nyoj Thepartialsumproblem（DFS）

[align=center]Thepartialsumproblem[/align]

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. 

Input

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).

Output

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!”.

Sample Input

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

Sample Output

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

[align=left]ps：深搜也要讲方法的，我临时想的一个搜法，果断超时。。[/align]
[align=left]超时代码：[/align]

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int n,k,flag,x;
int a[22];
void dfs(int s)
{
if(abs(s-k)>x)
return ;
if(s==k)
flag=1;
if(flag)
return ;
for(int i=0; i<n; i++)//这一点很关键，用此法无故多了好多分枝。。
{
if(a[i])
{
int m=a[i];
a[i]=0;
dfs(s+m);
a[i]=m;
}
}
}
int main()
{
while(~scanf("%d",&n))
{
memset(a,0,sizeof(a));
int i,y1=0,y2=0;
for(i=0; i<n; i++)
{
scanf("%d",&a[i]);
if(a[i]>0)
y1+=a[i];
else
y2+=a[i];
}
scanf("%d",&k);
if((k>0&&y1<k)||(k<0&&y2>k))
printf("Sorry,I can't!\n");
else
{
flag=0;
x=abs(k);
dfs(0);
if(flag)
printf("Of course,I can!\n");
else
printf("Sorry,I can't!\n");
}
}
return 0;
}

ps：吃一顿饭回来，想到了以前的部分和问题，然后用那种方法试一试，，竟然A了


AC代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int n,k,flag,x;
int a[22];
int dfs(int i,int sum)
{
if(abs(sum-k)>x)
return 0;
if(sum==k)
return 1;
if(i==n)
retur
4000
n sum==k;
if(dfs(i+1,sum))//少了好多分枝。。
return 1;
if(dfs(i+1,sum+a[i]))//
return 1;
}
int main()
{
while(~scanf("%d",&n))
{
memset(a,0,sizeof(a));
int i,y1=0,y2=0;
for(i=0; i<n; i++)
{
scanf("%d",&a[i]);
if(a[i]>0)
y1+=a[i];
else
y2+=a[i];
}
scanf("%d",&k);
if((k>0&&y1<k)||(k<0&&y2>k))
printf("Sorry,I can't!\n");
else
{
x=abs(k);
if(dfs(0,0))
printf("Of course,I can!\n");
else
printf("Sorry,I can't!\n");
}
}
return 0;
}

ps：深搜剪枝剪枝，还是要多想想，找到最优的剪枝方法。尽量少用循环。。