marzipan's blog

Good Bye 2023
By marzipan, 12 months ago, translation, In English

Hello, Codeforces!

I am happy to invite you to our Good Bye 2023, which will be held at Dec/30/2023 17:50 (Moscow time). The round will be rated for all the participants.

The tasks were created and prepared by 74TrAkToR, zwezdinv, OR_LOVe, marzipan, platelet.

We would like to thank everyone who helped us a lot with round preparation:

During the round you will need to solve 8 problems. You will have 2 hours to solve them.

Score distribution: 250—750—1250—1500—2000—2750—3750—(2750+1750)

We wish you good luck and high rating!!!

UPD https://codeforces.net/blog/entry/124060

UPD1:Tutorial was published

Announcement of Good Bye 2023
  • Vote: I like it
  • -4617
  • Vote: I do not like it

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12 months ago, # |
  Vote: I like it +118 Vote: I do not like it

i hope for a very good contest before 2024

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12 months ago, # |
  Vote: I like it +98 Vote: I do not like it

As a tester, I wish you have fun (and do not rage at the end of the year) Good luck!

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12 months ago, # |
  Vote: I like it +230 Vote: I do not like it

As a tester, I will not write any more rounds this year

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12 months ago, # |
  Vote: I like it +2 Vote: I do not like it

Thank you

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12 months ago, # |
  Vote: I like it +77 Vote: I do not like it

Lets hope aliens attack in 2024 and teach us how to P = NP before destroying us

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I think they would give us a quantum computer instead

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    12 months ago, # ^ |
      Vote: I like it -15 Vote: I do not like it

    Why depend on aliens do it yourself instead of posting sh*t online

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    12 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Let's hope aliens attack in 2024 and give us a better round than this one before destroying us

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

Good Bye 2K23 and Welcome 2K24! Hope that the last round of this year will be more enjoyable :)

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12 months ago, # |
← Rev. 2   Vote: I like it +6 Vote: I do not like it

Hoping to finally cross 2200!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I hope it will be my last nice and impressive round in this year

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

Wish everyone a happy new year 2024!!!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

goodbye 2023

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Hopefully 2024 will be good lol

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

The contest name is so excellent that I simply couldn't resist participating in it:)

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

So the final contest for the year 2023, hope for everyone to get positive delta

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Good Bye 2023..!! Wishing everyone a Happy New Year. and hope for a very good contest before 2024. Happy Coding.

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12 months ago, # |
  Vote: I like it -16 Vote: I do not like it

After bad performance on the last div 4, I hope to reach pupil in this contest

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I hope that this round will be lucky for all of us!!!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

bye 2023 , hi 2024 !

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I hope that this round will be lucky for all of us!!!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Any idea about the difficulty of the contest ?? Like similar to div2 div3

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12 months ago, # |
← Rev. 2   Vote: I like it -26 Vote: I do not like it

Why so many downvotes?

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    12 months ago, # ^ |
      Vote: I like it -16 Vote: I do not like it

    Did u practiced on some other platform before as i can See u were already good starting with your first contest

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12 months ago, # |
  Vote: I like it +2 Vote: I do not like it

as a tester, i can wish you only good luck

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Div 2??

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    12 months ago, # ^ |
      Vote: I like it -11 Vote: I do not like it

    Looking at the previous year contest it looks like it will be a Div 2

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12 months ago, # |
  Vote: I like it +1 Vote: I do not like it

Hoping to reach cyan before year end :)

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12 months ago, # |
  Vote: I like it +25 Vote: I do not like it

What will be the score distribution?

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Hi 2024 good bye almost prime 2023.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

What's the overall difficulty of the tasks?

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

good bye! <3

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

is the difficulty of the tasks similar to Div 2?

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    No it'll be div.1 + div.2

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    12 months ago, # ^ |
      Vote: I like it +2 Vote: I do not like it

    The first 6 problems are like Div.2 A-F. The last two are harder problems, like Div. 1 E and F.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Goodbye 2023 and Hello 2024... I hope that this year will be an end to all your problems and difficulties and that this year will be a year full of joy and happiness for everyone. Happy New Year everyone 2024.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

looking forward to it

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12 months ago, # |
  Vote: I like it +1 Vote: I do not like it

hope to have fun at the last contest of the year

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

i hope, i'll reach cyan in this contest

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12 months ago, # |
  Vote: I like it +1 Vote: I do not like it

Happy New Year everyone! Wishing happiness and joy for everyone in 2024, hope everyone gets positive delta in the last contest of the year!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Hoping to end the year with + delta

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12 months ago, # |
  Vote: I like it +37 Vote: I do not like it

Wow so E is hard. Gotta lock in

also why only 2 hours

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    12 months ago, # ^ |
      Vote: I like it +15 Vote: I do not like it

    also why only 2 hours

    Relatively easy problems? That would be my guess

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12 months ago, # |
  Vote: I like it +6 Vote: I do not like it

I hope it will be a good end to this year. Good luck to everyone

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I am expecting to reach the pupil! Let's see what happens in the last contest of 2023!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I promised last week that I won't play CF anymore before my final-exam. But...Hey guys,that's the last time in this year...

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Hope to have a better time in 2024!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I need +70 or I lose the bet to pk_27. :(

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12 months ago, # |
  Vote: I like it +1 Vote: I do not like it

wow so many problems but i only wish to reach pupil after this round

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I hope to achieve a score of 1600 in the new year.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

marzipan did the score distribution for problem E and G changed ?

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

hope i can solve 2 of this:(

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Hoping for a doable D

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

bye 2023!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I don't have courage to participate good bye and hello contest as I lost 200 rating last time.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Based on the score distribution, can we expect that H1 is approximately as hard as F?

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12 months ago, # |
  Vote: I like it +24 Vote: I do not like it

Looks like a math-force.

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12 months ago, # |
  Vote: I like it +19 Vote: I do not like it

74TrAkToR for excellent coordination and useful advices. Bro, it is advice not advices. Advice is an irregular plural.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

As a participant, I hope everyone will be happy in 2024!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

May this contest help me in reach pupil

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12 months ago, # |
  Vote: I like it +5 Vote: I do not like it

Goodbye 2023!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Always great to end the year with a contest!

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

Why is the SCORE distribution changing every couple hours?

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

At the end of the year, I want to have my best contest !!

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12 months ago, # |
  Vote: I like it +1 Vote: I do not like it

Hope it ends my slump. So the year enda on a Good note.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

as a participant, i say: Happy New Year!

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12 months ago, # |
  Vote: I like it +14 Vote: I do not like it

it's not sponsored by near?

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Sad to say....but this gonna be my last contest for this year:(

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

i wish everyone good luck ;)

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Good Bye 2023!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Good Bye 2023

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12 months ago, # |
  Vote: I like it +6 Vote: I do not like it

Goodbye 2023,Or as they say in China: "再见2023".

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12 months ago, # |
  Vote: I like it +8 Vote: I do not like it

The last round will be good for those who are trying hard for the whole year. Good luck to everyone.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Chilling contest, chilling day.

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12 months ago, # |
  Vote: I like it +12 Vote: I do not like it

Hope no queueforces like yesterday

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Usually that's only a problem in Div.3/4 and not in Div.1+2, but we'll see

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Happy new year in advance, everyone!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I don't wanna go back to being a newbie, please :!!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Good luck to everyone. This will be my first ranked contest

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Good luck!

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12 months ago, # |
  Vote: I like it -15 Vote: I do not like it

is this like a div2?

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12 months ago, # |
  Vote: I like it +23 Vote: I do not like it

It's a pity that it was postponed for 5 minutes(

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

guys，what happened? why is it put off？

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Good luck for us!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Good Luck!!

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12 months ago, # |
  Vote: I like it +55 Vote: I do not like it

1 refresh cost me 10 min... lol :)

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12 months ago, # |
  Vote: I like it +8 Vote: I do not like it

Hope the queue is not long today.

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12 months ago, # |
  Vote: I like it +36 Vote: I do not like it

is the contest postponed for severval minutes?

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Is the contest postponed for 10 minutes?

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12 months ago, # |
  Vote: I like it +85 Vote: I do not like it

Good Bye 2023+10minutes

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Hope I get some ratings :D

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Delayed ??

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12 months ago, # |
← Rev. 3   Vote: I like it +47 Vote: I do not like it

Out of curiosity, are the delays due to registration issues again?

I'm pretty sure I registered ~45 mins ago, but when I reopened CF around 20 mins I wasn't registered (and had to register again).

EDIT: Managed to reproduce the issue, opening the list of registered friends causes me to be unregistered even without clicking the unregister button on that page.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I also opened my list of registered friends. Didn't happen to me though.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I registered and entered the contest. After clicking "submit" on problem A, I found that there are no submission logs, and the submit button disappeared. After that I tried to click "submit code", but it said that I was not registered for this contest. I have to wait until 15:00 UTC for additional registration before submitting my code.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

why 22:45

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

good luck!

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

Good bye 2023 :>

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

April fool 2023? XD

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12 months ago, # |
  Vote: I like it +157 Vote: I do not like it

Looks like 2023 isn't quite ready to leave us yet

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12 months ago, # |
  Vote: I like it +7 Vote: I do not like it

The Last Ride of the Year...

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12 months ago, # |
  Vote: I like it +2 Vote: I do not like it

If I turn cyan today, it would be an actual "happy" new year :)

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

How to get rid of "Checking if the site connection is secure"?

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Good bye 2023(((

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12 months ago, # |
  Vote: I like it -9 Vote: I do not like it

Aren't we 2018 now? T_T

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I thought this problem was solved already back in 2018. But it still exists in 2023 but with different language :P

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    12 months ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    bro just lost from Nott'm Forest he lost track of time

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12 months ago, # |
  Vote: I like it +35 Vote: I do not like it

And yet another delay xD

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12 months ago, # |
  Vote: I like it +44 Vote: I do not like it

Delayforces

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12 months ago, # |
  Vote: I like it -6 Vote: I do not like it

Idk, but today I don't feel like giving contest X(

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12 months ago, # |
  Vote: I like it +52 Vote: I do not like it

one refresh cost me 5 more minutes

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

why still late for 5 minutes

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12 months ago, # |
  Vote: I like it +12 Vote: I do not like it

Why 5 more minutes postponed

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12 months ago, # |
  Vote: I like it +2 Vote: I do not like it

Are you kidding me? -_-

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12 months ago, # |
  Vote: I like it +17 Vote: I do not like it

Delayforces!

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12 months ago, # |
  Vote: I like it +24 Vote: I do not like it

With the repeated delays, I am starting to doubt the quality of the problems.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I guess its more likely to be a problem with platform issues such as judge stability (i.e, high risk of a large queue).

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    12 months ago, # ^ |
      Vote: I like it +35 Vote: I do not like it

    Now, it looks like the concerns were spot on

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Good luck!

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12 months ago, # |
  Vote: I like it +47 Vote: I do not like it

What happened? Do the contest authors not want to say goodbye to 2023? Rescheduled again(

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Why delayed?

