Here is my solution for O

[robert-king]

Here is my solution for O

[SamZhangQingChuan]

@SamZhangQingChuan how does your solution to O work? :)

I don't remember clearly. Looks like work out all the reachable states first and then do gaussian elimination.

[rversteegen]

Hi, thanks, I'll have a closer look at that.
"tried to go backwards"? What's TLE? (Edit: oh, time limit exceeded)

The trick for K is to decompose vertical and horizontal folds, handling them separately and then multiplying. For example if at a certain position the foil has been folded over 5 times when only folds along a horizontal axis are considered, and 3 times when only vertical folds are considered, that's a thickness of 15.
My solution for K stores the thickness of the foil as a list of thickness-length segments, like 5cm of thickness 2, 2cm of thickness 1. That's stored as [[2,5],[1,2]] in the R and C variables. Processing a fold then just requires splitting this list of segments in half (possibly splitting one segment) and combining the halves. Simple in theory, but easy to make off-by-one mistakes.

[rversteegen]

Also, in fold() in K.py, R1 and R2 are the two halves of the segment list, and when I iterate past the end of R1 or R2 I treat it as a virtual segment of length 1e9 thickness 0.

[SamZhangQingChuan]

Thank you for the explanation. Do you know what the time complexity is? I feel like it is O(2^n) but this bound might be too loose.

[rversteegen]

Running time for a single fold operation is linear in the number of segments (as I described above). Each fold can increase the number of segments by at most one (and can reduce it). So overall the worst case number of segments is number of folds + 1 and the worst case running time is O(V^2 + H^2) in number of horizontal and vertical folds. And there are only at most 20 folds!

[robert-king]

@SamZhangQingChuan how does your solution to O work? :)