专题一搜索 H - Prime Ring Problem

1. 题目
   A ring is composed of n (even number) circles as shown in diagram.
   Put natural numbers 1, 2, . . . , n into each circle separately, and thesum 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 thering beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the aboverequirements.
   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
2. 思路
   题目不难，对于每个数字，搜一下与它相加为质数的数就可以了，用dfs寻找每个分支，可以先预处理一次质数表（我没这么干），当然要是不嫌麻烦也可以直接预处理一次1到16的邻接表
   本题逆天的地方在于离谱的格式要求，相当严格，每行数字最后不能有多余空格，每组数据后面要空一行，最后一组数据后面还不能有回车，它题目里面还没提示，以至于很多交这题的同学未通过数达到了惊人的-8-9-10-11，最多的甚至有-14
3. 代码
   

   #include<cstdio>
   #include<cmath>
   #include<stack>
   #include<cstring>
   #include<algorithm>
   using namespace std;
   int vis[18],n,ch[18];
   
   int prime(int n)
   {
   	for(int i=2;i<=sqrt(n);i++)
   	{
   		if(n%i==0)return 0;
   	}
   	return 1;
   }
   
   void dfs(int x)//x:环里现在有几个数
   {
   	if(x==n)
   	{
   		if(prime(ch[x]+1))
   		{
   			for(int i=1;i<n;i++)
   			{
   				printf("%d ",ch[i]);
   			}
   			printf("%d\n",ch[x]);
   		}
   	}
   	else
   	{
   		for(int i=2;i<=n;i++)
   		{
   			if(vis[i]==0&&prime(i+ch[x]))
   			{
   				ch[x+1]=i;
   				vis[i]=1;
   				dfs(x+1);
   				ch[x+1]=0;
   				vis[i]=0;
   			}
   		}
   		//return ;
   	}
   }
   
   int main()
   {
   	int q=1;
   	while(scanf("%d",&n)!=EOF)
   	{
   		if(q==1){
   			printf("Case %d:\n",q);
   		}
   		else
   		{
   			printf("\nCase %d:\n",q);
   		}
   		q++;
   		memset(vis,0,sizeof(vis));
   		memset(ch,0,sizeof(ch));
   		if(n%2==1)
   		{
   			printf("\n");
   			continue;
   		}
   		ch[1]=1;vis[1]=1;
   		dfs(1);
   	}
   	return 0;
   }