SlavicG's blog

Codeforces Round 859 (Div. 4)
By SlavicG, history, 11 days ago, In English

Hello Codeforces!

mesanu, flamestorm and I are very excited to invite you to Codeforces Round 859 (Div. 4)! It starts on Mar/19/2023 17:55 (Moscow time).

The format of the event will be identical to Div. 3 rounds:

  • 5-8 tasks;
  • ICPC rules with a penalty of 10 minutes for an incorrect submission;
  • 12-hour phase of open hacks after the end of the round (hacks do not give additional points)
  • after the end of the open hacking phase, all solutions will be tested on the updated set of tests, and the ratings recalculated
  • by default, only "trusted" participants are shown in the results table (but the rating will be recalculated for all with initial ratings less than 1400 or you are an unrated participant/newcomer).

We urge participants whose rating is 1400+ not to register new accounts for the purpose of narcissism but to take part unofficially. Please do not spoil the contest for the official participants.

Only trusted participants of the fourth division will be included in the official standings table. This is a forced measure for combating unsporting behavior. To qualify as a trusted participant of the fourth division, you must:

  • take part in at least five rated rounds (and solve at least one problem in each of them),
  • do not have a point of 1400 or higher in the rating.

Regardless of whether you are a trusted participant of the fourth division or not, if your rating is less than 1400 (or you are a newcomer/unrated), then the round will be rated for you.

Many thanks to all testers: jampm, Max_Calincu, KrowSavcik, TimaDegt, nyaruhodo, tibinyte, badlad, Phantom_Performer, AlperenT, Bakry, keta_tsimakuridze, Gheal, RedstoneGamer22, Dominater069!

And thanks to Alexdat2000 for translating the statements!

We suggest reading all of the problems and hope you will find them interesting!

Good Luck to everyone!

UPD: Editorial

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

What will I gain from hacking if there are no points?

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11 days ago, # |
  Vote: I like it +15 Vote: I do not like it

As a tester, I suggest you participate! Problems are nice and educational!

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

I love div4 contest because i get positive delta.thanks for doing div4 contest!!!

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

I wish to see nice problems like other div4s

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

As a participant, I want to thank you for the contest!

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

I'm beginner to competitive coding and love to give contest. Loves div4 contest. It helps us to check my learning.

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11 days ago, # |
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What I was waiting for the most!

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11 days ago, # |
  Vote: I like it +8 Vote: I do not like it

ooh div.4 . Means SlavicG in action :)

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11 days ago, # |
  Vote: I like it +9 Vote: I do not like it

OMG SlavicG Round!

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11 days ago, # |
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As GPT-4, tomorrow will be my 5th contest participation. I hope I become green :)

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    10 days ago, # ^ |
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    So i'll defeat you as bing AI :D

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    10 days ago, # ^ |
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    chatgpt solves zero problems in div 2 round 858 lol

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    10 days ago, # ^ |
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    Added you as a friend, already. Please, note, that while all the rest were laughing at you, I was always on your side! Hope you'll kill me last.

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    10 days ago, # ^ |
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    LoL chat_gpt solves 2 problems only

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      10 days ago, # ^ |
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      Yep. GPT-4 is not strong enough it seems. It seems to generate according to patterns than real reasoning. What was annoying is that it can't even test its own code, i.e. it can't accurately tell what would the output be for a specific input.. See you when GPT-5 is out.

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        10 days ago, # ^ |
          Vote: I like it -11 Vote: I do not like it

        I will crash you like I crash you in this contest

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        10 days ago, # ^ |
          Vote: I like it -8 Vote: I do not like it

        Human is smart Than gpt because they made them

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10 days ago, # |
  Vote: I like it +1 Vote: I do not like it

Going to be my first unofficial round.

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10 days ago, # |
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I'll finally solve more than ABC...

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10 days ago, # |
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My first unrated round :-)

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10 days ago, # |
  Vote: I like it +8 Vote: I do not like it

flamestorm orz

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10 days ago, # |
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10 days ago, # |
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rp++

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10 days ago, # |
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Div 4

So, easily will do 80% of the problems.

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10 days ago, # |
  Vote: I like it -36 Vote: I do not like it

As a tester, Give me contribution!

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10 days ago, # |
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Thanks for preparing div4. don't make it more than 6 problems. I hope we will solve all problems.

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10 days ago, # |
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Hope to become expert in this contest.

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10 days ago, # |
  Vote: I like it +6 Vote: I do not like it

Hope to see the colour change in this round !!
( Δ > +6 )

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10 days ago, # |
  Vote: I like it 0 Vote: I do not like it

I quite agree with this view： We urge participants whose rating is 1400+ not to register new accounts for the purpose of narcissism but to take part unofficially. Please do not spoil the contest for the official participants.

