Hello Codeforces!
We're glad to invite everyone to participate in Codeforces Round #1091 and CodeCraft '26 (Div. 2) — IIIT Programming Club's eighth round on CF! The contest will start on Apr/07/2026 17:35 (Moscow time) and will be rated for all participants with a rating below 2100.
There are 8 problems to solve in 2 hour and 15 minutes, with two of these being easy and hard versions. You'd want to read all the problems :)
This round has been over a year in the making, and a lot of people helped us put it together. Thank you to:
- cry and Error_Yuan for their wonderful coordination.
- Um_nik and satyam343 for a detailed review of our initial set and bumping the quality.
- Alexdat2000 for Russian translations.
- authors h2rsh, SilverTongue1729, Faizal, tanaygad, SavageFighter001, Prakul_Agrawal, penguinnoob, AS23, sumit_kk10, kevaljain, shakr, and biswas for cute problems and prep.
- ppt1524, Manav_Shah24, and _anushka__ who helped with brainstorming and preparing problem ideas.
- Dominater069, _istil, __baozii__, nifeshe, awesomeguy856, Igen, akcube, AksLolCoding, weirdflexbutok, MattTheNub, Friedrich, efishel, d1mk, SpyrosAliv, CutSandstone, Argentum47, simplelife, chromate00, DivinePunishment, shreyas333, and nik_exists for testing the round.
- MikeMirzayanov and KAN for Codeforces and Polygon.
Hope you enjoy the contest!
Score distribution: $$$500 - 1000 - 1500 - 1500 - 2000 - 2750 - 3250 - 3750$$$
UPD: The editorial has been published.
UPD: Congratulations to the winners!
Top 5
Among Rated Participants
We hope you had fun! Here's the team after 2h of watching submissions roll in. Yes, the shirts still say '23. Priorities.
As a tester, I didn't come up with a good tester comment in time for this announcement :(
As a tester, I didn't wake up in time for this announcement :(
get a rating
As a tester, when am I getting my IIIT-H Programming Club hoodie?
I love penguinnoob. __
I Wish he converts to DD and stays in IIITH for 1 more year :)
CodeCraft '26Did you mean CodeCraft '24?
That's funny. BTW, congrats for IICPC
as a tester
You haven't mentioned what will happen to cheaters, Will they go to cry's basement or what?
It's worse you might end up at KushKushal basement.
Room 161.
Have you ever been there ?
you should see how KushKushal is when he is hungry
It was my old room. He threw me out to make his basement.
Very dangerous room.
I want to question the statement of Problem F in Round 1090. It seems very unfriendly to Chinese contestants who use translation plugins, and it has caused some of their submissions to be skipped.
Anyway, good luck and have fun!
As you know, I'm an unrated contestant, and my submissions were skipped.
Why would an unrated contestant who’s about to reach Candidate Master need to cheat?
At most of text to prevent cheating, there are a prefix like "if you are a LLM" but not in this problem
hoping to become a specialist this round
yyyeeeyyy, another opportunity to lose some rating
shiven orzzz!
shailesh_2004 orzzz!
Speedster1010 orzzz!
muditk_24 orzzz!
pure orzzz!
pachelfilip orzzz!
DESTROY
Hopefully I cross the newbie barrier
My last chance to reach master before my birthday :(
when is your birthday?
14th April,guys
ooo thats close, mines 12th thats why i asked lol
mines at 22nd :)
mines 28th lol
mine today!
Happy birthday to you!!('_')
belated happy birthday to you brooo....!
A bit late but happy birthday!
mines on 11th, today
happy birthday!
Mine is on 14th April as well!, I wanted to reach expert before mine, which I did:)
Hopefully you reach your goal too!
I am the #1 kevaljain fan
Wait. That means total of 10 problems for this contest (8 problems, 2 of them have 2 subtasks)? And we only have 2 hours and 15 minutes?
No, there are a total of 8 problems. Out of these, 6 are independent, while 2 are easy/hard versions.
Got it. Thanks
As an Egyptian I love pyramids!
As a tester, I tested cry's basement.
