Considering that d[i][j] indicate the number of i bricks divided into j columns steps.So if we cut away the downmost brick of every steps, and we can get d[i][j] = d[i-j][j-1](if the 1st step only have one brick) + d[i-j][j](if the 1st step have more than one brick).
Thus q = d[n][2->max_j]

really smart!
how can you get that dude?

Hello,

Thank you for your hint, it was very useful for me

 and yes, your solution is awesome!

really nice idea, i like it.

good job, man

But how does this transition ensures that the next column always has more bricks than the current column?