IOI 2024 Team Russia

• Daria Grekova (arbuzick) — 1st time at IOI, 1 attempt left

• Ivan Piskarev (Qwerty1232) — 2nd time at IOI (IOI 23 Gold), 0 attempts left

• Gimran Abdullin (bashkort) — 2nd time at IOI (IOI 23 Gold), 1 attempt left

• Petr Losev (green_gold_dog) — 1st time at IOI, 1 attempt left

WHY IS THERE EVEN AN OPTION TO ASK QUESTIONS IF YOU AREN'T GOING TO FUCKING ANSWER THEM???

Now PinkieRabbit was skipped and got a -220 rating. Glad to see that.

Ukraine's team:

• Andrii Smutchak(xGaz_) — 2nd time at IOI(IOI 23 Silver), 0 attempts left
• Ihnat Zharikhin(Ignut) — 1st time at IOI, 0 attempts left
• Denys Tereshchenko(TheQuantiX) — 1st time at IOI, 0 attempts left
• Illia Permiakov(160cm) — 1st time at IOI, 0 attempts left

Japan's team:

• Shogo Amacho (PCTprobability) — 2nd time at IOI, 0 attempts left
• Katsuki Ohta (shiomusubi496) — 1st time at IOI, 1 attempts left
• Ryotaro Hayashi (Forested) — 1st time at IOI, 0 attempts left
• Yuya Hirasawa (hirayuu_cf) — 1st time at IOI, 3 attempts left

China

56 300s.

This has been talked about before and the general consensus seems to be that while these last problems of Div. 2 contests may be hard, they are also too standard to be suitable for a Div. 1 contest.

Greetings from our team! Share your team photos too!

P.S: Guess who is who :)

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 :)

why put B in contest, nice ACD problems though

Vietnam's team:

• Nguyen Huu Tuan (ahihi1234) — 1st time, 2 attempts left
• Pham Cong Minh (socpite) — 1st time, 0 attempts left
• Pham Ngoc Trung (hotboy2703) — 1st time, 0 attempts left
• Hoang Xuan Bach (bachbeo2007) — 1st time, 1 attempt left

I think this was recently skipped too. He was mentioned as master in the round announcement.

[Unofficial rankings]

Please add your scores in this spreadsheet

the hero we do not deserve

No commment.

D is created only to give pain and trauma :skull:

Asia-Pacific Informatics Olympiad

Asia-Pacific Graphs Olympiad

Anyway, I think problems were cool

#include <iostream>
#include <stack>


using namespace std;

// Calculate the fp32 sum sequence like {s:1,2,3}
double calculateFp32(stack<double>& nums) {
    float currResultSingle = 0;
    while (!nums.empty()) {
        currResultSingle += static_cast<float>(nums.top());
        nums.pop();
    }
    return static_cast<double>(currResultSingle);
}

int main() {
    stack<double> s({1.0e-10, 1.0e-10, 1.0e0});
    double answer = calculateFp32(s);
    printf("%.20lf\n", answer);
}

This code prints 1.00000000020000001655 under "GNU G++17 7.3.0", and prints 1.00000000000000000000 under "GNU G++20 13.2". In addition to that, if we add a printf like printf("value: %.20lf\n", nums.top()); currResultSingle += static_cast<float>(nums.top());, the result changes to 1.00000000000000000000 under "GNU G++17 7.3.0".

well-balanced div3 round, GL for all participants 🏆

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

For me personally, the latest div3 tasks were very useful when I was a specialist/expert. And I think that only by solving them, I was able to quickly improve from 1500 to 1900 (you can look at the graph and make sure).

Therefore, I think that it is necessary to look for such tasks where it will be possible to pump your skills quickly with a minimum of effort. And one of these tasks for beginners is div3/div4, just get out of your comfort zone.

Some may be standard for IGMs and LGMs, but I don't think it holds true for most Masters and low reds. Many, not all, Div 2 rounds are definitely suitable for a "Div 1.5" and the coordinator should be able to decide this.

This is in meme territory.

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

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

bump

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

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)

How is your IOI?

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 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

Umanity will win EJOI 2024

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

how is your codeton 6?

