JeevanJyot's blog

Invitation to CodeChef Starters 89 (Rated till 5-stars) — 10th May
By JeevanJyot, 18 months ago, In English

We invite you to participate in CodeChef’s Starters 89, this Wednesday, 10th May, rated till $$$5$$$-stars coders (i.e. for users with rating $$$< 2200$$$).

Time: 8:00 PM — 10:00 PM IST

Note that the duration is $$$2$$$ hours. Read about the recent CodeChef changes 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 for all users for 1 day as soon as the contest ends, after which they will be available only to Pro users.

We’re hiring! If you’d like to work at CodeChef as a Learning Content Creator, click here.

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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18 months ago, # |
  Vote: I like it +11 Vote: I do not like it

I have noticed on recent Starters contests some are rated for 5 stars, some for 6 stars, some for all. Is there any pattern/way to check this for upcoming contests?

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

    Actually, there is no specific pattern. The number of stars till which the contest should be rated is decided based on the setting panel's opinion and the testers' feedback.

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18 months ago, # |
  Vote: I like it +3 Vote: I do not like it

How to solve EVIL_INF ? I tried to do some DSU stuff but couldn't get it to AC

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

    DP, first just find for every Y what they will take if their X is missing. Then just go over Y in decreasing order and take dp[ current ] = dp[ takes_from ] + 1 ; answer is max over dp — 1, since the last person is not evil, because he is left with nothing to take.

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

    It's simple greedy, sort all people by B[i](their arrival time) ,now iterate from last to first, suppose we are at person x and demon took his takeout then dp[x]=1+dp[id],where id is the minimum a[i] person x can steal. dp[i]-answer when i takeout is taken. To calculate id, we can use a priority queue.

    Here's my implementation- Code

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18 months ago, # |
← Rev. 2   Vote: I like it +26 Vote: I do not like it

Today i helped a demon.

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

when will ratings be updated?

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