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.

n (0 < n < 20).

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.

6
8

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
要求按字典序输出素数环。。DFS爆搜，不用剪枝。。
一开始用 next_permutation()函数纯暴力判结果TLE了。。
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#define L long long
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=32;
bool pri[100],vis[100];
int a[35],n;
void init()
{
	memset(pri,1,sizeof(pri));
	pri[0]=0;pri[1]=0;
	for(int i=2;i<=100;i++)
	{
		if(pri[i])
		{
			for(int j=2;j*i<=100;j++)
				pri[j*i]=0;
		}
	}
}
void dfs(int cur)
{
	if(cur==n&&pri[a[n-1]+a[0]])
	{
		printf("%d",a[0]);
		for(int i=1;i<n;i++)
			printf(" %d",a[i]);
		puts("");
		return ;
	}
	else
	{
		for(int i=2;i<=n;i++)
		{
			if(!vis[i]&&pri[i+a[cur-1]])
			{
				a[cur]=i;
				vis[i]=1;
				dfs(cur+1);
				vis[i]=0;
			}
		}
	}
}
int main()
{
	init();
	int cas=1;
	a[0]=1;
	while(scanf("%d",&n)!=EOF)
	{
		memset(vis,0,sizeof(vis));
		printf("Case %d:\n",cas++);
		vis[1]=1;
		dfs(1);
		puts("");
	}
	return 0;
}