China

56 300s.

Team Slovakia:

• Eliška Macáková (prvocislo) — 4th time at IOI, 0 attempts left

• Viliam Gottweis (viliamgott) — 1st time at IOI, 2 attempts left

• Martin Šindelář (IMackerI) — 1st time at IOI, 0 attempts left

• Patrik Prítrský (pritrskypatrik) — 1st time at IOI, 1 attempt left

Surprisingly, each of us has 6 letters in the first name and 8 letters in the surname. Our IOI statistics page is going to look super neat :)

Greetings from our team! Share your team photos too!

P.S: Guess who is who :)

This is in meme territory.

why put B in contest, nice ACD problems though

Croatia:

• Jakov Celin (celin) — 1st time at IOI, 0 attempts left
• Dino Hadžić (DinoHadzic) — 1st time at IOI, 3 attempts left
• Lovro Tunjić (ltunjic) — 1st time at IOI, 2 attempts left
• Nikola Vujica (nvujica) — 1st time at IOI, 1 attempt left

Sorry for repeating this for the third time, but I think I have not received any solid objections yet. Why can't we make Div. 2 partially rated for Div. 1 contestants? For example, it might mean that, if you are not in Div. 2, your rating change is lower: specifically, if you are rated $$$2100+d$$$, your rating change is multiplied by $$$e^{-d/400}$$$.

• What if the problems in Div. 2 are too standard for Div. 1? We can let the coordinator decide: every Div. 2 can either be partially rated or unrated for Div. 1 (also, the constant $$$400$$$ can be changed).
• This modification clearly makes the Codeforces rating not zero-sum, but as far as I know it's already not zero-sum.

About some contests I've coordinated:

• Codeforces Round 906 (Div. 2) was initially Div. 2 only, and it was in queue from more than 2 years. I noticed that 1890F - Game of Stacks was too good (and hard) to be used in Div. 2 only, and I asked the authors if they had some Div. 1 problems. During these 2 years, the authors had actually invented some Div. 1 problems and we made Codeforces Round 906 (Div. 1). If we had no Div. 1 problems, the contest could have safely been partially rated for Div. 1. In fact, 1890E2 - Doremy's Drying Plan (Hard Version) is rated 2600 and 1890F - Game of Stacks is rated 3000.
• If we replaced problem D with something less standard, Codeforces Round 917 (Div. 2) was suitable to be partially rated.

Of course, this can be generalized to Div. 3 and Div. 4.

[Unofficial rankings]

Please add your scores in this spreadsheet

Team Cyprus:

• Theofanis Orfanou (Theo830) (4th time, 1 attempt left)
• Antreas Knais (Trumling) (2nd time, 1 attempt left)
• Antreas Leonidou (ALeonidou) (2nd time, 0 attempts left)
• Mariza Paspalli (mariza_CY) (1st time, 4 attempts left)

It's clear as day that he was banned for contribution farming. Sadly, reporting cheaters wasn't his only blogs/comments activity.

No commment.

Guys I totally did not write a random solution for C and got either only st3 correct or both st2 and st3 correct, and in the end wrote a sol for st1 specifically!!!

There were no dependencies set !!!

Asia-Pacific Informatics Olympiad

Asia-Pacific Graphs Olympiad

Anyway, I think problems were cool

well-balanced div3 round, GL for all participants 🏆

G was pretty similar to this problem: 1526C2 - Potions (Hard Version)

Just too complicated

Movable rated boundaries make much more sense than this partial rated thing. But the issue with movable rated boundaries is its very subjective.

Sweden's team:

• Theodor Beskow (TheodorBeskow) (2nd time at IOI, 0 attempts left)
• Anastasiia Saranchina (qwusha) (1st time at IOI, 2 attempts left)
• Victor Vatn (goben) (4th time at IOI, 0 attempts left)
• Sofie Fu (slushtre) (1st time at IOI, 1 attempt left)

Good job by Mike and Co. Clearly he was farming contribution with cringe blogposts and comments.

In this round, there were three problems based on my ideas: 1974B - Симметричное кодирование, 1974C - Красивые пары троек, 1974F - Игра с отрезанием.

