Блог пользователя hmehta

Автор hmehta, история, 5 лет назад, По-английски

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!

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5 лет назад, # |
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How come there's unrated registrants in 2A?

User name:             osmanys
Rating (highest):      Not Rated
Rated events:          0
Member since:          Oct 17, 2016
Country:               Chile
Coder type:            Professional
Default language:      C++
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5 лет назад, # |
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How many participants advance to TCO19 Round 3?

This link doesn't seem to work

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5 лет назад, # |
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Unable to login to the web arena :(

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5 лет назад, # |
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Are there T-shirts for this year's contestants?

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    5 лет назад, # ^ |
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    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).

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5 лет назад, # |
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You missed tourist in https://tco19.topcoder.com/algorithm/round-4 (he is the winner of the first stage).

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    5 лет назад, # ^ |
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    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! :)

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      5 лет назад, # ^ |
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      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?

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5 лет назад, # |
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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 $$$O(N)$$$.

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5 лет назад, # |
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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