aryanc403
aryanc403
  • Видео 119
  • Просмотров 336 768

Видео

How to bruteforce game theory problems using CF Edu problem
Просмотров 1,9 тыс.4 месяца назад
How to bruteforce game theory problems using CF Edu problem
Codeforces Educational Round 169 Solution Discussion | ABCDEF
Просмотров 3,2 тыс.4 месяца назад
Codeforces Educational Round 169 Solution Discussion | ABCDEF
How to install atcoder library
Просмотров 2,4 тыс.5 месяцев назад
How to install atcoder library
TheOneYouWant AMA
Просмотров 10 тыс.8 месяцев назад
TheOneYouWant AMA
TheOneYouWant AMA Trailer
Просмотров 3,8 тыс.9 месяцев назад
TheOneYouWant AMA Trailer
#Atcoder #ABC 345 "E - Colorful Subsequence" Editorial (By #AIR #1)
Просмотров 1,1 тыс.9 месяцев назад
#Atcoder #ABC 345 "E - Colorful Subsequence" Editorial (By #AIR #1)
#Codeforces round 930 "Bitwise Operation Wizard" Editorial
Просмотров 1,2 тыс.10 месяцев назад
#Codeforces round 930 "Bitwise Operation Wizard" Editorial
#Codeforces round 930 "Shuffle Party" Editorial
Просмотров 1,2 тыс.10 месяцев назад
#Codeforces round 930 "Shuffle Party" Editorial
#Codeforces round 930 "Pokémon Arena" Editorial
Просмотров 59010 месяцев назад
#Codeforces round 930 "Pokémon Arena" Editorial
#Codeforces think-cell round 1 C. "Lexicographically Largest" editorial
Просмотров 2,7 тыс.10 месяцев назад
#Codeforces think-cell round 1 C. "Lexicographically Largest" editorial
#Codeforces think-cell round 1 A. "Maximise The Score" editorial
Просмотров 39110 месяцев назад
#Codeforces think-cell round 1 A. "Maximise The Score" editorial
#Codeforces think-cell round 1 B. "Permutation Printing" editorial
Просмотров 1,3 тыс.10 месяцев назад
#Codeforces think-cell round 1 B. "Permutation Printing" editorial
#StopPostContestHacking How to play cat and mouse game ft leetcode contests (and a shadow ban)
Просмотров 87410 месяцев назад
#StopPostContestHacking How to play cat and mouse game ft leetcode contests (and a shadow ban)

