done

как тестер скажу что этот контест высокого, как бутылка на фото, уровня

thanks, fixed

there is no contradiction in this, in any case, if we take any path of minimum length, this will be the upper bound

In the Input format: "Each of the following $$$q$$$ lines of each set of test case contains three integers $$$a$$$, $$$b$$$ and $$$k$$$ $$$(1≤a,b≤n, k≥1)$$$ — the next question. It is guaranteed that any path from vertex a to vertex b contains at least k edges" this is enough for upper bound

will be fixed soon

Однако неожиданное предложение, я с радостью его приму. Как смотришь на то чтобы пойти играть в Counter Strike 2 через минут 30?

Молодец Тагир, так держать! Уже несколько недель слежу за твоим прогрессом, ты не перестаешь удивлять!

Привет, Тагир! Желаю тебе успехов на codeforces.com и новых достижений! Удачи!

if this round didnt have samples, nothing would have changed

no, but you can dm me and ask for solution of any problem you want

omg, how i haven't seen this gold from perforator yet, thank u so much haZZlek bro <3

увы нет, ведь можно использовать абсолютно идентичную идею

только вместо дерева Фенвика используется дерево отрезков

задача D подозрительно похожа на задачу 1946F - Никто не нужен

it is not working that way lol

yes, it seems that they just generated 300+ random tests and thought "hmmm, maybe on some test the square will get a time limit")))

так less_equal компаратор вообще сломанный, например вот, лучше юзать less<pair<int, int>> вместо less_equal, второй параметр пары будет просто количество элемента и его можно например в мапчике поддерживать

bro 3rd time at EGOI and 3 attempts left is crazyyy

alexchist orz

you are right! fixed

правильно, туда его, пусть идет решение читает

actually one of our testers wrote this solution and it can be implemented easily without any handling any corner cases click

thanks! fixed it

thanks! fixed

added

in F we have $$$O(n \log^3 n)$$$ solution that works in about 6.5 seconds with proper optimization, so in order to cut it we made TL 6s and big $$$n$$$ and $$$q$$$ limit (to mention our $$$O(n \log^2 n)$$$ solution works in 1.6s without extra input optimizations and stuff like pragmas)

probably fixed now

ауф

поплачь хз

bro that joke got me dying never joke again pls

bro look at your profile picture, i can find it googling "cute авы стим аниме", you can't be taken seriously XDD just go chill somewhere, no need to litter here

bro go f yourself idk, f toxic

every cycle has length $$$\le \sqrt {2 \cdot 10^5}$$$, and the average lenght is $$$\frac{\sqrt {2 \cdot 10^5}}{2} = \sqrt {5 \cdot 10^4}$$$, so i think it will be something like $$$\sqrt {2 \cdot 10^5}! \cdot \sqrt {5 \cdot 10^4}$$$

a complete graph on $$$\sqrt {2 \cdot 10^5}$$$ vertices with about $$$2 \cdot 10^5$$$ edges, number of simple cycles will be something like $$$\sqrt{2 \cdot 10^5}!$$$ i guess

thanks!

норм вообще с кайфом

The last problem have a small idea, so maybe you should try it)

I have a contest, but all statements are in Russian, I will translate it soon with GPT: https://mirror.codeforces.com/contestInvitation/c2f6e93cf495684485aa4281001d484abb2f824b

ты считаешь что авторы не смогли бы отсечь перебор за 2^30 * 15? мне кажется они не пытались особо, можно было бы сделать n и m на 1 больше и tl поднять в два раза, при этом перебор бы уже работал намного хуже. или сделать отрицательные числа и тогда легче было бы намного строить тесты на которых перебор работает долго или неправильно

куб в C2 я тоже не понимаю как не смогли отсечь, даже если там условно n^3 / 27 то это все равно 5e9, я не понимаю как такое смогли допустить, сделали бы n <= 7000 и подняли бы ml в два раза то уже было бы намного лучше, 1.5e10, уже даже с рандомными тестами это бы не зашло

в чем же он не прав друг? тесты действительно были ужасные и заходило то что не должно, в C1 рандом а в C2 куб, кажется такого быть не должно. возможно слово "везение" нужно заменить на "умение загонять то что не должно заходить", это максимум. понятно что сильные участники и так прошли, но кто-то загнал то что не должно заходить из-за слабых тестов и поэтому прошел, забрав проход у людей с таким же баллом(кроме того что у них не зашел куб на 100), только наифавшим больше подгрупп в D

Красава раскрыл злой умысел автора

Полностью согласен! Нет буллингу серых!

Wow, thanks a lot, that's really creative!

Can you please try mine

it was until yesterday)

good bye! <3

Согласен

Брат я в тебя верю у тебя все получится главное просто частички понабирать по задачам и все выходишь на рег на чиле лютейшем

1900A - Заполните водой

UPD: теги убрали(((

хз подумай

bro your profile picture is insane

автор я твой фанат

this comment really lives up to your profile picture

as a tester, this round is awesome

насколько я понял,bleeding хотел сказать что ты написал полный бред

я согласен, пожалуйста donovilia2007 не пиши больше ничего, не позорься

wow that is insane, i know another example btw

bro your nickname is awesome

I dont know how to do it in general, but there is a post that compares your rating graph with some others person rating graph, you can see it here

wow, that is insane coincidence, i believe that they are not cheaters and this chance 0.1337228% is just a coincidence and also when they solved B first on the same minute and then A on the same minute too i believe this is just a coincidence too, just a coincidence, just a coincidence

ternary search does not work because: 1. too slow (log^3) 2. wrong because the function of the answer is not not strictly increasing

