I did not learn anything from Grzegorz

Where does acquaintanceship end and friendship begin, and isn't any individual related to any participant country (i.e. by citizenship) "interested in the outcome" of their delegates?

I think you didnt reply where you needed to

I agree

Very cool blog! I think (or at least hope, the details are kinda funky) this problem can be solved with this technique, maybe pushing it to the very edge.

Infocup committee has a hobby of unnecessarily obscuring the SC process, and for that reason results will appear much later

Why bother posting on (international based) CF and encouraging people not to use AI when you only give statements in chinese? It takes like 10 minutes to machine-translate a problemset.

Would you "especially appreciate" just as much some hard tasks that can't warrant a much more easy solution for 75 or so points?

Note that we will not be accepting problems from authors who are still attending high-school to avoid potential conflicts of interest.

Wishful thinking. Also ;(

these were the hints:

Systests = Pretests.

Kid named joke:

Why are they asking the guy named "soulless" for mercy?

Postpone el clasico

bleahbleah

jmyszka?

Constructive Olympiad in Informatics.

Cauta-ti de lucru.

Nice blog! I feel as though the warning is a bit superfluous..

I was opening a lot of accepted solutions to problem D, as I had already managed to hack a solution.

First mistake. Karma.

Where even were you last year man :(

The series doesn't converge. This is an abuse of notation brought on by the analytic continuation of the Riemann Zeta Function, which normally is well defined only for values pf n respecting Re(n)>1. As such, the only valid assumption that can be taken is that there is strictly more than one real problem in the problem set.

Very nice!

I think <20 deserves more points than it actually does (with regards to optimized 30 bits which leads to like 55).

I should delete this account...

Why does JOI every time have such mean grading criteria? What was the point of having such a slow ascent from ~60 bits to ~20 bits in fortune telling?

Same with all indian colleges, MikeMirzayanov please don't help them and don't resolve this.

bunp

Upvoted.

Is this same guy his single-serving friend?

Oh true. Who would've thought thinking before speaking is a good idea.

I think for the Mo problem you can use the second idea listed here (i.e. replace the segment tree which admits online O(log)/O(log) for sqrt decomposition which admits O(1)/O(sqrt) update/query)

MIHAI PATRASCU MENTIONED 🇷🇴 🇷🇴 🇷🇴 🇷🇴 🇷🇴 🇷🇴 FIRST EVER COMMUNICATION PROBLEM IN IOI 🇷🇴 🇷🇴 🇷🇴 🇷🇴 🇷🇴 🇷🇴 WHAT EVEN IS LR BINARY SEARCH 🇷🇴 🇷🇴 🇷🇴 🇷🇴 🇷🇴

Inlove hingaria

Are you hungarian?

How can you conceive splitting the E in this economy?

Propaganda

CF Round 804 problem B

You are welcome. Happy new year!

Codeforces indeed has emojis 👺

Thank God. Now Codeforces is a more secure place

Ok

Although I might certainly never be close to winning one, can you consider issuing the prizes in some other stablecoin?

Is this a real life?

Only after ~73 top 500 performances do you have the same chance of not winning a t-shirt as you have the chance of winning a t-shirt with one top 500 performance 😓

There is also a deterministic solution in $$$2 × log_2(N * 2)$$$, which uses the same means to determine degree, only that we'd query for how many neighbours have some bit on. Then, with this information, each leaf could only point to one other node, so we could get all the edges by removing leaves one by one.

Origin of problem D

Here apparently

If you were to put money on a match then you'd now be really interested in that match.

Gambling

And what site might be this? (I can't access the izho.kz and recently I started getting a frontpage about some PS Cloud Services)

E-scooters

Down

When will it be officially announced ;(?

When is IZhO

Please reschedule, it's my birthday

One really should find their own. But I think these help greatly.

First step would probably be to close codeforces in this search.

No one needs to solve thousands of problems, just the right ones.

Ah nvm im clown you're refering to blog problem

https://mirror.codeforces.com/contest/786/submission/289123812 ?

How are third type updates handled?

