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
1 /*代码一：构造法*/
2
3 //Memory Time
4 //188K 16MS
5
6 #include<iostream>
7 #include<cmath>
8 using namespace std;
9
10 int main(int i)
11 {
12 int n; //皇后数
13 while(cin>>n)
14 {
15 if(!n)
16 break;
17
18 if(n%6!=2 && n%6!=3)
19 {
20 if(n%2==0) //n为偶数
21 {
22 for(i=2;i<=n;i+=2)
23 cout<<i<<‘ ‘;
24 for(i=1;i<=n-1;i+=2)
25 cout<<i<<‘ ‘;
26 cout<<endl;
27 }
28 else //n为奇数
29 {
30 for(i=2;i<=n-1;i+=2)
31 cout<<i<<‘ ‘;
32 for(i=1;i<=n;i+=2)
33 cout<<i<<‘ ‘;
34 cout<<endl;
35 }
36 }
37 else if(n%6==2 || n%6==3)
38 {
39 if(n%2==0) //n为偶数
40 {
41 int k=n/2;
42 if(k%2==0) //k为偶数
43 {
44 for(i=k;i<=n;i+=2)
45 cout<<i<<‘ ‘;
46 for(i=2;i<=k-2;i+=2)
47 cout<<i<<‘ ‘;
48 for(i=k+3;i<=n-1;i+=2)
49 cout<<i<<‘ ‘;
50 for(i=1;i<=k+1;i+=2)
51 cout<<i<<‘ ‘;
52 cout<<endl;
53 }
54 else //k为奇数
55 {
56 for(i=k;i<=n-1;i+=2)
57 cout<<i<<‘ ‘;
58 for(i=1;i<=k-2;i+=2)
59 cout<<i<<‘ ‘;
60 for(i=k+3;i<=n;i+=2)
61 cout<<i<<‘ ‘;
62 for(i=2;i<=k+1;i+=2)
63 cout<<i<<‘ ‘;
64 cout<<endl;
65 }
66 }
67 else //n为奇数
68 {
69 int k=(n-1)/2;
70 if(k%2==0) //k为偶数
71 {
72 for(i=k;i<=n-1;i+=2)
73 cout<<i<<‘ ‘;
74 for(i=2;i<=k-2;i+=2)
75 cout<<i<<‘ ‘;
76 for(i=k+3;i<=n-2;i+=2)
77 cout<<i<<‘ ‘;
78 for(i=1;i<=k+1;i+=2)
79 cout<<i<<‘ ‘;
80 cout<<n<<endl;
81 }
82 else //k为奇数
83 {
84 for(i=k;i<=n-2;i+=2)
85 cout<<i<<‘ ‘;
86 for(i=1;i<=k-2;i+=2)
87 cout<<i<<‘ ‘;
88 for(i=k+3;i<=n-1;i+=2)
89 cout<<i<<‘ ‘;
90 for(i=2;i<=k+1;i+=2)
91 cout<<i<<‘ ‘;
92 cout<<n<<endl;
93 }
94 }
95 }
96 }
97 return 0;
98 }