hdu 1258 从n个数中找和为t的组合

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

 1 # include <cstdio>
 2 # include <cmath>
 3 # include <iostream>
 4 # include <cstring>
 5 # include <algorithm>
 6 using namespace std ;
 7 int t,n,flag;
 8 int num[14],save[14];
 9
10 bool cmp(int x , int y)
11 {
12     return x > y ;
13 }
14
15 void dfs(int i,int sum,int count){
16     if(sum>t)
17         return ;
18     if(sum==t){
19         for(int j=0;j<count-1;j++){
20             printf("%d+",save[j]);
21         }
22         printf("%d\n",save[count-1]);
23         flag=1;
24         return;
25     }
26     int tag=-1;
27     for(int k=i;k<n;k++){
28         if(num[k]!=tag){
29             save[count]=num[k];
30             sum+=num[k]; //每次的和
31             tag=num[k];  //保留当前的数，能避免重复  假如是 4 2 2  前一个2已经被排除 则不用考虑后一个2
32             dfs(k+1,sum,count+1);
33             sum-=num[k];
34         }
35     }
36 }
37
38 int main(){
39    // freopen("in.txt","r",stdin) ;
40     while(scanf("%d%d",&t,&n)!=EOF){
41         if(n==0)break;
42         for(int i=0;i<n;i++){
43             scanf("%d",&num[i]);
44         }
45         flag=0;
46         sort(num,num+n,cmp) ;
47         printf("Sums of %d:\n",t);
48         dfs(0,0,0);
49         if(!flag){
50             printf("NONE\n");
51         }
52
53     }
54     return 0;
55 }