hdu 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28064    Accepted Submission(s): 12487

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

 

Source

Asia 1996, Shanghai (Mainland China)

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
using namespace std;
int prime[40],vis[40];
int a[40],n;

int dfs(int x)
{
    if(x==n && !prime[a[n]+a[1]])
    {
        for(int i=1;i<=n;i++)
        {
            if(i==1)
            printf("%d",a[i]);
            else
            printf(" %d",a[i]);
        }
        printf("
");
    }
    else
    {
        for(int i=2;i<=n;i++)
        {
            if(!vis[i] && !prime[i+a[x]])
            {
                vis[i]=1;
                a[x+1]=i;
                dfs(x+1);
                vis[i]=0;
            }
        }
    }
}

int main()
{
    int k=0,j;
    while(cin >> n)
    {
        memset(prime,0,sizeof(prime));
        memset(vis,0,sizeof(vis));
        k++;

        prime[1]=1;
        for(int i=2;i<=n*2;i++)
        {
            if(!prime[i])
            for(j=i+i;j<=n*2;j+=i)
            {
                prime[j]=1;
            }
        }
        printf("Case %d:
",k);
        a[1]=1;
        dfs(1);
        printf("
");
    }
    return 0;
}