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.
 
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