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authorAnthony Wang2022-03-20 19:17:36 -0500
committerAnthony Wang2022-03-20 19:17:36 -0500
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- 1632D: Wow at least I can still solve 2000 rating problems, since it's this difficulty, you can assume it has a reasonably easy and greedy solution, and then some binary search or two pointers to speed things up
- ~~1650G: Mostly BFS but the overcounting is very tricky to avoid, but you can think about what the longer paths look like to find a way to only count them once, basically when they diverge from the short paths~~
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+- 1604F: Don't use binary search because it's cancerous to code, loop over every power of two instead which is much easier