Пожалуйста, подпишитесь на официальный канал Codeforces в Telegram по ссылке https://t.me/codeforces_official. ×

Блог пользователя MiteshAgrawal

Автор MiteshAgrawal, история, 7 лет назад, По-английски

Hello Codeforces Community!

I invite you all to take part in HackerEarth's December Easy, scheduled on the 1st of December at 22:00 IST.

Problems have been set by me(paradox1) and tested by satyaki3794. We are grateful to HackerEarth admin r3gz3n for his help.

You will be given 5 algorithmic problems to solve in 3 hours. Partial scoring will be used (you get points for passing each test case). Although the contest is targeted towards beginners, we hope even experienced problem-solvers find one or two problems to be interesting. The contest is rated and prizes will be awarded to the top 3 beginners(i.e. Programmers with a rating of 1600 or less before the challenge starts).

I hope you have fun solving the problems. You can discuss problems on this thread after the contest.

Good luck and have fun.

  • Проголосовать: нравится
  • +18
  • Проголосовать: не нравится

»
7 лет назад, # |
  Проголосовать: нравится 0 Проголосовать: не нравится

I think the problems were a little bit harder than November easy .

»
7 лет назад, # |
  Проголосовать: нравится +3 Проголосовать: не нравится

how to solve " Eerie Planet " ?

  • »
    »
    7 лет назад, # ^ |
      Проголосовать: нравится 0 Проголосовать: не нравится

    Kindly, check the editorials here. There are two solutions explained in the editorial and there is a discussion regarding third approach in comment section.

»
7 лет назад, # |
  Проголосовать: нравится 0 Проголосовать: не нравится

_The contest is rated and prizes will be awarded to the top 5 beginners(i.e. Programmers with a rating of 1600 or less before the challenge starts)._ This say that the prizes will be given for top 5 beginners but the website says only for top 3. Which is true?

»
7 лет назад, # |
  Проголосовать: нравится 0 Проголосовать: не нравится

Auto comment: topic has been updated by paradox1 (previous revision, new revision, compare).