 1 1 1 2 2 2 1 1 2 2 1 2 2 2 1 2 2 1 2 2\n1",
"output": "100"
},
{
"input": "100\n3 3 1 2 3 3 1 3 3 2 2 2 2 1 2 3 2 1 2 2 2 2 3 2 1 3 3 3 2 1 3 1 2 1 1 2 2 3 2 2 3 1 1 3 1 2 1 3 3 1 1 3 1 3 2 3 3 2 2 2 2 1 1 1 2 1 1 2 1 1 1 1 1 3 2 2 1 3 1 1 3 1 2 2 1 3 1 1 1 1 2 2 2 3 2 2 3 1 1 3\n1",
"output": "72"
},
{
"input": "100\n2 1 3 4 1 1 4 1 3 2 1 4 4 4 4 4 3 2 1 1 2 2 1 3 3 1 1 1 2 3 4 3 1 1 1 4 2 2 2 2 4 1 2 4 2 2 4 3 3 4 1 2 4 1 3 4 1 2 1 2 1 3 3 2 1 1 4 2 1 3 3 2 3 4 1 2 2 4 2 1 4 3 4 3 1 4 3 1 2 3 3 3 2 4 1 1 4 1 2 3\n1",
"output": "55"
},
{
"input": "100\n5 1 3 1 2 3 2 5 5 2 5 1 1 4 1 1 3 5 3 3 3 3 4 4 3 5 4 1 1 3 1 4 2 5 2 5 4 2 3 5 1 3 5 5 5 2 2 5 1 4 1 5 1 5 1 3 3 2 2 4 3 2 1 4 2 5 4 1 2 1 4 3 3 5 4 3 5 5 1 2 4 1 4 2 1 1 2 5 3 3 4 1 3 3 3 5 4 1 1 1\n1",
"output": "41"
},
{
"input": "100\n1 7 8 10 9 4 2 1 6 5 10 6 3 1 10 1 8 4 3 1 7 4 3 7 4 9 1 3 3 5 10 3 7 10 10 10 3 6 2 8 1 3 3 6 2 8 3 7 8 3 4 1 6 4 4 2 10 6 2 10 10 1 7 8 8 1 9 8 7 8 5 2 5 9 2 5 7 10 3 9 8 3 9 4 3 8 6 8 2 8 9 6 7 10 7 9 6 4 4 8\n1",
"output": "24"
},
{
"input": "1\n1\n1",
"output": "1"
},
{
"input": "1\n1\n1000",
"output": "1"
},
{
"input": "1\n849\n1",
"output": "1"
}
] | 1,695,401,858
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 31
| 0
|
N = int(input())
lst = list(map(int,input().split()))
T = int(input())
lst.sort()
count = 0
for i in range(len(lst)):
l, r = 0, len(lst)-1
while l<r:
m = (l+r)//2
if lst[i] - lst[m] == 1:
count+=1
break
elif lst[i] - lst[m] > 1:
l+=1
else:
r-=1
print(count)
|
Title: Fly, freebies, fly!
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Everyone loves a freebie. Especially students.
It is well-known that if in the night before exam a student opens window, opens the student's record-book and shouts loudly three times "Fly, freebie, fly!" — then flown freebie helps him to pass the upcoming exam.
In the night before the exam on mathematical analysis *n* students living in dormitory shouted treasured words. The *i*-th student made a sacrament at the time *t**i*, where *t**i* is the number of seconds elapsed since the beginning of the night.
It is known that the freebie is a capricious and willful lady. That night the freebie was near dormitory only for *T* seconds. Therefore, if for two students their sacrament times differ for more than *T*, then the freebie didn't visit at least one of them.
Since all students are optimists, they really want to know what is the maximal number of students visited by the freebie can be.
Input Specification:
The first line of the input contains integer *n* (1<=≤<=*n*<=≤<=100), where *n* — the number of students shouted "Fly, freebie, fly!" The second line contains *n* positive integers *t**i* (1<=≤<=*t**i*<=≤<=1000).
The last line contains integer *T* (1<=≤<=*T*<=≤<=1000) — the time interval during which the freebie was near the dormitory.
Output Specification:
Print a single integer — the largest number of people who will pass exam tomorrow because of the freebie visit.
Demo Input:
['6\n4 1 7 8 3 8\n1\n']
Demo Output:
['3\n']
Note:
none
|
```python
N = int(input())
lst = list(map(int,input().split()))
T = int(input())
lst.sort()
count = 0
for i in range(len(lst)):
l, r = 0, len(lst)-1
while l<r:
m = (l+r)//2
if lst[i] - lst[m] == 1:
count+=1
break
elif lst[i] - lst[m] > 1:
l+=1
else:
r-=1
print(count)
```
| 0
|
|
343
|
B
|
Alternating Current
|
PROGRAMMING
| 1,600
|
[
"data structures",
"greedy",
"implementation"
] | null | null |
Mad scientist Mike has just finished constructing a new device to search for extraterrestrial intelligence! He was in such a hurry to launch it for the first time that he plugged in the power wires without giving it a proper glance and started experimenting right away. After a while Mike observed that the wires ended up entangled and now have to be untangled again.
The device is powered by two wires "plus" and "minus". The wires run along the floor from the wall (on the left) to the device (on the right). Both the wall and the device have two contacts in them on the same level, into which the wires are plugged in some order. The wires are considered entangled if there are one or more places where one wire runs above the other one. For example, the picture below has four such places (top view):
Mike knows the sequence in which the wires run above each other. Mike also noticed that on the left side, the "plus" wire is always plugged into the top contact (as seen on the picture). He would like to untangle the wires without unplugging them and without moving the device. Determine if it is possible to do that. A wire can be freely moved and stretched on the floor, but cannot be cut.
To understand the problem better please read the notes to the test samples.
|
The single line of the input contains a sequence of characters "+" and "-" of length *n* (1<=≤<=*n*<=≤<=100000). The *i*-th (1<=≤<=*i*<=≤<=*n*) position of the sequence contains the character "+", if on the *i*-th step from the wall the "plus" wire runs above the "minus" wire, and the character "-" otherwise.
|
Print either "Yes" (without the quotes) if the wires can be untangled or "No" (without the quotes) if the wires cannot be untangled.
|
[
"-++-\n",
"+-\n",
"++\n",
"-\n"
] |
[
"Yes\n",
"No\n",
"Yes\n",
"No\n"
] |
The first testcase corresponds to the picture in the statement. To untangle the wires, one can first move the "plus" wire lower, thus eliminating the two crosses in the middle, and then draw it under the "minus" wire, eliminating also the remaining two crosses.
In the second testcase the "plus" wire makes one full revolution around the "minus" wire. Thus the wires cannot be untangled:
In the third testcase the "plus" wire simply runs above the "minus" wire twice in sequence. The wires can be untangled by lifting "plus" and moving it higher:
In the fourth testcase the "minus" wire runs above the "plus" wire once. The wires cannot be untangled without moving the device itself:
| 1,000
|
[
{
"input": "-++-",
"output": "Yes"
},
{
"input": "+-",
"output": "No"
},
{
"input": "++",
"output": "Yes"
},
{
"input": "-",
"output": "No"
},
{
"input": "+-+-",
"output": "No"
},
{
"input": "-+-",
"output": "No"
},
{
"input": "-++-+--+",
"output": "Yes"
},
{
"input": "+",
"output": "No"
},
{
"input": "-+",
"output": "No"
},
{
"input": "--",
"output": "Yes"
},
{
"input": "+++",
"output": "No"
},
{
"input": "--+",
"output": "No"
},
{
"input": "++--++",
"output": "Yes"
},
{
"input": "+-++-+",
"output": "Yes"
},
{
"input": "+-+--+",
"output": "No"
},
{
"input": "--++-+",
"output": "No"
},
{
"input": "-+-+--",
"output": "No"
},
{
"input": "+-+++-",
"output": "No"
},
{
"input": "-+-+-+",
"output": "No"
},
{
"input": "-++-+--++--+-++-",
"output": "Yes"
},
{
"input": "+-----+-++---+------+++-++++",
"output": "No"
},
{
"input": "-+-++--+++-++++---+--+----+--+-+-+++-+++-+---++-++++-+--+--+--+-+-++-+-+-++++++---++--+++++-+--++--+-+--++-----+--+-++---+++---++----+++-++++--++-++-",
"output": "No"
},
{
"input": "-+-----++++--++-+-++",
"output": "Yes"
},
{
"input": "+--+--+------+++++++-+-+++--++---+--+-+---+--+++-+++-------+++++-+-++++--+-+-+++++++----+----+++----+-+++-+++-----+++-+-++-+-+++++-+--++----+--+-++-----+-+-++++---+++---+-+-+-++++--+--+++---+++++-+---+-----+++-++--+++---++-++-+-+++-+-+-+---+++--+--++++-+-+--++-------+--+---++-----+++--+-+++--++-+-+++-++--+++-++++++++++-++-++++++-+++--+--++-+++--+++-++++----+++---+-+----++++-+-+",
"output": "Yes"
},
{
"input": "-+-+-++-+-+-",
"output": "Yes"
},
{
"input": "-+-++-+-",
"output": "Yes"
},
{
"input": "-+-++-+-+-",
"output": "No"
},
{
"input": "++-+-+-+-+--+",
"output": "No"
},
{
"input": "+++---",
"output": "No"
},
{
"input": "+-+-+-+-+--+-+-+-+-++--++--+",
"output": "Yes"
},
{
"input": "+-+-++",
"output": "No"
},
{
"input": "-++--+--+++-+-+-+-+-",
"output": "No"
},
{
"input": "+---+-+-",
"output": "No"
},
{
"input": "+-+--+-+",
"output": "Yes"
},
{
"input": "+++---+++---",
"output": "No"
},
{
"input": "-+++++",
"output": "No"
},
{
"input": "-+-+-+-+-+-+-++-+-+-+-+-+-+-",
"output": "Yes"
},
{
"input": "-+++--",
"output": "No"
},
{
"input": "+---+",
"output": "No"
},
{
"input": "-++",
"output": "No"
},
{
"input": "-+--+-",
"output": "Yes"
},
{
"input": "+---++--++",
"output": "No"
},
{
"input": "+++-",
"output": "No"
},
{
"input": "--+++",
"output": "No"
},
{
"input": "++-+",
"output": "No"
}
] | 1,547,694,224
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 62
| 248
| 1,228,800
|
t = input()
#-+-++--+
stack = []
for i in t:
try:
if stack[-1] == i:
stack.pop()
else:
stack.append(i)
except:
stack.append(i)
if len(stack) == 0:
print("Yes")
else:
print("No")
|
Title: Alternating Current
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Mad scientist Mike has just finished constructing a new device to search for extraterrestrial intelligence! He was in such a hurry to launch it for the first time that he plugged in the power wires without giving it a proper glance and started experimenting right away. After a while Mike observed that the wires ended up entangled and now have to be untangled again.
