题目链接：HDU 5800

题面：

To My Girlfriend

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 737    Accepted Submission(s): 292

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

∑i=1n∑j=1n∑k=1n∑l=1n∑m=1sf(i,j,k,l,m)(i,j,k,laredifferent)

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

Author

UESTC

Source

2016 Multi-University Training Contest 6

题意：

吐槽下题意，很难懂。求的是，给定n个数，其中选定若干数，这若干数的权值和为m，且这些数中没有下标为i,j的数，有下标为k,l的数的集合个数。

解题：

dp[i][j][s1][s2]，代表的是前i个物品，总权值为j，已有s1个必选，s2必不选的方案数。那么对于当前一个状态，它有四种转移状态。

1.选中当前的，增加权值，增加必选个数。

2.选择当前的，增加权值，不增加必选个数。

3.不选中当前的，不增加权值，增加不必选个数。

4.不选中当前的，不增加权值，不增加必选个数。

因为i，j可以互换，l，k也可以互换，故而最后方案数乘以4即为所求。

代码：

#include <iostream>
#include <iomanip>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define mod 1000000007
#define LL long long
#define p(a) printf("%d\n",a)
using namespace std;
unsigned dp[1005][1005][3][3];
int a[1005];
int main()
{
    int t,n,s,tmp;
	LL ans;
	scanf("%d",&t);
	while(t--)
	{
        ans=0;
		scanf("%d%d",&n,&s);
		for(int i=1;i<=n;i++)
		  scanf("%d",&a[i]);
		memset(dp,0,sizeof(dp));
		dp[1][a[1]][0][0]=1;
		dp[1][a[1]][1][0]=1;
		dp[1][0][0][0]=1;
		dp[1][0][0][1]=1;
		for(int i=2;i<=n;i++)
		{
			for(int j=0;j<=s;j++)
			{
               tmp=j+a[i];
			   if(tmp<=s)
			   {
				   for(int k=0;k<=2;k++)
				   {
					   dp[i][tmp][0][k]=(dp[i][tmp][0][k]+dp[i-1][j][0][k])%mod;
					   dp[i][tmp][1][k]=(dp[i][tmp][1][k]+dp[i-1][j][0][k])%mod;
					   dp[i][tmp][1][k]=(dp[i][tmp][1][k]+dp[i-1][j][1][k])%mod;
					   dp[i][tmp][2][k]=(dp[i][tmp][2][k]+dp[i-1][j][1][k])%mod;
					   dp[i][tmp][2][k]=(dp[i][tmp][2][k]+dp[i-1][j][2][k])%mod;

				   }
			   }
			   for(int k=0;k<=2;k++)
			   {
				   dp[i][j][k][0]=(dp[i][j][k][0]+dp[i-1][j][k][0])%mod;
				   dp[i][j][k][1]=(dp[i][j][k][1]+dp[i-1][j][k][0])%mod;
				   dp[i][j][k][1]=(dp[i][j][k][1]+dp[i-1][j][k][1])%mod;
				   dp[i][j][k][2]=(dp[i][j][k][2]+dp[i-1][j][k][1])%mod;
				   dp[i][j][k][2]=(dp[i][j][k][2]+dp[i-1][j][k][2])%mod;
			   }
			}
		}
		for(int i=1;i<=s;i++)
			ans=(ans+dp[n][i][2][2])%mod;
		printf("%lld\n",ans*4%mod);
	}
	return 0;
}