Poj 3239 Solution to the n Queens Puzzle

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1.Link：

http://poj.org/problem?id=3239

2.Content：

Solution to the n Queens Puzzle
Time Limit: 1000MS Memory Limit: 131072KTotal 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
80
Sample Output
5 3 1 6 8 2 4 7
Source
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 }

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