Q-Prime Ring Problem

最新推荐文章于 2022-01-24 12:03:09 发布

TimEckel

最新推荐文章于 2022-01-24 12:03:09 发布

阅读量290

点赞数

CC 4.0 BY-SA版权

分类专栏：搜索

本文链接：https://blog.youkuaiyun.com/TimEckel/article/details/72591097

搜索专栏收录该内容

8 篇文章

订阅专栏

本文介绍了一种构建素数环的算法实现，该算法通过深度优先搜索（DFS）完成从1到n的数字排列，使得环上任意相邻两数之和为素数。使用预处理的素数表简化判断过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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. 这里写图片描述

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
我就感觉我初始化表的时候那里好像错了，但是ac了没找出来，题意给你一个数n，要构成一个素数环，这个素数由1-n组成，它的特征是选中环上的任意一个数字i，i与它相连的两个数加起来都分别为素数，满足就输出。使用素数打表的方式简化素数判定。

#include<stdio.h>  
#include<string.h>  
int prime[] = {0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0};  
int visited[21];    
int value[21];  
int n;  

void DFS(int step){  
    int i;   
    if(step == n+1){  
        if(prime[value[step - 1] + 1]){  
            for(i = 1;i <= n;i++){  
                printf("%d%c",value[i],i==n?'\n':' ');  
            }  
        }  
    }  
    else{  
        for(i = 2;i <= n;i++){   
            if(!visited[i] && prime[i + value[step-1]]){  
                visited[i] = 1;  
                value[step] = i;  
                DFS(step+1);  
                visited[i] = 0;  
            }  
        }  
    }  
}  

int main(){  
    int caseNum = 1;  
    while(scanf("%d",&n) != EOF){  
        memset(visited,0,n);  
        printf("Case %d:\n",caseNum);   
        value[1] = 1;  
        DFS(2);  
        printf("\n");  
        caseNum ++;  
    }  
    return 0;  
}