1 1
1 1
3
100 1 0
100 1 1
0 1 0

answer: 1 0

27 5
O-O-O-O-O-O-O O-O-O-O-O-O-O
|\ \| | | |/| |\| | | |/ /|
O-O O O O O O-O O O O O O-O
|/ / /| |\ \| |/ /| |\ \ \|
O-O-O-O-O-O-O O-O-O-O-O-O-O

Answer: Just a picture

Anti-hash tests are added. Please choose mod = 2^64 and strange p in case of being hacked.

Maybe you can find out why you are wrong in this test if you WA on 7.

3 4
aaaa
aaaa
aaaa

answer:
3
1 1
1 2 (or other correct coordinates)

sorry for my bad English

Why output is 2 , I think it should be 22 , because the worm started from the first sheet of vol 1 and goes until last sheet of vol 2 . so it's path is 10+1+1+10 = 22

3
a
a
a

Answer: Impossible

Предусмотрена ли функция удаления аккаунта? Каким образом это можно сделать?

The following changes have been made to the problem:
- The checking program strictly checks that the initial formatting is preserved, for example it rejects outputs with additional empty lines before or after the text.
- Missing limitations added to the problem statements:
    - The words in the text to correct are no longer than 16 characters.
    - The lines in the text to correct are no longer than 80 characters.
    - The cumulative length of all lines doesn't exceed 10000 characters.
- New tests have been added.
- Time limit has been reduced to 0.5 sec.
- All solutions have been rejudged. 44% of authors have lost their accepted solutions.

1. Input: 7122 2119
2. Input: 7122 2120

почему если сначала
for k..
 for (i=n-1; i >= 0; i--)
       inc()

то ва12
а если
for(i = n-1; i >=0; i--)
  for k...
     inc()
то АС?
надеюсь кто-нибудь понял..

Edited by author 08.07.2022 14:59

Can be solved using two DFS calls and tree rerooting in O(N) while maintaining a DP array where DP[node] = sum for all paths from node to all the nodes in its subtree (considering 1 as the initial root). Tree rerooting can be used to calculate the sum for all nodes.

Why is there such a big difference between these two compilers and how do I know if I have solved this problem, it just feels like I wrote really dirty code and it somehow worked.

i dont have any idea why wa5..

Edited by author 02.07.2022 22:06

g++ 9.2 - mle 7
visual c++ - AC

2 11 12
aaaa
bb
ccc
ddddddddddd
eeeeeeee
ff
jjjjj
hhhh
iiiii
jjjjjjjjjj
kk
lll

Correct answer: 4

2 11 10
aaaa
bb
ccc
ddddddddddd
eeeeeeee
ff
jjjjj
hhhh
iiiii
jjjjjjjjjj

Correct answer: 3

2 100 12
aaaa
bb
ccc
ddddddddddd
eeeeeeee
ff
jjjjj
hhhh
iiiii
jjjjjjjjjj
kk
lll

Correct answer: 1

1 11 12
aaaa
bb
ccc
ddddddddddd
eeeeeeee
ff
jjjjj
hhhh
iiiii
jjjjjjjjjj
kk
lll

Correct answer: 7

2 13 8
aaaa
bb
ccccc
ddddddddddddd
eeeeeeee
ffff
jjjjj
hhhhhhh

Correct answer: 2

1 1 1
a

Correct answer: 1

The problem test case is:
3 2
3
1 1
3 2
1 2
but is it right??
My observation finds:
3 2
3
1 1
3 2
2 1
plz help me???

Is there any mathematical algorithm or i need to check every way mages can go

This firstly happens in 8th test.

a,b=map(int, input().split())
if a==1:
    print(2)
    exit()
if (a*2)//b<a*2/b:
    print(a*2//b+1)
else:
    print(a*2//b)

#include<stdio.h>
int main(){



int testCase,i,j,l;

scanf("%d",&testCase);

for(l=0;l<testCase;l++){
  int k=0,count;
  scanf("%d",&count);
int pnumber=0;

for(j=2;k<count;j++){

int flag=0;
for(i=2;i*i<=j;i++){
    if(j%i==0){
        flag=1;
        break;
    }
}

if(flag){

}else{
    pnumber=j;
    k++;
}

}
printf("%d",pnumber);

}

return 0;

}