Hello, Codeforces!
We are pleased to announce the resumption of the Global Rounds. Thanks to XTX Markets for supporting the initiative! In 2025, we will hold 3 such rounds. The series results will take into account the best 2 participations out of 3.
On Sep/20/2025 17:35 (Moscow time) we will host Codeforces Global Round 29 (Div. 1 + Div. 2). 
Codeforces Global Round 29 marks the first round in the 2025 series of Codeforces Global Rounds. These rounds are open and rated for everyone.
The prizes for this round are as follows:
- The top 30 participants will receive a t-shirt.
- 20 t-shirts will be randomly distributed among participants ranked between 31 and 500, inclusive.
The prizes for the 3-round series in 2025:
- In each round, the top-100 participants get points according to the table.
- A participant's final score will be the sum of the points they earned in their 2 highest-placing rounds.
- The top 20 participants across the series will receive sweatshirts and placement certificates.
We extend our gratitude to XTX Markets for supporting the global rounds initiative in 2025!
The problems were written and prepared by misteg168, FelixMP, Pablo-No, danx, MeGustaElArroz23, Clan328 and me.
We would also like to thank:
Coordination: Um_nik
🍅 Tomato testing: ikaurov, prvocislo, NemanjaSo2005, Merkurev, tfg, TheScrasse and WLZ
🍌 Banana testing: rlidon2006, Esomer, Sergio_2357, KiKoS, Dalek_of_Rivia and -firefly-
🫐 Blueberry testing: AngrySeal, Lucia_Aparicio, chromate00 and eloipages
🥒 Cucumber testing: Kalufas
🥥 Coconut testing: dlos
MikeMirzayanov and KAN for the great platforms
Round Information:
- Duration: 3h
- Number of problems: 9
UPD: Score distribution: $$$500 - 1000 - 1500 - 2000 - 2500 - 3000 - 4500 - 5500 - (4000 + 3500)$$$
UPD2: In this round, some of the problems will feature interactive visualizers that can help you to play around with small test cases. These are not part of the problem statements, although we tried our best to make sure that their behaviour correctly follows it. You can investigate examples or your own tests in more details with them. Note the following:
the visualizers use javascript in your browser to work; you don't need to install anything additional;
we won't be able to help with the visualizers during the round.
Picture of authors (and some testers) (and some non testers): 
We eagerly anticipate your participation!
UPD3 Congratulations to all the winners!!!
UPDT 4: Editorial is out: https://mirror.codeforces.com/blog/entry/146633
Auto comment: topic has been updated by BernatP (previous revision, new revision, compare).
the goats themselves are running a round! lfg.
Is it rated?
As a tester, I liked problems from this round more than from CAMA, which had really nice problems. Orz authors.
that hurts bro
Nemanja, you absolute galaxy-brained mastermind… normally your words descend from Mount Olympus like divine wisdom itself. But this time… oh boy. This round is cool, sure… but saying it’s better than CAMA is like saying a candle outshines the sun.
CAMA isn’t just better… it’s a whole separate lifeform, it’s peak evolution, it’s the final boss of awesomeness. CAMA >>>>>>>>>>>>>>>>>>>>>>>>>> reality itself.
oh wow !!!
I love how the colors of testers are named with a fruit .. very creative.
Spanish round omg :O
A round __baozii__ is not testing :,(
As a tester of teamscode, I'm excited to see danx in the author list.
I'm always excited,but after reading your comment, my excitement just got… more excited. Not quite as excited as that time I slept next to an almost-naked Esomer, but still dangerously close. Long, happy and exciting life to you and TeamsCode!
If there was a cyan tester, which food would it be?
this is why there's no cyan tester
Ikr :( it's sad
Where is __baozii__?
__baozii__ wants to reach IGM.
As a tester, Barça is the best team in the world (find me in the photo) :)
Madrid is better tho, statistically.
As a setter, I'm dyslexic
As a tester, i told all my friends to participate as the round is really high quality and has lots of nice problems!
as danx boyfriend, Esomer is the sexist (and sexiest) tester in the world, only cause Dalek_of_Rivia has not tested yet
i interested part of coder gang but my codeforces rating is low
__baozii__ Lost HIS Testing Streak
__baozii__ [user:wants] [user:TO] Reach igm
the colors for eggplant and blueberry emojis are swapped wrt the cf role colors and its driving me insane
We discussed that for a while. It turns out it depends on the browser
[deleted]
Excited for Global Round 29! Thanks to all authors, testers, and coordinators for their hard work.
