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.

 

Inputn (0 < n < 20). 
OutputThe 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 思路:DFS遍历所有结果
AC Code:
#include<iostream>
#include<cstring>
using namespace std;
int num[];
int vis[];
int n;
bool isPrime (int k) {
for (int i = ; i * i <= k; i++)
if (k % i == )
return false;
return true;
}
void dfs (int k) {
if (k > n) {
if (isPrime (num[] + num[n])) {
for (int i = ; i < n; i++)
printf("%d ", num[i]);
printf("%d\n", num[n]);
}
return ;
}
for (int i = ; i <= n; i++) {
if (vis[i] == false) {
if (isPrime (num[k - ] + i)) {
num[k] = i;
vis[i] = true;
dfs (k + );
vis[i] = false;
}
}
}
return ;
}
int main() {
int sum = ;
while(~scanf ("%d", &n)){
printf ("Case %d:\n", sum++);
memset (vis, false, sizeof(vis));
num[] = ;
vis[] = true;
dfs();
putchar('\n');
}
}

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