HDU 1016

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

附代码

#include <iostream>
#include <cstring>
using namespace std;
int n,a[25],book[25];
bool isPrime(int x)
{
	if(x <= 1)
		return 0;
	for(int i = 2;i*i<=x;i++)
		if(x % i == 0)
			return 0;
	return 1;
}
void dfs(int x)
{
	if(x == n + 1)
	{
		for(int i = 1;i<=n;i++)
		{
			if(isPrime(a[1] + a[n]))
			{
				if(i == n)
					cout<<a[n]<<endl;
				else
					cout<<a[i]<<" ";
			}
		}
		return;
	}
	else
	{
		for(int i = 2;i<=n;i++)
		{
			if(book[i] == 0)
			{
				if(isPrime(a[x-1] + i))
				{
					a[x] = i;
					book[i] = 1;
					dfs(x+1);
					book[i] = 0;
				}
				else
					continue;
			}
		}
	}
}
int main()
{
	int cnt = 1;
	while(cin >> n)
	{
		memset(a,0,sizeof(a));
		memset(book,0,sizeof(book));
		a[1] = 1;
		cout<<"Case "<<cnt++<<":"<<endl;
		dfs(2);
		cout<<endl;
	}
}