Привет, Codeforces!
Благодаря поддержке Neapolis University Pafos, продолжается серия образовательных раундов. Университет предлагает получение степени бакалавра в области компьютерных наук и искусственного интеллекта со стипендиями JetBrains. Получите передовые навыки в области искусственного интеллекта и машинного обучения, которые подготовят вас к востребованным техническим карьерам. Любопытно? Ознакомьтесь с учебной программой прямо сейчас. Доступно ограниченное количество стипендий. Не упустите свой шанс учиться в Европе бесплатно!
В 15.08.2024 17:35 (Московское время) состоится Educational Codeforces Round 169 (Rated for Div. 2).
Этот раунд будет рейтинговым для участников с рейтингом менее 2100. Соревнование будет проводиться по немного расширенным правилам ICPC. Штраф за каждую неверную посылку до посылки, являющейся полным решением, равен 10 минутам. После окончания раунда будет период времени длительностью в 12 часов, в течение которого вы можете попробовать взломать абсолютно любое решение (в том числе свое). Причем исходный код будет предоставлен не только для чтения, но и для копирования.
Вам будет предложено 6 или 7 задач на 2 часа. Мы надеемся, что вам они покажутся интересными.
Задачи вместе со мной придумывали и готовили Адилбек adedalic Далабаев, Иван BledDest Андросов и Максим Neon Мещеряков. Также большое спасибо Михаилу MikeMirzayanov Мирзаянову за системы Polygon и Codeforces.
Данный раунд частично пересекается по задачам с личным соревнованием летней школы Саратов-2024. Если вы участвовали в этом соревновании, то воздержитесь от участия в раунде.
Удачи в раунде! Успешных решений!
UPD: Разбор опубликован
Hope to reach candidate Pupil after this one
Best of luck
best of luck
Best of luck.
BledDest orz
hopefully expert after this one
good luck mate
good luck
why is this comment at -12???
idk either lmao, it is what it is
because this is cf
coz he's probably a cheater,see his submissions
if you accuse me of cheating, I suppose you have evidence?
.
Look through every single one of my submissions, and you can see I ALWAYS write long long int. Any single time I use 64-bit integer, I always type long long int. You just picked a random submission and the most random reason and decided to go with it? cope and seethe newbie, maybe focus on actually solving some problems instead of throwing baseless accusations.
So maybe you debug every question using ai,who knows!Also I am actually pupil,I didn't know multiple accounts are not allowed on cf,Hence kept this one only for practicing.
yea sure, I also use ai to type my code when I practice as well right? you're a complete clown, stop embarrassing yourself. oh and since when is AI considered cheating?
Independence day Contest !
All the best to all. Hope you all get +100.
I need 130 :) but 100 still good and will make me happy
thanks
How to gain rating in EDU Round? I've droped the rating at almost every the EDU round.
during the contest, if you see that your position is too low to gain rating, it is best to solve another problem
based
What do you mean?
Noob doubt: What's the difference between Education rounds vs Div 3 vs Div 4 ?
educ ≈ div2 and div4 < div3 < div2
I'm glad that there are no tester comments.
How to Hack someone's solution?? Shayan bro i think you should make a video on this topic!! I tried many times to hack someone.. but always ( Unsuccessful Attempt )
You should give test with high strength. Here is an example: 274841071
https://mirror.codeforces.com/contest/1999/hacks/1066931
If you can't understand this code, it doesn't matter. You just need to know what he is doing. In this code, he's using Substring all the times which is $$$O(|s|)$$$ in time. It will get a TLE by a test with high strength. Thus, I give a test like this:
1, and then 200000?s, and then ana.Unaccidentally, he is hacked by me.
You just used generator to generate worst case for s right?? But how did you figure out it'll give TLE..
How to practice for this?? I am very bad to figure out the time and space complexities
You can search the submissions in “Status”. And sort it by execution time at the end of the page. And get to the last page. The submissions there are almost TLE so that maybe TLE after you hack them.
What generator do you use?
Usually I use C++’s output. You can use what you familiar with.
if the test isn’t that long, just input the test by text is enough.
There you've wrote string s of length 2*10^5, how did you do that, and how to generate those big test cases?
There is a “Generate Input” Botton when you hack. Press it and paste your code there.
Example:
I tried this multiple times but never got successful Attempt
Try more, you’ll gain from it.
What’s more , see which problem is hacked most. It may mean that it is easier to hack than other problem.
Sorry I’m a beginner here, how are edu rounds different from normal contests ?
PS: Got it, thank you!
You can't hack during the contest, but you can EVERYONE hack after contest.
I am a total beginner and have no idea what hacking means in this context. Like changing the answer or sth? Where should I start from
https://mirror.codeforces.com/help
Read about hacks.
In some contests like Div. 2 Round, there is a Codeforces Format .
Maybe in most websites, it will give you the result at once when you submit. But if Codeforces, there is only a set of tests called Pretests.
It just means that this is just the preliminary test which shows you if the code can run normally. If you passed pretests, you can lock this problem and hack participants who pasts pretests in your room, by giving tests with high strength to make them failed passing this problem.
P.S. If you lock the problem, you can't resubmit any more until the contest ends.
To get more information, you can read this blog
https://mirror.codeforces.com/blog/entry/456
In Div. 3 , Div. 4, and EDU Round, it is EX-ICPC Format. There will be a 12-hour-long phase at which you can hack everyone after the contest.
Different from Div.1 , Div.2 , Div.1+2 contests, it is an EX-ICPC Format.
Similar to Div.3 and Div.4 contests. You can see participants rated and unrated separately in the standings after system testing. You can view last Div.3 contest as an example: Codeforces Round 966 (Div. 3)
And view this round to see the probable difficulty of each problem. Educational Codeforces Round 168 (Rated for Div. 2)
standings: View rated participants rated separately by pressing the
Both divisions vbotton at the rightmost of your screen. Then cancel the tick ofShow unofficialone.Thanks to all who contributes to this contest. Hope I can go back to specialist after this one!
hoping for the same!
Hope for 1300+ rating
good luck (you did well in the last contest congrats)
Let's go specialist in this one
I want to go back to Newbie...
My first contest as specialist. Good luck everyone.
Bet it or not, I'm going to reach Master this round.
Good luck to you. Good luck to everyone.
Not happening, ahahaha
I didn't say I bet it. haha.... ha... Failure is common.
Nevermind. I'll be still there next round.
How can you reply me just minutes before the contest, are you watching me? Thank you for your watching stuttering
Maybe I'm going to change my handle into
Lmao_Foxif this happens too many times.hopefully expert after this one
I will get CM
Congratulations! At least it's legal in Singapore since recently.
nice way of coming out of the closet
Hoping for a positive delta!
In Educational Round dose speed of solving question matters or only number of questions solved during the contest matters and does not depend even on the rating of the question solved??
