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 :)

This is in meme territory.

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

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 !!!

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

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

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.

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.

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

A well known saying in China: "Instead of beating the problem, I only need to beat the author." The person who said it got accepted in C with $$$n=75$$$, and his algorithm's easy to hack.

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)

он реально шарит в этом

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

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

Ukraine Team on EJOI2024:

• Maksym Shvedchenko(Umanity),
• Kseniia Strelbytska(ksu_),
• Daria Perekopska(perekopska.d),
• Dmytro Strakhal(dimastrakhal)

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.

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

go fuck yourself

Sorry I was wrong.

hopefully :)

Will find out in a week

See you at ioi24!

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!

Beautiful Contest...

I Loved Solving the problems...

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

Another Vladosiya round!!!

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.

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

"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.

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

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 $$$[x-a_1, \dots, x-a_n]$$$.

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]])

The mirror is currently running.

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

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)

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 $$$2$$$.

💀

If you took 100 random people, there would be a good chance all of them would be fully demotivated within 6 months. If you are talking about a highly intrinsically motivated person that has a passion in problem solving that can handle intense grinding for years (and the pain) plus good mindset and practice strategies, that person has a good shot at becoming yellow. This kind of person is definitely not "regular" and what I mentioned has nothing to do with IQ. There are even a lot of CF users that wouldn't fit any of these catagories (looking at the do CP for jobs people).

A very hard contest.

Kevin114514 got 115.

Awesome problems!:)

I had the $$$O(n \sqrt{n \log n})$$$ solution during the contest but I didn't want to code it since the TL was 1s and $$$n$$$ was up to $$$10^5$$$, why weren't there any subtasks rewarding that solution especially that there aren't that many subtasks to think of in P2 and P3 :))

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).

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

when will be results publish?

in 4 seconds, we should have atmost 4*(10^8) iterations not 10^32.you had done mathematical error.

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.

How many orange ones do you know that you were able to draw such a conclusion?

If you're assuming 1 second = $$$10^8$$$, then 4 second is $$$4 \times 10^8$$$, not $$$10^{32}$$$. $$$10^{32}$$$ is 1000000000000000000000000 times larger than $$$10^8$$$.

my bad , sorry for wrong calculation , lol all these days i was doing wrong calculation .

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

Even if they all reported him

Really ? In the screenshot above, 3 mentioned were capable of reporting :) peaked at Candidate at least.

What about the fact that one of them runs a 5K member TG group ? so you think all these reports won't suffice to trick the system into fake read only mode and content removal ?

It seems there is also other reason for banning, some spammy posts as per Vladosiya, but I honestly feel it is somehow an exaggerated move to ban cuz of this ?

My solution is O(n^{1.75}) and it passed this problem.

Are all submissions tested against the test cases from successful hacks at the end of the hacking phase?

u r right,coz 56 is not true.

It should be 58(seems that there is an international participant from THU high school,so 59 in China site).

Turkey looks promising.

Lol, non-sense. so ask a doubt for VS code on MAC via Hawkwatch, and solve it through heap ? Then I (bajelega) re state what was in heap comment so people can benefit (after comment removed probably due to read only mode) ?

Please, attend the logic class.

Why completely NEGLECT the acts of the biggest cheater in india with 5K Telegram group tho and haven't said a word about ?

sus

lol youre just making it speedforces

The key word here is consistently.

but i can ensure you most of >=1800-rated participants can pretty consistently solve 2100-rated problems

What's 'consistently'? If you mean solving them like 20% of the time is consistent then I have no more words. In the recent 5 of the rounds you participated in, there were 4 2400-rated problems and you solved only 1 of them, and 1 out of 2 2300-rated problems. This isn't consistent at all and that is why your performance averages at a little above 2200 and not 2400. Same goes for 1800-rated participants and if they can consistently solve 2100-rated problems then they should get to 2100+ in a few rounds for sure.

I agree that there are some inflation for the top problems of each division, but not as much as you stated. The problem ratings, at least for low-mid problems, are working as nearly as intended.

standings — example.

Do you claim all people who solved E1 are fake-CMs/Es and secretly have 2500 rating?

Your point is invalid.

I did a few. I enjoyed them. So definitely worth giving them a try.

Here is how I solved C.

for (X = u; X <= 25 000 000; X += v)

for (X = u_final; X <= 1e18; X += (product))

Also I think it should specified that it was a random interactor since it is more fair to have actual information about the checker

I ate a lot of chocolate while testing the round.

I think codeforces is better than both of them. I hate UVA, cringe.

I did some of SPOJ it is nice place to practice concepts and classical things. Here is some sheet I had used to solve on SPOJ before,

http://problemclassifier.appspot.com/

Also, check this you'd find replies by ICPC finalists.

There, https://youkn0wwho.academy/topic-list , you can find problems from the 3 websites! CF, SPOJ and UVA

That's good username, ain't gonna lie.

