Привет, Codeforces!
Я рад пригласить вас поучаствовать в Codeforces Round 959 при поддержке NEAR (Div. 1 + Div. 2), который пройдет в 18.07.2024 17:35 (Московское время). Вам будет предложено 8 задач и 2 часа на их решение. Он будет рейтинговым для всех участников.
Тема раунда — компьютерные игры!
Задачи для раунда готовили EJIC_B_KEDAX, zwezdinv, green_gold_dog, molney, azureglow и Sokol080808.
Мы от всей души хотим поблагодарить всех, кто оказал помощь в подготовке этого раунда:
Нашего координатора 74TrAkToR за полезные советы и помощь в подготовке задач!
Qwerty1232 за красную заинтересованность в тестировании раунда.
turmax за красно-чёрное тестирование раунда.
makrav, arbuzick, Anonymous_Noob, Hyperbolic за красное тестирование раунда.
ivan.alexeev, alenenok, Kihihihi, azureglow, pakhomovee, plagues, IzhtskiyTimofey, FelixArg, XaRDKoDblCH, meowcneil, AndreyPavlov, djm03178, MylnikovNikolay за жёлтое тестирование раунда.
Mr_Ell, krigare, marzipan, TimVen74, coder8080, MichsSS, PieArmy, tvladm, Vladosiya, bashem за фиолетовое тестирование раунда.
Algolagon, PoDuReMaN, zarubin, IgorA, leoper, kovir_aleksey_korroy, Kosya, Vamperox за синее тестирование раунда.
Sonya_2009, ishaandas1 за зелёное тестирование раунда.
MikeMirzayanov за прекрасные системы Codeforces и Polygon.
Я также поздравляю green_gold_dog с днём рождения и желаю ему удачи на IOI.
Этот раунд проводится при поддержке NEAR!
NEAR была основана в 2017 году Ильёй Полосухиным, одним из создателей технологии трансформеров, и Александром Скидановым как попытка создать систему, которая бы решала задачи по спортивному программированию. О том, что получилось тогда, можно почитать здесь.
В итоге NEAR сделала большой поворот на 180° и начала работу над протоколом блокчейна, который запустила в 2020 году.
В этом году NEAR сформировала новую лабораторию, NEAR.AI, задача которой — создание будущего, в котором технологии искусственного интеллекта открыты и доступны всем, а не контролируются небольшим количеством мега-корпораций.
Одним из направлений работы является обучение моделей мыслить рационально, и задачи по спортивному программированию — это отличное окружение для этой задачи. В этом контексте NEAR приглашает всех участников Codeforces с рейтингом 1400 и выше помочь нам описать решения задач по спортивному программированию. Мы хотим описать решения большого количества задач разной сложности и платим за это сравнительно большое количество NEAR. Присоединиться к системе можно по этой ссылке.
Спасибо NEAR за предоставление следующих призов победителям:
- 1-е место: 512 Ⓝ;
- 2-е и 3-е места: 256 Ⓝ каждому;
- с 4-го по 7-е место: 128 Ⓝ каждому;
- и так далее;
- с 512-го по 1023-е место: 1 Ⓝ каждому.
Разбалловка: $$$500-1000-1250-2000-2000-2500-2750-3750$$$
Удачи!
UPD: Разбор
UPD2: Поздравляю победителей:
I hope to become expert after this contest ^_^
Good luck everyone and have fun <3
I hope to become expert too. Good luck to us
I'm crying with you D;
Seeing 74TrAkToR as coordinator is quite scary, change my mind.
Relax, it's just another 74's redemption round. I believe he'll make it great.
what redemption? he has never apologized
Are you for real?
codeforces.com/blog/entry/124060
Are you for real? This one doesn't count
Are you realy?
In that blog, he didn't really mention why the round is disliked, thus I doubt he learnt from his mistakes
bro what?
https://youtu.be/15HTd4Um1m4?si=S86ETP2IxGkXFS5Z
ok. I'm wrong.
skill issue
hey, the round is either gonna be good or it's gonna cause a funny reaction from the community
it seems like it's going to be the latter
That's weird. I thought that A-D were pretty average — good problems, but maybe a lot of people didn't like them.