Speed matters. Rank is sorted first by number of solved problems, then time (10 minutes penalty for 1 wrong submission)
That means solving A and solving G are the same. It just calculates the total number of your solved problems and then the time.
I don't care what rating I will have I just solve tasks for fun
InShaAllah, I will get +100 today
.
Hope to solve four problems!
c < b
Great D but C < B.
was C really that easy?
it was easy I still got 6 WA's cause i didnt read "total" increase <=K
ahh, that changes a thing or two
or three
gotta be honest the difficulties of problems from A->D is too low, compared to normal div2 rounds or just edu rounds. I choked this one but generally speaking, i think this contest was not so good
so what the hell are you supposed to do for E (nims and games are kinda like the opposite of my skillset lol)
Find grundy numbers using SPF.
I don't know why it was named "not a nim problem". I solved it as a nim problem. Case analysis on small numbers showed that the nimber of a pile is basically the "rank" of its smallest prime factor (where a prime having rank k means it's the kth prime number, except that the rank of 2 is 0.)
Video editorial for D: Colored Portals
Here's a very fancy implementation of D: Colored Portals using Bitmask tricks that was introduced in D: Cases from a recent CF round. I also have a video editorial for the old problem.
Let's represent all portals as bitmasks of length 4. Then, a path between 2 nodes is possible if and only if the bitwise AND of both the numbers is not zero.
Assume $$$x \lt y$$$. On a number line, we know that the shortest path between 2 points is the straight line connecting them. So, if $$$a[x] \& a[y] \neq 0$$$, then, the answer is $$$|x - y|$$$. Otherwise, we need to jump to a portal that differs in exactly one color (i,e, the bitwise AND should contain exactly 1 set bit). From there, we can jump to $$$y$$$ directly.
Define $$$ldp[i]$$$ to be the nearest portal to the left that differs in exactly one color. You can populate it from left to right by maintaining a map of the latest bitmasks seen so far. Similarly, you can populate $$$rdp$$$.
There are only 3 possible paths. We take the best one amongst them.
i got the solution down immediately, but i could not implement it at all, fk me
I also got the solution, but my implement is not as good as the method above so... I don't know where I went wrong (I use binary search btw)
i tried using a map and then i also tried bs and some other stuff and i just couldnt get it. unlucky :(
276681877
You can look at my case work. I am laughing at my own code.
i shouldve just quit trying to do it nicely and just do that. good job ngl
Nice work
oh well I notice where I went wrong, need to reverse the array and position to binary search to get the correct answer. Not searching the y+1 or x+1 or (x+y)//2 will do
UPD: after fix the bs bug, my solution got TLE in test 6
In the hindsight, just maintain ldp and rdp will do me a favor more than bs
UPD: Resolved TLE, it's list.insert(0, list) which I think O(1) but it's O(n)
And the trend of 4k+ solves on D in Edu Rounds continues...
This D is only of Div2C level at most.
whoops
Why would you ever call a nim problem — $$$\texttt{Not a Nim Problem}$$$
Like I really don't understand, because it is funny?
When I read the name, I though that it is a bad idea to include "Nim" in the name of the problem, because when you search for nim tasks, this one would show up.
When though of the solution, I became even more confused. It was a nim problem. So why the author is telling me it is not true?
I thought that the name was part of the statement. I started to doubt myself that I got something wrong. Turns out everything is fine, nim works. Downvoted the contest when I saw "Accepted" verdict.
And also this statement:
Why you say "not not A" instead of just " A "
We can rewrite it like this:
This sentence is so confusing that even an announcement was made, where it was stated that
If it is so confusing, maybe the statement should be fixed? I don't understand why the announcement was made, and this sentence was not changed.
Lol, bro, relax
Agree with the title, just why?
Oh bro, is it a nim problem?
I immediately got idea for single pile, thought of using nim for n independent piles, but then I saw the problem name and thought its going to be some special case where nim doesn't work. And spent whole contest trying to solve without nim lol. I should understand nim better.
this nim thing is unfamiliar to me. any suggested resources ?
I really think you do not need to know that at specialist rank.
You can read the wikipedia article on sprague-grundy theorem and Nim Sum. I think they are pretty well written.
Agreed, read it wrong due to the not not thing and thus couldn't do it in time.
bruh C question due to "total" K wasted my 1 hour and 5 WA anyone else also faced same issue?
Every page from problem to submit takes so much time to load. It's infuriating.
for the i 1 its grundy number is 1
for the i 2 its grundy number is 0
for the i 3 its grundy number is 2
for the i 4 its grundy number is 0
for the i 5 its grundy number is 3
for the i 6 its grundy number is 0
for the i 7 its grundy number is 4
for the i 8 its grundy number is 0
for the i 9 its grundy number is 2
for the i 10 its grundy number is 0
for the i 11 its grundy number is 5
for the i 12 its grundy number is 0
for the i 13 its grundy number is 6
for the i 14 its grundy number is 0
for the i 15 its grundy number is 2
for the i 16 its grundy number is 0
for the i 17 its grundy number is 7
for the i 18 its grundy number is 0
for the i 19 its grundy number is 8
for the i 20 its grundy number is 0
for the i 21 its grundy number is 2
for the i 22 its grundy number is 0
for the i 23 its grundy number is 9
for the i 24 its grundy number is 0
for the i 25 its grundy number is 3
for the i 26 its grundy number is 0
for the i 27 its grundy number is 2
for the i 28 its grundy number is 0
for the i 29 its grundy number is 10
for the i 30 its grundy number is 0
for the i 31 its grundy number is 11
for the i 32 its grundy number is 0
for the i 33 its grundy number is 2
for the i 34 its grundy number is 0
for the i 35 its grundy number is 3
for the i 36 its grundy number is 0
for the i 37 its grundy number is 12
for the i 38 its grundy number is 0
for the i 39 its grundy number is 2
for the i 40 its grundy number is 0
for the i 41 its grundy number is 13
for the i 42 its grundy number is 0
for the i 43 its grundy number is 14
for the i 44 its grundy number is 0
for the i 45 its grundy number is 2
for the i 46 its grundy number is 0
for the i 47 its grundy number is 15
for the i 48 its grundy number is 0
for the i 49 its grundy number is 4
for the i 50 its grundy number is 0
for the i 51 its grundy number is 2
for the i 52 its grundy number is 0
for the i 53 its grundy number is 16
for the i 54 its grundy number is 0
for the i 55 its grundy number is 3
for the i 56 its grundy number is 0
for the i 57 its grundy number is 2
for the i 58 its grundy number is 0
for the i 59 its grundy number is 17
for the i 60 its grundy number is 0
for the i 61 its grundy number is 18
for the i 62 its grundy number is 0
for the i 63 its grundy number is 2
for the i 64 its grundy number is 0
for the i 65 its grundy number is 3
for the i 66 its grundy number is 0
for the i 67 its grundy number is 19
for the i 68 its grundy number is 0
for the i 69 its grundy number is 2
for the i 70 its grundy number is 0
for the i 71 its grundy number is 20
for the i 72 its grundy number is 0
for the i 73 its grundy number is 21
for the i 74 its grundy number is 0
for the i 75 its grundy number is 2
for the i 76 its grundy number is 0
for the i 77 its grundy number is 4
for the i 78 its grundy number is 0
for the i 79 its grundy number is 22
for the i 80 its grundy number is 0
for the i 81 its grundy number is 2
for the i 82 its grundy number is 0
for the i 83 its grundy number is 23
for the i 84 its grundy number is 0
for the i 85 its grundy number is 3
for the i 86 its grundy number is 0
for the i 87 its grundy number is 2
for the i 88 its grundy number is 0
for the i 89 its grundy number is 24
for the i 90 its grundy number is 0
for the i 91 its grundy number is 4
for the i 92 its grundy number is 0
for the i 93 its grundy number is 2
for the i 94 its grundy number is 0
for the i 95 its grundy number is 3
for the i 96 its grundy number is 0
for the i 97 its grundy number is 25
for the i 98 its grundy number is 0
for the i 99 its grundy number is 2
for the i 100 its grundy number is 0
Yes
I could generate grundy numbers, but didn't know how to extend them to 10^7. Small hint please ?