In short, you can't see the activities which were aimed at farm that I meant, but I started to notice them before he delete it.

(For more obvious evidence, you can compare his activity it to much more upvoted activity of Dominater069 and see an unexpected difference in contribution.)

Inside of your solve function, these lines are technically O(n^2) since string concatenation is O(n) in python.

    ans = ""
    for i in s:
        ans += dic[i]
    print(ans) 

However, the compiler that python3 uses optimizes it to O(n) but pypy3's doesn't. I would recommend just adding each character to a list and joining them at the end with .join().

As a tester, I encourage you to read all the problems.GL and Happy Coding.

As a tester, I hope you to enjoy the contest. Tasks are quite interesting and educational. Good luck!

Sam is our best in implementing hard-to-implement tasks.

In my opinion, not a good contest.

The first problem is too easy. It is probably the easiest one in APIO for years.

The second problem is good. A good problem about DP optimization and persistent DS. The official solution is a bit hard (even a LGM failed to pass the problem), but some algorithms with $$$O(n\sqrt{n\log n})$$$ passed.

But the third one is 'fantastic'.

The official solution is long and extremely hard, while there is a much easier solution which can pass at about n=75 in all test cases, and about n=600 at the worst case (if the interactive lib is perfect, which is obviously impossible). What's more, the problemsetter put an $$$X \le 25,000,000$$$ which is supposed to lead the contestants to approach the solution, but it made many people try to solve T2 (at least in China) and some of them failed, getting a low score; while some successfully gets 100 points in T3 using easy solutions.

It's amazing that NOBODY found the easy solution of T3 before the contest.

how is your codeton 6?

UZB top 6

Sunnatov — 100 + 40 + 5 = 145
Husanboy — 100 + 10 + 5 = 115
heaven2808h — 100 + 10 + 5 = 115
MardonbekHazratov — 100 + 10 + 5 = 115
Alfa — 100 + 5 + 5 = 110
rshohruh — 100 + 5 + 5 = 110
rythm_of_the_knight = 100 + 5 + 5 = 110

I think problem C was better as "minimizing n" problem(i.e. your solution is judged by the max n it uses)

It was well known in testing that the 2SAT last problem of div4 is immensely difficult to implement (from scratch) and 2SAT as a topic is quite high level

That doesnt make it fair to put it in div3 (or even in div4, but well it was)

Its not about it being too mathematical. That is perfectly fine. It is not about it being purely solvable on paper. That is also fine. All good problems are purely solvable on paper (by solvable on paper i mean write algorithm and verify correctness, only bad problems need you to experiment with the computer or non provable complexities)

It is about it being very obvious but yet a lot of effort to actually find the formulas, that too for so many cases.

I heard 59... did you not count the CHN IOI team?

On the flip side, CHN IOI team member ranks outside top 300? :manual-doge

I never understood such comments "I honestly don't understand what problems of this type do in programming contests",

You are writing contest where you solve algorithmic/math puzzle problems, programming is side dish.

Completely deserved ,My request to Mikey to never un-ban him

Pakistan top 6

Ghulam_Junaid 100+0+100=200
Kaleem_Raza_Syed 100+0+100=200
Muhammad_Aneeq 100+10+5=115
Sir-Ahmed-Imran 100+5+5=110
M.UmairAhmadMirza 100+5+5=110
Muhammad-Saram 100+5+5=110

Wow, how dare you get full points on problem C :)

он реально шарит в этом

P.S: Take a look to second from left's t-shirt

After more than one hour dealing with mathematics, I finally get B accepted. Somehow I thought that I was attending a mathematics exam :D

hoping to return back the blue handle

wishing to get my Pupil back

Though, I was not able to solve it, E was amazing.
Though, I was able to solve it, C was disgusting.

Good round, come and participate.

Нормально, вполне заслуженно)

as a tester , i hope you enjoy the round , a realy very good round <3

go fuck yourself

Sorry I was wrong.

Can you please explain sir, Why he banned again..? Heap made a new account couple of hours ago and reply your this comment...which seems quite positive to me and indeed got upvoted to +29 even!