I think C is a very standard problem. Similar ideas appear in many problems such as ARC169B.
Finding weird tricks to solve C sometimes gets old, so maybe having a more standard problem from time to time is not so bad. You are free to disagree, ofc.
praying for 1434 rating
assuming you don't cheat
1434 rating isnt that much...
Finally not 3h Div1 but 9 tasks in 150min, I hope the round is not too speedrun.
8 tasks in 120 min
It changed, I saw it was 9 tasks yesterday.
Actually now it's more speedrun than original (personally, I feel that 74 likes 8 tasks/2 hours combined round)
This time you are right!!!
Hello, I am writing this comment, possibly hoping for officials at NEAR to see.
Recently people on Baekjoon Online Judge (BOJ) have seen many DMs on Codeforces, apparently related to the payout for NEAR.AI. They have been asking for solutions (assumably they annotated the tasks themselves but did not have solutions) to many tasks on BOJ. I am gently asking you to remove the BOJ platform from the automated payouts for the AI. Here are a few reasons why.
The BOJ Rules state this is bannable offense.
Through investigation, the users at BOJ had found out that the BOJ account sahera5474 is related to verifying the solution for the payout (though, we did not know exactly what kind of service it was at that time). This kind of automation, not only is condemned by the users, but is also bannable offense according to the rules.
The following is part of what is stated on the rules of the platform (link):
Clearly, this kind of behaviour is considered cheating on the platform, and is bannable offense.
Most tasks on Baekjoon are not worth any payout for solutions, as they have the solutions publicly available.
For whoever does not know how the Baekjoon platform is maintained, the website contains the following kinds of tasks.
For case 1, most such contests contain official editorials, so the solutions are publicly available. For case 2 it is almost the same as case 1, but you will likely have to search outside the platform for the publicly available solution. Most cases where the solutions are not publicly available belong to case 3, but recently case 3 appears much less than case 1 or 2 in quantity. Therefore, an automated payout for training the AI on tasks of case 1 or 2 would make no sense.
Conclusion
Even if one suggests the two points above might not hold for this specific case, it is common sense to consider the DMs from the payout assignees to be considered as fraudulent use of both the Baekjoon platform and your payouts. Therefore, I am publicly asking to remove the BOJ platform from automated payouts and possibly the training data.
Hi, it's not the solutions, but the step-by-step way people solve them that we are after.
Apologies for violating your ToS, as of right now all the integrations with Baekjoon that we have are turned off, there should hopefully be no folks reaching out to you going forward.
Thank you for the reply! I have outlined in your DMs what could be possibly done to fix the issues in the long term. If you don't mind, please give it a look and tell how you think about it.
As a tester, I want comments again.
As a participant, how much NEAR for free solutions?
It's near $$$998244353$$$.
Iranian Rials?
seems markdown is not working
In NEAR, orange is yellow!
Hope to reach 1800 in this round
You can do it
he can do it
Hope the competition goes smoothly! :)
Do I really need coderforces-near-connect-account setting? It is tooooo complicated. Built-in wallet doesnt work and I even need certificate from goverment for proving where I live!
As a tester, I
Hope it's a good round with strong pretests!
Happy birthday green_gold_dog!
As a tester, I hope that everyone will enjoy solving the contest and find a wonderful problems. I wish everyone good luck)
Scoring distribution when?
sigmaforces
As a tester, I can say that problems are very interesting and I wish everyone good luck
What is N currency?
I also wonder
https://coinmarketcap.com/currencies/near-protocol/
Why is there no number for this round? Sequential Codeforces Round numbering would be more convenient in many ways. Also, there can be numbering for NEAR rounds since this is not the first round sponsored by this company.
Update: the round has number 959 now. Okay.
Hope it doesn't become a speedforces :3
9 problems for 150 minutes ???
Is there any reason why most of the contests score distribution follows GCD(score distribution)=50 or 100??
Actually, the GCD is always $$$250 k$$$ ( $$$k$$$ is a positive integer ) because of the rule.
