Problem Description
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

題意：輸入 正整數n，把整數1到n組成一個環，使得相鄰兩個整數之和均為素數，輸出時從整數1開始逆時針排列，n<=16。

思路：使用dfs，每次填數時判斷是否符合條件即可。

程式碼為：

#include<iostream>
#include<cstdio>
#include<string.h>
using  namespace std;
int n,q;
int a[20],b[32]={0,1,0,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1},c[20];
void dfs(int cur)
{
    if(cur==n&&b[a[0]+a[cur-1]])
    {
        cout<<a[0];    //符合輸出條件
 

        for(int i=1;i<n;i++)
        {
            cout<<" "<<a[i];
        }
        cout<<endl;
    }
    else for(int i=2;i<=n;i++)
    {
        if(!c[i]&&b[i+a[cur-1]])
        {
            a[cur]=i;
            c[i]=1;
            dfs(cur+1);
            a[cur]=0;
            c[i]=0;
        }
    }
}
int main()
{
    q=1;
    while(cin>>n)
    {
        if(q>1)     //控制空行的輸出，必須符合，否則錯誤。
            cout<<endl;
        memset(a,0,sizeof(a));
        memset(c,0,sizeof(c));
        a[0]=1;
        cout<<"Case "<<q++<<":"<<endl;
        dfs(1);
    }
    return 0;
}