nyoj 927 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!

#include <stdio.h>
#include <stdlib.h>
int n, arr[22], sum, vis[22], ok, count;
const char *sam[] = {"Sorry,I can't!\n", "Of course,I can!\n"};

int cmp(const void *a, const void *b){
  return *(int *)a - *(int *)b;
}

void DFS(int k)
{
	if(count == sum)
	{
		ok = 1; return;
	}
	for(int i = k; i < n; ++i)
	{
		if(i && arr[i] == arr[i-1] && !vis[i-1]) //cut
		continue;
		if(count > sum && arr[i] > 0) return; //cut	    
		count += arr[i];
		vis[i] = 1;
		DFS(i + 1);
		if(ok) return;
		count -= arr[i]; 
		vis[i] = 0;
	}
}

int main()
{
    while(scanf("%d", &n) == 1)
    {
	    for(int i = 0; i < n; ++i)
		{
	      scanf("%d", arr + i);
	      vis[i] = 0;
	    }
	    scanf("%d", &sum);
	    qsort(arr, n, sizeof(int), cmp);
	    count = ok = 0; DFS(0);
	    printf(ok ? sam[1] : sam[0]);
   }
  return 0;
}