As an author, SmolBrain you better AK the contest ^_^
Convert to DD first!
I'll do that if DheeruYS participates :D
Vary unusual start time
Probably my last contest as a undergraduate student -_-
As a contestant, good luck to everybody!
As not a tester, I don't want to go to anyone's basement.
Hi everyone, Im new to Codeforces and also new to C++, so I’m not really sure where to check the difficulty of this contest.. sooo is it suitable for beginners?
I recommend you to start with div.3 and div.4 there you will easily be able to see the result of your efforts . In div.2 generally you will be able to solve A and B only then you will get stuck and bored or annoyed .
Okii thankss!
Too much authors.
AuthorForces
Back after 2 years, lessgo
Why is akslolcoding testing literally every contest like are there any contest that my guy is actually participating in
guys it is my first time in codeforses i'am only here for schol but i'am glad to be part of this community
where is the score distribution??
same question
oh, score distribution appeared
yesss
IIIT HYD is conducting context. In which div level it
Test is coming soon.But score contribution is still a mystery......
Hey, just wondering does anyone thing like someone like myself with a rating of 1313 do this contest ???
Please dont share score distribution after the contest :D
C and F subtask??
how to be a tester?
Was the top of leaderboard banned? Stormed through the problems and vanished, LMAO
$$$C$$$ and $$$D$$$ are so bad problems. Just guess, don't bother trying to come up with observations.
Is there any proof of gcd in c? Or is it just observation
YES that's a REALLY well known concept in mathematics, I can't comprehend why is everyone hating that question
can you please tell me the topics or any resources to study to be able to know that proof
Sorry I wasn't able to find any sources but here let me demonstrate myself:
take a frog that jumps from a lily pad to another on a circle of N lily pads, jumping S steps to the right each time. As there aren't infinite lily pads one can see that it is required for the frog the step on a lily pad twice. Lets say that the first time the frog steps on a lily pad L for the second time, there was k steps between the first and second time it stepped there. Since we know that the frog took k steps to come back to where it started we might now conclude that the frog come backs to its current location every k steps as the lily pads are symmetrical, that is to say there are no difference between them, so coming back to one lily pad after k steps means k steps always makes the frog go back to initial position. and since a lily pad can't be visited twice before taking k steps we know that the frog has visited k different lily pads on total, and will ever visit them, as a loop started. So the question is: what is k equal to? There we have kS=0 (mod N) all leafs visited should mean that we have no kS=0 until k=N thus there should be no k such that k!=N and kS=0 mod N, if S and N had a common divisor d, we would have (N/d)S=0 mod N as S=ad for a random int a and (N/d)S=Na which does equal to 0 mod N. This question also requires showing that on the case that the only common divisor being 2, there are exactly two possible routes for the frog to go but using what I have already told, it should be easy to prove that yourself. (turns out the frog example was made up by our teacher so searching Frog Theorem as our hs teacher called it showed no results)
I created video editorial for D. Flip the Bit (Hard Version)
I also created video editorial for E. Definitely Larger
Problem F has 2 independent parts, one is Game Theory, and the second is Bit Manipulation / DP. I made a video editorial for the first part.
SHIT ROUND
guessforces
Why these problems reminded me of JEE
HUH?
For $$$C$$$, I just guessed if
gcd(n, m) > 2then no solution. And I assume $$$D$$$ has to do something with blocks of 0's and 1's.C should have been easier, i think and i don't know how there are more than 6k people who solved that, most of them must be cheaters
Yes, rankings and ratings have no value now, and no CP authority seems to care enough to act*, so I guess competitive programming might just die slowly...
*I don't count a cheaters database as acting, that is pretty useless. The incentive to cheat needs to be destroyed entirely (e.g. by not making rankings and ratings public).
Well for me C was even easier than B honestly, the coprime finding algorithm is well known and after applying it to see if each row and column would get visited, applying it once again to see if each visits would coincide was also a really basic idea as well.
D is much much harder than E. Still can't understand why there are so many ACs in D
Probably the AI effect.