The device is powered by two wires "plus" and "minus". The wires run along the floor from the wall (on the left) to the device (on the right). Both the wall and the device have two contacts in them on the same level, into which the wires are plugged in some order. The wires are considered entangled if there are one or more places where one wire runs above the other one. For example, the picture below has four such places (top view):
Mike knows the sequence in which the wires run above each other. Mike also noticed that on the left side, the "plus" wire is always plugged into the top contact (as seen on the picture). He would like to untangle the wires without unplugging them and without moving the device. Determine if it is possible to do that. A wire can be freely moved and stretched on the floor, but cannot be cut.
To understand the problem better please read the notes to the test samples.
Input Specification:
The single line of the input contains a sequence of characters "+" and "-" of length *n* (1<=≤<=*n*<=≤<=100000). The *i*-th (1<=≤<=*i*<=≤<=*n*) position of the sequence contains the character "+", if on the *i*-th step from the wall the "plus" wire runs above the "minus" wire, and the character "-" otherwise.
Output Specification:
Print either "Yes" (without the quotes) if the wires can be untangled or "No" (without the quotes) if the wires cannot be untangled.
Demo Input:
['-++-\n', '+-\n', '++\n', '-\n']
Demo Output:
['Yes\n', 'No\n', 'Yes\n', 'No\n']
Note:
The first testcase corresponds to the picture in the statement. To untangle the wires, one can first move the "plus" wire lower, thus eliminating the two crosses in the middle, and then draw it under the "minus" wire, eliminating also the remaining two crosses.
In the second testcase the "plus" wire makes one full revolution around the "minus" wire. Thus the wires cannot be untangled:
In the third testcase the "plus" wire simply runs above the "minus" wire twice in sequence. The wires can be untangled by lifting "plus" and moving it higher:
In the fourth testcase the "minus" wire runs above the "plus" wire once. The wires cannot be untangled without moving the device itself:
|
```python
t = input()
#-+-++--+
stack = []
for i in t:
try:
if stack[-1] == i:
stack.pop()
else:
stack.append(i)
except:
stack.append(i)
if len(stack) == 0:
print("Yes")
else:
print("No")
```
| 3
|
|
548
|
A
|
Mike and Fax
|
PROGRAMMING
| 1,100
|
[
"brute force",
"implementation",
"strings"
] | null | null |
While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string *s*.
He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly *k* messages in his own bag, each was a palindrome string and all those strings had the same length.
He asked you to help him and tell him if he has worn his own back-bag. Check if the given string *s* is a concatenation of *k* palindromes of the same length.
|
The first line of input contains string *s* containing lowercase English letters (1<=≤<=|*s*|<=≤<=1000).
The second line contains integer *k* (1<=≤<=*k*<=≤<=1000).
|
Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise.
|
[
"saba\n2\n",
"saddastavvat\n2\n"
] |
[
"NO\n",
"YES\n"
] |
Palindrome is a string reading the same forward and backward.
In the second sample, the faxes in his back-bag can be "saddas" and "tavvat".
| 500
|
[
{
"input": "saba\n2",
"output": "NO"
},
{
"input": "saddastavvat\n2",
"output": "YES"
},
{
"input": "aaaaaaaaaa\n3",
"output": "NO"
},
{
"input": "aaaaaa\n3",
"output": "YES"
},
{
"input": "abaacca\n2",
"output": "NO"
},
{
"input": "a\n1",
"output": "YES"
},
{
"input": "princeofpersia\n1",
"output": "NO"
},
{
"input": "xhwbdoryfiaxglripavycmxmcejbcpzidrqsqvikfzjyfnmedxrvlnusavyhillaxrblkynwdrlhthtqzjktzkullgrqsolqssocpfwcaizhovajlhmeibhiuwtxpljkyyiwykzpmazkkzampzkywiyykjlpxtwuihbiemhljavohziacwfpcossqlosqrgllukztkjzqththlrdwnyklbrxallihyvasunlvrxdemnfyjzfkivqsqrdizpcbjecmxmcyvapirlgxaifyrodbwhx\n1",
"output": "YES"
},
{
"input": "yfhqnbzaqeqmcvtsbcdn\n456",
"output": "NO"
},
{
"input": "lgsdfiforlqrohhjyzrigewkigiiffvbyrapzmjvtkklndeyuqpuukajgtguhlarjdqlxksyekbjgrmhuyiqdlzjqqzlxufffpelyptodwhvkfbalxbufrlcsjgxmfxeqsszqghcustqrqjljattgvzynyvfbjgbuynbcguqtyfowgtcbbaywvcrgzrulqpghwoflutswu\n584",
"output": "NO"
},
{
"input": "awlrhmxxivqbntvtapwkdkunamcqoerfncfmookhdnuxtttlxmejojpwbdyxirdsjippzjhdrpjepremruczbedxrjpodlyyldopjrxdebzcurmerpejprdhjzppijsdrixydbwpjojemxltttxundhkoomfcnfreoqcmanukdkwpatvtnbqvixxmhrlwa\n1",
"output": "YES"
},
{
"input": "kafzpsglcpzludxojtdhzynpbekzssvhzizfrboxbhqvojiqtjitrackqccxgenwwnegxccqkcartijtqijovqhbxobrfzizhvsszkebpnyzhdtjoxdulzpclgspzfakvcbbjejeubvrrzlvjjgrcprntbyuakoxowoybbxgdugjffgbtfwrfiobifrshyaqqayhsrfiboifrwftbgffjgudgxbbyowoxokauybtnrpcrgjjvlzrrvbuejejbbcv\n2",
"output": "YES"
},
{
"input": "zieqwmmbrtoxysvavwdemmdeatfrolsqvvlgphhhmojjfxfurtuiqdiilhlcwwqedlhblrzmvuoaczcwrqzyymiggpvbpkycibsvkhytrzhguksxyykkkvfljbbnjblylftmqxkojithwsegzsaexlpuicexbdzpwesrkzbqltxhifwqcehzsjgsqbwkujvjbjpqxdpmlimsusumizizpyigmkxwuberthdghnepyrxzvvidxeafwylegschhtywvqsxuqmsddhkzgkdiekodqpnftdyhnpicsnbhfxemxllvaurkmjvtrmqkulerxtaolmokiqqvqgechkqxmendpmgxwiaffcajmqjmvrwryzxujmiasuqtosuisiclnv\n8",
"output": "NO"
},
{
"input": "syghzncbi\n829",
"output": "NO"
},
{
"input": "ljpdpstntznciejqqtpysskztdfawuncqzwwfefrfsihyrdopwawowshquqnjhesxszuywezpebpzhtopgngrnqgwnoqhyrykojguybvdbjpfpmvkxscocywzsxcivysfrrzsonayztzzuybrkiombhqcfkszyscykzistiobrpavezedgobowjszfadcccmxyqehmkgywiwxffibzetb\n137",
"output": "NO"
},
{
"input": "eytuqriplfczwsqlsnjetfpzehzvzayickkbnfqddaisfpasvigwtnvbybwultsgrtjbaebktvubwofysgidpufzteuhuaaqkhmhguockoczlrmlrrzouvqtwbcchxxiydbohnvrmtqjzhkfmvdulojhdvgwudvidpausvfujkjprxsobliuauxleqvsmz\n253",
"output": "NO"
},
{
"input": "xkaqgwabuilhuqwhnrdtyattmqcjfbiqodjlwzgcyvghqncklbhnlmagvjvwysrfryrlmclninogumjfmyenkmydlmifxpkvlaapgnfarejaowftxxztshsesjtsgommaeslrhronruqdurvjesydrzmxirmxumrcqezznqltngsgdcthivdnjnshjfujtiqsltpttgbljfcbqsfwbzokciqlavrthgaqbzikpwwsebzwddlvdwrmztwmhcxdinwlbklwmteeybbdbzevfbsrtldapulwgusuvnreiflkytonzmervyrlbqhzapgxepwauaiwygpxarfeyqhimzlxntjuaaigeisgrvwgbhqemqetzyallzaoqprhzpjibkutgwrodruqu\n857",
"output": "NO"
},
{
"input": "rbehjxpblnzfgeebpkvzznwtzszghjuuxovreapmwehqyjymrkmksffbdpbdyegulabsmjiykeeqtuvqqyxlitpxjdpwmqtlmudqsksgwqekvwfjdsggzajcpsyserkctpbajgzdbiqaekfaepnecezdzamqszpwfvhlannszgaiewvcdnnvzhblmuzjtqeyjcqjqoxatavavokyxokuxwuqueskktxnxraihnqovrfykpzsyjmrhqsvbobzsnfqwvdwatvjxaubtiwhgvrqwjhgfnauqegqmcwnaruinplmtmxhvohhqgvcatsdqiwkvslhqzqevdgdhpijjevuyuitnixdjhwlmjsstmugownjdhvxueg\n677",
"output": "NO"
},
{
"input": "a\n3",
"output": "NO"
},
{
"input": "aa\n2",
"output": "YES"
},
{
"input": "abcaabca\n2",
"output": "NO"
},
{
"input": "aaaaaaaaab\n5",
"output": "NO"
},
{
"input": "aabbb\n3",
"output": "NO"
},
{
"input": "abcde\n5",
"output": "YES"
},
{
"input": "aabc\n2",
"output": "NO"
},
{
"input": "aabcaa\n3",
"output": "NO"
}
] | 1,438,423,633
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 4
| 46
| 0
|
def isPlaindrome(expression):
if(len(expression) <= 1):
return True
else:
if(expression[len(expression)-1] == expression[0]):
return isPlaindrome(expression[1:(len(expression)-1)])
else:
return False
s = input()
k = int(input())
l = len(s)//k
count = 0
b = False
i = 0
while i < len(s):
b = isPlaindrome(s[i:i+l])
if(b):
count += 1
i += l
if count == k:
print("YES")
else:
print("NO")
|
Title: Mike and Fax
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string *s*.