For even numbers, grundy is always 0. For 2, it is obvious because you can only take 1 and the grundy of 1 is 1. For the rest, because player is not allowed to take 2, then $$$g(2)=0$$$ would always be the mex of all options
For prime numbers, the grundy is $$$k$$$ where $$$k$$$ is the order of prime numbers (e.g. $$$g(3) = 2$$$, $$$g(5) = 3$$$ etc). This is because all prime numbers $$$p$$$ can have options of $$$1,2,3,\ldots,p-1$$$. Therefore the mex is always $$$k$$$ because it hasn't existed yet
For the rest, the grundy would be smallest prime factor of that number. Because it would be the smallest number that doesn't exist for all its transition (hence, it similar to the nature of mex). The other number would exist because if the smallest prime factor is $$$p$$$, then all numbers between $$$1, 2, \ldots, p-1$$$ can be taken of that pile. Hence, all numbers between $$$0, 1, 2, \ldots, g(p-1)$$$ can be exist as transition moves and won't appear as mex.
Thank you very much for explanation.
I could see the first and second observations from the brute-force that I had written.
But I couldn't deduce the third observation :( . otherwise I would have solved this.
Anyone who is interested in how to generate grundy numbers here, please refer to this.
~~~~~~
grundy[0] = 0; grundy[1] = 1;
~~~~~~~
Don't worry it took me 90 minutes to get it haha. F was much more straightforward today, so it's a good thing I wasn't afraid to skip.
small doubt, can we generate all the primes till 10^7 within the given time-limit ? ( never generated primes till 10^7, so asking ).
We need to find the what is the order of the given prime number, and we can't do that unless we have all the prime numbers within 10^7.
Do we need to implement SegmentedSieve ? or normal sieve is sufficient ?
fonmagnus
I think if we skip the even numbers, then we can do it for $$$5 \times 10^6$$$ numbers and should be fine (it's kinda close to TL but I guess that's one way to prune it)
Sure, will try later.
I wish, the pile sizes were within 10^5 range.
then I would have solved this question in 25-30 minutes :P .
yes we can. my submission ran in 406ms and i am running sieve till 1e7
yes, use linear sieve
Here, you forgot to include a link.
Normal sieve, if implemented correctly, runs in O(n*log(log(n))), which is sufficient (my sol passed with it).
Don't use " NOT NOT " in your statements ever again please
And yes, I googled E, sue me (I suspect many others did or have seen this problem before anyway)
This makes me wonder how many people thought of E during the contest by themselves instead of just googling. Because I couldn't think of anything after more than an hour (sad noises**)
I have spent most of the time going back and forth: "This is coprime nim. No, this isn't coprime nim. What if I take only divisors? But I can take any non coprime number. Wait, this is coprime nim, gcd should be equal to 1"
In regards to coprime nim (and not divisor nim, non-coprime nim and other variations) I've come up only with the most obvious stuff on my own:
say my name
Fck_cheaters
I'm curious if E is a classic problem of SG function. I don't know how to do it, but I have a vague feeling that it can do it.
what is SG function?
Sprague-Grundy function, super useful for nim-like problems.
Consider treating the game process as a DAG, taking the SG function value of a node as the mex of the SG function values of its child nodes.
I just generated all values of SG and guessed, then I realized that $$$g(x)$$$ is equal to the smallest prime factor of $$$x$$$ ($$$g(x) = 0$$$ if $$$x$$$ is even))
I come back to pupil again, thanks to problem B : )
C < A < D < B
Maybe I'm salty for not solving E, but is it ok to put a somewhat well known problem that apparently has a solution on the internet to a contest?
I am not a big fan of E myself but its edu round so the authors are somehow justified
Should such a contest be rated for anyone at all? C,D,E all were kind off not div2 level. If someone had a day where they did B and D fast enough (I couldn't, maybe thats why i am putting this out), E was a walkover.
Lovely problems thanks, problem B and D were a bit implementation heavy unless I just missed the obvious solution and did something unnecessarily long.
I think that D need a trick like bitmasks or else it would be so much pain
I got WA test 2 in D until the contest is over... unable to fix the code in time
binary search the indices would be enough no need for bitmasks
OMG! same avt. best of luck
yeppiee, u too
look at mine 276671836
Why this code (https://mirror.codeforces.com/contest/2004/submission/276666440) shows strange behavior when all the array elements are equal, for example:
5 4
3 3 3 3 3
Shouldn't the answer be 3 here?
Also i tried printing the final array after all the operations done, and in this case it is 3 3 3 1 0, but according to code none of the elements should change. Where i am going wrong?
Update: There was problem with my keyboard language setting due to which a special character was in the input instead of 3. But the solution is still wrong i think.
The variable ind will still bea -1 after your operations on nums, adding an another verification may help.(Though the solution is still wrong as you don't need to adjust all the numbers)
Wasted so much time on B lol, though D was very enjoyable!
How B ?
If you ever feel stupid, I got +7 penalties on D despite solving it in 20 minutes due to writing this function:
small piece of advice.next time create character set. and put characters of both the strings inside the set.
If the set size is less than (a.size() + b.size()) , means there is an overlap <3 .
I had the same issue and couldn't figure it out in contest lol. 276656805
IMAGE
i got also got penalty on D, because of this: wrong submission: 276659462 correct submission: 276671836
can any one tell me what is the wrong in my code in (d) or if i missed some cases may be 276679658
My sol for D looks like abomination 276680579
at least you solve it
R u blind? He didn't.
oh my god i was think he solve it
He did later.
