Program C 暴力求解

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

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int is_prime(int n)
{
  for(int k= 2; k*k <= n; k++)
    if(n% k== 0) return 0;
  return 1;
}

int n, A[50], 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("\n");
  }
  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("\n");
        printf("Case %d:\n", ++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;
}