I agree. In my country the competition was only about 15 points because of easy problem A and hard BC. Maybe we have skill issue, but it's still bad to be able to get A in the first 20 minutes and struggle to get anything else the rest of the contest

Great contest, but I think that problem A could have been written in a better way: there are too many people who did it after 10 mins, meaning that the statemenet was not clear at all. Something like "an app can only be placed on a single screen" would have been enough. I was (wrongly) thinking of application on multiple screens, so that, e.g. 4 screens could contain 15 y app. Hope next time I'll be sharper in statement comprehension

Well I tried to optimize your code, eliminated the use of one map in this submission 261926879. But it still gave TLE.

I initially also got a TLE during the contest on test case 14, that is because I did not notice the upper bound over all the test cases was on sum of hi and not m.

Thanks for the contest ! Good Problems !

Anyone else used Segment Trees on problem G?

It's a bit of an overkill, but it was my first idea and it worked.

bro come on you haven't even deleted comments from your code while copying and pasting from chatgpt ahahahhahaha

Unofficial Editorial for some more problems.

Problem F

void solve() {
    cin >> n;
    vl a = R(n);
    sor(a);
    if(a.ft==a.bk) {
        print(0); return;
    }
    ll extra = 0;
    rep(i, n) {
        if(a[i]>=10 || a[i]%2) {
            a[i]=(2*a[i])%10;
            extra++;
        }
    }
    debug(a)
    sor(a);
 
    if(a.ft==a.bk) {
        print(extra); return;
    }
    if(a[0]==0) {
        print(-1); return;
    }
    ll res = 1e18;
    debug(extra)
    auto func = [&](int val1, int val2) {
        if(val1==val2)return 0;
 
        if(val1==2) {
            if(val2==4)return 1;
            else if(val2==8)return 2;
            else return 3;
        } else if(val1 == 4) {
            if(val2==2) return 3;
            else if(val2==8)return 1;
            else return 2;
        } else if(val1==8) {
            if(val2==6)return 1;
            else if(val2==4) return 3;
            else return 2;
        } else {
            if(val2==2)return 1;
            else if(val2==4)return 2;
            else return 3;
        }
    };
    for(int i=2; i<=8; i+=2) {
        ll step = 0;
        rep(j, n) {
            step+=func(a[j], i);
        }
        chkmin(res, step+extra);
    }
 
    print(res==1e18?-1:res);
 
}

Problem K

pair<int, int> dfs(const vector<vector<ll>> &tree, int node = 0,
                   int previous = -1, int length = 0) {
    pair<int, int> max_path = {node, length};
    for (const int &i : tree[node]) {
        if (i == previous) { continue; }
        pair<int, int> other = dfs(tree, i, node, length + 1);
        if (other.second > max_path.second) { max_path = other; }
    }
    return max_path;
}
 
void solve() {
    cin >> n >> k;
    vvl g = RG(n, n-1);
 
    int s1 = dfs(g).fi, s2 = dfs(g, s1).fi;
 
    auto f = [&](int i)->vi {
        vi dis(n);
        dis[i] = 0;
        queue<int> q; q.push(i); vi vis(n, 0); vis[i] = 1;
        while(!q.empty()) {
            int j = q.front(); q.pop();
            for(auto i: g[j]) {
                if(!vis[i]) {
                    vis[i] = 1;
                    q.push(i);
                    dis[i] = dis[j]+1;
                }
            }
        }
        return dis;
    };
 
    vi a = f(s1), b = f(s2);
 
    rep(i, n) {
        if(a[i]>k&& b[i]>k){
            yesnoc(1);
            return;
        }
    }
    yesnoc(0);
}

Problem J

    cin >> n;
    ll res = 1;
    debug(n)
    for(int i=2;i*i<=n;i++) {
        debug(i)
        if(n%i==0) {
            ll c = 0;
            while(n%i==0) c++,n/=i;
            debug(c)
            res*=(2*c+1);
        }
    }
    if(n>1)res*=3;
    print(res);

On the technical committee end, I saw a whole bound of n^{1.5}logn get by in between 200-400ms.

2 and 3 both could have had much harder subtasks: trying to get people to put up apio24p2hard and apio24p3hard on DMOJ at the moment...

Indeed! Heap__OverFlow.'s comment in reply to mod Vlad, was removed and banned despite it was at +31. Something very fishy looks to be happening even 34z12000's comment in reply to mod Vladosiya was removed too but without a readonly mode.

Going off of what WeaponizedAutist said, anyone with a passion for solving math/coding problems, on average, has high mathematical ability. You aren't gonna enjoy solving problems if you're no good at it. If you solve thousands of problems, chances are that you're intelligent. This is partially why many masters/GMs have solved so many problems relative to other users: they didn't quit because they enjoy it, and they enjoy it because they are good at it. Practice, of course, plays some role in attaining master+ rank too.

Your comment is also kind of beside the point. Hypothetically, if you forced 100 people with average IQs to grind codeforces for a few years, there is a good chance that none of them would be master.

wala????

XD You are right. I missed that. Thanks

Bro is proud of ChatGPT solutions (Does look like that with all those comments, solution time and weird mistakes in testcase 1? Idk if that's even legal, correct me if I am wrong)