He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly *k* messages in his own bag, each was a palindrome string and all those strings had the same length.
He asked you to help him and tell him if he has worn his own back-bag. Check if the given string *s* is a concatenation of *k* palindromes of the same length.
Input Specification:
The first line of input contains string *s* containing lowercase English letters (1<=≤<=|*s*|<=≤<=1000).
The second line contains integer *k* (1<=≤<=*k*<=≤<=1000).
Output Specification:
Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise.
Demo Input:
['saba\n2\n', 'saddastavvat\n2\n']
Demo Output:
['NO\n', 'YES\n']
Note:
Palindrome is a string reading the same forward and backward.
In the second sample, the faxes in his back-bag can be "saddas" and "tavvat".
|
```python
def isPlaindrome(expression):
if(len(expression) <= 1):
return True
else:
if(expression[len(expression)-1] == expression[0]):
return isPlaindrome(expression[1:(len(expression)-1)])
else:
return False
s = input()
k = int(input())
l = len(s)//k
count = 0
b = False
i = 0
while i < len(s):
b = isPlaindrome(s[i:i+l])
if(b):
count += 1
i += l
if count == k:
print("YES")
else:
print("NO")
```
| 0
|
|
519
|
B
|
A and B and Compilation Errors
|
PROGRAMMING
| 1,100
|
[
"data structures",
"implementation",
"sortings"
] | null | null |
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
|
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
|
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
|
[
"5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n",
"6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n"
] |
[
"8\n123\n",
"1\n3\n"
] |
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
| 1,000
|
[
{
"input": "5\n1 5 8 123 7\n123 7 5 1\n5 1 7",
"output": "8\n123"
},
{
"input": "6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5",
"output": "1\n3"
},
{
"input": "3\n1 2 3\n3 2\n2",
"output": "1\n3"
},
{
"input": "10\n460626451 802090732 277246428 661369649 388684428 784303821 376287098 656422756 9301599 25720377\n277246428 388684428 661369649 460626451 656422756 802090732 9301599 784303821 376287098\n376287098 802090732 388684428 9301599 656422756 784303821 460626451 277246428",
"output": "25720377\n661369649"
},
{
"input": "3\n796067435 964699482 819602309\n964699482 796067435\n964699482",
"output": "819602309\n796067435"
},
{
"input": "3\n374054998 726316780 902899520\n902899520 726316780\n726316780",
"output": "374054998\n902899520"
},
{
"input": "3\n168638990 939116221 323703261\n168638990 323703261\n168638990",
"output": "939116221\n323703261"
},
{
"input": "3\n77 77 77\n77 77\n77",
"output": "77\n77"
},
{
"input": "3\n84 30 9\n9 84\n9",
"output": "30\n84"
},
{
"input": "6\n5 4 3 3 5 5\n3 5 5 4 3\n3 5 4 3",
"output": "5\n5"
},
{
"input": "4\n1 5 7 8\n1 5 7\n1 5",
"output": "8\n7"
}
] | 1,667,767,418
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 33
| 265
| 13,824,000
|
"""
https://codeforces.com/problemset/problem/519/B
"""
_ = input()
a = sorted([int(x) for x in input().split()])
b = sorted([int(x) for x in input().split()]) + [-1]
c = sorted([int(x) for x in input().split()]) + [-1, -1]
for i, z in enumerate(b):
if z != a[i]:
print(a[i])
break
for i, z in enumerate(c):
if z != b[i]:
print(b[i])
break
|
Title: A and B and Compilation Errors
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed *n* compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared — the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input Specification:
The first line of the input contains integer *n* (3<=≤<=*n*<=≤<=105) — the initial number of compilation errors.
The second line contains *n* space-separated integers *a*1,<=*a*2,<=...,<=*a**n* (1<=≤<=*a**i*<=≤<=109) — the errors the compiler displayed for the first time.
The third line contains *n*<=-<=1 space-separated integers *b*1,<=*b*2,<=...,<=*b**n*<=-<=1 — the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains *n*<=-<=2 space-separated integers *с*1,<=*с*2,<=...,<=*с**n*<=-<=2 — the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output Specification:
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Demo Input:
['5\n1 5 8 123 7\n123 7 5 1\n5 1 7\n', '6\n1 4 3 3 5 7\n3 7 5 4 3\n4 3 7 5\n']
Demo Output:
['8\n123\n', '1\n3\n']
Note:
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
|
```python
"""
https://codeforces.com/problemset/problem/519/B
"""
_ = input()
a = sorted([int(x) for x in input().split()])
b = sorted([int(x) for x in input().split()]) + [-1]
c = sorted([int(x) for x in input().split()]) + [-1, -1]
for i, z in enumerate(b):
if z != a[i]:
print(a[i])
break
for i, z in enumerate(c):
if z != b[i]:
print(b[i])
break
```
| 3
|
|
832
|
A
|
Sasha and Sticks
|
PROGRAMMING
| 800
|
[
"games",
"math"
] | null | null |
It's one more school day now. Sasha doesn't like classes and is always bored at them. So, each day he invents some game and plays in it alone or with friends.
Today he invented one simple game to play with Lena, with whom he shares a desk. The rules are simple. Sasha draws *n* sticks in a row. After that the players take turns crossing out exactly *k* sticks from left or right in each turn. Sasha moves first, because he is the inventor of the game. If there are less than *k* sticks on the paper before some turn, the game ends. Sasha wins if he makes strictly more moves than Lena. Sasha wants to know the result of the game before playing, you are to help him.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1018, *k*<=≤<=*n*) — the number of sticks drawn by Sasha and the number *k* — the number of sticks to be crossed out on each turn.
|
If Sasha wins, print "YES" (without quotes), otherwise print "NO" (without quotes).
You can print each letter in arbitrary case (upper of lower).
|
[
"1 1\n",
"10 4\n"
] |
[
"YES\n",
"NO\n"
] |
In the first example Sasha crosses out 1 stick, and then there are no sticks. So Lena can't make a move, and Sasha wins.
In the second example Sasha crosses out 4 sticks, then Lena crosses out 4 sticks, and after that there are only 2 sticks left. Sasha can't make a move. The players make equal number of moves, so Sasha doesn't win.