Lol, u r also blind xD since he didn't
Not during contest, but he did :)
Doesn't count :D
thank you
sorry, wrong submission
Bro what, 800 python lines...
276679322 Mine is not too far off xD , just that mine TLE'd
Lmao i just brute forced the first 100 grundy numbers and stared at the screen for 45 mins until i observed pattern
what is the pattern in this problem?
for even its 0 and for odd it is same as the grundy number of its smallest prime factor
Oh god, I've just realized that I misread the statement completely LOL. There was two "not" in the statement @@.
How do you solve D??
if x1 and x2 dont share a colour, then you want to find the closest city to the right of x1 and the closest city to the left of x1 that share only 1 colour, then you calculate which of the 2 paths is better. im so salty i couldnt implement this, but this is the solution i think
i was think if you find the closest city to minimum one between of them you can find the answer and i think this do same idea you mention but i got wrong 2
I added a 13 minute video editorial. Hope this helps.
For each pair query, there are three cases you have to consider:
Case 1: The two portals are not opposites (i.e. they share a color)
In this case, you can just directly teleport with a cost of their distance in the array
Case 2: The two portals are opposite (they do not share a color), but between them there is a portal color combo which is different from both
In this case, our cost is also just their distance in the array b/c you can just teleport to the element between them to get from one to the other
Case 3: The two portals are opposite (they do not share a color) and there is no element between them which has a different color combo than either one
In this case, we have to calculate the shortest distance from the left of the leftmost query point to an element different from it which is not its opposite and likewise for the rightmost query point to the right instead. We then take the minimum of these distances we call mindist which naturally gives the total cost for the query to be 2*mindist + the distance between the query points. If there is no such minimum distance (i.e. there is no way to get between the query points at all) then we instead output -1.
One can implement the third case by iterating forward and backward through the array for each color combo to fill an array of pairs calculating the index of the nearest element that is not equal and not opposite to its color combo both to the left and right.
can you tell me what's wrong in this solution for C? https://mirror.codeforces.com/contest/2004/submission/276677983
my code for C could pass the sample test cases but i kept getting WA after submission :(
i wasted 1 hr on it lmao, was C really so easy?
The problem is that if the array is initially of odd length, then you create a phantom element equal to 0 and further consider it as the original one and, accordingly, you can add values to it, which cannot be done, in fact, it does not exist.
Test when the code is not working correctly:
1
3 1
1 5 5
Program output: 0
Correct answer: 1
According to the logic of your program, element 0 is created and the array turns from [1, 5, 5] into [0, 1, 5, 5]. After that, in order to get the optimal answer, your code adds 1 to 0 and you get an array [1, 1, 5, 5]. Thus, the result will be equal to 0.
It turned out that we optimized the response using a non-existent element.
In a situation where the array is of odd length, you need to use another method — immediately add the minimum value in the array to the answer and then remove it from the array, thereby making it an even length
wow i wasn't expecting an answer so well explained lol. thanks a ton, i understand it now.
honestly i feel i could easily be cyan but i always mess up small things like this, for example related to the range of inputs, such as writing >x instead of >=x for some loop, etc. and realizing that after wasting 25 minutes on that. since you're blue do you have any suggestions for me?
The best practice will be to solve implementation tasks outside the contest and set a goal to pass the task the first attempt. That is, spend a lot of time checking all the key points of the code and be 99% sure that all the places are implemented correctly.
In the contest itself, you can:
1) Write a naive solution for small values and compare the answers received with the code that you are going to send
2) Manually check the boundary values, experiment with the boundaries (try to change > to >= or even to <, try to change the data type (if an overflow occurs))
3) Check each part of the code and intermediate values separately (whether the array was built correctly, whether the answer is updated correctly at each step, whether the necessary variables are used everywhere, etc.)
Usually these steps are enough, if not immediately to send a solution without bugs, then after an incorrect attempt to quickly find where it can be the error and fix it
Is there a $$$ O(n^2) $$$ solution for F?
just use hashmap
https://mirror.codeforces.com/contest/2004/submission/276639528
too tight constraints for D. This is a QlogN and not passing? [EDIT SOLVED . I was declaring the vector again]
O(N+Q) is possible (via previous/next tables for all six pairs, for example), so this may be a TLE intentionally
But there will be only 4 pairs possible
https://mirror.codeforces.com/contest/2004/submission/276683926
with vectors+lower bound tle.
Do you mean to say the complexity is not O(qlogn) but O( (n+q)logn) ?
I meant to say that an O(n+q) solutions exist, so I'm not sure if O(qlogn) is intended to pass. As @rahulmysuru7 pointed out, your code runs even slower, so it's not relevant to you, I was just guessing what might be the reason
Ohh I see. I will think of N+Q solution too.
Thanks for the response <3
@mahendraakshansh ~~~~~ for(int i=0; i<4 ;i++){ set ss = mp2[vv[i]]; .... ~~~~~
yeah ,you fked up here. you are declaring and extracting the entire set inside here. making your code O(qlog(n) * n)
fck me thank you so much.. I am a noob
E is too educational even for educational round
not sure why this is TLE?
https://mirror.codeforces.com/contest/2004/submission/276681345
B is the hardest out of the first 5 tasks for me 😭 Also, how tf does E have like 1400 ACs? Do so many people know sprague grundy now?
apparently its a well known problem with solution on the internet, im just surprised so many ppl managed to actually implement D.
B was also hardest for me. I wasted 20 minutes on casework before just (correctly) guessing the pattern. E was googlable.
I remember seeing a cf blog about game theory a few weeks ago that explained how to solve Nim problems, so that might be why
so, How B? Help me please !
I just bruted for all bridges [i, i+1] upto 100 in the end
Break the problem in several cases.
ref : (https://mirror.codeforces.com/contest/2004/submission/276649235)
Case 1: When there is no overlap ----> only blocking one gate will work.
Case 2: When there is an overlap at exactly one point ----> Block two gates.
Case 3: When both the intervals are same ---> Answer will be difference between them.
Case 4: When all the 4 points given are distinct ---> answer will be diff b/w overlap region + 2
Case 5: When only 3 of them(l, r, L, R) are distinct ----> answer will be diff b/w overlap region + 1
Analyse all the cases in order on pen & paper you will get the idea.
thanks everyone
should I feel proud that the problem I found easiest was hardest for Candidate Master (even thought I could only solve till C) :///
no, as they have a room temperature iq
looks inside
a Nim problem
Why? Just, why?
Fr I was not even thinking around the lines of NIM after looking at the title. 1400+ accepted submissions for that is crazy
One small step for man, a giant fall for my rating.
