my_tempo's blog

By my_tempo, history, 3 years ago, In English

Problem Link
I think it would be optimal to fill zeros starting from (m,m) and then at a gap of 2*m-2 in each row(because we will cover that much squares of size m*m).
So the minimum zeros required would be around n/(2*m-1)
Same will happen for columns, So the answer should be somewhere around (n/(2*m-1))^2 after just taking care of n%(2m-1) remaining cells.
But on the editorial, It is saying that minimum number of zeros should be (n/m)^2, Can someone tell where am I going wrong?

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3 years ago, # |
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A picture here is very instructive (below with N=10 and M=3).

Here it's easier to just think about this as starting from the top left corner and placing MxM tiles with a white square in the bottom right. Here the 10th row and column don't fit, but since the white square has a reach of M-1, it doesn't matter. For that reason it follows that formula can just be floor(N/M)^2.

The way you want to tile it forgets about some regions: