TCO19 Algorithm Round 2A and Parallel Rated SRM is scheduled to start at 21:00 UTC -4 on May 22, 2018. Registration is open for the Round in the Arena or Applet and it closes 5 minutes before the match begins, so make sure that you are all ready to go.
Click here to see what time it starts in your area.
Algorithm Round 2 Qualified Competitors (https://tco19.topcoder.com/algorithm/byes/, https://tco19.topcoder.com/algorithm/round-1a and https://tco19.topcoder.com/algorithm/round-1b) are eligible to compete in Round 2A. Those who have already qualified for Round 4 (https://tco19.topcoder.com/algorithm/round-4) from online stages are not eligible to compete.
All the best!
How come there's unrated registrants in 2A?
How many participants advance to TCO19 Round 3?
This link doesn't seem to work
Top 200 advance (positive score needed, but that shouldn't come up).
I've reported the broken link. In the meantime, this works: https://tco19.topcoder.com/algorithm/algorithm-ways
Unable to login to the web arena :(
Also can't login to web arena
Are there T-shirts for this year's contestants?
Everyone with a positive score in one of the Round 3 matches gets a T-shirt (and also everyone who made it past round 3 in a different way).
You missed tourist in https://tco19.topcoder.com/algorithm/round-4 (he is the winner of the first stage).
So we basically created the list stage wise
Stage 1 — tourist qualified — next 10 made it to round 4 (which included Petr)
Stage 2 — Petr qualified — next 10 made it to round 4 (which included you)(marked Petr in Stage 1 Round 4 as Qualified for Finals)
and so on! :)
By the way, I heard that some people got "you got a bye for R4!" email. I want to make sure that I haven't missed any emails from you — Will we get invitation letters to finals after R4?
Yeah, Finalists will receive the emails after Stage 4 is finalized for the visa travel and stay details.
Thank you for the information!
For 1000-point problem, we can solve in $$$O(A^3)$$$ ($$$A$$$ is the maximum pile size) by dynamic programming, if the grundy number of $$$n$$$-stone pile is $$$O(n)$$$.
However, I have not yet proved that. Is the hypothesis true? If so, is there any idea how to prove?
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Edit: It seems like I got the proof. Seeing hos.lyric's source code, the grundy number will be like $$$g = (0, 1, 2, 4, 3, 5, 6, 8n, 8n+2, 8n+1, 8n+5, 8n+7,...)$$$ where $$$n=1, 2, ...$$$. This yields $$$g_A \leq 1.6A$$$.
Also, this means that you can solve the 1000-points problem in $$$O(N)$$$.
In case anyone is interested in unorthodox solution to the 250 — was too sleepy to figure out the intended solution, so I googled and implemented this: https://en.wikipedia.org/wiki/Sparse_ruler#Wichmann_rulers