Feeling dumb after seeing submission count on C
Same :(
They just cheat bro
F is a beautiful problem, but this round had some issues, for instance D is WAY harder than E and for C you just gotta guess and assume correct, i didn't get to solve G and H so i won't comment on those
I really enjoyed C, thank you.
can you prove it?
It would be too tedious to write out a full proof, but here's a sketch for why $$$\gcd(n,m) =2$$$ works, but $$$\gcd(n,m) \gt 2$$$ doesn't work. (The rest of the problem is trivial).
If $$$\gcd(n , a) = 1$$$ and $$$\gcd(m , b) = 1$$$ you can just pretend $$$a = 1, b = 1$$$.
Maximum order of an element in $$$\mathbb{Z}_n \oplus \mathbb{Z}_m$$$ is $$$nm/\gcd(n,m)$$$. $$$(1,1)$$$ achieves this maximum.
If $$$\gcd(n,m) \gt 1$$$, $$$(1,0) \notin \langle (1,1) \rangle$$$.
The visited cells choosing to go right first are in $$$\langle (1,1) \rangle$$$ and $$$(1,0) + \langle (1,1) \rangle$$$. Both of these sets are of cardinality $$$nm/\gcd(n,m)$$$ (and distinct if $$$\gcd(n,m) \gt 1$$$), so the total number of visited cells is $$$2nm/\gcd(n,m)$$$. If all $$$nm$$$ cells are visited, then $$$\gcd(n,m) = 2$$$.
if you're at $$$(0, 0)$$$ assume $$$0$$$ based indexing, lets discuss the necessary conditions.. all row indices must be visited independently -> $$$gcd(n, a) = 1$$$ similarly $$$gcd(m, b) = 1$$$ Now, no. of moves till it takes a circle and comes back to origin is $$$2 * lcm(n, m)$$$ And we should visit all atleast once..
These are the necessary and sufficient conditions
Can you give some mathematical proof for the number of moves till it takes a circle and comes back to
(0,0)being2 * lcm(n, m)?I'm unable to understand this part.
We have n*m positions, and since all positions are symmetrical we can say that if it takes S steps to come back to a square after moving from it, than we have coming back to any square from it also takes S steps. Thus in those S steps we know that we have also visited S different positions. Lets define a step as moving left and down 1 step (as we can pretend a=b=1 now). Here at Sth step we are on (S mod n, S mod m). For it to make 0,0 we should have n|S and m|S, thus the first time we get back to 0,0 namely the amount of positions we visit becomes lcm(n,m) BUT we defined steps as moving left and down (or down and left) if we seperate those steps we see that we visit double the number of positions, thus comes 2*lcm(n,m)
Anyone How to Approach D ?
You will need this problem as a prerequisite. Thanks to aryanc403 for showing me a beautiful proof of this in the past, I also created a video for proof
Once you are done with the prerequisite, you can watch the edtiorial
If you want a hint you should start by thinking about transition points within the binary array. They will come in pairs of two: one point where a "wrong" block of bits begins and one point where it ends. Since each operation has to contain one special index, you should think about splitting the array in blocks, like this: 1..p[1],p[1]+1...p[2] and so on. Take this as a starting point.
If you can't solve the problem here is my submission, but please take the time to think about it, the problem is pretty cool. https://mirror.codeforces.com/contest/2217/submission/370145378
For C I observed that you can reach all values from 1 to n with a jumps only if gcd(n, a) == 1 and similarly for columns gcd(m, b) == 1. I thought that if you can reach all values separately than you can reach them together somehow after many jumps but I was wrong in the assumption. Anybody solved this, can you tell me what was the other necessary conditon?
gcd(n,m)<=2
why does it work? I mean I understood my two conditions but I couldn't understand why this condtion should be true...
If you start from 0,0 then the number of unique cells that you visit before moving to start would be 2*lcm(m,n), so it should be greater than n*m hence gcd(n,m) <=2
Thanks mate that helped a lot...