No.

well done dude very good joke!!!

poor how are you going to solve the problems of 1300 implementation((( i hope you wont die while you write 20 lines of code

Russia was not allowed to make it interesting to watch the first 4 places?

div4

i have my point of view, u have yours, so its ok i guess that we are just discussing, but if u dont want to talk anymore, thats ok, good luck

"You are just helping my point :) you sent a GM lebel problem and a LGM level problem, which was exactly my point" — those problems were in div2 and some people solved it. "Congrats i guess, you find stuff trivial this master found hard" — i dont see anything wrong with it. "Also congrats on learning guesses dont work always, just like it shud be. If you had some actual problem solvimg skills instead of complaining, you wud have solved div1A fast instead of guessing and rage quitting" — i wasnt even raging, idk were u get it. I am 100% sure i would solve div1A, but i dont want to solve problems if I dont get positive emotions from solving them

u want to say that problems are ok and i am just crying that i couldnt solve it, no, i dont care about it, i just dont like the problems thats all, if you think that contest feedback should be only positive, i am so pity for you

"how exactly do you intend to find problems which have 5 difficult steps in div2. Your hopes are quite unreasonable in a 2hr long contest unless youre GM+" a lot of div2E-div2F with quite few contest solves have 3-4 big steps in its solution, for example 1797F - Li Hua and Path or 1805F2 - Survival of the Weakest (hard version).

"Just because they are trivial to you doesnt mean they arent multistep" — thus every problem can be called multistep, for me those problems have 1 big step and then they are easy

maybe for u they are multistep, but i see only 1 big idea in every problem

"In B and C, tell me exactly which idea felt to you like this?" — B is trivial; i wasnt able to solve C, when i got wa2 i threw the contest in the trash, but i dont really remember(fortunately) what i was writing, i think some peace of shit with no proof, oh no, with "its easy to see".

div1B has two subtasks, one is note that if k is odd than answer is 1(k = 2 is actually trivial case), and one is for even k >= 2

in div1C as for me key idea is "Observe that for every u, there are only 2 possible different values for fu,w", after noticing this fact problem is trivial

i dont care about my rating so i don't hate contests because i lose rating on them

also i dont care about people opinion because the crowd is the crowd, the crowd does not have its own opinion, they see -100 on the post and immediately put a dislike without even reading it

as for me, greedy problems are complete shit, because i think its not ok when the hardest part of a problem is to guess one fact(almost always without proof) and then write a few lines of simple code that magically work

combinatorics problems are complete shit because its just math formula but you have to code it

i think that cp problems are something different where you can come up with a solution step by step, not just using one idea, its not interesting to sort through ideas and hope that the next unproven fact will get AC, its interesting to come up with a solution step by step, first n^3, then n^2, then n*log while using several ideas

in this contest unfortunately i didnt see any such problem, maybe div1D or div1E are interesting, but these tasks were not in div2

1. I was talking about div2
2. ABC — greedy, D1D2 — combinatorics is math if u dont know
3. E — optimization is based on "note that" => greedy

100+ people fail to look at the tags

do the authors know any topics except greedy algorithms and mathematics?

damn you're right

what else can you expect from anime fans XD

I don't trust your nickname

Правда что задача D1B Мясник была названа так, чтобы отдать дань уважения nn русскому репу, а именно в честь трека Мясник который вышел за день до раунда(https://music.yandex.ru/album/25431536)

I know, but in fact, the number of different elements added was very small, O(logw), so it should still work fast enough, in O(1)

$$$n log^3$$$ with unordered_map constant got TL in E(((

D: For every index $$$i$$$, let's see all such indexes $$$j$$$, that there is exists such subarray of $$$a$$$ $$$[l; r]$$$ of size $$$k$$$, that $$$i$$$ — $$$l$$$ = $$$r$$$ — $$$j$$$; for index $$$i$$$ let's call all such positions $$$j$$$ good.

We can see, that for all odd $$$i$$$ good positions are odd too, for all even $$$i$$$ good positions are even.

For every index $$$i$$$ we have to count $$$z$$$ = number of non-equal to $$$a_i$$$ elements on good positions, to do this we can count $$$x$$$ = number of good positions and $$$y$$$ = number of equal to $$$a_i$$$ elements on good positions, $$$z$$$ = $$$x$$$ — $$$y$$$.

Let's build two Merge Sort Trees, one for elements on odd indexes and one for elements on even indexes.

The number of good positions is just some simple math, the number of equal to $$$a_i$$$ elements on good positions is just one Merge Sort Tree query (if $$$i$$$ is odd, query to first MST, otherwise to the second).

Answer is (sum of all $$$z$$$ for all $$$i$$$) / $$$2$$$.

Time complexity: $$$O(n$$$ * $$$log$$$ ^ $$$2)$$$.

My solution: 199667931

lol

smart boy

I hope someday I will also become a specialist

speedforces

Б

чел хорош

omg i read it just 3 weeks ago

very interesting story, thanks!!!

Здравствуйте! У меня на сегодняшнем раунде упала задача D(div 2) на 13 тесте, хотя 13 претест она прошла. После контеста отослал такой же код, и задача зашла. Посылки: 148606610 и 148579156 Можете пожалуйста перетестировать

Good contest and very good balance, ths

ou yeah, another trash round, bad pretests C + no balance

Very good editorial and problems as usual. Ths

if this is really the case, then I think it's worth making the round unrated

It was necessary to try to make all three qualifying rounds of the technocup as terrible as this one

Good problems and contest) Nice problem B btw

trash round bad pretests C + no balance