Don't know why I hate L and R problems which have O(1) solution. Just hate the casework! P.S Talking about problem B
Damn, my programming skill is so bad that I couldn't implement C. The first test cases in problems are weak and doesn't help to find even some edge cases. E has no explanation for how Bob wins in the third case. Next year will be our year.
if u ever feel stupid then just know that I exist : wasted my whole contest on C only to realize it was total increase<=k, not k for every element
The question of double negative sentences is very unfriendly to foreign contestants, which seriously affects the questioning, and I feel very angry about this
Can someone pl look what is wrong in my implementation of D.
Link
if it_low==pos.end() you should try finding the last position <= x
D was actually easier to implement than I thought after I realized you could iterate over the colors and only check when you found an overlap for both endpoints. Then just binary search on the nearest index.
A, B, C, D, F is fucking easy, especially E is just grundy number which is not suitable for an edu round!
fucking rooms
For problem E, I found out quite early that I had to build grundy numbers for all possible number.
But I completely forgot all the theories. Almost all the time that I took for solving E was trying to figure out grundy again and for reading the question wrong and solving a wrong problem first. :3
can anybody point out whats wrong with the below idea for D ? (since failing testcase 2 is not visible till end of hacking) Submission: 276681155
Now for queries that have cities from one of the disconnected type we try three options. (WLOG assume y > x).
Find some city in range [x,y] that does not exist in the same disconnected group of cities as the queried (x, y). If found successfully then answer is again simply abs(x-y).
Find the closest city to the left of x and the *closest city to the right of y that are not a part of the disconnected group. Then answer is abs(x — x') + abs(x' — y) OR abs(y — y') + abs(x — y'). Take the minimum of the two.
If nothing is found in 1 or 2 then all cities lie in the same disconnected group and so the answer is -1. (remembering that we already checked for direct edge between the queried cities as the very first step).
Sounds right to me.(You mean "closest city to the right of y", not "largest city to the right of y", right?)
yeah i wrote that by mistake. I mean closest missing in the prefix of x and closes missing in the suffix of y. Looks like I effed up the code then. If you can take a look at it please ? (I assure you its written cleanly as much as possible.)
Sorry, can't spot a bug immediately. Maybe there's some bug in the implementation detail in finding the closest city to the left or right? Some off-by-one error, maybe?
if i understood it properly, that is the solution. the hard part of this shitty problem was the implementation
I don't like problem B and E (because I had a hard time with them, obviously). B is a boring classification, and for E you only need to print out the table of SG function values and find the pattern. TBH I didn't feel the joy of solving Cf problems in these two problems.
just now I was searching for D solution on youtube and saw problem A,B,C ,E and F were uploaded there during contest
my submission for problem b gave some negative number for first submission- https://mirror.codeforces.com/contest/2004/submission/276574322
and same solution got accepted after a few minutes-
https://mirror.codeforces.com/contest/2004/submission/276595377
if it is some kind of error from system then please recalculate penalty for this problem for me
I don't think they can be called "same solution".
The calc function in the first submission calls R itself in calculating R, which is undefined behavior
hmmmm whats wrong with my O(qlogn) solution for Problem D that got TLE on sixth test case? I realized that set::end is O(n) so i precomputed it in the
whatvector, but still didn't work. all other ops are O(1) or O(logn)submission
Use s.lower_bound instead of lower_bound(s)
Former has O(logn) and latter O(n)
yo what the heck! i had no idea. thank you so much. do you know why? same function yet big difference in time complexity.
You can see here set time complexity
didnt got D in time cause wasted 1 hour on fckn annoying B, for what do shit like that? B in hour, C in 10 mins)
Can someone help me?Your text to link here...
bro whattt
I am getting tle on test 5 can anyone help 276686377 even if the soln is qlog(n)
comlexity of your sol is n * q * log(n) see this accepted code using your code 276705976
i have no idea why my D is failing, i found the indices of all 6 types of values closest to the left and right and did accordingly am i missing something. solution
You have made exactly the same mistake that I made.
You need to take,
UPDATE :
That is exactly the issue. I copy-pasted your code, and changed one line. Proof :
https://mirror.codeforces.com/contest/2004/submission/276688327
dude i am crying......, i was burning my blood for this :(
Happens, don't worry. Happened with me also, I got 2 wrong penalty submissions for this.
E wasn't new...
projecteuler560
This is allowed during Educational rounds. It's one reason why they aren't rated for Div 1.
Source for this please
https://mirror.codeforces.com/blog/entry/21496
I don't think anything in the blog implies you can take exactly the same problem from somewhere else.
It's not exactly the same problem.
Project Euler asks you to calculate the number of losing positions with up $$$n$$$ stones in $$$k$$$ piles, while this round asked you to evaluate the winner for just 1 given position.
Looks to me like the CodeForces problem is strictly easier, except of course Project Euler has no fixed time limit.
It's true that the crucial insight behind the two problems is the same, so it's not completely new, but again, that's not unusual for Educational rounds and as far as I'm aware that's intentional.
Fair enough
awoo Why for Div2+ rounds you keep giving problems where cheating is extremely simple? No implementation, but only one critical observation which is extremely easy to pass from a cheater to another cheater.
Since you are giving feedback to Awoo, regarding the contest. I would like to chime in.
I didn't google a shit during the contest. usually I don't. For me, E was really good problem. except for one part, extra BOILER PLATE IMPLEMENTATION for sieve.
I would like to understand what was the motivation to keep pile sizes 10^7 instead of 10^5.
The problem would have been same, if the pile sizes were within 10^5. The core-crux of the solution logic would be same. generating prime till 10^7 and then doing extra implementation things, were more of boilerplate code.
Maybe so you can't completely bruteforce the Grundy numbers and actually have to come up with the "Grundy number of x is the Grundy number of its least prime divisor"? That would make sense in my opinion.
yeah, makes sense.
What problems are you talking about? C? For B and D, implementing a solution correctly is IMO much harder than coming up with it. For E, you still need to know about Grundy numbers and not screw up calculating for the small piles. And pretty much every single A is easy to implement, because it's supposed to be really easy, so there's not much organizers can do with that.
I spoke about problems D and E in particular. Ok, please tell me if these solutions for B and D look hard to implement. Do they?
I didn't realize B is just bruteforce-able like that and was thinking of an O(1) per query solution. I agree doing this is easy. As for D, I don't think your implementation is particularly easy, and looking at the comments, there's a bunch of people with the correct idea that failed to implement it, so I stand by it being difficult to implement.
B in O(1) per query. Need to block full intersection, additionally extending it by one on each side if needed.
In Problem D, I can't find my mistake, but got wrong answer on test 2. 276676154. and why every one use lower bound to find closest city from x and y? bcz vector was sorted and we can check it O(1)
Not that great of a contest IMO. A and C are fine, but B is just annoying casework, D is basically just implementation, and while I enjoyed E, it's googleable.
In B u can only check if 'i' is in one segment and i+1 in the other segment for every i in [1,99] then add 1 to the answer, there's no need for casework.
