hdu 1016 Prime Ring Problem

Prime Ring Problem

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

Total Submission(s): 14880 Accepted Submission(s): 6787

[/b]

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

#include<iostream>
 #include<mem.h>
 #include<stdio.h>
 using namespace std;
 bool b[20];
 int a[20];
 int n;
 bool prime(int p)
 {
     int i;
     if(p==3)
     {
         return true;
     }
     for(i=2;i<=p/2;i++)
     {
         if(p%i==0)return false;
         else if(i==p/2)
         {
             return true;
         }
     }
 }
 void DFS(int k)
 {
  if(k>n && prime(a
+a[1])==true)
  {
      for(int i=1; i<n; i++)
     printf("%d ",a[i]);
         printf("%d\n",a
);
         return;
  }
  for(int i=2; i<=n; i++)
  {
   if(b[i]==false&& prime(a[k-1]+i)==true)
   {
    b[i]=true;
    a[k]=i;
    DFS(k+1);
    b[i]=false;
   }
  }
 }
 int main()
 {
     int t=1;
  while(~scanf("%d",&n))
  {
   printf("Case %d:\n",t++);
   if(n%2==0)
   {
      memset(b,0,sizeof(b));
      memset(a,0,sizeof(a));
      a[1]=1;
      DFS(2);
   }
   printf("%s\n","");
  }
 }