HDU1016 素数环

最新推荐文章于 2020-07-04 22:25:58 发布

Fly_Eason

最新推荐文章于 2020-07-04 22:25:58 发布

阅读量327

点赞数

CC 4.0 BY-SA版权

分类专栏：回溯法文章标签： HDU

本文链接：https://blog.youkuaiyun.com/Fly_Eason/article/details/79712242

回溯法专栏收录该内容

3 篇文章

订阅专栏

本文介绍了一种基于回溯法的算法实现，旨在解决特定的排列组合问题：将1到n的自然数填入由n个圆圈组成的环中，使得任意相邻两个圆圈内的数字之和均为素数。通过递归深度优先搜索策略完成数字的有效放置，并最终输出所有可能的排列方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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:
4 3 2 5 6
6 5 2 3 4

Case 2:
2 3 8 5 6 7 4
2 5 8 3 4 7 6
4 7 6 5 8 3 2
6 7 4 3 8 5 2

import java.util.Scanner;

public class Main 
{
	static int n;
	static int arr[] = new int[16];
	static boolean vis[] = new boolean[16]; //记录是否使用该数字
	static int N=0;
	static boolean falg = true;
	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);
		while(cin.hasNextInt()){
			n = cin.nextInt();
			N++;
			arr = new int[16];
			for (int i = 2; i <= n; i++) {
				vis[i] = true;  //设置为每一个数都未使用过
			}
			arr[0] = 1;
			falg = true;
			DFS(1);
			System.out.println();
		}
	}
	private static void DFS(int cur) {
		
		
		if(cur==n && judge(arr[0]+arr[cur-1])){
			if(falg){
				System.out.println("Case "+N+":");
				falg = false;
			}
			
			for (int i = 0; i < n-1; i++) {
				System.out.print(arr[i]+" ");
			}
			System.out.print(arr[n-1]);
			System.out.println();
			
		}else{
			for (int i = 2; i <= n; i++) {
				if(vis[i] && judge(arr[cur-1]+i)){ //判断此时的数是否被使用了 并且判断该数字相加是否为素数
					//System.out.println(" cur"+cur+ " "+i);
					arr[cur] = i;
					vis[i] = false; //标记此时这个数已经被使用
					DFS(cur+1);
					vis[i] = true;  //取消标记 供下一种情况使用
				}
			}
		}
	}
	//判断此时相加的数是否为素数
	private static boolean judge(int num) {
		int i;
		for (i = 2; i <=num; i++) {
			if(num%i==0) break;
		}
		if(i>=num){
			return true;
		}else{
			return false;
		}
		
	}
	
}