hdu 1258 从n个数中找和为t的组合 (DFS)

题意：首先给你一个t，然后是n，后面输入n个数，然后让你求的是n个数中和为t的序列总共有多少种，把他们按从左到右的顺序输出来。

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 <cstdio>
 # include <cmath>
 # include <iostream>
 # include <cstring>
 # include <algorithm>
 using namespace std ;

 int a[] ;
 int save[] ;
 int  n , t ;
 bool flag ;

 bool cmp(const int &x , const int &y)
 {
     return x > y ;
 }

 void dfs(int count , int L , int pos)
 {
     if (L > t)
         return  ;
     if (L == t)
     {
         for(int j=;j<count-;j++)
         {
             printf("%d+",save[j]);
         }
         printf("%d\n",save[count-]);
         flag =  ;
         return  ;
     }
     for (int i = pos ; i < n ; i++)
     {
         save[count] = a[i] ;
         dfs (count+,L+a[i] , i+) ;
         while (a[i] == a[i+])
             i++ ;
     }

 }

 int main()
 {
     //freopen("in.txt","r",stdin) ;

     while (scanf("%d %d" , &t , &n) != EOF)
     {
         if (t ==  && n == )
             break ;
         int i ;
         for (i =  ; i < n ; i++)
         {
             scanf("%d" , &a[i]) ;
         }
         sort(a,a+n,cmp) ;
         flag =  ;
         printf("Sums of %d:\n",t);
         dfs(,,) ;
         if (flag == )
         {
             printf("NONE\n") ;
         }
     }

     return  ;

 }