Автокомментарий: текст был обновлен пользователем FelixArg (предыдущая версия, новая версия, сравнить).

Auto comment: topic has been updated by FelixArg (previous revision, new revision, compare).

Manifesting +ve delta for most.

Not that 6 — 7 problems joke again.

hope to become expert this round

I wish it a successful contest better than edu185！

edu185 is the worst edu I've ever participated.

Hope to become candidate master！！！

edu plz save cf again

if I solve 2 problems I celebrate, if I solve 3 I check if it`s a dream.

if i solve 3 question it soo good for me but if i have solved 4 question it will dream

hope dream come true

hopefully I don't reach newbie again after this contest

excited to try it out ;D

Why are Educational Div 2 contests turning into Div 3 day by day?

<3 lets do this!

Div-2 contest with Div-3 leaderboard loading i reckon

cl_crosshairstyle 4

cl_crosshairsize 2.5

cl_crosshairthickness 1

cl_crosshairgap -3

cl_crosshair_drawoutline 0

cl_crosshairdot 0

cl_crosshaircolor 1

cl_crosshairalpha 255

cl_crosshairusealpha 1

Aoje

2.

cl_crosshairstyle 4; cl_crosshairsize 3;

cl_crosshairthickness 1; cl_crosshairgap -5;

cl_crosshairdot 0; cl_crosshair_drawoutline

0; cl_crosshaircolor 2; cl_crosshairusealpha

1; cl_crosshairalpha 255;

cl_crosshair_t 1;

cl_crosshair_recoil 0; cl_fixedcrosshairgap -5

I hope I can solve min 2 problems today.

anyone on? lets do this :D feels so excited, need an exit.

gvil upi

queueforces

hoping to cross 1000 pls!!

The contest didn't go very well, partly because certain key points need to be clearly highlighted so that participants notice them, and partly because the instructions were worded incorrectly. Please correct this.

The math and the overflow problem in problem D, also the geometry in problem E cooked me so bad. Not really enjoying this round.

For E i thought of doing hexagonal packing. We can find the nearest circle centre and check if current point lies inside it. Hexagonal packing has ~90% efficiency, since we are forced to only take integer point, i think it would drop to ~89% which problem asks. I couldn't get through the rounding issue though.

i just hated B

How E? I can only solve if the percentage is $$$78.5\%$$$ instead of $$$89\%$$$.

A and B is more like GuessForces. On the other hand, C is great. In fact, I think this is the best C I've ever met.

E was worst experience ever

For F how do we compare 2 strings in less than 0(N). I mean we can use hashing but that gets hacked so what's the other way?

why F only pass with dc3

Problem $$$E$$$ is cool. My solution:

The essence of this problem is that we need to use circles of a fixed radius to cover the entire plane such that the coverage exceeds 89%.

A simple construction is to place circles in a regular grid:

$$$x = i \cdot 2r, \quad y = j \cdot 2r$$$

However, this results in a coverage fraction of $$$\pi/4 \approx 78.5\%$$$, which is insufficient for the required 89%.

A better construction is a hexagonal-like tiling using integer grid approximation:

• For odd $$$i$$$: $$$x = i \cdot \sqrt{3} r, \quad y = 2j \cdot r$$$

• For even $$$i$$$: $$$x = i \cdot \sqrt{3} r, \quad y = (2j+1) \cdot r$$$

This arrangement ensures the circles touch but do not intersect and achieves a coverage fraction higher than 89%.

Coverage Analysis

• Each circle has area $$$\pi r^2$$$.
• The hexagonal-like placement approximates a triangular lattice, whose area coverage fraction is:

$$$ \pi / (2 \sqrt{3}) \approx 90.7\% \gt 89\%$$$

Implementation Strategy

• For each point, check the four nearest candidate circles: top-left, top-right, bottom-left, bottom-right.
• Include the point in the circle where it lies within the radius.
• Collect all such circles to satisfy the coverage constraint.

This avoids enumerating the entire grid and works efficiently even for large $$$n$$$.

Hence, this construction guarantees enough coverage.

Had a solution to E that requires no randomization (aside from that which is given to us in the input generation :)).

Start by approximating the bounding box as being between the bottom left corner and top right. Then just make the direct hexagonal arrangement in row order and discard all circles that don't contain anything.

Hexagonal arrangement covers ~90% of the box. Every point is then uncovered with probability $$$\leq 0.1$$$. Using the Normal approximation for Binomial random variable then yields that the probability of failure (i.e. at least $$$1101$$$ uncovered points) is on the order of $$$4 \cdot 10^{-4}$$$, which (with 40 tests) has an overall success probability of just under $$$99\%$$$.

