### tibinyte's blog

By tibinyte, history, 3 months ago,

Hello, Codeforces! Or, as we like to say in Romania: Nu îmi amintesc să fi adresat întrebări, Codeforces!

I am glad to finally invite you to participate in CodeTON Round 3 (Div. 1 + Div. 2, Rated, Prizes!), which will start on Nov/06/2022 17:35 (Moscow time). You will be given 8 problems and 2 hours and 30 minutes to solve them. It is greatly recommended to read all the problems, statements are short and straight to the point.

I would like to thank:

Scoring Distribution: 500-750-1250-1750-2250-2500-3250-3500

The editorial has been published here!

And here are our winners!

And here is the information from our title sponsor:

Hello, Codeforces!

We, the TON Foundation team, are pleased to support CodeTON Round 3.

The Open Network (TON) is a fully decentralized layer-1 blockchain designed to onboard billions of users to Web3.

Since July, we have been supporting Codeforces as a title sponsor. This round is another way for us to contribute to the development of the community.

The winners of CodeTON Round 3 will receive valuable prizes.

The first 1,023 participants will receive prizes in TON cryptocurrency:

• 1st place: 1,024 TON
• 2–3 places: 512 TON each
• 4–7 places: 256 TON each
• 8–15 places: 128 TON each
• 512–1,023 places: 2 TON each

We wish you good luck at CodeTON Round 3 and hope you enjoy the contest!