True, didn't realize that doing contest. Doesn't really change my opinion about quality of the problem though, as the fastest solution still suffers from this, and I personally dislike when there's a fast solution that's more complicated than a (significantly) slower one, yet the slower one still passes everything.
I liked problem B. There are different ways to solve it, some of which don't require a lot of case work.
You complain that problem D is mostly implementation, but what's wrong with that? It's a programming contest, not a math contest.
Going by your username, you probably love math problems, but if all problems are math-based other users will complain about “MathForces”.
And problem E was much more math-heavy, so there was something for everyone. In my view the authors struck a good balance between math and code.
So, many $$$n^3$$$ solutions passed in F.
An easy way to fix it would have been to have $$$1 \le a_i \le 2 \cdot 10^4$$$, and decrease the TL to $$$3$$$s maybe. I had to use map because it was not possible to have a global vector of size $$$4 \cdot 10^8$$$. For the suggested constraints, one can simply use a global vector of size $$$8 \cdot 10^7$$$, and find $$$f(a[l,r])$$$ in $$$O(1)$$$ by traversing over $$$(l,r)$$$ in the increasing order of $$$r-l$$$.
$$$O(n^2)$$$ using your idea 276838722
some leaked codes that yall best be on lookout for
All of them has the useless check in their code for problem E lol.
In Problem D, I got wronganswer on test 2. I can't find my mistake.can any one help me? 276676154, and why everyone use lower bound to find the closest point from x and y, cz the vector are sorted and we can find it O(1).
update : problem solved
Can someone help me understand what I missed in
Problem D?Submission: 276664135
Idea:
If there is a Portal with same color available on source and target, then, it is
target - source.If not, then :
Let's say for example, source is BG and target is RY, then, I check for the minimum value of below logic with these 4 portal pairs — BR, BY, GR, GY. (i.e., trying to see if at-least 1 color matches). If no suitable answer, then return
-1.If any of these is present inbetween source and target, then, it is
target - source.Else if, it is present after target index, then,
(foundRight - source) + (foundRight - target).Else if, it is present before source index, then,
(target - foundLeft) + (source - foundLeft).Else, INT_MAX
But four more cases are possible- RB, RG, YB, YG.
You need to take the minimum of these two, not just the first one that's possible.
(Let's say you want to go from BG at 4, to RY at 6, and there are BR cities at 3 and 9, then clearly it's better to go via the city on the left at 3. But if BR cities are at 1 and 7 instead, then it's better to go via the city on the right at 7.)
edit: Also remember to handle the case where target < source. You can simply swap the indices in that case, since distance is symmetric, but if you forget, some of your math might not work out correctly unless you put some abs() calls in there.
That's exactly what was missed. Thank you very much for your reply.
Can't believe I missed this. (When initially reading the problem statement I thought of this scenario to take minimum of the two. But, later somehow forgot after starting to code, and got deviated from it).
(I did take into account the target<source scenario. I just didn't mention it in my comment above.)
I tried so hard to see the pattern from this sequence, which is the grundy value in the case we picked $$$y$$$ such that $$$gcd(x, y) \neq 1$$$. After contest I read the statement one more time and... F!!!!!!!!!!
For problem c:
276686828
shows wa..
But- 276688213
Shows ac with same logic...
N.B.: All variables and array in long long data type
But why?? Is there any difference for min() function??....
a[j+1] changes in the first code ...
Thanks.. I have got this....
yea xD i made the same mistake if you check my submissions.
Why would you refer to the operation in G as "process", when the most reasonable meaning of this word is a sequence of operations, not exactly one.
Petition for action on copied task E by recognizing it as unrated or in other possible ways.
The authors directly violated one of the obligations of the authors of the round: copied the task from an existing one, replacing some words that were not essential in solving the problem with others.
If you vote for this post, it is similar to digitally signing this petition. Let MikeMirzayanov and the authors see us and take action.
Thank you.
На русском:
Петиция о принятии мер насчет скопированной задачи E путём признания его нерейтинговым или другими возможными способами.
Авторы напрямую нарушили одно из обязательств авторов раунда: скопировали задачу с уже существующей, заменив некоторые несущественные в рамках решения задачи слова на другие.
Если вы голосуете за этот пост, это аналогично цифровому подписанию данной петиции. Пусть MikeMirzayanov и авторы нас увидят и примут меры.
Спасибо.
The problem statement is simple enough for the problem setter to come up with it independently and just failing to find it online (which would mostly be a failure of the coordinator). Still a big issue, but one of negligence, not of malice.
I can't see why my submission for C is failing:
I would really appreciate if someone could help
276684985
In the "increase to make i = i+1 for all i" part, in "k >= a[i + 1] — a[i]" state you should swap 2 lines in your if, otherwise k won't change at all.
I think it's the only problem.
totally missed that :(, thank you!
Compare and feel... 276693772 (I copied your solution I changed a little bit).
My best chance to reach orange but stuck at F for an hour :(
Erm, actually, yellow is different from orange.
you're right. I get them reversed.
Its difficult to reach expert without doing graph. I skipped trying D question because I thought it was from graphs, I tried E then.
It's just binary search.
Resolved TLE, it's list.insert(0, list) which I think O(1) but it's O(n)
Well, my D sol got hacked, but I have no idea how I could improve the time of it. It's QlogN and it barely passed during the contest, so I'm not sure if it's just not the intended solution, or if my code is unoptimized (thought I tried to speed it up during the contest and didn't find anything) : 276659141
dont use unordered map ever
It has only 6 entries though
oh didnt see
use
(auto &m: cnt)instead of(auto m: cnt).Sped it up 10x, thanks! Completely forgot about auto making copies, but I'm happy that at least I got the correct solution
From Problem D why 276671958 gives a TLE but when I replace mp with a vector of vector it passes?
Use
for (auto &it : mp)instead offor (auto it : mp). You are making copies ofiteverytime.need help in problem D, I keep getting TLE on test 12 (Python)... 276697598
My idea is binary search to get best choice for x on left and right (binary search 2 times) then minimize it.
UPD: Resolved TLE, it's list.insert(0, list) which I think O(1) but it's O(n)
I think this should be a pretty bad round. The first three questions are too simple, while the fourth question combines both simplicity and complexity. As for question E, it truly educated me：）
276687644
In probelm D, why does this give TLE on test 5, but when I changed unordered map to vector, it got accepted?
Just replace
for(auto it : mp)withfor(auto &it : mp).I don't know why I am getting wrong answer on C on this submission 276671135 and right answer on 276689047 Can someone please help me with this
a[i]+=min(k,a[i+1]-a[i]);changesa[i]and affects the followingk-=min(k,a[i+1]-a[i]);.Oh such a silly mistake , thanks for pointing it out
Made the same mistake.. took me 40 mins to figure out :(
Oh man, I was so frustrated coz of it that I left the contest midway.
