HDU-1016 Prime Ring Problem【素数环 打表+深搜】

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 55184    Accepted Submission(s): 24421

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
  
  

 把1-n的数放在一个两两相邻的环中 要求相邻两个相加是素数 首先将1-40的素数打表 后深度搜索所有情况输出

#include<bits/stdc++.h>
using namespace std;
int a[20],vis[20],isprime[45],n;
void get_prime()
{
    memset(isprime,0,sizeof(isprime));
    for(int i=2;i<8;i++)
    {
        if(!isprime[i])
            for(int j=i*i;j<45;j+=i)
        {
            isprime[j]=1;
        }
    }
}//素数打表，0是素数
int dfs(int step)
{
    if(step==n+1&&!isprime[a[n]+a[1]])
    {
        for(int i=1;i<n;i++)
        {
            printf("%d ",a[i]);
        }
        printf("%d\n",a[n]);
        return 0;
    }
    for(int i=2;i<=n;i++)
    {
        if(!vis[i]&&!isprime[i+a[step-1]])
        {
            a[step]=i;
            vis[i]=1;
            dfs(step+1);
            vis[i]=0;
        }
    }
}
int main()
{
    int k=1;
    a[1]=1;
    get_prime();
    while(scanf("%d",&n)!=EOF)
    {
        memset(vis,0,sizeof(vis));
        printf("Case %d:\n",k++);
        dfs(2);
        printf("\n");
    }
}