A - Prime Ring Problem （深度优先搜索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.

Inputn (0 < n < 20).
OutputThe 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
题大意是：输入正整数n，1—n组成一个环，相邻的两个整数为素数。输出时从整数1开始，逆时针排列。同一个环恰好输出一次，n (0 < n <20)
代码如下：

#include<iostream>
using namespace std;
int n;
int a[20],book[20];
//素数模板 
int sushu(int n)           
{
    if(n<2)  return false;
    for (int i=2;i*i<=n; i++)
    {
        if(n%i==0)
        return false;
    }
    return true;
}
void dfs(int step)
{
    //边界的处理 
    if(step==n&&sushu(a[1]+a[n]))  //递归边界。别忘了测试第一个数和最后一个数
    {
    {
        for(int i=1; i<n; i++)
            cout<<a[i]<<" ";
        cout<<a[n]<<endl;//空格的处理 
    }
    return ;//返回之前的一步(最近调用的地方) 
    }
        for(int i=2; i<=n; i++)
        {
            if(book[i]==0&&sushu(i+a[step]))   //如果i没有用过，并且与前一个数之和为素数
            {   //因为第一个已经存入,所以在这里a[step+1]
                a[step+1]=i;             //放入(该位置存入值) 
                book[i]=1;              //将值设为1，表示已不在手上 
                dfs(step+1);            //递归调用 
                book[i]=0;              //清除标记
            }
        }
        return;
}
int main()
{
    int t=0;
    while(cin>>n)
    {
        a[1]=1;          
        t++;
        cout<<"Case "<<t<<":"<<endl;
        dfs(1);
        cout<<endl;
    }
    return 0;
}