Hiya, could someone let me know why this solution for D is giving TLE? Is it the cuz of the lower bound calls on the sets? 276676038
I think that's because you use set, which will increase the complexity when inserting, and the order itself during traversal is already sorted, so if you replace set with vector, it can be accepted.
Yeah I completely forgot about the insertion complexity, I was way too focused on the lower bound part. There really was no need to use a set, such a silly mistake. But yeah thanks a bunch for the help!
My 311 ms B was hacked to tle. The pretests are awful.
Get a load of this guy, so many notorious coincidences.
First B solution got hacked, second one was FST'ed:
Codes for D are nearly identical:
...so on and so forth
I swear to god I never shared my code. I am thinking about someone hacking either my computer or codeforces server right now. Horrifying. I'm about to share this to my university group chat.
By the way I have recorded the whole coding process. It was all alone.
==============
Figured out, I shared my account once and now I have to change my password.
My solution for D is in O(n + q) (so optimal time) using 6 prefix and 6 suffix arrays that are precomputed, that tell the earliest/latest (depending on if its the prefix or suffix) occurence of some 2 letters combination in O(n), now if a[x] and a[y] share anything then it's just y-x, if not, iterate through all other letter combinations, then look them up in the precomputed arrays and get the result from that
Bet
cool! its same as jiangly's
Love problem F. Very simple implementation for a hard problem. Shows how elegant CP can be.
Can you share your approach?
The number of people having their D solution hacked is CRAZY (including me)
$$$O(\frac{1}{6}N^3)$$$ solution for problem F.
link : https://mirror.codeforces.com/contest/2004/submission/276598378
Is it right?
I'm curious why you mentioned a constant 1/6 explicitly.. is there any reason?
In my opinion,it's not the intended solution but obviously the problem setter wants to let $$$O(n^3)$$$ solutions pass.
Or there won't be awful constraints like $$$n\leq 2000$$$ and $$$5\rm s$$$ Time Limit which makes some $$$O(n^3)$$$ solutions with small constant like yours unhackable.
My solution was O(n^3), it didn't pass. Slightly less than that actually
At first, I thought what we do in problem E, is pick y stones in the pile which has x stones, guaranteed the gcd(x,y) is not equal to 1. As I do this problem what I thought for almost half an hour, I realize that the statement says:
It is not allowed to take from the pile a number of stones y such that the greatest common divisor of x and y is not equal to 1.
What was the idea for F?
How do I report suspicious submissions? I found several users that either have identical code to this comment or added useless snippets in their code (e.g. doing
ans ^= 23333;twice) to avoid detection.Why is the judging queue so long?
If the number of colors for D is not 4 but on the scale of $$$2\times10^5$$$, what will be the solution to the new problem? Can the new problem be solved with the given time limits?
In problem B, I implemented O(100^3) and give me TLE. I think my solution should be pass https://mirror.codeforces.com/contest/2004/submission/276617200
But there are at most $$$t=10^4$$$ testcases while no extra constraints on the sum of $$$l,r$$$ were declared. So your exact time complexity is $$$O(tl^3)$$$ in which $$$l$$$ is the length of the interval, and total operations is about $$$10^{10}$$$.
I made an Undirected graph for d and applied dijkstra for each query and ended up getting tle, D initially looked like a graph problem
its a lot simpler than that
i am trying to submit but getting in queue from so long... anyone knows why? the contest was over 12 hrs back
Here is my submission 276639740 for C but I got WA on test 4. Who can explain why??
int overflow. use long long.
Can someone please explain me this? My contest submission for Problem D was hacked https://mirror.codeforces.com/contest/2004/submission/276679036 But when I submit the same solution now, it gets accepted https://mirror.codeforces.com/contest/2004/submission/276743351
Also, I had 2 solutions (using map & unordered_map) which got accepted during the contest. What happens if one of them fails the systests? Is it the final submission that is tested or if any of them works then we get an AC? I was getting a TLE in D so I was trying different things and was not sure what was the best way to do this.
Thank you
if you have 2 subs in contest one hacked and one still AC. u will take points for the accepted sub by its submission time. (This for edu rounds only)
For D, why it works because it's probable that system tests have not been updated by hack tests yet. system testing didn't happen.
Why are the ratings not yet updated?
System testing is not completed yet.
Can anybody tell me what is wrong in my code for C problem.
Solution Link: https://mirror.codeforces.com/contest/2004/submission/276717142
Can anybody help me find the error in my code for problem C
Solution Link: https://mirror.codeforces.com/contest/2004/submission/276717142
Same mistake as this.
Note that
alice -= Math.min(alice, k);affectsk -= Math.min(alice, k);oh great, I am dumb
thanks a lot
F with $$$\mathcal{O}(n^3)$$$ complexity passed, but my solution in $$$\mathcal{O}(n^2 \log n)$$$ got TLE because I used "push_back" $$$2\times 10^6$$$ times. However, the problems are very nice, thanks for the problem providers!
TL47 on F, BLYAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAT
Ok, nvm, 4.968s / 5s xD ultra lucky.
Auto comment: topic has been updated by awoo (previous revision, new revision, compare).
It's very interesting to write from two accounts https://mirror.codeforces.com/submissions/keshavspam2800 https://mirror.codeforces.com/submissions/keshavgarg8800
This guy is the next LGM. Solving every question in <10 minutes. LOL keshavgarg8800
coming from a person who solved 10 questions in his entire life and he is at expert level
please have a life! Cheating in contests won't take you anywhere. your submissions have literally [................................................] these big variable names. clearly you have cheated!
I don't know exactly that maspy account was hacked or that is cheating but the submission 276630607 and 276683562 of Sparsh_Mittal are the same during this contest.
This submission 276683562 was done 14 minutes after the end of the contest when solutions were already available to everyone.
Hi everyone, why is the contest showing unrated for me even tho my rating is less than 2100. Any idea?
ratings weren't updated yet
I implemented an Nlog(N) solution for problem D (276652556), but since I coded it in Java, which is slower than C++, I got a Time Limit Exceeded error. This happened after the contest, putting me behind 2.5k more participants. Later, I submitted the same logic in C++ (276815660) with the help of ChatGPT, and it got accepted. This isn't the first time I've experienced this issue—same logic gives TLE in Java but passes in C++. I hope MikeMirzayanov can look into this and resolve the issue. I was close to becoming an expert after this contest, but due to the language (not the logic), I lost that chance.
if youre aware of this, why not just code in c++.. mike cant make java run faster
Mike can do something about this. On Codeforces, certain languages like Python often get a time limit multiplier (2x or even 3x) compared to C++. This allows users to code in other languages too.
As for why I don’t just switch to C++—I feel more comfortable coding in Java at the moment. Since Java is an available option for submission, it should be viable. If uniformity across languages can’t be ensured, then perhaps it’s better to remove Java as an option altogether.
