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
思路：
dfs演算法
程式碼：

#include<iostream>
#include<cstdio>
#include
 
<cstring>
#include<cmath>
using namespace std;
int cir[21];
int book[21]={0};
int prime(int num){
    int f=1;
    for(int i=2;i<=sqrt(num);i++){
        if(num%i==0){
            f=0;
            break;
        }
    }
    return f;
}//判別素數
void dfs(int n,int s){
    int f=0;
    if(s==n+1
 
&&prime(cir[s-1]+cir[1])==1){
        for(int i=1;i<n;i++){
            cout<<cir[i]<<" ";
        }
        cout<<cir[n]<<endl;
        return;
    }
    for(int i=2;i<=n;i++){
        f=prime(cir[s-1]+i);
        if(book[i]==0&&f==1){
            book[i]=1;
            cir[s]=i;
            dfs(n,s+1);
            book[i]=0;
        } 
    }
}
int main(void){
    int n,s=0;
    while(~scanf("%d",&n)){
        s++;
        memset(cir,0,sizeof(cir));
        cir[1]=1;
        if(n==1){
            printf("Case %d:\n",s);
            cout<<n<<endl;
            cout<<endl;
        }
        else{
            printf("Case %d:\n",s);
            dfs(n,2);
            cout<<endl;
        }
    }
    return 0;
}