翻译：n个数字（1,2,3...n）围成一个圈，要求相邻的两个数字之和是质数。题目要求根据给出的n，计算所有能够组成满足条件的圈的数字序列。

解题思路：

首先选择1，然后选择和1相加等于质数的数字，可能有很多种情况，例如（n=6的情况）：

1+21+41+6然后针对每种情况，再选择与第二个数的和为质数的数字，得到下面的序列

1+2+31+2+51+4+31+6+5直到所有的数字都选择完，把所有的组合列出即可。

 

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



#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
int a[50],f[50],ans=0,m,flag;
int sd[41];
int fun(int c)
{   int i;
    for( i=2;i<=sqrt(c);i++)
        if(c%i==0)return 0;
      if(i>sqrt(c))return 1;
}
void dfs(int c)
{

    a[1]=1;
    if(c>m&&sd[f[1]+f[m]])
    {
        if(!flag)
            {
                cout<<"Case "<<ans<<":"<<endl;
                flag=1;
            }
        for(int i=1;i<m;i++)
        {
            cout<<f[i]<<" ";
        }
        cout<<f[m]<<endl;
    }
    else
    {
        for(int i=2;i<=m;i++)
        {
            f[c]=i;
            if(!a[i]&&sd[(f[c-1]+f[c])])
            {
                a[i]=1;
             //   if(m%2==0)
                dfs(c+1);
                a[i]=0;
            }
        }
    }
}
int main()
{
    int n,a[100];
    while(cin>>m)
    {
        ans++;
        flag=0;
        memset(sd,0,sizeof(sd));
        for(int i=1;i<=40;i++)
            if(fun(i))sd[i]=1;
        memset(a,0,sizeof(a));
        f[1]=1;
        if(m%2==0)
        dfs(2);
        else
            cout<<"Case "<<ans<<":"<<endl;
        cout<<endl;
    }
    return 0;
}