1125 - Divisible Group Sums （DP）

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 tochoose any M of them. So you can choose in ^NC_Mways. 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 Nindicates 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 querieswill have to be answered based on these N numbers. Each of the next Qlines contains two integers D (0 < D ≤ 20) and M (0 < M≤ 10).

Output

For each case, print the case number in a line. Then foreach 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：给你n个32位有符号整数，Q个询问，每次询问从n个数中选出M个数相加能够整除D的方案数

题解：

简单的计数DP问题，首先先找最优子结构，dp[i][j][k]表示n个数的前i个数选j个，mod D等于k的方案数。

知道这个就可以通过三层for递推出结果了，最终答案为dp[n][M][0];

AC代码：

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
ll dp[205][15][25];
int a[205];


int main() {
    int T,n,q,cnt = 0;
    cin>>T;
    while(T--) {
        scanf("%d %d",&n,&q);
        for(int i = 1; i <= n; i++) {
            scanf("%d",&a[i]);
        }
        printf("Case %d:\n",++cnt);
        while(q--) {
            int D,M;
            memset(dp,0,sizeof dp);
            scanf("%d %d",&D,&M);
            dp[0][0][0] = 1;
            for(int i = 1; i <= n; i++) {
                dp[i][0][0] = 1;
                for(int j = 1; j <= M; j++) {
                    for(int k = 0; k < D; k++) {
                        dp[i][j][k] += dp[i - 1][j][k];
                        dp[i][j][k] += dp[i - 1][j - 1][(k - a[i] % D + D) % D];
                        //printf("%d %d %d %d\n",i,j,k,dp[i][j][k]);
                    }
                }
            }
            printf("%lld\n",dp[n][M][0]);
        }
    }
    return 0;
}