As a tester, I am sure this is what happened for many of you when you saw Global Round happening in 2025.
miguel
The goats are awasome
Can we have Orange Testing Instead of Banana
Banana is more hotting
orz danx
Much respect from Ethiopia
Auto comment: topic has been updated by BernatP (previous revision, new revision, compare).
__baozii__ fan are crying in this contest lol
Wishing everyone has a blast in this competition!:)
Visca!
Hoping that the small baby in my DP (display picture not dynamic programming ^_^) will smile till end of September :).
you didnt declared rating bro
$$$5500$$$ points? That's incredible bro
the visualisation feature is so sick....love it...hope it will be in future contests as well...
yeah!!!
lol what happened to problem G?
There was a lot of cheating on this problem, which is why there were many in-contest solves. Cheaters will be banned. We’re sorry if the high solve count during the contest misled anyone, but please understand that this isn’t something the authors, coordinators, or codeforces can control.
114514
how are many coconut handles able to solve G but not E and F *_*
The contest is roughly good and I love the format that most of problems have their visualizer... but what happened in the AC count of G?
There was a lot of cheating on this problem, which is why there were many in-contest solves. Cheaters will be banned. We’re sorry if the high solve count during the contest misled anyone, but please understand that this isn’t something the authors, coordinators, or codeforces can control.
Please hurry and ban them or I'm going to lose rating
I was so shocked this situation happens, this scenario can collapse the meaning of the standings dualing the contest totally... (should I use friend-feature to create "legit" standings then?)
i think it's gpt-able
A bunch of cheaters between rank 100 and rank 200. They are all newbies and they cannot solve E and F, even D, but they can solve G. Please check if their code is AI-generated
There was a lot of cheating on this problem, which is why there were many in-contest solves. Cheaters will be banned. We’re sorry if the high solve count during the contest misled anyone, but please understand that this isn’t something the authors, coordinators, or codeforces can control.
I understand. The problems are very good, and removing the cheaters is the responsibility of admin, not yours.
I am also going to holder my own div2 contest. (still need 5 or more testers), hope cheaters will not ruin my contest.
YES! They are cheaters!
Please ban all cheaters that solved G with ChatGPT
There was a lot of cheating on this problem, which is why there were many in-contest solves. Cheaters will be banned. We’re sorry if the high solve count during the contest misled anyone, but please understand that this isn’t something the authors, coordinators, or codeforces can control.
легенда
What had happened on G?
There was a lot of cheating on this problem, which is why there were many in-contest solves. Cheaters will be banned. We’re sorry if the high solve count during the contest misled anyone, but please understand that this isn’t something the authors, coordinators, or codeforces can control.
I can understand that,but please ban all of the cheaters.
Comment below if you solved G completely by yourself(like me ^.^)!!
^.^(but I have no time to complete F)
spent so long solving it that i didnt have enough time to implement ):
very nice math problem tho!
AI could pass G
please try to give the statement to AI before posting it to codeforces
A lot of cheaters might be detected due to this however not all
gambleforces
AIforces
D and C are very good problems
C was a very neat problem and I enjoyed solving it even if I got +8 in penalties.
I then got to D and got it in 20 minutes with a guessed solution :/
That's the same thing which happened to me. Except i got 4 WAs in penalty not 8. D was so much easier than C
problem C alone costed me almost 250 perf....
Dislike most problems. Hard and strange. D<A<B<C.
Almost solved F with some $$$\mathcal O(n\sqrt n)$$$ solution.
I'm losing hundreds of rating. Probably contribution as well.
I think it's more like an AtCoder round. My brain is not enough for this :(
I have no idea how people can enjoy today’s contest, especially the Div. 2 participants. The first four problems weren’t interesting at all. B and C felt like torture to me, and by the time I reached D I was completely exhausted
UPD: upsolved D, ok, that is good problem
I didn't hate B that much it was sort of ok to me, but C just sucks.
update: Problem F 339613954
-78 rating :(
AIforces
C>>>>>>>>>>>D
Is C was like a prefix sum question where you have to find the minimum distance of a 0 to another zero (left and right) and if the distance is greater than 2 then the answer is false otherwise it is true.
no
so can you explain me the solution that you have applied.
