I did not think of going in reverse. I get it now. Thank you

same. couldn't AC. frustrated.

bros personality fits his name

I do not think it is fair that some authors simply deny responsibility with the notion on that, "I set problems that I like, I do not care about AI". In the end, you're setting problems for people on this website to solve, and most people here are affected by the fact that there are cheaters. Regardless, the amount of hate that you're getting is not justified at all.

I think it's unfortunate that there are barely any (visible, at least) efforts to combat cheating. I strongly believe that one reason for this is that top-rated people are not affected by this as much as people that are lower rated. I also believe that a website, that has so many, extremely talented engineers can figure out some way to combat cheating. Codeforces needs to adapt to this new era of AI. It cannot remain the same.

Did it get better, or did you reach CM despite experiencing all that? My experiences are similar to you and I believe it is hindering my growth a lot.

excited for the only thing that makes me happy, good luck to all

yeah, can you share?

"Stretch the tree about 2 nodes" idea works very nicely in 979C — Kuro and Walking Route.

Did the authors mistakenly perform B^=C, C^=B, B^=C?

You stand by your name

A suggestion for manual review -- Anyone is able to report, but only trusted testers/problem setters/high rated people can review. Similar to how there is a stream button for only specific people on CF.

There may be many accounts reported, and not enough reviewers, but I think that's okay. Even if only few accounts get banned, that helps.

If we implement many systems (like the ones in this post) and each of them help to certain extent, I'm sure the problem of cheating will go down significantly.

This can be used to solve PizzaForces in a fun way.

Any integer N > 1 can be represented as N = 2*a + 3*b, where a >= 0 and b >= 0. Extending further:

=> N = 2*(a-x) + 3(b-x) + 2*x + 3*x

=> N = 2*p + 3*q + 5*r

If we write 2*p as 4*l, the only number that we lose is N = 2. That is, any number N > 2 is also of the form N = 3*q + 4*l + 5*r.

Which can be used to solve the above question.

what was the point of F

Edit: Fixed. Mistakenly wrote rep[i] instead of find(i) at line 84.

Can anyone tell me why my E fails (on only 3 test cases)?

fr. i keep confusing x with x-coordinate and not the x-th row in my mind

i honestly could not figure out A and just guessed it

I mean, he has only given one Div. 3 and solved 2 questions there. I would not call that smurfing

give me a referral :p

Very few people in this world are born with gifted intelligence. Then why can there such a vast difference between two people in the ability to think critically (/solve codeforces problems)? It's because of life experiences leading up to this point.

Intelligence = Experience + Genetics. Someone who has spent some part of their childhood/teenage playing chess, or doing maths, or solving puzzles like sudokus, or maybe preparing for a competitive exam/olympiad that has logical subjects like maths/physics is certainly going to be better at solving logical problems (or picking up the ability to solve logical problems) than someone who has not had these experiences.

In fact, research says that people who have a lot of hobbies/know a lot of skills (like art, sports, etc.) are more intelligent than those who do not.

So, it all boils down to your experiences until your current age + some amount of genetics.

What's up with C? Why is removing the highest degree node, updating degrees, and then removing the highest degree node again wrong?

This has been fixed. Please try now :)

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Just made my first new year submission. Hope it turns to happy new year soon!

One the authors already apologized (editorial / announcement comments, don't remember where), and I really don't think that it should become unrated — that would be very unfair to those who solved on their own and got positive delta :p

As a pupil, I did solve C

Can you explain further? Did you do this on pen and paper (seems like a daunting task), and if you did this via code — did you calculate S(P) for all of them manually?

patternforces

i am so dumb

how to do D

happy birthday! :)

This is amazing. Thank you!

i feel like i will have to build one myself

Wait till it happens to you and you'll understand

But why would programmers join Refact.AI to build something that takes their job?

Right, my bad. Anyways, I still think that it is a bad choice to put up questions whose 70% logic can be copied from an already-existing source on the internet.

If you had a little bit of brain, you'd try to think about what I'm inferring in my comment. The point I was making is that there's often a lot of cheating when popular pre-existing problems are put up in contests. That's all.

D was extremely annoying. It shouldn't have been a part of this contest, especially when spiral matrix (on leetcode) is such a well-known problem.

Very true. It's very hard to remain calm and "think simple" during contests. I always overcomplicate under pressure

It can be done in Nlog(N). Check the implementation above. You can store v[i] + v[i+1] in an array, then to find the first pair with ai + ai+1 > mx, just find the upperbound of mx in this array.

My approach:

It is sufficient to ensure that a1 + a2 > mx, where a1 and a2 are the two smallest elements of the array, and mx is the largest.

We can change the maximum element of the array. So, sort the array and iterate from the end. Let's stay mx1 = largest element, mx2 = second largest and so on...

For each mxi, we find the number of operations required such that a1 + a2 > mxi. To do that, we find the first pair ai and ai+1 such that ai + ai+1 > mx, and number of operations is simply i - 1, as we must change all elements smaller than ai.

The answer is min(op1, op2, ...), where opi = minimum number of operations for mxi

real. it's so stressful when you keep getting WAs lol

unfortunately the world doesn't revolve around India

excited

congratulations dude! hoping to get to (at least) specialist before next year (placements) :(

I started from the beginning. The order of the first and last elements do not matter, so put them in any order. Then, while filling the positions (l, r), I compared them to the elements right outside (l-1, r+1), the position I filled in the previous step. This works, but I do not know why.

