杭电1016Prime Ring Problem

最新推荐文章于 2024-07-20 14:45:00 发布

bluedream1219

最新推荐文章于 2024-07-20 14:45:00 发布

阅读量481

点赞数

CC 4.0 BY-SA版权

分类专栏：杭电题目文章标签：杭电 Prime Ring Problem10

本文链接：https://blog.youkuaiyun.com/bluedream1219/article/details/26625061

杭电题目专栏收录该内容

25 篇文章

订阅专栏

本文详细介绍了如何通过递归算法解决PrimeRingProblem，即在一个环形排列中，每个相邻的圆圈内填充从1到n的自然数，使得每个相邻数之和均为质数。通过实例演示了输入输出格式，并提供了关键代码片段。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25319 Accepted Submission(s): 11301

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

此题要用到递归

代码：

#include<stdio.h>
//此函数为判断素数
int prime(int m)
{
    int n = 0 ;
    for(int i = 2 ; i < m ; i++)
    {
        if(m % i == 0 )
            n++;
    }
    if( n==0 )
        return 1;
    else
        return 0;
}
//递归
int a[100];
int flag[100] = {0};
void DFS(int x,int n)
{
    int i = 0 , j = 0 , m = 0 ;
    for(i = 2 ; i <= n ; i++)
    {
        int m = 0 ;
        m = prime(a[x-1]+i);
        if(m == 0 || flag[i])//判断当两个数相加不是素数并且这个数被标记
           continue;
        a[x] = i ;
        flag[i] = 1;
        DFS(x+1,n);
        flag[i] = 0 ;
    }
    if(prime(a[x - 1]+1) == 1 && x == n)
    {
        for( i = 0 ; i < n-1 ; i++)
           printf("%d ", a[i]);
        printf("%d\n",a[n-1]);
        return ;
    }
}
int main()
{
    int n = 0 , b = 1 ;
    while(~scanf("%d",&n))
    {
        a[0] = 1 ;
        int x = 1 ;
        printf("Case %d:\n",b);
        DFS(x,n);
        printf("\n");
        b++;
    }
return 0;
}