ABCDEF | Atcoder Beginner Contest 361 Solution Discussion
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- čas přidán 5. 07. 2024
- ABCDEF | Atcoder Beginner Contest 361 Solution Discussion
My submissions - atcoder.jp/contests/abc361/su...
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good work
F is good, I used inclusion exclusion on kth root to solve it
With bit of precomputation, it got accept with 1ms runtime
Yeah, mine got accepted with 1ms runtime too. Though I only considered counting numbers of the form a^p, where p is a prime, so this reduced the task to checking 2,3, 5, 7, 11, ... , 59th power of numbers and applying inclusion exclusion on that
Won't there be (N+1) places where we can have 2 consecutive Dots.
Yes, we will have N+1 places.
I was able to solve ABCE . Any tips for solving problems with lower constraints? I am only able to solve O(n) and O(nlogn) ish solutions and just can't think when the constraints are lower
i am so dumb for D still
is G that hard?
I felt it was hard, but editorial makes it look easy.