3101: N皇后

3101: N皇后

Time Limit: 10 Sec  Memory Limit: 128 MBSec  Special Judge
Submit: 88  Solved: 41
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Description

n*n的棋盘,在上面摆下n个皇后，使其两两间不能相互攻击…

Input

一个数n

Output

第i行表示在第i行第几列放置皇后

Sample Input

4

Sample Output

2
4
1
3

HINT

100%的数据3<n<1000000。输出任意一种合法解即可

Source

题解：一道神(dou)奇(bi)的题目，传说中貌似有种O(N)构造N皇后解的方法，具体为啥貌似也查不到，求神犇给出证明orzorzorz（引自N皇后的构造解法）

一、当n mod 6 != 2 或 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)

然后就是码代码了= =

 

 /**************************************************************
     Problem:
     User: HansBug
     Language: Pascal
     Result: Accepted
     Time: ms
     Memory: kb
 ****************************************************************/

 var
    i,j,k,l,m,n:longint;
 begin
      readln(n);
      case n mod  of
           ,:begin
                    k:=n div ;
                    case (k mod )+(n mod )* of
                         :begin
                                for i:= to (n-k) div  do writeln(k+i*);
                                for i:= to (k-) div  do writeln(+i*);
                                for i:= to (n-k-) div  do writeln(k++i*);
                                for i:= to k div  do writeln(+*i);
                         end;
                         :begin
                                for i:= to (n-k-) div  do writeln(k+i*);
                                for i:= to (k-) div  do writeln(+i*);
                                for i:= to (n-k-) div  do writeln(k++i*);
                                for i:= to k div  do writeln(+*i);
                                writeln(n);
                         end;
                         :begin
                                for i:= to (n-k-) div  do writeln(k+i*);
                                for i:= to (k-) div  do writeln(+i*);
                                for i:= to (n-k-) div  do writeln(k++i*);
                                for i:= to (k-) div  do writeln(+*i);
                         end;
                         :begin
                                for i:= to (n-k-) div  do writeln(k+i*);
                                for i:= to (k-) div  do writeln(+i*);
                                for i:= to (n-k-) div  do writeln(k++i*);
                                for i:= to (k-) div  do writeln(+*i);
                                writeln(n);
                         end;
                    end;
           end;
           else begin
                if odd(n) then
                   begin
                        for i:= to (n-) div  do writeln(i*);
                        for i:= to (n+) div  do writeln(i*-);
                   end
                else
                    begin
                         for i:= to n div  do writeln(i*);
                         for i:= to n div  do writeln(i*-);
                    end;
           end;
      end;
      readln;
 end.