hdoj 1016 Prime Ring Problem

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

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

 

题目大意：

给出一个数n,将从1到n的数全排列，将相邻两个数互为素数且第一个数为1，第一个数与最后一个数也互为素数的排列进行输出

 

 #include <stdio.h>
 #include <string.h>
 int n;
 int vis[], a[];//数组vis记录改点是否被访问过，数组a记录符合条件的数
 int judge(int s)//判断s是否是素数
 {
     int flag = ;
     for(int i = ; i <= s/; i++)
     {
         if(s%i == )
         {
             flag = ;
             break;
         }
     }
     return flag;
 }
 void dfs(int s, int cnt)//利用深搜来解决该问题，cnt是将要往数组中放入第几个数，s是放入数组中的末端的数
 {
     int i, j;
     if(cnt  == n+ && judge(a[]+a[n]))//当放入数组中的数（cnt-1)等于n时，且数组第一个数与最后一个数也互为素数时，进行输出
     {
         for(j = ; j < n+; j++)
         {
             if(j != )
                 printf(" ");
             printf("%d", a[j]);
         }
         printf("\n");
         return ;
     }
     for(i = ; i <= n; i++)
     {
         if(judge(i + s) && !vis[i])//如果i与s互为素数并且i没有被访问过时，访问i,将i放入数组中
         {
             a[cnt] = i;
             vis[i] = ;
             dfs(i, cnt + );//i进行与它上一个数相同的操作，将要往数组中放入第cnt+1个数
             vis[i] = ;//返回后，要将i标记为未访问

         }
     }
     return ;
 }
 int main()
 {
     int num = ;
     while(~scanf("%d", &n))
     {
         printf("Case %d:\n", num++);
         memset(vis, , sizeof(vis));
         vis[] = ;//将1标记为已访问
         a[] = ;//将1放入数组中
         dfs(, );
         printf("\n");
     }
     return ;
 }