Блог пользователя shubhamgrg1000

Автор shubhamgrg1000, история, 20 месяцев назад, По-английски

We invite you to participate in CodeChef’s Starters 82, this Wednesday, 22nd March, rated till 6-stars Coders (ie. for users with rating < 2500). This contest is organized by Udyam’23, IIT BHU.

Time: 8:00 PM — 11:00 PM IST

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.

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.

Udyam'23 is the technical fest of the Electronics Engineering Society of IIT BHU, and this contest is conducted under its Competitive Programming and Web Development event, Devbits!!

Link to the Web Development Hackathon organized by Devbits.

So what are you waiting for? Visit our site, register for the event, and get a chance to win cash prizes worth Rs 20k+!!

Hope to see you participating. Good Luck!

  • Проголосовать: нравится
  • +51
  • Проголосовать: не нравится

»
20 месяцев назад, # |
  Проголосовать: нравится 0 Проголосовать: не нравится

Auto comment: topic has been updated by shubhamgrg1000 (previous revision, new revision, compare).

»
20 месяцев назад, # |
  Проголосовать: нравится -16 Проголосовать: не нравится

Excited!

»
20 месяцев назад, # |
  Проголосовать: нравится 0 Проголосовать: не нравится

... this contest is conducted under its Competitive Programming and Web Development event, Devbits!!

How these two pieces are connected together? Is Starters 82 some kind of qualification for the further hackathon or it's two separate things?

  • »
    »
    20 месяцев назад, # ^ |
      Проголосовать: нравится +19 Проголосовать: не нравится

    These are two separate events, but final winners will be decided on the basis of combined result of both events.

»
20 месяцев назад, # |
  Проголосовать: нравится +15 Проголосовать: не нравится

Would recommend participating the round. Problems are good in my opinion. The round might be slightly tougher than usual. It is recommended to read all the problems as some problems are equi-difficult (So pick according to your taste :P)

»
20 месяцев назад, # |
  Проголосовать: нравится +10 Проголосовать: не нравится
»
20 месяцев назад, # |
  Проголосовать: нравится +24 Проголосовать: не нравится

As a setter, I hope you'll have as much fun solving as we had in preparing :D

»
20 месяцев назад, # |
  Проголосовать: нравится +11 Проголосовать: не нравится

Waku waku :)

»
20 месяцев назад, # |
  Проголосовать: нравится +13 Проголосовать: не нравится

As a Problem Solver, I like all the problems (Special Treatment for Tree Game)...

»
20 месяцев назад, # |
  Проголосовать: нравится +16 Проголосовать: не нравится

Nice Problemset! What is the intended solution for GRIDMEET?

  • »
    »
    20 месяцев назад, # ^ |
      Проголосовать: нравится +19 Проголосовать: не нравится

    There are a total of N*N significant points where K people can meet. We can fix an x[i] (let's say X) and perform a scanline for all y[j] (let's say Y). After fixing X, points can only be located either low (y<Y) or high (y>=Y). We can start iterating from the smallest Y to the largest. During the scanline, points transition from high to low.

    To efficiently track the best K points, we can maintain two sets: one for the high points (s1) and one for the low points (s2). In set s1, we can store the points as {abs(X-x[id])+y[id], id}, and in s2, we can store them as {abs(X-x[id])-y[id], id}. Initially, all points are high, so the best K points are simply the top K elements of s1.

    As we move to higher values of Y, some points will shift from s1 to s2, and the iterators tracking the least good points in s1 and s2 can shift a total distance of the number of points shifted. Since the iterators can shift a total distance of N for any X, we can perform this operation in O(NlogN) time. We can iterate over all such values of X in O(NNlogN).