Prime Ring Problem

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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

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

思路：典型的深搜题，思想看代码

#include<stdio.h>#include<string.h>using namespace std;int a[20],vis[20],n;int prime(int m){    int i;    for(i=2;i*i<=m;i++)    if(m%i==0) return 0;    return 1;}//判断素数函数int dsf(int step){    int i;    if(step==n+1&&prime(a[n]+a[1]))//递归结束的条件，不要忘记判断最后一个数和第一个数的和是否是素数    {        for(i=1;i<n;i++)        printf("%d ",a[i]);        printf("%d\n",a[n]);        return 0;    }    for(i=2;i<=n;i++)//题目要求从一开始，所以这里从二开始循环    {        if(!vis[i]&&prime(i+a[step-1]))//判断当前的数是否被用过，并且与上一个数的和是素数        {            a[step]=i;            vis[i]=1;//每用一次就标记上            dsf(step+1);调用递归函数            vis[i]=0;回溯的时候清零，以便下次用        }    }}int main(){    int k=1;    a[1]=1;    while(scanf("%d",&n)!=EOF)    {        memset(vis,0,sizeof(vis));标记数组清零        printf("Case %d:\n",k++);        dsf(2);从二开始        printf("\n");    }    return 0;}

虽然题很水，但还是做了很长时间，这就是菜鸟的世界，你不懂。。。。。。

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