| 500
|
[
{
"input": "1 1",
"output": "YES"
},
{
"input": "10 4",
"output": "NO"
},
{
"input": "251656215122324104 164397544865601257",
"output": "YES"
},
{
"input": "963577813436662285 206326039287271924",
"output": "NO"
},
{
"input": "1000000000000000000 1",
"output": "NO"
},
{
"input": "253308697183523656 25332878317796706",
"output": "YES"
},
{
"input": "669038685745448997 501718093668307460",
"output": "YES"
},
{
"input": "116453141993601660 87060381463547965",
"output": "YES"
},
{
"input": "766959657 370931668",
"output": "NO"
},
{
"input": "255787422422806632 146884995820359999",
"output": "YES"
},
{
"input": "502007866464507926 71266379084204128",
"output": "YES"
},
{
"input": "257439908778973480 64157133126869976",
"output": "NO"
},
{
"input": "232709385 91708542",
"output": "NO"
},
{
"input": "252482458300407528 89907711721009125",
"output": "NO"
},
{
"input": "6 2",
"output": "YES"
},
{
"input": "6 3",
"output": "NO"
},
{
"input": "6 4",
"output": "YES"
},
{
"input": "6 5",
"output": "YES"
},
{
"input": "6 6",
"output": "YES"
},
{
"input": "258266151957056904 30153168463725364",
"output": "NO"
},
{
"input": "83504367885565783 52285355047292458",
"output": "YES"
},
{
"input": "545668929424440387 508692735816921376",
"output": "YES"
},
{
"input": "547321411485639939 36665750286082900",
"output": "NO"
},
{
"input": "548973893546839491 183137237979822911",
"output": "NO"
},
{
"input": "544068082 193116851",
"output": "NO"
},
{
"input": "871412474 749817171",
"output": "YES"
},
{
"input": "999999999 1247",
"output": "NO"
},
{
"input": "851941088 712987048",
"output": "YES"
},
{
"input": "559922900 418944886",
"output": "YES"
},
{
"input": "293908937 37520518",
"output": "YES"
},
{
"input": "650075786 130049650",
"output": "NO"
},
{
"input": "1000000000 1000000000",
"output": "YES"
},
{
"input": "548147654663723363 107422751713800746",
"output": "YES"
},
{
"input": "828159210 131819483",
"output": "NO"
},
{
"input": "6242634 4110365",
"output": "YES"
},
{
"input": "458601973 245084155",
"output": "YES"
},
{
"input": "349593257 18089089",
"output": "YES"
},
{
"input": "814768821 312514745",
"output": "NO"
},
{
"input": "697884949 626323363",
"output": "YES"
},
{
"input": "667011589 54866795",
"output": "NO"
},
{
"input": "1000000000000000000 2",
"output": "NO"
},
{
"input": "1000000000000000000 3",
"output": "YES"
},
{
"input": "1000000000000000000 4",
"output": "NO"
},
{
"input": "999999999999999 1",
"output": "YES"
},
{
"input": "17 4",
"output": "NO"
},
{
"input": "2 2",
"output": "YES"
},
{
"input": "1000000000000000 2",
"output": "NO"
},
{
"input": "12 4",
"output": "YES"
},
{
"input": "6 1",
"output": "NO"
},
{
"input": "2 1",
"output": "NO"
},
{
"input": "10000000005 1",
"output": "YES"
},
{
"input": "10000000000000009 2",
"output": "NO"
},
{
"input": "12457895452123 1",
"output": "YES"
},
{
"input": "999999999999999999 9",
"output": "YES"
},
{
"input": "1000000000000 3",
"output": "YES"
},
{
"input": "13099714659575475 6549849616514894",
"output": "NO"
},
{
"input": "100000000000000001 1",
"output": "YES"
},
{
"input": "825175814723458 324",
"output": "YES"
},
{
"input": "20 4",
"output": "YES"
},
{
"input": "100000176877 4",
"output": "YES"
},
{
"input": "100000 3",
"output": "YES"
},
{
"input": "946744073709551614 10",
"output": "YES"
}
] | 1,649,993,225
| 2,147,483,647
|
Python 3
|
WRONG_ANSWER
|
TESTS
| 2
| 30
| 0
|
import sys
input = lambda : sys.stdin.readline().strip()
##############################################################################
n,k = map(int,input().split())
print('YES' if n % 2 == 1 else 'NO')
|
Title: Sasha and Sticks
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
It's one more school day now. Sasha doesn't like classes and is always bored at them. So, each day he invents some game and plays in it alone or with friends.
Today he invented one simple game to play with Lena, with whom he shares a desk. The rules are simple. Sasha draws *n* sticks in a row. After that the players take turns crossing out exactly *k* sticks from left or right in each turn. Sasha moves first, because he is the inventor of the game. If there are less than *k* sticks on the paper before some turn, the game ends. Sasha wins if he makes strictly more moves than Lena. Sasha wants to know the result of the game before playing, you are to help him.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=1018, *k*<=≤<=*n*) — the number of sticks drawn by Sasha and the number *k* — the number of sticks to be crossed out on each turn.
Output Specification:
If Sasha wins, print "YES" (without quotes), otherwise print "NO" (without quotes).
You can print each letter in arbitrary case (upper of lower).
Demo Input:
['1 1\n', '10 4\n']
Demo Output:
['YES\n', 'NO\n']
Note:
In the first example Sasha crosses out 1 stick, and then there are no sticks. So Lena can't make a move, and Sasha wins.
In the second example Sasha crosses out 4 sticks, then Lena crosses out 4 sticks, and after that there are only 2 sticks left. Sasha can't make a move. The players make equal number of moves, so Sasha doesn't win.
|
```python
import sys
input = lambda : sys.stdin.readline().strip()
##############################################################################
n,k = map(int,input().split())
print('YES' if n % 2 == 1 else 'NO')
```
| 0
|
|
262
|
B
|
Roma and Changing Signs
|
PROGRAMMING
| 1,200
|
[
"greedy"
] | null | null |
Roma works in a company that sells TVs. Now he has to prepare a report for the last year.
Roma has got a list of the company's incomes. The list is a sequence that consists of *n* integers. The total income of the company is the sum of all integers in sequence. Roma decided to perform exactly *k* changes of signs of several numbers in the sequence. He can also change the sign of a number one, two or more times.
The operation of changing a number's sign is the operation of multiplying this number by -1.
Help Roma perform the changes so as to make the total income of the company (the sum of numbers in the resulting sequence) maximum. Note that Roma should perform exactly *k* changes.
|
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=105), showing, how many numbers are in the sequence and how many swaps are to be made.
The second line contains a non-decreasing sequence, consisting of *n* integers *a**i* (|*a**i*|<=≤<=104).
The numbers in the lines are separated by single spaces. Please note that the given sequence is sorted in non-decreasing order.
|
In the single line print the answer to the problem — the maximum total income that we can obtain after exactly *k* changes.
|
[
"3 2\n-1 -1 1\n",
"3 1\n-1 -1 1\n"
] |
[
"3\n",
"1\n"
] |
In the first sample we can get sequence [1, 1, 1], thus the total income equals 3.
In the second test, the optimal strategy is to get sequence [-1, 1, 1], thus the total income equals 1.
| 1,000
|
[
{
"input": "3 2\n-1 -1 1",
"output": "3"
},
{
"input": "3 1\n-1 -1 1",
"output": "1"
},
{
"input": "17 27\n257 320 676 1136 2068 2505 2639 4225 4951 5786 7677 7697 7851 8337 8429 8469 9343",
"output": "81852"
},
{
"input": "69 28\n-9822 -9264 -9253 -9221 -9139 -9126 -9096 -8981 -8521 -8313 -8257 -8253 -7591 -7587 -7301 -7161 -7001 -6847 -6441 -6241 -5949 -5896 -5713 -5692 -5644 -5601 -5545 -5525 -5331 -5253 -5041 -5000 -4951 -4855 -4384 -4293 -4251 -4001 -3991 -3762 -3544 -3481 -3261 -2983 -2882 -2857 -2713 -2691 -2681 -2653 -2221 -2043 -2011 -1997 -1601 -1471 -1448 -1363 -1217 -1217 -1129 -961 -926 -801 -376 -327 -305 -174 -91",
"output": "102443"
},
{
"input": "12 28\n-6652 -6621 -6471 -5559 -5326 -4551 -4401 -4326 -3294 -1175 -1069 -43",
"output": "49488"
},
{
"input": "78 13\n-9961 -9922 -9817 -9813 -9521 -9368 -9361 -9207 -9153 -9124 -9008 -8981 -8951 -8911 -8551 -8479 -8245 -8216 -7988 -7841 -7748 -7741 -7734 -7101 -6846 -6804 -6651 -6526 -6519 -6463 -6297 -6148 -6090 -5845 -5209 -5201 -5161 -5061 -4537 -4529 -4433 -4370 -4266 -4189 -4125 -3945 -3843 -3777 -3751 -3476 -3461 -3279 -3205 -3001 -2889 -2761 -2661 -2521 -2481 -2305 -2278 -2269 -2225 -1648 -1524 -1476 -1353 -1097 -867 -785 -741 -711 -692 -440 -401 -225 -65 -41",
"output": "-147832"
},
{
"input": "4 1\n218 3441 4901 7601",
"output": "15725"
},
{
"input": "73 26\n-8497 -8363 -7603 -7388 -6830 -6827 -6685 -6389 -6237 -6099 -6013 -5565 -5465 -4965 -4947 -4201 -3851 -3793 -3421 -3410 -3201 -3169 -3156 -2976 -2701 -2623 -2321 -2169 -1469 -1221 -950 -926 -9 47 236 457 773 1321 1485 1545 1671 1736 2014 2137 2174 2301 2625 3181 3536 3851 4041 4685 4981 4987 5145 5163 5209 5249 6011 6337 6790 7254 7361 7407 7969 7982 8083 8251 8407 8735 9660 9855 9957",
"output": "315919"
},
{
"input": "53 5\n-9821 -9429 -9146 -8973 -8807 -8801 -8321 -7361 -7222 -7161 -6913 -5961 -4877 -4756 -4753 -4661 -3375 -3031 -2950 -2661 -2161 -2041 -1111 -1071 -905 -697 -397 323 772 1617 1752 2736 2737 3201 3465 4029 4121 4463 4561 4637 4814 6119 6610 6641 6961 7217 7523 8045 8610 8915 9004 9265 9576",
"output": "92703"
},
{
"input": "1 1\n0",
"output": "0"
},
{
"input": "1 1\n10000",
"output": "-10000"
},
{
"input": "1 2\n-1",
"output": "-1"
},
{
"input": "2 1\n0 1",
"output": "1"
},
{
"input": "2 2\n-1 0",
"output": "1"
},
{
"input": "5 6\n-3 -2 -1 5 6",
"output": "15"
},
{
"input": "3 3\n-50 -10 30",
"output": "70"
},
{
"input": "4 4\n-100 -90 -80 1",
"output": "269"
},
{
"input": "4 3\n-3 -2 1 2",
"output": "6"
},
{
"input": "6 4\n-6 -3 -2 1 2 3",
"output": "15"
},
{
"input": "5 6\n-10 -9 -8 1 2",
"output": "28"
},
{
"input": "1 2\n1",
"output": "1"
},
{
"input": "2 2\n-1 3",
"output": "2"
},
{
"input": "4 7\n-3 -2 1 6",
"output": "10"
},
{
"input": "4 3\n-7 -6 1 5",
"output": "17"
},
{
"input": "4 4\n-5 -1 1 2",
"output": "9"
}
] | 1,696,263,260
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 4
| 92
| 0
|
n, k = map(int, input().split())
line = list(map(int, input().split()))
i = 0
while k:
if line[i] < 0:
line[i] *= -1
k -= 1
if k == 0:
break
elif line[i] >= 0:
if k % 2 == 1:
min_num = line.index(min(line))
line[min_num] *= -1
break
i += 1
s = sum(line)
if sum(line) < 0:
print(0)
else:
print(s)
|
Title: Roma and Changing Signs
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Roma works in a company that sells TVs. Now he has to prepare a report for the last year.
