Prime Ring Problem

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

题意：给你一个n，得到一个数组，1~n；让你写出一个素数环，要求相邻两个数（顺逆时针）和是素数，输出的时候第一个永远是一；

解题思路：按照递归求全排列的思想，从第2个开始递归一直到最后一位，并且首尾也是素数才算是搜索完成；

感悟：一开始我还以为得剪纸，但是只有20个数，一次就过了；

代码：

#include
#include
#include
#include
#define maxn 25
using namespace std;
int ans[maxn],visit[maxn],n;
int Prime(int a)
{
    for(int i=2;i<=sqrt(a);i++)
        if(a%i==0)
            return 0;
    return 1;
}

void dfs(int cur)
{
    if(cur==n&&Prime(ans[0]+ans[n-1]))//递归到最后一位，并且首尾也能是素数
    {
        for(int i=0;i
            printf("%d ",ans[i]);
        printf("%d\n",ans[n-1]);
    }
    else
    {
        for(int i=2;i<=n;i++)
        {
            if(!visit[i]&&Prime(i+ans[cur-1]))//这个数没用过并且相邻的是素数
            {
                ans[cur]=i;
                visit[i]=1;
                dfs(cur+1);
                visit[i]=0;
            }
        }
    }
}
int main()
{
    //freopen("in.txt", "r", stdin);
    int s=1;
    memset(visit,0,sizeof(visit));
    while(~scanf("%d",&n)&&n)
    {
        for(int i=0;i
            ans[i]=i+1;
        printf("Case %d:\n",s++);
        dfs(1);
        printf("\n");
    }
}