In the case of 1010101, the distance is 1 in between all zeroes, yet (if I understood correctly) there is no way to make the rabbits not jump. Either the first two face eachother and the last makes a jump, or the last 2 face eachother and the first makes a jump, unless I am mistaken.
for C you just had to do a simple reduction.
if there are consecutive 0's (empty pots), you can imagine there's a wall between them.
so
011001001is equivalent to three separate problems that are easier to solve0110,010, and01now for each subproblem, you just check the number of zeros in chains of 0101010s. if its odd and none of the zeros is on a border, then its impossible.
339557173
why consecutive zeros might lead to a wall between them ?
A: but they won't jump if there is a rabbit in that pot already
the main idea is that you solve the problem for groups of 0's that are not separated by more than one 1 indepandently , because when there are 2 consecutive 1's the rabbids from previous positions can not interact with rabbids in the next positions (this is the most important idea) . now when you have a group that we described earlier you can see that when the count of rabbids in that group is even you can always pair them up and none of them can escape , for example : 10->1<-0 and so on , now we can expand this to odd count of rabbids at one condition : "there are at least 2 consecutive rabbids in the group" , because we can do this : 10->1<-0(<-0)1 , so we pair (cnt-1(even)) and the last one can see in the direction of its blocked neighbour , although i don't have a formal proof of why it won't work otherwise i think it is pretty intuitive . you can refer to my implementation : submission
I saw that there were more AC in pG than pF, so I spent all my time to solve pG. Then my rank drop from 30 to 200...
I regret my decision.
rip baited
I did the same. Solve count can be inflated if AI can solve it, but I didn't really consider that.
Problem C and D should have been swapped.
Way to many cheaters on G
Just ban everybody below CM who solved G but not E or F. The chance of false positive isn't big anyways
lol. i bet there is going to be 0 false positive.
I can bet on the fact that many CMs (not all) could not have even interpreted what the problem is stating, leave alone coding the solution !!
worst contest(performance wise)for me -160 :(
Started 1:10 hours late thought i would cover it but f'ed whole contest. But will upsolve
Thanks for the round!
The problems are interesting and diverse enough.
The visualizers are a nice touch, hope to see more of these in the future.
To be honest, I really don't like C. There are too many edge cases and I got 8 Wrong Answers but still didn't solve
When one tries to solve a problem by exhausting "edge cases", and these turn out to be too many (i.e., not edge cases at all), it isn't the problem's fault.
I'm sorry I think i should describe it as "The edge cases are too hard to find". The problem is that I really don't know where did I do wrong and it becomes super annoying. There are a lot of test cases but only 1 or 2 of them are strong. You can observe that there are many people attemped the problem but did not get AC.
Nah, my point is, rather: if you have more than say 3 "edge cases", usually it's time to rethink the solution from the start, not to find 5+ more edge cases for the current solution.
There aren’t really edge cases tbh. There’s one important case: 101 and 10101. If there’s an even number of 0’s -> true, odd ->false. All other cases are trivial, a string of 1’s or 0’s is obviously true
When your approach is a bit off track then you'll get wrong answer. This is why I don't like this problem.
I guess this just depends on your solution, as I didn't have to add any extra condition for edge cases(there was just no edge case for me).
C was kinda a mindfuck but really the only case that mattered was the alternating case. Repeating 1s or 0s was trivial.
alternating which case can u tell pls
I did that onnly, can you check my code please
https://mirror.codeforces.com/contest/2147/submission/339586315
339573079 I figured this out but I still got WA on test 2
You’re not accounting for 001010101 case. You will find that case as false when should be true. You can’t just pull out 10101 blocks without accounting for the ends
bruh my code giving yes oly
My code give it YES because the
if g1 != "0" and g3 != "0":already covers it. It would be matched as0(0)(1010101)()Great round! Especially balanced difficulty of problems
why my C 339573079 got WA on test 2?