• +552

 » 3 months ago, # |   -8 Good luck in LeafTON round 3.
•  » » 3 months ago, # ^ |   +1 Omg Green round
 » 3 months ago, # |   +83 omg green round
•  » » 3 months ago, # ^ |   +12 omg green round
•  » » » 3 months ago, # ^ |   +29 omg green round
•  » » » » 3 months ago, # ^ |   +18 omg green round
•  » » » » » 3 months ago, # ^ |   +26 omg green round
•  » » » » » » 3 months ago, # ^ |   0 omg green round
•  » » » » » » » 3 months ago, # ^ |   +12 omg green round
•  » » » » » » » » 3 months ago, # ^ |   +12 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   0 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   +10 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   +7 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   0 omg green round
•  » » » » » » » » » 3 months ago, # ^ | ← Rev. 2 →   -17 (-_-)
•  » » » » » » » » » 3 months ago, # ^ | ← Rev. 2 →   -49 Broke the chain
•  » » » » » » » » » 3 months ago, # ^ |   -26 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   +3 What does "omg green round" mean? Thanks in advance and good luck to the participants!!
•  » » » » » » » » » 3 months ago, # ^ |   0 it means omg round with green problemsetter
•  » » » » » » » » » 3 months ago, # ^ |   +5 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   +6 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   +4 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   +4 omg green round ;)
•  » » » » » » » » » 3 months ago, # ^ |   +8 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   +8 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   +11 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   +6 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   +9 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   +5 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   -19 omg so many attempts to get some + contribution
•  » » » » » » » » » 3 months ago, # ^ |   +9 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   +11 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   +8 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   +8 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   -11 ahahahah
•  » » » » » » » » » 3 months ago, # ^ |   +8 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   +8 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   +7 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   +10 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   +7 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   0 omg green round
•  » » » » » » » » » 3 months ago, # ^ |   0 omg green round
•  » » » » » » » » » 2 months ago, # ^ |   0 omg green round
•  » » » » » » 3 months ago, # ^ |   +7 omg green round
•  » » » » » » » 3 months ago, # ^ |   0 rmg oreen gound
•  » » » » 3 months ago, # ^ |   0 Is it rated?
•  » » » » » 3 months ago, # ^ |   0 Yes, it's rated and mentioned in the title.
•  » » 3 months ago, # ^ |   0 omg green round
•  » » 3 months ago, # ^ |   +11 Please stop this nonsense, every time I read the round announcement I have to scroll down things like orz, as a tester this round is great, etc... And this thread is too much
•  » » » 5 weeks ago, # ^ |   0 omg green round :o
 » 3 months ago, # |   +31 As a tester, I am sure you will find some interesting problems whatever rating you have and I wish you good luck and high rating!
 » 3 months ago, # |   +38 First of all, as a tester, I tested. Secondly, I strongly recommend this round!!!
 » 3 months ago, # | ← Rev. 2 →   +12 As a tester, I'm already yellow, not purple!Upd: Good luck to all participants! tibinyte orz
 » 3 months ago, # |   +45 The contest's duration is 2.5 hours in blog but 2 hours in Current or upcoming contests?
 » 3 months ago, # |   +61 As a tester I suggest everyone to participate in this contest. Even I want to participate in it but I obviously can't :(I wish everyone good luck and a good experience
•  » » 3 months ago, # ^ |   +9 As a sad I can't participate in the round myself I'm a tester too!good luck everyone! hope you find some interesting problems and have a good time
 » 3 months ago, # |   +18 As a tester, Problem A is one of the problems of all time.
•  » » 3 months ago, # ^ |   +21 Okay I will start with B then :p
 » 3 months ago, # |   +4 GREEN>>CYAN
 » 3 months ago, # |   +39 statements are short and straight to the point. finally it's in the blog. Thanks
 » 3 months ago, # |   +5 As a newbie , it will be more fun to participant in a green author round . Best of luck all.
 » 3 months ago, # |   +12 As a tester I tested
•  » » 3 months ago, # ^ |   +87 As a tester, I haven't tested yet
 » 3 months ago, # |   +13 Leaf forces
•  » » 3 months ago, # ^ |   +15 I appreciate your dedication
 » 3 months ago, # |   +10 Excited for a green round :)
 » 3 months ago, # |   0 what is the starting difficulty range of the question ??