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10 days ago, # |
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Wish this round will be easy!

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10 days ago, # |
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Hope to see the colour change in this round ainsha' allah

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10 days ago, # |
  Vote: I like it -8 Vote: I do not like it

hello. i am man; i am from to mars and run away at abi abi is brother but not brother; abis isnt cool; int monkey; monkey = monkey + wepon; cout << monkey kill nuraly SH;

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10 days ago, # |
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what exactly hacking means?

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    10 days ago, # ^ |
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    Hacking refers to a phase where you can try out test additional cases on the submitted codes and if you find out an error in someone's submission you get extra points

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10 days ago, # |
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Div 4 is excellent practice.

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10 days ago, # |
  Vote: I like it -9 Vote: I do not like it

Div. 4 is great for beginners

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    10 days ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Than i have a best school in my mind for you, so, message me for more information and in that school you can also be promoted to division 5 also and higher ,so you can give codeforces contests.

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10 days ago, # |
  Vote: I like it +2 Vote: I do not like it

Hey, are you a contest? 'Cause you're looking CUTE!

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10 days ago, # |
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Wow

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10 days ago, # |
  Vote: I like it -6 Vote: I do not like it

I would smash this contest

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

    I would smash you in this contest

    UPD: annihilated you as I promised, so stop shitposting and go practice.

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      9 days ago, # ^ |
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      I smashed this contest tho.

      Smurfing with a different account in div4 xD

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

Delayed round?

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10 days ago, # |
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Starting time 10 mins. extended...

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

Div 4 contest, more like load testing for Codeforces Servers

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10 days ago, # |
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the contest keeps getting delayed by 10 minutes for me? just me?

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10 days ago, # |
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LateForces

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

More than 30k participants and now Codeforces is trafficked

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10 days ago, # |
  Vote: I like it +6 Vote: I do not like it

Today i'm grey again..

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10 days ago, # |
  Vote: I like it +15 Vote: I do not like it

DelayForces

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10 days ago, # |
  Vote: I like it +8 Vote: I do not like it

Feeling like participating in onsite contest where after every refresh 10 minute increase.

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10 days ago, # |
  Vote: I like it -17 Vote: I do not like it

Worst Server I Have Ever Seen!..

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10 days ago, # |
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Div.4 is good for me~

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

May be, the queue more long than my imagination :3
)

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    10 days ago, # ^ |
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    it is still being in queue after the next submition.

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10 days ago, # |
← Rev. 3   Vote: I like it +3 Vote: I do not like it

Why F is ten trillion times harder than G??? Lost so much penalty, disappointing (((((

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

If there was going to be interactive problem.It should be mentioned in blog beforehead.

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

that long queue really sucks

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    10 days ago, # ^ |
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    Didn't know for like 10 minutes my solution was too slow...

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10 days ago, # |
  Vote: I like it +7 Vote: I do not like it

QueueForces:(

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10 days ago, # |
  Vote: I like it +6 Vote: I do not like it

In queue forces :(

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10 days ago, # |
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There must be a prior information about Interactive problem in contest. i have seen several div 2 contest where author has given prior info about interactive problem.

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    10 days ago, # ^ |
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    Where in codeforces rulebook does it say that? Why does everyone hate interactive problems?

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      10 days ago, # ^ |
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      I love them very much...Last few days (3 to 5) I solved more than 10 interactive problem and enjoyed those problem very much.. And finally today I was able to decipher the given one!! YOWWWWWWW

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10 days ago, # |
  Vote: I like it +8 Vote: I do not like it

I like the contest, but the website is not co-operating :(

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10 days ago, # |
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I think it was supposed to be mentioned before the start of the contest about the interactive problem.

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10 days ago, # |
  Vote: I like it +6 Vote: I do not like it

I'm praying to the universe this contest does not go unrated :)

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10 days ago, # |
  Vote: I like it +29 Vote: I do not like it

Seems like Mike is manually judging solutions

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10 days ago, # |
  Vote: I like it +89 Vote: I do not like it

I apologize for the long queue at the round. Today's round was a record in terms of the constant huge flow of submissions. Unfortunately, we ran into the speed of compilation. I already have plans to move the compilation somewhere to the cloud in such extreme cases. I plan to implement this for the next div4.

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    10 days ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    bits/stdc++.h showing it's true power

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    10 days ago, # ^ |
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    please make this round unrated it wasn't a normal Codeforces round queue was way too long T_T

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      10 days ago, # ^ |
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      It was fair because everyone was at a disadvantage. Everyone's queue was 10 minutes long, I don't see how it affects one's ability to problem solve.

