D - Prime Ring Problem HDU - 1016 （深搜+素数打表）

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

分析：

意思是给你一个数n，要构成一个素数环，这个素数由1-n组成，它的特征是选中环上的任意一个数字i，i与它相连的两个数加起来都分别为素数，满足就输出

要全部搜索，所以用dfs，要判断是否是素数，并且看到数据最大是40.打表把1到40判断素数的结果存在数组里就最简单啦，而且高效

而素数打表有四种方法详见我另一篇博客

https://blog.youkuaiyun.com/qq_42079027/article/details/81410361

#include<stdio.h>
#include<string.h>
int prime[40]={0,1,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0};//素数打表，因为n最大是20，所以只要打到40
int n;
int book[21],a[21];
void dfs(int num)//深搜
{
   int i;
   if(num==n&&prime[a[num-1]+a[0]])  //满足条件了，就输出来
   {
       for(i=0;i<num-1;i++) 
           printf("%d ",a[i]);
       printf("%d\n",a[num-1]);
   }
   else
   {
       for(i=2;i<=n;i++)
       {
           if(book[i]==0&&prime[i+a[num-1]])//如果未被标记，且是素数 
           {
                   book[i]=1;//标记了
                   a[num]=i;//放进数组
                   dfs(num+1); //递归调用
                   book[i]=0; //退去标记                 
          }
       }
   }
}    
int main()
{
      int id=0;
      while(scanf("%d",&n)!=EOF)
      {
         id++;
         printf("Case %d:\n",id);
         memset(book,0,sizeof(book));
         a[0]=1;
         dfs(1);
         printf("\n");
      }
      return 0;
}