I got trolled by rand on B, it cost me about 20 minutes and a bunch of TLE submits, and i got ac once i switched to mt19937. Cautionary tale to never use rand, i guess i finally learned that the hard way lol
there was a construction so rand was really never needed
Funny story, there was an incredibly similar problem in the 2024 AMO:
so i used the solution to that one instead of looking for a construction
ironic that prior knowledge of a generalised version of the problem made me perform worse
can I solve problem C in DP??
dp[i][0] : ith cat can see left
dp[i][1] : ith cat can see right
i was thinking along that line, but could not come up with a solution.
https://mirror.codeforces.com/contest/2147/submission/339547978
Yes. See my submission, I used this approach
was trying similar approach
I did one dimensional top down DP.
https://mirror.codeforces.com/contest/2147/submission/339550657
I did use similar idea.
power=1 -> can see only one side power->2 can see both side (I merged all continuous 0's into single 0, now this 0 can cover both the sides)
Why did I1 and I2 only have 2 testcases
How to approach C ?
There is 2 solution , dp and greedy
339573079 Is there a edge case I missed?
what is the greedy way of doing ?
There are some edge cases for the first and last elements of the string. You can think of it yourself. For the general case, for any substring of the form 110...011, you can see that thus substring can never affect any other part of the string. So, basically, any substring that has 11 after and before it will be independent. We can remove these extra ones in the prefix and suffix of these substrings and get something of the form 0...0. For these type of strings, it is possible to arrange the rabbits if and only if there is either any substring of consecutive 0s in it, or the number of 0 is even. You should try to prove it yourself. If all independent substrings can be arranged correctly, the answer is yes, else no.You can also check my code
are you saying that you created different pools of substring such that to its left and right there is 11?after that you processed that substring and checked whether you can do what is asked to do in the question?
Yes. There are some edge cases for the first and last element. As an example, if the substring has the first or the last element in it, amd it’s 0, its also always possible to create an arrangement for that substring. Basically decompose into substrings that are independent, and find if all of them are possible.
oh ok ok.
also if there are consecutive zeros like 0000 we can ignore all the zeros in middle and take only the 00 right?this is what we will be doing in your approach right?
No. That substring will be possible as there are consecutive 0s. My solution and the author's is basically the same.
how to do using dp?
Let's say there is ask(int X) function which will tell us wether we can arrange rabbits from level X to n
Now suppose we are at level a, then there are multiple transition we can make
If pot[level] == 1 means there is grass then we don't have to worry about it and answer for level a is same as ask(a+1)
If pot[level] == 0 then either it's immediate right has to be rabbit or immediate left or if immediate right is grass then immediate right to right has to be rabbit , now we have to call ask function according to the transition we made
Ask me if you have any doubts or you can check my code
Thank you
This might seem difficult than others but it seems easier to reach this solution if you knew this. It can also be solved using simple Bipartite check for a graph.
Run the dfs and check for bipartite.(Start from all the vertices already having a fixed color/direction)
Note: There is no restriction on any zero that does not have any adjacent 1.(Could be in any direction)
Submission: 339613547
was G google searchable or chat gpt-able? so many of CM solved ABC and then G. what the hell?
I can bet on the fact that many CMs (not all) could not have even interpreted what the problem is stating, leave alone coding the solution !!
And people who are Grey, Green and Cyan should not even think about defending themselves !!
I was not even able to reach G , how did so many people solve it
you are ratist pro max
PS: I could not solve it, even though I knew that in $$$O(log(m))$$$ steps, the series converges :/
I am talking about those who have submitted the accepted solution !!
I know everything is not about rating but you seriously think problem G of a Global Round with so many Red testers (generally 3100 rated) can be solved by so many contestants who can't even solve E and F and directly attacked G..
Obviously: I explicitly mentioned not all.. There are always exceptions but here we are talking about larger chunk.
Google search only yields some special case on mod $$$5^k$$$ (many papers stated that they are looking for something like "freezing trailing digits in tetration") as what I got when I tried Googling. Also tried searching for papers on Google Scholar but get no solutions.
GPT is illegal in contest so I didn't try.
(can't even submit once on pG cuz I can't think of a solution lol, although spending more than half of my time and skipping D, E, F for it. Would be appreciated if someone can provide a solution here)
Well I tested it post-contest
pG can be solved by ChatGPT 5 Thinking (post-contest submission). You don't even need Pro so that might cause even more cheating issue.
I could not solve it, but it is project euler right? I remember that in $$$O(log(m))$$$ steps, we have the result for all tetrations independent of $$$n$$$.
