Hello Codeforces!

ijxjdjd and I are pleased to invite you to participate in Codeforces Round 728 (Div. 1) and Codeforces Round 728 (Div. 2) which will be held on Jun/25/2021 18:35 (Moscow time). **Please note the unusual timing**. Each division will have **5 problems** and **2 hours and 15 minutes to solve them**.

We would like to thank:

adedalic for amazing coordination of our round!

singleLOOP, skittles1412, penguinhacker, wxhtzdy, coderz189, dorijanlendvaj, qlf9, Nezzar, Tlatoani, MagentaCobra, Etherite, and prabowo for testing the round and providing invaluable feedback.

MikeMirzayanov for the amazing platforms Codeforces and Polygon.

You, for participating!

Read all of the statements carefully, and we hope you enjoy the problems! Wish you high ratings!

**UPD:** I would also like to thank Roberticey, golions, and 244mhq for testing.

**UPD:** Here's the score distribution:

Div 1: **500 — 1250 — (1500 + 750) — 3000 — 3500**

Div 2: **500 — 1000 — 1500 — 2250 — (2500 + 750)**

**UPD:** The editorial is out!

**UPD:**

Congratulations to the winners!

**Div. 1:**

**Div. 2:**

Omg purplecrayon orz. Can't wait to solve your geniosity problems

Omg purplecrayon orz. Can't wait to solve your geniosity problems

Omg purplecrayon orz. Can't wait to solve your geniosity problems

Omg purplecrayon orz. Can't wait to solve your geniosity problems

Omg purplecrayon orz. Can't wait to solve your geniosity problems

The comment is hidden because of too negative feedback, click here to view it

The comment is hidden because of too negative feedback, click here to view it

Don't steal my tricks. lol

Thanks for the blog, hope to see u in purple next time ;)

Whatta coincidence! The sequence of their profile picture is like Mocha bear is rotating milk bear in some order :o

that's the point... cf is tired of gyrating cats

Is it a coincidence that the first question in div 2 contained gyrating cats. I am not into vodoo stuff but this sh*t is deep.

well, purplecrayon is part of our gyrating cat club :P

Those are bears..? For so long I thought they were cats.

.

It's a bad sight that the first five comments on the blog have all these negative votes!

Suggest removing it, MikeMirzayanov

I hope everyone enjoys our problems and learns something new!

PurpleCrayon is a skilled programmer, and I'm excited to see the tasks he and ijxjdjd have prepared. Good luck to all participants, and hopefully we can see more crayon rounds in the future!

Maybe I get purple on PurpleCrayon's round

I feel really sad to say it won't happen but you might get blue in this round :P

BlueCrayon orz

What is orz?

It's a person bowing, 'o' is head, 'r' is arms and 'z' is folded legs.

So, if I say "someone" orz, this means I would like to shoot "someone"?

No, it means you pay respect to someone and admire them.

Why not? QWQmaster got from Specialist to CM in a recent contest and there are more such examples.

What is purple?

You are lucky to never find out.

afterall: What is yellow?

Come from real id

👉 👈

As a tester, give me ..

As a tester I can assure you that you will write purplecrayon orz after contest (before too)

Here. I can give you this latest striver meme.

SpoilerLol downvoting my greatness :(

waifu sauce please

Just search Rikka, she's famous.

Each division will have 5 problems and 2 hours and 15 minutes to solve them. This line scares me.

Why is the rating limit for Div.2 is 1899? Shouldn't it be 2099?

I think because this contest has Div 1 too.

Score distribution for this round will be really interesting I guess. 5 problems in 2:15 is very challenging

Damn, 5 problems in 2:15 hours. Seems like a Div-1.5 :)

If both divisions are 2 hours and 15 minutes why here the length is 2 hours?

Spoiler.

Don't tag them unnecessarily

Edit1: Lol you totally changed the main idea of your commentMind your own rating! :P

You could reply me instead of writing my name ....

As a tester, I can confirm problems are awesome! Good luck in the round!

but u said problems will be awesome!!!

Hope to be expert.

Just don't cheat again. GL;HF.

As someone purple did not allow test the round, give me contribution!

As someone purple also did not allow to test this round, give me contribution!

