Please share that glorious kompot with me!

as a participant, I look forward to it being one of the rounds of all time.

May God bless you with a positive delta in tomorrow's round!

No. IPL clashing with CF. Tell them to change the schedule xD

Only the brave coders can participate in a contest during an exam.

Me too

Finally reached pupil

Cf>RCB>Finals

It's probably interactive

https://en.wikipedia.org/wiki/Nikita_(given_name)

Exactly. Also a few youtube channels show the answers to the problems before the contest ends. It is unfair for those trying genuinely. I am not sure if there is any way to catch those who copied and those who tried on their own. Otherwise ratings of this contest will be false representation of ones capability.

I want the round to be unrated to dodge this -40 delta I'm about to get :(((

It's already hard to check for AI solutions and also it's very hard to come up with div2A which will be unsolvable by LLMs.

I think it's not feasible even now but in 6-12 months LLMs will be able to solve most existing div2A and div2B and some div2C so it will be almost completely impossible to prepare AI-resistant div2 contests.

I wasn't able to perform as well as others == The round needs to be unrated

$$$32$$$ is the number of bits, how you want to solve it?

Because you need to be smart enough to not use brute force, which took me some time to realise

Yes it's kind of dp, the main idea is that there are 2 options:

1. LCM of whole subset is larger than max(a) -> then answer is n
2. LCM of whole subset is <= max(a) <= 1e9 and then there are not that many possible LCMs, I don't have a proof but intuitively if you have a lot of different prime factors, then LCM quickly grows to be large and also that if you add a new number to a subset LCM either stays the same (does not increase amount of LCMs) or multiplies (so grows very fast).

So the solution is: check for option 1, if it's not the case just keep dp = map<LCM, max_subset_size> and add elements one by one (iterate over all previously calculated LCMs) and update dp

Balanced Ternary

x + 2x = 4x — x

$$$2^0+2^1+...+2^{n-1}=2^n - 1$$$

And yet knowing the solution of longest trip doesn't help you much

Basically, yes. You first calculate the LCM of all the numbers. If that number is greater than $$$10^9$$$ then the LCM cannot be inside of the array, so you just take all the numbers. Otherwise, you iterate over every divisor of the LCM (works in $$$\sqrt{LCM}$$$ time), check for each number in the array if it's divisible, and then return the maximum result. Here is my submission: Submission

Find LCM of list. If LCM > max element: answer is n.

Otherwise iterate through all factors of LCM.

If factor not in list. Keep track of cnt of items divisible by factor and LCM of said items.

Answer max(cnt of items for all factors not in list and where LCM(items) = factor).

• If LCM of array =/= largest value in array, answer is N
• Iteratively calculate LCM of array. If at any point exceeds 10^9, we know the above case is automatically satisfied.

Now, where LCM = largest value in array = L.

The LCM of the elements for our answer can be only factors of L. So, find all factors of L.

For each factor of L that does not appear in the original array, count how many maximum elements divide it. Ensure that LCM of these elements = this factor. If this is satisfied, find maximum across all count values.

complexity: O(N sqrt(max(A_i)))

i had a hashing soln but i couldn't submit due to small bug.
my idea is for each column compute all possible operation which can be made on i'th column  such that i'th column becomes special
there will be n such operation possible ops
pick the operation which occurs most of the times

Even I'm wondering how to D. I had some partial ideas , first of all take a particular column j , and make all it's elements zero , by choosing appropriate rows and inverting them. Then we can choose an i such that we will again apply inversion at the row i making this column good with 1 at row i. Now all the other columns , which had a 1 at i_th row and one more 1 at some other row will become good , but the main challenge seems to me is how to find them efficiently?

I saw your code and you have the right idea for small $$$m$$$.

For small $$$n$$$ you can just iterate over all the $$$n$$$-sized bitsets that make each column have exactly one 1, group them in a map, and return the bitset that maxes the result.

Code: 262803308

First convert x into binary form 10110100111.
After that fix all positions where you have 11.

if x is odd we can turn x divisible by 4 by either adding 1 or subtracting 1, and if x is divisible by 4 the next element can be 0 meaning there can always be 0 between 2 non zero elements

no, I didnt

many participants do cheat but the whole point of solving cp problems is to improve problem solving skills or to get dopamine boost after solving a problem which you don't get just by copy pasting code

Conest should be unrated

agree T_T

TBH this B = normal C, this C = normal D, this D = normal E. Tough round, but you need only 3 instead of 4 questions to get massive + delta.

I just submitted it after the contest my bad :)

My idea was something like this (I'm trying to prove it formally, but couldn't yet succeed) —

If the binary representation has $$$x$$$ digits, then my claim is that — in the given constraints, we can build the array in almost $$$x+1$$$ digits. I wrote a recursive function to implement this.

AC Submission: https://codeforces.com/contest/1977/submission/262782660

damn chat GPT is smarter than me

Is it enough to go through max(A) 's divisors only? I went through all divisors of all elements

How do you get the idea that if $$$n\mod2==1$$$, decide the next digit based on $$$n\mod4$$$ ??

Looks like there are many ways to solve C. Mine is exactly the same as your editorial lol.

262772972

Why do you care about rating? =)

WA is WA.

Ahh... bad luck. I used to have this feeling 10 years ago (feeling that such a small error should be allowed to rejudge, or why do I have to suffer for such a negligible mistake where my logic/thought process is correct), so I totally understand your frustration. Thanks for making me nostalgic.

Lesson: Error is error. 100% your fault. Own it, embrace it, learn from it, overcome it, grow from it.

Ok, so what I think is that people probably misunderstood you: You were saying that your solution was incorrect because of a small typo, but it got marked as correct during the contest, because of which you got a good rank which you feel like you didn't deserve, correct (instead, they assumed that you were saying that your solution got WA because of a small bug, and so you should get AC just because it's so small)? Well, I'm saying that you do deserve it: Just look above / below you, there are so many people with solutions that aren't actually correct for D, but they work on all the testcases. In the end, in competitive programming, what matters is passing all the tests, not whether your solution is actually correct (although for most problems with strong tests, those two are equivalent...). So, if you deserve your rating, you will keep it in the next contest, and if you don't, well you're gonna drop down next contest anyway, so it's fine :)

and getting three problem quickly is equivalent to master.

$$$a_i \le 10^9$$$

so how can the LCM be greater than the MOD and in the array at the same time?

Why didn't you just put a condition to check if lcm gets larger than 1e9 and output n in that case?

no, your solution can't be hacked since it is correct) you are searching for all possible LCMs only if LCM of elements is already in array, so it is less then $$$10^9$$$, and obviously all possible LCMs of any subsequence are divisors of global LCM so there are less than $$$\sqrt{10^9}$$$ possible LCMs hence your time complexity is $$$O(n \sqrt{maxA})$$$ (oh I see you find global LCM wrong way and solution already got hacked :(( )

The tests are weak indeed. My solution clearly fails this test case

1
9
2 3 6 7 1003 1003 14042 21063 42126

Answer should be 5

My prayers were not unheard , they increased 96, it was +75 before Happy happy to reach 1700