Комментарии

  • @TheJohnniePlays
    @TheJohnniePlays 35 минут назад

    My comment: Here is an idea to experiment on some future duels. Duel can be between two pairs of 2400+ rated users where for each round you choose 2 (up to 2400 codeforces difficulty rating) problems that are not solved by any of 4 duelists and then each pair split 2 problems so that each duelist solves only 1 problem without trying to solve the other one at all and then the main point is that each duelist can not write the solution for the problem player A solved but player A must collaborate with the other teammate (player B) in a way where the other teammate is using whiteboard and explaining all about the problem that this other teammate (player B) solved and giving instructions on what to write in order to solve the problem solved by the problem B. The same thing for problem solved by player A has to give instruction and explain all bits of the solution to player B. You could also have 3 streams separate where you have one main stream for your commentary and splitting between the other 2 streams where these 2 other streams each is for each pair of duelists with their commentary and screen share. Like the main points is in this solution explaining to the other teammate which could be interesting to see how top rated people come up with solutions. Maybe there doesn't even have to be any round but basically a set of problems that are not solved by any competitor like 5 problems or such and then each duelist marks the problem he is doing so the other one does not try to solve it and they have some time limit like 1h or 2h and are just trying to solve as many as possible. Also problems should be picked so they are not implementation heavy because that would be really hard I assume to code for the other person getting instruction, so maybe keep problems around 2000 cf rating difficulty or such so it keeps the stream quick paced and have something going at all time for majority of viewers to watch and learn something from the stream. This is just a rough idea with main focus on this solution explaining to the other teammate, I guess other people can give more suggestions and figure what could be done better and so. Anyway this ended up bit of a longer write up. Thanks for the streams. ================================================= Revised and more concise version of my comment by chatgpt: Here is an idea to experiment with for future duels. A duel could involve two pairs of 2400+ rated users. For each round, you could select two problems (with a maximum difficulty of 2400 Codeforces rating) that none of the four duelists have solved before. Each pair would then split the two problems so that each duelist solves only one problem without attempting the other. The twist is that no duelist is allowed to write the solution to the problem they solved. Instead, they must collaborate with their teammate. For example, player A explains the solution to the problem they solved to their teammate, player B, who codes the solution based on player A's explanation. The same applies in reverse-player B must explain and guide player A in coding the solution for the problem they solved. This format could be enhanced by having three separate streams: A main stream for your commentary, which alternates between the pairs. Two additional streams, one for each pair of duelists, with commentary and screen sharing. The main focus is on the solution explanation and collaboration between teammates, which could be fascinating to watch as it showcases how top-rated players approach problem-solving and communicate their ideas. You could also eliminate the concept of "rounds" and instead have a set of unsolved problems (e.g., 5 problems). Each duelist selects a problem to solve, ensuring no overlap, and the teams have a set time limit (e.g., 1 or 2 hours) to solve as many as possible. To keep the stream engaging, the problems should not be implementation-heavy, as this would make it difficult for one person to code based on another's instructions. Instead, problems around a 2000 Codeforces rating would likely be ideal, as they strike a balance between difficulty and fast pacing, keeping the stream interesting for viewers. This is just a rough idea focusing on the solution explanation aspect. I’m sure others can provide additional suggestions to improve this concept further. Thanks for the streams! ================================================= TLDR also by chatgpt lol chatgpt is the best: Proposed a duel format with two pairs of 2400+ rated users. Each duelist solves one problem (max 2400 rating) from a shared set of unsolved problems, then explains their solution to their teammate, who codes it based on the explanation. Focus is on collaboration and showcasing top players' problem-solving and communication skills. Suggested streams for commentary and screen sharing, a time limit (e.g., 1-2 hours), and selecting problems (~2000 rating) that aren't implementation-heavy to keep it engaging for viewers. Open to further suggestions to refine the idea.

  • @vishalmishra-ue4te
    @vishalmishra-ue4te 11 часов назад

    First time seeing dominator

  • @Flux-e4y
    @Flux-e4y 18 часов назад

    denominator to bhout neend mai lag raha hai😪

  • @devanshkhandelwal6722
    @devanshkhandelwal6722 18 часов назад

    Cool ❤

  • @accountaccount3840
    @accountaccount3840 21 час назад

    Nice duel ...

  • @dampdigits.69
    @dampdigits.69 21 час назад

    keep them coming!

  • @dampdigits.69
    @dampdigits.69 21 час назад

    Wow! This is so cool. Always wanted to see how Dominater solves problems

  • @shivanshkasaudhan4130
    @shivanshkasaudhan4130 23 часа назад

    Damn, the elimination of n colour states was so good. Thanks.. Please keep uploading videos of such hand-picked good questions.

  • @vasuaggarwal2334
    @vasuaggarwal2334 День назад

    Scrasse orz orz Dominator orz orz

  • @user-fm8jx
    @user-fm8jx День назад

    Cp is about to be chess....1v1

  • @aadarshdhiman7768
    @aadarshdhiman7768 День назад

    This is amazing to see🎉❤

  • @rayyankhan8525
    @rayyankhan8525 День назад

    wow this can be the future of cp🎉 looking forward to tourist vs jiangly 1v1 someday

  • @ishankdeep9823
    @ishankdeep9823 День назад

    I guess vMix can help to stream both the participants screen at the same time

  • @ishankdeep9823
    @ishankdeep9823 День назад

    Congrats 👏🎉

  • @studynewthings1727
    @studynewthings1727 День назад

    Congrats to both the participants.

  • @iamarpit.
    @iamarpit. День назад

    Congo.. shreyan for day 1 and best of luck for day 2

  • @saadman-t3j
    @saadman-t3j 4 дня назад

    ❤🎉

  • @DevGorasiya-z5e
    @DevGorasiya-z5e 6 дней назад

    Rank of dhirubhai ambani institute of information technology

  • @dreamer2387
    @dreamer2387 8 дней назад

    Very good explaination 👏👏

  • @allaboinadivakar1976
    @allaboinadivakar1976 10 дней назад

    When two legends met together💀💥

  • @throwawayuser9931
    @throwawayuser9931 11 дней назад

    when will an online mirror be available for practice?