As a participant, I hope my fellow comrade hbarp and I can become

purplein this PurpleCrayon blessed round, unlike last time. Wish us luck!! xDGood luck to you :)

As a tester, I can say that I enjoyed the problems. Good luck to everyone participating :)

I hope my colour change with a +ve delta :)

Unfortunately, most of IOI participants can't participate in this round :(

I think upcoming contest was little bit tricky..so,I hope u all are enjoying this contest and learns something new!

It's not started yet :)

Excited for the round, knowing the writers I have no doubt it will be great!

I wish problem statements would be as clean as

this announcementI wish statements would be as short and clear as the annoucement, too.

I wish that this contest has Graph theory and DP related problems !

sounds funny

Thats all you had to say

.

I will remember the contest start time properly this time, the previous contest div 2 which I thought usually to start at 20:05 but it was around 3pm RIP

PurpleCrayon is not even purple Sadge

"You, for participating!" This line hit hard :(

As a cyan, D seems not killable.So, speedforces :(

Really excited to participate in div2.. All the best everyone.

The scoring distribution for div2 is weird. I hope I get positive delta.

Can anyone tell what does (1500 + 750) mean in the score distribution? Does that mean the score of the question can be (1500 or 750) or does that mean the score would be 2250?

It means the problem has an easy version and a hard version. Solving just the easy version would give you a max 1500 score, solving both would give you a max 2250 score.

Why does the easy version has a higher rating than the hard version?

This is a fine move by the setters so that we lame programmers can feel better.

Why it shouldn't be?

The complete score is the score of the "hard version". Easy version should be considered as partial score for partial solution. If it contains significant part of the final solution, it is natural that is worth significant part of the final score.

Hope that I can become a crayon today ^_^

Is it logical that the second part of the last problem of div2 has the same score as in div1 (750)?

.

It's 38 mins dear

how to solve D when you are green?

How to solve D when you are blue ?

How to solve D when you are Cyan?

I am waiting for someone to say become red! xD

Respect the unusual timing as a usual one. That could work probably work. :D

Seems to solve current D, I need to be atleast 4000 rated :/

Well you are speaking facts. D is too OTZ levelled. T_T :(

So are we solving E2 after A?

Went with that approach at start. Only to return back sad faced after reading E1 fr 5 mins. T_T . Dumb me!!

I wish everybody an awesome codeforces round

gl

i think today i become pupil

F

Anyone in Div 2 feeling like they accidentally participated in Div 1?

Speedforces, redefined.

The game of hand speed.

The expected value to solve D by myself is 0.

Most difficult div.2 D ever?

Never seen such a big jump in difficulty from C to D

What is Normal Red?

Div 2 is definitely a speed force.

speedforces

I have no idea on 1B or 2D, wtf

Some relavant meme:

I want to give 200 upvotes T_T

E1 is like Atcoder grand contest math problem.

ive read all the problems and they are really nice good job!

Solved 3! Then saw D. WTF only 13 submissions!?

Started with D, read it & thought something, then saw A. WTF crossed 10,000 submissions!?

I actually never read D..just saw the submissions lol.

Forget about it

It's probably hard to solve problems when you are creating memes during the contest.

There must be more than 16k people creating memes, because only 15 people could solve the problem

I feel like this was speedforces but at the same time wasn't speedforces :)

So by Div 2 u meant Div 1.5

Good and challenging problems ! Good job but the difference of difficulties of problems between c and d was a lot !

Literally 15 mins on 1A and 2 hours on crying

Same pinch !!

Same here , 5 min and then nothing

Same ...

That sums it up :(

I hate probability.

Me in Div2 today:

First, solve three problems.

Second, look at the accepted number of problem D

Third, give up and go to the comment section.

DIV-2: When I solved B there were ig 200-300 people who solved C, When I solved C there were 2 people who have solved D. Got the idea about the question before reading it.

Can you share your approach to DIV 2 b ?

Iterated i from 1 to n-1; We have to find the all the possible (j>i) & (j<=n) which satisfy the given condition.

