lightoj1125_dp

http://lightoj.com/volume_showproblem.php?problem=1125

Given a list of N numbers you will be allowed to choose any M of them. So you can choose in ^NC_M ways. You will have to determine how many of these chosen groups have a sum, which is divisible by D.

Input

Input starts with an integer T (≤ 20), denoting the number of test cases.

The first line of each case contains two integers N (0 < N ≤ 200) and Q (0 < Q ≤ 10). Here N indicates how many numbers are there and Q is the total number of queries. Each of the next N lines contains one 32 bit signed integer. The queries will have to be answered based on these N numbers. Each of the next Q lines contains two integers D (0 < D ≤ 20)and M (0 < M ≤ 10).

Output

For each case, print the case number in a line. Then for each query, print the number of desired groups in a single line.

Sample Input	Output for Sample Input
2 10 2 1 2 3 4 5 6 7 8 9 10 5 1 5 2 5 1 2 3 4 5 6 6 2	Case 1: 2 9 Case 2: 1

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <ctime>
#include <queue>
#include <list>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;

int a[210], d[20], m[20];
LL dp[210][30][20];//dp[i][j][k]表示遍历2到第i个数，选了j个数，余数为k时的方案有多少
int main()
{
    int t, n, q;
    scanf("%d", &t);
    for(int ca = 1; ca <= t; ca++)
    {
        scanf("%d %d", &n, &q);
        for(int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        for(int i = 1; i <= q; i++)
            scanf("%d %d", &d[i], &m[i]);
        printf("Case %d:
", ca);    
        for(int l = 1; l <= q; l++)
        {
            memset(dp, 0, sizeof(dp));
            for(int i = 0; i <= n; i++)
                dp[i][0][0] = 1;
            for(int i = 1; i <= n; i++)
            {
                for(int j = 1; j <= m[l]; j++)
                {
                    for(int k = 0; k < d[l]; k++)
                    {
                        dp[i][j][k] += dp[i-1][j][k];
                        dp[i][j][((k+a[i])%d[l]+d[l])%d[l]] += dp[i-1][j-1][k];
                    }
                }
            }
            printf("%lld
", dp[n][m[l]][0]);
        }
    }
    return 0;
}