Vjudge Prime Ring Problem (dfs)

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. 

 

Input

n (0 < n < 20). 

Output

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. 

Sample Input

68

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

AC代码：

#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>using namespace std;int path[20+2];int vis[20+2];int n;bool isPrime(int t){for(int i = 2; i <= sqrt(t);i++){if(t % i == 0) return false;}return true;}void dfs(int cnt){if(cnt - 1 == n) {if(isPrime(path[1] + path[n])){printf("%d",path[1]);for(int i = 2; i <= n; i++)printf(" %d",path[i]);printf("\n");}    return ;}for(int j = 2; j <= n; j++){if(!vis[j] && isPrime(path[cnt-1] + j)){vis[j] = 1;path[cnt] = j;dfs(cnt + 1);vis[j] = 0;}}}int main(){int count = 0;while(scanf("%d",&n)==1){printf("Case %d:\n",++count);memset(vis,0,sizeof(vis));path[1] = 1;vis[1] = 1;for(int i = 2; i <= n; i++){if(isPrime(path[1]+i)){path[2] = i;vis[i] = 1;dfs(3);vis[i] = 0;}}printf("\n");}return 0;}