For this number at ith position. i+j can range from i+i+1(a) to i+n(b);

v[i]*v[j] = i+j.=> We have v[i] and range of i+j in hand. So what is the possible range of v[j]? ((a/v[i]) + (a%v[i]==0?0:1)) (c) to (b/v[j]) (d);

Now iterate from c to d (possible v[j]) and find whether the number in this range exist at respectively jth index, such that a[i]*a[j] = i+j;

Take sum.

Time complexity should be O(n). The range of c-d is maximum when a[i]=1 ~ n but since all numbers are different so if at next index a[i]=2 then range of c-d shuld be n/2 . Now n+ n/2 +n/3 + ... ~ O(n)

Probably sum of n + n/2 + n/3 + ... is O(N*logN), as harmonic series (without

nfactor in each summand) is equivalent to logarithm. But yeah, that is still AC :)Yes you are right. Thnx.

Am I the only that forgot numbers in B are distinct and wasted more than 20 min because of that

no

Yep. Me too. But it's my fault.

Seems like I'm getting dumber with each passing day, half of the world solved C in one go.

:) while I could not able to solve it.

:( another rough day for us

Same here :(

I wonder how 4k+ people solved C. Was it that easy?

for me it was pretty hard to understand the problem for first 10 minutes, then it went from graphs -> dp -> sorting and turn out to be easy problem lol

I get wa on both A and B :( ,same pinch bro,I also feel same

What should the answer be for the 3rd example testcase for div2D? I got 141/40 = 925000010 but it should be 500000007. What's the actual fraction that gives 500000007 though?

agh i think i figured it out.. I was computing the probability of marking X before Y, given the heights from the root of X and Y, but it really should have been the heights from the LCA of X and Y, not from the root. :(

3.5 but no idea what the actual fractions going into it were

I guess it's 7/2

Thanks for the contest! I really enjoyed it.

The maximum boring problem set in Div2. Three observation based problems, and D + E ridiculously difficult. The contest was practically over after 20 minutes.

ok ...i hate my life now

I entered 25 minutes later and after 35 minutes I left by topping in my room! This line describes the contest!

MEET AN EXPRIENCED & SHAMELESS CHEATER This is how Master_Jiraya bypasses Plagiarism testing.

i reported to codeforces and MikeMirzayanov about him from past 2 contest and he does cheating in it also and got plagiarism , thanks to u for upvote my comment so that he got punished . and today again he cheated in the contest , pls again upvote my comment ......

Master_Jiraya does cheating from starting and i reported about it to MikeMirzayanov and he got plag in last 2 rounds , he abused me in private chat becz i reported him https://ibb.co/JmhSwKL . guys show your support and again upvote my comment so he again got punished.

People like Master_Jiraya are spoiling the sport. I don't understand where would cheating take them in life. They will never get anywhere in life but always remain what they are i.e cheater. He should be banned from the platform as soon as possible . MikeMirzayanov sir pls ban him and skip his solutions .

his todays contest submission 120565019 120557120 , saw his submission timing and also see this dummy variables snippet

xx[3]++;xx[1]++;xx[3]--;xx[1]--;xx[5]++;xx[3]++;xx[1]++;xx[3]--;xx[1]--;xx[5]++; xx[3]++;xx[1]++;xx[3]--;xx[1]--;xx[5]++;xx[3]++;xx[1]++;xx[3]--;xx[1]--;xx[5]++; xx[3]++;xx[1]++;xx[3]--;xx[1]--;xx[5]++;xx[3]++;xx[1]++;xx[3]--;xx[1]--;xx[5]++; xx[3]++;xx[1]++;xx[3]--;xx[1]--;xx[5]++;xx[3]++;xx[1]++;xx[3]--;xx[1]--;xx[5]++;

Others Template; #include <>

Your Template: MEET AN EXPRIENCED & SHAMELESS CHEATER.. kedos123

And you submit it after every contest in announcement. Keepitup.

One more guy vadapavandchill check his submissions with obsfucated code 120565541 120557993

Can anyone explain Div2 problem C

..

On above that put that as Div2D :(

My logic for C, take the sum of differences of all pairs and subtract the difference of adjacent elements from it and print the answer with a negative sign. Why is this failing?

The answer is take the absolute differnce of each pair and then subtract maximum value from array d and then multiply with. 1 , The logic is that to minimise the cost you should add as many -ve no. You can and as less +ve no. You can , So for +ve no. you will take maximum no. As the sum of difference between adjacent sorted no. will lead to it and for -ve add All -ve no.s .

