素数环

Description

A ring is composed of n (even number) circles as shown in diagram. Put natural numbers  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 <= 16)

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.

You are to write a program that completes above process.

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

 因为起点已经确定为1；

只要用dfs回溯就行；

同是要注意使用vis[]数组判重；

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int is_prime(int x) {
  for(int i = 2; i*i <= x; i++)
    if(x % i == 0) return 0;
  return 1;
}

int n, A[50];
bool isp[50],vis[50];
void dfs(int cur) {
  if(cur == n && isp[A[0]+A[n-1]]){
    for(int i = 0; i < n; i++){
      if(i != 0) printf(" ");
      printf("%d", A[i]);
    }
    printf("
");
  }
  else for(int i = 2; i <= n; i++)
    if(!vis[i] && isp[i+A[cur-1]]){
      A[cur] = i;
      vis[i] = 1;
      dfs(cur+1);
      vis[i] = 0;
    }
}

int main() {
  int kase = 0;
  while(scanf("%d", &n) == 1 && n > 0){
    if(kase > 0) printf("
");
    printf("Case %d:
", ++kase);
    for(int i = 2; i <= n*2; i++) isp[i] = is_prime(i);
    memset(vis, 0, sizeof(vis));
    A[0] = 1;
    dfs(1);
  }
  return 0;
}