The partial sum problem （DFS）nyoj

The partial sum problem

时间限制：1000 ms  |  内存限制：65535 KB

输入

There are multiple test cases.
Each test case contains three lines.The first line is an integer N(1≤N≤20),represents the array contains N integers. The second line contains N integers,the ith integer represents A[i](-10^8≤A[i]≤10^8).The third line contains an integer K(-10^8≤K≤10^8).

输出

If Tom can choose some integers from the array and their them is K,printf ”Of course,I can!”; other printf ”Sorry,I can’t!”.

样例输入

4
1 2 4 7
13
4
1 2 4 7
15

样例输出

Of course,I can!
Sorry,I can't!

描述

One day,Tom’s girlfriend give him an array A which contains N integers and asked him:Can you choose some integers from the N integers and the sum of them is equal to K. 

 

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int sum,a[30],k,ok,n;//a数组用来储存每个数字，ok存放是否存在题目要求的组合

void dfs(int x)//深搜
{
    if(sum>k)//当sum比k大的情况，就不用再往深处搜索了
        return ;
    if(sum==k)//当sum==k，也就是加起来的和存在和k相等的时候，把ok赋值为1；
    {
        ok=1;
        return ;
    }
    for (int i = x; i <= n; i++){
        sum += a[i];//加上某一个数
        dfs(i + 1);//递归
        sum -= a[i];//把加上的这个数减去
    }
}

int main()
{
    while(~scanf("%d",&n)){
         for(int i=1; i<=n; i++){
            scanf("%d",&a[i]);
        }
        scanf("%d",&k);
        ok=0,sum=0;
        dfs(1);
        if(ok)
            cout<<"Of course,I can!"<<endl;
        else
            cout<<"Sorry,I can't!"<<endl;
    }
    return 0;
}