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12 months ago, # |
  Vote: I like it +10 Vote: I do not like it

Ig the contest will start untill 2024

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12 months ago, # |
  Vote: I like it +1 Vote: I do not like it

GoodBye 2023 typically

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12 months ago, # |
  Vote: I like it +59 Vote: I do not like it

20:05 No

20:15 No

20:20 No

20:23 XD

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I don't wanna use codeforces mirror but cloudflare keeps reloading the page when i verify :(

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

What the he*ll bores?

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12 months ago, # |
  Vote: I like it +1 Vote: I do not like it

I wanna cry(

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

Hope this is the last postponement...

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12 months ago, # |
  Vote: I like it +1 Vote: I do not like it

Why it always delay

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12 months ago, # |
  Vote: I like it +1 Vote: I do not like it

This is the most annoying thing that website can do

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

why did the time got rescheduled ?

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    12 months ago, # ^ |
      Vote: I like it -13 Vote: I do not like it

    This is my first comment on codeForces, lmao

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12 months ago, # |
  Vote: I like it +10 Vote: I do not like it

stress

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12 months ago, # |
  Vote: I like it +5 Vote: I do not like it

why is the time changing? I can't get on to the contest

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12 months ago, # |
  Vote: I like it +8 Vote: I do not like it

Delay no more please

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12 months ago, # |
  Vote: I like it +5 Vote: I do not like it

Codeforces ..its not april month XD

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12 months ago, # |
  Vote: I like it +8 Vote: I do not like it

Can we get a Belated Good Bye 2023 :D

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Delayforces pls! :)

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Goodbye "Good Bye 2023"

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12 months ago, # |
  Vote: I like it +36 Vote: I do not like it

I was 10 minutes late and then to my surprise, I logged in 5 minutes earlier!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Bruh why codeforces trolling?

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12 months ago, # |
  Vote: I like it +5 Vote: I do not like it

Delayforces

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12 months ago, # |
  Vote: I like it +11 Vote: I do not like it

delay until 2024?

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

last contest of this year

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12 months ago, # |
  Vote: I like it +24 Vote: I do not like it

5 minutes delay takes away the time about 0.3% of this year

→ Reply
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    12 months ago, # ^ |
      Vote: I like it -6 Vote: I do not like it

    8% of remaining 2023 into a dust. Only 22h remaining in 2023 at last.

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12 months ago, # |
  Vote: I like it +11 Vote: I do not like it

Today’s contest is sponsored by SBI

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

C'mon. Don't be angry. Let's focus on the contest.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Delay it one more time and It will be welcome 2024

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12 months ago, # |
  Vote: I like it +24 Vote: I do not like it

Everytime i refresh got 5 min penalty

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12 months ago, # |
  Vote: I like it +24 Vote: I do not like it

April Fools Day Contest 2023

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12 months ago, # |
  Vote: I like it +10 Vote: I do not like it

As a tester, I think this round will be amazing

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Seems like it will keep on being rescheduled till its actual year end for Mike. X0

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

22:50?(UTC+8)

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Why am I sweating in winters

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12 months ago, # |
  Vote: I like it +11 Vote: I do not like it

I could have played 1 more Valorant-ranked game.

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

thank you sir for this amazing platform I wish this platform should be forever .

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

welcome the coming of 2024!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

The delay was so long that contest become Hello 2024

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12 months ago, # |
  Vote: I like it +5 Vote: I do not like it

I was registered, but I can't submit because it says I'm not registered. Can someone please fix this ASAP?

→ Reply
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12 months ago, # |
  Vote: I like it +5 Vote: I do not like it

I was registered and now it is showing not registered and there is also not option for registration

→ Reply
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12 months ago, # |
← Rev. 7   Vote: I like it +5 Vote: I do not like it

Good Bye 2023 with LOL contest ....registered for it but showing i am not...:)

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12 months ago, # |
  Vote: I like it +8 Vote: I do not like it

Ending 2023 with another mathforces round this is amazing

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

the rooms are broken i cant find mine

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

Thanks for the contest, Finally Cyan :)

→ Reply
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12 months ago, # |
  Vote: I like it +143 Vote: I do not like it

Worst D I have ever seen even though I solved it.

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12 months ago, # |
  Vote: I like it +22 Vote: I do not like it

Now I'm really curious about the intended solution of H. Because it's 100000000000% not my solution.

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    12 months ago, # ^ |
      Vote: I like it +22 Vote: I do not like it

    there is an already available solution for this question on google.

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12 months ago, # |
  Vote: I like it +31 Vote: I do not like it

Biggest Mathforces of the year

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12 months ago, # |
  Vote: I like it +34 Vote: I do not like it

Good Bye my rating... See you in the next year

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12 months ago, # |
  Vote: I like it +8 Vote: I do not like it

Now I'll become a purple coder again :)

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12 months ago, # |
  Vote: I like it +98 Vote: I do not like it

How much did 74Tr paid to Mike ?

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12 months ago, # |
  Vote: I like it +47 Vote: I do not like it

I guess problem H is not an original question

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12 months ago, # |
  Vote: I like it +61 Vote: I do not like it

I have a message for the author of D.

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12 months ago, # |
  Vote: I like it +10 Vote: I do not like it

Mathforces

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12 months ago, # |
← Rev. 6   Vote: I like it +5 Vote: I do not like it

Sorry, guys. I was trying to report a streamer, but I didn't realize that my question would be visible to everyone.

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

All is well that ends well :( But It wasn't

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12 months ago, # |
  Vote: I like it +306 Vote: I do not like it

And the prize for the worst contest of 2023 goes to...

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12 months ago, # |
  Vote: I like it +14 Vote: I do not like it

Is H copied from somewhere or is it just easy? I dont remember the last time the last problem was solved so fast in a div1.

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12 months ago, # |
  Vote: I like it +8 Vote: I do not like it

how to solve E?

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    12 months ago, # ^ |
    ← Rev. 2   Vote: I like it +21 Vote: I do not like it

    Build segtree on euler tour.

    Now we do a dfs, and when are at node $$$u$$$ and have traversed all nodes in its subtree, for each node $$$v$$$ in its subtree, store only those nodes $$$v$$$ such that no ancestor of $$$v$$$ in subtree of $$$u$$$ has the same color as $$$v$$$. How to "store" such a node $$$v$$$? Simply add $$$+1$$$ on $$$[tin_v, tout_v]$$$ in the segtree (when deleting it, we will just add $$$-1$$$ on the same range) . It is equivalent to only considering highest occuring color on all paths to prevent overcounting.

    For node $$$u$$$, we can now find the optimal path when we fix $$$u$$$ as the lca. This is simply the product of the two maximum values of $$$(1 + max(tin_c, tout_c))$$$ accross all children $$$c$$$ of $$$u$$$. The answer is the maximum of this value across all $$$u$$$.

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12 months ago, # |
  Vote: I like it +8 Vote: I do not like it

How D?

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    12 months ago, # ^ |
      Vote: I like it +7 Vote: I do not like it

    For larger n its this pattern. X on the left, X**2 on the right.

    10000011 100000220000121 10000101 100002020010201 10000110 100002200012100 10001001 100020021002001 10001010 100020201020100 10001100 100022001210000 10010001 100200120020001 10010100 100202102010000 10011000 100220121000000 10100001 102010020200001 10100010 102010202000100 10100100 102012020010000 10110000 102212100000000 11000001 121000022000001 11000010 121000220000100 11000100 121002200010000 11001000 121022001000000 11010000 121220100000000

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    12 months ago, # ^ |
    ← Rev. 2   Vote: I like it +41 Vote: I do not like it

    Or you can simply construct these:

    1690000
9610000
1060900
9060100
1006009
9006001
1960000
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      12 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      might be simple question, but how do I know for sure these are perfect squares? Did you brute force for small n and guess the pattern? or is there some kind of proof for this?

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

        Observing the answers for <=7 will be enough

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        12 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        A0..n..0B^2 = (A*10^n+B)^2 = (A*10^n)^2 + 2(A*10^n)(B) + B^2

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        12 months ago, # ^ |
          Vote: I like it +67 Vote: I do not like it

        Notice that $$$(10^k+3)^2=10^{2k}+6\times 10^k+9=10000\ldots60000\ldots9$$$ and $$$(3*10^k+1)^2=9\times 10^{2k}+6\times 10^k+1=90000\ldots60000\ldots1$$$.

        So, the answer for something line $$$n=11$$$ can be:

        • $$$90000600001 = ((3\times10^5+1)\times 10^0)^2$$$
        • $$$90006000100 = ((3\times10^4+1)\times 10^1)^2$$$
        • $$$90060010000 = ((3\times10^3+1)\times 10^2)^2$$$
        • $$$90601000000 = ((3\times10^2+1)\times 10^3)^2$$$
        • $$$96100000000 = ((3\times10^1+1)\times 10^4)^2$$$
        • $$$10000600009 = ((10^5+3)\times 10^0)^2$$$
        • $$$10006000900 = ((10^4+3)\times 10^1)^2$$$
        • $$$10060090000 = ((10^3+3)\times 10^2)^2$$$
        • $$$10609000000 = ((10^2+3)\times 10^3)^2$$$
        • $$$16900000000 = ((10^1+3)\times 10^4)^2$$$
        • $$$19600000000 = (14\times 10^4)^2$$$ (a special case)
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        12 months ago, # ^ |
          Vote: I like it +1 Vote: I do not like it

        you can try observation. I think its hard in this case, but possible.

        I bruteforced for n<=9 and solved it by finding the pattern that way.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Jokes of assumption

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12 months ago, # |
← Rev. 2   Vote: I like it +76 Vote: I do not like it

H: https://math.stackexchange.com/a/1859668 ?

$$$\prod_{j=1}^r\,\frac{q^{j-1}\,\left(q^{m-j+1}-1\right)\,\left(q^{n-j+1}-1\right)}{\left(q^j-1\right)}\,$$$
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    12 months ago, # ^ |
      Vote: I like it +57 Vote: I do not like it

    https://math.stackexchange.com/questions/3259887/distribution-of-matrix-rank-over-finite-fields

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

      The most down-to-earth formula I found (and implemented) is here, at the bottom of page 19 (or 26 in PDF numeration): Thesis of Geoffrey Critzer

      $$$ f (n, p, k) = \frac{\left((p^n - 1) (p^n - p) \cdots (p^n - p^{k - 1})\right)^2}{(p^k - 1) (p^k - p) \cdots (p^k - p^{k - 1})} $$$

      Along with how to infer it in the few pages above (which can be skipped in contest time).

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        12 months ago, # ^ |
          Vote: I like it +30 Vote: I do not like it

        btw, i wondered for a while how to deal with "divided by zero" is issue. It seems that the constraint does not imply $$$p^k \not\equiv 1 \mod 998244353$$$?

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          12 months ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Interesting, I just assumed the values are "good".

          Perhaps system tests will say otherwise :) .

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            12 months ago, # ^ |
              Vote: I like it 0 Vote: I do not like it

            i asked the authors for clarification during contest, and was responded "no comments" as expected xD

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              12 months ago, # ^ |
                Vote: I like it 0 Vote: I do not like it

              That may mean they (will?) add such tests before system testing phase. Especially if the author solution avoids such divisions. Would explain the wait, too.

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          12 months ago, # ^ |
          ← Rev. 2   Vote: I like it +47 Vote: I do not like it

          It might not be an issue at all (and I think most contestants, myself included, kinda lucked out here).