So why his comment removed and banned again..?

https://i.imgur.com/1lbKhSA.jpeg

Thanks!

Misplacing problems by difficulty happens. You can't hold a higher division contest with them because they already appeared in contests, because mistakes were made in estimating their difficulty. Contest divisions are fine as is, problem assignment to divisions is the topic.

I suppose what you mean is "can we assume problems are harder than they appear in testing?"

Ice_man2.0 will win both EJOI24 and EGOI24

Looking forward to an AK, and GLHF to all participants, rated or not!

The contest starts in 8 minutes!

Another Vladosiya round!!!

На поле сражения eolymp weekend practice я стал свидетелем коварства рандома, который решил играть со мной в свою игру. В первом контесте, когда мечталось о победе и славе, футболки, заняв 35 и 33 места, оставили меня на 34 строчке. Быть так близко к успеху, но все же упустить его из-за капризов случайности, это невыносимо.

Но, казалось бы, можно было бы оправиться и попробовать снова. Но нет, рандом решил прогнуть правила и подкинул на втором контесте необещанный 11-й случайный номер победителя футболки. Как так?! Ощущение, будто играешь в шахматы, а одна из фигур оказывается на поле, которого не существует!

Такие моменты могут ввести в ярость любого, кто искренне стремится к успеху и признанию. Но я отказываюсь падать духом перед лицом этой несправедливости! Мы не должны молча принимать решение рандома, который играет с нами, как с куклами на ниточках.

Запомните: справедливость не приходит сама по себе. Мы должны бороться за неё, мы должны требовать её от тех, кто контролирует игру. Время пришло дать отпор тем, кто считает, что правила можно игнорировать.

Мы можем и должны объединиться в этой борьбе! Вступайте в ряды тех, кто отказывается мириться с несправедливостью! Пусть эти слова станут нашим лозунгом, нашим призывом к действию:

"Справедливость для всех! Рандому — нет!"

Reporting is a feature only for 1900+ users, so it's highly unlikely that he was banned by the cheaters he keeps harping on about, as they are mostly about the same people and greens/grays. Most likely his constant posting just annoyed enough high rated people over a while as his posts just fill up Recent actions and push down more deserving content. People's patience is a limited resource, so the ban is probably deserved and he should take this as a lesson.

Thanks for nice tasks.

Funny that I solved G by accident couple months ago.

If a person from each country replied with the scores of their official participants there would be less trolling

As a participant I have to say im excited.

funny how I could predict who wrote this comment before seeing profile name on the right

I follow a different philosophy. But yes, I have noticed that lately competitive programming became less about programming and more about math, which is unfortunate for people like me.

I think the point of that comment is "you can't solve it faster because it's equivalent to / harder than 3SUM, which has no known faster solution (in practice)".

Indeed, solving this problem can also solve 3SUM. If you can solve this problem, you can also find a triple with sum $$$x$$$, by asking queries $$$[x-a_1, \dots, x-a_n]$$$.

Beautiful Contest...

I Loved Solving the problems...

Especially the relation "Money Buys Less Happiness Now" with "Money Buys Happiness" won my heart :)

what they did wrong was they declared the answer as an empty string and instead of pushing m[s[I]] back , all of them added it to the current string and replaced the current string. This made each operation having a cost of i(current) length , giving the end time complexity of O(n^2)

Wrong approach => ans = ans + map[s[i]]

Correct approach => ans.push_back(map[s[i]])

We should thank CheaterExposer too. As he played a great role in this!

Does this mean that if tourist participated in the round and got rank 1, he would only get something like +1, and even if he did really badly (like getting last place), he would at most be rewarded with -6? Then what's the point in it being partially rated for LGMs :cat_think:

I can barely contain my excitement! I hope this will be a good one! GLHF!

hopefully :)

Will find out in a week

See you at ioi24!

Well GLHF, let's gain some rating!