 » 3 months ago, # |   -55 Well, it's obvious that tibinyte is a talented programmer, but he just managed to get negative deltas to become green to pretend to be weak. Also, he is just the blog sender, not one of the problem setters.So, I wish you guys get positive deltas & have fun!
•  » » 3 months ago, # ^ |   +97 As a tester, tibinyte is the main author of this round. Please do not spread misinformed opinions.
•  » » » 3 months ago, # ^ |   +20 I can confirm that.
•  » » 3 months ago, # ^ |   +27 Capital L
•  » » 3 months ago, # ^ |   +21 As a Tibinyte's friend I can tell you that he has been developing this round since long time ago and he's the main author of round. Please do not spread misinformed opinions. (and reach master)
•  » » 3 months ago, # ^ |   +8 As a Leafeon supremacist, I can confirm that this claim is baseless. Also green is the best color because Leafeon is the best Pokemon
•  » » » 3 months ago, # ^ |   0 vanilluxe is better :3
•  » » 3 months ago, # ^ |   +11 Okay I'm sorry for that, I apologize for this.
•  » » 3 months ago, # ^ |   +48 Personally, a round is the result of everyone's joint efforts. Whoever thinks of the idea, who thinks the solution, who proves the correctness of the algorithm, who perfects the statement, who creates the powerful test data, and who fixes the issue is actually unnecessary so clearly. What matters is our joint efforts and our friendship.
•  » » » 3 months ago, # ^ |   +13 LLLL LLLL LLLL LLLL LLLL LLLL LLLL LLLLLLLLLLLLLLLL LLLLLLLLLLLLLLLL 
•  » » 3 months ago, # ^ |   +3 I believe you had a huge surprise...
 » 3 months ago, # |   0 Good Luck everyone!!!
•  » » 3 months ago, # ^ |   0 Thank you!
 » 3 months ago, # |   -32 Is it rated?
•  » » 3 months ago, # ^ |   -30 As a predictor , It will be unrated.
 » 3 months ago, # |   +11 How do they'll send cryptocurrency? We have to have crypto wallet?
•  » » 3 months ago, # ^ | ← Rev. 2 →   +7 This Medium blog gonna help you creating one.
 » 3 months ago, # |   +4 As a tester I have nothing more to add than "Good luck!" and "May the God have mercy on you", because the problem setters surely don't know the definition of this word.
•  » » 3 months ago, # ^ |   +35 I assure you, tibinyte knows what mercy is, it is the only hero I can play on overwatch
 » 3 months ago, # |   0 Wish for positive delta Good luck everyone<3!
 » 3 months ago, # |   0 Time to become CM now ;)
•  » » 3 months ago, # ^ |   0 Sa nu iti fie cu suparare dar...
•  » » 3 months ago, # ^ |   0 yup, you can Shivu!
 » 3 months ago, # |   0 G Gaming
 » 3 months ago, # |   -6 Time for Remontad
 » 3 months ago, # |   +1 Oooh wow! 58 people contribute to making this round happen.
 » 3 months ago, # |   0 Aiming green in the green round... Wish me luck :)
•  » » 3 months ago, # ^ |   0 Solve A and B problems very quickly and you will become pupil. Also don't make any wrong answers. All the best bro!
 » 3 months ago, # |   +3 Wish me luck hope i become green after this contest.
 » 3 months ago, # |   +31 Hope I will get GM. :)
•  » » 3 months ago, # ^ |   +33 Oh.. you are close to 2400. Wish you luck for today's contest, hope you become GM..
•  » » 3 months ago, # ^ |   +23 Same D:
•  » » » 3 months ago, # ^ |   0 Frate, iti zic io, ajungi si tu (grand)master.Marinush
•  » » » » 3 months ago, # ^ | ← Rev. 2 →   +14 I have just realized that I read statement of problem E incorrectly, I thought that we can shift only whole string D:
•  » » » » » 3 months ago, # ^ |   +8 You too? Me too!11!!1!!!1
•  » » » » » » 3 months ago, # ^ |   +3 Yeah, so sad :(
•  » » 3 months ago, # ^ |   +11 Congratulations on reaching GM!!
•  » » » 3 months ago, # ^ |   0 Thanks!!
 » 3 months ago, # |   0 I wish good luck to all to this contest !!
 » 3 months ago, # | ← Rev. 2 →   -41 round need to be unrated, because it is round
 » 3 months ago, # |   +6 How do they'll send cryptocurrency? We have to have crypto wallet? someone can answer to this question plz!
•  » » 3 months ago, # ^ |   0 Create a wallet on this app; this'll give you a wallet addresshttps://tonkeeper.com/P.S. Checkout the TON username market on telegram : https://fragment.com/ where you buy and sell in TONs
•  » » » 2 months ago, # ^ |   0 I've created my wallet address and update my profile but did not receive the prize yet. Would you like to tell me the reason or what should I do to receive the prize?
•  » » 3 months ago, # ^ |   0 Yes, you'll have to provide your wallet address to receive prize.Here is the list of TON wallet applications: https://ton.app/wallets
•  » » » 3 months ago, # ^ |   0 Who should I provide my wallet address to?
•  » » » » 3 months ago, # ^ |   0 In CodeTON Round 2 there was a Codeforces message from System, containing link to form. So I think you should wait for that message (it will be visible in notifications panel).
