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.

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

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?

No, anyone can get the prizes.

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

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

Shitty problemset!

Really awesome problem set. Especially team division was really cool.

Can you please explain how to solve it.

Hope you won't host similar contests on codeforces.

Can we get an editorial for the contest?

how to solve C?

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.

Prepared this for my friendThe contest was really great! When will the editorial of the questions posted?

Editorials:## Winners!!

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

Ratings will change after the April Long contest is over.

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

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.

what is po fx for? btw seriously thanks a lot.

Po is modular exponentiation function

Ok thank you.

Happy to help:)

:)