On CF, if you solve a task during $$$k$$$-th segment of the round when the round is equally seperated into $$$120$$$ segments, you got (max points for the task) $$$\times$$$ $$$( 1 - \frac{k-1}{250})$$$ points (minus, Wrong Answer penalties), therefore, to make the score integer, the max score must be a multiple of $$$250$$$.
For further details, look at this page (just above of "Hacks") and this comment.
As a tester, I was expert when I tested so participate in the contest.
PieArmy can you explain more about your statement ? how does testing of a contest work , is it rated when you test?
Good luck everyone.
Спасибо!
How to change my account because I lose it
If you are referring to NEAR account, send me a DM. I'd need more details to be helpful here.
When would the rating rollback happen for the previous rounds?
As a participant, I hope cheaters don't ruin the contest. Let's ensure that everything is fair and done correctly.
I'm excited!
What are the score distributions for the problems?
As a tester, I can say that problems are very interesting and I wish everyone good luck
I hope I reach 1500 in this. Thanks PoDuReMaN and best of luck to all !!!
Hopefully I get back the rating I lost in last rounds
"Dear God, I come to you with a heavy heart, seeking your guidance and wisdom for those who resort to cheating in coding contests. Grant them the insight to understand the value of honesty and integrity, and the humility to seek success through their own efforts. May they realize that true achievement comes from diligence and skill, and may your wisdom guide them to choose the path of integrity in all their endeavors. Amen."
please give me positive delta
Good Luck!
Same to You
I hope to not become pupil again. I wish the problems to be interesting. I wish there were a round based on specific games and the problems have a story of the game. (Imagine, "Kratos has brought n deers and Atreus wants to eat them with maximum weight first...would be so awesome xDxDxD)
any tips to reach specialist like you? I mean you havent solved that many questions of range >1200/ 1300 but still you are a good specialist
build logic as fast as possible. moreover, i had done 230 questions on my previous id but had to stop it and made this new one. i was solving for quite a time but only from last year or so i started giving contests seriously. trust me problems <2000 are not hard. i strictly believe that you should try your best from your id only as you will be fooling urself. try to stay thinking fast, learning concepts, seeing code of top ones etc. have a zeal to learn. don't give up. choose discipline, not motivation. peace.
cout <<" GOOD LUCK EVERYONE "<< endl;
Two and a half hours is too short for round 1,2
What is "512 Ⓝ" mean? It's name of money type ?
it's the cryptocurrency NEAR. (1 NEAR = $6.15 for now)
another speedforces?
yes
I think if you solve A B C fast you will get expert performance
The number of problems changed from 9 to 8... So it is going to be speedforces (?) :(
Okay, I'm not fast enough :(. Solved ABCDF but ranked lower than 70% of the ABCDE solvers.
Hope @74TrAkTor doesn't spoil the contest
surely i wont throw this time
ok genuinely that was the most atrocious Cdiv2 i have ever touched
why?
really janky index nightmares
What? There was no issue with C at all.
might just be skill issue ig but it straight up took me an entire hour to debug
My implementation during the contest was pretty clean I think: 271215425
Can you explain the intuition behind problem C? Wasn't a good contest for me, could only solve AB.
;(
also i was 3 minutes too slow for D wow
nvm
i hope to become newbie after this contest, after long time.
think about CM
nice pfp lol
hope to reach 1900 today
whats penality for this round
I'm 909 rating should I join or I will lost alot rating ?
Join bro, let's lose rating together, don't worry about it, it doesn't matter, what matters is attempting & working on problems, if/when we get stronger, the rating will come. Enjoy.
best piece of words i heard on cp
Thank you for your suggestion I was going to increase the rating because thank God I solved b but unfortunately wrong answer in a
Lord Krishna had said , You have a right to perform your prescribed duties, but you are not entitled to the fruits of your actions. Never consider yourself to be the cause of the results of your activities, nor be attached to inaction.
hoping to get under 1k
Congratulations, Codeforces checked that I am human in just 5 minutes!
bad D
Problem C , statement's input is written The first line of each test case contains three integers n , x (1≤n≤2⋅105,1≤x≤109) — the number of mushrooms and the maximum toxicity level. But three integers are not there are just 2, is this a mistake or am i trippin?
