coins on chessboard, who has the winning strategy The game begins with empty $n\times n$ chessboard and a fixed number $m\in\{1,2,\dots,n\}$.
Two players are making moves alternately, each move is placing a coin on one empty square, each row and column can contain at most $m$ coins, the guy who cannot put a coin when he is to play, loses.
Who has the winning strategy?

In the original problem there was $n=2011$ and $m=1005$.
My solution:
The first guy wins. First move: a coin in the centre, then symetrical reflections of opponent's moves.

After solving the problem, I generalised it. 
My above solution works for all $n,m$ both odd.
If $n$ is even, then the second guy wins by symetrical reflections.
What about remaining cases?

The remaining case is $n$ odd, $m$ even. My first intuition was the second player wins but my proof fails.
The maximum number of coins that can fit on a $n \times n$ chessboard without making a $m+1$-alignment is $n*m$, which is even. 
If there are less than $n*m$ coins disposed, you can always find at least one row and one column with less than $m$ coins. If their intersection is free, you can play there. Alas we cannot be sure that this is the case, for instance with n=3, m=2:
| _ | _ | X |
| X | X | _ |
| X | X | _ |
Second player cannot play anymore.