Ban ALL those shitty cheater CMs. Completely delete their account and shadow ban their IPs. MikeMirzayanov
How to do C
I think G is a good question, but with so many AI passing through this question, these people, especially Indians, should be severely punished
Don't say Indians bro, Instead say cheaters. Lol even you also took a new account and participating in contests instead participate with your old account itself.
How to solve D??
https://mirror.codeforces.com/contest/2147/submission/339575538
I just guessed the solution, but it got pretests passed, the only issue I could think of when the testcase is something like, 1,1,2,2,3,3 but my code is passing that case as well, not sure what will happen after System testing is complete...
No one want to choose a even number $$$x$$$, because the other player can choose $$$x-1$$$, then you gain no points, or maybe losing some points.
Even numbers :- both will get same number of points no matter what
Odd numbers :- this is the game changer,because one will get (x/2) and other will get (x/2) + 1 times.Thus both will prefer odd numbers with max_freq. So just take <x,freq> pair sort it in desc order,on alice's turn she will get (x/2)+1 times the freq and bob will get (x/2) and on bob's turn he will get (x/2)+1 times and alice will get (x/2) times.
For even numbers :- both will get (x/2) times the freq.
Contrary to most people I solved C and E but not D lol. I would love some logic on D.
At least in my solution, you don't need to consider the case where a number changes into another number (i.e. 2 2 3 3 -> 2 2 2 2)
my approach was a bit simple.Create a vector of pair storing frequency and the number and then sort it in non increasing fashion. Now alice will start first, as he is starting first he will get upperbound of that number i.e frequency*(number+1)/2 and bob will get frequency*(number)/2. Now the only thing we need to understand is who will start picking first when the next number starts which is simple to find, if the prev number was odd and alice started then for the next number bob will start first and vice versa. For the even number the person who picked first prev will pick the next one first as well.
https://mirror.codeforces.com/contest/2147/submission/339586315
why is it failing
Thanks to the authors, nice round! There are a lot of cheaters on G, I hope you'll check them out. I was also confused by this, I passed a n^2 solution in F(TL 4), but decided to leave this task and switch to E because of the small number of OK on F. Although without mass cheating, I think a lot of people would have passed F.
my best
How to solve E?
First, we prove an important lemma: There exists an optimal set of changes such that every set bit in the OR is either set before any changes, or is a part of a suffix containing only set bits.
Assume otherwise. Consider an optimal solution, and consider the leftmost set bit in the final OR that wasn't set at the beginning. Then, by undoing some changes to a number with this bit, we can reach a number that has all smaller bits set, and by undoing every change to any other number, we get an OR that has all bits to the right of this one set and all original set bits set. So either all of them were set before, meaning the OR has the required form, or this new OR is also optimal and it has the required form. QED.
Now, for each possible suffix of ones, we calculate the minimum number of changes required to reach it. To do this, we can just go bit by bit starting with the leftmost one, if it's already set, do nothing, and if it isn't, change the number with the highest remainder modulo respective power of 2 (only these modulos matter now, as higher bits can't be unset by doing this and the required ones are already set). Calculate this for every bit, get the minimum number of changes for every suffix length, and then for each scenario, just pick the longest suffix of set bits that's possible. Time complexity is $$$O(n log^2(n))$$$.
But when we are calculating the minimum number of moves for a suffix, our past moves matter, because by adding to one of the numbers in the array you might lose one of the bits on the right.
I don't understand why your greedy approach for the last part works
Because you need to somehow get to that bit, so you need to change some number to set this bit, and changing any other number than the maximum after modulo can be simulated by changing the maximum to set this bit, then changing the other number to the maximum (actually its modulo), which takes the same amount of steps, so you don't lose anything by just changing the maximum.
I still didn't get the proof of greedy strategy to have all k bits in suffix set to 1 , setting only kth bit to 1 is trivial whichever has maximum modulo, needs least op but how does it work for suffix k bits?
oh, B is too hard for me. I only managed to solve it at 2:53
Same. I spent half an hour staring at it
I solved CD but was blocked on B forever lol
:>
You should write it on paper start from n like 1st term as n ,2nd as n-1 and so on till you reach the nth term as 1 then n+1th term as n and then again from 1 to n-1.
Or you could do that on visualiser (like I did).
That's just how some constructive problems are
Was G GPT-able? I saw that more people were solving G then F and decided to skip F, so it affected my result even discounting the people who passed me unfairly...