          By repeatedly factoring, you can see that $$$f(n, p, k)$$$ is

          $$$ \displaystyle \frac{(p^n - 1)^2 (p^{n - 1} - 1)^2 \cdots (p^{n - k + 1})^2}{(p - 1) (p^2 - 1) \cdots (p^k - 1)} $$$

          multiplied by some nonzero constant (everything happens in the field of numbers modulo $$$998\,244\,353$$$).

          I claim that one of two things is true: either there are no zeros or there are strictly more zeros in the numerator than in the denominator. Let $$$r$$$ be the smallest positive number such that $$$p^r = 1$$$. It's a well-known fact in number theory that if $$$p^m = 1$$$, then $$$r \mid m$$$.

          In the denominator, the first time a zero appears is the term $$$p^r - 1$$$. By that time, we have had all exponents $$$n - r + 1, \ldots, n$$$ in the numerator. That is $$$r$$$ consecutive values, one of them is divisible by $$$r$$$. Since the terms in the numerator are squared, it means that we have two zeros in the numerator among the first $$$r$$$ terms. Repeat the same argument for all blocks $$$[tr + 1, (t + 1)r]$$$ in the denominator, and the claim follows.

          Since there are more zeros in the numerator, if you go back to the integers you can see that the entire thing is a multiple of $$$998\,244\,353$$$.

          If you had a "typical" solution (e.g. incrementing $$$k$$$ and adding terms to the formula) and your code doesn't throw an error when calculating the inverse, then you could just pretend that no division by zero ever occurs and your solution would still be correct. Even if you had a different solution using some similar formula (maybe with additional pointless terms cancelling out), you can probably show that it is correct using a similar argument.

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            12 months ago, # ^ |
              Vote: I like it +10 Vote: I do not like it

            Thanks for your clarification! That's a cool proof.

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12 months ago, # |
  Vote: I like it +232 Vote: I do not like it

Was this contest made from the rejected problems of Good Bye 2013? :)

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I tried submitting C last minute but it lagged. Really annoying.

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12 months ago, # |
  Vote: I like it +109 Vote: I do not like it

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12 months ago, # |
  Vote: I like it +96 Vote: I do not like it

WTF H1 & H2???????

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12 months ago, # |
  Vote: I like it +15 Vote: I do not like it

ruined my last cp contest of year

I hate game problem so much like it's unreal

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12 months ago, # |
  Vote: I like it +70 Vote: I do not like it

OEIS moment... https://oeis.org/search?q=1+49+294+168+&language=english&go=Search

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12 months ago, # |
← Rev. 2   Vote: I like it +46 Vote: I do not like it

Does anyone use the same method as me in problem D?

Violent enumeration of numbers $$$1 \sim 575000$$$ squared, calculate the answer for $$$n \le 12$$$, and find that the number of answers for $$$n=11,12$$$ has exceeded $$$99$$$. Simply add an even number of zeros after these numbers.

Submission link

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    12 months ago, # ^ |
      Vote: I like it +18 Vote: I do not like it

    I did that as well, D feels more like an April first question than a div2. D

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I used the fact that 31, 13 and 14 work for $$$n = 3$$$ and extended it with zeros, explained by the following example for $$$n = 7$$$:

    $$$1300, 1030, 1003, 3100, 3010, 3001, 1400$$$ (all squared)

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    12 months ago, # ^ |
      Vote: I like it +37 Vote: I do not like it

    violent enumeration lol

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      12 months ago, # ^ |
        Vote: I like it +5 Vote: I do not like it

      truly the best translation of 'brute force'

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Anyone else lose time trying to find the n = 7 case for D? I was stuck for 30 minutes until I brute-forced it...

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12 months ago, # |
← Rev. 2   Vote: I like it +9 Vote: I do not like it

Problem H: Link (page 20)

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

how tf do u do B

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

    i returned lcm ,but if lcm<=b then multiply lcm by lowest prime factor of a/b and return. it worked for me hope it passes system tests

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

      prove?

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        12 months ago, # ^ |
          Vote: I like it +11 Vote: I do not like it

        The smallest divisor of n is one. Next is a prime number p. Next may be other prime number q, or p^2. In first case a = n/q, b = n/p. gcd(a,b)=n/(pq) and lcm(a,b)=ab/gcd(a,b)=n. In second case a = n/p^2, b = n/p. lcm(a,b)=n/p=b. p = b / a, n = b * p.

        As you can see, b/a is a prime number in second case. It is unnecessary to check it.

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12 months ago, # |
  Vote: I like it +87 Vote: I do not like it

will it be unrated because of H.

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12 months ago, # |
  Vote: I like it +91 Vote: I do not like it

OMG H1&H2, doesn't any testers noticed the task is already exist?

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Can you guys tell me which topics or questions I need to work on to build intuition for problems like problem B

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

The moon is beatiful, isn't it? But this contest is ...

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12 months ago, # |
  Vote: I like it +10 Vote: I do not like it

Did H leak?

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12 months ago, # |
  Vote: I like it +24 Vote: I do not like it

What is this D????

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12 months ago, # |
← Rev. 3   Vote: I like it +149 Vote: I do not like it

[deleted]

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12 months ago, # |
  Vote: I like it -16 Vote: I do not like it

How to solve B? Here is my code, it is getting wrong answer in 2nd test case......

void solve() { ll a, b; cin >> a >> b; if(a == 1) cout << b*b << '\n'; else if(a%2 == 0 || b%2 == 0) cout << b*2 << '\n'; else { bool f = 0; for(ll i = 3; i <= a; i+=2) { if(b%i == 0 && (b*i)%a == 0) { f = 1; cout << b*i << '\n'; break; } } if(!f) cout << a*b << '\n'; } }

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12 months ago, # |
  Vote: I like it +41 Vote: I do not like it

First 4 problems were really really bad. Bad problems for Good Bye 2023!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

very good guessforces, I have a good 2023!

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12 months ago, # |
  Vote: I like it +7 Vote: I do not like it

Squeezed H2 using $$$O(n\log^2n)$$$ FFT using AtCoder Library's implementation.

Hope that it doesn't FST.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Pretest and problem statement for A were very bad, seems like a FORCED problem.

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12 months ago, # |
  Vote: I like it +110 Vote: I do not like it

Why in hell someone need to verify if I'm a human or not when only 15s is remaining in the contest?

How frustating it is to not being able to submit because of verfication?

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    THIS. THIS HAPPENED TO ME. I could not submit my code for C b/c of this. I am malding rn

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

      Couldn't submit D. An already frustating contest wasn't frustating enough at the end of the year.

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12 months ago, # |
  Vote: I like it +11 Vote: I do not like it

Did anyone solve D without brute-forcing and guessing the pattern, if so, how did you come up with the logic?

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Yeah i guessed a pattern but ain't sure abt that

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    i did not guess the pattern i coded for n = 5, 7, 9 digits and got the pattern.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Couldn't solve it as I spent most of the time generating cases for higher n and verifying locally. I think the solution is to append "00" and insert "0" in between the strings {"169", "196","961"} for higher n. For example {"16900","10609","19600","10906","96100","90601"}

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      12 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      10906 isn't a perfect square...

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

      You can observe pattern by brute force.. see my code both brute force and pattern...

      Pattern Code
      Brute Force For pattern
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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

D was a pattern question right? for n=1 i can make 1 ** n=3 i can make 13 31 14 169 961 196** ** n=5 i can make 103 130 301 310 and 14xx kinda pattern?** tell plz or was it a dp question?-

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    you could make the solution by using a 1, a 6, a 9 and a bunch of 0's, and the rest you can guess

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    12 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    For n=13 there exists 99 numbers that have the same multiset of digits. For larger n, just add zeroes to the end, and for smaller n, just bruteforce

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Start with 169, 196 and 961.

    Now for the next n i.e. with two more digits simply add 00 to each, and you need only two more numbers. Those are 1x6x9 and 9x6x1, where x is (n-3)/2 0s.

    With this you can go on to find for any n.

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      12 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      How to prove that adding (n-3)/2 zeros between the numbers will make it a perfect square ?? can you please explain ?

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        12 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        It easier if you look at its root. 1x6x9 is obtained by squaring 1x3, where x denotes some non-negative number of zeros.

        However, for your question we can rewrite 1x6x9 as 1a + 6b + 9, where a and b denote some non-negative number of zeros.

        With x = (n-3)/2, we have a = n-1 and b = (n-1)/2, since n is odd, both a and b are even.

        1x6x9 = 1a + 6b + 9 = 1b*1b + 2*1b*3 + 3*3 = (1b + 3)^2 = (1x3)^2

        1x6x9 is a perfect square is proven by the existence of its integer root.

        Apologies if the notations seem a bit unconventional/confusing, hardcode some 0s inplace of x

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Yeah this is my solution, pattern like you described

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12 months ago, # |
  Vote: I like it +165 Vote: I do not like it

trash problem D, downvoted. :(

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12 months ago, # |
  Vote: I like it +130 Vote: I do not like it

D is the worst problem I have ever seen.

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12 months ago, # |
  Vote: I like it +5 Vote: I do not like it

oh i feel i got a slap instead of gift for the new year

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

Good Bye 2023, Good Bye my blue color T_T

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12 months ago, # |
  Vote: I like it +16 Vote: I do not like it

Why is the last problem of a Div1+2 round solved by as many as 140 participants? Do you have any idea?

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12 months ago, # |
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how E??

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12 months ago, # |
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a more difficult f

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12 months ago, # |
  Vote: I like it +129 Vote: I do not like it

Good Bye 2023 Bad Bye 2023 , worst contest of 2023

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12 months ago, # |
  Vote: I like it +19 Vote: I do not like it

At least thanks for E

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    12 months ago, # ^ |
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    How to Solve?

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

      During dfs we can maintain diff function for each vertex in subtree (to current vertex). If current vertex has value $$$x$$$ we need to add +1 for each vertex in its subtree except vertices already contating $$$x$$$. This vertices are united to subtrees so we can add +1 to whole subtree and add -1 to these subtrees. And after just choose two maximums and update answer.

      The rest is usage of rmq with range add and find subtrees where -1 will be added. There will be no more than $$$n$$$ such subtrees.

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        12 months ago, # ^ |
          Vote: I like it +8 Vote: I do not like it

        Thanks!

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        12 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        can you please elaborate on how are you adding +1 over a subtree? Are you using a euler tour ?

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

good bye rating round <3

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12 months ago, # |
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  • A done
  • B done
  • C done
  • D WTF + penalties -> DONE
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12 months ago, # |
  Vote: I like it +11 Vote: I do not like it

I have a headache after the contest. Can anyone else relate, or is it just me? ;_;

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    12 months ago, # ^ |
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    how about having headache before, during and after the contest ? :)

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12 months ago, # |
  Vote: I like it +16 Vote: I do not like it

I wasn't assigned to any of the rooms during the contest.

I checked my submissions' page (the one which can be opened in the "standings"), and there isn't any room information as it used to be in the first line.

As a result of that, I can't make any hacks (and I thinks my solutions can't be hacked by others too).

I checked the standings and saw many people with the same problem.

I asked about that during the contest and got replied with "no comment".

Will this happen in the next contest? or There's some rules about this that I didn't noticed?