There is penalties for wa on test1 ???
No.
There's no way, I fumbled on B and got humbled by C.
B fked me up too but check your messages!
Same just straight up couldn't solve B and C was a nightmare
I took 1 hour to prove my approach wrong and finally did it.
The fact that I could straight up ignore the tree structure in E is the funniest thing I've seen in a while hehe.
thanks for shit F and G
F btw : https://math.stackexchange.com/questions/1342269/picking-edges-from-a-connected-graph-so-that-any-vertex-is-incident-with-an-odd https://algorithmist.com/wiki/Euler_tour
Very very good set of problems! Thank you!
C : Hungry Games is a simple application of the famous interview problem Jump Game. I created a video editorial for today's C.
There is also a linear solution: for every l find first r such that $$$\sum\limits_{i=l}^{r} a_i > x$$$ (you can do it with 2 pointers), then calculate a $$$dp[i]$$$: number of segments with i as left border such that game on these segments will end with some none-zero score. $$$dp[i] = (right[i] - i) + dp[right[i] +1]$$$.
You can add a fictive $$$+\infty$$$ element to the end of array, also $$$dp[right[i] +1]$$$ should be added if only $$$right[i] < n$$$ (numeration from 1).
Can we do the same by going from $$$1$$$ to $$$N$$$ and calculating number of segments with i as right border?
Yeah, sure
Just keep in mind that array $$$right$$$ can not be easily transformed into $$$left$$$ (with analogical meaning), you need to find $$$left[i]$$$ explicilty with 2 pointers going from $$$n-1$$$ to $$$0$$$.
The rest actions are completely symmetrical.
how to be good at problem like C
I'm going down green again 💩
Just thought about Pigeonhole principle in problem D at the last minute of the contest.
explain please
Also thought of it, but couldn't code it in time. The basic idea is that you have n-1 operations at most, but n elements in the array. Hence, no matter what operation, it is guaranteed you get two elements which are divisible by the operation number. I couldn't really prove that this will extend from any operation to all operations, so a proof of that will be nice.
Resolved
For each $$$x$$$, it's easy to see that we only can pick two vertices $$$u$$$ and $$$v$$$ when they have the same modulo $$$x$$$. So what we can do after knowing this? Suppose we solve this problem in the reverse order of $$$x$$$, i.e iterate $$$x$$$ from $$$n - 1$$$ to $$$1$$$. Then we can blindly pick any two vertex $$$u$$$ and $$$v$$$ such that they are not picked in the previous steps and have the same modulo with the current $$$x$$$, and we are done. And the fact that this problem always have the answer $$$YES$$$.
So why does it work? Let's consider the first iteration when $$$x = n - 1$$$. Then, when we modulo all $$$a_i$$$ with $$$x$$$, we will have at most $$$n - 1$$$ different outcomes. And according to the Pigeonhole principle, if we have $$$n$$$ elements and $$$n - 1$$$ outcomes, there will be at least two $$$u$$$ and $$$v$$$ have the same modulo, so we choose those two $$$u$$$ and $$$v$$$. And the same procedure could be applied in the next iteration, as after the previous iteration, we will care about $$$n - 1$$$ vertices so we could apply the Pigeonhole principle again until $$$x = 1$$$.
Damn that's actually crazy. Thanks for writing this.
what was D?
At each stage i, take A[j] mod i for each j. Vertices with the same remainders can be connected by an edge. Pigeon hole principle guarantees that you can always find an edge that doesn't create a cycle.
I got the first part, but what is the pigeon hole principle? And how would you go about finding those edges?
Ok I forgot to mention that you have to reverse the process and add edges starting from stage $$$i = N - 1$$$.
