cry's blog

By cry, 4 months ago, In English

Wazzup Codeforcers!

sum, satyam343, and I are extremely ecstatic to invite you to Codeforces Round 965 (Div. 2) on Aug/10/2024 17:35 (Moscow time). You will be given 5 problems and 2 hours to solve them. 1 problem will have subtasks. This round will be rated for all participants with rating below 2100. We spent the most time cooking up this round than any other round, so it means a lot if you will participate.

We would like to orz the following individuals for making the contest possible:

Score Distribution: $$$500 - 750 - 1250 - 1500 - (1750 + 1750)$$$

UPD: We know there has been a lot of disapproval for problem C. We're sorry for it. Please view https://codeforces.net/blog/entry/132507

UPD: Editorial

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4 months ago, # |
Rev. 5   Vote: I like it +6 Vote: I do not like it

Wazzup guys! I encourage you to participate in CerealCodes for their Summer 2024 Contest.

What is CerealCodes?

Disclaimer: This round is not affiliated with CerealCodes.

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    4 months ago, # ^ |
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    cereauwucodes

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    4 months ago, # ^ |
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    I knew the CerealCodes round was going to be good once I saw cry there!

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      4 months ago, # ^ |
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      I will be cry ing because I can't participate :/

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        4 months ago, # ^ |
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        I will be Laugh ing because you can't participate :)

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          4 months ago, # ^ |
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          I solved around 7-8 800 rating problems and still couldn't solve a single problem in todays contest. Any tips on improving it or some prerequisite i don't know? I know a bit of STL but don't know how much is required. Please help

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            4 months ago, # ^ |
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            this is because the round is much more difficult than usual. number of solutions on c should be way higher. like that kind of a gap occurs at c-d problems most of the times, not on b-c.

            just try learing some simple logic, i recommend you participate in div.4 or div.3, those problems are much more beginner friendly

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            4 months ago, # ^ |
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            I would say the first problem today is maybe slightly harder than the first problem normally is.

            I would recommend just solving some more low level problems, you'll get the hang of it eventually.

            This was my approach for problem A today.

            If the points have to be centered on xc, yc then our points have to be symmetric in some way. An easy way to do this is just to put all our points in a line, like this:

            . . . c . . .

            If we have an odd number of points, then we should put a point in center, and then put half of the other points above it and half the points below it.

            If we have an even number of points, we do the same thing but leave the center blank.

            There's a bunch of ways to calculate where the other points go.

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    4 months ago, # ^ |
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    this will be a great round

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4 months ago, # |
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As a first-time tester, I am proud to finally be able to write an "as a tester..." comment :)

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As a tester.

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4 months ago, # |
Rev. 3   Vote: I like it +69 Vote: I do not like it

As a tester,

🦧

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4 months ago, # |
Rev. 5   Vote: I like it +34 Vote: I do not like it

🙈 🙉 🙊 🦍 🦧 🙈 🙈 🦍 🙊 🙊 🙊 🐒 🐒

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As a tester, I cried because of how beautiful the problems are

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    4 months ago, # ^ |
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    Aa a participant, I will cry because of how beautiful the problems are

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    4 months ago, # ^ |
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    As a participant, I cried because of the difficulty gap between B and C.

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4 months ago, # |
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As a tester, I will take the round for infinite delta.

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I think I'm the first non-tester to send a messages!!! Can't wait for another cry round!!!

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As a participant, I feel like the tasks will be very unpredictable.

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    4 months ago, # ^ |
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    skill issue

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    4 months ago, # ^ |
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    what does having subtasks mean ? Will we get partial points for solving the subtasks? I am new to this concept, can anyone explain please

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      4 months ago, # ^ |
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      its essentially 2 identical problems, except for some constraint or some important detail being changed, so that the 2nd version is harder than the first. sometimes you can just solve the harder version and submit the same code for both of the versions, but there are exceptions to that, for example this and this,

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4 months ago, # |
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cry orz

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Good luck to everyone!

P.S. It seemed that the author wrote"Russian"wrongly,instead,he wrote"russian",the first letter was a lowercase letter.

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This contest is the end of an era of Contest IDs with a 4-digit number beginning with 1. On to the 2xxx’s for IDs

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4 months ago, # |
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Did anyone note "coordinatORZ"?

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4 months ago, # |
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The round feels more solid with cry and sum as problem setters! :)

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4 months ago, # |
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how Grandmaster become cyan over 1 year? I'm genuinely curious. I'm talking about the coordinator of this round satyam343. is this level of degeneration really something that could happen or just troll?