How to solve B ? I don't know how it can be solved in o(n)

It's not O(n). Rewrite the eqn as aj = (i+j)/ai

Now, as aj is an integer, i+j must be a multiple of ai, you need to check only ai blocks in each case. Since ai is distinct, this is similar to prime sieve in terms of time complexity.

Let us start with the brute force approach..

You would initialise two for loops.. one from i (1 to n) and the inner one j (i+1 to n) .. iterating over all values and obtaining all such pairs. However, this would take order of n*n and is not feasible.

So, the key observation is it can be reduced to approximately nlogn per testcase, by only considering the valid indices that occur

afteri that are themselves multiples of a[i] , incrementing j by a[i] each time. The first such j for which i+j is a multiple of ai can be found by simple math (because ai*aj=i+j, so i+j has to be a multiple of ai).I feel more dumb now because i got the same idea but thought it would give TLE ; _ ;

It would indeed give tle if all elements of the array were not unique.. as it would now tend towards n*n for big n and small ai.

Based on what I observed in my room, there were 2 major ways to solve B: Either an $$$O(n\sqrt n)$$$ solution or $$$O(nlogn)$$$, which I thought was pretty cool.

$$$O(n\sqrt n)$$$:

First, create a map $$$where[n]$$$ for all $$$a[i]$$$ -> $$$i$$$. Now, iterate over all integers $$$i$$$ from 1 to $$$2n$$$. Get the factors of each integer $$$i$$$. Now, for each factor $$$f$$$, check if $$$where[f] + where[i/f]$$$ = $$$i$$$. If so, increment the answer.

$$$O(nlogn)$$$:

We iterate over every multiple $$$k$$$ of each $$$a[i]$$$ and check if there exists an index $$$j$$$ where $$$j > i$$$ and $$$j + i = k$$$. If so, check if $$$a[j] = k$$$. If so, increment the answer. This works since all $$$a[i]$$$ are distinct, so the time complexity is bounded by the harmonic series $$$n (1 + \frac {1}{2} + \frac {1}{3}...)$$$ which is well known to have an upper bound of $$$O(nlogn)$$$.

How to solve problem C ?

Sort the input as the order of pastures doesnt matter.

Then notice that more the roads between pastures more minimal the time will be.

Now create a difference array from the sorted array as they will correspond to our postive edges in graph.

Now we take make negative edges for every pair of pastures.

The sum of these will be the negative of sum of all subarrays of difference array(Call it NEG). This can be calculated in O(n) using contribution technique.

So final answer would be sum of positive edges(all elements in difference array) + NEG.

This is greedily the minimal ig. You can check my submission out. I feel it is neat.

Note that you need to arrange the vertices in such a way so we minimize the sum of positive edge, which can be done by sorting the array and taking minimum times to be a difference of two consecutive elements.

Eg:- $$$w x y z$$$ be the sorted array then $$$(x-w), (y-x)$$$ and so on will be the costs of edges which gives a path from patch 1 to all patches with minimum positive cost.

Now we can need to add negative backward edges to make the total sum of each cycle zero. Let us say the edges are $$$a,b,c,d,e$$$. Then the negative edges cost will be $$$N=(a+b+c+d+e) + (a+b+b+c+c+d+d+e) + (a+b+c+d+b+c+d+e) + (a+b+c+d+e)$$$. Each term corresponds to nullifying cycles of with 2,3,4 and 5 vertices respectively. Generalize the summation for $$$n$$$.

The answer would be $$$a+b+c+d+e-N$$$.

I really don't understand how you do it

Here the cost of each curved arc will be a negative value such that the total cost of its respective cycle becomes zero.

Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads.

can u explain where does the above statement play a role in your solution? As I thought that it doesn't allow more than one longest cycle to form in the graph.

"As I thought that it doesn't allow more than one longest cycle to form in the graph." ==> It does not allow a cycle with the total sum of edge weights < 0.

Ahhh... Thanks for clearing the doubt as it kept bugging me during the contest and couldn't ask the author the same thing during the contest thought it might violate any rules..

