Prime Ring Problem

题目链接
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

#include<iostream>#include<cstring>#include<cstdio>#include<cmath>using namespace std;int n, dis[20];bool visited[20];bool is_Prime(int a){    for(int i = 2; i <= sqrt(a); i++)        if(a % i == 0)            return false;    return true;}void dfs(int a){    if(a == n && is_Prime(1+dis[n-1])){//如果到达结尾,并且最后一个数和第一个数1能够构成素数        for(int i = 0; i < n; i++){            if(!i)                cout<<dis[i];            else                cout<<" "<<dis[i];        }        cout<<endl;        return;    }    else        for(int i = 2; i <= n; i++)            if(!visited[i] && is_Prime(i+dis[a-1])){                visited[i] = true;                dis[a] = i;                dfs(a + 1);                visited[i] = false;            }}int main(){    int sign = 1;    while(cin>>n){        cout<<"Case "<<sign++<<":"<<endl;        if(n == 1){//一个数字也可以构成环            cout<<"1"<<endl;            continue;        }        if(n&1){//&1的结果就是取二进制的最末位t,t==1是奇数,t==0为偶数            /*                奇数不行的原因是:                    1.奇数+奇数 = 偶数(不是质数)                    2.奇数+偶数 = 奇数(不能确定是否是质数)                    3.偶数+偶数 = 偶数(不是质数)                ps:如果n是奇数的话,定会导致奇数的个数比偶数的个数多一个                   换句话说,只有奇数和偶数一一对应的时候,才有可能出现质数,即奇数的个数==偶数的个数            */            cout<<"No Answer"<<endl;            continue;        }        dis[0] = 1;        memset(visited,false, sizeof(visited));        dfs(1);        cout<<endl;    }    return 0;}