nyoj927 The partial sum problem(dfs)

The partial sum problem

时间限制：1000 ms  |  内存限制：65535 KB

难度：2

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.

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!

题意：很简单，就是给你一个数组，和一个值k，是否能从数组中找到n个数之和等于k

思路：直接dfs，不过这个题好像时间有点限制，所以要一个简单的剪枝

Code:

//开始的思路不是很清晰，太盲目的做，因为初始化和回溯的问题wa了两次，然后改好之后又果断超时了，没有考虑剪枝和重复搜索的问题.
//看了一下别人的代码瞬间开窍了,这么简单的题目还wa真是醉了
#include <stdio.h>
#include <string.h>
int a[20],n,k,flage,vis[20];

void dfs(int now,int sum)
{
    int i;
    if(sum>=k)
    {
        if(sum==k)
            flage=1;
        return ;
    }
    else
    {
        for(i=now;i<n;i++)              //这里从now开始就行了，之前的思路是把所有的情况全部考虑了中间会出现重复的情况
        {
            if(vis[i]==0)
            {
                vis[i]=1;
                sum+=a[i];
                dfs(i+1,sum);
                if(flage)               //这里做了一个剪枝，如果已经可以找到，那么后面就不用找了
                    return;
                sum-=a[i];
                vis[i]=0;
            }
        }
    }
}

int main()
{
    int sum,i;
    while(scanf("%d",&n)!=EOF)
    {
        flage=0;
        sum=0;
        memset(vis,0,sizeof(vis));
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        scanf("%d",&k);
        dfs(0,0);
        if(flage)
            printf("Of course,I can!\n");
        else
            printf("Sorry,I can't!\n");
    }
    return 0;
}