That being said this might be one of the worst problems I've seen on a contest — completely unreasonable to expect contestants to prove that hexagonal covering actually covers 90% of the grid and hence this degenerates to a boring knowledge test/implementation task.

I made video editorial for the key idea behind D. Exceptional Segments.

Highly recommend using Geodeb library for geometry problems like todays E!

Why do unofficial contestants like Nachia appear in the official leaderboard?

QAQ I don't check ranges of the data in D. I'm so stupid.

I actually struggled more with the logic on B than C. I reached my 100 day streak today. Thanks for the round!

I wasted 30 minutes on problem E because I got WA just for having trailing spaces at the end of the lines.

Why so much downvotes?? It was a good contest indeed.

my solution to A was

let y/x be d

y = dx

a number smaller than y but divisible by x can be written as dx-x or y-x

numerator(z) = dx-x (because y-x will always be a multiple of x if y is a multiple of x and greater than x+1) denominator = x

(dx-x)/x = x(d-1)/x = d-1 which means dx-x will always be a multiple of x now let's check the second condition, which is "y is not divisible by z"

dx/dx-x dx/x(d-1) d/(d-1) and we know that 2 consecutive number have hcf of 1, so remainder is not 0 at all

thus we come to conclusion that, z = dx-x, or y-x

but we know that x<z<y, so if here our z or y-x , is not greater than x then it is breaking the order rule, so basically it isnt possible when y-x <= x or not greater than x

and our final solution is

if y-x >x yes, else no

Why can't I hack submissions on problem F?

Weirdest contest ever, should have unrated

Possible idea source of E: https://www.youtube.com/watch?v=hGa8_P8X3yY Shouldn't implement it on an Edu.E ...

Problem E is the worst one I have ever seen.

Problem E: WA on test 36!

Sometimes it's desparate as well even knowing $$$\dfrac{\pi}{2\sqrt 3}$$$

Please ban cheaters Mostafafied and moh-kaz333.

Problem E:
1. https://codeforces.me/contest/2225/submission/372033090
2. https://codeforces.me/contest/2225/submission/372034904

Problem G:
1. https://codeforces.me/contest/2225/submission/372048258
2. https://codeforces.me/contest/2225/submission/372043513

Their code is exactly the same.

i genuinely don't know why this got downvoted to hell, it was your typical div 2 contest. people care too much about rating.

I solved all 7 problems but got hacked because I used hash in F.

The only chance to get grandmaster for me gone.

when will the ratings be updated ?

when I solved A to D, I know I have chance to get my rating up, but when I readed E, I know I'd better go to sleep(I'm in china exactly midnight when I participant the contest).

Is hashing even suited for problem F? I used 2 moduli hash (as suggested in editorial). Still got hacked by Sugar_fan. Like is Z function the only suggested method for this, though I commonly used it but was feeling petty for my Hashing template left unused since long?

Update: Should we be careful in choosing the primes to prevent anti-hash hacks? I used random_generator() and got AC.

CF Rating update time nowadays:

is not it rated?

Is this contest rated ??

is this contest unrated ?

How am I going to get rated ? Because it has been nearly 24 hours since I participated in this contest, but I still haven’t been rated yet (I solved one problem)[contest:189 div 2]

today, codeforces mailed me that my solutions for this contest are skipped because structure of my code looks same as some other candidate, i didnt copy any code from anyone, im new to codeforces , hence reading editorials may be i unconsciously wrote in same style but its my own code, i solved on my own.

please restore my solutions as i deeply respect codeforces rules . it was my second contest and attempted it with so much courage and energy.

Hello Everyone, I got a plag for the problem E(covering points with circles) . I really surprised by seeing the solution. It similarly matched but I don't even understand how this even happens. Let me tell my approach for this problem and share the screenshots of my coding style. Due to this it showed that I am out of contest.

Based on the problem statement my understanding of the question is

We need Place circles of radius r with integer centers, non-overlapping (may touch) Each circle must contain ≥1 given point Cover ≥89% of all points Points are uniformly distributed in a rectangle whose area ≥ 10 × circle area

Since points are uniformly random, the fraction of points covered ≈ fraction of the plane covered by circles. We need ≥89%. So I thought that hexagonal packing is better approach.

