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): 41136    Accepted Submission(s): 18209

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>using namespace std;int n;int initial[21];int now[21];int isPrime(int x){  int prime_num[12] = {2,3,5,7,11,13,17,19,23,29,31,37};  for(int i = 0; i < 12;i++)    if(x==prime_num[i])  return 1;  return 0;}int dfs(int preNum,int findNum,int find){  if(!isPrime(preNum+findNum)){    return 0;  }  now[find] = findNum;  if(find==n && isPrime(findNum+1)){    for(int i = 1; i < n;i++)      cout << now[i] << " ";    cout << now[n] << endl;  }  initial[findNum] = 0;  int i;  for(i = 2;i <= n; i++){if(initial[i]!=0 && dfs(findNum,i,find+1))  break;  }  initial[findNum] = 1;  return 0;}int main(){  int count = 1;  while(cin >> n){cout << "Case " << count++ << ":" << endl;int i;for(i = 1;i <= n; i++){  initial[i]=i;}now[1] = 1;    if(n == 1){  cout << now[1] << endl;  continue;}for(i = 2;i <= n; i++){  dfs(1,i,2);}cout << endl;  }  return 0;}