When stage $$$i$$$ begins, you currently have $$$N - i - 1$$$ edges added. Let $$$rem[k]$$$ be a length $$$i$$$ array denoting the number of vertices that falls into the remainder of $$$k$$$. So clearly the sum of $$$rem[k]$$$ is equal to $$$N$$$. Now for contradiction, suppose that adding any valid edge to the DSU will create a cycle. This implies that for each group of vertices that falls into remainder $$$k$$$ bucket, these vertices form a connected component. This implies that there are $$$\sum_{k = 0}^{i - 1} rem[k] - 1 = N - i$$$ edges currently part of the DSU. but This is clearly a contradiction, as we started out with an assumption that there are currently $$$N - i - 1$$$ edges in DSU.
And for finding those edges, well, you just have to find two vertices within each remainder group that have different roots of the DSU.
Pigeonhole Theorem.
For any $$$m$$$ numbers, there exist at least two numbers with the same value modulo $$$m-1$$$.
So the answer is always yes.
I got this, but how to show that the 2 nodes with same modulo also weren't connected via some previous set of edges?
Suppose currently we have $$$k$$$ elements, and $$$x = k - 1$$$. After we choose two vertices $$$u$$$ and $$$v$$$ in the current iteration, then in the next iteration, we will only need to care about $$$k - 1$$$ elements as we can ignore $$$u$$$ or $$$v$$$, and the Pigeonhole Theorem could be applied again.
Do it backwards: start with stage $$$n-1$$$ and keep going until $$$1$$$, then reverse the answer later. At stage $$$x$$$, let's say vertices $$$a$$$ and $$$b$$$ are the same mod $$$x$$$. Connect $$$a$$$ and $$$b$$$, then erase one of them (i.e. ignore it for the rest of the problem). This way we are sure that the same set is not connected twice. At each stage we remove one element, and we start with an extra element ($$$n$$$ elements at stage $$$n-1$$$). So this means that at every stage we have an extra element. Hence, pigeonhole guarantees the existence of two elements with same mod in every stage.
Was D a variation of Hall's Marriage Theorem where there are more nodes on one part of the bipartite graph? I didn't have enough time for it, seems like a fun problem though :(
no it was just observing php
Yeah, I was searching for a way to do a pairing between the diffs and the nums from $$$1$$$ to $$$N-1$$$, but it seems you cannot go wrong with it...
I actually saw F before: https://codeforces.net/gym/101879/problem/C
Something something traktor bad something something
d hint?
I am not sure but I was trying with DSU and thought today's contest was 2.5hrs long :)
Are pretests = sys tests? I'm 90% sure that my solution for problem D is bs, but somehow it passed 30 pretests.
Actually the solution was fine, for $$$n$$$ numbers and $$$x = n - 1$$$ at least two numbers will always be equal modulo $$$x$$$, by removing 1 number then we get $$$n - 1$$$ numbers left and $$$x - 1$$$, so pigeonhole principle repeats itself.
I hope so, I am pretty sure my solution can be TLEd, but it passed pretests as well...
The most speedrun round I have participated. I feel they are too classic for D1.
Tags for D??
How come there are so less submissions for C?
bruh :(
cause B fked me up
How do you think and come up with the solution for B?
If $$$s$$$ only has
0, you cannot modify anything. Otherwise, every index from the first1to beyond can be modified at will.Once we simplify operation, we understand that in one operation, first $$$x$$$ elements of $$$s$$$ we can XOR with first any substring of $$$x$$$ elements in $$$s$$$, for any $$$x$$$ we chose..
Now if the first bit in $$$s$$$ is
1, we can clearly make any string $$$t$$$. (Remember that XOR with 1 flips a bit, and XOR with 0, it remains the same.)Now if the first bit in s is
0, but second bit is1, we can make any string $$$t$$$ whose first bit is also0.And so on.. So the final answer: the first position where
1appears in $$$s$$$ must be <= first position where1appears in $$$t$$$, if it possible to make $$$t$$$ from $$$s$$$.Solved ABCD, but I am below people that only solved ABC or ACD...
Great set of problems!!!! But Internet and "Confirm you are not a robot" costed me 5 minutes on A :(
what is the idea behind D?
