90% of Python solutions to A will get hacked or fail systests, since using the obvious ans += s[i] in a loop or similar will immediately balloon into O(n^2). My question is, why would one even need to consider this for a Div2A? It's not like it even adds to the complexity of the problem — the difficulty and algorithm would be the same for $$$|s| \lt 10^3$$$ or something. At the very least, have a pretest that's the trivial print(1) print('A'*200000) generator I wrote in 30 seconds, instead of leaving that to the hacking phase.

(To be fair, I'm pretty salty right now because I dropped from 2000 to 8000 after hacking my A. But I still think this is a pretty serious problem, especially for an educational round where problems have equal weighting... unless the goal was to educate me to not use Python ever again :/)

Update: Yep, I know that Python string concatenation is rather expensive now, you don't have to keep commenting that (although I appreciate everyone who responded :D). I guess the Educational Round succeeded in being educational! I still think that there should have been a strict $$$2*10^5$$$ pretest, but whatever, I'll take the L here. Now time to get back to Expert...