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.

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

wishing to get my Pupil back

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

the current format is more educational than competitive which i like for lower divisions and when u get to div 1 where competition starts then its fair for everybody i think divisions which target lower rated participants should focus more on the educational side than the competition side

In general, yes.

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

Good round, come and participate.

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.)

Please write suffix array template and never submit hashing solution again

In some cases, this actually seems useful for finding a balance between time and memory. For example, it seems that it is better to use flat_map when storing edges at the node of a trie.

As far as I understand this is just an implementation like vector<pair<Key, Value>> with sorted Keys. And it works great with cache.

Also TimDee and TimaDegt was skipped too. Probably MikeMirzayanov will post something.

hoping to return back the blue handle

Deducing this, std::flat_map, std::print

Wansur will win EJOI2024

According to regulation schedule, the contest window is over. But is it truly over? Are there anyone who didn't participate?

When will ranking come out

why do we need to wait 2-3 days

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.

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.

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 have some trouble understanding the given proof for the $$$O(1)$$$ solution for Problem A. I'll explain how I got that solution. Let us take an optimal solution (i.e., a solution with the max possible number of draws that corresponds to the given values of $$$p_1$$$, $$$p_2$$$, and $$$p_3$$$). The total number of games played is $$$\frac{p_1 + p_2 + p_3}{2}$$$. Let $$$a$$$, $$$b$$$, and $$$c$$$ denote the number of draws in the $$$A_1$$$ vs. $$$A_2$$$, $$$A_2$$$ vs. $$$A_3$$$, and $$$A_1$$$ vs. $$$A_3$$$ matches respectively, where the names of the players are $$$A_1$$$, $$$A_2$$$, and $$$A_3$$$. Since this is an optimal solution, the correct answer to the problem is $$$a+b+c$$$. Now, if it so happens that more than one of these players had wins (say, $$$A_1$$$ and $$$A_2$$$ both had nonzero wins — say, $$$k_1$$$ and $$$k_2$$$) — then, we could have repurposed them as $$$\min(k_1,k_2)$$$ more draws in $$$A_1$$$ vs. $$$A_2$$$ games, and one of them having $$$\mid k_2 - k_1 \mid$$$ wins instead, contradicting the optimality of our solution. Therefore, we may assume that only one of these players had wins (if any), say $$$w$$$ of them, in our optimal solution.

Suppose $$$A_3$$$ is the player who had all $$$w$$$ wins ($$$w \geq 0$$$) in our optimal solution. Then, $$$p_1 = a + c$$$, $$$p_2 = a + b$$$, and $$$p_3 = b + c + 2w$$$. Solving for $$$a$$$, $$$b$$$, and $$$c$$$, we get $$$a = \frac{p_1 + p_2 - p_3 + 2w}{2}$$$, $$$b = \frac{p_2 + p_3 - p_1 - 2w}{2}$$$, and $$$c = \frac{p_1 + p_3 - p_2 - 2w}{2}$$$. Verify that these values for $$$a$$$, $$$b$$$, and $$$c$$$ will be the same even if you assume $$$A_1$$$ or $$$A_2$$$ won all $$$w$$$ games! Note that we get integral solutions for $$$a$$$, $$$b$$$, and $$$c$$$, iff the number of odd numbers in $$${p_1, p_2, p_3}$$$ is even. If this is not the case, we output $$$-1$$$, and finish.

Now, thanks to $$$p_1 \leq p_2 \leq p_3$$$, $$$w = 0$$$ (i.e., all games are draws) leads to a valid (i.e., nonnegative integer) solution for $$$b$$$ and $$$c$$$, and a valid solution for $$$a$$$ only if $$$p_1 + p_2 \geq p_3$$$. That is, if $$$p_1 + p_2 \geq p_3$$$, the answer we must output is $$$\frac{p_1 + p_2 + p_3}{2}$$$. The case left to deal with is when $$$p_1 + p_2 < p_3$$$. To make the value of $$$a$$$ nonnegative, the minimal value of $$$w$$$ to be set (we want a minimal value for $$$w$$$ as we want maximum draws) is $$$\frac{p_3 - p_1 - p_2}{2}$$$. With this setting of $$$w$$$, we get a valid solution $$$a = 0$$$, $$$b = p_2$$$, and $$$c = p_1$$$, so that the answer is $$$a + b + c = p_1 + p_2$$$. I wrote this solution (sadly, after the contest) and it was accepted.

