ACM 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): 8651    Accepted Submission(s): 3877

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)

 

Recommend

JGShining

 

 

 

深搜，简单题：

看代码，还可以优化，尤其是判断素数的时候，可以用筛选法，或者直接列出40内的素数：

#include<stdio.h>
#include<cmath>
#include<string.h>
int a[21];
int used[21];
int n;
bool prime(int n)
{
    int m=(int)sqrt(n);
    int i;
    for(i=2;i<=m;i++)
      if(n%i==0)break;
    if(i>m)return true;
    else return false;
}   
void dfs(int k)
{
    if(k==n+1)
    {
        if(prime(a[1]+a[n]))
        {
            for(int i=1;i<n;i++)printf("%d ",a[i]);
            printf("%d\n",a[n]);
        }    
    } 
    else
    {
        for(int i=1;i<=n;i++)
        {
            if(used[i]==0&&prime(i+a[k-1]))
            {
                used[i]=1;
                a[k]=i;
                dfs(k+1);
                used[i]=0;
            }    

        }    
    }       
}     
int main()
{
    int i=1;
    while(scanf("%d",&n)!=EOF)
    {
        memset(used,0,sizeof(used));
        a[1]=1;used[1]=1;
        printf("Case %d:\n",i++);
        dfs(2);
        printf("\n");
        
    }    
    return 0;
}