Roma has got a list of the company's incomes. The list is a sequence that consists of *n* integers. The total income of the company is the sum of all integers in sequence. Roma decided to perform exactly *k* changes of signs of several numbers in the sequence. He can also change the sign of a number one, two or more times.
The operation of changing a number's sign is the operation of multiplying this number by -1.
Help Roma perform the changes so as to make the total income of the company (the sum of numbers in the resulting sequence) maximum. Note that Roma should perform exactly *k* changes.
Input Specification:
The first line contains two integers *n* and *k* (1<=≤<=*n*,<=*k*<=≤<=105), showing, how many numbers are in the sequence and how many swaps are to be made.
The second line contains a non-decreasing sequence, consisting of *n* integers *a**i* (|*a**i*|<=≤<=104).
The numbers in the lines are separated by single spaces. Please note that the given sequence is sorted in non-decreasing order.
Output Specification:
In the single line print the answer to the problem — the maximum total income that we can obtain after exactly *k* changes.
Demo Input:
['3 2\n-1 -1 1\n', '3 1\n-1 -1 1\n']
Demo Output:
['3\n', '1\n']
Note:
In the first sample we can get sequence [1, 1, 1], thus the total income equals 3.
In the second test, the optimal strategy is to get sequence [-1, 1, 1], thus the total income equals 1.
|
```python
n, k = map(int, input().split())
line = list(map(int, input().split()))
i = 0
while k:
if line[i] < 0:
line[i] *= -1
k -= 1
if k == 0:
break
elif line[i] >= 0:
if k % 2 == 1:
min_num = line.index(min(line))
line[min_num] *= -1
break
i += 1
s = sum(line)
if sum(line) < 0:
print(0)
else:
print(s)
```
| -1
|
|
257
|
B
|
Playing Cubes
|
PROGRAMMING
| 1,300
|
[
"games",
"greedy",
"implementation"
] | null | null |
Petya and Vasya decided to play a little. They found *n* red cubes and *m* blue cubes. The game goes like that: the players take turns to choose a cube of some color (red or blue) and put it in a line from left to right (overall the line will have *n*<=+<=*m* cubes). Petya moves first. Petya's task is to get as many pairs of neighbouring cubes of the same color as possible. Vasya's task is to get as many pairs of neighbouring cubes of different colors as possible.
The number of Petya's points in the game is the number of pairs of neighboring cubes of the same color in the line, the number of Vasya's points in the game is the number of neighbouring cubes of the different color in the line. Your task is to calculate the score at the end of the game (Petya's and Vasya's points, correspondingly), if both boys are playing optimally well. To "play optimally well" first of all means to maximize the number of one's points, and second — to minimize the number of the opponent's points.
|
The only line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of red and blue cubes, correspondingly.
|
On a single line print two space-separated integers — the number of Petya's and Vasya's points correspondingly provided that both players play optimally well.
|
[
"3 1\n",
"2 4\n"
] |
[
"2 1\n",
"3 2\n"
] |
In the first test sample the optimal strategy for Petya is to put the blue cube in the line. After that there will be only red cubes left, so by the end of the game the line of cubes from left to right will look as [blue, red, red, red]. So, Petya gets 2 points and Vasya gets 1 point.
If Petya would choose the red cube during his first move, then, provided that both boys play optimally well, Petya would get 1 point and Vasya would get 2 points.
| 500
|
[
{
"input": "3 1",
"output": "2 1"
},
{
"input": "2 4",
"output": "3 2"
},
{
"input": "1 1",
"output": "0 1"
},
{
"input": "2 1",
"output": "1 1"
},
{
"input": "4 4",
"output": "3 4"
},
{
"input": "10 7",
"output": "9 7"
},
{
"input": "5 13",
"output": "12 5"
},
{
"input": "7 11",
"output": "10 7"
},
{
"input": "1 2",
"output": "1 1"
},
{
"input": "10 10",
"output": "9 10"
},
{
"input": "50 30",
"output": "49 30"
},
{
"input": "80 120",
"output": "119 80"
},
{
"input": "304 122",
"output": "303 122"
},
{
"input": "500 800",
"output": "799 500"
},
{
"input": "900 1000",
"output": "999 900"
},
{
"input": "1 1000",
"output": "999 1"
},
{
"input": "997 9",
"output": "996 9"
},
{
"input": "341 678",
"output": "677 341"
},
{
"input": "784 913",
"output": "912 784"
},
{
"input": "57 888",
"output": "887 57"
},
{
"input": "100000 100000",
"output": "99999 100000"
},
{
"input": "10000 100000",
"output": "99999 10000"
},
{
"input": "9999 99999",
"output": "99998 9999"
},
{
"input": "12 100000",
"output": "99999 12"
},
{
"input": "9999 31411",
"output": "31410 9999"
},
{
"input": "12930 98391",
"output": "98390 12930"
},
{
"input": "98813 893",
"output": "98812 893"
},
{
"input": "99801 38179",
"output": "99800 38179"
},
{
"input": "831 69318",
"output": "69317 831"
},
{
"input": "99999 99997",
"output": "99998 99997"
},
{
"input": "74 99",
"output": "98 74"
},
{
"input": "159 259",
"output": "258 159"
},
{
"input": "245 317",
"output": "316 245"
},
{
"input": "947 883",
"output": "946 883"
},
{
"input": "7131 3165",
"output": "7130 3165"
},
{
"input": "11536 12192",
"output": "12191 11536"
},
{
"input": "25938 40897",
"output": "40896 25938"
},
{
"input": "81314 31958",
"output": "81313 31958"
},
{
"input": "294 83621",
"output": "83620 294"
},
{
"input": "64896 18105",
"output": "64895 18105"
}
] | 1,540,128,711
| 2,147,483,647
|
Python 3
|
RUNTIME_ERROR
|
TESTS
| 0
| 156
| 0
|
n=int(input().split())
m=int(input())
print(max(n,m)-1,min(n,m))
|
Title: Playing Cubes
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Petya and Vasya decided to play a little. They found *n* red cubes and *m* blue cubes. The game goes like that: the players take turns to choose a cube of some color (red or blue) and put it in a line from left to right (overall the line will have *n*<=+<=*m* cubes). Petya moves first. Petya's task is to get as many pairs of neighbouring cubes of the same color as possible. Vasya's task is to get as many pairs of neighbouring cubes of different colors as possible.
The number of Petya's points in the game is the number of pairs of neighboring cubes of the same color in the line, the number of Vasya's points in the game is the number of neighbouring cubes of the different color in the line. Your task is to calculate the score at the end of the game (Petya's and Vasya's points, correspondingly), if both boys are playing optimally well. To "play optimally well" first of all means to maximize the number of one's points, and second — to minimize the number of the opponent's points.
Input Specification:
The only line contains two space-separated integers *n* and *m* (1<=≤<=*n*,<=*m*<=≤<=105) — the number of red and blue cubes, correspondingly.
Output Specification:
On a single line print two space-separated integers — the number of Petya's and Vasya's points correspondingly provided that both players play optimally well.
Demo Input:
['3 1\n', '2 4\n']
Demo Output:
['2 1\n', '3 2\n']
Note:
In the first test sample the optimal strategy for Petya is to put the blue cube in the line. After that there will be only red cubes left, so by the end of the game the line of cubes from left to right will look as [blue, red, red, red]. So, Petya gets 2 points and Vasya gets 1 point.
