Problem 1608 - Calculation

Time Limit: 500MS   Memory Limit: 65536KB  
Total Submit: 343  Accepted: 96  Special Judge: No

Description

Today, Alice got her math homework again!

She had n integers, and she needed to divide them into several piles or one pile. For each pile, if the teacher could get S, by + or – operator, then Alice got 1 small red flower. Alice wanted to get as many flowers as possible. Could you help her? Just tell her the maximum number of flowers she could get.

Input

The input consists of several test cases.
The first line consists of one integer T (T <= 100), meaning the number of test cases.
The first line of each test cases consists of two integer n (n<=14), meaning the number of the integer, and S (0<= S<= 100000000), meaning the result which teacher wanted.
The next line consists of n integer a1, a2, …, an (0<= ai <= 10000000).
You should know a few cases that n is larger than 12.

Output

For each test case, output one line with one integer without any space.

Sample Input

2
5 5
1 2 3 4 5
5 5
1 2 3 8 8

Sample Output

3
2

Hint

Source


分析： 于是比赛时我又煞笔了。。。记忆化搜索T了，直接暴力DP反而更快。。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
using namespace std;
int f[1<<14],sum[1<<14],n,m,s,t,ans,a[15];
bool jud[1<<14];
int main()
{
    cin.sync_with_stdio(false);
    cin>>t;
    for(int T=1;T <= t;T++)
    {
        cin>>n>>s;
        int N=(1<<n)-1;
        memset(sum,0,sizeof(sum));
        memset(f,0,sizeof(f));
        memset(jud,0,sizeof(jud));
        for(int i=1;i <= n;i++) cin>>a[i];
        for(int i=N;i;i=(i-1) & N)
        {
            for(int j=0;j < n;j++)
            if(i & (1<<j)) sum[i]+=a[j+1];
            if(sum[i] == s) jud[i]=true;
        }
        for(int i=N;i;i=(i-1) & N)
        if(!jud[i])
        for(int j=i;j;j=(j-1) & i)
          if(sum[i]-2*sum[j] == s)
          {
            jud[i]=true;
            break;
          }
        for(int i=0;i <= N;i++)
        for(int j=i;j;j=(j-1) & i)
          if(jud[j]) f[i]=max(f[i],1+f[i-j]);
        cout<<f[N]<<endl;
    }
}