Prime Ring Problem

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

68

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

题意：按字典序输出素数环，素数环就是两两相邻的两个数相加 和为素数

#include<cstdio>#include<cmath>#include<cstring>int vis[20];int num[20];int n;int sushu(int x)//判断是否为素数{    int i,k=sqrt(x);    for(i=2;i<=k;i++)    {        if(x%i==0)            break;    }    if(i>k&&x!=1)//1不是素数        return 1;    return 0;}void dfs(int index){    if(index==n)    {        if(sushu(num[index]+num[1]))//输出答案        {            for(int i=1;i<n;i++)                printf("%d ",num[i]);            printf("%d\n",num[n]);        }    }    for(int i=2;i<=n;i++)    {        if(!vis[i]&&sushu(num[index]+i))        {            vis[i]=1;            num[index+1]=i;//注意这里自加用index+1;            dfs(index+1);            vis[i]=0;//回溯回去的时候把标记清除掉        }    }}int main(){    int Case=1;    while(~scanf("%d",&n))    {        memset(vis,0,sizeof(vis));        printf("Case %d:\n",Case++);        if(n%2&&n!=1)//n为奇数是不可能存在素数环的,所以可以直接跳过,加速;        {            printf("\n");            continue;        }        num[1]=1;        dfs(1);        printf("\n");    }    return 0;}