hdu 1016Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 47348    Accepted Submission(s): 20921

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

68

 

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2
#include<iostream>#include<string>#include<memory.h>#include<algorithm>using namespace std;int primer[40]={0,1,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0};int n,k=1,a[100],sum;int visited[100];void dfs(int m){if(m==n&&primer[a[m-1]+a[0]]==1)//找到最后一个数且首尾和为素数{   for(int i=0;i<n-1;i++)       cout<<a[i]<<" ";   cout<<a[n-1]<<endl;    }   else//没有找到最后一个    {        for(int i=2;i<=n;i++)//从2到n-1找{    if(visited[i]==0&&primer[i+a[m-1]]==1)//没有被访问过且和前一个的和为素数 {     visited[i]=1;//标记访问过a[m++]=i;//存入数据方便下次输出dfs(m);//递归找visited[i]=0;//返回2重新找 m--;//下次寻找 } }    }}int main() {while(cin>>n){   memset(visited,0,sizeof(visited));//标记状态  cout<<"Case "<<k<<":"<<endl;  k++;  a[0]=1;//数组保存一个素数环   dfs(1);//从1开始搜 （必须从1开始）   cout<<endl;}return 0;}