Is there any chance to pass Div.1 D by $$$O(n^2 \log n)$$$ solution...? I know $$$3e9$$$ per second sounds insane :(

Hello! Nice problems but the leveling of the questions was not very good. D Div2 was very hard

Div2D was so hard, only 18 people solved it.

I thought of doing something like this:

And inversion counting like that, but I didn't work out the details of how to figure that out.

Was the intended complexity $$$O(N^3)$$$?

ah, that was my idea too! I coded it up but got WA on the example pretest #3 :(

It might be possible to do it in $$$N^3$$$ but my solution was $$$O(N^3\text{log}N)$$$. The bottleneck was finding the $$$O(N^2)$$$ lca's for each pair of nodes given a fixed root.

I asked instead:

This leads to an $$$O(N^3)$$$ solution. However it took me 2 hours past the end of the contest to correctly work out how to implement it.

How to solve Div 1C/2E?

it seems that initial array is valid if and only if $$$\sum_{i=1}^k a_i \ge \sum_{i=1}^k (x+s_i)$$$ holds for all $$$1 \le k \le n$$$, where $$$s_i=\sum_{j=1}^{i-1} b_j$$$. But I have not submitted, so i have no idea if there is anything wrong or not.

Couldn't submit 1B by 10 seconds :(

Well I'm glad to see I wasn't the only one having trouble with 2D. It's a nice problem though, I think I have a solution but it will take me a few hours to implement it.

Biggest drop of my contest history. Well, worth a fall if I hope to perform well again!

Felt like the entire contest was me making little progress on C1 and secretly hoping the solve count would stop rising :P at least I did well in the speedforces race.

How to solve div-2 D ??

Anyone remember when galen_colin said that your rank can really depend on the time you take to solve the problems? Welp, it's even more severe this contest, being from positions 19 - 4269 (approx).

PurpleCrayon, I liked the problems a lot, but I think that most of Div. 2 participants will disagree and say that it was not very balanced, turning things into speedforces somewhat

Pure speedforces

c

Because you have UB, you're accessing index out bound, your array is not big enough

For heaven's sake. If you're writing code in c++ you should know about UB and consider it whenever something inexplicable happens

Check out time complexity for problem BIf anyone has better approach then you can add hereSpoilerCan you put it in a spoiler message, please?

The problems are pretty good,but the diffculty gap between Div2 C and Div2 D are too big,making this Div2 round a "speedforces" round,so to my opinion,not very enjoyable because the round is not very balanced. Again,the contest is great,but I think there are room for improvements.

how to solve B div2

Iterated i from 1 to n-1; We have to find the all the possible (j>i) & (j<=n) which satisfy the given condition.

For this number at ith position. i+j can range from i+i+1(a) to i+n(b);

v[i]*v[j] = i+j.=> We have v[i] and range of i+j in hand. So what is the possible range of v[j]? ((a/v[i]) + (a%v[i]==0?0:1)) (c) to (b/v[j]) (d);

Now iterate from c to d (possible v[j]) and find whether the number in this range exist at respectively jth index, such that a[i]*a[j] = i+j;

Take sum.

Time complexity should be ~ O(nlog(n)). The range of c-d is maximum when a[i]=1 ~ n but since all numbers are different so if at next index a[i]=2 then range of c-d shuld be n/2 . Now n+ n/2 +n/3 + ... ~ O(nlog(n))

For each element you can find the number of elements that have product less than 2*n that is max(i+j). By mathematics you can prove that total such pairs are not more that nlogn. Hence brute force will work under time limits.

Where are those cats with their generosity problems now?

The gap between div2-C and div2-D is too large so if u submit some wrong answers in the first three questions u get a bad rank...

First time to solve a problem as difficult as div1D!

This is the most difficult Div.2 problem D I have ever seen.

120609085 my sweet O(NQ) solution has passed... :)

I also wrote the same code, and I think it is hackable. I'm very happy that it passed!!

friends!!!!!

Hacked both of you :)

Nope, I was first.

Can somebody post the screen with Swistakk's comment about $$$O(n \cdot \sqrt{n} \cdot \log(n))$$$ vs $$$O(n^2)$$$, please?

Here we go.

