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

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

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;
}