Problem:
Exhaustive Search

Write a program which reads a sequence A of n elements and an integer M, and outputs "yes" if you can make M by adding elements in A, otherwise "no". You can use an element only once.

You are given the sequence A and q questions where each question contains Mi.

Input

In the first line n is given. In the second line, n integers are given. In the third line q is given. Then, in the fourth line, q integers (Mi) are given.

Output

For each question Mi, print yes or no.

Constraints

n ≤ 20

q ≤ 200

1 ≤ elements in A ≤ 2000

1 ≤ Mi ≤ 2000

Sample Input 1

5
1 5 7 10 21
8
2 4 17 8 22 21 100 35

Sample Output 1

no
no
yes
yes
yes
yes
no
no

Notes

You can solve this problem by a Burte Force approach. Suppose solve(p, t) is a function which checkes whether you can make t by selecting elements after p-th element (inclusive). Then you can recursively call the following functions:

solve(0, M)

solve(1, M-{sum created from elements before 1st element})

solve(2, M-{sum created from elements before 2nd element})

...

The recursive function has two choices: you selected p-th element and not. So, you can check solve(p+1, t-A[p]) and solve(p+1, t) in solve(p, t) to check the all combinations.

For example, the following figure shows that 8 can be made by A[0] + A[2].

Solution in JAVA (judge verdict: Accepted)

import java.util.HashMap;
import java.util.Map;
import java.util.*;
import java.util.Arrays;
class Main {
    public static void main(String[] args) {
       int M = 0;
       Scanner scner = new Scanner(System.in);
        int T = 0;
        int n = scner.nextInt();
        int[] S = new int[n];
        for (int i=0; i<n; i++) {
            S[i] = scner.nextInt();
        }
        int f = scner.nextInt();
        for(int i = 0; i<f; ++i){
         int b = scner.nextInt();
        
       
       
       if(sSearch(S,b)){
                 System.out.println("yes");
             }else{
                 System.out.println("no");
             }

         }
      }
         public static Boolean sSearch(int[] a, int b){
        
        Boolean res=false;
        
        for(int j=0; j<a.length; ++j){
            if(seSearch(a,b,j)){
                res=true;
                break;
                
                
               }
        }
        return res;
    }

    public static Boolean seSearch(int[] a, int b,int k){
   
    if(b==0){
            return true;
        }else if(k>=a.length){
            return false;
        }
        
        
    
Boolean res=seSearch(a,b,k+1)||seSearch(a,b-a[k],k+1);
return res;

}
}