Rayan 2024 Selection Round (Div. 1 + Div. 2, Rated, Prizes!)
By ArshiaDadras, 5 weeks ago, In English

👋 Salam, Codeforces!

We are thrilled to invite you to the Rayan Programming Contest 2024 - Selection (Codeforces Round 989, Div. 1 + Div. 2).

The Selection Round is a rated contest for Div. 1 + Div. 2, taking place on Nov/30/2024 17:35 (Moscow time). The problems for this round were crafted and prepared by ArshiaDadras, MohammadParsaElahimanesh, AmShZ, AmirrzwM, and Keshi.

The top 60 trusted participants (with a maximum of three from each country) will qualify for the onsite Final Round in Tehran, in Spring 2025. An additional quota is reserved for the host country. Hotel accommodation, local transportation, and meals will be provided to all finalists. More details about the final contest will be available on the Rayan website.

🏆 Selection Round Prizes:

  • The top 50 participants will receive T-shirts. 👕
  • Additionally, 50 random participants from ranks 51-1000 will also receive T-shirts.

🏅 Final Round Prizes:

  • 1st place: $15,000 💵
  • 2nd place: $10,000 💵
  • 3rd place: $5,000 💵
  • 4th–6th places: $2,000 each
  • 7th–10th places: $1,000 each

🙏 Special Thanks:

We would like to extend our deepest gratitude to:

📊 Score distribution: $$$500 - 750 - 1500 - 1500 - 2000 - (2000 + 2000) - (3000 + 3000) - 5000$$$

We look forward to seeing you both online and onsite!

Good Luck & Have Fun! 🎉

Note: Editorial just published here! 🚀🍿

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4 weeks ago, # |
  Vote: I like it +17 Vote: I do not like it

Auto comment: topic has been updated by ArshiaDadras (previous revision, new revision, compare).

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    4 weeks ago, # ^ |
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    Is there a separate category for Iranians, or will all participants be grouped together?

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      4 weeks ago, # ^ |
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      All participants are considered together. The only difference is that we have a separate quota (about 30–40) for Iranians, which will not be counted within the main quota.

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4 weeks ago, # |
← Rev. 2   Vote: I like it +43 Vote: I do not like it

As a tester, I didn't know I could have won $15,000 (just needed to beat a certain Belarusian...)

Y'all should give it a shot!!! Good luck.

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    4 weeks ago, # ^ |
      Vote: I like it +12 Vote: I do not like it

    Sorry man, but I have to admit that I'm kinda glad you didn’t know that 🫂

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    4 weeks ago, # ^ |
      Vote: I like it +21 Vote: I do not like it

    For the 1st time in my life, I found the tester's comment funny!

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4 weeks ago, # |
← Rev. 2   Vote: I like it +21 Vote: I do not like it

As an author, please give me T-shirt too.

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4 weeks ago, # |
← Rev. 2   Vote: I like it -114 Vote: I do not like it

I predict tourist will fall below 4000.

Edit after round: everyone who downvoted this should apologize to me in DMs.

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4 weeks ago, # |
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Nice! This must be one of the rounds of all time, I'm so excited to see my teachers' round ;)

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    4 weeks ago, # ^ |
      Vote: I like it +13 Vote: I do not like it

    Let me tell u sth about ArshiaDadras:

    I officially started competitive programming and programming classes with Arshia Dadras four years ago. He is a very capable and ethical teacher. Thank you, Arshia, for being so great <3

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      4 weeks ago, # ^ |
        Vote: I like it +15 Vote: I do not like it

      Thank you so much for your kind words and support, AmirParsa!

      It’s truly a pleasure to be remembered in such a thoughtful way. I’m also really happy to see that you’re continuing to improve your coding skills.

      Wishing you the best of luck in your journey, and I hope we get to meet in person soon!

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        4 weeks ago, # ^ |
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        Thank you, man. It's an honor for me to have grown under the guidance of professors like you. I'm looking forward to meet you in person too ;)

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4 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

"The top 60 trusted participants (with a maximum of three from each country)"

How will you enforce this?

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    4 weeks ago, # ^ |
      Vote: I like it +14 Vote: I do not like it

    We will start inviting participants from the top of the list while considering the country limit (a maximum of three participants per country) until we fill all 60 spots. Each invited participant will have a specific time frame to confirm their attendance. If someone declines or cannot attend, their spot will be given to the next eligible participant from the scoreboard who meets the criteria.

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4 weeks ago, # |
  Vote: I like it -38 Vote: I do not like it

Paying people that much to solve problems with already existing solutions..

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4 weeks ago, # |
  Vote: I like it +24 Vote: I do not like it

As a tester, this is the $$$4^{th}$$$ time in a row I found at least one problem in the round I tested really satisfying, and I hope people would feel the same.