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    4 months ago, # ^ |
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    look at his recent contests, seems like troll

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    4 months ago, # ^ |
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    He is just solving the hard questions or the questions he wants to solve. He has no intention of increasing or maintaining his rating

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Monkey doesn't have any pants

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4 months ago, # |
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WTF THATS ME

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4 months ago, # |
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Did satyam343 sell his account or what

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the best birthday present ever :"]

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i will reach gm

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Although the number of the round is 1998,it is the last contest with the id like "1xxx`.

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As an expert participant, I hope to reach back to CM.

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more contest noice

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milind0110 sir Orz

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Hope to reach Pupil 3rd time

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first unrated div 2 for me :D

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Not gonna lie your last div3 was probably the best div3 I have ever given. Decode and Bomb, both were beautiful problems.

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what is wazzup??

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Good luck and have fun! :)

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potato monkey

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Score distribution?

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Hopefully I get to add some blue to the green of my handle.

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Hoping to become cyan after the contest.

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Hoping for this to be my best round in 2024 :prayge:

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Excited for the contest. yeeeeee

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i hope one day i can proudly stand in the "Our testuwuers: " line owo

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2 Min silence for those who judged the coordinator on his current rating

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I'd be thrilled if i reach cyan after this round.

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    4 months ago, # ^ |
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    Awesome recent performance bro. I also hope to reach cyan this round.

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4 months ago, # |
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Wazzup

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18o3 and satyam343 Orz

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Just dont make us cry :')

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4 months ago, # |
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anyone miss my comments?

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Good luck, hope its not math forces again

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4 months ago, # |
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Good luck for everyone <3

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what is the point distribution?

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Good luck everyone !!!

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4 months ago, # |
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is it only me or codeforces is quite lagging recently

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I see satyam343. I'm shaking. Please no counting maths problems this time. I hate those. Please satyam343 We believe in u.

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    4 months ago, # ^ |
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    I read binary strings and permutations, I stayed out. Glad I did. Indian authors and their love for maths, binary strings, and permutations, etc.

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    4 months ago, # ^ |
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    I hope you don't mind me using this comment in problem C. I just found it funny :p

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As a participant I am excited .

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Any specific reason for the contest to be of 3 hours? Also, eagerly waiting for the point distribution to be updated.

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Spoiler
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hope i can solve at least 4 problem XD XD XD

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Counting is not fun.

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Hoping for delta 200++ round. been some time...

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    4 months ago, # ^ |
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    May you good luck!

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      Thanks !

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      4 months ago, # ^ |
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      first two problems, solved in 5 minutes.

      C problem solved in 1 hour 53 minutes. .

      3 times understood the problem completely wrong. Feeling so stupid. should have read the the test cases first.

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        4 months ago, # ^ |
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        how did u do C ?

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          4 months ago, # ^ |
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          Hint 1 :

          We need to go greedy. If we apply 'k' addition, we will apply it on one single element. To deduce this, you can try few test cases on pen-paper. find the current median, try to delete elements from left of the median, or right of the median. There also, we need to check if the array is EVEN or ODD size array. So many IF-ELSE :| . you will always find this EXCEPT one edge case ( below described)

          Hint 2 : Edge case handling. 3 4 1 1 100 1 1 1

          In above input, it is best to pick 100 and delete it, and then take median of rest of the array. Sort the values, and pick the last value, and then see, what is the best median you can get from remaining N-1 elements. For that I used binary search.

          Now, we have to take the MAX() value from HINT-1 approach , and edge-case result.

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As a contestant want to be Master, i wish i can be Master in this round

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cry orz

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Excuse me
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    4 months ago, # ^ |
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    There is no such rule or guideline that the score should be equal to estimated problem rating. It's just that they tend to be similar, but they don't necessarily need to be like that. Score distribution tends to be more exponential than problem rating, and therefore we sometimes have problems of 250 points and 6000 points depending on relative difficulty difference of problems, which is pretty far from their problem rating. Also in Div. 1/2 separated rounds 1A is usually scored 500 points but they're as hard as *1500-rated problems.

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thembululquaUwU

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interesting score distribution;GL&HF

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aaj speedforces hoga kya bc :D

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As a tester, all the problems are very good and no problem will make you cry

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At first glance the monkey looks like its showing middle fingers!