If Petya would choose the red cube during his first move, then, provided that both boys play optimally well, Petya would get 1 point and Vasya would get 2 points.
|
```python
n=int(input().split())
m=int(input())
print(max(n,m)-1,min(n,m))
```
| -1
|
|
60
|
A
|
Where Are My Flakes?
|
PROGRAMMING
| 1,300
|
[
"implementation",
"two pointers"
] |
A. Where Are My Flakes?
|
2
|
256
|
One morning the Cereal Guy found out that all his cereal flakes were gone. He found a note instead of them. It turned out that his smart roommate hid the flakes in one of *n* boxes. The boxes stand in one row, they are numbered from 1 to *n* from the left to the right. The roommate left hints like "Hidden to the left of the *i*-th box" ("To the left of *i*"), "Hidden to the right of the *i*-th box" ("To the right of *i*"). Such hints mean that there are no flakes in the *i*-th box as well. The Cereal Guy wants to know the minimal number of boxes he necessarily needs to check to find the flakes considering all the hints. Or he wants to find out that the hints are contradictory and the roommate lied to him, that is, no box has the flakes.
|
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=1000,<=0<=≤<=*m*<=≤<=1000) which represent the number of boxes and the number of hints correspondingly. Next *m* lines contain hints like "To the left of *i*" and "To the right of *i*", where *i* is integer (1<=≤<=*i*<=≤<=*n*). The hints may coincide.
|
The answer should contain exactly one integer — the number of boxes that should necessarily be checked or "-1" if the hints are contradictory.
|
[
"2 1\nTo the left of 2\n",
"3 2\nTo the right of 1\nTo the right of 2\n",
"3 1\nTo the left of 3\n",
"3 2\nTo the left of 2\nTo the right of 1\n"
] |
[
"1\n",
"1\n",
"2\n",
"-1\n"
] |
none
| 500
|
[
{
"input": "2 1\nTo the left of 2",
"output": "1"
},
{
"input": "3 2\nTo the right of 1\nTo the right of 2",
"output": "1"
},
{
"input": "3 1\nTo the left of 3",
"output": "2"
},
{
"input": "3 2\nTo the left of 2\nTo the right of 1",
"output": "-1"
},
{
"input": "3 0",
"output": "3"
},
{
"input": "2 2\nTo the left of 2\nTo the left of 2",
"output": "1"
},
{
"input": "5 0",
"output": "5"
},
{
"input": "6 2\nTo the right of 2\nTo the left of 4",
"output": "1"
},
{
"input": "8 6\nTo the right of 3\nTo the left of 4\nTo the left of 4\nTo the right of 1\nTo the right of 5\nTo the left of 6",
"output": "-1"
},
{
"input": "459 16\nTo the right of 9\nTo the right of 148\nTo the left of 406\nTo the left of 335\nTo the right of 240\nTo the left of 409\nTo the left of 365\nTo the left of 426\nTo the left of 416\nTo the right of 247\nTo the left of 284\nTo the right of 187\nTo the right of 134\nTo the right of 128\nTo the right of 98\nTo the left of 412",
"output": "36"
},
{
"input": "1000 1\nTo the right of 117",
"output": "883"
},
{
"input": "1000 0",
"output": "1000"
},
{
"input": "1 0",
"output": "1"
},
{
"input": "2 0",
"output": "2"
},
{
"input": "1 1\nTo the right of 1",
"output": "-1"
},
{
"input": "1 1\nTo the left of 1",
"output": "-1"
},
{
"input": "2 1\nTo the right of 2",
"output": "-1"
}
] | 1,581,214,198
| 2,147,483,647
|
PyPy 3
|
OK
|
TESTS
| 54
| 374
| 1,433,600
|
n, m = input().split(' ')
n = int(n)
m = int(m)
hints = {}
max_right = None
min_left = None
if m > 0:
for i in range(m):
do_nothing = False
line = input().split(' ')
hint_position = int(line[4])-1
if line[2] == 'left':
hint_box = hint_position - 1
if min_left is None or hint_box < min_left:
min_left = hint_box
hint_pointer = -1
else:
hint_box = hint_position + 1
if max_right is None or hint_box > max_right:
max_right = hint_box
hint_pointer = 1
hints[hint_position] = hint_pointer
value_1 = hints.get(max_right)
if max_right is not None and min_left is not None:
if max_right <= min_left:
boxes = 0
for i in range(max_right, min_left+1):
value = hints.get(i)
if value is None:
boxes +=1
if boxes == 0:
print(-1)
else:
print(boxes)
else:
print(-1)
elif max_right is None:
if min_left < 0:
print(-1)
else:
print(min_left+1)
else:
if max_right > n-1:
print(-1)
else:
print(n-max_right)
else:
print(n)
|
Title: Where Are My Flakes?
Time Limit: 2 seconds
Memory Limit: 256 megabytes
Problem Description:
One morning the Cereal Guy found out that all his cereal flakes were gone. He found a note instead of them. It turned out that his smart roommate hid the flakes in one of *n* boxes. The boxes stand in one row, they are numbered from 1 to *n* from the left to the right. The roommate left hints like "Hidden to the left of the *i*-th box" ("To the left of *i*"), "Hidden to the right of the *i*-th box" ("To the right of *i*"). Such hints mean that there are no flakes in the *i*-th box as well. The Cereal Guy wants to know the minimal number of boxes he necessarily needs to check to find the flakes considering all the hints. Or he wants to find out that the hints are contradictory and the roommate lied to him, that is, no box has the flakes.
Input Specification:
The first line contains two integers *n* and *m* (1<=≤<=*n*<=≤<=1000,<=0<=≤<=*m*<=≤<=1000) which represent the number of boxes and the number of hints correspondingly. Next *m* lines contain hints like "To the left of *i*" and "To the right of *i*", where *i* is integer (1<=≤<=*i*<=≤<=*n*). The hints may coincide.
Output Specification:
The answer should contain exactly one integer — the number of boxes that should necessarily be checked or "-1" if the hints are contradictory.
Demo Input:
['2 1\nTo the left of 2\n', '3 2\nTo the right of 1\nTo the right of 2\n', '3 1\nTo the left of 3\n', '3 2\nTo the left of 2\nTo the right of 1\n']
Demo Output:
['1\n', '1\n', '2\n', '-1\n']
Note:
none
|
```python
n, m = input().split(' ')
n = int(n)
m = int(m)
hints = {}
max_right = None
min_left = None
if m > 0:
for i in range(m):
do_nothing = False
line = input().split(' ')
hint_position = int(line[4])-1
if line[2] == 'left':
hint_box = hint_position - 1
if min_left is None or hint_box < min_left:
min_left = hint_box
hint_pointer = -1
else:
hint_box = hint_position + 1
if max_right is None or hint_box > max_right:
max_right = hint_box
hint_pointer = 1
hints[hint_position] = hint_pointer
value_1 = hints.get(max_right)
if max_right is not None and min_left is not None:
if max_right <= min_left:
boxes = 0
for i in range(max_right, min_left+1):
value = hints.get(i)
if value is None:
boxes +=1
if boxes == 0:
print(-1)
else:
print(boxes)
else:
print(-1)
elif max_right is None:
if min_left < 0:
print(-1)
else:
print(min_left+1)
else:
if max_right > n-1:
print(-1)
else:
print(n-max_right)
else:
print(n)
```
| 3.90383
|
598
|
A
|
Tricky Sum
|
PROGRAMMING
| 900
|
[
"math"
] | null | null |
In this problem you are to calculate the sum of all integers from 1 to *n*, but you should take all powers of two with minus in the sum.
For example, for *n*<==<=4 the sum is equal to <=-<=1<=-<=2<=+<=3<=-<=4<==<=<=-<=4, because 1, 2 and 4 are 20, 21 and 22 respectively.
Calculate the answer for *t* values of *n*.
|
The first line of the input contains a single integer *t* (1<=≤<=*t*<=≤<=100) — the number of values of *n* to be processed.
Each of next *t* lines contains a single integer *n* (1<=≤<=*n*<=≤<=109).
|
Print the requested sum for each of *t* integers *n* given in the input.
|
[
"2\n4\n1000000000\n"
] |
[
"-4\n499999998352516354\n"
] |
The answer for the first sample is explained in the statement.
| 0
|
[
{
"input": "2\n4\n1000000000",
"output": "-4\n499999998352516354"
},
{
"input": "10\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10",
"output": "-1\n-3\n0\n-4\n1\n7\n14\n6\n15\n25"
},
{
"input": "10\n10\n9\n47\n33\n99\n83\n62\n1\n100\n53",
"output": "25\n15\n1002\n435\n4696\n3232\n1827\n-1\n4796\n1305"
},
{
"input": "100\n901\n712\n3\n677\n652\n757\n963\n134\n205\n888\n847\n283\n591\n984\n1\n61\n540\n986\n950\n729\n104\n244\n500\n461\n251\n685\n631\n803\n526\n600\n1000\n899\n411\n219\n597\n342\n771\n348\n507\n775\n454\n102\n486\n333\n580\n431\n537\n355\n624\n23\n429\n276\n84\n704\n96\n536\n855\n653\n72\n718\n776\n658\n802\n777\n995\n285\n328\n405\n184\n555\n956\n410\n846\n853\n525\n983\n65\n549\n839\n929\n620\n725\n635\n303\n201\n878\n580\n139\n182\n69\n400\n788\n985\n792\n103\n248\n570\n839\n253\n417",
"output": "404305\n251782\n0\n227457\n210832\n284857\n462120\n8535\n20605\n392670\n357082\n39164\n172890\n482574\n-1\n1765\n144024\n484545\n449679\n264039\n5206\n29380\n124228\n105469\n31116\n232909\n197350\n320760\n136555\n178254\n498454\n402504\n83644\n23580\n176457\n57631\n295560\n59704\n127756\n298654\n102263\n4999\n117319\n54589\n166444\n92074\n142407\n62168\n192954\n214\n91213\n37204\n3316\n246114\n4402\n141870\n363894\n211485\n2374\n256075\n299430\n214765\n319957\n300207\n493464\n39733\n52934\n81193\n16510\n15..."