Is there going to be a hacking period?

why downvotes?

Multiple reasons for you getting downvotes.What I don't like about observation problems that it is too easy to cheat and too difficult to catch the cheaters

And this was like all the problems were observation based

Even though I kind of tired of cheaters topics as well

How to solve D? Can anyone explain? Never seen such a big difficulty jump between C and D.

How to prove the correctness of problem C's solution?

Auto comment: topic has been updated by PurpleCrayon (previous revision, new revision, compare).the round wasn't balanced :(

how to solve D?

Problem B Detailed Explanation

Very enjoable contest! Very enjoable not math tasks! Very enjoable raiting everybody've gained! Very enjoable enjoy i've felt. Good job! Good job! Good job!

Did anyone noticed weird template of this guy tampa (Rank 1 in today's div 2 contest)

I think he is simply using the atcoder library templates.

Diego nice

I never understood how it happened

Do you mean that you don't know what "Unexpected verdict" means?

Yes. And I don't know, why it happened on my hack

You've hacked the model solution (it got TLE or RE, probably TLE here).

Oh, realy nice!

Well, I'm not that sure, it should not happen if everything would be prepared correctly, but shit happens. And yep, for you it's nice :P

Maybe "shit" is too much actually. I that guess it means that one of the solutions marked as a correct one got TLE, it doesn't mean that the main one got it.

One of slowest solutions we considered correct, to be precise. But it doesn't belittle your achievement, congrats!

Hey Radewoosh please tell me how to solve div-2 D (div-1 B)

Sure. Read the statement, then start thinking about the solution.

EDITOr read the editorial.:'/

This contest which had no corner cases:

This round, Will be rated?

Yes, you can see the rating changes on the standings page.

Tranks you :)

MikeMirzayanov PurpleCrayon I received a plagiarism notice. But I don't know on what basis I got the Plagiarism. I dont even know him and A problem was so simple that one can have exact same code since we only need to swap the elements adjacent element and swap the last 2 element if it is odd.. Many could have thought the same solution .. Please Look into this.. I wouldn't consider it a plagiarism! I have given like more than 50 contest and why would i do this kind of thing !! His code is same as mine because thinking was same and it was an easy problem . Please Look into this ..

This is what I got from you guys !

Attention!

Your solution 120540551 for the problem 1541A significantly coincides with solutions phoenix_slayer/120540551, vedant.gupta.cer20/120587528. Such a coincidence is a clear rules violation. Note that unintentional leakage is also a violation. For example, do not use ideone.com with the default settings (public access to your code). If you have conclusive evidence that a coincidence has occurred due to the use of a common source published before the competition, write a comment to post about the round with all the details. More information can be found at http://codeforces.com/blog/entry/8790. Such violation of the rules may be the reason for blocking your account or other penalties. In case of repeated violations, your account may be blocked.

Did you get any solution from system yet? phoenix_slayer

No not yet.. But I got my rating change!! But still there's a warning which I would like to get removed from them!! Like A question always have same kind of submissions and don't know why they plagiarised!! I don't think they will correct it!!

You atleast got your rating changed bro!!

...,,..

...

You can say you didn't cheat and can ask for not to give any punishment. But why you will be in huge pain and stress just because of loosing some rating?.do you think life is that much cheap??

You have participated in the contest using two accounts and this is forbidden by the rules.

If kritarth2121 is actually some imposter rather than your alt account, then the obvious code similarity is very suspicious. What are the chances to have similarly named accounts submitting similar solutions just accidentally?

Code obfuscation is also forbidden by the rules. Even though this is not actively enforced yet.

.....

Yes, must be a coincidence that as soon as kritarth2121 solution to problem B and C got accepted, you submitted it after that with your additional useless functions.

I am removing this comment now . I don't want to give him more mental stress but i am hoping that he will never cheat again.

Nowdays codeforces solutions are available in net during contest. Someone are getting ideas from live streaming and someone are getting direct code. I don't know what do you want to achieve by cheating? Rating? Show some courage. The problem solvers who are practicing hardly they won't get a correct rating for your cheating during the contest.

