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题意:
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
注意每个数只能用一次,且不能出现重复的等式
题解 :dfs深搜
1 #include<stdio.h> 2 int a[20],ans[20]; 3 int sum,n,flag; 4 void dfs(int x,int count,int m) 5 { 6 int last,i; 7 if(m>sum) 8 return; 9 if(m==sum) 10 { 11 flag=1; 12 for(i=1;i<count;i++) 13 { 14 if(i==count-1) 15 printf("%d\n",ans[i]); 16 else 17 printf("%d+",ans[i]); 18 } 19 } 20 last=-1; 21 for(i=x;i<=n;i++) 22 { 23 if(last!=a[i]) 24 { 25 ans[count]=a[i]; 26 last=a[i]; 27 dfs(i+1,count+1,m+a[i]); 28 } 29 } 30 return; 31 } 32 33 int main() 34 { 35 int i; 36 while(~scanf("%d %d",&sum,&n)) 37 { 38 if(sum==0&&n==0) 39 break; 40 for(i=1;i<=n;i++) 41 scanf("%d",&a[i]); 42 flag=0; 43 printf("Sums of %d:\n",sum); 44 dfs(1,1,0); 45 if(!flag) 46 puts("NONE"); 47 } 48 return 0; 49 }
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原文地址:http://www.cnblogs.com/awsent/p/4269503.html