特殊数
特殊数
| Time Limit: 1000MS | Memory Limit: 65535KB |
| Submissions: 83 | Accepted: 29 |
Sample Input
1
Sample Output
1
解析:
这也算是一道水题,就用简单的深搜即可得到结果,见如下代码:
# include<stdio.h> # include<string.h> int vis[12]; int n,leap; void DFS(int index,int num)//index用来记录现在的位数,num用来记录index-1位的值 { int i; if(leap)return; if(index==n+1) { printf("%d\n",num); leap=1; return; } for(i=1;i<=9;i++) { if(leap)break; if(vis[i]==0&&(10*num+i)%index==0) { vis[i]=1; DFS(index+1,10*num+i); vis[i]=0; } } } int main() { scanf("%d",&n); leap=0; memset(vis,0,sizeof(vis)); DFS(1,0); return 0; }
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