},
{
"input": "1\n16",
"output": "74"
},
{
"input": "60\n536870912\n536870911\n536870913\n1000000000\n999999999\n1\n2\n3\n4\n268435456\n268435455\n268435457\n536870912\n536870911\n536870913\n1000000000\n999999999\n1\n2\n3\n4\n268435456\n268435455\n268435457\n536870912\n536870911\n536870913\n1000000000\n999999999\n1\n2\n3\n4\n268435456\n268435455\n268435457\n536870912\n536870911\n536870913\n1000000000\n999999999\n1\n2\n3\n4\n268435456\n268435455\n268435457\n536870912\n536870911\n536870913\n1000000000\n999999999\n1\n2\n3\n4\n268435456\n268435455\n268435457",
"output": "144115186196807682\n144115186733678594\n144115186733678595\n499999998352516354\n499999997352516354\n-1\n-3\n0\n-4\n36028796079439874\n36028796347875330\n36028796347875331\n144115186196807682\n144115186733678594\n144115186733678595\n499999998352516354\n499999997352516354\n-1\n-3\n0\n-4\n36028796079439874\n36028796347875330\n36028796347875331\n144115186196807682\n144115186733678594\n144115186733678595\n499999998352516354\n499999997352516354\n-1\n-3\n0\n-4\n36028796079439874\n36028796347875330\n36028796347875..."
},
{
"input": "13\n1\n19\n31\n19\n19\n92\n74\n69\n32\n32\n91\n42\n73",
"output": "-1\n128\n434\n128\n128\n4024\n2521\n2161\n402\n402\n3932\n777\n2447"
},
{
"input": "1\n16383",
"output": "134176770"
},
{
"input": "16\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100\n100100",
"output": "5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908\n5009792908"
},
{
"input": "1\n414234",
"output": "85794061921"
},
{
"input": "1\n414232",
"output": "85793233454"
},
{
"input": "3\n414231\n414231\n414231",
"output": "85792819222\n85792819222\n85792819222"
},
{
"input": "1\n121",
"output": "7127"
}
] | 1,620,905,043
| 2,147,483,647
|
Python 3
|
OK
|
TESTS
| 21
| 62
| 6,963,200
|
def is_power_of_2(n):
return (n & (n - 1)) == 0
def solve(n):
sum_1_to_n = (n*(n+1))//2
sum_powers_of_2 = 0
i = 1
while i <= n:
if is_power_of_2(i):
sum_powers_of_2 += i
i *= 2
return sum_1_to_n - (2 * sum_powers_of_2)
if __name__ == '__main__':
t = int(input())
while t > 0:
n = int(input())
result = solve(n)
print(result)
t -= 1
|
Title: Tricky Sum
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
In this problem you are to calculate the sum of all integers from 1 to *n*, but you should take all powers of two with minus in the sum.
For example, for *n*<==<=4 the sum is equal to <=-<=1<=-<=2<=+<=3<=-<=4<==<=<=-<=4, because 1, 2 and 4 are 20, 21 and 22 respectively.
Calculate the answer for *t* values of *n*.
Input Specification:
The first line of the input contains a single integer *t* (1<=≤<=*t*<=≤<=100) — the number of values of *n* to be processed.
Each of next *t* lines contains a single integer *n* (1<=≤<=*n*<=≤<=109).
Output Specification:
Print the requested sum for each of *t* integers *n* given in the input.
Demo Input:
['2\n4\n1000000000\n']
Demo Output:
['-4\n499999998352516354\n']
Note:
The answer for the first sample is explained in the statement.
|
```python
def is_power_of_2(n):
return (n & (n - 1)) == 0
def solve(n):
sum_1_to_n = (n*(n+1))//2
sum_powers_of_2 = 0
i = 1
while i <= n:
if is_power_of_2(i):
sum_powers_of_2 += i
i *= 2
return sum_1_to_n - (2 * sum_powers_of_2)
if __name__ == '__main__':
t = int(input())
while t > 0:
n = int(input())
result = solve(n)
print(result)
t -= 1
```
| 3
|
|
412
|
B
|
Network Configuration
|
PROGRAMMING
| 900
|
[
"greedy",
"sortings"
] | null | null |
The R1 company wants to hold a web search championship. There were *n* computers given for the competition, each of them is connected to the Internet. The organizers believe that the data transfer speed directly affects the result. The higher the speed of the Internet is, the faster the participant will find the necessary information. Therefore, before the competition started, each computer had its maximum possible data transfer speed measured. On the *i*-th computer it was *a**i* kilobits per second.
There will be *k* participants competing in the championship, each should get a separate computer. The organizing company does not want any of the participants to have an advantage over the others, so they want to provide the same data transfer speed to each participant's computer. Also, the organizers want to create the most comfortable conditions for the participants, so the data transfer speed on the participants' computers should be as large as possible.
The network settings of the R1 company has a special option that lets you to cut the initial maximum data transfer speed of any computer to any lower speed. How should the R1 company configure the network using the described option so that at least *k* of *n* computers had the same data transfer speed and the data transfer speed on these computers was as large as possible?
|
The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=100) — the number of computers and the number of participants, respectively. In the second line you have a space-separated sequence consisting of *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (16<=≤<=*a**i*<=≤<=32768); number *a**i* denotes the maximum data transfer speed on the *i*-th computer.
|
Print a single integer — the maximum Internet speed value. It is guaranteed that the answer to the problem is always an integer.
|
[
"3 2\n40 20 30\n",
"6 4\n100 20 40 20 50 50\n"
] |
[
"30\n",
"40\n"
] |
In the first test case the organizers can cut the first computer's speed to 30 kilobits. Then two computers (the first and the third one) will have the same speed of 30 kilobits. They should be used as the participants' computers. This answer is optimal.
| 1,000
|
[
{
"input": "3 2\n40 20 30",
"output": "30"
},
{
"input": "6 4\n100 20 40 20 50 50",
"output": "40"
},
{
"input": "1 1\n16",
"output": "16"
},
{
"input": "2 1\n10000 17",
"output": "10000"
},
{
"input": "2 2\n200 300",
"output": "200"
},
{
"input": "3 1\n21 25 16",
"output": "25"
},
{
"input": "3 2\n23 20 26",
"output": "23"
},
{
"input": "3 3\n19 29 28",
"output": "19"
},
{
"input": "100 2\n82 37 88 28 98 30 38 76 90 68 79 29 67 93 19 71 122 103 110 79 20 75 68 101 16 120 114 68 73 71 103 114 99 70 73 18 36 31 32 87 32 79 44 72 58 25 44 72 106 38 47 17 83 41 75 23 49 30 73 67 117 52 22 117 109 89 66 88 75 62 17 35 83 69 63 60 23 120 93 18 112 93 39 72 116 109 106 72 27 123 117 119 87 72 33 73 70 110 43 43",
"output": "122"
},
{
"input": "30 13\n36 82 93 91 48 62 59 96 72 40 45 68 97 70 26 22 35 98 92 83 72 49 70 39 53 94 97 65 37 28",
"output": "70"
},
{
"input": "50 49\n20 77 31 40 18 87 44 64 70 48 29 59 98 33 95 17 69 84 81 17 24 66 37 54 97 55 77 79 42 21 23 42 36 55 81 83 94 45 25 84 20 97 37 95 46 92 73 39 90 71",
"output": "17"
},
{
"input": "40 40\n110 674 669 146 882 590 650 844 427 187 380 711 122 94 38 216 414 874 380 31 895 390 414 557 913 68 665 964 895 708 594 17 24 621 780 509 837 550 630 568",
"output": "17"
},
{
"input": "40 1\n851 110 1523 1572 945 4966 4560 756 2373 4760 144 2579 4022 220 1924 1042 160 2792 2425 4483 2154 4120 319 4617 4686 2502 4797 4941 4590 4478 4705 4355 695 684 1560 684 2780 1090 4995 3113",
"output": "4995"
},
{
"input": "70 12\n6321 2502 557 2734 16524 10133 13931 5045 3897 18993 5745 8687 12344 1724 12071 2345 3852 9312 14432 8615 7461 2439 4751 19872 12266 12997 8276 8155 9502 3047 7226 12754 9447 17349 1888 14564 18257 18099 8924 14199 738 13693 10917 15554 15773 17859 13391 13176 10567 19658 16494 3968 13977 14694 10537 4044 16402 9714 4425 13599 19660 2426 19687 2455 2382 3413 5754 113 7542 8353",
"output": "16402"
},
{
"input": "80 60\n6159 26457 23753 27073 9877 4492 11957 10989 27151 6552 1646 7773 23924 27554 10517 8788 31160 455 12625 22009 22133 15657 14968 31871 15344 16550 27414 876 31213 10895 21508 17516 12747 59 11786 10497 30143 25548 22003 2809 11694 30395 8122 31248 23075 19013 31614 9133 27942 27346 15969 19415 10367 8424 29355 18903 3396 6327 4201 24124 24266 22586 724 1595 3972 17526 2843 20982 23655 12714 18050 15225 2658 7236 27555 13023 729 9022 17386 2585",
"output": "8122"
},
{
"input": "100 1\n199 348 489 76 638 579 982 125 28 401 228 117 195 337 80 914 752 98 679 417 47 225 357 413 849 622 477 620 487 223 321 240 439 393 733 660 652 500 877 40 788 246 376 723 952 601 912 316 598 809 476 932 384 147 982 271 202 695 129 303 304 712 49 306 598 141 833 730 946 708 724 788 202 465 951 118 279 706 214 655 152 976 998 231 487 311 342 317 243 554 977 232 365 643 336 501 761 400 600 528",
"output": "998"
},
{
"input": "80 50\n15160 6853 20254 11358 19535 27691 2983 31650 9219 11833 32053 31695 21511 4320 4384 24843 1454 31543 18796 13815 1546 27926 16276 14315 12542 25370 24890 29647 3584 17867 12446 15072 19852 30207 16361 7964 5343 398 10837 31114 9252 12767 15098 22562 32637 31823 8160 12658 6422 19142 12448 6765 7373 868 31712 24856 23251 29200 8159 16144 27165 4308 13652 12502 4183 7961 3032 26855 8687 12263 24319 7722 19460 30700 29806 1280 21141 25965 25550 26881",
"output": "12448"
},
{
"input": "50 16\n16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16",
"output": "16"
},
{
"input": "100 1\n16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16",
"output": "16"
},
{
"input": "100 2\n16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16",
"output": "16"
},
{
"input": "100 100\n16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16",
"output": "16"
},
{
"input": "100 99\n16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16",
"output": "16"
}
] | 1,397,838,477
| 1,077
|
Python 3
|
OK
|
TESTS
| 34
| 62
| 0
|
[n, k], speed = map(int, input().split()), sorted(map(int, input().split()), reverse = True)
print(speed[k - 1])
|
Title: Network Configuration
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
The R1 company wants to hold a web search championship. There were *n* computers given for the competition, each of them is connected to the Internet. The organizers believe that the data transfer speed directly affects the result. The higher the speed of the Internet is, the faster the participant will find the necessary information. Therefore, before the competition started, each computer had its maximum possible data transfer speed measured. On the *i*-th computer it was *a**i* kilobits per second.