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12 months ago, # |
  Vote: I like it +16 Vote: I do not like it

mathforces

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12 months ago, # |
  Vote: I like it +18 Vote: I do not like it

If the test data of G isn't wrong, I will start to learn Russian.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

1916B - Two Divisors

I don't think problem B was a good one.How can we determine that input values a=6, b=8 are not valid?

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    12 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    I think they expected us to assume that the input has a valid answer.

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      12 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      this is hilarious

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        12 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        i checked and it didnt say give -1 if the inputs are wrong therefore it must be valid in every case (or at least thats what you have to assume)

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    If x is divisible by 6 and 8, then it is also divisible by 24. However, that means 12 is a larger factor, so a and b are not the largest factors.

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    12 months ago, # ^ |
      Vote: I like it +18 Vote: I do not like it

    The author has stated that a, and b will be provided in the test cases such that it always leads to a valid x. I will quote the statement — "It is guaranteed that a, b are the two largest divisors for some number 1≤x≤10^9. "

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      12 months ago, # ^ |
        Vote: I like it -24 Vote: I do not like it

      I understand sir. but that's not how you write a problem. the condition(1<=a<b<=10^9) says it is possible to have input such as a=6,b=8;a=4,b=8... So, as an ordinary human being, anyone can think of it. No one is going to make a research on which kind of inputs are possible and which are not to solve B. Therefore it's not a good problem.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It is already stated in the problem that they are divisors of x. So a valid answer must exist.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Because they have mentioned in the problem that a and b are the largest divisors of x with the condition a<b<x. You will not be able to find any x for a=6, b=8 which will follow the condition

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I can feel you, I myself missed that line where it is mentioned that a & b are given such that they will always be the largest divisor of x. They should have mentioned it above not in the input section.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

for problem B, if a=6, b=8, what is X ?

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    12 months ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    24

    EDIT: Wrong, it will divide 12!

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      12 months ago, # ^ |
      ← Rev. 2   Vote: I like it 0 Vote: I do not like it

      But 24's biggest divisor is 12, not 8. The question does not guarantee input is valid !

      nvm i missed reading a line

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    its not vaild input i guess. No ans exist for this imo.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    a = 6 and b = 8 aren't valid. It is not always the case that there is a number x such that a and b are the two biggest divisors for all a and b.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    there is not solution that's why it said:

    It is guaranteed that a , b are the two largest divisors for some number 1≤x≤109.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It isn't possible to have a number with 8 and 6 as it's greatest divisors

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12 months ago, # |
  Vote: I like it +6 Vote: I do not like it

This round quality compliments my experience of the entire year,garbage.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

The contest made me realize my skill level

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12 months ago, # |
  Vote: I like it +19 Vote: I do not like it

First time I ever feel need to downvote round. Terrible set.

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12 months ago, # |
  Vote: I like it +7 Vote: I do not like it

Lets see if we can make this blog the most downvoted blog in 2023 as a prize for the most trash round in 2023

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12 months ago, # |
  Vote: I like it +192 Vote: I do not like it

Someone might like B and C or not, we can argue. E was nice. BUT D IS OBJECTIVELY THE WORST PROBLEM THAT WAS EVER MADE IN THE HISTORY OF THIS PLANET.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Ended the last contest of the year with a terrible performance :( Hoping to do much better next year! Happy New Year to all!

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12 months ago, # |
  Vote: I like it +17 Vote: I do not like it

Not gonna lie, problem D made me think for little bit.

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12 months ago, # |
  Vote: I like it +249 Vote: I do not like it

The quality of the contest doesn't deserve the title "Good Bye 2023".

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    12 months ago, # ^ |
      Vote: I like it +31 Vote: I do not like it

    More like "Goodbye good quality problems".

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12 months ago, # |
  Vote: I like it +29 Vote: I do not like it

These types of contest along with CloudFlare Feature is quite annoying

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12 months ago, # |
  Vote: I like it +23 Vote: I do not like it

problem D is definitely something

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

    D told us look question carefully, you can find the answer in the first line itself

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    12 months ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    can you tell me what is bad about D, i never solved a D so i dont know how normally Ds are

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      12 months ago, # ^ |
      ← Rev. 2   Vote: I like it +2 Vote: I do not like it

      Start by generating a sequence using 13, 31, and 14 for n = 3.

      For n ≥ 5, calculate the solution for k = n — 2. Multiply these numbers by 10. Then, add two more numbers: 100...003 and 300...001. After finding those numbers for n, square them and you get the answer.

      I don't know why but it works.

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12 months ago, # |
  Vote: I like it +58 Vote: I do not like it

Weak pretest of A and very easy H1 and H2. Not a good round. I hope Hello2024 can be better.

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12 months ago, # |
← Rev. 2   Vote: I like it +190 Vote: I do not like it

74TrAkToR Could you please OEIS the sequence before you use a "several-integer-input-problem" in contest next time?

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12 months ago, # |
  Vote: I like it +148 Vote: I do not like it

i actually want to quit cf after this round

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    12 months ago, # ^ |
      Vote: I like it -16 Vote: I do not like it

    i actually want to quit life after this round

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12 months ago, # |
  Vote: I like it +7 Vote: I do not like it

Editorial should be fun, no way this was the intended solution for D right?

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

What's the intuition behind B? It felt confusing to me. I'm really bad at math :(

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12 months ago, # |
  Vote: I like it +172 Vote: I do not like it

Typical 74TrAkToR round :)

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Am I the only one who actually solved H instead of googling the answer?

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12 months ago, # |
  Vote: I like it +59 Vote: I do not like it

Actually, the ending of 2023 would be much better without this "awesome" contest. Thank you very much for ruining everyone's expectation.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Yes, I love you too.

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12 months ago, # |
  Vote: I like it +40 Vote: I do not like it

Today's contest was the greatest I have ever seen. Spectacular problems from start to finish. The problem D was exceptionally good. Having to hardcode and look at some 1s 6s and 9s truly is the epitome of contest problems. I definitely didn't have a headache and no, I won't cry myself to sleep.

Thanks for making 2023 even more bearable for me!

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12 months ago, # |
  Vote: I like it +76 Vote: I do not like it

Hacking overflow solutions in A was an interesting task in itself. You basically had to find at most $$$5$$$ numbers less than or equal to $$$2023$$$ such that their product is of the form $$$k\times 2^{32}+x$$$ where $$$k$$$ is any positive integer and $$$x$$$ is a factor of $$$2023$$$ (i.e. $$$x\in {1,7,17,119,289,2023}$$$).

The smallest number satisfying this is $$$3\times 2^{32}+289 = 12884902177=7\times 691\times 1489\times 1789$$$. When this product overflows in the hacky solutions, the value becomes $$$289$$$ which is a factor of $$$2023$$$ and they print $$$\texttt{"YES"}$$$ instead of $$$\texttt{"NO"}$$$. I really hope such tests are present in the system tests.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    There's no overflow if you use long longs right?

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    12 months ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    You could also print five numbers whose product is $$$2^{32}$$$, which would overflow to $$$0$$$ and cause a runtime error in many solutions which then attempt to modulo by $$$0$$$.

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      12 months ago, # ^ |
        Vote: I like it +5 Vote: I do not like it

      Well, that works too and is much easier to find. I was so focused on getting WA that I didn't think about this, lol.

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    12 months ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    For solutions with a test like 2023 % product == 0, another idea to hack them is to use $$$2^{32} - 7 = 3 \cdot 37 \cdot 167 \cdot 223 \cdot 1039$$$. (I kept making Wolfram Alpha factor numbers around $$$2^{32}$$$ and finally found this!)

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      12 months ago, # ^ |
        Vote: I like it +8 Vote: I do not like it

      And yes I've just realized $$$2^{32}$$$ works just fine because it causes zero division...

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    You can just write test like this:

    1
4 1
1024 1024 1024 1024

    And this product will be equal to 0 (with integers).

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    They put so much of brains in A and no time was left to prepare D so they put some garbage in there.

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

    Finally the round where you can hack someone in easier problems! ++

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12 months ago, # |
  Vote: I like it +34 Vote: I do not like it

F is https://www.luogu.com.cn/problem/P9394 by ix35 and H is very similar to 1603F - October 18, 2017

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12 months ago, # |
  Vote: I like it +12 Vote: I do not like it

By the way I think I have a funny solution to D. I noticed through brute forcing that for 11 digits there is > 100 different numbers, so i just copypasted the best sequence of numbers for every number of digits up to 11: https://codeforces.net/contest/1916/submission/239681879

I also got TLE on pretest 1 for trying to calculate it instead of copypasting, lol

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I have the same solution (but I didn't notice n has to be odd, so I did the same for 12 digits too)

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

    Great solution, tbh after seeing this solution I feel problem D was not bad at all XD. Also there have been similar problems like this before on cf as well, but ig for most people doing cp now this might have been new, hence the angry reaction. XD

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

What is special about testcase 15?

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12 months ago, # |
  Vote: I like it +319 Vote: I do not like it

The worst contest in 2023.

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12 months ago, # |
  Vote: I like it +30 Vote: I do not like it

2:00 Hours only wasn't enough

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    12 months ago, # ^ |
      Vote: I like it +51 Vote: I do not like it

    my theory is that they didn't have problems that were hard enough for strong contestants so they just shortened the length of the contest to artificially increase the difficulty

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

How to do B?

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I know there are another solutions and in fact I don't think this is the intended. But the one that I've used is this one:

    Okay you know that $$$a <b$$$. They are the biggest divisors of $$$x$$$, if we don't count $$$x$$$. So it means that $$$b \cdot e = x$$$, where $$$e$$$ is the smallest divisor of $$$x$$$. So if we find $$$e$$$, we can know $$$x$$$. So the key is how to find $$$e$$$. We can be sure that either is in the decomposition into prime numbers of $$$a$$$ or $$$b$$$. This is because if $$$e^{2]$$$ divides $$$x$$$, $$$e$$$ is going to divide $$$b$$$ too, but if not, $$$e$$$ is going to be divided by $$$a$$$.

    So you can just iterate over primes until we reach one that divides $$$a$$$ or $$$b$$$.

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      12 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      If a is a factor of b, than the answer is b*b/a, otherwise the answer is lcm of a and b. So yeah, I don't think your solution is intended)

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        12 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Yes hahaha, in fact i've got tle on test 5. I don't know why, but maybe it was just not suposed to get accepted

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          12 months ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Because there can be a pair of a and b which are co-prime means a and b can be such larger that while interacting for the smallest prime factor of their LCM can lead to tle better is you can mention a condition before interacting if hcf(a, b)=1 return their product. In this case you should also need to mention a condition for a=1

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12 months ago, # |
  Vote: I like it +1 Vote: I do not like it

I doubt my mathematical skills after the contest

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12 months ago, # |
  Vote: I like it +128 Vote: I do not like it

I saw some blogs asking something like "what is the worst problem in 2023". Seems like voters have been tricked by the author of problem D!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I couldn't even solve A, so I gave up

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12 months ago, # |
  Vote: I like it +34 Vote: I do not like it

What a disaster this was.

