nyoj 927 The partial sum problem

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The partial sum problem

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!

#include <stdio.h>
int e[25];
int imt;
int num;

int dfs(int step,int count)//深度搜索 这里还用到了动态规划的思想 01背包问题
{
    if(step==num)
        return count==imt;
    if(dfs(step+1,count))
        return 1;
    if(dfs(step+1,count+e[step]))
        return 1;
    return 0;
}

int main()
{

    int i,j,k;
    
    while(scanf("%d",&num)!=EOF)
    {
        for(i=0;i<num;i++)
        {
            scanf("%d",e+i);
        }
        scanf("%d",&imt);
        
        if(dfs(0,0))
            printf("Of course,I can!\n");
        else
            printf("Sorry,I can‘t!\n");
    }
    
    return 0;
}

刚开始写的时候一直超时，很郁闷，搜索了下，发现用深搜加上动态规划的思想便可以解决~，很棒的一种方法~

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nyoj 927 The partial sum problem

标签：style   blog   color   os   art   for   

原文地址：http://www.cnblogs.com/ltwy/p/3841278.html