Prime Ring Problem

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#include<stdio.h>#include<string.h>int prime[40]={0,1,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0},n;//素数打表，因为n最大是20，所以只要打到40int visited[21],a[21];void dfs(int num)//深搜{   int i;   if(num==n&&prime[a[num-1]+a[0]])  //满足条件了，就输出来   {       for(i=0;i<num-1;i++)           printf("%d ",a[i]);       printf("%d\n",a[num-1]);   }   else   {       for(i=2;i<=n;i++)       {           if(visited[i]==0)//是否用过了           {               if(prime[i+a[num-1]]) //是否和相邻的加起来是素数               {                   visited[i]=-1;//标记了                   a[num++]=i;//放进数组                   dfs(num); //递归调用                   visited[i]=0; //退去标记                   num--;               }           }       }   }}    int main(){      int num=0;      while(scanf("%d",&n)!=EOF)      {         num++;         printf("Case %d:\n",num);         memset(visited,0,sizeof(visited));         a[0]=1;         dfs(1);         printf("\n");      }      return 0;}

时间限制: 2000ms内存限制: 32768KBHDU ID: 1016
64位整型: Java 类名:
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题目描述
A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.

Note: the number of first circle should always be 1.

输入
n (0 < n < 20).
输出
The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order.

You are to write a program that completes above process.

Print a blank line after each case.
样例输入
6
8
样例输出
Case 1:
1 4 3 2 5 6
1 6 5 2 3 4

Case 2:
1 2 3 8 5 6 7 4
1 2 5 8 3 4 7 6
1 4 7 6 5 8 3 2
1 6 7 4 3 8 5 2
来源
Asia 1996, Shanghai (Mainland China)

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