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    I don't know why but looking at this monkey is so relaxing

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      is it still relaxing after the contest ? The monkey was the silence before the storm.

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I thought this is a LeetCode contest after looking at these problem titles

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    I couldn't think of a good title for this problem, so I decided to learn from LeetCode. — Sun Tzu, The Art of War

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      its well known leetcode is the best competitive programming site, we should all learn from it

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        4 months ago, # ^ |
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        its well known CODEFORCES is the best competitive programming site. We are learning from it. xD

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      If it ain't broke, don't fix it. -Sun Tzu, The Art of War

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UnbalancedForces!

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4 months ago, # |
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unbalancedforces!

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nice 2000 rating third lmao

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what is this problem balancing

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mathforces , worst contest

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Did they accidentally give us a Div1 instead of Div2.

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    4 months ago, # ^ |
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    not really only Div2C was off in difficulty. Acted more like a D.

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    there goes my rating. Got outplayed in problem C so hard I don't even want to read D and E anymore

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speedforces!

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Horrible coordination, score dist. is shit.

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problem rating: 800 800 2000 .... so on. Just let me reach pupil pls..

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    4 months ago, # ^ |
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    800, 900, 1800, 2000, 2100, 2500 if I were to estimate.

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as a Newbie, Pupil, Specialist and new experts the contest duration was 20 minutes or less

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    Yeah A+B took me 5 mins, C and D took almost an hour each.

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-_-

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If you can't make a div 2 round just dont.

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    4 months ago, # ^ |
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    If you can't come up with your actual id then don't comment :)

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unbalanced-forces, contest should be standard div 2

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i am not giving the contest currently but this guy is streaming and giving out the solution for the contest

can you do something about this

i can rarely solve 3rd in div 2 this mf is giving them away

mf madarchod https://www.youtube.com/watch?v=dDZyHLJVHaM

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    4 months ago, # ^ |
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    Well the most you can do is not advertise cheating telegram groups.

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Is it just me or there is a problem with same solution to problem B on Codeforces? Also,you liked your social credits joke?Now you have earned your negative credit from me.Hope you liked it.

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4 months ago, # |
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bad contest problem c was too hard for normal div 2

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    4 months ago, # ^ |
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    I feel like it is okay, but requires a lot of time to implement, if one doesn't know certain tricks. I'd like to have a full day on it, breaks included, not just 2 hours ;D ...and still not sure my idea would work, maybe this is all just a ramble of a low-skill.

    Of course, given the amount of people who solved, 1250 on rating distribution doesn't make any sense. Looking forward to the editorial...

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if you can't offer normal div2 problems, please don't do it

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    4 months ago, # ^ |
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    I personally didn't like their Div 3 either. "Counting is fun", huh. But math guys will argue.

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4 months ago, # |
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DIV2 C ???

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Contest too hard, me sed. Back to newbie here I go.

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I've just become Specialist last Div. 4 Round. Again, I'll be Pupil, or even Newbie.

What Div.2 C??? Wish not this again.

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    4 months ago, # ^ |
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    The comment removed because of Codeforces rules violation
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    One's demise is another one's triumph. Someone would have the rating. Maybe one specialist guy, who solved A and B fast enough (not me though, 0:11 A). Let's cheer for them this time.

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When participants like neal , aryanc403 got stuck on 3rd problem How am I going to get motivation to solve that .

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Really only 1250 points for problem C?? should be 1500 considering difficulty!

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Too much harder than Div.2 before!

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stupid contest

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Man the C was so hard it took my soul with it .

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Gonna need a 3 mo break from cp after this

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E1 is just B(altic)OI 2022 islands on array instead of tree.

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wtf is that C?

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problem C score should be at least 1750 imo

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Can someone please explain, why in in 4th sample Bessi loses if she starts at 2nd and 3rd islands?

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Median again.

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    4 months ago, # ^ |
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    this problem C crush my soul, I can solve it if b was all 1. But not like this

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      Came back after skimming others thoughts/code. splitting array a into b=0 and b=1 only actually help... Well, need to take a hit and learn tho

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why this approach (https://codeforces.net/contest/1998/submission/275607879) for problem A is wrong, any counter test cases?

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4 months ago, # |
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:(

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Why do I get WA4 in D? 275609322

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May be, i am noob or C is a bad problem xD

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    4 months ago, # ^ |
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    I'd say the rating distribution was not okay, but also yeah, we are noobs.