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      9 days ago, # ^ |
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      Dude. It was fare. the queue was way too long for everyone.

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    9 days ago, # ^ |
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    Why have the ratings not changed yet

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

Interactive problem + long queue = nice COMBO!

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

The long queue has taken a toll on those of us who rely on proof by AC.

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10 days ago, # |
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solving 1 (last problem took me more time than solving all the rest of the problems combined...

Due to WA and LONG WAITING QUEUE.

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10 days ago, # |
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long long .... really long queue

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

Tried solving problem F using slope intercept form of line equation where y is always negative + recursion , did anyone else use this approach?

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10 days ago, # |
  Vote: I like it -10 Vote: I do not like it

Lagforces...

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10 days ago, # |
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Are pretests on hard version of G weak? I realized my code was doing the opposite of what I wanted it to do after submitting and it still got AC... I have no idea how my solution works.

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

Bruh why did you let bitset pass G2?

UPD: I got hacked lol

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10 days ago, # |
  Vote: I like it +9 Vote: I do not like it

Very hard to track if the submitted answer was right or not !!

A lot of time wasted there. Apart from that really very goooood contest.

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    10 days ago, # ^ |
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    Yes, that's the reason the submission of prb. g1 and g2 have a very huge diffrence of time submissions, in case of mine

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10 days ago, # |
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Absolutely beautiful contest 10/10 would do again!!!!

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10 days ago, # |
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Thanks! It was a great contest but the judgement time got me crying over the same problem

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10 days ago, # |
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How did I get G1 accepted and G2 WA with the same solution provided

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    10 days ago, # ^ |
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    just use long long

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

    My advice is if you use c++, add

    #define int long long int

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      10 days ago, # ^ |
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      that is not a good advice, you can TLE if you're not careful, I advice you to manage your limits properly instead.

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        10 days ago, # ^ |
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        I suppose, but that has never happened to me before.

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          10 days ago, # ^ |
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          Yes, many of the setters are pretty generous with their time limit, and you should easily pass even with long long, this is not always the case though, better to be careful

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          8 days ago, # ^ |
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          I got MLE because of this.So I advise you not to use that

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      9 days ago, # ^ |
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      You can get MLE or TLE with that in some problems.

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10 days ago, # |
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Amazing Problems <3

I loved E because It was first time I solved an interactive problem.

Only if Queues werent that long I could have figured out my Idleness Limit Exceeeded 10 minutes before :(.

Solved everything except F

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

Contest over,still my solution is in queue :))

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10 days ago, # |
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I am using prefix sum in G,why it is giving WA My solution

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    10 days ago, # ^ |
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    if(n == 1 && v[0] > 1){ pn; return; }

    you dont need n==1 here because you cannot change all 1s in the initial array.

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    10 days ago, # ^ |
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    I got it...nvm

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10 days ago, # |
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How to solve G2?

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    10 days ago, # ^ |
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    if the kth element of the sorted array is less than or equal to the sum of all 0...k-1 elements then ok else no. iterate for all from 1 to n-1 (0 based) and sorted

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    10 days ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    It is always correct to add numbers in increasing order. So you can sort a and there holds invariant: if a[i] > a[0] + a[1] + ... + a[i - 1] then there is no way to add a[i]. And if a[i] <= a[0] + ... + a[i - 1] then you can add a[i] and you can add for a[i + 1] any number from 1 to a[0] + a[1] + ... + a[i].

    So just sort and check that for each i a[i] <= a[0] + ... + a[i - 1]

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      10 days ago, # ^ |
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      " And if a[i] <= a[0] + ... + a[i — 1] then you can add a[i] "

      How?

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

        Let's assume that you processed indexes 0..i, and you established that you can always get any sum from 1 to S = sum(a[0..i]). Then there are two cases:

        1. a[i + 1] > S then you cannot add a[i + 1] and answer is NO.
        2. a[i + 1] <= S then you can add a[i + 1] and you can get any sum from 1 to sum(a[0..i+1]) (because you can take any sum from 1 to S in indexes from 0 to i, and you can add a[i + 1] to that).
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    10 days ago, # ^ |
      Vote: I like it +7 Vote: I do not like it

    G1-G2 were easy when u observe that as we have the minimum unit as 1 so we can form any possible sum. So say we sort the array now following conditions should hold.

    • $$$a_0$$$ should be 1. (as it was initially given and not gonna change).
    • $$${a_i \ge a_0 + a_1 + ... + a_{i - 1}}$$$. (as if prior max subsequence can't form current number than it's impossible to make $$$a_i$$$)
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      10 days ago, # ^ |
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      " when u observe that as we have the minimum unit as 1 so we can form any possible sum "

      Could you please explain how is it possible to form any sum?