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4 weeks ago, # |
  Vote: I like it +34 Vote: I do not like it

I love Persian contests!

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    4 weeks ago, # ^ |
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    Why do you like Persian contests?

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      3 weeks ago, # ^ |
        Vote: I like it +9 Vote: I do not like it

      Because most of them are about their history, and Persian history is very good.

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4 weeks ago, # |
  Vote: I like it +14 Vote: I do not like it

As a tester I wish good luck to everyone!

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4 weeks ago, # |
  Vote: I like it +35 Vote: I do not like it

As a tester, I can confirm that I tested this round.

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4 weeks ago, # |
  Vote: I like it +20 Vote: I do not like it

As a tester, I liked to test my first round! :)

PS: If you win $15,000, please give me the T-Shirt, hahaha

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4 weeks ago, # |
  Vote: I like it -19 Vote: I do not like it

Final Round Prizes: 1st place: $15,000 2nd place: $10,000 3rd place: $5,000 4th–6th places: $2,000 each 7th–10th places: $1,000 each

^^^

How much will you win? They're really rich...

(This is my first comment on CodeForces so if you think I'm wrong please politely scold me)

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    4 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    If you look at their website, the onsite part is sponsored by a university and from what I can tell an entire ministry, and these prizes are for the onsite part. This round is only a selection round before the onsite one.

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      3 weeks ago, # ^ |
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      I understand, thank you. By the way, the prize money for the competition held by Huawei last time was also as high as this

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4 weeks ago, # |
  Vote: I like it +3 Vote: I do not like it

as a persian, i can confirm i ain't winning shit

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    4 weeks ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    But I guess we'll have someone from Iran in the top 10... P.S. Let’s bet lunch on it!

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    4 weeks ago, # ^ |
      Vote: I like it -8 Vote: I do not like it

    as a persian, i also confirm i'm not winning anything

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4 weeks ago, # |
  Vote: I like it +5 Vote: I do not like it

W salam

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4 weeks ago, # |
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I'm quite interested that, is there any special about the sponsor? why does this contest appears on the top bar?

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4 weeks ago, # |
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niceeeeeeeeeeeeeeee

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4 weeks ago, # |
  Vote: I like it +37 Vote: I do not like it

tourist will destroy this contest to earn money

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4 weeks ago, # |
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As a participant, I will participate!

BTW, what watch are you wearing, sir ?! ArshiaDadras

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4 weeks ago, # |
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Do you arrange this every year?

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    4 weeks ago, # ^ |
      Vote: I like it +13 Vote: I do not like it

    We’re planning to do that, but this is Rayan’s first year!

    We need to evaluate the feedback and reception from participants before making a clearer decision.

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      4 weeks ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Thanks.. It will be great. Can I ask you the meaning of "Rayan"

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        4 weeks ago, # ^ |
          Vote: I like it +33 Vote: I do not like it

        “Rayan” is derived from the first part of the Persian word “Rayaneh,” which means “computer.”

        Fun fact: If someone asks us, “What is the event name?” we would reply, “It’s Rayan.” In Persian, we could say this as “Rayan eh,” which follows the same pattern as saying, “It’s cold” -> “Sard eh” or “It’s good” -> “Khoob eh.” Interestingly, this sounds exactly like the word “Rayaneh”!

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4 weeks ago, # |
  Vote: I like it -6 Vote: I do not like it

hope to reach expert this round. any recommendations?

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4 weeks ago, # |
  Vote: I like it -14 Vote: I do not like it

this round will make tourist get back to gm i guess. lol

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4 weeks ago, # |
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Fantastic ! nice job Iran. Good luck everyone.

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4 weeks ago, # |
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and maybe the final prize in the next year

Bcz problems for this year have been leaked and the winners are already decided $$$??$$$

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4 weeks ago, # |
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I am new to codeforces but still I try.

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4 weeks ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

As a Participant, I can confirm that I am not a tester.

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4 weeks ago, # |
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I'm just looking at the rewards knowing I won't be even close to top 10 :/

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    4 weeks ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    Why not? At the very least, you’d get to enjoy a wonderful trip to Iran, stay in one of the best hotels here for free, and experience some amazing hospitality! (if you can make it to the top 60 😁)

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4 weeks ago, # |
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Hey Tourist, I accept donations here ಥ_ಥ

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4 weeks ago, # |
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Excited for my first contest as a specialist!

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4 weeks ago, # |
  Vote: I like it +4 Vote: I do not like it

if there are $$$50$$$ russians above me and all of them refuse to go to the onsite, will i be invited?

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    4 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Yes, as I said here! (if you'll be one of the next candidates to invite)

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      4 weeks ago, # ^ |
      ← Rev. 2   Vote: I like it +1 Vote: I do not like it

      if someone sets their country to Gibraltar will you somehow check that they are indeed from Gibraltar?

