素数环（简单搜索）

 

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36338    Accepted Submission(s): 16024

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

 

 

其中n如果是除1的奇数的话就不存在素数环，在测试数据含有大量奇数时加以判断可节省时间

 

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int a[50],b[21],test[21];
int n;
void prime()
{
    int i,j;
    for(i=2;i<=50;i++)
    {
        for(j=2;j<=i/2;j++)
            if(i%j==0)
                break;
        if(j==(i/2+1))
            a[i]=1;
        else
            a[i]=0;
    }
}
void dfs(int s)
{
    int i,j,k;
    if(s==n+1&&a[b[n]+1])
    {
        for(i=1;i<n;i++)
            cout<<b[i]<<" ";
        cout<<b[n]<<endl;
    }
    else for(i=2;i<=n;i++)
        if(a[b[s-1]+i]&&!test[i])
        {
            b[s]=i;
            test[i]=1;
            dfs(s+1);
            test[i]=0;
        }
}
int main()
{
    b[1]=1;
    prime();
    int i,j;
    while(cin>>n&&n)
    {
        cout<<"Case "<<j<<":"<<endl;
        memset(test,0,sizeof(test));
        dfs(2);
        cout<<endl;
        j++;
    }
}