Separate things properly. If PinkieRabbit really is a cheater, he should be punished accordingly. And that doesn't invalidate the good deeds he had done, if any. Those two are completely independent topics, and none would outshadow the other.

You can't use your good past as an excuse for your present wrongdoing. And vice versa, your present good deeds might earn people's respect, but they don't fully delete any sins in your past — at least their marks for remembrance and atonement will always stand.

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

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?

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

он реально шарит в этом

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

Thank you!

My post contest discussion stream

Also, can someone hack my solution for G?
Gentle ping maroonrk because of this

Cool. Nice to see justice being served in both cases. Thanks MikeMirzayanov!

What if someone creates a website where everyone enters their results?

Good and interesting problems, thanks for the good round!

Great problems!

Most of this blog is copy-pasted from an uncredited geeksforgeeks article.

*800 *1000 *1300 *1600 *1900 *2100 is more true for div2 than *800 *1300 *1700 *2100 *2500 *3000. Div2 is rated for ~1899, and no 1899 can solve *3000(except for future GM).

Difficult div2 often feels like div1.5, which may be good for M or CM, but is disastrous for E. Creating difficult well-made task is very difficult, and I think it is impossible for the rated participants of div2 to solve *2500 or harder task. Why not save the most difficult task for div1? Wouldn't it be satisfactory for everyone if put in a moderately easy problem(yes, around *1000~*1500) instead?

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!

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

I thought that was the point of educational rounds, to learn to be able to use techniques and algs that you wouldn't be able to use on problems like game theory.

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.

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.

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

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

Hope to ak!

This time I solved problem F much faster than usual(maybe 15 minutes), because I have met a similar one, https://codeforces.com/contest/650/problem/D , during my virtual participation.

This makes me believe again that hard work will pay off sooner or later.

When will every country participated and I can discuss problems?

Yeah, so lets take an example, lets say the array is this -

idx: 1 2 3 4 5 6
-----------------
  p: 5 1 6 2 4 3
  q: 2 6 1 5 3 4
-----------------
     7 7 7 7 7 7

Here $$$p[2] = 1$$$, so we try to make values at odd indices local maximums.

The values at odd indices are $$${5, 6, 4}$$$.

Since we are going to iterate over increasing values, we go in the order $$$4 \rightarrow 5 \rightarrow 6$$$.

So we first swap(q[pos[1]], q[pos[4]]) to get -

idx: 1 2 3 4 5 6
-----------------
  p: 5 1 6 2 4 3
  q: 2 3 1 5 6 4
-----------------
     7 4 7 7 10 7

Then we perform swap(q[pos[1]], q[pos[5]]) to get -

idx: 1 2 3 4 5 6
-----------------
  p: 5 1 6 2 4 3
  q: 3 2 1 5 6 4
-----------------
     8 3 7 7 10 7

Then we finally perform swap(q[pos[1]], q[pos[6]]) to get -

idx: 1 2 3 4 5 6
-----------------
  p: 5 1 6 2 4 3
  q: 3 1 2 5 6 4
-----------------
     8 2 8 7 10 7

And we are done!

Will we play volleyball again? :D

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

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

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

hopefully :)

Will find out in a week

See you at ioi24!

I think it is related to the number of read-only modes, and that's why I say it was unfair to ban him! because there had been fake read-only modes caused by high rated cheaters, he reported (master and cand).

the reason said by vlad, is

It's clear as day that he was banned for contribution farming.

The checker seems deterministic, why would the order of operations change?

Could you share this situation or describe how to reproduce it? I don't think it should be legal in this code in standard C++, at least with IEEE 754-compliant FP arithmetic.

Could you share this test? float->double->float roundtrip should be lossless.

Edit: The example yosupo provided answers all my questions; "GNU G++17 7.3.0" doesn't strictly adhere to IEEE 754.