poj 3844 Divisible Subsequences

一道简单的组合数学题，我们把从1~m的和的余数相同的放在一个集合里面，那么他们任意一个相减就会是得到余数为0；那么也是d的倍数；这里要注意的是当余数为0的集合那么可以从中任意选2个与1个，这个题会超出int型：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include<cstring>
#include<vector>
#include<string>
#define LL long long
using namespace std;
LL mod[1000024];
LL Solve( int n )
{
    LL ans = mod[0];
    for( int i = 0; i < n ; i ++ )
    {
        if( mod[i]>1 )
        {
        ans += (mod[i]*(mod[i]-1))/2;
        }
    }
    return ans;    
}
int main(  )
{
    int num,sum,n,MOD;
    int T;
    while( scanf( "%d",&T )==1 )
    {
        while( T-- )
        {
            int sum = 0;
            memset( mod , 0 , sizeof( mod ) );
            scanf( "%d %d",&MOD,&n );
            for( int i = 0 ; i < n ; i ++ )
            {
                 scanf( "%d",&num );
                 sum = (sum + num)%MOD;
                 mod[sum]++;    
            }
            printf( "%I64d\n",Solve( MOD ) );
        }
    }
    return 0;
    //system( "pause" );
    return 0;
}