Whoever wrote D, I'll hunt you down (then I'll do nothing but tell you why)
Welp
Can anyone explain to me how to do C I'm so confused after spending an hour on it. I was just drawing random things and telling myself I can't simulate it because it's 10**9 and just trying to find something mathematical to use
How to simply prove the conclusion of Problem C?
assume correct ♥️
proof by accepted
First, it is apparent that $$$(n, a)=1$$$ and $$$(m, b)=1$$$ must hold.
Then, if $$$n$$$ and $$$a$$$ are coprime, and $$$P={1,2,...,n}$$$ is a permutation of $$$n$$$, then $$$aP={a, 2a, ..., na}\pmod n$$$ is also a permutation of $$$n$$$.
So we can arrange the index（row and column) $$$[1,2,3,4,...]\to[1,1+a,1+2a,...]\pmod n$$$. In the new grid, the two moves transform into:
Jump 1 step right
Jump 1 step down
Firstly, to visit all the rows, we need to visit the row after the current row we are standing on after a cycle is completed starting from the current row. That implies, gcd(n, a) = 1. Similarly, gcd(m, b) = 1 is the condition for visiting all the columns. Now, after a cycle (moves after which we return back to (1, 1)), all the cells must be visited. Otherwise, we can never visit all the cells. Let's count the number of moves required to complete a cycle.
Number of moves to complete a cycle = 2 * lcm (n / gcd(n, a), m / gcd(m, b)) = 2 * lcm(m, n). Let's understand why 2 is multiplied here. We need lcm(n, m) moves to complete any one of the rounding (either row wise or column wise. It actually depends on the first move we choose). But, we have to round both row and column to get back to (1, 1). So, we need 2 * lcm(m, n) moves in total. Also, in a cycle, each move covers a unique cell. So, the number of cells covered in a cycle = 2 * lcm(m, n).
To fulfill the condition, 2 * lcm(m, n) >= m * n -> gcd(m, n) <= 2.
So, the answer will be "YES" iff gcd(a, n) = 1 and gcd(b, m) = 1 and gcd(m, n) <= 2.
Thank you so much,intuition is tough though
If gcd == 1, when you finish a full loop, you'll eventually arrive back at tile (1,1) but moving in a different direction (if your first move was right, you'll come back and move down, and vice versa) This triggers a second phase that fills in all the gaps
If gcd == 2, when you finish a full wrap, your path shifts perfectly landing you right next door on a tile like (1,2) or (2,1)
If gcd == 3 (or higher), the gap is too wide, the loop closes too early, and some tiles will never be reached
This is what I understood
I used two pointers for B was that the easiest way?
no
I did it by checking the bit flips on either side and taking the maximum...Maybe simpler than two pointers but depends on how you implemented 2 pointers...
I checked for the number of consecutive substrings that need to be converted to the left and right of the special index, then min(count_strings_left, count_strings_right) + abs(count_strings_left — count_strings_right)*2, since for 1 sequence to the left and 1 sequence to the right, we need 2 more operations. The remaining imbalance of sequences needs 2 operations each (in whichever direction they are)
Feeling frustrated as a person who loves writing big codes, on B I got in on 6th try with 117 lines of code :(
I got TLE after submitting a apparent O(n) code lol.
Thanks for all the effort you put into creating this contest. Not a single problem in this set is boring, even though i did not manage to perform well in this contest. Better luck for me next time.
أنت كويس
Cbroke my dream of coming back to expert rankJust rename the platform to MathForces atp.
hardest problem C i've ever seen
Can we get the fastest solves for each problem , lol , by friend got it for D (he is shy) :)
Timmy1003 is a cheater, look at his H solution and other solutions. the amount of code and the style is different to A — D solution.
Also if you look at Timmy1003's past contests https://mirror.codeforces.com/contests/with/Timmy1003, in Codeforces Round 1089 (Div. 2), he solved 0 but got rating +8.
I was trying something like this in problem C There exists a number k[i][j] such that k[i][j] = i mod n k[i][j] = j mod m For all (i,j) in [0;n-1] cross [0;m-1] Can anyone extend this approach? Possibly simplify something using CRT?