Previous to this, I tried swapping each position and comparing whether disturbances decrease. I compared l with both neighbors (l+1, l-1) and r with (r+1, r-1). I got a score out of 4. If swapping decreases this score, then swapping is optimal. Else it's not. This did not even pass the sample test cases. Any idea why?

Great contest, but sadly D is literally on leetcode

C++ comparator should define the < sign. Using this, we automatically get the other two cases: x > y = !(x < y) and x == y = !(x < y) && !(x > y)

Thanks! This knowledge is life-changing

There is an edge case when number of inversions in both orders is equal. In such a case, putting the pair with the largest number on the right gets AC. I don't know why though

Brother, the code you've written and the logic that you say are different. The code is correct, but the logic that you've written in your comment is not

pretests failed

Can anyone tell me why my Div. 2 C doesn't work? The idea: Pair (a,b) comes before (c,d) if at least two of these inequalities hold: a <= c, a <= d, b <= c, b <= d. Else, pair (c,d) comes before (a,b)

I made a comparator with this logic and sorted the pairs. Doesn't work

Your rating directly relates with your rank, and not the number of problems you solve. Today, your rank was exactly what is expected at your rating, hence no rating change. Yesterday, your rank corresponded to -10. Read more: CF Rating System

Are boarding schools common in your country? Are they strict?

Yeah true actually, that part is not clear from the statement. It's ambiguous

I think C was pretty clear, but that clarification for B was important. I personally believe it should've been included in the statement

Love the round timeline! It's really fascinating to learn how contests are prepared, and how much time and efforts it takes. Thank you :)

same, wasted 1.5 hours figuring out what was wrong

genuine question — how are scores for problems chosen? does the score indicate difficulty of problems with respect to each other?

bro is making life harder for no reason

funny how almost similar comments but the better color dude has much higher upvotes

Thanks! Increasing my understanding of this question comment by comment

i will never recover from B's trauma

Hi, was searching for exactly this, but the link is broken. Can you please share another link?

Same question

All of this while averaging ~10k rank in previous contests. Truly marvelous.

Yeah, but it also went from 50 to 89 in like less than an hour

Only problems that you submit while a contest is going on / system testing is going on have that sort of delay. The contest yesterday was smooth, and even system testing has been really fast today

they did warn us about the 10 minute penalty

Saw my first ever interactive problem, panicked, but eventually did it when around 3 minutes were left. Feels good.

How did you go from having 6k rank to 476 and then 11?

We have similar solutions, mine passed. I did not use a set though, just vectors

bro is faster than the flash

Please understand that the definition of fun is different for different people. If, for some people, fun means getting a good rank and having a good rating -- then regardless of what you say AI IS ruining competitive programming for them.

And if you think their definition of fun should match yours, then I don't know what to say

This has been said multiple times but I'm suggesting again: why not put phone number verification (one number = one account) as well as impose strict bans on accounts whose solutions have been skipped in, say, more than 3 contests?

This is not full proof by any means, but we should make cheating as difficult as possible. I'm confident most cheaters are not much motivated and will give up after 2 or 3 phone numbers get banned.

False positives will happen, but overall outcome can be positive.

I'm seeing negligence and over-reliance on AI all around me. University and school students in my country have started to heavily rely on LLMs to cheat. There is cheating in coding competition, hiring tests everywhere. I liked when people wrote code and built things by themselves, but no one is doing that anymore.

The advent and growth of AI makes me immensely sad, but this is the future people are building for themselves. It's depressing to me.

bro can't read

Nevermind. Understood

The function that we're minimizing is abs(quadratic). Imagine the graph of a quadratic — on taking absolute value, the part below x-axis goes above x-axis. Now, the minima occurs at one of the roots, because everything else is positive. Graph is symmetric around both roots, so we need to check one of them only.

However, the root may not be an integer. Say, the root of our quadratic is equal to 1.22. You need to check at both 1 and 2 for minima because your range, in this question, is integers only. That is why he takes min(f(i), f(i+1)). At least, this is my understanding of the solution.

In D, how do we prove that there will be no other cases? Also, can someone share how did they arrive at those 3 points (last case)?

I eventually realized that those 3 points made a right triangle during contest, but it took me a long time.

Agreed. Making it unrated when m1, m2, m3 were working perfectly fine seems unfair.

was having the best performance of my career and then i heard "unrated" :(

Typo in the editorial of B, it should be first character of the 'second' line

lmao literally, B was an absolute guess

This is what you're looking for

I don't understand. The time limit given in questions is 'per test case', then how do we analyze multiple test cases? Shouldn't O(n) for l = 1, r = 10^5 always pass in 1 second?

This is what I did too, although I ended up writing extremely messy code.

I don't think you've seen a lot of Div 4 As, because today's was certainly not as easy as this.

i swear to god B took more time than A,C,D,E

real

While phone number may be easily available, I do support this. Having this is certainly better than not having this.

Alright, thanks

Pardon my lack of understanding, but when you say overflow, are you referring to "mid" or something else? If it is mid, shouldn't writing mid = low + (high-low)/2 be enough?