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12 months ago, # |
  Vote: I like it +16 Vote: I do not like it

so much maths

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12 months ago, # |
  Vote: I like it +22 Vote: I do not like it

I feel like crying after this contest

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12 months ago, # |
  Vote: I like it +316 Vote: I do not like it

I'm not super shocked by authors setting a purely mathematical problem for 4500pts, thinking it's very difficult. I'm shocked that so many testers went through it and didn't try to Google it. Even just looking at the statement you feel like this must've been an original problem only in the 1800s.

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12 months ago, # |
  Vote: I like it +33 Vote: I do not like it

cringe D

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12 months ago, # |
  Vote: I like it +73 Vote: I do not like it

with all respect , round was terrible

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12 months ago, # |
  Vote: I like it +13 Vote: I do not like it

problem D is so weird.

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12 months ago, # |
  Vote: I like it +604 Vote: I do not like it

I'm gonna go and say (controversial opinion time) it might be worth making this contest unrated. The balance of top ranks is completely screwed by H since it's worth so many points and takes so little effort to find, or even a surprisingly small amount of effort to figure out. In addition, the gaps between number of solutions for other problems are huge.

This is a bad situation to be in as a contest organiser, where you're either discarding a bad contest even though you don't need to (results are not invalid, they just suck), or you're not discarding a contest even though the results suck.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

The most memorable contest throughout 2023, in a way

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12 months ago, # |
  Vote: I like it +11 Vote: I do not like it

disaster

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12 months ago, # |
  Vote: I like it +14 Vote: I do not like it

just learnt that guess forces are also welcomed by codeforces.

in D just look at number 31

31^2, FOR SERIES OF 3 ANAGRAMS 31^4 for series 5 anagrams.

and so on.

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12 months ago, # |
  Vote: I like it +6 Vote: I do not like it

what's wrong with D, I was stuck at B.

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12 months ago, # |
  Vote: I like it +59 Vote: I do not like it

2023's worst contest DELETE H!!!!!!!!!

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12 months ago, # |
  Vote: I like it +21 Vote: I do not like it

Can we officially call this MathForces?

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12 months ago, # |
  Vote: I like it +13 Vote: I do not like it

I hated C so much

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12 months ago, # |
  Vote: I like it +68 Vote: I do not like it

ABC meh

D I can't evaluate it coz it is AD-HOC.Personally I dislike it.

E a bit classical,but not bad.

F I used a random algorithm and passed it easily...

H Bro why it could exist in a codeforces round in nearly 2024???????

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Please explain me how to solve B. I'm out of breathe after thinking for so long how to solve B.

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12 months ago, # |
  Vote: I like it +18 Vote: I do not like it

I see A got hacked for many people. Is it because of integer overflow?

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    12 months ago, # ^ |
      Vote: I like it +18 Vote: I do not like it

    i think so

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    12 months ago, # ^ |
      Vote: I like it +18 Vote: I do not like it

    yes it is. that is how some people got lots of hacks

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    You are right.Some people multiply all the numbers in the b array and it exceeded the range of int.

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    12 months ago, # ^ |
      Vote: I like it +5 Vote: I do not like it 
    1
5 1
64 64 64 128 128

    This hacks a few people who used int. The product(2^32) overflows to 0 and could cause RE through division by 0 error

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12 months ago, # |
  Vote: I like it +29 Vote: I do not like it

Before Contest... : D

After Contest... : |

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12 months ago, # |
  Vote: I like it +380 Vote: I do not like it

Ah darn, I forgot that we're supposed to google problems instead of trying to solve them, what a newbie mistake by me

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    12 months ago, # ^ |
      Vote: I like it -15 Vote: I do not like it

    That's what I thought seeing people say that H is already online. That shouldn't have happened, but still, I think googling it instead of solving it is cheating.

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      12 months ago, # ^ |
        Vote: I like it +33 Vote: I do not like it

      it's not against the rules, however, the problem that everyone expects to be the most difficult shouldn't be solved by a simple google search though, serious oversight from the authors

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Sadly I gave up googling too soon :(

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    lol... I guess they will make the round unrated

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12 months ago, # |
  Vote: I like it +88 Vote: I do not like it

Have been testers and coordinator awake when they were doing their job?

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12 months ago, # |
  Vote: I like it +16 Vote: I do not like it

Solved D 10 seconds after the contest ended. That was a great lesson by the end of the year to look less at the scoreboard.

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12 months ago, # |
  Vote: I like it +9 Vote: I do not like it

I really think this is the worst contest of all time. What the hell are these problems????

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12 months ago, # |
  Vote: I like it +10 Vote: I do not like it

why pretests of problem A were so weak...Get fst qwq

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12 months ago, # |
  Vote: I like it +38 Vote: I do not like it

Bad ending for 2023 ...

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I guess it does make me want to say goodbye to 2023 harder

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Solved A,B,C.

Could not understand D.

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

After participating in this contest, I felt like a failure

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Should not it be unrated ??

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12 months ago, # |
  Vote: I like it +143 Vote: I do not like it

aGreE weLl

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    12 months ago, # ^ |
      Vote: I like it +49 Vote: I do not like it

    I thought people were overreacting when saying that this round was gonna be bad because traktor was coordinating. I was wrong.

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12 months ago, # |
  Vote: I like it +16 Vote: I do not like it

Here is the test case for hacking problem A:

1
5 5
1024 1024 1024 2 2

Anyone who didn't use long long will have a product of 0, which will result in RTE when checking 2023%(product of array).

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12 months ago, # |
← Rev. 2   Vote: I like it +219 Vote: I do not like it

What happened with the statements of this round? The statement of problem B is just incorrect (x is still a divisor of x), the word "parent" is used in two different ways in problem E, and problem F is so hard to read. I think testers were supposed to actually READ the statements.

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    12 months ago, # ^ |
      Vote: I like it +29 Vote: I do not like it

    It's also possible testers read the statements, told that the were unclear and the setters ignored it. This happened to me once as a tester so wouldn't be surprised.

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12 months ago, # |
  Vote: I like it +21 Vote: I do not like it

I was participating in an online round previously and there were many beginners there I remember they were shocked that the round was online and they said "oh it should be a test for googling skill". After problem H I think I am a beginner and they were masters actually :).

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12 months ago, # |
  Vote: I like it +77 Vote: I do not like it

Positive topic: I like E! At once it seems very hard task, but it's actually good training of data structures and Euler Tour Technique. (though TL=1s is too tight)
(but H makes everything into trash, so sad)

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12 months ago, # |
  Vote: I like it +49 Vote: I do not like it

You can simply find solution of H by searching example output "1 49 294 168" in OEIS.

https://oeis.org/search?q=1+49+294+168+&language=english&go=Search

I think this round should be unrated

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12 months ago, # |
← Rev. 3   Vote: I like it +18 Vote: I do not like it

tired. time to quit. this game is just so fxxked up.

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12 months ago, # |
  Vote: I like it +26 Vote: I do not like it

You can get the answer of H1 if you copy the example to oeis.Why can it happened in an codeforces round???Don't you guys try finding answer when testing???

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Thank you for positive delta

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12 months ago, # |
  Vote: I like it +55 Vote: I do not like it

74TrAkToR = -delta

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12 months ago, # |
  Vote: I like it -18 Vote: I do not like it

It's a pity that H was easily googlable, but guys, have some respect to the authors, problems except H were alright (as I see by the comments D is arguable, but for me it is not that bad). Also I think that H is not even made by initial authors and was taken from proposed Div.1 problems (which explains that platelet is an author whereas other authors are all from Lyceum Num. 31 in Chelyabinsk).

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12 months ago, # |
  Vote: I like it +5 Vote: I do not like it

thx for this round,but i think i waste plenty of time in these "excellent" problems,i should play genshin impact instead.

by the way,does 74TrAkToR play genshin impact?i think so because these problems are so "excellent" that only genshin players can make these.

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12 months ago, # |
← Rev. 2   Vote: I like it +56 Vote: I do not like it

Imagine if H1 and H2 were switched up and presented as D1 and D2! I bet a whole bunch more would've nailed them with googling -_-

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

fun" math force

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    12 months ago, # ^ |
      Vote: I like it +23 Vote: I do not like it

    Being mathforces was the least of the problems on this round

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12 months ago, # |
  Vote: I like it +18 Vote: I do not like it

Though hating the fact that H1 is googlable, I'm now aware of the existence of oeis :v

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12 months ago, # |
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what if I hack my own submission

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12 months ago, # |
  Vote: I like it -18 Vote: I do not like it

I hope this contest is rated, because this is the first time I solve F in 1:57. I don't know what happened in H, but I will be very upset if it is unrated.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

It wasn't the best round I've ever written.: 1) The authors delayed the round by 15 minutes (they had 2 months to prepare, but they couldn't do everything on time) 2) Task H is on the Internet, which is unfair to all participants. However, there are also positive aspects — the first 5 tasks were very interesting (not including d)

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    12 months ago, # ^ |
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    Ok, what's wrong with D?

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      12 months ago, # ^ |
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      What aspect of problem solving does problem D test?

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        12 months ago, # ^ |
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        Crytical thinking maybe?

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          12 months ago, # ^ |
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          If there was some logical path by which you could arrive to the solution then sure, but don't see how brute forcing small values tests any of that.

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            12 months ago, # ^ |
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            If it was this simple, why didn't you solve it?

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            12 months ago, # ^ |
              Vote: I like it 0 Vote: I do not like it

            Maybe because you still have to get to that idea and it takes quite a time?

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        12 months ago, # ^ |
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        you can construct answer from i to i+2

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      12 months ago, # ^ |
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      this is a task to build an example in which you almost don't have to think, but do the implementation

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        12 months ago, # ^ |
          Vote: I like it +10 Vote: I do not like it

        You have to get to that idea. If it was that simple, you could have solved it

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12 months ago, # |
  Vote: I like it +35 Vote: I do not like it

Very Good.That's like my terrible 2023.

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12 months ago, # |
  Vote: I like it +15 Vote: I do not like it

How to prove or disprove that $$$\prod_{i=1}^{n}p^{i+k}-1$$$ is divisible by $$$\prod_{i=1}^{n}p^{i}-1$$$ for all $$$n,k>0$$$? I thought division by 0 would be an issue in H, but if that's true then every time division by 0 happens the answer is 0 anyways.

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    12 months ago, # ^ |
      Vote: I like it +20 Vote: I do not like it

    The quotient is equal to the q-binomial coefficient which is an integer.

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    12 months ago, # ^ |
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    Or, alternatively, you can check the degree of any complex root of 1 in these polynomials.

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    12 months ago, # ^ |
      Vote: I like it +2 Vote: I do not like it

    If n is 7168, p is 573817, and k is 3584, division by zero will occur, but the answer will be nonzero.

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      12 months ago, # ^ |
      ← Rev. 5   Vote: I like it +44 Vote: I do not like it

      My code says that the answer is 0. Here's why division by 0 should cause the answer to be zero if what I wrote in the comment above is true(which it is, as the other 2 replies to it show):

      The answer is $$$\prod_{i=1}^{r}\frac{p^{i-1}(p^{n-i+1}-1)^2}{p^i-1}$$$. For $$$k=n-r$$$, product of $$$p^{i+k}-1$$$ is divisible by the product of $$$p^i-1$$$. So, that expression is an integer multiplied by $$$\prod_{i=1}^{r}p^{i-1}(p^{n-i+1}-1)$$$. The latter is a multiple of the product of $$$p^i-1$$$, so if the product of $$$p^i-1$$$ is 0, then it's a multiple of 0, so the entire answer is 0.

