1125 - Divisible Group Sums （DP）

最新推荐文章于 2016-09-15 11:47:21 发布

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本文详细介绍了使用计数DP方法解决给定整数集合中选取若干个数使其和能被特定数整除的问题。通过三层递推实现最优解计算，并提供了AC代码示例。

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

10 2

5 1

5 2

5 1

6 2

Case 1:

Case 2:

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;
}