 » 3 months ago, # |   +12 Will score distribution be announced briefly before the beginning of the round?
•  » » 3 months ago, # ^ |   +17 It will be posted along with the editorial as to not spoil anything about the round
 » 3 months ago, # |   +1 Hope that I will be Pupil after this round
•  » » 3 months ago, # ^ |   -8 you have not submitted a single problem coward
•  » » » 3 months ago, # ^ |   0 Sorry I am completing my assignment for my cllg .As Tommorow is the last day
•  » » » 3 months ago, # ^ |   0 And Finally Somebody who noticed me
 » 3 months ago, # |   +4 Score distribution?
•  » » 3 months ago, # ^ |   -18 it is unrated
 » 3 months ago, # |   0 Hopefully it will be great contest because of short statements and straight to the point.
 » 3 months ago, # |   0 In the previous CodeTON rounds, I lost my rating. Hope Today I will gain plus point. Love to see short statements and straight to the point.
 » 3 months ago, # |   -8 let's go
 » 3 months ago, # |   -8 time to + 100
•  » » 3 months ago, # ^ |   +2 Haha prediction
 » 3 months ago, # |   -9 When Life gives you a match with Zimbabwe... Let's be India! All the best for the Contest!
 » 3 months ago, # |   0 Good Luck!!!!!!!!!
 » 3 months ago, # |   -49 Please, don't make rounds again. Thanks.
 » 3 months ago, # |   +10 Last few contests gave me cancer
 » 3 months ago, # |   -16 What is the conversion ration TON to INR ??
 » 3 months ago, # |   0 How to solve D?
 » 3 months ago, # |   +14 How to optimize E from O(n^2) to O(n) or O(nlogn) ?
 » 3 months ago, # |   -10 Good problems, very good D and E was easy
•  » » 3 months ago, # ^ |   -7 How to solve D? please!! QAQ
•  » » 3 months ago, # ^ |   +1 whats the logic for D ?
•  » » » 3 months ago, # ^ |   -6 There arent any logic just print("Hello world")
 » 3 months ago, # |   -8 How to calculate the number of numbers from 1 to M coprime to N ?
•  » » 3 months ago, # ^ |   +17 Factorize N, and then calculate how many numbers between 1 and M are divisible to any of the number factorized using the Inclusion-Exclusion formula:https://www.geeksforgeeks.org/inclusion-exclusion-principle-and-programming-applications/
•  » » 3 months ago, # ^ |   0 Can be done Using Inclusion Exclusion For Example Let prime factors of N be 2,3,7,5 Then ans = M — M/2 — M/3 — M/5 — M/7 + M/(2*3) + M/(2*5) .... — M/(2*3*5) — M/(2*3*7) ..... + M/(2*3*5*7) Given the range of Number we are working on this problem , There will be max 10 distinct prime factors of N so we can generate all 2^10 combinations and get the answer
•  » » » 3 months ago, # ^ |   0 I did this and I TLEed on test 12??
•  » » » » 3 months ago, # ^ |   0 You factorize every $v_i$ and it obviously exceeds time limit $O(n \sqrt{m})$. You only needed to find prime factors of $\frac{v_{i - 1}}{v_i}$ because there are $O(\log m)$ such elements greater than $1$.
•  » » » » 3 months ago, # ^ |   0 There may be a large number of a[i]=a [i+1]. In this case, you only need to do: ans=ans*[m/a[i]],which only do O(log n) times calculations.
•  » » 3 months ago, # ^ |   0
•  » » 3 months ago, # ^ |   0 https://www.acwing.com/blog/content/19417/using principle of inclusion and exclusion.
 » 3 months ago, # |   -8 Took me 40 minutes to search and read about Mobius function(Problem D). Not sure if there is any solution that doesn't utilize it.
•  » » 3 months ago, # ^ |   +5 Use the Inclusion-Exclusion Principle.
•  » » » 3 months ago, # ^ |   0 Yeah, I thought about that one too. But I believe that I won't be able to implement in time.
•  » » » 3 months ago, # ^ |   0 I thought that would maybe time out. Took me an eternity to get the Mobius solution to work.
•  » » » » 3 months ago, # ^ |   0 It won't be time out. Think about this, this principle works on a[i]/a[i+1], and the product of them is a[1]/a[n] which isn't more than 1e9.
•  » » » » 3 months ago, # ^ |   0 It won't time out because, you will apply factorization and inclusion exclusion stuff only when $a[i] != a[i-1]$. The prefix gcd can only change $O(log(m))$ times, because everytime the gcd changes, it has to reduce at least by a factor of $2$.
 » 3 months ago, # |   0 What was logic for C , Solved D but was not able to process C's logic.
•  » » 3 months ago, # ^ |   0 Consider the xor of the two arrays, you will notice something ...
•  » » 3 months ago, # ^ |   +4 Some useful observations are: If every index is equal/unequal in both strings, then it's possible. When an operation is performed, either bit in string a, or bit in string b changes, so if all of the index were equal in both, then it becomes unequal, and vice-versa. You just have to make all index 0 in string a, let's say. So, string b can be all 0's or all 1's. If all 0's then we're done. If not, then just add 3 extra operations: (1,1) (2,N) (1,N)
 » 3 months ago, # |   0 better luck next time for all ****