There was a youtube/telegram leak.
After reading a complaint from LGM about AI, I became sure 100% that we are cooked.
Poor you, pal! many did the same. I hope this round will be unrated so it wont affect the try-hards too much!
Guessforces!
what special property does G have that it was solvable by GPT despite being so hard!?
Solving maths would consume less tokens from AI. the precision will get higher as the result
Is there a method to come up with a solution for constructives like B? I could only do trial and error.
Great problems tho :)
though I guessed by intuition , each i from 1 to n-1 will have distance 2i , and n will have distance n. So the sequence would be like n , n-1 , n-2 , ..... , 1 , n , 1 , 2 , 3 , ..... n-1
Woahh, thats way simpler than the bs construction I used, thanks xD
it took me an hour to come up with this
You can try printing all answers using brute force, and then try to see the pattern.
:)
Abolish combined rounds (to clarify: I enjoyed problems, but problem G situation is crazy, I regret solving it over F based on standings:/)
Problem F is really beautiful, thanks!
Now let's test GPT 5 Pro, which is granted for the participants of ICPC WF 2025!
G seems really GPT-able. chat
idk it's actually correct, but it passes the samples and some tests compare with AC submissions.
Thank you for sharing that (jokes aside). I see extremely similar and very suspicious solutions in my friend standings and this is really helpful for me to know.
Bro, actually gpt-5 PRO solves anything, H solution, idk why cheaters don't get TOP1 during the contests(I am not very familiar with modern technologies, so there is a chance that chatgpt googled the solution, but it says it didn't and I tend to believe it)
I like contest problems even though I am cooked, excited to know the solutions
going to lose lot of rating :( couldnt come up with a solution for B after staring at it for an hour and then -8 (wrong answer on testcase 2) for C.
Thank you for the round! B felt harder than usual but D and E were very cool.
I enjoyed working on G although I wasn't close to solving, sad to see that it was trampled by cheaters.
I actually find B very fascinating... Indeed, it took me way too long. But in retrospect, even though I did find a viable pattern, I know there was a more methodical approach. Particularly, I remember stumbling onto the fact that you could stack even or odd numbers onto each other. Additionally, that most numbers could be placed at a distance of 2x from each other, and not just x. Put the two and two together and you get a trivial pattern. For some reason, I didn't internalize that second point enough and didn't try to draw a conclusion like that. So I guess for future reference, look for more properties instead of looking for patterns?
Hi fellow 199!
Yeah it was a skill issue on my part. I remember staring at something of the form 53113542x24 for 30 minutes without any thoughts on how to vacate the x, but during the time I could've sought other constructions. If I tried writing a brute force I also could've solved sooner.
In general, maybe it's an issue of "too much thinking, not enough solving" ;(
omg, I didn't even realize xd
btw, weren't you also at the wf in Baku? Ain't that an interesting coincidence...
Why are we unable to see hidden pretest yet??
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Editorial?
Can anyone help me debug the code for Problem C. I used greedy, first fixing the direction of rabit for those pots where they dont have any other option like 11010 or 01011, then find if I can find the pair for them and then for rest of the rabbits which have options chose them greedily for left or right. Left those on the border or consecutive 0s as they always find a direction to not jump: https://mirror.codeforces.com/contest/2147/submission/339602741
204 solves currently for G. Excited to see how many solves after removals.
Now 48 solves. Impressive!
"UmM, aCtuaLly, I DidN't cHeaT! gIvE mE My rAtinG!!"
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Wasn’t d easier than c?
Auto comment: topic has been updated by BernatP (previous revision, new revision, compare).
For D, in this tc:
answer should be 5 2 right? I see multiple AC submissions that give 4 3.
its 4, 3 only.
can anyone please tell me why my solution for E couldn't work.
Problems were really Nice , don't know why there were that many G submissions though
I know many people are hurt by what happened with G situation and I understand the disappointment. I would like to ask to not direct any hate towards the authors though. As a tester, I can confirm that they spent a lot of time working on the round and making the problems for you to enjoy. And they are even dedicating more time to catch as many cheaters as possible. They are more upset about the situation than any of us.
So please, don't give cheaters all the attention and praise the authors for the work they put in to make this round possible!
is problem B == Leetcode 1718 ?