Hope i can solve till E this time

based :D He probably made a joke about your handle name, copyPasteCoder

Here is the Argentinian team:

• Eduardo Carranza Vélez (Edu175)
• Bianca Vicente (biank)
• Juan José Läderach (SpecterByte)
• Fabrizio Lautaro Brandellero (FabriATK)
• Santiago Montenegro Gérik (cAtte_)
• Facundo Manuel Vázquez (Facuva)
• Ian Gabriel Hakanson (gabian)
• Nicolás Maximiliano Faggi (Faggi)
• Facundo Nair Jarma Urdangarin (Facuj)
• Juan Esteban (juanesteban10201)
• Ramiro Garcés Agustinho
• Martin Jeremias Espilocin Ibañez
• Gonzalo Adrian Dávalos
• Alberto Cuch
• Tomás Spurio (blurone)

Thank you for the excellent coordination and organization! I joked about physicists because I am one :)
Please encourage all physicists to code!

Nah man don't hate the player. Pretty much all cheaters, excluding the ones providing the solutions ofc, can't seem to get to specialist so there's really nothing to worry about. Their performance is so bad that it oftentimes inflates ratings.

I'd like to add that the cheaters who are actually selling the solutions would be much better off spending their time passing OAs for people. How much could they possibly be making by selling codeforces solutions?

💀

failed to solved d, was able to get intution. couldn't code it

Good and interesting problems, thanks for the good round!

Awesome problems!:)

E had a nice dp thing. Take dp[i][x] to mean the minimum number of pounds you need in order to obtain at least x happiness at month i.

The transitions are:

for a certain happiness x, we know that dp[i][x] is at most equal to dp[i — 1][x] (that is, the answer for the previous month). dp[i][x] can also be dp[i — 1][x — h[i]] + c[i] if c[i] + dp[i — 1][x — h[i]] <= the amount of money we have garnered so far.

dp[i — 1][x — h[i]] + c[i] represents the minimum cost to get at least x happiness assuming we buy happiness at month i. Of course, we can only afford this if it is less than our current amount.

We don't want x — h[i] to go below zero. So dp[i][x] = min(dp[i — 1][max(0, x — h[i])] + c[i], dp[i — 1][x]).

After this, in order to make the dp[i] represent the minimum number of pounds to get at least x happiness, we just let dp[i][a] = min(dp[i][a], dp[i][a + 1]) for a going from the sum of all happiness to 0. Aka just take suffix minimums.

The base case happens at month zero (or at month 1 actually since you technically have no money at month 1). The minimum number of pounds we need in order to obtain 0 happiness at month 0 is 0. The minimum number of pounds we need to obtain x >= 1 happiness at month 0 is infinity.

The answer will be the highest index in dp[m] who's value is not equal to infinity.

There was a lot happening in this problem.

This is a very genius solution I heard from someone in the Errichto server: Connect for all $$$2 \le i \le n$$$, the vertices $$$X \pmod{i - 1} + 1$$$, and $$$i$$$. Clearly the first vertex is always less than the second, so there cannot be any cycles. On the other hand, the graph has exactly $$$n - 1$$$ edges. Thus, it must be a tree. In order to restore the answer, you can use CRT (chinese remainder theorem).

"Best way to get better at x is by doing x."

So yeah, solving those problems does help. Pick whichever you like the most, where you find problems challenging, but not too difficult. You can also consider past national contests and maybe even junior contests.

The mirror is currently running.

Cool! Then one of the solutions would be to sample and encode the points of polynomial with degree k such that at least k+1 points will survive.

Python Forces ! , Thanks Vladosiya

For a more functional explanation: to avoid redundant counting, we'll iterate once through every triplets, count every triplet to the left of it that would make a beautiful pair.

