Same here, I feel like our individual hard work has finally paid off. Good Luck for purple. :D

Wow! You have a nice participation graph — you are the first person I see that participated in 1970. How was it back then? :)

same :)

I guess vovuh has betrayed div3, Curious why he is staying loyal to educational rounds xd

Somewhat relevant meme:

SupaHotFire oTZ

They be like:

downvote him, guys.

You will be given 7 problems and 2 hours to solve them.

yea div3 after some times

Me too!

No, you will still have 2 hours to solve the problems. Penalty time is just a counter used as a tie-breaker for contestants with the same number of solved problems. The less penalty you have, the better.

Better Practice for you Tho...Lol

Didn't seem that hard. It was okay according to Div. 3

Nope :(

Why?

Why? It was my idea of the problem. I found this problem to be a rare case of not a math puzzle, but rather a programming problem (and not on standard algorithms). I think it is okay. The only issue: we didn't guess its correct position in the set.

[name changed]

the name before: cmii02

B->A

Weird flex, but okay

Congrats. We based on the results of the testers. I am glad that the round was quite suitable for most of div3 participants.

it wasn't that easy for me :(

Well, to be more precise

Ranking by the amount of time I spent + penalty time

D(44) > C(31) > G(28) > E(19) > F(8) > A(5) > B(4)

logic? or how you decide that x gifts can be made?

If we take $$$N$$$ gift sets of type 1 ($$$a$$$ red, $$$b$$$ blue) and $$$M$$$ gift sets of type 2, then we want the maximum $$$N + M$$$ constrained by $$$Na + Mb \leq x$$$ and $$$Nb + Ma \leq y$$$, which are two lines. The optimal point is at the intersection of those two lines. So we get an $$$\mathcal O(1)$$$ solution.

Submission

Same with me.

Except that I quickly decided that I'm too stupid for math and started to write binary search from the start

I solved G with a logic/math solution similar to neal's first ac submission (he resubmitted with binary search uh oh) which is $$$\mathcal{O}(1)$$$. I'm not confident that it is correct (bcz I suck at proofs) but it seems to intuitively make sense. My general approach was as follows:

Assign $$$\min(x, y)$$$ to $$$x$$$ and $$$\max(x, y)$$$ to $$$y$$$. Then, attempt to balance $$$x$$$ and $$$y$$$ by subtracting $$$\min(a, b)$$$ from $$$x$$$ and $$$\max(a, b)$$$ from $$$y$$$ until either $$$x$$$ and $$$y$$$ are balanced or $$$x$$$ has been exhausted. Then, evenly subtract the rest from $$$x$$$ and $$$y$$$ by alternating $$$a$$$ and $$$b$$$ (starting with $$$\min(a, b)$$$ for $$$x$$$ and $$$\max(a, b)$$$ for $$$y$$$). My solution code is below.

pls hack this or smth

Does anyone else have this solution?

can you give some hint for F

What is wrong with D? Probably, you just have bad implementation. My solution is short and clear, no corner cases. You see it is part of the programming skill to write in such a way that there are no corner cases and huge code.

G is also my problem. I think it is good) Ternary search, probably, is wrong in this problem.

What's wrong with case analysis? The fastest solution for D runs in 30 ms, how can you say the TL is was strict?

But it would be nicer if the TL was 4-5 seconds or if ${T} was decreased to 1000 from 10000. This would allow solutions in slower languages to pass comfortably, without compromising the core idea.

It can be tempting to say a round isn't very good because you found it difficult, or what you tried didn't work. A good round has a range of problems of different difficulties, not 7 easy problems that anyone can get. I think this round was good, because it had interesting, and in some cases unusual, problems, which were varied in difficulty.

I struggled on G too because I tried ternary search. I got it wrong, and then had a different idea which worked. That's part of the fun.

D I also got wrong initially because I missed a crucial pruning insight. Again, that's part of the fun.

There are only two cases to consider to get both numbers a and b to the same number c: if operations are too much or little...

longlong

1. binary search

binary search for the good range of all numbers >= a[i] excluding a[i] for each i

just sort it and use lower_bound and upper_bound for l — vec[i] and r — vec[i] for each index to the right

Binary Search

You can also use a 2 pointer approach.

I don't think so

I precomputed the primes till $$$\sqrt{10^9}$$$ and then iterated over it code

Use wheel factorization.

Store all prime numbers upto $$$4e4 (sqrt(1e9))$$$ there are about $$$3000$$$ of them, then check only these numbers atlast check if there is a prime factor greater than $$$4e4$$$ (there can be atmost 1).

First, we need primes <= sqrt(1e9), we have approximately 4000 primes You can do it by sieve. Then you can use those primes to factorize. Like this: for prime in primes, while(num % prime==0) { num/=p; power++;} do it until prime < sqrt(num). In the end if num > 1 then it is also a prime. Max count of prime can be approximately 13 for a number < 1e9. For each prime power count can be at max 28. So it will pass.

We need store 3 first and 3 last character of each string, then merge them easily

Registered yesterday as an Alt you mean. Well, I'd guess it's because you're not a Trusted Participant, but you must have read that from the announcement. But I guess you missed it.

Because long long eats up memory stack, you are not supposed to use long long everywhere, use it only when you suspect an overflow.

My condolences.

Do you even understand what you're saying?

Around O(2 * (sqrt(n) + log2(n))) for each test case.

If there is no need in long long use int because long long is slower

Task 2 (Vepar)

Your code got hacked due to Arrays.sort() function which in worst case have O(n^2) time complexity.

Instead of long long use int

Got accepted in my case

missed the condition when a and b are equal and k!=1. My bad :(

use int instead of long long.I faced same situation

When K=1, only when a%b or b%a is zero, the answer is yes.

I think every error after the first test case counts as WA submission.

System testing hasn't started yet. First that'll happen and after it's finished the ratings are updated within the next hour or so.