HDU 1016 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

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<string.h>
#define ll long long
using namespace std;
char a[10005];
int prime[100];
int vist[100];
int b[100];
int n;
void init()
{
    memset(prime,0,sizeof(prime));
    prime[1]=1;
    for(int i=2;i<=10;i++)
    {
        if(!prime[i])
        {
            for(int j=i*i;j<=100;j+=i)
            {
                prime[j]=1;
            }
        }
    }
   /* for(int i=0;i<20;i++)
    {
        if(!prime[i])
            cout<<i<<" ";
    }*/
    memset(vist,0,sizeof(vist));
}
void dfs(int num)
{
    //printf(" %d",x);
    if(num==n&&!prime[b[num-1]+b[0]])
    {
        for(int i=0;i<=num-1;i++)
        {
            if(!i) printf("%d",b[i]);
            else printf(" %d",b[i]);
        }
        printf("\n");
    }
    else
    {
        for(int i=2;i<=n;i++)
        {
            if(!vist[i])
            {
                if(!prime[i+b[num-1]])
                {
                    vist[i]=1;
                    b[num++]=i;
                    dfs(num);
                    vist[i]=0;
                    num--;
                }
            }
        }
    }
}
int main()
{
    int num=0;
    while(~scanf("%d",&n))
    {
        num++;
        init();
        b[0]=1;
        printf("Case %d:\n",num);
        dfs(1);
        printf("\n");
    }
    return 0;
}