HDU 5800 To My Girlfriend DP

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

\[\sum_{i=1}^{n}\sum_{j=1}^{n}\sum_{k=1}^{n}\sum_{l=1}^{n}\sum_{m=1}^{s}f(i,j,k,l,m)\quad (i,j,k,l\quad are\quad different) \]

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 n, s (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
2
4 4
1 2 3 4
4 4
1 2 3 4
Sample Output
8
8

**题意：**给出n个数，求权值和为m，没有下标i，j，有下标k，l的数集合个数 **思路：**主要是如何描述状态，权值大小，必不选取和必选取的状态的描述可以类似背包，dp[i][j][k][l] 代表选取到i个元素和为j，使用了k个必选名额，l个必不选名额 那么状态可以有四种转移方法 $$ \begin{equation}\label{Dirichlet}dp[i][j][k][l]+=\begin{cases} dp[i-1][j][k][l], & \text{ 不选当前的数，亦不作为必不选的数 }\\\ dp[i-1][j-a[i]][k][l], & \text{ 选当前的数，不作为必选的数 } \\\ dp[i-1][j-a[i]][k-1][l], & \text{ 选当前的数，作为必选的数 } \\\ dp[i-1][j][k][l-1], & \text{ 不选当前的数，但作为必}\color{#F00}{不}\text{选的数} \end{cases}\end{equation} $$

/** @Date    : 2016-11-09-17.37

  * @Author  : Lweleth (SoungEarlf@gmail.com)

  * @Link    : https://github.com/

  * @Version :

  */

#include <stdio.h>

#include <iostream>

#include <string.h>

#include <algorithm>

#include <utility>

#include <vector>

#include <map>

#include <set>

#include <string>

#include <stack>

#include <queue>

#define LL long long

#define MMF(x) memset((x),0,sizeof(x))

#define MMI(x) memset((x), INF, sizeof(x))

using namespace std;



const int INF = 0x3f3f3f3f;

const int N = 1e5+2000;

const int mod = 1e9 + 7;

int a[1010];

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



LL solve(int &n, int &s)

{

    MMF(dp);

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

    //dp[1][0][0][0] = 1;

    LL ans = 0;//返回值要用LL...

    for(int i = 1; i <= n; i++)

    {

        for(int j = 0; j <= s; j++)

        {

            for(int k = 0; k <= 2; k++)

            {

                for(int l = 0; l <= 2; l++)

                {

                    dp[i][j][k][l] += dp[i-1][j][k][l];

                    dp[i][j][k][l] %= mod;

                    if(j >= a[i])

                    {

                        dp[i][j][k][l] += dp[i-1][j-a[i]][k][l];

                        dp[i][j][k][l] %= mod;

                    }

                    if(k > 0)

                    {

                        if(j >= a[i])

                            dp[i][j][k][l] += dp[i-1][j - a[i]][k-1][l];

                        dp[i][j][k][l] %= mod;

                    }

                    if(l > 0)

                    {

                        dp[i][j][k][l] += dp[i-1][j][k][l-1];

                        dp[i][j][k][l] %= mod;

                    }

                }

            }

            if(i == n)

            {

                ans += dp[i][j][2][2];

                ans %= mod;

            }

        }

    }

    return (ans * 4 + mod) % mod;

}







int main()

{

    int T;

    scanf("%d", &T);

    while(T--)

    {

        int n, s;

        scanf("%d%d", &n, &s);

        for(int i = 1; i <= n; i++)

            scanf("%d", a + i);

        printf("%lld\n", solve(n, s));//输出LL型？

    }

    return 0;

}