There will be *k* participants competing in the championship, each should get a separate computer. The organizing company does not want any of the participants to have an advantage over the others, so they want to provide the same data transfer speed to each participant's computer. Also, the organizers want to create the most comfortable conditions for the participants, so the data transfer speed on the participants' computers should be as large as possible.
The network settings of the R1 company has a special option that lets you to cut the initial maximum data transfer speed of any computer to any lower speed. How should the R1 company configure the network using the described option so that at least *k* of *n* computers had the same data transfer speed and the data transfer speed on these computers was as large as possible?
Input Specification:
The first line contains two space-separated integers *n* and *k* (1<=≤<=*k*<=≤<=*n*<=≤<=100) — the number of computers and the number of participants, respectively. In the second line you have a space-separated sequence consisting of *n* integers: *a*1,<=*a*2,<=...,<=*a**n* (16<=≤<=*a**i*<=≤<=32768); number *a**i* denotes the maximum data transfer speed on the *i*-th computer.
Output Specification:
Print a single integer — the maximum Internet speed value. It is guaranteed that the answer to the problem is always an integer.
Demo Input:
['3 2\n40 20 30\n', '6 4\n100 20 40 20 50 50\n']
Demo Output:
['30\n', '40\n']
Note:
In the first test case the organizers can cut the first computer's speed to 30 kilobits. Then two computers (the first and the third one) will have the same speed of 30 kilobits. They should be used as the participants' computers. This answer is optimal.
|
```python
[n, k], speed = map(int, input().split()), sorted(map(int, input().split()), reverse = True)
print(speed[k - 1])
```
| 3
|
|
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"
}
] | 1,693,371,864
| 2,147,483,647
|
Python 3
|
COMPILATION_ERROR
|
TESTS
| 0
| 0
| 0
|
a=input("")
b=input("")
c=a.length()
i=0
j=c-1
for i in range(c//2):
swap(a[i], a[j])
i++
j--
if (a==b):
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
a=input("")
b=input("")
c=a.length()
i=0
j=c-1
for i in range(c//2):
swap(a[i], a[j])
i++
j--
if (a==b):
print("YES")
else:
print("NO")
```
| -1
|
459
|
B
|
Pashmak and Flowers
|
PROGRAMMING
| 1,300
|
[
"combinatorics",
"implementation",
"sortings"
] | null | null |
Pashmak decided to give Parmida a pair of flowers from the garden. There are *n* flowers in the garden and the *i*-th of them has a beauty number *b**i*. Parmida is a very strange girl so she doesn't want to have the two most beautiful flowers necessarily. She wants to have those pairs of flowers that their beauty difference is maximal possible!
Your task is to write a program which calculates two things:
1. The maximum beauty difference of flowers that Pashmak can give to Parmida. 1. The number of ways that Pashmak can pick the flowers. Two ways are considered different if and only if there is at least one flower that is chosen in the first way and not chosen in the second way.
|
The first line of the input contains *n* (2<=≤<=*n*<=≤<=2·105). In the next line there are *n* space-separated integers *b*1, *b*2, ..., *b**n* (1<=≤<=*b**i*<=≤<=109).
|
The only line of output should contain two integers. The maximum beauty difference and the number of ways this may happen, respectively.
|
[
"2\n1 2\n",
"3\n1 4 5\n",
"5\n3 1 2 3 1\n"
] |
[
"1 1",
"4 1",
"2 4"
] |
In the third sample the maximum beauty difference is 2 and there are 4 ways to do this:
1. choosing the first and the second flowers; 1. choosing the first and the fifth flowers; 1. choosing the fourth and the second flowers; 1. choosing the fourth and the fifth flowers.
| 500
|
[
{
"input": "2\n1 2",
"output": "1 1"
},
{
"input": "3\n1 4 5",
"output": "4 1"
},
{
"input": "5\n3 1 2 3 1",
"output": "2 4"
},
{
"input": "2\n1 1",
"output": "0 1"
},
{
"input": "3\n1 1 1",
"output": "0 3"
},
{
"input": "4\n1 1 1 1",
"output": "0 6"
},
{
"input": "5\n1 1 1 1 1",
"output": "0 10"
},
{
"input": "5\n2 2 2 2 2",
"output": "0 10"
},
{
"input": "10\n2 2 2 2 2 2 2 2 2 2",
"output": "0 45"
},
{
"input": "3\n2 2 2",
"output": "0 3"
},
{
"input": "3\n3 3 3",
"output": "0 3"
},
{
"input": "2\n10000000 100000000",
"output": "90000000 1"
},
{
"input": "5\n5 5 5 5 5",
"output": "0 10"
},
{
"input": "5\n3 3 3 3 3",
"output": "0 10"
},
{
"input": "6\n1 1 1 1 1 1",
"output": "0 15"
},
{
"input": "2\n5 6",
"output": "1 1"
},
{
"input": "10\n1 1 1 1 1 1 1 1 1 1",
"output": "0 45"
},
{
"input": "10\n1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000",
"output": "0 45"
},
{
"input": "4\n4 4 4 4",
"output": "0 6"
},
{
"input": "7\n1 1 1 1 1 1 1",
"output": "0 21"
},
{
"input": "11\n1 1 1 1 1 1 1 1 1 1 1",
"output": "0 55"
},
{
"input": "8\n8 8 8 8 8 8 8 8",
"output": "0 28"
},
{
"input": "2\n3 2",
"output": "1 1"
}
] | 1,660,243,380
| 2,147,483,647
|
PyPy 3-64
|
OK
|
TESTS
| 58
| 155
| 27,545,600
|
t=int(input())
l=list(map(int,input().split()))
maxi=max(l)
mini=min(l)
count1=l.count(maxi)
count2=l.count(mini)
if mini==maxi:
print(maxi-mini,count1*(count1-1)//2)
else:
print(maxi-mini,count1*count2)
|
Title: Pashmak and Flowers
Time Limit: None seconds
Memory Limit: None megabytes
Problem Description:
Pashmak decided to give Parmida a pair of flowers from the garden. There are *n* flowers in the garden and the *i*-th of them has a beauty number *b**i*. Parmida is a very strange girl so she doesn't want to have the two most beautiful flowers necessarily. She wants to have those pairs of flowers that their beauty difference is maximal possible!
Your task is to write a program which calculates two things:
1. The maximum beauty difference of flowers that Pashmak can give to Parmida. 1. The number of ways that Pashmak can pick the flowers. Two ways are considered different if and only if there is at least one flower that is chosen in the first way and not chosen in the second way.
Input Specification:
The first line of the input contains *n* (2<=≤<=*n*<=≤<=2·105). In the next line there are *n* space-separated integers *b*1, *b*2, ..., *b**n* (1<=≤<=*b**i*<=≤<=109).
Output Specification:
The only line of output should contain two integers. The maximum beauty difference and the number of ways this may happen, respectively.
Demo Input:
['2\n1 2\n', '3\n1 4 5\n', '5\n3 1 2 3 1\n']
Demo Output:
['1 1', '4 1', '2 4']
Note:
In the third sample the maximum beauty difference is 2 and there are 4 ways to do this:
1. choosing the first and the second flowers; 1. choosing the first and the fifth flowers; 1. choosing the fourth and the second flowers; 1. choosing the fourth and the fifth flowers.
|
```python
t=int(input())
l=list(map(int,input().split()))
maxi=max(l)
mini=min(l)
count1=l.count(maxi)
count2=l.count(mini)
if mini==maxi:
print(maxi-mini,count1*(count1-1)//2)
else:
print(maxi-mini,count1*count2)
```
| 3
|
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