Upvote if you think that the penultimate problem of a prized CF round having almost 200 solves is a great idea
Man, the limits in C were pretty harsh. This got me a MLE: https://codeforces.net/contest/1994/submission/271247958. And this one got me TLE for some reason: https://codeforces.net/contest/1994/submission/271261706
Both solutions exceeded the problem's limits despite being O(N^2), I guess my rating is plunging once again hehe.
Usually $$$n \leq 2 \cdot 10^5$$$ indicates that $$$O(n^2)$$$ solutions will not work. There was a $$$O(n\log{n})$$$ solution for the problem.
:(
mine
Great answer, thanks for sharing!
I went through hell and came back while trying to solve B.
Anyways, great problem set! Thanks for the round EJIC_B_KEDAX zwezdinv green_gold_dog molney azureglow Sokol080808 and all testers!
G = https://codeforces.net/contest/1670/problem/F
Do the coordinator and the problem setter believe that DIV 1 + 2 equals DIV 3 ?
It seems like 1 + 2 = 3.
I had a very bad experience.The network is terrible.codeforces keep verifying me “are you a human?”. And the statement on m1.codeforces.com is somewhat not available.So I can only read the problems on my phone and solve the problems.After about half an hour am I able to visit the page normally.And now and then it needs verifying and the page just stuck there for minutes,which is quite annoying....
Will my solution for C be accepted if its O(n^2)?
Nope, C is le funny problem
Unless you use bitsets in your program, this is impossible
Fucking Speedforces :(
Didn't want to participate because I didn't sleep well last night, read problem E in bathroom, realized solution was simple, turns out I was one of the first to get accepted lol.
Solution is just count occurrences of bits of sizes of trees, if a bit occurs twice you can turn on every other bit below it. The shape of the trees is not important
how did you reach this conclusion ? ( in the last line ? )
You can get any size removing leaf by leaf. Then, if you have more than one tree with the i-th bit on, you can remove leaves from other trees to turn on all the bits less than i.
Hii, Suppose I have two trees of size 8, Both trees are of the form: Root is 1 , and 7 other nodes attached to root 1. 1st tree will give me 8, Can you please tell how will you generate 4,2,1 using the second tree? I can generate either (4 and 1) or (2 and 1) but not all three simultaneously.
You just take a leaf from one of the trees then You have 7 and 8, 7 | 8 -> 15
Thanks,got it.
You can always delete some unimportant leaves, so if the size of some tree is $$$N$$$, you can get a tree with any size less than $$$N$$$. If you reach a tree size of $$$2^m$$$, by removing 1 more leaf you can set to $$$1$$$ all bits less than $$$m$$$.
Too bad I spent a lot of time figuring out if my solution for D is correct or not, E wasn't actually difficult at all.
Someone just ping me when the editorial is out I'm only able to solve A,B and tried D but TLE :) , As my first contest solving 2 questions is fine ? :')
Harder E: https://codeforces.net/contest/1935/problem/E
c was wild
I had to reevaluate my whole life while solving B and C today.
But overall nice problem set!
In problem B, what will be the answer for this testcase : 0110
0101
Yes.
Apply operation $$$(2, 3)$$$ to make it
0100, then $$$(3, 4)$$$ to make it0101.Thanks! I didn't even consider overlapping segments in the contest.
speedforce
E got WA on pretest 11 and don't know why :(
I got the pretest passed and even I don't know why
When I was finally able to open the very first problem statement (which was problem A, but I wanted to start with B or C), the timer already showed more than one minute passed since the start of the contest. During this time I tried to use
m2.codeforces.com, but it told me there is no access to the statements. The main site was either displaying the connectivity error message or asking me to prove that I am a human being.+1
same issue
I just can't solve problems anymore xD
;!
Super easy solution to H: 271267941
You learn m because $$$(X - a - b) \text{ mod } m = (((X - a) \text{ mod } m) + (m - b)) \text{ mod } m = m - b$$$.
To form string for query 3, first subtract 1 from the base value by reducing the first character by 1, then subtract something between a and 2a. To do this, just consider how many multiples of every power of P you must subtract to subtract a. If it is at most 24, subtract that straight from the letter. Otherwise, subtract 1 from the next letter. This works because P <= 2 * 25, and every letter is subtracted at most once from the carry, and at most by 24 directly.