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        12 months ago, # ^ |
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        Oh, you are right, I forgot that I need to multiply further. My apologies. Thank you for your detailed explanation, I understand now.

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        12 months ago, # ^ |
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        Hey, can you share some resources where I can understand the maths behind this? Please. I have been stuck on this problem for so long.

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      12 months ago, # ^ |
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      Does this mean that FST will happen?

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

What a trash round in any sight and what a trash year I've passed.

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12 months ago, # |
  Vote: I like it +56 Vote: I do not like it

Does anyone find it strange? Where is k?

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    12 months ago, # ^ |
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    lol, in the first line of the statement they have already mentioned $$$k$$$

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      12 months ago, # ^ |
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      If you don't want to write k here, then you shouldn't explain k here.

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    12 months ago, # ^ |
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    yeah the statements are full of weirdness

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12 months ago, # |
  Vote: I like it +118 Vote: I do not like it

Never going to compete in codeforces rounds again in 2023.

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12 months ago, # |
← Rev. 3   Vote: I like it +15 Vote: I do not like it

500+ downvotes in just 30 mins after the contest ended, insane

UPD: 800+ now (40 mins), let's keep going!!!

UPD: 1k+ reached in under 1 hour!

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12 months ago, # |
  Vote: I like it +18 Vote: I do not like it

Well problem D managed to capture how my 2023 went. I had falshbacks while trying to find the patterns.

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12 months ago, # |
  Vote: I like it +11 Vote: I do not like it

"Save the best for last"

I guess it did not apply this year.

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

Bad contest!!!!!!

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12 months ago, # |
← Rev. 2   Vote: I like it +339 Vote: I do not like it
  • A: no int32 hack case (12884902177, if you wonder)
  • C: write brute to find rule
  • D: write brute to find rule
  • E: 1 second??
  • F: c+p bipolar orientation
  • H: c+p oeis (I didn't even tried to c+p sample, but like wtf)

I don't really mind tho, who cares about CP these days

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    12 months ago, # ^ |
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    I agree. Who cares about CP.

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    12 months ago, # ^ |
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    LOL, 1 second E is insane

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    12 months ago, # ^ |
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    Wow, Now I get how people find the pattern (write brute to find the rule). Thanks!

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    12 months ago, # ^ |
      Vote: I like it -76 Vote: I do not like it

    bro if you are an lgm and you can't solve task c after 50 minutes of the contest when there are around 3000 solves, then maybe the issue is not with the task.

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      12 months ago, # ^ |
        Vote: I like it +51 Vote: I do not like it

      yeah the issue is def me being an lgm but why i have positive delta here

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    12 months ago, # ^ |
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    Can you please elaborate A? What does "no int32 hack case" means?

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      12 months ago, # ^ |
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      Some of the code involves taking products of all numbers in int32. If you put the factorization of above number into the input, such code will produce an invalid solution.

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    12 months ago, # ^ |
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    I just brute forced C, didn't need to find the idea.

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12 months ago, # |
  Vote: I like it +16 Vote: I do not like it

If this round goes rated i will never come to cf again and also will encourage people to leave codeforces

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12 months ago, # |
  Vote: I like it +52 Vote: I do not like it

I know really think that this round should be UNRATED

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    12 months ago, # ^ |
      Vote: I like it +21 Vote: I do not like it

    i think instead of this we should post like #UNRATED THIS!

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

How does sum of n*n remain less than 1e5 if length is upto 99 in D?

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

Is this the intended solution for problem E?

Calculate for all vertices number of different values $$$a_v$$$ on path to the root. We can do it using one dfs with a set.

Now we want to calculate answer for the root. Actually, for all pairs of vertices, such that their lca is root. Look at subtrees of root. Either one of the vertices is root, and other one is in subtree, or both of them are in different subtrees. So we want to know max on subtrees and out of them 2 max values.

But what if answer is not root? Go to the child of the root. Now we want to recalculate array $$$a$$$. Assume value of root is $$$x$$$. We want to do $$$-1$$$ for all vertices in subtree, if there is no vertice with value $$$x$$$ on path betreen them, and the root. So we iterate over all vertices with values $$$x$$$ in subtree and do $$$+1$$$ on that subtree. But we don't want to do it for vertices, which has another vertice with value $$$x$$$ on their path to the root. So do the following. Do euler tour. For all values calculate positions of vertices with such values. Now we do vector lower_bound on segment of subtree on value $$$x$$$. Do $$$+1$$$ on found segment (using segment tree) and jump to the end of the found segment. Do lower_bound again until end of subtree. Now we changed the root and ready to answer for this one as lca.

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    12 months ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    I think that should work but I got TLE on pretest 15 for some reason for O(nlogn) :( seems like the limit is tight. Very frustrating :|

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      12 months ago, # ^ |
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      This solution passed in 0.8s. Looks time limit is evil here. I used only vector lower_bound in "heavy" part of code and no sets and maps.

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12 months ago, # |
  Vote: I like it +175 Vote: I do not like it

An $$$O(n \sqrt n)$$$ solution for G: (I didn't write the code and I doubt whether it can pass due to huge constant factors, but possibly this could inspire provable AC solutions?)

If the length of the resulting path $$$(u, v)$$$ is less than $$$B = \sqrt{n}$$$, the GCD along the path $$$(u, v)$$$ has to be at least $$$\frac{w_e}{B}$$$ for every edge $$$e$$$. Therefore, we can calculate the number $$$\frac{w_e}{j}$$$ for each edge $$$e$$$ and $$$j \in [1, B]$$$ and work on the connected components formed by edges containing the same number. This should be $$$O(nB)$$$ (and actually smaller than that because $$$j$$$ divides $$$w_e$$$) with careful implementation.

If the length of the resulting path $$$(u, v)$$$ is at least $$$B = \sqrt{n}$$$, consider the path $$$(u_0, v_0)$$$ that covers the middle $$$\frac{1}{3}$$$ of the path $$$(u, v)$$$. $$$(u_0, v_0)$$$ has length at least $$$\frac{1}{3}B$$$. Since we can select a vertex set $$$S$$$ of size $$$O(\frac{n}{B})$$$ from the tree such that every path of length $$$\frac{1}{3}B$$$ passes through an element of $$$S$$$, $$$(u_0, v_0)$$$ must go through a vertex in $$$S$$$, denoted by $$$w$$$. Suppose $$$\gcd(u, w) = x$$$ and $$$\gcd(w, v) = y$$$. Then, $$$\frac{x}{y}$$$ can only be one of $$$1, 2, \frac{1}{2}$$$ (or $$$(u, w)$$$ or $$$(w, v)$$$ would have a larger value than $$$(u, v)$$$). By enumerating $$$w \in S$$$ and computing the GCD over each path starting from $$$w$$$ in linear time (this might require special properties of this problem and is a bit tricky, but easy with an extra log factor), the process for finding the optimal long path $$$(u, v)$$$ can be done in $$$O(\frac{n^2}{B})$$$ time.

Disclaimer: I am strongly against setting competitive programming problems without provable (possibly aided by computer) AC solutions.

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    12 months ago, # ^ |
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    Unrelated, what do the equations in your profile picture signify?

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Problem A is probably the "best" problem in Codeforces. Thank you 74TrAkToR!!!!!!

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    12 months ago, # ^ |
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    It's UNBELIVEABLE that you can pass the pretests without using long long 239642528

    How shit the pretests.

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12 months ago, # |
  Vote: I like it +48 Vote: I do not like it

I think problem D is actually a nice problem.

First, solve it for small values and list all possibilities. It is a common enough step to arrive at a solution, no guessing involved.

Listing the multisets of digits which appear at least n times, we get:

multisets and their counts

Now, it can be seen that n = 1 is a special case. For the others, the common part is the digits 1, 6, and 9 with some zeroes. It is now within reach to explicitly construct such solutions.

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    12 months ago, # ^ |
      Vote: I like it +70 Vote: I do not like it

    What you mentioned is exactly why I think it's a bad problem. It's just "find the pattern in small cases" which requires no thinking at all.

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      12 months ago, # ^ |
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      Perhaps I regard a more diverse set of activities as "thinking" then.

      Anyway, we can agree to disagree on that, and move on :) .

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        12 months ago, # ^ |
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        You arent thinking. Thinking requires motivation. You are just hoping for a pattern, and you got lucky (i also did but i hate it because of that)

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    12 months ago, # ^ |
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    so you encourage guessing from just an observation? how do you prove this?

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      12 months ago, # ^ |
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      How do I prove... what? That 100...0600...09 is a square of 100...03 and the like? The number of zeroes can be arbitrary, as long as it is the same. We just have to have no carries in the multiplication.

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        12 months ago, # ^ |
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        I didn't even notice this till you mentioned it, I just saw some patterns with zeros, 1, 6, and 9, and coded something. Another reason why the problem is not as elegant as you think.

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          12 months ago, # ^ |
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          My point is this: "the problem can be solved without guessing". No more, no less.

          It is also true that the problem can be solved with some guessing. That's a bit unfortunate. However, my view is that the no-guessing solution takes the same time, and has the benefit of being reliable.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    You can also bruteforce the squares of $$$10^{(n-1)/2}+i$$$ for all small $$$i$$$ until you find the first such set, and store only all the relevant values of $$$i$$$ in your solution, so it doesn't exceed the submission size limit. The full solution then just does $$$O(N^2)$$$ addition and square. No need for brain.

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      12 months ago, # ^ |
        Vote: I like it +45 Vote: I do not like it

      Oh, I love when there are various approaches to a problem, like a mathematician's approach and a programmer's approach! Contestants with different backgrounds can then play to their strengths, and still solve the problem. Perhaps some would then call each other's solutions "no-brainey" and "too mathy", respectively. Okay, good for them.

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      12 months ago, # ^ |
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      What is the maximum value of $$$i$$$? I thought you will be checking too many values

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        12 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        $$$10^8$$$ is definitely enough, and you'll likely stop before that. Besides, your program can take 10 minutes to calculate that, and you can write another solution in the meantime.

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12 months ago, # |
  Vote: I like it +76 Vote: I do not like it

How to become a coordinator?

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12 months ago, # |
← Rev. 2   Vote: I like it +59 Vote: I do not like it

Hi.

Of course, there was an unpleasant incident with the problem today. In fact, the problem can be effortlessly googled.

You, of course, are shocked. You, of course, think that the round should be unrated.

You're wrong right. Here's why.

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12 months ago, # |
  Vote: I like it +8 Vote: I do not like it

make this UNRATED!!!!!

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12 months ago, # |
  Vote: I like it +18 Vote: I do not like it

Editorial when?

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12 months ago, # |
  Vote: I like it +20 Vote: I do not like it

Will system testing happen today?

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12 months ago, # |
  Vote: I like it -8 Vote: I do not like it

Can someone provide any hints for E, I was thinking about HLD but couldn't progress further

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12 months ago, # |
  Vote: I like it +2 Vote: I do not like it

My 2023 is just like this round.TRASH FROM BEGINNING TO END.