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        4 weeks ago, # ^ |
          Vote: I like it +18 Vote: I do not like it

        I hope we don’t encounter these types of cheaters in the top 60. But don’t worry—we’ll verify their passports during the invitation process to catch any discrepancies.

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4 weeks ago, # |
  Vote: I like it +14 Vote: I do not like it

As an Iranian, this really makes me proud!

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    4 weeks ago, # ^ |
    ← Rev. 3   Vote: I like it +14 Vote: I do not like it

    And I would like to thank the medalists who designed the questions and all the people involved and hardworking in this event!

    I think it is a good thing that the famous programmers of Codeforces come to our country.

    I hope to see the tourist here!

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4 weeks ago, # |
  Vote: I like it +41 Vote: I do not like it

The same day as chinese NOIP! Hoping we all have good performance in the both round!

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4 weeks ago, # |
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is it for rating

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4 weeks ago, # |
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May everyone get their expected rating plus this round<3

GG everyone!!

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4 weeks ago, # |
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Is this round rated?? as normal cf contest

participation is enough for registration or i have to register on website or something??

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    4 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It's like a normal rated contest in CF. Also, participation is enough for registration and you don't need to register on the website or somewhere else.

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4 weeks ago, # |
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Great

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4 weeks ago, # |
  Vote: I like it +7 Vote: I do not like it

Time Conflict with ICPC Regionals QAQ

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4 weeks ago, # |
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im new to codding but still im going to Participate hopefully i receive a t-shirt

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3 weeks ago, # |
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looking forward!

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3 weeks ago, # |
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Salaam too.

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3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Hope to cross 1100 again this time!

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    3 weeks ago, # ^ |
    ← Rev. 2   Vote: I like it -10 Vote: I do not like it

    Hey..I looked at your profile..you have been doing cp for more than 2 years now but still not able to improve much even after being so consistent..with this much pracitce you should have touched pupil by now..Mayb you are doing something wrong.. Try solving cses that helped me..also you can practice usaco guide silver level prblms and learn techniques on prefix/suffix sum,two pointers etc..Also try atcoder contests and gym prblmes..Good luck buddy

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      3 weeks ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Thank you, buddy, for your suggestion. I’ve already solved many problems on CSES and am familiar with concepts like prefix sums, two-pointers, binary search, basic graphs, and recursion. I’ve also participated in a few Atcoder contests but haven’t practiced USACO or GYM problems yet— planning to work on those. Previously, I lacked consistency. So, recently I have been consistent and hopefully, I will reach pupil soon. I will try my best. Again buddy thanks for your suggestion. I am getting more inspiration to reach pupil now.

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        3 weeks ago, # ^ |
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        You're most welcome.. I would suggest to go through the topics of usaco silver level and try to solve the problems of those topics like bs, prefx sum etc. from there first..(some of those problms are really nice but you may need to create accounts on multiple sites :)) Maybe then you can try out the gym problems.. (Also dont spend too much time on a problem if you cant solve it on first try.. Go to the next ones then come back agaim here.. I used to spend too much time on a prblm but thats not worth it but just a waste of time for most of the times) Hope for the best.. See ya

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3 weeks ago, # |
  Vote: I like it +14 Vote: I do not like it

As a tester, I wish everyone good luck! :)

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3 weeks ago, # |
  Vote: I like it -16 Vote: I do not like it

teams or no?

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3 weeks ago, # |
  Vote: I like it +22 Vote: I do not like it

Rayan Gosling Round

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3 weeks ago, # |
  Vote: I like it +3 Vote: I do not like it

Wish tourist back 4000++ once more.. yo yo

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3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

rainboy orz (Problem H).

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3 weeks ago, # |
← Rev. 2   Vote: I like it +42 Vote: I do not like it

Seems orzdevinwang will become tourist soon!
...Why does the sentence sounds that strange.

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3 weeks ago, # |
  Vote: I like it +6 Vote: I do not like it

I said D > C;

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3 weeks ago, # |
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What is the case in E where a solution exists for both k and n being odd and k < n?

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    3 weeks ago, # ^ |
      Vote: I like it +11 Vote: I do not like it

    in all cases, a solution will exist if n and k are odd except for k=1 and k=(n-1)!

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    3 weeks ago, # ^ |
      Vote: I like it +3 Vote: I do not like it 
    1
5 3

    YES
1 2 3 4 5 
3 4 5 1 2 
5 3 1 4 2
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      3 weeks ago, # ^ |
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      how do you construct the solution if k=3 and n>k?