  • @vyomgoyal3125
    @vyomgoyal3125 11 дней назад

    Bhai can't we solve D question using DP? If yes, can you explain and give some hints?

    • @gaurishojha1202
      @gaurishojha1202 11 дней назад

      Hi, I solved using dp. first count the frequency of each equal groups elements like string=110111000 freq = 2 1 3 3 and we can observe that each group greater than 2 freq reduces to 2 and adds count of the removed element. our final freq array contains only elements 1 and 2. we can run recursion and then memo it. cases if a single element of a parity is surrounded by diff parity then we can remove that middle single element and merge both group and reduce that group upto atmax 2 and treat removed as profit. why dp? if we have multiple single elements adjacent how to choose which one give more profit. below is the psedo code of diff operation. int f(freq, prev, i){ ans = f(freq, v[i], i + 1); if(i + 1< freq.size and freq[i] = 1){ nextPrev = prev + freq[i + 1]; reduces = nextPrev - 2; nextPrev -= reduces; ans = max(ans, reduces+ 1 + f(freq, nextPrev, i + 2)); } return ans; }

    • @vyomgoyal3125
      @vyomgoyal3125 11 дней назад

      @@gaurishojha1202 Damn bhai, thanks. Understood upar upar se. Will try it once later today. Thanks man. I just wasted a lot of time on this question thinking in a completely wrong direction 😭😭😭😭

  • @jashanpreet.753
    @jashanpreet.753 11 дней назад

    Can you please do a discussion stream for ICPC Kanpur 2024 regionals.

  • @Flux-e4y
    @Flux-e4y 12 дней назад

    Orgasm aa jata hai CP dekh ke🥵🥵🥵

  • @Shivam_smraj
    @Shivam_smraj 12 дней назад

    Dominater069,

  • @priyangshumanna3231
    @priyangshumanna3231 12 дней назад

    Oh dominator!

  • @acrsahil5631
    @acrsahil5631 13 дней назад

    You teach really well but why you write what you are saying like a whole sentence it is not required !!!! <3

  • @sudhadevi6692
    @sudhadevi6692 15 дней назад

    Very Clear Explanation of all questions bhaiya ❤

  • @ashokdaukiya3211
    @ashokdaukiya3211 16 дней назад

    Why cant we just calculate min sum and max sum of sub array from entire array? Why we are taking two separate part.

    • @aryanc403
      @aryanc403 15 дней назад

      1 -1 100 -1 1 Min sum is -1 and max sum is 100 But not all values between -1 and 100 are achievable.

  • @saibharathalam6402
    @saibharathalam6402 16 дней назад

    in the 3rd question we should check for first 30 primes but by mistake u checked for first 30 numbers while proving it is greater than 1e18

    • @aryanc403
      @aryanc403 15 дней назад

      Yes, for my proof it should be first 30 primes rather than first 30 nos. In practice its difficult to find counter testcase even for first 10 nos.

  • @theepicman8160
    @theepicman8160 16 дней назад

    Thumbnail has wrong problems

  • @Nisikto
    @Nisikto 16 дней назад

    Which drawing tool are you using?

  • @AkashKumar-kl4xd
    @AkashKumar-kl4xd 16 дней назад

    So damn good explanations

  • @HarshitKumar-dj4ev
    @HarshitKumar-dj4ev 16 дней назад

    Bro, can u give some direction about xor getting 0 in 1 + log 50

  • @cdrrjt
    @cdrrjt 16 дней назад

    Sir i am very new to codeforces , but have decent knowledge of dsa, but one thing.i don't understand is how much question should i aim for like for div 4 should i try upsolving all questions and same for div 2 , should i upsolve the first 3.. i mean my rating is just 1100 rn.. so can you please tell.. btw i really like your approaches and way of teaching

    • @aryanc403
      @aryanc403 13 дней назад

      You can check yt403.com/discord -> roadmap-to-red-in-2-mins channel.