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

Problem E is really good, but the time limit is super, super tight.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I don't think E is really good. It was an extremely standard problem, I am a low master was able to find a solution in several minutes.

    It would be a great problem for an Edu round, but not for the Good Bye.

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      12 months ago, # ^ |
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      can we solve E without recursion?bcoz java and python sucks at recursion

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      12 months ago, # ^ |
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      Can you please explain your solution?

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12 months ago, # |
← Rev. 4   Vote: I like it +15 Vote: I do not like it

"Good Bye 2023" change to "RIP 2023"...

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12 months ago, # |
  Vote: I like it +12 Vote: I do not like it

A very excellent experience of this round that literally mirrors my life in 2023: pain, sickness, bad luck and helplessness. Thank 74TrAkToR for fxxking hurting me again.

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12 months ago, # |
  Vote: I like it +11 Vote: I do not like it

You should stop coordinating!!

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12 months ago, # |
  Vote: I like it +54 Vote: I do not like it

My great-grandmother can do better than 74TrAkToR as a tester.

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12 months ago, # |
  Vote: I like it -11 Vote: I do not like it

marzipan path to lowest contribution :/

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12 months ago, # |
  Vote: I like it +77 Vote: I do not like it

marzipan is not to blame

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

    im not blaming him.... I just feel bad for him considering all the downvotes to the contest. Other than the math and the copied problem, it wasn't too bad.

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

The worst contest in 2023

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12 months ago, # |
  Vote: I like it -8 Vote: I do not like it

I wonder if the weakness of pretests is an accident or if the authors made them weak on purpose?

I am very disappointed in this competition.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

As a gray tester, this contest will be the BEST!

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12 months ago, # |
← Rev. 2   Vote: I like it +1 Vote: I do not like it

Why the responsibility for coordinating such a crucial contest is given to somneone who has a great history in coordinating rounds

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12 months ago, # |
  Vote: I like it +7 Vote: I do not like it

Great! Now i have to read more retarded comments and blogs of xlpg0317

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12 months ago, # |
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If there were various issues with the contest's problems, can the system test still be started?

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Thanks for MathForces. I hope this terrible round doesn't spoil your mood. Happy New Year!

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12 months ago, # |
  Vote: I like it +1 Vote: I do not like it

2023 deserved a better ending.

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12 months ago, # |
  Vote: I like it +8 Vote: I do not like it

I wonder if this blog could actually reach a thousand downvotes. :)

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

And the prize for the worst contest of 2023 goes to...

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12 months ago, # |
  Vote: I like it -6 Vote: I do not like it

The worst round I've seen in half a year, and that's 74

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I need a explanation for why is 10 not a valid answer for a = 1 and b = 5 and 18 same for a = 3 and b = 9

Question B. Note: i may have misread the question or something cuz i need an actual explaination for this. like I dont get what i thought wrong first.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Because greatest divisors of $$$10$$$ are $$$2$$$ and $$$5$$$. Similarly for $$$18$$$, are $$$6$$$ and $$$9$$$.

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      12 months ago, # ^ |
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      OHHH i get it. I read the fking question wrong.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    If X was 10, a and b would be 2 and 5, not 1 and 5. If X was 18, a and b would be 6 and 9, not 3 and 9.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Because largest divisors of 10 are 2 and 5, not 1 and 5. For 18, largest divisors are 9 and 6, not 9 and 3.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I have got my answer. I misread the damned question before.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    for 10; 5 and 2 are the two largest factors
    for 18; 9 and 6 are the two largest factors

    I got a WA on Test Case 2 because of that ....

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12 months ago, # |
  Vote: I like it +2 Vote: I do not like it

Never let him cook again

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12 months ago, # |
  Vote: I like it +39 Vote: I do not like it

Key lesson:

Spoiler
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12 months ago, # |
  Vote: I like it -17 Vote: I do not like it

That was a very nice round Especially d , very good problem pro ❤❤

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12 months ago, # |
← Rev. 3   Vote: I like it -7 Vote: I do not like it

I hope the contest should be rated.

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12 months ago, # |
  Vote: I like it +13 Vote: I do not like it

Now I understand why this round didn't get supported by NEAR.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

can we attempt for hack after contest is over and if yes, how?

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12 months ago, # |
← Rev. 2   Vote: I like it -11 Vote: I do not like it 
Your code here...
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

void solve() {
    int n, k;
    cin >> n >> k;

    int pro = 1;
    int temp;
    for (int i = 0; i < n; i++) {
        cin >> temp;
        pro *= temp;
    }

    if (2023 % pro == 0) {
        cout << "YES" << endl;
        cout << (2023 / pro) << " ";
        for (int i = 0; i < k - 1; i++) {
            cout << 1 << " ";
        }
        cout << endl;
    } else {
        cout << "NO" << endl;
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int t;
    cin >> t;

    while (t--) {
        solve();
    }

    return 0;
}

my solution of question A was accepted at time of contest but now showing Runtime error at test 8 why??

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12 months ago, # |
  Vote: I like it +22 Vote: I do not like it

Goodbye 74TrAkToR

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12 months ago, # |
  Vote: I like it +10 Vote: I do not like it

Round must go Unrated!!

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12 months ago, # |
  Vote: I like it +12 Vote: I do not like it

Folks kindly chill! There has been a mistake and the host has acknowledged it. What else he can do? Most of us are here to learn something new and all of us atleast learnt something in today's contest, even if it was very small. We are not here to criticise someone, afterall mistakes do happen and its a very very very tough task to come with new problems with such high complexity. Lets not turn toxic and appreciate the positives, thats what the whole cp community is about. Nothing like all your years hardwork was dependent on this 1 contest... or is it? JUST A CONTEST!

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Up, just safe critics. But we wished for a well done contest for the end of the year :(

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    12 months ago, # ^ |
      Vote: I like it +67 Vote: I do not like it

    There has been a mistake

    Oh, there have been so many mistakes you probably can't even count them. Here's just a few:

    • Weak pretests in A
    • Incorrect and self-contradicting usage of "divisor" in the statement of B
    • Incorrect and self-contradicting usage of "parent" in the statement of E
    • F has appeared previously
    • G doesn't have a proven solution
    • Author's solution of G is incorrectly implemented
    • H is too easy
    • H is in OEIS
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12 months ago, # |
  Vote: I like it +69 Vote: I do not like it

my new steps for solving problems :

1-google the problem

2- put it on OEIS

3- try some random solutions

4- think

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    12 months ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    You can still skip step 4 and get the same result

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12 months ago, # |
  Vote: I like it +6 Vote: I do not like it

Make this unrated pls

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    why the heck do u want it unrated ? u couldn't even solve B today

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      12 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      maybe thats the reason? The weak pretests harmed alot of people

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Can someone be my friend. and help me get out of being a newbie :(

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12 months ago, # |
  Vote: I like it +1 Vote: I do not like it

goodbye cyan, hello green

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12 months ago, # |
  Vote: I like it +10 Vote: I do not like it

where is the excellent co-ordination mentioned in the blog?

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

Problem D is why I will be switching from Competitive Programming to Linguistics Olympiad.

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12 months ago, # |
  Vote: I like it +1 Vote: I do not like it

Petition: Make round rated only for users with positive rating delta, just for new year's sake

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12 months ago, # |
  Vote: I like it +6 Vote: I do not like it

"high-quality testing and valuable advice"

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

Why yet not it is unrated ?

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Please don't make the round unrated (re-judge the submissions of G again instead).

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12 months ago, # |
← Rev. 3   Vote: I like it -6 Vote: I do not like it

.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Let's not downgrade ratings to save the holiday vibe :)

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12 months ago, # |
  Vote: I like it +13 Vote: I do not like it

Got FST(tle) in B on tc6. Why make such a weak pretests? Whatever solution we think of, apply it to problems like A and B. At least let us know if our solution doesn't pass the time limit!

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12 months ago, # |
← Rev. 3   Vote: I like it +16 Vote: I do not like it

Good Bye 2023 with 2023 downvotes?

UPD 2 : surpassed 2023. Please upvote!

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12 months ago, # |
  Vote: I like it +1 Vote: I do not like it

Can someone try to uphack my solution in E? I think that my solution is O(N^2) in worst case.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Looking at 2023, few contest where I performed good, but most of the times I struggled lot in these maths constructive problems. I hope in 2024 I reach CM

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Oh hooray, now there are (at least) 3000 FSTs on A. The cherry on top.

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12 months ago, # |
← Rev. 2   Vote: I like it +1 Vote: I do not like it

Goodbye my rating! :D

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12 months ago, # |
  Vote: I like it +18 Vote: I do not like it

I understand that people are complaining, I don't understand why people that cannot solve E,F,G and H complain the loudest though. Should be kinda okay to make the Contest rated for Div2 only.

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

    it's feels nice to get onto the bandwagon when you have f'ed up really bad today :sob::sob:

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

    Admittedly alot of the time I get annoyed it is in bad faith but the large amount of complaints should hopefully ensure these issues do not happen again (especially considering the history of past contests like the one where div2c is NP-hard).

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      12 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      ohh absolutely. But in that case, the minus should not go to marzipan, but 74TrAkToR

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      12 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      that was a nice video of you eating an onion

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      12 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      what does onion taste like

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    yeah same thing

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

I just wanna say sorry for all the negative contrib due to traktor :skull:

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12 months ago, # |
  Vote: I like it +12 Vote: I do not like it

Gee I love Googleforces

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12 months ago, # |
← Rev. 4   Vote: I like it +17 Vote: I do not like it

marzipan is now the most downvoted CF profile and thats just sad, since the bad quality of the round doesn't seem to be his fault

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

what will be answer for Problem B, if a=30 and b=210? will answer exist under problem's constraints?

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    this is not a valid case, as no such x would exist. the prime factors of a either have to be a subset of b, or the 2 sets(prime factors of a and prime factors of b) have to differ by at most one element. here, both these conditions are not satisfied for a and b, so they are not a valid input

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Same wondering about a = 4 & b = 5, till I returned back to the problem and read it again :)

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    12 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    70 would also be a factor

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Worst D question, I wasted more than 1 hour on that problem.

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12 months ago, # |
← Rev. 2   Vote: I like it +1 Vote: I do not like it

2023 is NEAR

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12 months ago, # |
  Vote: I like it +19 Vote: I do not like it

Meanwhile me who had a bad contest today watching this round get unrated:

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I'll do suicide, i can't solve problem B still now, I'm a loser. i can't do anything in my life. my friends aee progressing

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I'm 20 months into Codeforces and I haven't been able to solve it, don't worry You're good

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      12 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      hey thank you so much for appreciating! I'll do better in sha Allah

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    You need to solve harder problems. You solved 37 problems rated higher than 900 in 11 months. That's ~0.11 problems rated higher than 900 per day.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    You need to solve higher rated problems. Div2B is around 1100-1300 rating range. Start with the 900 rating problems, gradually increase the rating as you get comfortable and faster at solving them.

    And don't get discouraged. We all start at zero. Just reflect on what you need to work upon. You'll eventually get there.

    :)

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      12 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      thank you so much for your concern :') i'm actually so much drepressed, I'll start solving >=1000 problem then. let's see I'll update about this to you later

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

long long

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Can anyone explain how to solve C please?

