hmehta's blog

By hmehta, history, 6 months ago, In English,

The first Algorithm Competition Online Rounds of the 2019 Topcoder Open has arrived! Round 1A will be held on Saturday April 20, 2019 at 12:00 UTC -4.Registration is open for the round and closes at 11:55 UTC -4, April 20 2019.

Did you win an Automatic Berth a.k.a Bye for Round 1? Check out the list of members who won an automatic berth to Round 2 here.

How many will qualify? The 750 highest scorers from each Round 1 will win a spot in Round 2 of the Algorithm Competition. Note: To be eligible to advance from an Online Round Match, the Competitor must finish the Match with a point total greater than zero.

There will be one more Round 1 — Round 1B on Wednesday, May 1, 2019 at 07:00 UTC -4.

Best of luck to you in the Arena! -Topcoder

Update: Match Summary and Editorials: https://www.topcoder.com/blog/2019-topcoder-open-algorithm-round-1a-editorials/

 
 
 
 
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6 months ago, # |
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I think you need to mention positive scores are needed to qualify..

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    6 months ago, # ^ |
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    Added as a note! thanks :)

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    6 months ago, # ^ |
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    And it also looks like the only thing need mentioning! If you have positive score, you will be in top 750...

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6 months ago, # |
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Just received an email saying I’ve won a bye to Round 2. Will there be another email for those who got to Round 4 (or even to the Finals) through the SRM point system in one of the 4 online stages? I imagine people who advanced through that rule won’t be eligible to participate in Round 2 despite the bye, right?

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    6 months ago, # ^ |
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    Yeah, members who have qualified for round 4 from stages 1,2 or 3 will receive an email about the same(being not eligible) after stage 3 ends and before round 2A. The ones qualifying for Round 4 from stage 4 will be notified later when stage 4 ends.

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6 months ago, # |
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    6 months ago, # ^ |
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    Hey Golovanov399, We generally don't reveal writers in tournament rounds especially TCO Rounds :)

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    6 months ago, # ^ |
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    That would be me.

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      6 months ago, # ^ |
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      Bonjour! Loved your sin(t) codeforces rating graph! another maxima expected after few contest. :)

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      6 months ago, # ^ |
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      Thanks for the very nice problems, reachable to lower-rated coders also but requiring some careful thought (which I botched in both the 500/1000!!).

      Very enjoyable, and my failures were excellent reminders of general principles like "use integer for float problems whenever possible" :-)

      Will there be parallel rounds for 1B and 1C?

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6 months ago, # |
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Is it rated?

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6 months ago, # |
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Will this count towards the points for TCO regionals?

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6 months ago, # |
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UPD: Found it. :/

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6 months ago, # |
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1000 point problem's precision issues were frustrating, otherwise the round was good!

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    6 months ago, # ^ |
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    That's because the intended way was to avoid any floating-point number calculations until the very end?

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      6 months ago, # ^ |
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      I wanted to do that but don't the bounds mean it can overflow 64-bit integers?

      If the percentages are set to 99, you end up (almost) doubling up to 100 times, which seems like it should be too much.

      Maybe the idea is that such cases will never arise because they don't have a unique answer... but it doesn't seem obvious that this is true.

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        6 months ago, # ^ |
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        Spoiler
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          6 months ago, # ^ |
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          You're right — I forgot to think about the average when I was doing the estimations.

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    6 months ago, # ^ |
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    Indeed, the intended solution was to use integers only and since I expected many people wouldn't do that, I expected the problem to provide plenty of challenge opportunities. Apparently, creating a test wasn't that easy, thus most solutions failed on the System Test.

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    6 months ago, # ^ |
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    I managed to pass this problem by calculating everything in cents, so only integers needed. To round a real number to the nearest int, I used int(x + 0.5).

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      6 months ago, # ^ |
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      I had a separate divide function that rounded numbers fully in integers.

      auto divide = [](li a, li b){ 
          if (a % b * 2 >= b) 
              return a / b + 1; 
          return a / b; 
      };
      
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6 months ago, # |
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Auto comment: topic has been updated by hmehta (previous revision, new revision, compare).

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6 months ago, # |
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So, what is best solution for 250 problem?

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    6 months ago, # ^ |
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    Just brute force all possibilities.

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      6 months ago, # ^ |
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      I did it, but it was very long solution. It can be done recursively or iteratively, but I saw solution in 10 rows. I didn't understand that using <<

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        6 months ago, # ^ |
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        OK, thank you I understood using mask