lightoj1125：Divisible Group Sums（类01背包）

1125 - Divisible Group Sums

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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

题意：从N个数选M个使他们的和能被D整除，求方案数。

思路：dp[i][j]表示选i个数，他们的和取余D的值为j的方案数，dp[i][j] += dp[i-1][(D+ j-(a[k]%D)%D)]

# include <stdio.h>
# include <string.h>
int main()
{
    int a[203], t, T, i, j, k, D, M, N, n, Q, num;
    long long dp[11][21];
    scanf("%d",&T);
    for(t=1; t<=T; ++t)
    {
        scanf("%d%d",&N,&Q);
        for(i=1; i<=N; ++i)
            scanf("%d",&a[i]);
        printf("Case %d:\n",t);
        while(Q--)
        {
            scanf("%d%d",&D,&M);
            memset(dp, 0, sizeof(dp));
            dp[0][0] = 1;
            for(i=1; i<=N; ++i)
                for(j=M; j>0; --j)
                {
                    num = a[i]%D;
                    for(k=0; k<D; ++k)
                        dp[j][k] += dp[j-1][(k-num+D)%D];
                }
            printf("%lld\n",dp[M][0]);
        }
    }
    return 0;
}