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Consider the sum value of all numbers. If you merge two number with the same parity, the sum stays the same. If you merge two number with the opposite parity, the sum decrease by 1.

    So the first player would merge the two number with the same parity (always possible) and the second would merge numbers with opposite parity (if possible). Also note that after merging two numbers you get an even number. The first player would take this advantage. He knows that the second player can decrease the sum only if there are odd numbers, so he would take two odds and merge them if possible (instead of two evens).

    Now, we know that the number of odd numbers is what matters, take mod 3 of the number of the odd number (mod 3 because if the first player take two of them and the second take one of them, it would decrease by 3 after both player played).

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12 months ago, # |
  Vote: I like it +8 Vote: I do not like it

What is the idea behind E problem?

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12 months ago, # |
  Vote: I like it +8 Vote: I do not like it

https://codeforces.net/contest/1916/submission/239707511
This is my submission for E problem. I had just did 100 times bruteforce until I get an optimal solution. My code is wrong.. but still it gets accepted by the judge.

I wanted to know the exact approach for this E problem.. Anybody explain pls.
Also try to hack my submission . It would be a great help.

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

    DFS the tree. From the leaves to the top, maintain the numbers of the color by using segment tree, then calc the max subtree and the second max subtree.

    For a node u, we get the lowest ancester that has same color. We may calc one color for multiple times, so reduce it on the ancester. Note that we only do operations like interval +1 or interval -1, and we work O(n) times. So the complexty is O(n log n).

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    12 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    this is amazing! need moar tosters!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

man , i am a bit sad . Was really close to get both problem A and B right , and i could have gotten D too , i see that trick but ran out of time :(( the pretest in A is really weak ngl

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12 months ago, # |
  Vote: I like it +25 Vote: I do not like it

Can anybody clarify why this round is rated?

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    12 months ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    I am surprised myself. The real issues were problem F(harder solutions exist: https://loj.ac/p/3176), G(had no valid solution when the contest ran) and H(solvable with oesi: https://oeis.org/A286331)

    • Maybe not enough people solved problem F&H that way to make it unrated.
    • Maybe its still up to debate
    • Maybe there is a debate to make it rated for Div2 only (there were minor issues with Problem A and D, but not enough reason to make it unrated imo)
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12 months ago, # |
  Vote: I like it +3 Vote: I do not like it

is it rated ?

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

i think it was a good contest with weak pretests sad to see all this hacks

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

i cheese cbrt for b

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

It would be nice if this post reaches 2024 downvotes

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12 months ago, # |
← Rev. 2   Vote: I like it -49 Vote: I do not like it

This amount of downvotes! Weak test! Weird problem D.... Two last problems are available in online. Still rated contest? Mike has no feeling for us. He doesn’t even care for our opinion.

Everyone? Time to leave codeforces. Its not a cp platform anymore.

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12 months ago, # |
← Rev. 2   Vote: I like it +202 Vote: I do not like it

Happy New Year!

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

[ Deleted ]

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12 months ago, # |
  Vote: I like it +45 Vote: I do not like it

It seems that nobody mentioned it before, so: in problem F there is a randomized solution. 239664519. The idea is that we building random spanning tree using dfs (dfs-tree) and looking for subtrees of size n1 or n2. If such subtree found, it is the answer. Repeat the procedure for all roots ~20 times. I have no real understanding why it works. Maybe someone can hack..

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

A ....weak pretests (no overflow handling lol) B ....statement is wrong (b<x and 1<=x<=1e9) D H ....googlable G ....wrong author solution F ....stolen Very bad contest

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12 months ago, # |
  Vote: I like it +17 Vote: I do not like it

Editorial.Please.

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12 months ago, # |
  Vote: I like it +17 Vote: I do not like it

I used to think that 74TrAkToR is a good problem setter because of good problems in Round 905. But I was wrong!!!!!!

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12 months ago, # |
  Vote: I like it +8 Vote: I do not like it

Wake up to check the problems I missed hours ago, only to find this... Worst Goodbye contest ever!

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

Worst contest ever! Make this unrated! It's okay if the contest was having small problems but the contest was delayed 2 times, the site keep using cloudfare, first 4 problem was really really bad (unimaginable) and leaked a problem that can be easily searched.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

I think I should read the last few questions and OEIS them first next time.

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12 months ago, # |
  Vote: I like it +32 Vote: I do not like it

IMO, problem E time constraint should've been at least 2 seconds, considering 3 * 1e5 input size.

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

All are saying B a bad problem. But when i submitted accepted one solution around 1hr 10min (by changing approach to gcd) after getting TLE 3 times, then around 10k-11k users have already understood & solved it, which i felt unusual, and also in problem A. i vaguely remembered that around 10k users have already passed pretest in 15-20min. when i submitted:( And yes, also i didn't handled that int overflow in A, i have got Runtime error in main test :(

upd: i liked problem C, i just now got it accepted in one go, happy ending 2023:)

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    a problem can be bad and easy at the same time. tbh i dont have any probllem with this competition cuz im using my fun account so i dont have to take things serius (solve some then leave) and ofc this is new account so i gained elo. i dont think people have huge problems with problem B as its only problem was not clarifyng enough that invalid input cannot happen (some toddlers need to learn how to understand texts). The real problem was with H which was an existing probllem that you could google. (sorry for bad english not my first language)

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

Can anyone please tell why this solution for Problem A fails. And it got accepted without the if (multiply > 2023) condition in the for loop?

#include<bits/stdc++.h>
using namespace std;
#define ll long long 
 
void solve(){
    int n, k;
    cin >> n >> k;
    ll multiply = 1;
    for (int i=0; i<n; i++){
        ll temp;
        cin >> temp;
        multiply *= temp;
        if (multiply > 2023){
            cout << "NO" << endl;
            return;
        }
    }
    
    if (2023 % multiply != 0){
        cout << "NO" << endl;
        return;
    }
    else {
        cout << "YES" << endl;
        cout << (2023 / multiply) << " ";
        k--;
        while (k > 0){
            cout << 1 << " ";
            k--;
        }
        cout << endl;
    }
}
 
int main() {   
    int testCases;
    cin >> testCases;
    while (testCases--){
        solve();
    }
    return 0;
}
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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Because you are not completing the input, whenever the product goes beyond 2023. Never return something in the input loop.

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Because you return in the loop too early and ended up not reading all the numbers.

    Consider the case below:

    2
5 1
2023 2023 2023 2023 2023
1 1
1

    Your code will treat the second 2023 as the $$$n$$$ for Case#2

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    you should read all the inputs in one task before output and handle next

    void solve(){ int n, k; cin >> n >> k; ll multiply = 1; for (int i=0; i<n; i++){ ll temp; cin >> temp; multiply *= temp; }

    if (2023 % multiply != 0){
    cout << "NO" << endl;
    return;
}
else {
    cout << "YES" << endl;
    cout << (2023 / multiply) << " ";
    k--;
    while (k > 0){
        cout << 1 << " ";
        k--;
    }
    cout << endl;
}

    }

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12 months ago, # |
  Vote: I like it +16 Vote: I do not like it

So where is the tutorial?

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    12 months ago, # ^ |
      Vote: I like it +32 Vote: I do not like it

    They just argue about who should publish the editorial, because that unlucky guy will have to face -3000 contribution too :)

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      12 months ago, # ^ |
        Vote: I like it +36 Vote: I do not like it

      Ah, now I understand why some contests have no editorial for long time.

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      12 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Now I can see Tutorial Good Bye 2023 (en) on the contest page. But when I click on it, I see

      Can't find such topic

      What the f*cking sh*t is it???

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        12 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        That was a joke tutorial by someone and it got removed

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12 months ago, # |
  Vote: I like it +18 Vote: I do not like it

I think I,47TrAkToR can do better than 74TrAkToR as a tester.

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12 months ago, # |
  Vote: I like it +8 Vote: I do not like it

Is it hard for you to create the Editorial ?

Spoiler
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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

So where is the tutorial?

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Tutorial please!

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

please help me . when I get pupil rank

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

waiting for the editorial please

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Python users !! Can E be solved using recursion?If yes,then how and how do python users solve recursive problems as python can do only limited recursions (not enough for CP).Advise needed! look at my submissoin:my E solution

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    12 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Time limit here is tight even for C++. Probably, it is impossible to get intended solution on python, which pass time limit. (However, probably, it is possible to submit something hacky.)

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

lets not hope hello 2024 will not be as terrifying as this

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

ending the year in green

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12 months ago, # |
  Vote: I like it +33 Vote: I do not like it

fun fact.

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12 months ago, # |
  Vote: I like it +2 Vote: I do not like it

Does anyone know what is the lowest contribution on CF ever? This guy might have made a history

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Since there is no editorial, can anyone explain E ? I would be very grateful.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Sorry for the rudeness, but I won't write any more rounds from 74tractor

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

It should be unrated. I wasted a lot of time because of this round. And...... Where is Tutoria ??? It's a bad ending of 2023.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Why is hack gets Unexpected verdict?

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

Everyone :-3231 downvotes

Mike : what a beautiful round ! it should be rated.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Good one

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

i need help?how to understand C?my idea is wrong,emmm,please give me a idea

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Can someone please share resources for H problem? I can't understand how the permutation and combination work in the maths of matrix problems.

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12 months ago, # |
  Vote: I like it +21 Vote: I do not like it

Oh, what a magical number!

-3333

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12 months ago, # |
← Rev. 2   Vote: I like it -17 Vote: I do not like it

Why everyone is disapointed about this contest.If you did not know that 2023^5>10^12 then it is your problem.At minimum you can use calculator to see it.Be realistic and attentive.And I think it was good contest to see your math skills.

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

Goodbye,$$$ 2023 $$$

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

We hope hello 2024 be better ❤️

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Sorry, I clicked downvote by mistake

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

A somewhat disappointing leave for 2023. I hope the welcome, Hello 2024, would be more favorable than this.

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

This round describes 2023 very well

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12 months ago, # |
  Vote: I like it +34 Vote: I do not like it

Why did they allow such a crazy man(74TrAkToR) to be responsible for such an important contest?

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12 months ago, # |
  Vote: I like it +8 Vote: I do not like it

oeisforces

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Mathforces af

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12 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Everyone is complaining about Good Bye 2023 and problems D and further, while I (a Div.3) didn't even make it that far.

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12 months ago, # |
  Vote: I like it +1 Vote: I do not like it

Good Bye 2023

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12 months ago, # |
  Vote: I like it +51 Vote: I do not like it

monument to 4k. congrats!

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

Are you waiting for -5k to make this round unrated?

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12 months ago, # |
  Vote: I like it +8 Vote: I do not like it

is this the most downvoted post ever?

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11 months ago, # |
← Rev. 2   Vote: I like it +1 Vote: I do not like it

Really Nice Contest.

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11 months ago, # |
← Rev. 2   Vote: I like it +1 Vote: I do not like it

Same Situation as i can only solve two questions.

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8 months ago, # |
  Vote: I like it 0 Vote: I do not like it

poor marzipan he just invited everyone to the contest, and now he is the worst contributor ever.

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6 months ago, # |
  Vote: I like it +2 Vote: I do not like it

Why so disliked?

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