524 Prime Ring Problem

A ring is composed of n (even number) circles as shown in diagram. Put

natural numbers 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 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

68

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

题意：

       给出一个偶数N，需要将1到N这N个数等距排列在一个圆周上，使得相邻两个数的和总是一个素数，将所有的解（镜面对称的两个解视作不同解）都按顺时针顺序打印出来。

 

输入：

       正整数N。

输出：

       所有解。

分析：

使用深搜的方式进行枚举即可。从1开始打印解。

1 #include
      
        2 #include
       
         3 #define MAX_N 50 4 using namespace std; 5 int n,A[MAX_N] = {
    1},ispe[MAX_N],vis[MAX_N]; 6 void dfs(int cur){ 7     if(cur == n && ispe[A[0] + A[n - 1]]){ 8         for(int i = 0; i < n; i++) 9             i ? printf(" %d", A[i]) : printf("%d", A[i]);10         printf("\n");11     }else for(int i = 2; i <= n; i++)12         if(!vis[i]&& ispe[i + A[cur - 1]]){13             A[cur] = i;14             vis[i] = 1;15             dfs(cur + 1);16             vis[i] = 0;17         }18 }19 int main(){20     for(int i = 2; i <= 50; i++) ispe[i] = 1;21     for(int i = 2; i <= 50; i++)22         for(int j = i + i; j + i <= 50; j += i)23             ispe[j] = 0;24     int kase = 0;25     while(cin >> n){26         if(kase++)27             printf("\n");28         printf("Case %d:\n",kase);29         dfs(1);30     }31     return 0;32 }