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        3 weeks ago, # ^ |
          Vote: I like it +8 Vote: I do not like it
        Spoiler
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          3 weeks ago, # ^ |
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          ah ok i get the solution i just assumed that if k=3 and n>k it was impossible

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3 weeks ago, # |
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nice pics SIR

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3 weeks ago, # |
  Vote: I like it +17 Vote: I do not like it

C was so annoying to solve. Forgot one edge case and spent all my contest time on it, leaving no time for D :(

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    3 weeks ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Actually, C seemed difficult to me not because of solving the problem, but because of the implementation. What edge case did you get stuck on? I can share my solution if you want.

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      3 weeks ago, # ^ |
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      Yes, the implementation was also painful. But I solved it at the last minute :')

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3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

How to solve E?

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    3 weeks ago, # ^ |
    ← Rev. 2   Vote: I like it +21 Vote: I do not like it
    Spoiler
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3 weeks ago, # |
  Vote: I like it -68 Vote: I do not like it

Sooo nice E! Though it contains some casework, there is a beautiful construction method.

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    3 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Can you share it with us common folk?

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      3 weeks ago, # ^ |
        Vote: I like it +9 Vote: I do not like it

      We treat $$$n=1$$$, $$$k=1$$$, $$$k=n!$$$, $$$k>n!$$$ first.
      Let's define $$$i$$$ th permutation as the permutation when we call next_permutation for $$$(1,2,\dots,n)$$$, $$$i-1$$$ times.
      First we construct for even $$$k$$$. We can just take $$$i$$$ th permutation $$$p$$$ and $$$(n!)+1-i$$$ th permutation $$$q$$$ together. (actually, $$$p_i = n+1-q_i$$$)
      For even $$$n$$$, if $$$k$$$ is odd, the answer is No (because $$$k \times \frac{(n+1)n}{2}$$$ isn't divisible by $$$n$$$)
      Now, we construct, for odd $$$n$$$, $$$k=3$$$. This can be done by this method (as an example, let $$$n=7$$$).

      • $$$(1,2,3,4,5,6,7)$$$
      • $$$(4,5,6,7,1,2,3)$$$ (left shift $$$\frac{n-1}{2}$$$ times)
      • $$$(7,5,3,1,6,4,2)$$$ (make sum = $$$3 \times \frac{n+1}{2}$$$)

      At last, we construct, for general odd $$$k$$$. We ignore following pairs:

      • $$$(1,2,3,4,5,6,7)$$$ and $$$(7,6,5,4,3,2,1)$$$
      • $$$(4,5,6,7,1,2,3)$$$ and $$$(4,3,2,1,7,6,5)$$$
      • $$$(7,5,3,1,6,4,2)$$$ and $$$(1,3,5,7,2,4,6)$$$

      At the current state, we can take $$$\frac{n!}{2} - 3$$$ pairs. After that, just add $$$k=3$$$. Then, we can construct for $$$3 \le k \le n!-3$$$.

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        3 weeks ago, # ^ |
          Vote: I like it +3 Vote: I do not like it

        tysm

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        3 weeks ago, # ^ |
          Vote: I like it +1 Vote: I do not like it

        wow! That left shift (n-1)/2 is insane, Any inspiration to come up with that?

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          3 weeks ago, # ^ |
            Vote: I like it +54 Vote: I do not like it

          I saw this problem and immediately coded brute force for $$$n=5$$$ and $$$k=3$$$ and the first one that came out was that construction

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          3 weeks ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          It comes by try & error so I can't tell exact step, but I tried to match $$$(1,4,7)$$$ first, suddenly cames up my construction.

          Maybe $$$(1,2,3,4,5,6,7)$$$ + (shift) leads the difference of the sum of two permutation $$$+2, +2, \dots, +2, -n+2, +2, \dots, +2$$$ (and under odd $$$n$$$, this works) is the decent way?

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          3 weeks ago, # ^ |
          ← Rev. 3   Vote: I like it 0 Vote: I do not like it

          I didn't manage to solve E during contest (I was really close, but I missed a few simple cases, like n = 1 and forgot about the cap for evens). I did get something very close. But basically once you realise that you can do all evens up to n! you can come up easily with "maybe if I can figure out a solution for 3 then it will work". And then you realise that you are looking for something that once you add 1 2 3 ... n it makes the same thing, you get the idea that you make something where the sum increases by 1 every time. I still had some trouble, but then I thought it would be easier if I just do every 2.

          So for every 2 I could do for 7:

          4 5 6 7

          1 2 3 4

          that would add up to 5 7 9 11

          and then the rest complete the sequence. So my own solution for 7 was:

          4 1 5 2 6 3 7

          1 5 2 6 3 7 4

          7 6 5 4 3 2 1

          Then it's somewhat obvious that 4 = 7 / 2 + 1. Then it was easily expandable as start for n / 2 + 1 for the first one.

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    3 weeks ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    why is this getting downvoted

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      3 weeks ago, # ^ |
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      Because the problem E is neither nice, nor beautiful.

