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

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

Solution

Code

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47  #include #include #define maxn 50 int is_prime[maxn], a[maxn], n, vis[maxn]; void get_prime(int n) { memset(is_prime, 0, sizeof(is_prime)); is_prime[0] = is_prime[1] = 1; for (int i = 2;i if (!is_prime[i]) { for (int j = 2;j*i 1; } } void dfs(int step) { int i; if (step == n + 1 && !is_prime[a[n] + a[1]])//结束 { for (i = 1; i printf("%d ", a[i]); printf("%d\n", a[n]); return; } for (i = 2; i if (!vis[i] && !is_prime[i + a[step - 1]]) { a[step] = i; vis[i] = 1; dfs(step + 1); vis[i] = 0;//回溯 } } } int main(int argc, char const *argv[]) { int cas = 1; get_prime(maxn); while (~scanf_s("%d", &n)) { memset(vis, 0, sizeof(vis)); printf("Case %d:\n", cas++); a[1] = 1; dfs(2); printf("\n"); } return 0; }