 » 3 months ago, # |   0 Tough contest for me :((
 » 3 months ago, # |   -6 How to solve problem D ?
 » 3 months ago, # |   -8 How to round number in D?
•  » » 3 months ago, # ^ |   0 If you think of the number I think of, you have to round it down :)
 » 3 months ago, # |   0 Is it ObservationForces now?
 » 3 months ago, # | ← Rev. 2 →   +25 Upvoted the contest for the story behind Problem E. I couldn't focus on the Problem, because I had to show the story to my fiancée and my brother and needed to hype it. Beautiful Daemon Targaryen. I would upvote more if I could.
•  » » 3 months ago, # ^ |   0 I still didn't watch episode 10 of HOD :(
 » 3 months ago, # |   0 Did anyone solve D using Euler's totient function?
 » 3 months ago, # | ← Rev. 2 →   +10 I used about 10 minutes to think about the conclusion of problem E but used more than an hour on coding. Anyway, I think this round is the best round I've ever participated.UPD: after I solved problem D I got rk20. :) (though my final rank is 200 :(
 » 3 months ago, # |   0 omg green round
 » 3 months ago, # |   -13 Literally dying atm. Could get a grasp on D (Yes, I did think of inclusion-exclusion here and there but did not find a way to preprocess the values of the mobius function. Did I want to copy paste an overkill $O(n^\frac{2}{3})$ method to preprocess prefix sums of the mobius function? Hell no.) but couldn't finish. C felt hard, like, very hard. Might be even convinced that C > D if C was after D.
 » 3 months ago, # | ← Rev. 2 →   +8 is it the approach for problem c ?  If A^B == 0000 or 11111 then answer exists  Then we shift all ones to left side and zeros to right for A and do same operations for B also. which makes A as 00001111 and B as 11110000 something like this 
•  » » 3 months ago, # ^ |   0 The answer is only possible if for all i either a[i]=b[i] or a[i]!=b[i]. Just make all a[i]=1,you'll observe that all b[i] become either 0 or 1.If they are all 0 just do 1,n so all a will be 0 and b will remain 0,else if all b are 1 then do (1,1) (2,n)
•  » » 3 months ago, # ^ |   0 Yes. I did operations on $[i,n]$ for all $i$ such that $b_i$ and $b_{i-1}$ are different. Then $b$ consists only of $1$s or only of $0$s. The same goes for $a$. Then you just need to look at $a_1$ and $b_n$ (they did not get changed yet) an accordingly make everything $0$.
•  » » 3 months ago, # ^ |   0 Let me Explain by example Example 1: A -> 110010, B -> 001101 (Initially Given) A -> 001110 B-> 001110 (1-4 index operation) A -> 000001 B-> 111110 (2-4 index operation)  Example 2: A -> 10101 B-> 01010 (Initially Given) A-> 01101 B-> 01101 (1-2 index operation) A-> 00011 B-> 11100 (1-3 index operation) I think the above approach was harder for implementing atleast to me. I guess making all equal to one was simpler than this. Instead of making x 0's and y 1's as above.
 » 3 months ago, # |   0 I always lose rating in div 1+ div2 contest any suggestions…?
•  » » 3 months ago, # ^ |   +1 I guess, in div 1 + div 2 contests it is more important to solve the earliest problems fast enough (since there will be more people solving the same amount of problems, relatively speaking). Interestingly, for me it is the other way round: I always win rating in the joint contests...
 » 3 months ago, # |   +8 How to solve E ?
•  » » 3 months ago, # ^ | ← Rev. 5 →   +11 Take a segment. Let's call $C$ the number of closing brackets and $O$ the number of opening brackets. Let's call $M$ the minimum value the prefix sum of the segment reaches (Opening bracket is $+1$, closing is $-1$). The cost of a segment then is $max(0, O-C)+min(0, M)=max(0, O-C)+M$. Explanation$M$ is the amount of mismatched brackets. We can use a rotation to fix one of them. If we have opening brackets left to distribute, we can also use them.$max(0, O-C)$ comes from the need to balance the amount of both, opening and closing brackets. But we only need to calculate more openings, because more closing ones is already taken into account by $M$.You can calculate both parts of this sum for all segments idependently by iterating and doing confusing calculations. See 179633595 the two blocks after long long ans=0;. For me it was easier to visualise the bracket sequence as a graph of the prefix sum and then look at area on the graph which both of the terms contribute.
•  » » » 3 months ago, # ^ |   +8 Very clean code and solution, thank you! I knew that the solution is very clean, but did not have the skill to find it tho.
 » 3 months ago, # | ← Rev. 2 →   0 I think Problem B is based more on implementation side rather than observation.
•  » » 3 months ago, # ^ | ← Rev. 2 →   0 my sol : just find longest substring with same character let k be the longest length substring and x=no of 0's and y=no of 1's in whole stringour ans is max of (k*k , x*y)