  • @codequest-yt
    @codequest-yt 16 дней назад

    For problem C, for finding max sub arr sum and min subarray sum for left and right of the unique number why will kadane algorithm not work? Thanks in advance <3

  • @A57278
    @A57278 16 дней назад

    is G hard for u?

    • @aryanc403
      @aryanc403 16 дней назад

      Yes.

    • @A57278
      @A57278 16 дней назад

      @@aryanc403 read red in 2 mins roadmap maybe helpful

  • @vampconnoisseur
    @vampconnoisseur 17 дней назад

    What is the point of using INF and -INF in "E. Best Price" inside the impPoints vector

    • @aryanc403
      @aryanc403 16 дней назад

      I have a variable defined at top INF = 1e18. Lets say its just a fancy way to write a big no.

    • @vampconnoisseur
      @vampconnoisseur 16 дней назад

      @@aryanc403 No i'm not talking about the type definition, im saying why is it necessary to have 1e18 and -1e18 in the beginning and end of the array

  • @multiversesyndrome2299
    @multiversesyndrome2299 17 дней назад

    Sir very thankful to you for your post contest discussions . Really helps !!! . Please keep doing it .

  • @1moreredcoder
    @1moreredcoder 17 дней назад

    @demoralizer welcome back

  • @AMANMISHRA-vd7bj
    @AMANMISHRA-vd7bj 17 дней назад

    my only inspiration of competitive programming ,

  • @theorist_2
    @theorist_2 18 дней назад

    Which os u use

  • @gopaljee6063
    @gopaljee6063 18 дней назад

    thank you so much sir

  • @manasXDlol
    @manasXDlol 18 дней назад

    can you share the software you use for these annotations

  • @Sha-256-rath
    @Sha-256-rath 18 дней назад

    Lord Aryan with the solutions🙇‍♂🙇‍♂🙇‍♂🙇‍♂

  • @than0s869
    @than0s869 18 дней назад

    can any one answer me? how to prove greedy approach? please

    • @aryanc403
      @aryanc403 18 дней назад

      codeforces.com/blog/entry/133289

    • @than0s869
      @than0s869 18 дней назад

      @@aryanc403 thanks

  • @abhirajmandal9782
    @abhirajmandal9782 18 дней назад

    cant we use binary search for prpblem E

  • @vaishnavplays203
    @vaishnavplays203 18 дней назад

    please add timestmaps for each problem

  • @vaibhavtiwari6030
    @vaibhavtiwari6030 18 дней назад

    hi bro i am new to cp , how to build these concepts to solve problem like i was trying to do with python but unable to solve tried few codechef contest able to solve only first question butttt any advice you can give

  • @PranjalChabra
    @PranjalChabra 19 дней назад

    my approach for B is as follows : we need to find OR of of range [L,R], lets says there is only 30 bits in the number at max ; then we will see if any number in the range of L,R can set the Result as 1 if not then in result it will be 0, next question is how we see that ? for lets say i th bit we want to see this; then we will set all the bit of lesser significance from ith to 0 and set i th bit as 1 in L if the this number exist the range then that bit will be set 1 in result ; now we need to prove if this number does not exist in the range then no number with this bit set will be in the range , so what we did ? we set the bit which was 0 in L as 1 and made all the bits with lesser significance as 0 hence resulting number will be bigger than the L hence if it doesnot exist in the range it will be bigger then the R, hence you may argue there may be a number in the range with i bit set which is smaller then number , that can only happen if we set bit of higher significance which was 1 as 0 but if we do that then it less than L hence no othen number can exist in the range ..... code is : void solve(long long n, long long m) { long long number = max(0ll, n - m); long long orignalNumber = max(0ll, n - m); long long base = max(0ll, n - m); long long j = 1; for (int i = 0; i <= (long long)(log2(n + m) + 1); i++) { long long reqNumber = ((base >> i) << i) | j; if ((reqNumber <= n + m) && (reqNumber >= (n - m))) { number = (number | j); } j = j << 1; } cout << number << endl; } time complexity is ig log(max(n+m)) cause in the result at max bits can only be set till the bits of significance present in n+m