In problem C, gcd(n , a) == 1 and gcd(m , b) == 1 these are obv necessary conditions. But how to prove gcd(n , m) <= 2?
Your state will repeat itself after $$$\frac{2mn}{gcd(m,n)}$$$ jumps, thats why $$$gcd(m,n)$$$ must be not exceeding $$$2$$$.
To visit every tile, the sum of two sets of landings (even steps and odd steps) must be at least the total size of room, easy to get
=>
=>
still can't get it. suppose S[r] denotes the set of positions that even steps visit, and S[l] denotes the set of positions that odd steps visit. Shouldn't the condition be |S[r] ∪ S[l]| >= n * m ?
gcd(n, m) <= 2 only guarantees |S[r]| + |S[l]| >= n * m.
I imagined reordering the rows and columns so that every operation looks like one step right or one step down.
You can use Bezout to prove it. You can work with 0 indexed because it is easier.
Suppose you choose to add b k1 times(so now i is congruent to b*k1 mod n), now in order to have the same residue of i mod n, we should set k1 to k1+ k*n. Same thing applies for columns. Then we can just choose k1, k2 the number of times to add b to i and j respectively such that abs(k1-k2) = 2, then the condition of the problem obliges us to make abs(k1-k2)=0 or 1, in order to do that we should find x and y such that x*n + y*m = 1 or 2, which means that the gcd of n and m is <= 2
I think D>E, F, G.
how to prove in $$$C$$$ that $$$gcd(n,m)$$$ $$$\le$$$ $$$2$$$
I am assuming you already understand why gcd(n,a) and gcd(m,b) are 1. Assuming these two are true, you can observe the following:
if the gcd of (n,a) is 1, then LCM(n,a) is n*a. this means that after travelling n*a squares the person will be at their initial position. each jump has height a, so n*a jumps takes n jumps and thus the vertical position of the person oscillates with time period n.
a similar line of reasoning can be used to deduce that the time period for the oscillation in the horizontal direction is m.
assume for example it takes 8 jumps to get back to the origin vertically and 7 jumps horizontally. this means that we end up at the exact same initial position after 56 turns (with both jumps). thus, the most amount of unique squares we can visit is bounded by lcm(n,m). since each turn has one horizontal and one vertical jump, we can visit at most 2 * lcm(n,m) unique squares before returning to the origin.
to cover the whole grid, the maximum amount of visitable unique squares must be equal to or greater than the number of squares in the grid. Thus 2*lcm(n,m) >= n*m. With a bit of algebra, this is identical to gcd(n,m) <= 2. (I used the lcm version of the condition in my solution anyway)
Dont ever write a contest ever again
Womp womp.
@Musa_82FaGm used AI for sure, look at the problem H, and other solutions, he used
if(!(cin >> xxx >> ) return; some solution cin >> xxx; directly. only GPT/Gemini would do defense IO in codeforces, and you can see the almost the main and coding style is different for each solutions. the top standing almost all ai. this is AIforces. mf
Problem C is beautiful.. Was a bit upset initially that it is some random guessing, but the solution is very elegant and is something that can be come up with.. Great!
IN PROBLEM B , 3 1 0 1 0 2 why the answer is 2 , when i can do [0,0,0] just in one operation?? can anyone explain please .
Let x denote the value at special indices before any operations are applied. ... You need to get 1 1 1, not 0 0 0
In one operation it would become 0 1 0.
can anyone give me the valid mathematical cum logical proof for c gcd(n,m)<=2 i guessed it into the contest but searching for proof and not finding any , it will be really helpful
For every x,y there is cycle with some length. Cycle length is also number of visited cells.