•  » » 3 months ago, # ^ | ← Rev. 2 →   0 I don't think so... This was just 10 lines of code solve :: String -> Int solve str = max count (ind * ind) where x = length $filter (=='0') str y = length$ filter (=='1') str count | x > 0 && y > 0 = x * y | x > 0 = x * x | otherwise = y * y ind = maximum $length <$> group str 
 » 3 months ago, # |   +65 E was nice. Until the last 40 minutes, I thought that the first operation can be only applied to the entire string. That led to a much more complicated problem. :(
•  » » 3 months ago, # ^ |   0 Wow... I was thinking and thinking with that misunderstanding of the first operation for the whole contest. I got to read the statement a bit more carefully.
•  » » 3 months ago, # ^ |   0 Me too, I have written two brute-forces to understand that it's not the case...
•  » » 3 months ago, # ^ |   0 And I realized only after reading your comments that we can apply to any substring :(
 » 3 months ago, # |   0 How to solve problem E? The most I can do is a brute force O(n^3) solution.
 » 3 months ago, # |   0 Great contest!Congrats to the authors!Cool problem D, unfortunately I didn't solve it.I'm waiting for editoiral!Thanks!
 » 3 months ago, # |   +2 Me After Solving A & B :')
 » 3 months ago, # |   -10 Can anyone please explain how to solve problem C with intution?
•  » » 3 months ago, # ^ |   0 Consider A^B. Every time we conduct an operation A^B would be XORed by 111...1. So for a YES case, A^B would either be 000...0 or 111...1 after whatever operations. Now you just need to make A to 000...0. In the end, B is either 000...0 or 111...1. Then it's easy to make a few more operations to get job done.
•  » » 3 months ago, # ^ |   0 Hint1Observe that answer is only possible if both the string is the same or if we can get b after inverting each character of a. Hint2After each operation, you can get either the same string or an inverted version of each other. TutorialMake every character of a 1. Then you get b either in the form of 00...0 or 11...1. Now can simply perform an operation (1, n) for the former case and (1, 1) + (2, n) for the latter. Solutionfrom sys import stdin input = stdin.buffer.readline def func(): ans = list() flag1 = True # same flag2 = True # inverted for i in range(n): if a[i] != b[i]: flag1 = False if a[i] == b[i]: flag2 = False if not flag1 and not flag2: print('NO') return for i in range(n): if a[i] == '0': ans.append((i+1, i+1)) l = len(ans) if flag1: l += 1 if l % 2 == 1: ans.append((1, 1)) ans.append((2, n)) else: ans.append((1, n)) print('YES') print((len(ans))) for i in ans: print(*i) for _ in range(int(input())): n = int(input()) a = input().decode().strip() b = input().decode().strip() func() 
•  » » » 3 months ago, # ^ |   0 im not able to understand the hint1 and hint 2,can you please explain by taking examples?
 » 3 months ago, # |   0 Why the time limit of G is 3s... My FFT solution passes. It does not seem to be the intended solution.
 » 3 months ago, # |   +149 Congratulations to conqueror_of_tourist for not only living up to their username, but also becoming LGM with Python!
•  » » 3 months ago, # ^ |   +8 you also achieved the goal of your username with python only.
 » 3 months ago, # |   +29 Countforces
•  » » 3 months ago, # ^ |   +10 OMG bicsi the legend
•  » » 3 months ago, # ^ |   -6 can you explain C's logic
 » 3 months ago, # |   0 I wasn't able to solve A, with just looking at the problem and understanding it, I saw that in the correct cases, position 0 had an 1 or position 1 had a 2, so intuitively I went on and did an algorithm that checks that case. Too bad for me it was just when it started with 1.But can someone explain why is it when it starts with 1? I did it so if it's 1 or 2 at the first 2 positions, i.e. arr[0] == 1 or arr[1] == 2 then yes, else then "No".
•  » » 3 months ago, # ^ |   +1 before u completely look at my answer read the question once for clarity..... constraint : i < j < k 1st position element has to smaller than kth position element element at 1st position cannot be changed.... but using 1st position you can swap any positions of j and k .so the smallest possible element that can take 1st position must be one .
•  » » 3 months ago, # ^ |   0 Because when it starts with 1 you will always be able to make it sorted, suppose you have 1,5,4 you just select i=1, j= 2 and k=3 so that you can swap 5,4, similarly for every other wrong pair you just can choose those pair as j and k and by selecting i as 1 you are ensuring the 2nd operation which is swap
•  » » » 3 months ago, # ^ |   0 But then, what happens with 2,4,3 for example? You can sort it, but at the start there is a 2.
•  » » » » 3 months ago, # ^ |   0 But since there is no 1 at first place, it is not possible to sort the list, so ans for this is No, one thing 2.3,4 this input is not possible, read carefully the list is permutation, that means you will always have 1,2,...n