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        3 weeks ago, # ^ |
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        how you rate a problem can be subjective, however, why would you take it out onto someone else who's not involved in setting it?

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3 weeks ago, # |
← Rev. 2   Vote: I like it +20 Vote: I do not like it

Nothing could be worse than losing rating on my birthday :(

Edit: problems are nice, I choked on H with some cases like 1 3 105 63 45 and what I missed is just enumerating the GCD of all picked numbers...

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    3 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    happy birthday! :)

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    3 weeks ago, # ^ |
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    Happy Birthday!

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    3 weeks ago, # ^ |
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    Happy birthday tho!

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    3 weeks ago, # ^ |
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    Well a gain of 0 is better than any loss I suppose. Happy birthday!

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      3 weeks ago, # ^ |
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      Yeeeees! I’m so happy that codeforces give me this non-negative delta as my birthday gift, and my solution for H works after fixing the bug. Thank you! OvO

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3 weeks ago, # |
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My God!

I don't know; maybe as an Iranian I would have preferred not to take these internal disputes to CF and resolve them by ourselves!

Of course, I definitely didn't want a few one-sided people talking about it, and I'm happier that both sides' opinions were presented.

But in the end, the beauty of the previous comments was lost in this kind of discussion and debate....

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3 weeks ago, # |
  Vote: I like it +92 Vote: I do not like it

AT 2:53 OF CONTEST TIME THE DIFFERENCE BETWEEN TOURIST AND ORZDEVINWANG WAS 32 POINTS. TOURIST NEEDED TO HACK SOMEONE AND HE MANAGED TO DO SO 8 SECONDS BEFORE THE END OF THE CONTEST!!!!!! BUT ORZDEVINWANG SOLVED G2 3 MINUTES BEFORE THE END OF THE CONTEST, SECURING THE FIRST PLACE!!!!! WHAT A DRAMA OMGOMGOMGOMGOMGOMGOMGOMGOMGOMGOMGOGMOG

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3 weeks ago, # |
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The problems were nice, though I feel like I solved something very similar to F a long time ago on Codeforces (I didn't have time to solve it though, I spent too long on implementing D and E). Also, they have very nice pictures :D Do you think that they are AI generated? They are very intricate but I don't see any obvious AI artifacts. Otherwise, big respect to the artist who made them.

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3 weeks ago, # |
  Vote: I like it +21 Vote: I do not like it

orz orzdevinwang. Also IF orzdevinwang didn't solve G2, then tourist would have won by doing a successful hacking attempt at the last minute, that's crazy.

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3 weeks ago, # |
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Is it true that in H the max size of the answer is 3? And does it have randomized solution?

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    3 weeks ago, # ^ |
      Vote: I like it +18 Vote: I do not like it

    The max size of the answer is 6. let $$$x=2\cdot3\cdot5\cdot7\cdot11\cdot13$$$, then $$${\frac{x}{2},\frac{x}{3},\frac{x}{5},\frac{x}{7},\frac{x}{11},\frac{x}{13}}$$$ is a valid set

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    3 weeks ago, # ^ |
    ← Rev. 4   Vote: I like it +14 Vote: I do not like it

    Here is the rough sketch for H.

    This is how we can characterize all possible linearly independent sets $$$a$$$.

    First find a set of distinct primes $$$p_1,p_2,\ldots,p_k$$$ and let the gcd of all the $$$a$$$ be $$$g$$$. We will write $$$P = \prod p_i$$$ as the product of all the primes here.

    Then we can construct a linearly independent set $$$a_1,a_2,\ldots,a_k$$$ where $$$a_i$$$ is a multiple of $$$\frac{gP}{p_i}$$$ but not $$$gP$$$.

    For each linearly independent set, you can find a corresponding $$$(g,p_1,p_2,\ldots,p_k)$$$ that satisfy the above constraint.

    Then after brute forcing everything, you realize the the number of possible $$$(g,p_1,p_2,\ldots,p_k)$$$ with $$$k \geq 3$$$, $$$g \mid P$$$ and $$$\frac{gP}{\min(p_i)} \leq 10^5$$$ is at most 500k or something. So you brute force everything. I had to use a bunch of constant opts to pass because my code is shit.

    Checking answer is at least $$$2$$$ can be done in $$$O(n \log n)$$$.

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3 weeks ago, # |
  Vote: I like it +31 Vote: I do not like it

What's with the constraint of A? I see a lot of people passed it with brute forcing, but they're quite close to the time limit, while I got TLE and I suspect it's because I used long long instead of int.

I don't see why it couldn't be much higher so that no brute force solution could pass, or be much lower so that every brute force solution could pass.

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3 weeks ago, # |
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Why are there few pretests on problem D and E? I'm worried that my solution won't pass the system tests.

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3 weeks ago, # |
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Problem C was nice (although very standard dfs problem). D wanted me to kms.

