Prime Ring Problem

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

题解：数据量小，深搜。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;

bool visited[50];
bool prim[100];
int pre[100];

void prime()
{
	memset(prim,true,sizeof(prim));
	prim[2] = false;
	for(int i = 2;i <= 99;i++)
	{
		for(int j = 2 * i;j < 100;j += i)
		{
			prim[j] = false;
		}
	}
}

void print(int x)
{
	if(pre[x] == x)
	{
		printf("1");
		return;
	}
	print(pre[x]);
	printf(" %d",x);
}

void dfs(int x,int t,int n)
{
	if(t == n - 1)        //找到最后一个需要判断前驱和还有和1的和是不是素数
	{
		if(prim[x + pre[x]] && prim[x + 1])
		{
			print(x);
			printf("\n");
		}
		return;
	}
	visited[x] = true; //表示该数目前可行
	for(int i = 2;i <= n;i++)
	{
		if(!visited[i] && prim[i + x]) //前面没有出现过并且可行
		{
			pre[i] = x;
			dfs(i,t + 1,n);
		}
	}
	visited[x] = false;    //该数字遍历完了，标记为没访问过
}

int main()
{
	int n;
	prime();
	int t = 1;
	memset(visited,false,sizeof(visited));
	while(scanf("%d",&n) != EOF)
	{
		printf("Case %d:\n",t++);
		for(int i = 1;i <= 60;i++)
		{
			pre[i] = i;
		}
		dfs(1,0,n);
		printf("\n");
	}
	
	return 0;
}