•  » » » » 3 months ago, # ^ |   0 Array is a permutation so there must be a 1
•  » » » » » 3 months ago, # ^ |   0 I didn't know what was a permutation, that was the problem....Thank you very much guys! Next time i hope to do better, meanwhile i'll be learning and doing problems.
 » 3 months ago, # |   0
 » 3 months ago, # |   0 can someone explain C logic
 » 3 months ago, # |   +21 Problem D is similar to this problem which I proposed to codechef before.
 » 3 months ago, # |   -25 Sorry, but I didn't like this round at all:A: just one simple observationB: just one simple observationC: just one simple observationD: just find number of coprime with n not greater than m (testers in comments, is it interesting? Really?)E: closed this problem as soon as I've read. Yet another "count some *** for all subsegments"Can say nothing about F+, because didn't readNo, don't misunderstand me, each problem is good. But as set of problems in one contest...
•  » » 3 months ago, # ^ | ← Rev. 5 →   +38 I'm sorry that you found the first 2 problems of the contest(which are supposed to be easy) too easy for a candidate master / master, I will talk to tibinyte and kindly ask him to give out some smart and well hidden minimum cost maximum flow with a lot of observations and implementation details for problems A and B in his future rounds. As of C, consider it a simple observation for you, judging by the counter of solves it was just as hard as it had to be and I also needed some time to figure it out(but I might also have some skill issues so ignore me if you want). I really liked problem D because of its "smooth" solution that included some simple math and observations, so I found it interesting and I am sorry that this does not fit your view. Your argument about problem E is probably by far the dumbest thing I read on the internet today and this really means something as I've seen a lot of dumb memes today. Anyway, I am sorry that you didn't enjoy the round and wish you good luck in the next ones.
•  » » » 3 months ago, # ^ | ← Rev. 2 →   +3 First and foremost, LSecond, this is not how you argue, this is how you correctly argue this is satire, by all means the purpose is not to offend anyone Dear contestant,Please don't worry if you had to little to no observations to pass some problems that were set to not cater to you. In a Codeforces round, the mechanics employed by some problem do not by any means determine the quality of any problem*. Apologies if this does not fit your view.Thanks for the feedback,Tester of Codeton Round 3 *: (although, admittedly, it can determine the quality of the round as you stated-- I won't argue with that because I am not in the mood for jejune disputes)
•  » » » 3 months ago, # ^ |   +3 you found the first 2 problems of the contest too easy for a candidate master / master Where have I said this? consider it a simple observation for you It's my point: this is "one observation problem". You either find this observation->"ez ac" or you don't find this observation at all->you can't solve problem. This argument is applied to A and B I really liked problem D It's your personal opinion and I've shared mine. Your argument about problem E It's not argument, it's my personal opinion. I don't find problems which require some counting on all subsegments interesting. I didn't solve D incontest because I'm dumb, but if I solved D, I would skip E.The dumbest thing is "This contest is the most bad/good/... contest ever" comments without any details. I tried to share my opinion about whole problemset why I didn't like this contest. And seems that you misunderstood my initial comment. Once more: every problem is good. But you don't find problemset good if every problem is about, for example, geometry, do you?
•  » » » » 3 months ago, # ^ | ← Rev. 5 →   +8 "you found the first 2 problems of the contest too easy for a candidate master / master" — I just wanted to point out that it is a normal fact that you found them easy because you have a higher rating than the people that should find them challenging, I didn't want to say that you mentioned this, because you didn't."It's my point: this is "one observation problem". You either find this observation->"ez ac" or you don't find this observation at all->you can't solve problem. This argument is applied to A and B" — I find this quite common in the modern day competitive programming where problems are more and more based on tricky observations, I wouldn't say that I love it, but we should adapt to these kind of problems if we want to do CP, and if not, as we say in Romania, "ayaye"."It's your personal opinion and I've shared mine." — I respect that."I don't find problems which require some counting on all subsegments interesting." — I get that, but in a contest(especially in a serious one with a higher stake like OI or