Prime Ring Problem

最新推荐文章于 2019-11-23 11:20:08 发布

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本文介绍了一个关于环形排列的问题，要求将1到n的自然数填入由n个圆圈组成的环中，使得相邻两个圆圈内的数字之和为素数。文中提供了一段C++代码实现该问题的解决方案，采用深度优先搜索算法完成这一过程。

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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 <cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
int a[25];
bool vis[25];

int n;
bool prime(int k){
	int i;
	for(i =2 ; i * i < k+1 ; i++){
		if(!(k%i))
			return false;
	}
	return true;
}

void dfs(int cur)//深度搜索；
{
	if(cur == n && prime(a[n-1]+a[0])){
		for(int i = 0 ; i < n; i++)
			printf("%d ",a[i]);
		printf("\n");
	}
	for(int j = 2  ;j <= n ; j++){
		if(!vis[j]){
			if(prime(a[cur-1]+j)){
				vis[j] = true;
				a[cur] =j;
				dfs(cur+1);
				vis[j] = false;
			}
		}
	}
}

int sum = 1;
int main(){
	while(scanf("%d",&n)!=EOF){
		a[0] = 1;
		printf("Case %d:\n",sum++);
		dfs(1);
		printf("\n");
	}
	return 0;
}