IceKnight1093's blog

Invitation to CodeChef Starters 101 (Rated till Div-2) — 20th September
By IceKnight1093, 14 months ago, In English

We invite you to participate in CodeChef’s Starters 101, this Wednesday, 20th September, rated till Division-2 (ie. for users with rating < 2000).

Time: 8:00 PM — 10:00 PM IST

Read about the recent judge migration here.

Joining us on the problem setting panel are:

Written editorials will be available for all on discuss.codechef.com. Pro users can find the editorials directly on the problem pages after the contest. The video editorials of the problems will be available only to Pro users.

Also, if you have some original and engaging problem ideas, and you’re interested in them being used in CodeChef's contests, you can share them here.

Hope to see you participating.

Good Luck!

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Comments Comments (15)
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14 months ago, # |
  Vote: I like it +1 Vote: I do not like it

While migrating servers no one was creating problems?
I thought we can have multiple rounds which will be rated for all as there was enough time to create problems which can rated for all.

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    14 months ago, # ^ |
      Vote: I like it +17 Vote: I do not like it

    If you think creating problems is easy, then why you don't create one?

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14 months ago, # |
  Vote: I like it 0 Vote: I do not like it

The Input Format in Split and ADD is wrongly mentioned.
The line of the test case contains, $$$N, L$$$ and $$$R$$$, but in the problem input format, $$$N$$$ and $$$K$$$ are mentioned!

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14 months ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Can we solve PIARQ using Mo's algorithm in $$$\mathcal{O}(n\sqrt{n})$$$ time?

It seems from the TL that we can, but my submission gives TLE.

Edit: I guess I should have set the block size to $$$600$$$, instead of $$$400, 500$$$. Bruh.

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

    Let nxt[i] and prv[i] be the next larger and previous smaller to i. Answer is max value of (prv[i]-l+1 + 1 + r-nxt[i]+1) over all i such that l <= prv[i] <= nxt[i] <= r. You can solve this offline using segtree.

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      14 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Thanks for another approach I can try that, though I still don't get why it TLEs.

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    14 months ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    Me too thought of MO in contest but the time constraints were 1.5sec it might pass for 3sec or 4sec!

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      14 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I mean, I could have agreed with you, if the same thing hasn't passed after the contest, that too in 1 sec.

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        14 months ago, # ^ |
          Vote: I like it +5 Vote: I do not like it

        Can you please provide a link to the identical submission that you had made during the contest?

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          14 months ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Sorry for the confusion, but by same thing I mean the same Mo's algorithm, I didn't mean the same submission. I have changed a variable after the contest.

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14 months ago, # |
  Vote: I like it +31 Vote: I do not like it

A quick request: For future contests, please announce which divisions will be rated with more than eight hours' notice. Many contestants' decisions to participate could reasonably depend on whether the round will be rated, not just because they care about their rating but because (a) the fact that a round is rated for Div. 1 signals that there will be challenging problems targeted at high-level contestants and (b) rated rounds attract a higher-quality competition pool. Providing over a week's notice like Codeforces generally does for Div. 1 rounds would be ideal, but at the very least, contestants should be able to go to sleep the night before a contest knowing whether or not they should expect to participate in a rated round the next day.

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14 months ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

deleted

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