HDU 1258 Sum It Up

http://acm.hdu.edu.cn/showproblem.php?pid=1258

题意：给你一个数t作为最后等式的和，再给你n个数作为等式的加数，每个加数最多只能使用一次，要求输出所有满足条件(加数从大到小输出)的等式，并且不能重复

Sample Input
4 6 4 3 2 2 1 1
5 3 2 1 1
400 12 50 50 50 50 50 50 25 25 25 25 25 25
0 0

Sample Output
Sums of 4:
4
3+1
2+2
2+1+1
Sums of 5:
NONE
Sums of 400:
50+50+50+50+50+50+25+25+25+25
50+50+50+50+50+25+25+25+25+25+25

#include <iostream>
#include <algorithm>
#include <string>
#include <map>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
//#include <ctime>
#include <ctype.h>
using namespace std;
#define LL __int64
#define inf 0x7fffffff
#define M 15

map<string, int> m;
int num[M], ans, n;
bool vis[M], flag;
char res[105];

void dfs (int x, int sum, int k)
{
	int i;
	if (sum == ans)
	{
		res[k] = 0;		//+结束符，然后判是否重复
		if (m[res] == 0)
		{
			flag = true;
			m[res] = 1;
			printf ("%d", (int)res[0]);	//转换成整数输出
			for (i = 1; i < k; i++)
				printf ("+%d", (int)res[i]);
			printf ("\n");
		}
		return ;
	}
	for (i = x + 1; i < n; i++)		//由于是递增，所以从x+1开始
	{
		if (vis[i] || sum + num[i] > ans)
			continue;
		vis[i] = true;
		res[k] = num[i];
		dfs (i, sum+num[i], k+1);
		vis[i] = false;
	}
}

int main()
{
	int i;
	while (scanf ("%d%d", &ans, &n), (ans||n))
	{
		printf ("Sums of %d:\n", ans);
		m.clear();
		for (i = 0; i < n; i++)
			scanf ("%d", num+i);
		flag = false;
		for (i = 0; i < n; i++)
		{
			res[0] = num[i];	//直接当num[i]是ASCⅡ码使用，方便map记录重复
			memset (vis, false, sizeof(vis));
			vis[i] = true;
			dfs (i, num[i], 1);
		}
		if (!flag)
			puts ("NONE");
	}
    return 0;
}