When I got 10 I didnt account for cyclic shifts :(

my mistake, when I divided 2024 by 50 I got 400

for 15, how do you solve in 39+eps moves a graph which is like a star, but every edge is a 49-node chain?

When will results be published?

Is this like IGO?

Because its retarded. Deque needs like 2 kilos on its own to prepare for a lot of the shenanigans it can maintain (i.e. random access in constant time). All its subderivates also have the same thing (i.e. queue and stack). So try your best not to use them if you have a lot of them instantiated in parallel.

Should we stop him while ae have the chance?

Don't add context if you do not know any.

I will get politically tried and persecuted if I post on main

what do you want to hear?

It seems as though this was the last day (because its 15th October 12:28 UTC).

Hey! I called him pedantic first >:((

When will I get my first bouquet of flowers?

What are you trying to imply :(

Damn, I wanted to upvote but missclicked downvote. Sorry man.

❌

Can you predict mine please

Sedmoklasnikut orz

I love Albania, greetings from Romanian brothers 💕

Bump.

No? One has a — and the other not.

You are missing the point, as most pedantics do. And in doing so, you will fail to see why it is this life you hold so dear is worth fighting for. Pray tell, when did a pretty painting ever bring end to famine? Never. Yet we still do it, although it has zero practical applications. Makes you think..

Wait until you hear about a domain called math

Second problem (of second day) looks like it might be from atcoder..

No. But check this page for more such details.

Google Cross Product:

The thing about concavity is that if you go one way, by turning left, one time you might end up turning right. This is because in a convex polygon, all adjacent vertices vectors are traversed by angle while traversed in some order. Concave polygons do not respect this, the order of the adjacent vertices vectors have their angle spawned randomly across the unit circle. As such, to adjust from one adjacent vector to the next, you might have to rotate both clockwise and counter-clockwise (assuming you rotate only with <180 degrees). If you use both, then it means that it is not convex.

No; During a Div1/Div2 round, all participants under 1900 participate in Div2 and all those above participate in Div1. Candidate masters can also participate in Div2 only rounds, which have an upperbound of 2100. But because Div1s always have a Div2 along with them, it is no use (and more harmful) to let people of lesser rating participate there in the stead of Div2s.

Andrei_ierdnA and RaresFelix will win IOI24! gold in my stead

Have a look at the headlock here

Sir, wrt to blogs, this is a 40% shitposting website, 40% spam website and 20% actual CP website. Probabilistically speaking, so very few people ever add anything meaningful to the conversation ;)))

Thank you for your innovative idea.

I clearly am doing something wrong while writing code. My question is whether anyone can tell me what :pleading_face:

Personal story:

Today I met a most wonderful person and had a very beautiful interaction with them. They told me "The weather sure is hot, right?", to which I responded with "Yeah".

In fact, there is a problem the same as this one: link

CodeCodeCodeCodeCodeCodeCodeCodeCodeCodeCodeCodeCode CodeCodeCodeCodeCodeCodeCodeCodeCodeCodeCodeCodeCode CodeCodeCodeCodeCodeCodeCodeCodeCodeCodeCodeCodeCode CodeCodeCodeCodeCodeCodeCodeCodeCodeCodeCodeCodeCode

Do the visible rankings already take into account the age limit?

Ok time to add something meaningful to the conversation.

Although the trick detailed can be easily dismissed as unrolled queue-like undoing, the problem linked poses the same question but for 2D spaces. And so I would like to detail this scaling:

The trick here, which can be applied in the 2D case, is to assign some reference points for which we will calculate some information which will be necessary and sufficient for completing an entire query. For triangles, given an interiour reference point, we can divide the triangle by drawing lines parallel to the axis through the given reference point, and as such we divide our query to require only information for a: trapeze in NW/SE quadrants, simple cornered triangle in NE, and rectangle in SW, all computable in linear time wrt the dimensions required for a query, which is of the order of $$$k^2$$$ (sufficiency?)

Furthermore, to assure the necessity? of the construction, we can calculate only for gridpoints which have both their coordinates divisible by $$$\frac{k}{2}$$$; as such, any query has a reference point within it.

The complexity is, therefore, clearly linear.