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satyam343 bro played with the emotions of his biggest fan lol!

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https://qoj.ac/problem/3511

Problem E1, E2 was on JOISC.

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    4 months ago, # ^ |
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    what's this site? qoj? thank you. (there is no intro in their website)

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One more median problem and I am killing myself :D

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С sucks. Its true rating seems to be pretty higher than it's stated. Can't imagine what was on contest creators' mind when making this distribution

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ok yay you did it hope you didnt wa on test 2

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4 months ago, # |
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C solution:

Consider two cases:

I) Greedily choose the largest number that has its $$$b = 1$$$ and spend all your $$$k$$$ on it.

II) Binary search on the largest $$$\operatorname{median}(a)$$$ you can make by the given $$$k$$$, then calculate the answer for this case.

The answer of the problem would be the maximus answer of two cases described above.

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    4 months ago, # ^ |
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    In test case 4: how answer : 13

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      4 months ago, # ^ |
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      all operations go into 4, and then median is 5 after you remove the 8, 8+5=13

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      4 months ago, # ^ |
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      Spend all $$$k$$$ on that $$$4$$$, so it becomes $$$8$$$. Then take that $$$8$$$ and new median which is 5. So the answer would be $$$8 + 5 = 13$$$.

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    4 months ago, # ^ |
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    I realized this, but i couldnt implement finding the larget median, how do you do that?

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    4 months ago, # ^ |
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    can you provide your solution ?

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    4 months ago, # ^ |
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    Yeah but how exactly do you apply binary search in the 2nd case? Arent there too many posibilites to consider?

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    4 months ago, # ^ |
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    in II in case of odd length our median is (sz+1)/2-1 right ?

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4 months ago, # |
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i hate your biggest fan.

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4 months ago, # |
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from collections import *
for _ in range(int(input())):
  x,y,k=[int(i) for i in input().split()]
  x_sum=x*k
  y_sum=y*k
  if k%2==0:
    print(0,0)
    k-=1
  c=1
  while k>1:
    print(-c,-c)
    print(c,c)
    c+=1
    k-=2
  print(x_sum,y_sum)

why this not working for A

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    4 months ago, # ^ |
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    Maybe if k%2==0---->if k%2==1

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    4 months ago, # ^ |
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    if xc,yc=0,0 and k is even you will print two times the point (0,0)

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      4 months ago, # ^ |
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      nice observation man now i got the answer

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      4 months ago, # ^ |
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      from collections import *
      for _ in range(int(input())):
        x,y,k=[int(i) for i in input().split()]
        x_sum=x*k
        y_sum=y*k
        if x_sum==0 and y_sum==0 and k%2==0:
          while k>0:
            print(-c,-c)
            print(c,c)
            c+=1
            k-=2
          continue
        if k%2==0:
          print(0,0)
          k-=1
        c=1
        while k>1:
          print(-c,-c)
          print(c,c)
          c+=1
          k-=2
        print(x_sum,y_sum)
      

      it Got Accepted,so that was the only case my code failed

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4 months ago, # |
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5 4
7 5 2 5 4
0 0 1 0 1

13

???

Why in problem C 4th example it's 13 ?

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    4 months ago, # ^ |
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    You spent all 4 op on 4 then you get: 2 5 5 7 8. The answer=8+5=13

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      4 months ago, # ^ |
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      thanks. you are right

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      4 months ago, # ^ |
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      I just realized that I was solving a different problem.

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        4 months ago, # ^ |
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        soon you will realise that Gojo in your DP isn't real :)

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    4 months ago, # ^ |
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    Apply operation on the last element 4th times it becomes 8. Removing last element the median will be 5. So 5 + 8 = 13

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    4 months ago, # ^ |
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    If you use all your operations on a[4] (initally 4), it will become 8.

    The rest of the array is [7, 5, 2, 5]. Median is 5. So, answer is at least 13.

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    4 months ago, # ^ |
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    pick 4 and spend k = 4 on it. The median is 5 (2, 5, 5, 7). So answer is 4 + 4 + 5 = 13

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4 months ago, # |
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After this contest, I will never look at score distribution again.

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4 months ago, # |
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Nice, well-balanced contest. Also felt like "real" competitive programming (too many contests feel like math these days).

Well maybe B-C gap is a bit too large but C-D-E1-E2 is well-balanced imho

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4 months ago, # |
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Oh my God, I only have 15 minutes to do problem E1, and because I was in such a rush, I didn't realize that the complexity of my code is $$$O(n*log^3n)$$$. Maybe if I convert the segment tree to a sparse table, I'll get AC.

