题目大意：给你N个数，从中挑选M个，使得M个数的和能被S整除

解题思路：用dp[i][j][k][mod]表示前i个数中选出了j个，除数为k，余数为mod的有多少种情况，接着就可以转移了，分两种情况，选和不选进行转移
这里有个坑点，就是给的数有可能是负数，所以要先转正

#include <cstdio>
#include <cstring>
typedef long long LL;
const int N = 210;

//dp[i][j][k][l],表示从前i个种，选择j个，除数为k，余数为l的个数
LL val[N], dp[N][10][21][20], num[N][22];
int n, q, cas = 1;

void init() {
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= n; i++)
        scanf("%lld", &val[i]);
    for (int i = 1; i <= n; i++)
        for (long long j = 1; j <= 20; j++)
            num[i][j] = (val[i] + j * 100000000000) % j;

    memset(dp, 0, sizeof(dp));

    for (int k = 1; k <= 20; k++) 
        dp[1][1][k][num[1][k]] = 1;

    for (int j = 2; j <= n; j++)
        for (int k = 1; k <= 20; k++) {
            dp[j][1][k][num[j][k]]++;
            for (int l = 0; l < 20; l++) 
                dp[j][1][k][l] += dp[j - 1][1][k][l];
        }

    for (int i = 2; i <= 10; i++)
        for (int j = i; j <= n; j++) 
            for (int k = 1; k <= 20; k++)
                for (int l = 0; l < 20; l++) {
                    //选
                    dp[j][i][k][(num[j][k] + l) % k] += dp[j - 1][i - 1][k][l];
                    //不选
                    dp[j][i][k][l] += dp[j - 1][i][k][l];
                }
}

void solve() {
    int d, m;
    printf("Case %d:\n", cas++);
    while (q--) {
        scanf("%d%d", &d, &m);
        printf("%lld\n", dp[n][m][d][0]);
    }
}

int main() {
    int test;
    scanf("%d", &test);
    while (test--) {
        init();
        solve();
    }
    return 0;
}