So, my intuition is rows at y = j·d_v, even rows have centers at x = 2r·i, odd rows offset by r (so x = 2r·i + r). For non-overlap, adjacent-row neighbors must be ≥ 2r apart: r² + d_v² ≥ 4r², so d_v ≥ r√3. So,For each point, we need to find its nearest hex lattice center (4 candidates — floor/ceil of row and column indices). If it is within distance r, we should mark that center and need to print all marked centers. And the time complexity for this is O(n). So, I don't know why I faced plag for this. I did it on my own and I really can't understand why this happened. So, because of this will my rating and account will be affected? Or else will there be any further enquiry? Please restore my solution as I deeply respect codeforces rules.

Hello, regarding the plagiarism warning for 2225E: both referenced accounts belong to me. I submitted the same solution from another account to check whether it would be accepted before submitting on my main account. I understand this violates the rules and I won’t do this again. Sorry for the issue.

Hi first time getting flagged here but I'll try my best to explain my thought process for problem E . I used hex lattice packing w random offset to maximize coverage. This is a standard geometric construction for covering points, it naturally leads to similar coordinate transforms and local neighbor checks, which may explain the resemblance to other submissions... also used a personal template which used a fixed seed for randomization that I use (mt19937 rng(Leet)) generally in practice. I use sublime text as an ide, pc's too potato to handle both my browser and vsc. I did not copy another participant’s code during the contest. I will be more careful to ensure complete independence and perhaps employ other concepts during future contests. 372019544

@psgawa is it still possible to hack the solution of F if we use single hashing with modulo as 2^61-1 and using randomized base ??

I was flagged for plagiarism in 2225D, I think this is a mistake since my solution doesn't match with those flagged.here is the link to my submission :

my submission

some of the other flagged ones are as follows :

flagged 1

flagged 2

flagged 3

flagged 4

flagged 5

None of these submissions match my code or structure completely. flagged 4 is not even cpp code. I believe this is a mistake and request you to please verify and correct this mistake.

FelixArg, BledDest, basalov_yurij, MikeMirzayanov

Subject: Regarding plagiarism detection on problem 2225D [Submission: 372013367] Dear Codeforces Team, I hope you are doing well. I am writing regarding the plagiarism notice I received for my submission to problem 2225D (Submission ID: 372013367). I would like to respectfully clarify that I solved this problem entirely on my own. I did not use any unauthorized resources, nor did I share my code with anyone. The approach I used is based on a well-known mathematical property of XOR prefix sums, specifically the pattern involving ( k \mod 4 ), which is commonly used to optimize XOR computations. Because this technique is standard, it is possible that my implementation appears similar to others who applied the same logic. Additionally, I used a std::map to maintain frequency counts and iterated through the ranges ((0, x-1)) and ((x, n)), which I believe is a natural and straightforward approach for solving this type of counting problem. I take the integrity of competitive programming very seriously and always strive to follow the rules of the platform. I kindly request you to review my submission again. I am fully willing to explain any part of my solution in detail if needed. Submission link: https://codeforces.me/contest/2225/submission/372013367 Handle: shubhachakma Thank you for your time and consideration. Sincerely, Shubha Chakma

Hello FelixArg BledDest basalov_yurij MikeMirzayanov Please read full

This evening, I got shocked after getting plagiarism mail after giving 190 contests on Codeforces in this contest in question D(Exceptional Segments).

Just want to put some points from highest to lowest priority: - I took the idea to solve the subproblem of this problem from blog: https://www.geeksforgeeks.org/dsa/calculate-xor-1-n/ which does not violates the rules of codeforces. - It is stated in the mail that solution with ID: 372046384 is matching with others. But this solution of mine was not even accepted. The accepted code is with ID: 372047912. - I was already having a pathetic contest(solved C problem with a lot of struggle) and was already expecting a big negative delta then for which purpose I will cheat in problem D !!! - I code on antigravity code editor which gives some free credits of claude opus. The moment I got solution in my head with help of that blog. I directly explained the solution to my inbuilt claude and it wrote the solution. Now I don't know how the heck it wrote wrong modulo value in it's solution due to which I got that wrong ans on pretest 4 (the code which you caught). But quickly I caught that thing and changed modulo value then submitted and it got accepted(there is nothing said about this code in your mail). - Never thought that after genuinely grinding through all these years and transitioning from an average student to a CM and not stopping even after reaching CM (most of the Indians stop) would later result me to such a rubbish plag tag :) **** MikeMirzayanov please look into my matter as fast as possible. If you want any further proofs then ask me I am ready to cooperate.

I think I know why.I would't use AI translator anymore....

i am writing an appeal for the warning for submission of D problem.. I checked the matched submission that were given to me.. most of the solutions didn't match with my submission.. that is very disappointing for me for such a contest as i am new to competitive programming as well as very demotivating for me.. Please check out the matter my submission: 372032649 for 2225D.