Where is contest's rate Lol

Your solution 120588348 for the problem 1541A significantly coincides with solutions jatin3028/120548283, Saitama1227/120548701, TheBeardedCoder/120549565, yog_111/120550555, Ali_salloum/120550818,

Atom_Bomb/120559071, phoenix639/120560190, sanskar00/120562876, Astro_Aditya/120563489, reshma_27/120569500, Dilip24/120573780, Shalika22/120582653, Farhan_meb/120588348, Paavan_parekh/120595630, Dlip/120597081. Such a coincidence is a clear rules violation. Note that unintentional leakage is also a violation. For example, do not use ideone.com with the default settings (public access to your code). If you have conclusive evidence that a coincidence has occurred due to the use of a common source published before the competition, write a comment to post about the round with all the details. More information can be found at http://codeforces.com/blog/entry/8790. Such violation of the rules may be the reason for blocking your account or other penalties. In case of repeated violations, your account may be blocked.Hello, so i received a message saying my solution for a got matched with some random people. I am quite surprised to realize how could you judge a straight forward a problems solution as plag as the solution was change the indexing with its near index and that's it. And you can also check my ip or somehow if i have multiple accounts at contest time. This is totally unfair as i have solved b,c and didn't get plag but got you judged i copied some easy problem a. Please check this matter out. Got -184 which is quite a big deal for me. MikeMirzayanov. I dont even know any of the mentioned persons.

Seeking your kind attention!!! MikeMirzayanov I hope you will look into this matter. About submitting an easy problem so late, i was busy solving C and B first, this is why i solved A at last because B and C mattered to me much as A was straight forward problem and i knew everyone who participated the contest could solve it with 1 less than 1 month of experience. Please, Check this out.

No brother.. Rather say that,you got the solution of B and C first and you made a slight change so that plag detector Won't catch you and at last you got A and you didn’t change it much. Lol.have some brain before cheating.

Do you even know me bro that you are so sure about your accusation? WHO basically needed solution for A? Kindly check my profile and then say that, i finally got solution for A!!!!! and then didnt change it!!!! Kindly stop creating new accounts and scamming please. This community is becoming full of scammers day by day!

Your solution for C and B is exactly same as the leaked code(you know but wont confess)and you just changed a little bit. I saw you just started doing good in last few days.Why you needed to cheat in this circumstances.you could hold a little bit for your rating to continue in expert rather than cheating. Such a shame

So you saying i copied b,c and changed the codes. But didn't bother to change code of A? Makes sense? Who would be this dumb bro? I literally have no idea what leaked code you are talking about tho. Ideas can match of a solution!

This is why we should record while participating a contest as a emergency proof. Gonna do it from next rounds though!

Sorry to say but you submitted all three with a delay of 8 and 7 mins, that clearly shows you got the answer and just uploaded them after few changes but unluckily you got plag. in A.(But you will not accept it)

Yes bro, you live in my house, so you can predict everything what happened during contest time

Were you using ideone or pastebin?

No. I didn't use that.

So poisonous mathforces round!

Seeking your attention MikeMirzayanov PurpleCrayon !

I got the message that my solution 120541662 significantly coincides with solution 120540725 of user Sky200 whom I don't even know. The similarities in code are because code was short and for the case of odd, "3 1 2" was given in the sample case so, I have used same the user also have used the same. The problem was quite straightforward and almost everyone implemented using two loops. I haven't copied any single line of code from anyone.

I have participated in 60+ contests and solved 400+ problems. I think our thinking was the same and because of short code, they are having similarities.

Even Official Editorial followed the exact same approach!!

Kindly check it again.

Thanks!

Seeking your attention MikeMirzayanov PurpleCrayon!

I got a message from the system stating that me and FranSkarica have copied in some way the solution of 1541A - Pretty Permutations. But I do not even know the person and did all my code locally. So there is no chance of that person coping my code or me copying his. The reason it has been caught in this type of test is because the logic is extremely simple (div-2 A) and most people would have got the same logic. The code is also very small which could lead to higher probability of this being a coincidence. I have particpated in 100+contests and solved around 1000 problems, I think I will definitely be able to solve A problem by myself. I also see that many have faced the same issue in the comments section. Kindly have a look at all the requests and do what is necessary

Same thing happened with me

I have submitted two solution to problem div2.B.My logic for this one is

j can written as x*arr[i]-i. I look for all possiblej<=n. In first version I comparei+j==(arr[i]*arr[j])and it is giving wa on pretest 3 where as in 2nd version of solution I writearr[j]==x, here if you multiply both side witharr[i]then it becomearr[j]*arr[i]=x*arr[i]which is equal toi+j,both of them are equivalent then why in first submission it is giving WA on test 3. 1st submission giving WA on pretest 3 2nd submission getting AcceptedInt overflow is the reason. In the 1st submission, arr[i]*arr[j] is greater than 2^32.

