hmehta's blog

TCO18 Beijing Parallel Rated SRM and Editorials
By hmehta, history, 7 years ago, In English

TCO18 Beijing Parallel Rated SRM

Round Editorials are now published — https://www.topcoder.com/blog/tco18-beijing-regionals-round-editorials/

TCO18 Beijing Parallel Rated SRM is scheduled to start at 03:00 UTC -4 on May 26, 2018. Click here to know what time it starts in your area.

You will be able to register for the SRM in the Arena or Applet 4 hours prior to the start of the match. Registration closes 5 minutes before the match begins, so make sure that you are all ready to go.

Don’t know how to compete in Topcoder SRMs?

Check out this guide to successfully compete in an algorithm match.

You can compete using:

  • Topcoder Web Arena(Beta) - Please watch this video for step by step guide.
  • Topcoder Java Applet - You can refer to this guide here to set up the applet. (Note that those who have Java 8 installed on their machine will see a security issue — You will have to add Topcoder in security exceptions in Java Control Panel. Please refer to the details in the guide here)
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Comments Comments (18)
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7 years ago, # |
  Vote: I like it +2 Vote: I do not like it

Bumping up.

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7 years ago, # |
  Vote: I like it +8 Vote: I do not like it

Reminder: The contest starts in 4 hours. So, registeration has started.
If you are not an onsite participant, you can register "Fun Round Beijing" not "Beijing Round".

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    7 years ago, # ^ |
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    The Round begins in 40 mins! Hope to see most of you competing! All the best! :)

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7 years ago, # |
  Vote: I like it +16 Vote: I do not like it

How to solve 1000?

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    7 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    To make it easier, let's assume a = 0.

    There is a naive upper bound for the number of rows: , the left hand side is the sum of all rows, and the right hand side is the minimum sum of all columns. The inequality gives , and in fact we can always construct a solution of .

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Could please explain how to get second inequality and achieve second upper bound?

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        7 years ago, # ^ |
          Vote: I like it +5 Vote: I do not like it

        The second inequality comes directly from the first one.

        The main idea of the construction is to fill the columns with [0, 1, 2, ..., r - 1] and [r - 1, r - 2, ..., 1, 0] alternatively.

        The tricky part is when the number of columns is odd. In this case, we can construct a special case for three columns and expand to the case of any odd columns.

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7 years ago, # |
← Rev. 2   Vote: I like it +8 Vote: I do not like it

For the 500ptr, there is an alternate solution that does not involve the 2 DP states for each i. Find all underscores that can be converted to dots (leading underscores can't be converted, underscores followed by a digit can't be converted).

For a group of consecutive underscores of size K, the no. of ways is FK + 2, where Fi denotes the ith Fibonacci number. The final answer is the product of each group's ways.

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7 years ago, # |
  Vote: I like it +15 Vote: I do not like it

When are the ratings going to be updated?

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    7 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    Ratings will be updated in 7-8 hours from now! We had an issue with Poland Round Ratings — Fixing that currently! -Ratings will be updated after that :)

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Ratings have been updated :)

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