Not exactly, in B you must use two copies of 1, meanwhile in this problem 1 must only show up once. The problem you linked also does not allow distances to be multiples of the number, and you must output the lexicographically largest array possible rather than B which does not have that requirement.
I hope that authors will always be given pre-tests like these.
Are the tests for Problem D correct? My solution passes the maximum test, but gets a time limit on the smaller test.
UPD It's really anti-hash test?
Well, it was a hack that was added to the testset after the contest (which is a standard practice). I don't really know what we should have done differently in your opinion. Have it in pretests in the first place? Don't add the hack to the tests?
Good answer, but it would be better if you also good checked the tasks in the GPT before the round.
Recur + memo approach for C := 339626295
Thanks for the round:
unordered_map. A few days earlier I spent a long time on ABC 418E wheremapTLE’d butunordered_mappassed. I studied it superficially and used it in D today, which cost me the problem. Now I’ll study it thoroughly.The main thing isn’t the delta, but persistence in understanding concepts, starting from the simplest ones. Today my “persistence delta” grew a lot thanks to the contest.
Thanks to this contest because I've become a pupil now. The problems were delightful. And, C was just awesome. I found a greedy solution for C. Most of my friends wrote a DP solution for C.
What do you call Specialists?
will cheaters be removed from the standings and ratings recalculated?
We will try. It is hard to prove they cheated but I can ensure we will ban as many as we can
It seems that the authors have banned many cheaters, and the top 100-200 no longer looks the same as before
Spent all my time trying B, then I heard some coconuts solved G :D
I was just scrolling through the top ranks and the number of obvious cheaters is unreal. It's not just G! They clearly used GPT to solve A,B,C,D as well. Someone has said that GPT-5-Pro could even solve H. At this point, we need chess.com level detection of cheaters to make these contests fun again. Feeling really bad for the setters, testers, and honest participants. So many greys have outscored IGMs :(
Decided to upsolve I1 since I didnt look at the contest until a lot of time already passed, so I just looked at the hard problems that looked feasible to solve. It was actually a really cool problem. Interestingly enough, this problem shares a technique with a problem I already knew before.
The "make the first jump size really large, then make the next jumps only +1 larger for as long as possible" approach is very similar to one of the approaches in NOI.PH 2020 Finals 1B with the "very large values" padding so that the problem goes from unique subarray sums to unique pairs of subarray size and sum. The only major difference was that I1 needed you to go up/down to make sure you can return to the start for the next point skip size.
I1 ends up having a very similar walk possible by looking at the number of points skipped since it was effectively the same as "making subarrays that have different sizes always have different sums", since the gap between points can be treated similarly to subarray sums, or prefix sum:points skipped :: subarray sum:jump distance, except it only added a pathing requirement solved by going up/down (again, the small jump between the up/down segments is trumped by the points skipped amount so it doesn't matter)
I wonder if there's other problems that use a similar approach of using a big padding to split the problem into two or more parts, effectively allowing a single value to be a set of n values in a mixed radix system.
My DP solution to C: 339573488
I was comparing with what I did at (i-2) to calculate for the ith state, Here the second state if:
0: Means a plant
1: Rabbit placed left looking
2: Rabbit placed right looking
But as it is dependent on (i-2), I calculated manually first all starting [0][0], [0][1], .. [1][2] 24 states for initialization. How to think of better dp states or how to modify solution in better way.
Tried to solve F during the contest and afterwards — it seems that my "DCP-style" solution in $$$Nlog^2N$$$ is too slow and gets 5s to work (and $$$\ge$$$ 256Mb memory) (checked with changing TL / ML in mashup)
Well, at least it works! (Now it's time to check the authors' approach)
My rating suffered from skipping D, E and from not being able to participate unrated..
This was actually the first solution we came up with.
Try to think of another way of doing a segment tree such that you can erase ranges, I believe you have solved the first part of the problem.
Thanks for the contest. The Contest was amazing!!!
My solution for problem 2147C - Rabbits is 339594760. Time Complexity O(N).
Will there be rollback?
https://mirror.codeforces.com/blog/entry/146629 Somewhere in this blog one of the authors said that the score will be recalculated
Congratulations to tshirts winners! In a few weeks you will be contacted via private messages with instructions to receive your prize.
As usual, we used the following two scripts for generating random winners, seed is the score of the winner.
ORZ, Spanish CP goats!!!!!