hdu 5800 计数dp

To My Girlfriend

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 929    Accepted Submission(s): 359

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 modulo109+7 (both Liao and Guo like the number).

 

Sample Input

24 41 2 3 44 41 2 3 4

 

Sample Output

88

               加维以便计数

#include <bits/stdc++.h>using namespace std;typedef long long ll;const long long mod=1e9+7;int  dp[1005][3][3];int num[1005];int main(){    int n,t,s;    scanf("%d",&t);    while(t--)    {        memset(dp,0,sizeof(dp));        scanf("%d %d",&n,&s);        for(int i=1;i<=n;i++)        {            scanf("%d",&num[i]);        }        dp[0][0][0]=1;        for(int i=1;i<=n;i++)        {            for(int j=s;j>=num[i];j--)            {                for(int a=2;a>=0;a--)                {                    for(int b=2;b>=0;b--)                    {                        dp[j][a][b]+=dp[j-num[i]][a][b];                        dp[j][a][b]%=mod;                        if(a!=0)                        {                            dp[j][a][b]+=dp[j-num[i]][a-1][b];                            dp[j][a][b]%=mod;                        }                        if(b!=0)                        {                            dp[j][a][b]+=dp[j][a][b-1];                            dp[j][a][b]%=mod;                        }                      //  cout<<"i="<<i<<" j=="<<j<<" a=="<<a<<" b="<<b<<" value="<<dp[j][a][b]<<endl;                    }                }            }            for(int j=num[i]-1;j>=0;j--)            {                 for(int a=2;a>=0;a--)                {                    for(int b=2;b>=0;b--)                    {                        if(b!=0)                        {                            dp[j][a][b]+=dp[j][a][b-1];                            dp[j][a][b]%=mod;                        }                      //  cout<<"i="<<i<<" j=="<<j<<" a=="<<a<<" b="<<b<<" value="<<dp[j][a][b]<<endl;                    }                }            }        }        ll ans=0;        for(int i=1; i<=s; i++) ans = (ans + dp[i][2][2])%mod;        cout<<(ans*4)%mod<<endl;    }    return 0;}