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    4 months ago, # ^ |
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    $$$O(n * log^2n * log \max)$$$ passes with iterative segment tree

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      4 months ago, # ^ |
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      Maybe my implementation is bad because I wrote around 100 lines in a very rushed time (< 10mins). Unlucky for me this time.

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    4 months ago, # ^ |
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    I also had the same complexity and did not realize it at first. I ended up adding a bunch of heuristics and it passed pretests in 2 seconds (out of 4), still may fail systests though. I am not sure if those heuristics improved asymptotics. I kept segment tree.

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      4 months ago, # ^ |
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      I didn't even have time to check the verdict of my submission because the time was up :<

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4 months ago, # |
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worst contest i ever participated in. i acidentally sent my solution where arrays were too small and i got RTE on test 5 and i placed 6000 instead of 3000 because of that. its a skill issue, but still crazy

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    4 months ago, # ^ |
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    Messed up with output on A, found it's failed on 1 pretest only after ten minutes :D Kinda feel the same...

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4 months ago, # |
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If this is the level at which C is going to come i don't think i will ever reach expert

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4 months ago, # |
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Almost submitted C, found the error in the last minute but had 5 secs left to submit..

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4 months ago, # |
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I was ready to solve C before the 1 hour mark, maybe even solve D and get back to expert. Let's just say I was too naive hahaha

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    4 months ago, # ^ |
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    as someone who did that, it would place you closer to CM in ratings.

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      4 months ago, # ^ |
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      Oh wow, yeah... this Div2 is so broken, some past few contests I solved A. B, C too slowly and couldn't get D, my final placement for those contests is lower than this one where I only solved A and B lmao. Codeforces contests can be really random sometimes...

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4 months ago, # |
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The comment removed because of Codeforces rules violation
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4 months ago, # |
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can't do C :(

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4 months ago, # |
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Has anyone observed in the last 20 min, that accepted submission in C increased by 50% from 1K to 2K?

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4 months ago, # |
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Felt like this div2 had 2 As and 4Es....

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4 months ago, # |
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when is rating roll back?

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4 months ago, # |
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Yaaaay I solved C lol! I'm so happy!

Just choose one of two options:

  1. Greedily choose the largest a with b = 1 and use all k on it.
  2. Use all k to make median as big as possible (binary search).

max(ans1, ans2) is the answer.

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4 months ago, # |
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Happy to solve C, get the logic in 20 min but take an hour to complete code.

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4 months ago, # |
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Interesting round. The solutions end up being so simple you almost feel ashamed for spending so much time on the problem lol.

Also, C might be more difficult than what I've seen in the previous rounds.

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4 months ago, # |
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What were all these testers doing?

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    4 months ago, # ^ |
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    Making Fun of us.... ok yay you did it hope you didnt wa on test 2 (4 test cases)

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4 months ago, # |
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C was a long implementation, atleast for the approach that I thought T_T

imho C shouldn't be this long implementation based problem, but other than that cool round, infact I love how C boils down to a simple solution

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4 months ago, # |
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terrible contest, i dont really see your balance setting on Problem C. Maybe you need more preparation on testing your problem. BIG DISAPPOINTMENT

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4 months ago, # |
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I finished C at the last minute literally, so I hardly had time to check my answers in detail. I hope it will pass the system tests :I

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4 months ago, # |
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use 15 minutes to think about C and use 2 hours to code and debug it

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4 months ago, # |
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Can anyone tell how binary search can be used to find the max median in C?

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    4 months ago, # ^ |
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    Possibly use the same idea as the problem D of the last contest.

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    4 months ago, # ^ |
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    OK... so here is the my idea for the whole C...

    1) Sort the array $$$a$$$(Try binding with $$$b$$$ to make coding easier);

    2) Assume the index of median of $$$a$$$ is $$$i_{mid}$$$, the median of $$$a$$$ is $$$median(a)=MED(a)$$$.