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    3 weeks ago, # ^ |
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    how to solve that?

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      3 weeks ago, # ^ |
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      C or D?

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        3 weeks ago, # ^ |
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        C

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          3 weeks ago, # ^ |
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          I applied DFS here. Firstly, i marked all the non '?' boundary cells bad if they were actually making the pointer out of bound (for example, if a[0][0] = 'U', this is bad cell). Now, I just applied dfs from an unvisited cell. For the question mark, we can replace it with either of the 'L', 'R', 'D', or 'U'. Hence, if any of the four gives me a good answer, I'll take that, otherwise I'll mark that cell as bad. For the explicit cells, like a[i][j] = 'L', it will be marked based on dfs(i, j - 1).

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      3 weeks ago, # ^ |
      ← Rev. 7   Vote: I like it 0 Vote: I do not like it

      Problem C (my solution)

      1. Model each maze square as a node in a graph and connect them using the info given
      2. Realize that for each connected component of the graph, whatever node u start from will always give u the same end point

      Start with your answer = n*m. Perform a DFS from each node and find the end state of the BFS from each node (either an infinite cycle, a question mark node, or an escape). If a node leads to an escape, subtract 1 from your answer. For the question marks, check each adjacent grid square. If the destinations of all adjacent squares are out of bounds / escape, subtract 1 again from your answer. Done! My submission at 294079080. Reply if you have more questions.

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        3 weeks ago, # ^ |
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        Never thought it would be a graph problem, thanks for your explanation :P.

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    3 weeks ago, # ^ |
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    I was able to solve D, but chocked on C :(

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      3 weeks ago, # ^ |
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      I too choked on C, but figured out that I was making a silly mistake. I was running if-else loops for the '?' marked cells instead of four separate if loops. This costed me whole hour lmao. For D, I was trying to first put all the 0's in correct positions, then 1 and 2's will eventually become easy to swap (I failed miserably).

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3 weeks ago, # |
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How to solve F? I cant find any clear approach to solve this problem

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    3 weeks ago, # ^ |
    ← Rev. 3   Vote: I like it +11 Vote: I do not like it

    Firstly, for simplicity, redefine the special conditions to be in terms of the number of rubies/sapphires in your satchel (not chest). Then add dummy states $$$(0, 0)$$$ and $$$(n, m)$$$ for convenience (we won't double value at these though).

    Define dp states to be $$$dp[i]$$$, which is the total sum of value of your satchel across all ways you can reach the state defined by the $$$i$$$-th condition.

    To solve this dp you will firstly need to define some ordering on the conditions so that when evaluating some dp state, all the states that it depends on have already been evaluated. Just sort by $$$x + y$$$.

    Then, the transition from state $$$i$$$ to state $$$j$$$ will correspond to the ways you can go from state $$$i$$$ to state $$$j$$$ without visiting any other special condition state in between.

    Computing the number of ways you can go from state $$$i$$$ to state $$$j$$$ (where $$$i$$$ occurs before $$$j$$$ in the topological ordering) without visiting any intermediate special condition state is easy and can be done using inclusion exclusion in $$$O(n^3)$$$. Call this $$$w[i][j]$$$.

    Also define the total number of ways for us to go from state $$$i$$$ to $$$j$$$ to be $$$g[i][j]$$$. This can be computed in $$$O(1)$$$ using simple combinatorics.

    Define the addition in value for going from state $$$i$$$ to state $$$j$$$ without any intermediate doubling to be some $$$h[i][j]$$$.

    Then, the dp formulation for state $$$i$$$ is:

    $$$dp_i = 2 \cdot \sum{dp_j * w[j][i] + g[0][j] * w[j][i] * h[j][i]}$$$

    (where $j$ occurs before $$$i$$$ in the topological sort)

    All states will be evaluated in $$$O(n^2)$$$.

    The final answer is the value of the dp state corresponding to the the state $$$(n, m)$$$ divided by the total number of ways to go from $$$(0, 0)$$$ to $$$(n, m)$$$, which is just $$$\binom{n + m}{n}$$$.

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      3 weeks ago, # ^ |
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      Can you explain more about how to calculate $$$w(i, j)$$$ using inclusion exclusion, I still didn't get this part.

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        3 weeks ago, # ^ |
        ← Rev. 2   Vote: I like it +5 Vote: I do not like it

        $$$w[i][i] = 1 \; \forall \; i$$$.

        $$$w[i][j] = g[i][j] - \sum_{k} {w[i][k] \cdot g[k][j]}$$$ (here $$$i$$$ occurs before $$$j$$$ and $$$k$$$ occurs between $$$i$$$ and $$$j$$$ in the ordering)

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3 weeks ago, # |
  Vote: I like it +6 Vote: I do not like it
meme on C
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3 weeks ago, # |
  Vote: I like it +68 Vote: I do not like it

I liked F. However E is bad.