ICPC) you can't just say that you don't like a problem, you just have to try to solve it because you wanted to take part in that competition(I will assume that no one can force you to take a contest if you do not want that)."I didn't solve D in contest because I'm dumb" — I know this is a bad feeling, but you probably just missed out on something, it happened to all of us, no need to be so harsh about it."The dumbest thing is "This contest is the most bad/good/... contest ever" comments without any details" — I agree that this is usually toxic and people tend to say that after a round that got them positive delta, but my opinion as a tester was just that the round was good and recommended others to try it."But you don't find problemset good if every problem is about, for example, geometry, do you" — No, I do not. I also recommend that you spend some time analysing the problems with editorials and realize that they are not just some combinatorics / dp problems, but also require some clever observations. Also, A-C are not counting at all and E and H are not combinatorics / dp(as I know).Anyway, I think we discussed more about this than it was necessary. The contest is done and others will come.
•  » » » » » 3 months ago, # ^ |   +6 Thank you for discussion!Just one correction: in ICPC you have wider set of topics to choose from and what is more important you have teammates. This allow you to skip/delegate some problems if you don't like it more often, than in personal contests.
•  » » 3 months ago, # ^ |   0 and you could just solve simple observations problems didn't notice something?
 » 3 months ago, # |   0 Why is my output on the code I wrote when run on my compiler different from the one used in the contest ?
 » 3 months ago, # |   0 Great contest
 » 3 months ago, # |   -16 as newbie downvote me
 » 3 months ago, # |   0 Tourist in 10th place wow
 » 3 months ago, # |   +10 Ratings updated preliminarily. We will remove cheaters and update the ratings again soon!
•  » » 3 months ago, # ^ | ← Rev. 3 →   0 Hope updated ratings would changed my rating by 2 points
•  » » » 3 months ago, # ^ |   0 Hope updated ratings would change my rating by 54 points
 » 3 months ago, # |   +3 input : 4 1000000000 60 30 1 1output: 595458194how? can anyone explain?
 » 3 months ago, # |   +3 Personally, I think this E is very good. I can use BIT or segment tree, or even deduce the expression O(n) directly. I really like this problem
 » 3 months ago, # |   0 Hey, is it possible to pass problem B with python?
 » 3 months ago, # |   +1 in problem D if m was <= 1e6 , could we solve the problem using Möbius Function? something similar to this problemhttp://acm.hdu.edu.cn/showproblem.php?pid=1695
•  » » 2 months ago, # ^ |   0 It's Möbing time!
 » 3 months ago, # |   0 How to claim TON cryptocurrency after winning in this contest?
 » 3 months ago, # |   -12 My friend rating is 1110 and rank in contest is 9000 by solving only problem and by rating is 926 and got 8200 rank by solving 2 questions...But the worst things my rating decreases only by 6 and my rating decreases by 28....How is it even possible.....Please help anyone
•  » » 3 months ago, # ^ |   0 You get more rating points than you should first 5 or 6 contests you participate in, which is most likely to be the situation with your friend.
 » 3 months ago, # | ← Rev. 2 →   0 I see that in recent contests the questions don't get difficulty tags!! Why is it so? Anyone?
 » 3 months ago, # | ← Rev. 3 →   +11 Alsalam Alykom,My solution for A,B is skipped and i swear i didn't use ideone or another account or anything its my clear solution ! I swear to god i didn't cheat why something like that happen to me ? how to avoid this MikeMirzayanov tibinyte
 » 3 months ago, # |   +47 How do I receive my money?
 » 3 months ago, # |   +65 Where is my ton please?
 » 3 months ago, # |   +68 How can I receive my money?
 » 2 months ago, # |   +25 Has anyone received the prize yet?
 » 2 months ago, # |   +11 Has anyone received their TON yet? The deadline was supposed to be on Nov 26.
•  » » 2 months ago, # ^ |   +22 Hi. The wallets were collected and sent to TON. Of course, they need several days to process the data.
 » 2 months ago, # |   +11 Is the team still in the process of sending prizes? Or is it already finished and I didn't receive anything due to some mistake from my side?
•  » » 2 months ago, # ^ |   +10 I didn't receive it either.
•  » » » 2 months ago, # ^ |   +10 same, still havent got anything
 » 2 months ago, # | ← Rev. 2 →   +31 I just received my prize a minute ago. Maybe you guys will receive it soon
•  » » 2 months ago, # ^ |   0 I also just received the prize :)
•  » » » 2 months ago, # ^ |   0 I also just received the prize :)
 » 3 weeks ago, # |   0 Will this round send NFT trophies? Like the kind of NFT trophy in round 1 and round 2.