Description

The eight queens puzzle is the problem of putting eight chess queens on an 8 × 8 chessboard such that none of them is able to capture any other. The puzzle has been generalized to arbitrary n × n boards. Given n, you are to find a solution to the n queens puzzle.

Input

The input contains multiple test cases. Each test case consists of a single integer n between 8 and 300 (inclusive). A zero indicates the end of input.

Output

For each test case, output your solution on one line. The solution is a permutation of {1, 2, …, n}. The number in the ith place means the ith-column queen in placed in the row with that number.

Sample Input

8
0

Sample Output

5 3 1 6 8 2 4 7

解题思路：

一、当n mod 6 != 2 或 n mod 6 != 3时：

[2,4,6,8,...,n],[1,3,5,7,...,n-1] (n为偶数)

[2,4,6,8,...,n-1],[1,3,5,7,...,n ] (n为奇数)

二、当n mod 6 == 2 或 n mod 6 == 3时

(当n为偶数,k=n/2；当n为奇数,k=(n-1)/2)

[k,k+2,k+4,...,n],[2,4,...,k-2],[k+3,k+5,...,n-1],[1,3,5,...,k+1] (k为偶数,n为偶数)

[k,k+2,k+4,...,n-1],[2,4,...,k-2],[k+3,k+5,...,n-2],[1,3,5,...,k+1],[n] (k为偶数,n为奇数)

[k,k+2,k+4,...,n-1],[1,3,5,...,k-2],[k+3,...,n],[2,4,...,k+1] (k为奇数,n为偶数)

[k,k+2,k+4,...,n-2],[1,3,5,...,k-2],[k+3,...,n-1],[2,4,...,k+1],[n ] (k为奇数,n为奇数)

(上面有六条序列。一行一个序列，中括号是我额外加上的，方便大家辨认子序列，子序列与子序列之间是连续关系，无视中括号就可以了。第i个数为ai，表示在第i行ai列放一个皇后；... 省略的序列中，相邻两数以2递增。)

#include
  
   
#include
   
     using namespace std; int main(int i) { int n; //皇后数 while(cin>>n) { if(!n) break; if(n%6!=2 && n%6!=3) { if(n%2==0) //n为偶数 { for(i=2;i<=n;i+=2) cout<