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 = ;
const int mod = 1e9+;
int a[maxn];
int dp[maxn][maxn][][];
int n,s;  

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