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 (1 < n < 17).

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

提示：

用printf打印输出。

#include<iostream>

#include<stdio.h>

#include<queue>

using namespace std;

int ans[];

bool mark[];

bool check(int a){
    
    if ((a%2 == 0||a%3==0||a%5==0)&&a!=2&&a!=3&&a!=5)//因为输入的n<17，所以相邻数字加和小于32，合数的因子就235 
    return false;
    return true;
}

int n;

void dfs(int cnt){

    if (cnt>  )

    if (check(ans[cnt-]+ans[cnt-])==false)

    return;

    if (cnt==n)

    if (check(ans[cnt-]+ans[])==false)

    return;

    if (cnt == n){

        printf("");

        for (int i=;i<n;i++){

            printf(" %d",ans[i]);

        }

        printf("\n");

    }

    for (int i=;i<=n;i++){

        if (mark[i]==false){

            mark[i]=true;

            ans[cnt]=i;

            //cnt++;

            dfs(cnt+);

            mark[i]=false;

        }

    }

}

int main (){

    int c=;

    while (cin>>n){

        c++;

        for (int i=;i<=n;i++)

        mark[i]=false;

        ans[]=;

        mark[]=true;

        printf("Case %d:\n",c);

        dfs();

        printf("\n");

    }

    return ;

}