lightoj 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.

题意：给你n个数，提出q的问题，问你从n个数取出m个求和能被d整除的有几个。

我们设一下dp[i][j][mod]表示取到第i个数，取了j位数余数为mod的方案数。显然我们可以得到递推公式,

dp[i][j][mod]+=dp[i-1][j][mod](这一位数不取）or dp[i][j][((mod+a[i])%d+d)%d]+=dp[i-1][j-1][mod]。

((mod+a[i])%d+d)%d这样取余是为了处理负数的出现。

#include <iostream>
#include <cstring>
using namespace std;
int a[220];
long long dp[220][30][30];
int main()
{
    int t;
    cin >> t;
    int ans = 0;
    while(t--) {
        ans++;
        int n , q;
        cin >> n >> q;
        for(int i = 1 ; i <= n ; i++) {
            cin >> a[i];
        }
        cout << "Case " << ans << ":" << endl;
        for(int i = 0 ; i < q ; i++) {
            int d , m;
            cin >> d >> m;
            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 ; j++) {
                    for(int l = 0 ; l < d ; l++) {
                        dp[i][j][l] += dp[i - 1][j][l];
                        dp[i][j][((l + a[i]) % d + d) % d] += dp[i - 1][j- 1][l];
                    }
                }
            }
            cout << dp[n][m][0] << endl;
        }
    }
    return 0;
}