Exhaustive Search dfs

最新推荐文章于 2021-05-06 16:55:45 发布

ujn20161222

最新推荐文章于 2021-05-06 16:55:45 发布

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本文介绍了一个经典的子集求和问题，通过递归的方法来判断是否能从给定序列中选出若干个数使得它们的和等于指定的目标值。文章详细解释了递归函数的工作原理，并提供了一段C++代码实现。

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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 M_i.

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 (M_i) are given.

Output

For each question M_i, print yes or no.

Constraints

n ≤ 20
q ≤ 200
1 ≤ elements in A ≤ 2000
1 ≤ M_i ≤ 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].

#include<bits/stdc++.h>

using namespace std;

const int N(30);
int a[N];

int n;

bool dfs(int i,int m){
    if(m==0)return 1;
    if(i>=n)return false;
    return dfs(i+1,m-a[i])||dfs(i+1,m);
}


int main(){
    //freopen("in.txt","r",stdin);
    cin>>n;
    for(int i=0;i<n;i++)
    scanf("%d",&a[i]);
    int q;
    cin>>q;
    for(int i=0;i<q;i++){
        int m;
        cin>>m;
        if(dfs(0,m))
        puts("yes");
        else puts("no");
    }
    //fclose(stdin);
}