Clear explanation!
With more and more coordinators like 74TrAkToR who would place the G at G, I'm sure it will soon become a trend for users from all rating ranges to solve problems A ~ G in Div 1+2!!!!!!!!! Very refreshing speeeeeeedy round, love from Botswana
UPD: The trend also appeared in past Div2 contests like 1891F.
Nice round!
I got in top 1023 places but I'm not specialist yet. How can I gain my prize? Haven't NEAR thought about people that have <1400 rating??
Try participating from your main account next time.
Yes I am sorry for using my alt account (to make it seem less severe, the first round I'm taking I'm having a headache so I decided not to risk it and thought rounds with money must be participated with this account eversince). But actually the thing I'm most concerned with is there may be other main <1400 rating accounts who have missed the prize.
Having multiple accounts is against the rules. You should be banned on your both accounts. cc: MikeMirzayanov
Snitch.
I recommend you to let Mike ban my friends first: They have higher rating which may cause more harm codeforces.
Damn.
nice hoping you did it by urself
I spent one hour trying to know how to find the Euler circuit but failed, and I saw this problem at least twice before... This round should not be a division 1 and should be an educational round. Additionally, Codeforces was intermittently down during the contest.
In my opinion, understanding the Euler circuit problem is important for problem F, but not all of it.
Thanks for the amazing test cases in problem D.
Got the idea of A in first 5 mins but it took me 30mins to implement it I am such a dumb:(
fapipotato orange when (v_v")
hope for positive delta and road to lagenry grandmister!!1! (^o^)/
It wasn't until I saw the "int foo" in the tourist's E-question code that I realized I was just a fool.
WTF？？？？Extortion？？？
bro is a menace
i can't stop laughing
LMAOOO Grand Master Oogway blackmailing
when you use 100% of your brain
bro tries to blackmail
(Problem F)
I think that the constraints on the input should be written not only in the problem statement but also in the Input section.
You're right. Sorry, missed that when editing the statements. The validator checks that, of course.
I'm sorry,but I want to say that it's a rubbish contest because the rank depends on the speed,not the knowledge you have.In the other words,if you have enough knowledge but little speed,then you will fail in this contest.But OI and things like that needs a lot of thinking and the knowledge is more important than speed.This contest can't train contestants to think or to learn.So I think it's a bad contest.Hope that author can have more high-quality contests!
womp womp
Can someone suggest me some source for
questions on codeforces so i can practice and learn??? I'm really afraid of the topic and want to overcome that fear.
cses problemset on dp
I finally became blue :)
Can I work for Near using this url? https://acade.studio/ https://acade.studio/cf/~~~/42b3dc3613063549da80c2054da288280814be413611e67fafb53798ea43095f/1924/dd557c66e72750997f1962df175f063ac20910ca334354bfda8244a477b75814
If I understand your question correctly, the link you pasted is generally the right way to onboard to Acadé Studio (modulo the right handle instead of ~).
Feel free to ping me via DMs if something doesn't work.
Hi i got rating (0) does that mean its a +0 or its a glitch :(
+0
bruh i think i should make a new id this id could be a cursed
You know that is a rule violation right? If you only care about rating you will just keep on making new accounts learning nothing, I'm sure that's not your goal??
You don't want to be like zh0ukangyang, right? Then don't make alt accounts.
F: There weren't any extreme cases such that
Some bad implementation of Eulerian Circuit (including my contest submission) got uphack, so if you're afraid I suggest you to resub to check.
The problems are so known that even chat gpt can solve it
About C, I think this question is enlightening:
https://atcoder.jp/contests/arc169/tasks/arc169_b
Why the time for 1+2 is 2h. I nearly passed G but I had no time at last.
Yoo ,C was a good educational question.Got to learn something new.Thanks
I like TrAkToR:)
It seems a lot of people didn't like this round, but I liked the round. The cloudflare issue wasn't that bad for me. Apparently there were repeated problems, but I hadn't seen them, and I thought F was interesting.