Became blue by courtesy of speedforces.

Seeking your attention MikeMirzayanov PurpleCrayon!

I got the message that "Your solution 120588348 for the problem 1541A significantly coincides with solutions jatin3028/120548283, Saitama1227/120548701, TheBeardedCoder/120549565, yog_111/120550555, Ali_salloum/120550818, Atom_Bomb/120559071, phoenix639/120560190, sanskar00/120562876, Astro_Aditya/120563489, reshma_27/120569500, Dilip24/120573780, Shalika22/120582653, Farhan_meb/120588348, Paavan_parekh/120595630, Dlip/120597081. Such a coincidence is a clear rules violation. Note that unintentional leakage is also a violation" But i don't know any of them at all!.

The similarities in code are because code was short and for the case of odd, "3 1 2" was given in the sample case so, I have used same the user also have used the same. The problem was quite straightforward and almost everyone implemented using two loops. I haven't copied any single line of code from anyone. For this i got -184 which is a really big downfall for me.

I have participated in 200+ contests and solved 800+ problems. I dont think i am unable to solve problem A. I think our thinking was the same and because of short code, they are having similarities.

Even Official Editorial followed the exact same approach!!

Kindly check it again.

Thanks!

Seeking your attention MikeMirzayanov PurpleCrayon

This is regarding a mail I received for plagiarism in my first solution [ GordeyZav/120537624, pranav_mehta/120551724 ], I would just say that 1st solution was itself very intuitive and I believe a ton of users would have used the same logic, thereby the logic being same should not be a deciding factor, now if the plagiarism is because I used same iterator "i" and variable "n" then I believe the basic coding books and online tutorials are to be blamed, I don't really care about the rating or stuff, however getting a plagiarism badge on you when you didn't cheat isn't something anyone would want.

You got your rating changed bro?

nop

You didn't even get any penalties bro?

People get penalties in the second or third time when they are plagarised. In the first time simply the participant becomes unrated. Rating not gonna down.As your rating got down,obviously you have previous history of coincidence / plagiarism

Hi!

I received a punishment from the system which claimed that I had cheated with romanychev, but I didn't.

As the system says:Attention!

Your solution 120543231 for the problem 1541A significantly coincides with solutions romanychev/120536673, sepehrking/120543231. Such a coincidence is a clear rules violation. Note that unintentional leakage is also a violation. For example, do not use ideone.com with the default settings (public access to your code). If you have conclusive evidence that a coincidence has occurred due to the use of a common source published before the competition, write a comment to post about the round with all the details. More information can be found at http://codeforces.com/blog/entry/8790. Such violation of the rules may be the reason for blocking your account or other penalties. In case of repeated violations, your account may be blocked.

As you see the codes generally have similarities, (which has became accidently) but they are not the same!!

And why should a question with a simple algorithm be copied!

And why should I do an wrong submission when I have copied it.

And why should I not give other solutions meanwhile he hasn't submitted them (he hadn't accepted any other problem except the one which has been claimed as an cheat) and had helped me.

Dear MikeMirzayanov. Please do something because it has been a misunderstanding, and it's too bad to lose the chance of becoming an expert just because of a misunderstanding.

Expert Again :)

A friend of mine got 0 rating change in this contest, he had 1WA and 1AC for problem A. He was at rating 761 and is still at 761 after the contest. It is not showing as out of the contest for him. On the contest page of his profile, it comes as 1 solved and 0 rating change.

What could be the reason for this?

Probably because his expected performance, based on his rating, matched with his actual performance during contest, his rating remained the same. You can always use webbrowser extensions, such as CF-predictor, to see what your actual performance is during the contest.