HDU - 1016 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
问题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1016
问题简述：要求输出从1到n的之间相邻的数相加为素数的序列。（第一个和最后一个也要满足要求）
问题分析：判断素数+dfs
AC通过的C++语言程序如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include <algorithm>
#include<string.h>
#include<stdio.h>
using namespace std;
const int maxn = 100005;
int num[maxn];
bool vis[maxn];
int n;
bool prime(int x)
{
	for (int i = 2; i*i <= x; i++)
	{
		if (x%i == 0) return 0;
	}
	return 1;
}
void dfs(int x)
{
	if (x == n && prime(num[n - 1] + 1))
	{
		cout << 1;
		for (int i = 1; i < n; i++)
		{
			cout << " " << num[i];
		}
		cout << endl;
	}
	if (x == n) return;
	for (int i = 2; i <= n; i++)
	{
		if (!vis[i] && prime(num[x - 1] + i))
		{
			vis[i] = 1;
			num[x] = i;
			dfs(x + 1);
			vis[i] = 0;
		}
	}
}
int main()
{
	int z = 1;
	while (cin >> n)
	{
		memset(num, 0, maxn);
		memset(vis,0, maxn);
		vis[1] = 1;
		num[0] = 1;
		cout << "Case " << z << ":" << endl;
		dfs(1);
		cout << endl;
		z++;
	}
	return 0;
}