2016暑假多校联合---To My Girlfriend

2016暑假多校联合---To My Girlfriend

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

 

思路：令dp[i][j][s1][s2]表示前i个物品填了j的体积，有s1个物品选为为必选，s2个物品选为必不选的方案数(0<=s1,s2<=2)，则有转移方程dp[i][j][s1][s2] = dp[i - 1][j][s1][s2] + dp[i - 1][j - a[i]][s1 - 1][s2] +dp[i - 1][j - a[i]][s1][s2] + dp[i - 1][j][s1][s2 - 1]，边界条件为dp[0][0][0][0] = 1，时间复杂度O(NS*3^2)；

 

代码如下：

#include<bits/stdc++.h>  
using namespace std;  
#define LL long long  
const int maxn = 1005;  
const int mod = 1e9+7;  
int a[maxn];  
int dp[maxn][maxn][3][3];  
int n,s;  

int main()  
{  
    int T;  
    scanf("%d",&T);  
    while(T--)  
    {  
        memset(dp,0,sizeof(dp));  
        scanf("%d%d",&n,&s);  
        for(int i = 1;i<=n;i++)  
            scanf("%d",&a[i]);  
        dp[0][0][0][0]=1;  
        for(int i = 1;i<=n;i++)  
        for(int j = 0;j<=s;j++)  
        for(int s1 = 0;s1<=2;s1++)  
        for(int s2 = 0;s2<=2;s2++)  
        {  
            dp[i][j][s1][s2]=(dp[i][j][s1][s2]+dp[i-1][j][s1][s2])%mod;  
            if(j>=a[i])  
            dp[i][j][s1][s2]=(dp[i][j][s1][s2]+dp[i-1][j-a[i]][s1][s2])%mod;  
            if(s1>0&&j>=a[i])  
            dp[i][j][s1][s2]=(dp[i][j][s1][s2]+dp[i-1][j-a[i]][s1-1][s2])%mod;  
            if(s2>0)  
            dp[i][j][s1][s2]=(dp[i][j][s1][s2]+dp[i-1][j][s1][s2-1])%mod;  
        }  
        LL ans = 0;  
        for(int i = 1;i<=s;i++)  
          ans = (ans+dp[n][i][2][2])%mod;  
        printf("%lld
",ans*4%mod);  
    }  
}