bayleef's blog

[HackerRank] Ad Infinitum 18 — Math Programming Contest
By bayleef, history, 8 years ago, In English

Hi!

Ad Infinitum is back with 18th Edition, a contest restricted to mathematics domain, held on HackerRank.

Register at https://www.hackerrank.com/infinitum18

The contest commences on 9th June 15:30 UTC. You are allowed to enter the contest anytime, for tie breaking the timer will start when you view the first challenge, which allows you to start late at your convenience, but once started try to finish as fast as possible to be on top of the leaderboard :)

It's a 2 days contest with 7 problems from easy to hard Scoring Distribution: 10 20 30 40 55 80 100

Top 10 on leaderboard gets Cool HackerRank T shirt.

Detailed editorials will be available by the end of contest :) I suggest you to try all challenges and at the end of contest understand the solutions.

GL&HF

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Comments Comments (15)
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8 years ago, # |
← Rev. 2   Vote: I like it -10 Vote: I do not like it

for tie breaking the timer will start when you view the first challenge

Please remove or change this.

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    8 years ago, # ^ |
      Vote: I like it +10 Vote: I do not like it

    As far as I know, this line is true for Ad Infinitum 18.

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      8 years ago, # ^ |
        Vote: I like it -13 Vote: I do not like it

      I meant you should start the timer for a user from the contest start, not when he/she opens the first problem. I think this is going to be disadvantageous for honest users unless I have misinterpreted the rule.

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        8 years ago, # ^ |
        ← Rev. 2   Vote: I like it +13 Vote: I do not like it

        Dishonest users don't win contests. But the rule is somewhat strange of course.

        UPD Great rule, I think.

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          8 years ago, # ^ |
          ← Rev. 2   Vote: I like it 0 Vote: I do not like it

          I believe that starting time from opening first statement is good, because it make contest convenient for all timezones and holiday schedules. The same works for SNSS for example.

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          8 years ago, # ^ |
            Vote: I like it +6 Vote: I do not like it

          apparently this link shows that dishonest users does win contests. (at least on hackerearth)

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        8 years ago, # ^ |
        ← Rev. 2   Vote: I like it +38 Vote: I do not like it

        First of all, I'm not a HackerRank representative, so here is my own point of view.

        Cheaters always gonna cheat. But the common observation is cheaters usually do not have great coding skills — probably that's why they start cheating. So, to ignore cheaters one should just improve in solving problems (which is a good goal overall).

        On the other hand let's look at some honest skilled user from a bad time zone. Let's assume it's like 3 AM in user's town when the contest starts. That's what I call disadvantageous.

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          8 years ago, # ^ |
            Vote: I like it -12 Vote: I do not like it

          I completely understand. But this is a rated contest and low rated coders who participate honestly will face issues due to this. Maybe consider a score based ranklist?

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            8 years ago, # ^ |
              Vote: I like it +18 Vote: I do not like it

            No, this contest doesn't affect main rating, for math competitions there is separate rating. And haven't seen cheaters on the top.

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8 years ago, # |
  Vote: I like it +3 Vote: I do not like it

one more divisor exploration!

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

Please unlock solutions for the last problem

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8 years ago, # |
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The problems were nice. Thanks!

Can someone please explain the formula from the editorial for the last problem?

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    8 years ago, # ^ |
    ← Rev. 7   Vote: I like it +24 Vote: I do not like it

    Say the of a people gets 2xi + 1 candies, of b people gets 2yi candies, and of c people gets zi candies.

    Then the answer is

    Now using

    and

    ,

    you can see that the answer is the coefficient of xn in

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

Where can we find the official solutions? Can anyone tell me please?

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