hdu 5800 To My Girlfriend (dp)

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

24 41 2 3 44 41 2 3 4

 

Sample Output

88

 

思路：dp[i][t][k][j]; 0到i个数 和为j t个必取 k个必不取

解为 dp【n】【2】【2】【1~s】

#include <set>#include <map>#include <stack>#include <queue>#include <deque>#include <cmath>#include <vector>#include <string>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define L(i) i<<1#define R(i) i<<1|1#define INF  0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-9#define maxn 10010#define MOD 1000000007long long ans;int T;const int mod = 1e9 +7;int n,s;int a[1010];int dp[1010][3][3][1010];int main(){    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&s);        for(int i=1; i<=n; i++) scanf("%d",&a[i]);        memset(dp,0,sizeof(dp));        dp[0][0][0][0]=1;        ans=0;        for(int i=1; i<=n; i++)        {            for(int j=0; j<=s; j++)            {                for(int t=0; t<=2; t++)                    for(int k=0; k<=2; k++)                    {                        dp[i][t][k][j] = dp[i-1][t][k][j];                        if(j>=a[i])                        {                            dp[i][t][k][j] = (dp[i][t][k][j] + dp[i-1][t][k][j-a[i]]) % mod;                            //choose                            if(t>0)                                dp[i][t][k][j] = (dp[i][t][k][j] + dp[i-1][t-1][k][j-a[i]]) %mod;                        }                        // no choose                        if(k>0)                            dp[i][t][k][j]  = (dp[i][t][k][j] + dp[i-1][t][k-1][j]) %mod;                        }            }        }//        for(int i=1; i<=n; i++)//        {//            for(int j=0; j<=s; j++)//                for(int t=0; t<=2; t++)//                    for(int k=0; k<=2; k++)//                    {//                        printf("dp[%d][%d][%d][%d] = %d\n",i,t,k,j,dp[i][t][k][j]);//                    }//        }        for(int i=0; i<=s; i++)            ans =(ans + dp[n][2][2][i])%mod;        printf("%I64d\n",ans*4 %mod);    }    return 0;}