(1) I believe it made the contest unfair that one sample testcase was silently added. What you should have done is to make a clarification as an annoucement instead.

(2) The essential part had appeared multiple times — IMO 2009 Shortlist C2 and UTPC2023 and possibly many others. The "distinct" part does not add fun (perhaps personal opinion).

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3 weeks ago, # |
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who can tell me why runtime error pleaseYour text to link here...

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3 weeks ago, # |
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294071265 this gives runtime error on pretest 8 for problem C . Can anyone tell what is the reason . I have not seen similar verdicts in any other solutions , so I am curious what caused this error ?

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    3 weeks ago, # ^ |
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    have you considered what would happen if dfs(i, j) went out of bounds?

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      3 weeks ago, # ^ |
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      How can this even happen , for all border elements which are overflowing i have assigned visited =1 and verdict=-1 initially only. Can you explain or give example

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3 weeks ago, # |
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How to solve D ?

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    3 weeks ago, # ^ |
    ← Rev. 3   Vote: I like it +1 Vote: I do not like it

    What I did was iterate until the columns are sorted. On each iteration I:

    1. Swap the 1 that is furthest to the left with the 0 that it furthest to the right until all 0s are to the left of every 1.

    2. Swap the 2 that is furthest to the left with the 1 that is furthest to the right until all 1s are to the left of every 2.

    I did have to use a set to store the indexes otherwise it would TLE.

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    3 weeks ago, # ^ |
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    If there are only $$$0$$$'s and $$$1$$$'s or only $$$1$$$'s and $$$2$$$'s, just swap the values in the way Leongg wrote.

    If the array contains all three numbers, then you must put all the $$$2$$$'s to the rightmost part of the array or (if more than half elements in the array are $$$2$$$'s) put all the $$$0$$$'s to the leftmost indices. If you have to swap $$$1$$$ and $$$2$$$ or $$$0$$$ and $$$1$$$, then just swap them. Otherwise you must do an intermediate swap involving any (for example, first to the left) $$$1$$$ in the array. If there were $$$2$$$, $$$1$$$, $$$0$$$ at some indices, after these two swaps you will have $$$1$$$, $$$0$$$, $$$2$$$ on these respective indices.

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3 weeks ago, # |
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I can't imagine how many people were blocked by $$$n=m=1$$$ in $$$C$$$ and $$$n=k=1$$$ in $$$E$$$...

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    3 weeks ago, # ^ |
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    why is n=m=1 in C an edge case? I didn't consider this and it seems fine

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      3 weeks ago, # ^ |
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      if you have all ?-s in the grid, then n=m=1 is the only case you will have 0 as answer. otherwise the answer is n*m always.

      but still depends on the implementation obviously, i failed it though

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    3 weeks ago, # ^ |
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    Wrong answer on test 4 ...

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    3 weeks ago, # ^ |
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    I realized right after the end of the contest is that the wrong with my code for $$$E$$$ is that I didn't handle the case where $$$n = 1$$$

    it really hurts 😢

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3 weeks ago, # |
  Vote: I like it +16 Vote: I do not like it

A regular request to increase the memory limit to allow solutions in python. Going from https://mirror.codeforces.com/contest/2034/submission/294035798 to https://mirror.codeforces.com/contest/2034/submission/294041682

was neither insightful nor fun.

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3 weeks ago, # |
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BRO I passed A with 999ms whats on about the tight TL

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3 weeks ago, # |
← Rev. 2   Vote: I like it -10 Vote: I do not like it

GreedyForces

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3 weeks ago, # |
  Vote: I like it +37 Vote: I do not like it

idk how many CF rounds I have to do to stop wasting half an hour on finding N = K = 1 testcases on E and similar :v

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    3 weeks ago, # ^ |
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    If you have a hard time figuring that out, then what am I supposed to do? I missed two test cases in E....

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      3 weeks ago, # ^ |
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      & I used "int n" instead of "long long n" and couldn't debug it though i had 1 hours left. :) sadlyf.

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3 weeks ago, # |
← Rev. 2   Vote: I like it +26 Vote: I do not like it

I FST-ed on A :( Why half allow O(t*a*b) solution

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3 weeks ago, # |
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loved the statements, especially the characters. look forward to seeing this competition continue in the coming years!

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3 weeks ago, # |
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TLE on A system testing because of define int long long

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3 weeks ago, # |
  Vote: I like it -19 Vote: I do not like it

Loved the images :)

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3 weeks ago, # |
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гослинг?

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3 weeks ago, # |
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I still do not understand why I got TLE on A and B in the system testing while the other similar answers were getting accepted...

