1.Link：
http://poj.org/problem?id=3239
2.Content：
Solution to the n Queens Puzzle
Time Limit: 1000MS Memory Limit: 131072K Total Submissions: 3459 Accepted: 1273 Special Judge 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
0Sample Output
5 3 1 6 8 2 4 7Source
POJ Monthly--2007.06.03, Yao, Jinyu
3.Method:
一开始用8皇后的方法，发现算不出来。
只能通过搜索，可以利用构造法，自己也想不出来构造，所以直接套用了别人的构造公式
感觉没啥意义，直接就用别人的代码提交了，也算是完成一道题目了
构造方法：
http://www.cnblogs.com/rainydays/archive/2011/07/12/2104336.html
一、当n mod 6 != 2 且 n mod 6 != 3时，有一个解为：
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)
(上面序列第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。下同)
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)第二种情况可以认为是，当n为奇数时用最后一个棋子占据最后一行的最后一个位置，然后用n-1个棋子去填充n-1的棋盘，这样就转化为了相同类型且n为偶数的问题。
若k为奇数，则数列的前半部分均为奇数，否则前半部分均为偶数。
4.Code：
http://blog.csdn.net/lyy289065406/article/details/6642789?reload
/*代码一：构造法*/ //Memory Time
//188K 16MS #include<iostream>
#include<cmath>
using namespace std; int main(int i)
{
int n; //皇后数
while(cin>>n)
{
if(!n)
break; if(n%!= && n%!=)
{
if(n%==) //n为偶数
{
for(i=;i<=n;i+=)
cout<<i<<' ';
for(i=;i<=n-;i+=)
cout<<i<<' ';
cout<<endl;
}
else //n为奇数
{
for(i=;i<=n-;i+=)
cout<<i<<' ';
for(i=;i<=n;i+=)
cout<<i<<' ';
cout<<endl;
}
}
else if(n%== || n%==)
{
if(n%==) //n为偶数
{
int k=n/;
if(k%==) //k为偶数
{
for(i=k;i<=n;i+=)
cout<<i<<' ';
for(i=;i<=k-;i+=)
cout<<i<<' ';
for(i=k+;i<=n-;i+=)
cout<<i<<' ';
for(i=;i<=k+;i+=)
cout<<i<<' ';
cout<<endl;
}
else //k为奇数
{
for(i=k;i<=n-;i+=)
cout<<i<<' ';
for(i=;i<=k-;i+=)
cout<<i<<' ';
for(i=k+;i<=n;i+=)
cout<<i<<' ';
for(i=;i<=k+;i+=)
cout<<i<<' ';
cout<<endl;
}
}
else //n为奇数
{
int k=(n-)/;
if(k%==) //k为偶数
{
for(i=k;i<=n-;i+=)
cout<<i<<' ';
for(i=;i<=k-;i+=)
cout<<i<<' ';
for(i=k+;i<=n-;i+=)
cout<<i<<' ';
for(i=;i<=k+;i+=)
cout<<i<<' ';
cout<<n<<endl;
}
else //k为奇数
{
for(i=k;i<=n-;i+=)
cout<<i<<' ';
for(i=;i<=k-;i+=)
cout<<i<<' ';
for(i=k+;i<=n-;i+=)
cout<<i<<' ';
for(i=;i<=k+;i+=)
cout<<i<<' ';
cout<<n<<endl;
}
}
}
}
return ;
}