Есть ли скрипт, позволяющий сравнивать пользователей в виде таблицы по задачам с "+"("АС") - решена, "-"(WA5, TL12 ...) - не решена, " " - не пробовал? Или что-то в этом духе? Простите, за мои плохие знания английского языка.
Is there a script that allows you to compare users in the form of a table for tasks with "+" ("as") - solved, "- "(WA5, TL12 ...) - not solved, "" - not tried? Or something like that? Sorry for my poor English skills.

Edited by author 04.06.2009 17:47

Edited by author 04.06.2009 17:47

It can help you:
Test:
1 13
0 -10 -7 -10 1 -5 6 -10 -8 -8 -7 -10 1
Answer: 6

Edited by author 04.04.2018 15:40

Use printf("%.0lf\n",  floor(answer + .5)); for output answer ;)

I don’t understand what is the mistake here. Help pls
              string[] val = Console.ReadLine().Split(' ');

               double[] k = new double[10000];

              for(int i = 0; i < Convert.ToInt32(val[1]); i++)
               {
                   string s = Console.ReadLine();
                   k[Convert.ToInt32(s) - 1]++;
               }
               double swap = 0;
               for(int i = 0; i < k.Length-1; i++)
               {
                   for(int j = i + 1; j < k.Length - i - 1; j++)
                   {
                       if (k[j+1] > k[j])
                       {
                           swap = k[j+1];
                           k[j+1] = k[j];
                           k[j] = swap;
                       }
                   }
               }
               for (int i = 0; i < Convert.ToInt32(val[0]); i++)
               {

                   Console.WriteLine(string.Format(CultureInfo.InvariantCulture, "{0:N2}", (100 * k[i]) / Convert.ToDouble(val[1]))+"%");
               }

In python 3.6 extra print() in the end of the program gives you WA2. Therefore you should output
"8
 1 3 4 2 0 1 3 2"
instead of
"8
 1 3 4 2 0 1 3 2
 "

a = int(input())
b = int(input())
print(a+b)

Try this:
#X#
#OO
XOX

15 17
1 2
2 3
3 4
4 5
5 6
6 14
14 15
15 9
9 7
7 5
7 8
8 13
13 12
11 12
10 11
10 4
1 6
1 12 15
ans: 3

#include<bits/stdc++.h>
using namespace std;
void prin ( int i)
{
    int j;
    for ( j =1 ; j <= i ;j++)
    {
        cout << "sin(" << j;
        if ( j == i){
        for (j = 0 ; j<i ; j ++)
            cout <<")";
        break;}
        else {
        if (j%2)
            cout << "-";
        else cout << "+";
        }
    }
}
int main()
{
    int i,j,n;
    cin >> n;
    int k = n;
    for (i = 1; i < n ;i++)
    cout << "(";
    for ( i = 1 ; i <= n ; i++,k--)
    {
        prin(i);
        if (i!= n)
            cout << "+"<<k<<")";
    }
    cout << "+1\n";
    return 0;
}

I spent two hours finding this stupid mistake. Do not be like me, do everything with a margin.

4 4 2 3
1 2
1 3
2 4
3 4

ans: 6

12 16 2 30
1 2
2 3
3 1
4 5
5 6
6 4
7 8
8 9
9 7
10 11
11 12
12 10
1 4
1 7
4 10
7 10

ans: 24

6 8 2 30
1 2
2 3
3 1
4 5
5 6
6 4
1 4
2 4

ans: 12

I checked two accepted solution But They are giving different output in same test case.
AC code 1:  https://paste.ubuntu.com/p/RwtFH2s4xK/
AC code 2:  https://paste.ubuntu.com/p/vmrXrdcJN6/

Tese case :
25 28
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
10 11
11 12
12 13
13 14
14 15
15 16
16 17
17 18
18 1
5 19
19 20
20 21
21 22
22 23
23 24
24 25
25 14
15 22
6 22
1 10 21

Output in code 1 : 2
Output in code 2 : 3

Answer should be : 2

5
1 2 1
2 3 1
3 4 1
4 5 1
7
+ 1
+ 3
+ 5
- 3
+ 4
- 1
- 5

ans:
0
2
4
4
4
1
0

I can't figure out where my mistake is. Can someone give me some tests?

If it is impossible to find out the distance from C to D, then write "Impossible." and not "Impossible"

Seems an easy problem, but I can't understand it...

test it your self

after deleting this code, please
explain me why WA twoalias[animal]inbox[youknow]ru

#include <stdio.h>
#include <memory.h>


struct elt
{
    int c;
    short i;
};

elt T[100001];

int main()
{
    memset(T,0,sizeof(T));
    int N;
    scanf("%i",&N);
    short i;
    for (i=1;i<=N;i++)
    {
        int S,C;
        scanf("%i %i",&S,&C);
        if (C>T[S].c)
        {
            T[S].c=C;
            T[S].i=i;
        }
    }
    int j,ch=0;
    for (j=0;j<100001;j++) if (T[j].c>0) ch++;
    printf("%i\n",ch);
    for (j=0;j<100001;j++) if (T[j].c>0) printf("%i ",T[j].i);
    return 0;
}

Edited by author 29.08.2006 20:03