A: https://mirror.codeforces.com/contest/2034/submission/294007119 B: https://mirror.codeforces.com/contest/2034/submission/294033976

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3 weeks ago, # |
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https://mirror.codeforces.com/contest/2034/submission/294091655

please someone help me in figuring out whats wrong in my code?

please!

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    5 hours ago, # ^ |
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    Fixed in 297729300

    The issue is that when this is executed:

    if(s[j]=='0') {cnt++; j++;}

    j may be out of bounds.

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      84 minutes ago, # ^ |
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      yea i realised it later that night! just missed continue; in my code

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3 weeks ago, # |
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Hoping to get my first t-shirt(among Randomly choosen 50 participants)

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3 weeks ago, # |
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This is my code for question A but it has not been accepted and showing TLE . How is this possible. I have reviewed my code thoroughly and believe it is correct .The code should definitely work as the constraints are — (1≤a,b≤1000) and (1≤t≤100). Many with same logic have their solutions accepted. I kindly request re-evaluation of my submission.

Submission Id — 294008048

#include <bits/stdc++.h>
using namespace std;
#ifndef ONLINE_JUDGE
#include "Debugtemplate/template.hpp"
#else
#define debug(...)
#define debugArr(...)
#endif
#define int long long
#define pii pair<int,int>
#define vpii vector<pair<int,int>>
#define vi vector<int>
#define vvi vector<vector<int>>
#define pb push_back
#define mp make_pair
#define F first 
#define unmp unordered_map<int,int>
#define S second 
#define all(x) x.begin(),x.end()
#define sortasc(x) sort(all(x))
#define sortdes(x) sort(x.rbegin(),x.rend())
#define f(i,ind,n) for(int i=ind;i<n;i++)
#define PI 3.1415926535897932384626
const int mod = 1000000007;

void solve(){
    int a,b;
    cin >> a >> b;
    for(int i=min(a,b);i<=a*b;i++){
        if(i>=a || i>=b){
            if(i%a==i%b){
                cout << i << endl;
                return;
            }
        }
    }
}
#undef int
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
// freopen('input.txt','r',stdin);
// freopen('output.txt','w',stdout);
    int t;
    cin >> t;
    while(t--){
        solve();
    }
    return 0;
}
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    3 weeks ago, # ^ |
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    Your solution works in $$$O(tab)$$$, with the constraint the worst case is $$$10^3\times 10^3\times 10^2=10^8$$$.

    However, the % operator has a large constant factor, which might case TLE in this case.

    #define int long long also slow down your code a little bit.

    PS: your code runs for about 2700ms on AtCoder (usually faster than codeforces) in about a $$$8\times 10^7$$$ total product, which is within the constraint. It would definitely TLE in the worst case here.

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    3 weeks ago, # ^ |
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    problem con see hi

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3 weeks ago, # |
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The fact that the glorious history of Iran and amazing myths were mentioned in the text of the questions made me proud that I was born in Iran. I hope this beautiful work will be repeated later.

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3 weeks ago, # |
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unable to view the editorial

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3 weeks ago, # |
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hello everyone do you know how to get job because i was recently graduated i don't know what to do

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3 weeks ago, # |
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This contest gave me 120+ delta and made me a Specialist <3 standings

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3 weeks ago, # |
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can anybody give me the id of this contest

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3 weeks ago, # |
  Vote: I like it +9 Vote: I do not like it

vulestamenkovic orz

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3 weeks ago, # |
  Vote: I like it +18 Vote: I do not like it

Nice problems! F2 and H are nice, even though I wasn't able to debug the latter in contest time :(

On a side note: Why insert the image in the middle? It randomly divides the statement into two parts, which can be really annoying when checking to see if you have missed something.

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3 weeks ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Runtime Error

AC

can someone explain why the first one gets Runtime error while the second code gets accepted?

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3 weeks ago, # |
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C is too implementation heavy.

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    3 weeks ago, # ^ |
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    idk its pretty standard implementation, you probably just need to do more dfs/bfs problems

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3 weeks ago, # |
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When will we know if we got tshirts or not?

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2 weeks ago, # |
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Today I received this message:

Spoiler

This is unfair decision. The only similar parts of this submissions: mine, studboy — is parts with sets, which is not the main logical part. Main ideas looks quite similar, but ways of solving are different. Hope this misunderstanding will be fixed.

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    2 weeks ago, # ^ |
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    Hello, Timophey!

    Detecting similarities between codes and marking them as potential cheating cases is not something handled by the problem authors. In fact, we aren’t informed when these checks are performed or who ends up being flagged as a result.

    This process is part of the system’s automated operations. I believe reaching out to KAN or contacting the appropriate individuals in charge might help resolve this situation for you.

    I’m sorry that I’m unable to assist directly with this matter.

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43 hours ago, # |
  Vote: I like it 0 Vote: I do not like it

best problems

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