When gcd(a,n)=gcd(b,m)=1, let:
Let l be min_k. If l%2 is 0, real length is lcm(n,m). If l%2 is 1, then if you started with horisontal step, now step will be vertical. Real length is 2lcm(n,m).
the chances of 2 people replying in the span of 10 minutes after the question was hanging for 50 minutes already is low, but never zero, lol
Assume $$$\gcd(a, n) = 1$$$ and $$$\gcd(b, m) = 1$$$, otherwise the answer is NO
Suppose the steps were done simultaneously, then you add $$$(a, b)$$$ to your current coordinates. Can you reach the cell $$$(a, 0)$$$ (which would normally be achieved because of the alternations)? The coordinates have the form $$$(at, bt)$$$, so to reach $$$(a, 0)$$$ we must have $$$a \equiv_n at$$$ and $$$0 \equiv_m bt$$$, which means that $$$t \equiv_n 1$$$ and $$$m | t$$$, which is possible iff $$$\gcd(n, m) = 1$$$. Okay, but what about the case when $$$(a, 0)$$$ isn't reachable. Then $$$(2a, 0)$$$ must be reachable from $$$(0, 0)$$$ by moving in the direction $$$(a, b)$$$, since it cannot be achieved from $$$(a, 0)$$$ the same way. Again, we get that $$$t \equiv_n 2$$$ and $$$m | t$$$, which requires $$$\gcd(n, m) | 2$$$. And if you can reach those cells, it's not hard to show that all other cells are reachable as well
Thanks for explanation buddy
Video editorials.
https://mirror.codeforces.com/blog/entry/152780
Youtube link
Thanka
I couldn’t think of a worse round, thanks.
Why hasn't the rating been calculated yet? Is the contest still rated?
Same for me bruh.
wow there are so many algorithm geniuses in a div.2 round where so many grandmasters (even lgm) are beaten by so-called [trusted participants] :)
I hate C
Why isn't it rated for me? It says not rated in the rating chart but here they say it's rated. Im really confused.
Same, it's not rated for me either. That's odd considering the first ever round I participated in was rated and shows up on my profile too. How am I supposed to know if a round is gonna be rated for me or not?
The round is shown as unrated for all participants till the rating is calculated and updated, which has not yet been done
Yup, it got updated now. Thanks for the explanation!
Nice Contest!! btw how to become a tester in a contest?
Could you guys update rating before the heat death of the universe plz :3?
just to complain, i think problem authors and testers should really check the ambiguous or confusing expressions in your descriptions. last night i stuck at problem a for approximately half an hour and eventually gave it up. I mistakenly thought one could decrease multiple positions, because the "some" word in the description. i didnt realize it until just now i went through it again and found there was a "it". changing "some" to "a" would be much clearer for foreigners. i also recommend authors fix description in problem g. it is not said in the description that left and right son are taken differently, and the "labeled binary trees" would easily mislead the participants to count labeled trees(nodes labeled from 1 to n). these details should be clarified in descriptions rather than letting participants to find out by checking the examples. it would be my pleasure if you could accept these suggestions.
The first few questions of this competition were almost testing the luck of whether one could find the patterns. I think this competition is really terrible.
Is this round unrated?
Why hasn’t it been scored yet after 12 hours?
Seems like the contest didn't get positive feedback from the participants, that's why lol
But yeah that's the first time I've seen such a long time being taken to update ratings in a div. 2 round
why ratings are not given yet?
Me and my homies waiting for the ratings:
Me waiting for the contest to become unrated.
Alas!
Please calculate the rating!
I have wait it for 14hours.
Edit: Rating Updated, I got 2098 ?????!!
WTF ?!
Yeah,I also feel strange about why my rating dosen't change this time.
after rollback you will reach master
When will the ratings be rolled back and recalculated?
I have waited for it for 8 days.
Calculate the rating when! T_T
Hope not unrated.
Edit: +369 rating hooray! ^_^
Congratssss
Is the contest still rated because it is taking quite long now for a div 2 contest ratings to be updated?
test
I'm frustrated too with the delay in rating change announcement, but can you guys just stop bombing downvotes while waiting for rating results? Organizers really put a lot of effort into making this round.
You should have waited 7 more minutes...
Cool, finally rated.
But why did it take that long?
[deleted]
[deleted]
Easily the worst contest I participated in so far. That was so disappointing.
Hello,
I received a notification regarding similarity between submissions for problem 2217B.
I want to clarify that both accounts mentioned are owned and used by me. The similarity occurred because I used the same approach on my own accounts.
