yes , no impossible cases are present.

Так с гугл-переводчиком понятно

and in the 4

Try this test:
1 5 2
to
be

answer: 1

Edited by author 25.09.2015 16:26

And this

input:
091

output:
109

and

092
119

0920
1019

Edited by author 22.12.2006 19:08

Try this and look for the time

010000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

Should give

100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

Edited by author 18.07.2007 05:06

My program give AC on your tests, but give WA#4. Please give me tests #4 or more tests. Thank.

Test 59 look like:

1000000000000000000000000

Answer: -1

There seem to be some non-ASCII characters in the test cases. If the input is loaded byte by byte, then the result should be output byte by byte as well.

I used Rust and handled the bytes as chars, which were stored as 4-byte unicode, but mistakenly output the result by chars instead of bytes.

8*n-7 <--- integer overflow

t = int(input())
for i in range(t):
    n, k = [int(i) for i in input().split()]
    m = n // k
    q = n % k
    s = (m * (n - m) * (k - q) + (m + 1) * (n - m - 1) * q) // 2
    print(s)

#распределяем максимально равномерно и "деремся" команда из m против всех оставшихся + команда из m + 1 против всех оставшихся и, поскольку учли каждый бой по два раза (для обоих участников), делим пополам =)

thx for test

my code passed yours test but shows WA1

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int N,a[55],b[55],i,c[55],j,d[55],t,e[55],m,n;
    unsigned long long x,y,p,arr[55];
    cin>>N;
    for(i=0;i<N;i++){
        cin>>a[i]>>b[i]>>c[i]>>d[i];
        e[i]=i;
    }
    cin>>m>>n;
    for(i=0;i<N;i++){
        x=max(a[i]-m,m-c[i]);
        if(x<0){
            x=0;
        }
        y=max(b[i]-n,n-d[i]);
        if(y<0){
            y=0;
        }
        arr[i]=x*x+y*y;
    }
    for(i=0;i<N;i++){
        for(j=i+1;j<N;j++){
            if(arr[i]>a[j]){
                p=arr[i];
                arr[i]=arr[j];
                arr[j]=p;
                t=e[i];
                e[i]=e[j];
                e[j]=t;
            }
        }
    }
        for(i=0;i<N;i++){
            cout<<e[i]+1<<" ";
        }

    return 0;
}

There are many more combinations. You also forgot 9

n, m = map(int,input().split(' '))
voices = []
p = ['p']
h = 0
for i in range(m):
    voices.append(int(input()))

for i in range(n):
    p.append(0)

for i in voices:
    for c in range(n+1):
        if i == c:
            p[c] += 1

for i in range(1, len(p)):
    h = p[i] / m * 100
    print(f'{h:.2f}%')



This is your code that prints the right answer. Python 3 uses BANKER'S rounding while the problem asks you to the MATHEMATICAL rounding. I used 'f-string' to perform MATHEMATICAL rounding. Anyways, the solution is not fast enough - try to come up with some speed-ups.

Sorry!
I understood my mistake.