    When no operation can be done($$$k=0$$$), the answer should be $$$max(a)+MED(a)$$$;This can be shown that:

    2a) Choose the last element, we get the score $$$p_{max}=max(a)+MED(a)$$$;

    2b) At index $$$i<=i_{mid}$$$,we select $$$a_i<=MED(a)$$$ and delete it , the median may change to the $$$a_{i_{mid+1}}$$$. But after all the answer will be $$$a_i+a_{i_{mid}+1}$$$. Since $$$a_i<=MED(a)$$$ and $$$a_{i_{mid+1}}<=max(a)$$$, This is not the answer;

    2c) At index $$$i>=i_{mid}$$$, deletion won't change $$$MED(a)$$$, so $$$p_i=a_i+MED<=p_{max}$$$;

    3) Then we are allowed to do operation. Obviously the answer will just get larger. Since the answer is contributed by two parts: one element and one median, the intuition is that:

    3a) we increase the element;

    3b) we increae the median;

    3c) we increase both the element and median;

    4) Assume after optimal operation, the final array become $$$a^\prime$$$.Then we find that:

    4a) 3c) will never be the optimal solution.(Little hard to find tho...). From $$$p_{max}=max(a^\prime)+MED(a^\prime)$$$, we know that if we increase the $$$max(a^\prime)$$$ by 1, $$$p_{max}$$$ will also increase by 1. At the same time there is no assurance that increase some element in the array $$$a$$$ will change $$$MED$$$ (if changed, increase at most 1). Just to make it clear that increase the $$$max(a^\prime)$$$ is more effective.

    4b) So if we only change the $$$max(a^\prime)$$$, we bet all the operation on it. This can be done by increase one available element in $$$a$$$ to $$$a+k$$$ and calculate the answer with $$$O(n)$$$ or even $$$O(1)$$$;

    4c) if we only change $$$MED$$$, $$$max(a^\prime)=max(a)$$$ will be fixed and won't change ever.

    Then all we need to do is to find the maximum reachable $$$MED$$$ in the $$$n-1$$$ size $$$a[:-1]$$$ within $$$k$$$ operations.

    One possible solution is that Binary search the $$$MED$$$ through this way:

    Given the target $$$MED$$$, we iter through the $$$n-1$$$ elements and count all the element smaller than $$$MED$$$ while recording the operation needed if the element can be changed to $$$MED$$$. After the iteration if we find there are more than $$$int(n/2)$$$ elements smaller than $$$MED$$$, then we should use the operation to decrease the number until less than $$$int(n/2)$$$.

    Obviously we should change the largest element as possible to reduce the operation needed.

    Finally if $$$op>k$$$ or $$$cnt$$$ can't be reduced to $$$int(n/2)$$$, $$$MED$$$ is unreachable. Otherwise reachable.

    Wish my explaination could help you. If my solution is wrong just hack it..

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4 months ago, # |
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I don't understand why my first solution do not pass. Maybe problem with C#.

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4 months ago, # |
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Whoever created C deserves capital punishment

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4 months ago, # |
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Didn't solve C, too difficult for me (`o`)=3

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4 months ago, # |
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Please let us know if an author is indian. I will skip contests which have indian authors. PS : I'm Not tryna be racist, but indian authors like too much maths and observational trick problems.

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    4 months ago, # ^ |
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    isn't it the case with russian and chinese authors too?

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    4 months ago, # ^ |
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    Satyam is only indian author, and i dont believe he has a single problem in the contest

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      4 months ago, # ^ |
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      my bad. I apologize.

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      4 months ago, # ^ |
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      It's clearly written--

      "Our coordinatorz satyam343 for his gracious coordination and donation of his own problems to this contest."

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4 months ago, # |
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so we're in copper

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4 months ago, # |
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finally get true purpose of the author posting ape pic

Spoiler
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4 months ago, # |
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Most imbalanced and disappointing round ever...

Any idea for Problem — C

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    4 months ago, # ^ |
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    key observation:

    if b[i] = 1 then there is no difference between in action (increasing value itself vs increasing median) so we take a[i] + k + (median after remove a[i])

    if b[i] = 0 then we need to take a[i] + (median after remove a[i] then use k operations to maximize the median) and it can be shown in case of b[i] = 0 we only need to consider one case where a[i] is max

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      4 months ago, # ^ |
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      Thanks for the good observation, even so after this we have to solve median problem by using k to divide accumulation result of binary search on (a[i] having b[i]==1) to have the final answer though

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4 months ago, # |
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cool round FST on A :)

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4 months ago, # |
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I struggled so much on B when all people did it in less than 5 minutes it took 30 minutes how are you guys able to guess this thing in a seconds ? can you tell me what did you think when reading the problem

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    4 months ago, # ^ |
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    Wrote a brute force solution and found a pattern.