I think people deserve second chances. Let's support 74TrAkToR !
Couldn't agree more! Even though most of these problems have shown up on leetcode and code.org (jk haha)
Thanks for decent quality contest.
It's really a great SpeedForces Round.
Here's my thought during this round.
I have some problem with my near account. I didn't copy my Private Key or Seed Phrase and now I can't login. Is there any solution or who do I need to find for help?
tourist is back
tourist is back
tourist is back
tourist is back
tourist is back
tourist is back
Can anyone hack or explain my solution in problem C?
i don't know why this pass. TwT
271325693
Stuck in B forever Gang
Very good trush round!This made my brain quickly rotate.
I should've won 1 NEAR and after the contest I opened my NEAR account and it shows the amount to claim is 0?
You would receive message from System when near are distributed
OK, thank you
Downvoted
I should've won 2 NEAR and after the contest I opened my NEAR account and it shows the amount to claim is 0?
The prizes for this round have not been entered into the system yet.
It takes several days after the round to filter out all the duplicate submissions and finalize the standings.
Oh,thanks.
for problem 3 and I solved the question with a good amount of time and also got a wrong submission . But codeforces saying my code is same as of someothers . Just check it : My code :https://codeforces.net/contest/1994/submission/271254358 Others code ( given by codeforces) : https://codeforces.net/contest/1994/submission/271237089 How it could be simillar I agree the intution is quite simillar. And its very simple to be . Please recheck it and give me back my rating .
(There are many one Who have also the same code as the others but they didn't got any plagrism) Also Got an solution https://codeforces.net/contest/1994/submission/271241875 which is as same as the others but this account did't got an plagrism . I appriciate the thing of plagrism it should be more trict but at first it should be more accurate.If giving plagrism please check it out carefully. My Submission : 271254358 Othhers Submission : 271237089 One's Submission same as others but didn,t get plagged : 271241875
Same with me I submitted after a lot of Wrong answer at almost closing time of the contest on my own.But I received a mail that mine coincides with a lot others.
Bro . Thats Good that they are checking but its wrong that giving plag to the genuine code
Dear Codeforces Team,
I am writing regarding submission 271253205 in contest 1994. I have been flagged for plagiarism, but I would like to emphasize that my code and the flagged submission may share similar intuition, but they are not identical.
I have invested significant effort in solving this problem, and my last submission was just before the contest ended. I understand the importance of maintaining integrity on Codeforces and respect your responsibility in ensuring fair play. However, I kindly request you to review my submission separately from others that have been flagged.
I am willing to discuss any concerns or provide additional clarification if needed. Please reconsider the decision and avoid banning my ID outright. I am confident in my abilities to regain my rating fairly, but I hope to resolve this misunderstanding promptly.
Thank you for your attention to this matter.
Hello, [submission:271243589][submission:271238016], i have got email from CF that my solution is copied from the other solution. I want someone to please check and help me what thing i copied, as i have not done any kind of cheating. i am new to this platform and getting this email is discouraging. How to stop it and avoid in future if someone can guide me , and what mistake i had done in the solution that seems to be copied, i am not getting that point. Its was question with general answer. Please someone guide me and help. And request CF to remove cheating tag from my solution.
Subject: Clarification on Solution Similarity and Account Suspension
Dear Codeforces Team,
I received a notice about my recent solution (ID: 271257872) matching another submission. I want to clarify that, while the approach may be similar, I used my own code snippets and did not copy any solutions. After two unsuccessful attempts, I submitted my original solution. Please note that my other two submissions, which have no allegations of copying, were also skipped.
Additionally, my old ID (Anas_006) was suspended for a similar issue. Could you please inform me of the expected time for its reinstatement?
Thank you for your understanding and assistance.
Best regards,
suplex_city
https://youtu.be/T1PWmRblsyA?si=R_S3Nl1vaHUMm_0W bro save cp you copy code from youtube we can see and you mail cf team! i think you know you had cheat but in case you want proufes please inform me i will happy to give you proufes:)