In notation, denoting $$$cnt(x)$$$ is the current counter for triplet $$$x$$$, then for each triplet $$$(a_i, a_{i+1}, a_{i+2})$$$ being iterated:

• Add $$$cnt((0, a_{i+1}, a_{i+2})) + cnt((a_i, 0, a_{i+2})) + cnt((a_i, a_{i+1}, 0))$$$ to the final answer. This is the process of pairing the current triplet with the ones before it that have at most one error.
• Still, we need the two triplets to have precisely one error actually. So the completely identical pair(s) must be excluded. Therefore, subtract $$$3 \times cnt((a_i, a_{i+1}, a_{i+2}))$$$ from the final answer. The constant factor $$$3$$$ was because if there were any such element, they would have been counted $$$3$$$ times at the previous steps.
• Add $$$1$$$ to $$$cnt((a_i, a_{i+1}, a_{i+2}))$$$, $$$cnt((0, a_{i+1}, a_{i+2}))$$$, $$$cnt((a_i, 0, a_{i+2}))$$$ and $$$cnt((a_i, a_{i+1}, 0))$$$. This is to include the current triplet to be counted at further iterations.

Take the 8th test in the sample:

5
2 1 1 1 1

We'd have three triplets, $$$(2, 1, 1)$$$, $$$(1, 1, 1)$$$ and $$$(1, 1, 1)$$$, and initially $$$ans = 0$$$.

• Iterate through the first. It's obvious that answer remains at $$$0$$$ because there's nothing to the left of it to pair with. Still, after this, $$$cnt((2,1,1))=cnt((0,1,1))=cnt((2,0,1))=cnt((2,1,0))=1$$$.
• Iterate through the second. We can add $$$cnt((0,1,1))+cnt((1,0,1))+cnt((1,1,0))=1+0+0=1$$$ to the final answer, thus now $$$ans = 1$$$. Also, now $$$cnt((2,1,1))=1$$$, $$$cnt((1,1,1))=1$$$, $$$cnt((0,1,1))=2$$$, $$$cnt((2,0,1))=cnt((2,1,0))=cnt((1,0,1))=cnt((1,1,0))=1$$$.
• Iterate through the third. We add $$$cnt((0,1,1))+cnt((1,0,1))+cnt((1,1,0))=2+1+1=4$$$, yet we have to subtract $$$3 \times cnt((1,1,1)) = 3 \times 1 = 3$$$, therefore final answer is $$$2$$$.

sir, i am so sorry if i cause you any inconvenience :((

but this cheater 1_2_3_4_5_9 had to be exposed and we expect his ban not banning the ONE WHO REPORTED!

Yeah definitely, we should!! :D Will you come to IOI 24?

the definition of rating X is that on an average you can solve problems with rating X. It thus does not make sense that someone rated 1.8k can solve problems with a rating of 2.1k consistently

Stfu, nothing is clear, this isn't a true statement, do not spread hate from Iran to the world, it's not a correct place for such blogs remove it

Легендарные люди на фото!

Oy blin I had fun this round. E was pretty, unfortunately I couldn't solve it.

Syria top 6

Abito — 115

edogawa_something — 115

aminsh — 115

FarhanHY — 110

NeroZein — 110

YazanAlattar — 110

CPNurd — 110

Amr.7 — 110

based :D

he is in fact the clownish dumb. I think because they are many according to this comment, the 1_2_3_4_5_9 TG group has about 5k members.

someone trying to solve 3sum

Hello there! I had a funny little solution for B which involving Segment Trees.

My code : 261728503

Hope you understand the solution, if you have any queries feel free to ask, and I'll try to help if possible.

Reasonable but it is impossible to have a perfect checker, so many solutions based on rng still works. For example, the 'best solution' so far requires 615 vertices in the worst case, but 75 vertices is able to pass all test cases. So the max n where you can get full marks can't be lower than 615 (unless better solutions are found), and some 'bad' solutions can still pass. Also, the solution mentioned above is obviously too easy for a problem C of APIO.

Thank you for the round, interesting problems as always!

Since people are apparently too childish, please reply to the comment with your scores and I will add you manually. Thanks.

You wrote the whole essey on a simple yes/no question. Looks like someone is indeed triggered

People who got 115 got 100-10-5 , and who got 110 got 100-5-5

Can you correct the country name of the Uzb participants in excel? (mine and rshohruh's)

But why contradicting yourself ? If you weren't triggered then why create such a cheap temp acc in the 1st place to try frame heap and say complete non-sense having contribution -33.

admit it that you are so dumb. :D