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      that's what I did but no one did this I think

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    4 months ago, # ^ |
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    well if you just rotate it once, it works. kinda just intuitive i guess

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    4 months ago, # ^ |
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    My observation is that changing one of the pair can make its sums different, then consider what about shift more numbers.

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4 months ago, # |
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FST on A, whyyyy?

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4 months ago, # |
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i think i can understand the meaning of cry now lol

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4 months ago, # |
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Submitted C in last few seconds,I have even no time to check it:(

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4 months ago, # |
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WTF literally FST on A

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4 months ago, # |
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now i can understand the meaning of cry

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4 months ago, # |
  Vote: I like it +47 Vote: I do not like it

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4 months ago, # |
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Cry made me cry today

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i got FST on A, this was my second div 2 contest and i was so happy i solved 2 this time :\ (i solved only 1 last time) and btw why was there such a huge difficulty gap between B and C? damn

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4 months ago, # |
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Hi, can somebody please explain why my submission gets WA? 275619036

When I change the line

int lo=0,hi=1e18;

to

int lo=0,hi=1e10;

I get accepted, but I don't see why that would cause problems? (I define int as long long)

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    4 months ago, # ^ |
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    I think the variable used to be overflowing.

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    4 months ago, # ^ |
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    a better way to get a hi value without any overflow problems. create a function $$$ f(x) $$$ which return true if x is a valid option in binary search or not. (you cannot create it if you want but it looks cooler and nicer). then do this

    int lo = 0, hi = 1;
    // f(x) is of form 0000011111 then do the following
    while(!f(hi)) hi *= 2;
    // f(x) is of form 111110000 then do the following
    while(f(hi)) hi *= 2;
    

    This ensures we get a valid hi value without running into overflows

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4 months ago, # |
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cry definately made me cry lol....Didn't expected to get back to pupil after grinding 1600 rated problems rigorously for past one month.....

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4 months ago, # |
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Why was it that after being hacked, the correct code in front of me was skipped

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4 months ago, # |
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This code

This code of mine was wrong so I resubmitted because this code fails for: 1 10 0 20

Don't know how this code later got accepted. Please check into this.

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The number of correct submissions going from ~1000 to 2000+ in the last 20 minutes totally not suspicious?

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hard C.

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4 months ago, # |
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Do we have editorial for this?

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4 months ago, # |
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Aside from the difficult C, I think the description for E was also a little too complicated. I mean, this kind of problem would be much more understandable if the legend has a more reasonable story behind it, rather than using all the mathematical formulas to be extremely formal.

For example, this problem can be easily modeled as each $$$a_i$$$ denoting a power of the $$$i$$$-th character and it can absorb either the $$$i-1$$$-th or the $$$i+1$$$-th character's power if the $$$i$$$-th one has greater or equal power. Then the question is to simply find all the ones that can possibly survive to the end. In this way it's easy to reason why this process has to happen, and is much more simple than explaining this process with an array and a set.

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    4 months ago, # ^ |
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    The description of E is what makes it the last problem lol.

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275593102 Why does it fail? (Problem A)

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    4 months ago, # ^ |
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    Because your code is printing the same point again for some testcase. The goal is to print k distinct points.

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4 months ago, # |
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I got destroyed by this contest. Still I must admit that problems were interesting.Thank you for the round.

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I really hate it when it gives pretests passed but wrong answer later,just demolished my rank.

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4 months ago, # |
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"All wars is deception" Sun Tzu, The Art of War

and ofc... "Whatever you do, never reveal all your secrets in a Youtube video.. you fool!" Sun Tzu, The Art of War

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4 months ago, # |
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That C was something...

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So many W5 in A.

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A had very weak pretests (my code had some flawsobviously and pretests just didn't include x=y cases where my code failed) Anyways I have learnt not to trust on pretests (and sample test cases obviously XD)

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4 months ago, # |
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If i registered for a contest, but unable to participate in that, Will it affect my rating ?

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4 months ago, # |
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Can someone help me why 275686807 is failing on test 15 ? I'm trying to remove each element. If the removed element corresponds to b[i] = 1 then adding k to the removed element. And if b[i] is 0 at that index then binary searching on the median to maximize it with k.

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4 months ago, # |
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How beautiful the problems are.

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4 months ago, # |
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Hi[user:cry],

I got plaged in this round at problem C. I don't know how. I use to give Claude AI my codes to rewrite it and make it not readable but keep the logic same. I use to do this after my solution was hacked by someone a while ago. My friend suggested to make the code less readable. But I can't be plaged for that. Please look closely at this code.

