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

 

简单DFS，不多说什么了

 

#include <stdio.h>
#include <string.h>
int n,prime[50],a[50],vis[50];

void isprime()
{
    int i,j;
    for(i = 0;i<50;i++)
    prime[i] = 1;
    prime[0] = prime[1] = 0;
    for(i = 2;i<50;i++)
    {
        if(prime[i])
        for(j = i+i;j<50;j+=i)
        prime[j] = 0;
    }
}

void dfs(int step)
{
    int i,j;
    if(step == n+1 && prime[a[n]+a[1]])//结束条件
    {
        for(i = 1;i<n;i++)
        printf("%d ",a[i]);
        printf("%d\n",a[n]);
        return ;
    }
    for(i = 2;i<=n;i++)
    {
        if(!vis[i] && prime[i+a[step-1]])//此数未用并且与上一个放到环中的数相加是素数
        {
            a[step] = i;
            vis[i] = 1;
            dfs(step+1);
            vis[i] = 0;
        }
    }
}

int main()
{
    int cas = 1;
    a[1] = 1;
    isprime();
    while(~scanf("%d",&n))
    {
        memset(vis,0,sizeof(vis));
        printf("Case %d:\n",cas++);
        dfs(2);
        printf("\n");
    }

    return 0;
}