This is false for official rounds.
that is like saying you can't win an F1 race because you used a 2nd hand honda civic, there's a reason why almost everyone here codes in cpp but i'm sure you could've done something to optimize it in java itself, try looking it up online
I get your point, but the analogy isn't quite right. Codeforces supports multiple languages, so it's fair to expect some parity. I've already optimized my Java code (276816388), and it still runs close to the TLE (just 79ms away, might give tle if we run it again). On the other hand, unoptimized logic in Python (276837719) is giving runtime close to C++ (276815660) because of better time scaling, even though Python is slower than Java. So, even a 3rd hand Honda Civic can win if the rules are well-structured.
I notice that you are using the standard Java Scanner class. This is generally not recommended. Here is your first solution with my modified version of Kattio, 276843722.
I used to be a Java enjoyer for competitive programming. And I still enjoy using Java for other tasks. But it only took me a week to start using C++ comfortably (at least in CP). It is not an undue burden to expect the contestants to be mindful of what language they use.
Two reasons I wouldn't want some sort of time scaling:
People have gone to fairly high places on this website with alternative languages, such as Java. Worry about improving your problem solving skills, because at our level, the language choice really doesn't matter that much.
Why are the ratings still not updated in the platform?? It's over 24 hrs!!
They will probably update rating for this round after recalculating the rating changes of the last div1+2 round. Lot of cheaters were skipped but rating hasn't changed yet
still unrated....
wait
I swear to god if i wake up and ratings still not updated
congratz on the positive delta :)
Well My solution got a plag warning. But seriously I haven't copied from any source but I Googled to read a few sources to optimise my solution as it failed to pass all the test cases I implemented in my IDE. Any suggestions on how to make our code more unique and plagiarism-free are highly appreciated.
Stop cheating.
I never thought about creating unique solution and never got skipped verdict.
And also explain these lame excuse to those who have never coded in their life.
Your previous 2 contests were skipped. You are clearly cheating lol.
Dear Codeforces Team,
I have received a notification that my solution for problem 2004E (submission ID: 276630140) significantly coincides with the solutions of other participants. I would like to clarify that my approach was independently developed using well-known concepts from combinatorial game theory, specifically the Sprague-Grundy theorem, which is a standard method for solving problems of this nature.
My solution is based on calculating Grundy numbers to determine the outcome of the game. This technique is well-documented and is commonly used in problems involving impartial games, such as Nim. Given that these concepts are fundamental and widely taught, it is likely that multiple participants arrived at similar solutions independently.
For reference, here are some sources that explain the theory behind my approach:
Sprague-Grundy Theorem on CP-Algorithms Game Theory: Sprague Grundy Theorem on GeeksforGeeks Nim Game and Grundy Numbers — TopCoder I assure you that I did not engage in any form of code sharing or plagiarism. My work was done independently, and I am committed to upholding the integrity of the competition. If needed, I am open to providing further details or discussing the specific steps of my solution to demonstrate its originality.
Thank you for your understanding
Coincidentally, your solution and the leaked solution both have this useless check
LOL AnkushRaj Someone leaked solution so that guys like you would be able to trace easily
Attention!
Your solution 276582979 for the problem 2004B significantly coincides with solutions sushi_666/276582979, Farhan.WaheedSS/276587437, AryanSeth/276637757, parshiv_k/276649425. 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://mirror.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 I got a message about my code being plaged in B. I dont know any of the other three remotely, and by looking at the submission time I hope its evident that my code isnt something like a copied code. I also didnt use IDEONE so thats out of question. This was one of the best contests I ever had so please remove the plag check if you belive Im innocent which I can guarantee you that I am. I went to pupil after a long struggle and dont wanna lose the rating for something I didnt even do. Please awoo help me with this one.Thank YOu
The submissions are the EXACT SAME, except for random newlines that don't do anything.
Can someone Help me in D problem, why am i getting a tle at test 6, here is my solution link
https://mirror.codeforces.com/contest/2004/submission/276709586
Replace
for (auto it : msi)withfor (auto& it : msi).Thanks a lot bro, it worked, but why there was a tle? and what difference does the & play? also when we need to use auto & and when is it not necessary? can you provide some clarification please..i was just using auto it for all my answers till now, is it working because auto it was creating copy of itself in O(n) and hence giving tle, and auto& it is passing it by reference and hence don't give tle?
The
autoversion copies the entire vector for each iteration, in the same way as pass-by-value function arguments.Your solution 276648513 for the problem 2004D significantly coincides with solutions davit_tsibadze/276637520, morphinecode/276648513.
I want to clarify that this is merely a coincidence, I don't even know the other person, we don't even belong to the same country, I wasn't using any third party code editor and this isn't even a case of leak as he is the only person whose code shares similarities.
I want to point out that there are significant syntactical differences between both solutions, it is just the binary search part that matches and since the binary search was repeated 6 times so it caused a lot of same syntax.
I wish Mikemizayanov, awoo, and the authors adedalic, BledDest and Neon to take a look at it and solve this issue.
I am writing in response to the email regarding my solution (ID: 276675706) for problem 2004E, which has been flagged for significantly coinciding with other users' solutions. I would like to provide some context regarding how my solution was developed.
For problem E, for E, my solution vedu1004/276675706 is not even close with yash3003/276594150 , as both are in different languages and code structure is also different.
For problem D, I asked ChatGPT to generate a helper function. Given that the approach is relatively straightforward, it is possible that another user who used a similar tool might have ended up with code that appears similar to mine.
I want to clarify that I did not share my code with anyone, nor did I receive code from others. The code I submitted was based on my own work and public resources that were available before the contest.
I understand the importance of adhering to the rules, particularly the clause that states solutions and test generators must use source code completely written by the participant, except for code published or generated using tools available before the round. My intention was not to violate any rules, and I hope this explanation clarifies the situation.
I wish @Mikemizayanov, @awoo, and the authors @adedalic, BledDest and Neon to take a look at it and solve this issue.
I am writing in response to the email regarding my solution (ID: 276675706) for problem 2004E,and 2004D which has been flagged for significantly coinciding with 1 other user solutions. I would like to provide some context regarding how my solution was developed.
For problem E, both my code vedu1004/276669741 and other user code yash3003/276669660 are in different language and the code structure is also very different.
For problem D, I asked ChatGPT to generate a helper function. Given that the approach is relatively straightforward, it is possible that another user who used a similar tool might have ended up with code that appears similar to mine.
I want to clarify that I did not share my code with anyone, nor did I receive code from others. The code I submitted was based on my own work and public resources that were available before the contest. I wish @Mikemizayanov, @awoo, and the authors @adedalic, BledDest and Neon to take a look at it and solve this issue.
Another incredibly obvious cheater. Vast majority of the code from the submissions in your post, including, but not limited to, the entire while(query_count--) loop, is exactly the same except for variable names. Codeforces plagiarism checker is smart, and trying to outsmart it like this is futile.