I was asked to post comment here to help me with my plag. Can someone please help me with this? I have been giving contest from this Id to put the rating on my resume, but in the notification it was mentioned that I may get blcoked for this plag. If this happens then I won't be able to give contests again before my placements and it will not be fair.

I have already put this Id in the application form of several important companies and if my offer got revoked just because of this mismatch then my career will be adversely affected.

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4 months ago, # |
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Dear Codeforces Team,

I hope this message finds you well. I'm writing with a heavy heart regarding the recent issue flagged for my solution to problem 1998D. The notification states:

*"Your solution 275630271 for the problem 1998D significantly coincides with solutions [user:razz_11]razz_11/275627442, his submission

EclipseWraith/275630271."my submission

I want to clarify that the solution I submitted was entirely my own work, and I have no connection whatsoever with the other participants involved.

With a contest involving around 30,000 participants, isn’t it plausible that solutions with similar logic might emerge, especially when we all are tackling the same problem? It’s disheartening to be penalized for something that’s beyond my control. My code is not even close to the ones it’s being compared to, yet I’m facing the possibility of penalties that could significantly affect my time, efforts, and rating.

I understand the importance of maintaining fairness and integrity in these competitions, but I assure you that my intentions were honest, and I have always followed the rules. It’s upsetting to see my hard work and dedication being questioned due to circumstances that I didn’t create or contribute to.

I request you to please reconsider the decision and review my submission carefully. I’m confident that you will find no malintent on my part. This issue not only affects my rating but also demotivates me from participating in future contests.

MikeMirzayanov,Vladosiya, I know you might receive many such messages, but this truly affects participants like me who put in a lot of time and effort to compete. If this issue cannot be rectified, it may lead to many users, including myself, considering a shift to other competitive programming platforms where such occurrences are handled with more nuance and fairness.

Thank you for your understanding.

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

satyam343 sum cry

Could you guys consider reviewing my C solution once again please? While I saw the similarities and won't deny their existence, you can check my unofficial submissions to see the Binary Search on answer codes I have been writing before which also match the syntax used in that particular program. The only difference is using auto instead of my normal loop (which I had used a day before on some spoj and leetcode problems). Rest is just similar logic for binary search because that's how I have been writing binary search till now.

I have been accused for similar submission in this one particular problem before too however that's not the actual case since it's just the nature of the problem that gave similarities.

The template used is unique to me as well. Also, I have never got any skipped solution before this (never cheated before) and you can see my rating graph also looks normal with certain ups and downs unlike those of cheaters who go straight up. There have been contests where I have solved just 1 problem due to poor temperament as well which wouldn't be the case if I cheated my way through here. I am aware of the fact that past records don't really prove that a person can't cheat in future but they might help you confirm if my growth is organic. I'd request you to review it again in an unbiased way (Not calling you biased but from what I have observed people on this platform jump to conclusion too quick once someone is accused, not realising the difference between accused and convict).

I am also ready if you decide to get my account banned if genuinely found guilty. Another thing I would like to bring to your notice is that solution got a lot of similar submissions. Now while this might seem an evidence for the counter, I would want you guys to notice that only 2000 people solved it in the contest. So was it like 60-70% people cheated in that? And if it leaked at mass scale then why did it have just 2000 submissions while other leaked codes have 10k+? Again not a statement just an open ended question.

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

Dear Admin,

I have received the notification regarding the significant similarity between my solution and those of other participants for problem 1998C. I would like to address this issue as follows:

Independent Work: I developed my solution independently during the contest without any external assistance. I did not engage in any form of collaboration, nor did I share my code on any public or private platforms before or during the contest.

Common Approach: The approach I used in my solution, which involves binary search and sorting to find the optimal value, is a well-known technique for such problems. It is possible that other participants arrived at a similar approach independently, leading to the resemblance in our code.

Commitment to Fair Play: I am fully committed to adhering to the rules and spirit of fair play in competitive programming. If there has been any inadvertent error on my part, I sincerely apologize and am willing to cooperate fully with any investigation.

Request for Review: I kindly request that you review the situation thoroughly. If there is any evidence that points to a common source or method that might have led to this coincidence, I would appreciate knowing the details so that I can understand how this occurred.

Thank you for your attention to this matter. I am ready to provide any additional information or clarification if needed.