hdu 5800 To My Girlfriend (dp)

最新推荐文章于 2018-10-06 19:35:21 发布

原创最新推荐文章于 2018-10-06 19:35:21 发布 · 203 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一种使用动态规划解决复杂组合问题的方法，具体为计算特定条件下物品子集的组合数量。通过定义状态转移方程并采用递推的方式，有效地解决了问题，并提供了完整的代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Problem Description

Dear Guo

I never forget the moment I met with you.You carefully asked me: "I have a very difficult problem. Can you teach me?".I replied with a smile, "of course"."I have n items, their weight was a[i]",you said,"Let's define f(i,j,k,l,m) to be the number of the subset of the weight of n items was m in total and has No.i and No.j items without No.k and No.l items.""And then," I asked.You said:"I want to know

\sum i = 1 n \sum j = 1 n \sum k = 1 n \sum l = 1 n \sum m = 1 s f (i, j, k, l, m) (i, j, k, l a r e d i f f e r e n t)

Sincerely yours,
Liao

Input

The first line of input contains an integer T

(T≤15) indicating the number of test cases.
Each case contains 2 integers

(4≤n≤1000,1≤s≤1000). The next line contains n numbers:

a1,a2,…,an

(1≤ai≤1000).

Output

Each case print the only number — the number of her would modulo

109+7 (both Liao and Guo like the number).

Sample Input

Sample Output

8
8

思路：dp[i][t][k][j]; 0到i个数和为j t个必取 k个必不取

解为 dp【n】【2】【2】【1~s】

#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define L(i) i<<1
#define R(i) i<<1|1
#define INF  0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-9
#define maxn 10010
#define MOD 1000000007

long long ans;
int T;

const int mod = 1e9 +7;
int n,s;

int a[1010];

int dp[1010][3][3][1010];


int main()
{
    scanf("%d",&T);

    while(T--)
    {
        scanf("%d%d",&n,&s);

        for(int i=1; i<=n; i++) scanf("%d",&a[i]);

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

        dp[0][0][0][0]=1;
        ans=0;

        for(int i=1; i<=n; i++)
        {
            for(int j=0; j<=s; j++)
            {
                for(int t=0; t<=2; t++)
                    for(int k=0; k<=2; k++)
                    {
                        dp[i][t][k][j] = dp[i-1][t][k][j];

                        if(j>=a[i])
                        {
                            dp[i][t][k][j] = (dp[i][t][k][j] + dp[i-1][t][k][j-a[i]]) % mod;

                            //choose
                            if(t>0)
                                dp[i][t][k][j] = (dp[i][t][k][j] + dp[i-1][t-1][k][j-a[i]]) %mod;
                        }

                        // no choose
                        if(k>0)
                            dp[i][t][k][j]  = (dp[i][t][k][j] + dp[i-1][t][k-1][j]) %mod;
                        }
            }
        }


//        for(int i=1; i<=n; i++)
//        {
//            for(int j=0; j<=s; j++)
//                for(int t=0; t<=2; t++)
//                    for(int k=0; k<=2; k++)
//                    {
//                        printf("dp[%d][%d][%d][%d] = %d\n",i,t,k,j,dp[i][t][k][j]);
//                    }
//        }

        for(int i=0; i<=s; i++)
            ans =(ans + dp[n][2][2][i])%mod;

        printf("%I64d\n",ans*4 %mod);
    }
    return 0;
}