The partial sum problem

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

41 2 4 71341 2 4 715

样例输出

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

   一开始只想到超时的代码，一直不知道原来for循环里可以这么奇妙，看来是对dfs不够熟练。。。。

超时代码：

 #include<iostream>#include<cstring>#include<cstdio>using namespace std;int bo[25];int a[25];int k,n,f;void dfs(int t){if(f||t>k)     return; if(t==k)f=1; for(int i=0;i<n;++i){if(!bo[i]){bo[i]=1;dfs(t+a[i]);bo[i]=0;}}}int main(){while(~scanf("%d",&n)){f=0;memset(bo,0,sizeof(bo));for(int i=0;i<n;++i){scanf("%d",&a[i]);}scanf("%d",&k);dfs(0);if(f){printf("Of course,I can!\n");}else{printf("Sorry,I can't!\n");}}return 0;}        

AC代码：

#include<iostream>#include<cstring>#include<cstdio>using namespace std;int a[25];int k,n,f,t;void dfs(int x){if(t==k)f=1;if(f||(t>k&&t>0))return;for(int i=x;i<n;++i){t+=a[i];dfs(i+1);t-=a[i];}}int main(){while(~scanf("%d",&n)){f=0;t=0;for(int i=0;i<n;++i){scanf("%d",&a[i]);}scanf("%d",&k);dfs(0);if(f){printf("Of course,I can!\n");}else{printf("Sorry,I can't!\n");}}return 0;}