杭电oj 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23458    Accepted Submission(s):
10465

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

 

　　这一题的意思是输入一个数n，然后让你用1~n的数围成一个圈，但是每相邻的两个数的和必须为素数。

　　　下面是代码：

 #include <iostream>
 #include <cmath>
 using namespace std;
 #define MAX 22
 int a[MAX];    //标记数组
 int b[MAX];
 int n;
 bool prime(double n)
 {
     int i, m;
     m = (int)sqrt(n);
     for (i=; i<=m; i++)
         if ((int)n%i == )
             return false;
     return true;
 }
 void dfs(int i)
 {
     int j;
     if (i>=n)
     {
         if (prime(double(b[n-]+)))    //判断最后一个和第一个数的和是不是素数
         {
             cout<<b[];
             for (j=; j<n; j++)
                 cout<<" "<<b[j];
             cout<<endl;
         }
     }
     else
     {
         for (j=; j<=n; j++)
         {
             if (a[j] || !prime(double(b[i-]+j)))    //a[j]已经被加入到圈中或者相邻两个数和不是素数，则continue
                 continue;
             a[j] = ;    //如果j已经加入圈中,则标记为1
             b[i] = j;
             dfs(i+);
             a[j] = ;
         }
     }
 }
 int main()
 {
     int i=, j;
     while (cin>>n)
     {
         memset(a,,sizeof(a));        //全置为0
         cout<<"Case "<<i++<<":"<<endl;
         b[] = ;
         dfs();
         cout<<endl;
     }
     return ;
 }