shikhar7s's blog

Invitation to ICM Technex 2021 — IIT (BHU) Varanasi (Rated for Div. 2 and Div. 3)
By shikhar7s, history, 4 years ago, In English

Hello everyone,
I would like to invite you all to the International Coding Marathon 2021, under the banner of Technex'21, IIT (BHU) Varanasi.

Technex is the annual techno-management fest of Indian Institute of Technology (BHU), Varanasi organized from April 2-6, 2021.

International Coding Marathon is the flagship event of Byte The Bits, the set of programming events organized under Technex. It will take place on 6th April, 8pm — 10:30pm IST. The contest will be held on Codechef.
Click here to go the contest page.

The problemset has been prepared by the_nightmare, CoderAnshu, DenOMINATOR, rivalq, loud_mouth, _nitr0_, sharabhagrawal25 and me, shikhar7s.
I would like to thank jtnydv25, Ashishgup, fmota and amnesiac_dusk for their invaluable help in testing and preparation of the contest.

Some of our previous contests :
ICM Technex 2017 and Codeforces Round 400 (Div. 1 + Div. 2, combined)
ICM Technex 2018 and Codeforces Round 463 (Div. 1 + Div. 2, combined)
ICM Technex 2019
ICM Technex 2020
Participants will have 2.5 hours to solve 8 problems. The scoring will be ICPC style. Contest will be rated for Div. 2 and Div. 3 participants, but it has exciting problems and great prizes for everyone!

Prizes!
Global 1st — worth 10k
Global 2nd — worth 7k
Global 3rd — worth 5k
India 1st — worth 4k
IIT BHU (overall 1st) — worth 3k
IIT BHU 1st year — worth 1k

Good luck everyone! Hope to see you on the leaderboard.

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Comments Comments (25)
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4 years ago, # |
  Vote: I like it +39 Vote: I do not like it

Thanks for the contest, this one fills the gap for some of us left by Codeforces Break right now.

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4 years ago, # |
  Vote: I like it +44 Vote: I do not like it

When I try to sign up on the website,

It asks for the year and college. Does it mean the prizes are only for college students?

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4 years ago, # |
  Vote: I like it +27 Vote: I do not like it

Can we expect video editorials by COPS IIT BHU? You guys explain really well.

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4 years ago, # |
  Vote: I like it +48 Vote: I do not like it

Hey everyone, we hope you enjoyed the contest. Please provide us with your valuable feedback.

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

how to solve C?

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    4 years ago, # ^ |
      Vote: I like it +38 Vote: I do not like it

    Suppose that a given polygon has $$$n$$$ sides of length $$$1$$$. Then, using the Law of Cosines, we find that the polygon generated from its midpoints will have side-length equal to $$$\frac{1}{2^2} + \frac{1}{2^2} - 2 \cdot \frac{1}{2^2} \cdot \cos \left(\pi - \frac{2\pi}{n} \right) = \frac{1}{2} \left( 1 - \cos \left( \frac{(n-2)\pi}{n} \right) \right).$$$

    Since these two polygons are similar, we can find the ratio of their areas by squaring the ratio of their side-lengths. From here, knowing that the fraction of points visible to the first player is $$$1$$$ and knowing the ratio of the areas between any two successive shapes, we can compute the answer using the formula for the sum of an infinite geometric series.

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    4 years ago, # ^ |
      Vote: I like it +39 Vote: I do not like it
    Prepared this for my friend
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4 years ago, # |
← Rev. 3   Vote: I like it 0 Vote: I do not like it

The contest was really great! When will the editorial of the questions posted?

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4 years ago, # |
  Vote: I like it +35 Vote: I do not like it

Editorials:

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

Winners!!

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

Is the contest not rated? rating didn't change?

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

    Ratings will change after the April Long contest is over.

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

How to solve ICM0005 (Team Division)? I couldn't understand the editorial.

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

    I will explain the way i solved it.

    Sort according to (Skill, Count) in ascending order. For each index i from 1 to N. consider ith Score as the min possible Score that is in Set B. So, You will get only one way of arranging i.e. (In Set A — indexes 1 to i-1 because if any index j < i is in set B means jth value will be the min and not ith, In Set B — indexes i to N because if any j > i is in set A, jth value can't divide ith value). So while calculating for each index i you need Prefix Lcm of i-1 and Suffix Gcd of i to solve this.

    link to my code.

    Though it is not a complete editorial, it might give you idea to solve.

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