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): 17618    Accepted Submission(s): 7937

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
 
生成素数环，用dfs
 
AC代码：
#include <iostream>#include <cstring>#include <cmath>using namespace std;int vis[25],a[25],vis_p[60],n;void dfs(int cur){    if(cur==n&&!vis_p[a[0]+a[n-1]])    {      for(int i=0;i<n-1;i++)      cout<<a[i]<<" ";            //最后一个数后面不能输出空格      cout<<a[n-1];      cout<<endl;    }    else    for(int i=2;i<=n;i++)    {        if(!vis[i]&&!vis_p[i+a[cur-1]])        {            a[cur]=i;            vis[i]=1;            dfs(cur+1);            vis[i]=0;        }    }}int main(){    int m = sqrt(50);    memset(vis_p,0,sizeof(vis_p));     for(int i=2;i<m;i++)     if(!vis_p[i])      for(int j=i*i;j<=50;j+=i) vis_p[j]=1;     int c=0;     a[0]=1;     while(cin>>n)     {         memset(vis,0,sizeof(vis));         c++;         cout<<"Case "<<c<<":"<<endl;         dfs(1);         cout<<endl;     }     return 0;}