hdu1016.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

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <cstring>using namespace std;int n;int a[20],vis[20];bool isprime(int n)//判断素数{    if(n<2)return false;    for(int i=2;i*i<=n;i++)    {        if(n%i==0)        {            return false;        }    }    return true;}void dfs(int cur)//cur表示个数{    if(cur==n&&isprime(a[1]+a[n]))//记得判断最后一个和第一个的和是不是素数    {        for(int i=1;i<n;i++)        {            printf("%d ",a[i]);        }        printf("%d\n",a[n]);    }    else    {        for(int i=2;i<=n;i++)        {            if(!vis[i]&&isprime(i+a[cur]))            {                a[cur+1]=i;                vis[i]=1;//标记                dfs(cur+1);                vis[i]=0; //搜索完清除标记            }        }    }}int main(){    int cas=0;    while(scanf("%d",&n)!=EOF)    {        memset(vis,0,sizeof(vis));        a[1]=1;        if(cas++)        {            printf("\n");        }        printf("Case %d:\n",cas);        dfs(1);    }    return 0;}