杭电 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 25187    Accepted Submission(s): 11246

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)

 

注意边界  还有首尾推断是否符合条件   

#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#define M 25
using namespace std;
int vis[M],n,ans[M],t=0;
int sum(int a,int b)//判定素数的函数
{
     int k=0,sum,i;
    sum=a+b;
        for(i=2;i<sum;i++)
            {
                if(sum%i==0)
                    {
                        k=1;
                     return 0;
                    }
            }
            if(!k)
                 return 1;
}

void dfs(int x)
{
    int i;

    if(x==n&&sum(ans[0],ans[n-1]))//递归边界   判定首尾是否符合条件
         {

             for(i=0;i<n;i++)
               {

                if(i==0)
                    cout<<ans[0];
                else
                   cout<<" "<<ans[i];

               }
               cout<<endl;
         }
    else
      {
        for(i=2;i<=n;i++)
           {
                if(!vis[i]&&sum(i,ans[x-1]))
                  {
                       {
                           ans[x]=i;
                           vis[i]=1;//标记用过的数
                           dfs(x+1);
                           vis[i]=0;//还原用过的数
                       }
                 }

           }
      }
}
int main()
{
    int i,j;
    while(cin>>n)
    {

        t++;
        memset(vis,0,sizeof(vis));
        cout<<"Case "<<t<<":"<<endl;
            vis[1]=1;ans[0]=1;
            dfs(1);
            cout<<endl;//每一个例子之间有一个空行
    }

    return 0;
}