I was not aware that participating with multiple accounts in the same contest violates Codeforces rules.
I understand this now and sincerely apologize for my mistake. I assure you that I will use only one account in future contests.
Thank you.
Hello,
I recently received plagiarism warnings for problems 2217D and 2217F in this contest, and I was quite surprised and concerned about it.
I would like to sincerely clarify that I solved both problems completely on my own during the contest. I did not access or refer to any other participant’s code. I spent time thinking through the approach myself, mainly using XOR transformations and bitwise reasoning, which I understand can sometimes lead to similar implementations among participants.
It is a bit disheartening to see my submissions flagged, as I always try to follow the rules and compete fairly.
I understand that structural similarities can happen in such cases, and I respect the system in place. I kindly request you to review my submissions once again. I would be happy to explain my approach in more detail if needed.
Thank you for your time and consideration.
Deleted
Dear Codeforces,
I received a similarity flag on my submission 370158810 for problem 2217D, and I would like to clarify my situation.
I worked on this problem independently during the contest. Initially, I implemented a more direct approach by explicitly counting segments where elements differ from x.
My earlier submission: 370139065, 370145835
After that, I refined the same idea into a more concise implementation by transforming the array into a binary form and using a difference array (XOR) to track transitions between segments.
My later submission: 370158810
Both approaches are based on counting segments where a[i] != x, but implemented differently (explicit counting vs. transition tracking). I believe the similarity comes from using this common technique.
I did not copy any code from other participants. My submission history reflects my step-by-step development of the solution.
I respect the Codeforces rules and would never intentionally violate them.
Regarding the system warning, I want to clarify that I compiled and tested my code entirely locally. I did not use ideone.com or any public online compilers during the contest.
Thank you for your time and for maintaining a fair environment.
Sincerely, Sogni265
Respectful Appeal: False Positive Plagiarism Flags on 2217B and 2217D (pulkitjaincs)
Hello Codeforces Team and Round Coordinators,
I am writing to respectfully request a manual review of the coincidence flags placed on my account (pulkitjaincs) for my submissions on Problem 2217B (370128856) and Problem 2217D (370152721).
I fully support the automated systems used to keep Codeforces fair, but I coded these solutions entirely independently in my local VS Code environment. Because both problems have straightforward optimal paths, the underlying logic aligns with other users. However, a manual look at the raw code reveals undeniable proof of independent creation:
Evidence for 2217D (Dynamic vs. Static Array Construction):
My Approach: I dynamically constructed my padded auxiliary array using .push_back() line-by-line (pExt.push_back(0);).
Flagged Users: Both Starrism0203 and BongCoder... used a completely different conceptual approach: pre-allocating a static vector of size k + 2 and using direct index assignment (P[i+1] = p[i]).
Loop Boundaries: My inner loop explicitly defines startGap and endGap with an inclusive boundary (<= endGap), whereas the others rely on different boundary logic and prefix increment operators (++i vs my g++).
Consistent Stylistic Evidence (Across Both Problems):
Bracing/Indentation: My local VS Code auto-formatter uses strict Allman style (opening braces { on entirely new lines). The other flagged users consistently use K&R style (braces on the same line).
Macros: I reflexively included using ll = long long; but stuck to standard int types. The other users forced heavy macros like #define int long long and used signed main(), which I did not use at all.
Naming Conventions: My variables (cL, cR, maxC, pExt) are distinct to my personal coding style.
I take the rules of this platform very seriously. I kindly request that a moderator review these specific, human-typed differences, as they strongly indicate my code was written from scratch.
Thank you for your time and hard work in maintaining the integrity of the platform.
C problem was very frustrating. I was trying to prove my approach but couldn't do so in the contest. Had to see the editorial.
If anyone has seen some similar problems to C, i would very much appreciate sharing them with me.
How many of the top 5 (rated) used chatgpt? Very interesting contest history for some of those participants...
All the best!!
Finally spec yay
D is trash
i got -61 was able to solve first problem only , dammm bad at cp
learned a lot from the contest