Prime Ring Problem

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

6
8

Sample Output

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

析：

题目大衣：一个由n个数组成的圆圈，相邻的两个数的总和为素数，最后一个与第一个相加为素数。且第一个数为1.
利用回溯法，见紫皮书194页例题

代码

#include<iostream>
#include<string.h>
#include<math.h>
using namespace std;
int a[20],vis[20];
int n;
int isp(int x) //判断是否为素数
{
    int i;
    if(x<2)
        return 0;
    for(i=2;i<=sqrt(x);i++)
    {
        if(x%i==0)
            return 0;
    }
    return 1;
}
void dfs(int s)//深搜
{
    int i;
    if(s==n&&isp(a[0]+a[n-1]))// 递归边界，测试第一个数和最后一个数
    {
        for(i=0;i<n-1;i++)
            cout<<a[i]<<" ";
        cout<<a[n-1]<<endl;
    }
    else
    {
        for(i=2;i<=n;i++)
        {

            if(!vis[i]&&isp(i+a[s-1]))//如果i没有用过，并且与前一个数之和为素数
            {
                a[s]=i;
                vis[i]=1;//设置使用标记
                dfs(s+1);
                vis[i]=0;//清除标记
            }
        }
    }
}
int main()
{
    int t=1;
    while(cin>>n)
    {
        memset(vis,0,sizeof(vis));
        a[0]=1;
        cout<<"Case "<<t++<<":"<<endl;
        dfs(1);
        cout<<endl;
    }
    return 0;
}