hdu 5800 To My Girlfriend(2016 Multi-University Training Contest 6——DP)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5800

To My Girlfriend

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

Total Submission(s): 908    Accepted Submission(s): 348

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

 

题目大意：

给n个数，求：

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

i，j为必选数的序号，k，l为必不选数的序号，m是背包体积。

解题思路：

定义dp[i][j][k][l]表示到第i个为止，和正好是j，有k个必选，l个必不选的方法数。

那么分析一下，只有四种情况：选，不选，必选，必不选。

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

需要注意的是，i和j可以交换，k和l也可以交换，所以最后的结果要乘以4。

详见代码。

//dp[i][j][k][l]表示到第i个为止，和正好是j，有k个必选，l个必不选的方法数

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int Mod=(1e9+7);
#define ll long long
int dp[1010][1010][3][3],a[1010];

int main()
{
    int t;
    scanf("%d",&t);
    while (t--)
    {
        int n,s;
        ll sum=0;
        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++) //循环每一位数
        {
            for (int j=0; j<=s; j++) //循环每一个重量
            {
                for (int k=0; k<=2; k++) //k个必选 
                {
                    for (int l=0; l<=2; l++) //l个必不选
                    {
                        dp[i][j][k][l]+=dp[i-1][j][k][l];//不选
                        dp[i][j][k][l]%=Mod;
                        if (l>0)
                        {
                            dp[i][j][k][l]+=dp[i-1][j][k][l-1];//必不选
                            dp[i][j][k][l]%=Mod;
                        }
                        if (j>=a[i])
                        {
                            dp[i][j][k][l]+=dp[i-1][j-a[i]][k][l];//选
                            dp[i][j][k][l]%=Mod;
                        }
                        if (k>0&&j>=a[i])
                        {
                            dp[i][j][k][l]+=dp[i-1][j-a[i]][k-1][l];//必选
                            dp[i][j][k][l]%=Mod;
                        }
                    }
                }
            }
        }
        for (int i=0; i<=s; i++)
        {
            sum+=dp[n][i][2][2];
            sum%=Mod;
        }
        printf ("%lld\n",(sum*4)%Mod);
    }
    return 0;
}