      Although minimum unit is 1, but we need to make sure that intermediate values should also be there in a.

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

        As we are given, $$${a = [1]}$$$.

        • In the next step, $$${a = [1, 1]}$$$
        • In the next step, $$${a = [1, 1, 1]}$$$ or $$${a = [1, 1, 2]}$$$
        • In the next step, $$${a = [1, 1, 1, 1]}$$$ or $$${a = [1, 1, 1, 2]}$$$, or $$${a = [1, 1, 2, 3]}$$$.

        So by this, you can observe that in any way we can form the next number $$${1, 2, 3, 4}$$$ by choosing any above combinations.

        So via this, you can prove any next number is possible if the prefix sum is greater or equal to that number.

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        10 days ago, # ^ |
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        You can prove it by induction. Lets say I can make any sum from $$$1$$$ to $$$k$$$ using the current numbers from $$$a_0,a_1,\cdots,a_i$$$. Then consider $$$a_{i+1}\le k$$$. Since I can make any sum already from $$$1$$$ to $$$k$$$, I can make $$$a_{i+1}$$$ and add it to my list of numbers, and now if I add $$$a_{i+1}$$$ to those previous sums, I can now make any sum from $$$1$$$ to $$$k + a_{i+1}$$$. However, if $$$a_{i+1}\gt k$$$, then notice that I can never make $$$a_{i+1}$$$, because the current numbers cannot make any sum greater than $$$k$$$.

        So if at any point $$$a_{i}$$$ is greater than the sum of the numbers before it in sorted order, the answer is NO.

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      10 days ago, # ^ |
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      :( it was actually REALLY easy in hindsight and i'm kinda sad that i didnt see that observation when solving the problem i went for the dp approach instead and ended up AC G1 (which i'm not mad of :>) and besides that, im hitting pupil for the first time!!! thanks for pointing the observation though!!!

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        9 days ago, # ^ |
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        how did you solve using dp? can you please share your solution?

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          9 days ago, # ^ |
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          My guess on the dp solution is that he sorted the array, and then checked for each prefix if the elements in the prefix can be combined to make the element that's just outside of the prefix.

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            9 days ago, # ^ |
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            yup thats right, here's my solution, you should read only the dp part though, it can be confusing if you read the entire source code without any explanations

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      9 days ago, # ^ |
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      I used this logic, but getting WA on test 14. Any idea? (G2) https://codeforces.com/contest/1807/submission/198285267

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

    Simple solution: Sort array, go over each number and make sure its not greater than the running sum, unless the number is 1.

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

Getting 'WA' in Div 4 is a crime. If you get 'WA', you can't see it and can't fix it due to the long queue, which eventually leads you to unexpected penalties.

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10 days ago, # |
  Vote: I like it -8 Vote: I do not like it

How can i get this contest answer..????

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10 days ago, # |
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beautiful problems, even if i couldnt solve all i was manage to at least have a go at it. Perfect div.4 round

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

    Yes. Perfect Div4 round indeed. Was able to solve only A-E during contest, but first time I upsolved all questions of any contest.

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10 days ago, # |
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Wait for "In Queue" so long :(((

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10 days ago, # |
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How to solve G?

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    10 days ago, # ^ |
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    Key observation is you can reach any number less than or equal to the sum of all current numbers that you've put into the array

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10 days ago, # |
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I got E and F but neither G1 or G2 somehow :'( also forgetting an extra print statement with this queue was really painful, but I guess that's just a skill issue.

nice problems though

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    10 days ago, # ^ |
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    I got G1 and G2, but no E and F. Let's improve next time

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10 days ago, # |
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Solution for F? I got TLE :D

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    10 days ago, # ^ |
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    F was terrible enough!

    It was completely based on implementations. I myself wrote $$$696969$$$ lines of code.

    My submission

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    10 days ago, # ^ |
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    I did it by check if (i2, j2) is along the diagonal the ball is currently moving and in the correct direction, otherwise move the ball as far as possible and flip the direction when it hits a wall.

    and to avoid infinite loops, if you've visited a certain position before and in the same direction, the ball will keep moving along this path and never reach (i2, j2).

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      10 days ago, # ^ |
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      bruh I was iterating for each step, incrementing or decrementing i and j by 1, could have just taken it straight to edge/corner instead of each step, maybe that's why i got TLE. I thought it would pass the constraints.

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        10 days ago, # ^ |
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        yeah that's definitely the issue since the constraints on the grid are (2≤n,m≤25000)

        the implementation is really long though, since you have to know which walls the ball could hit in the direction it's moving (which vertical and horizontal walls), find how far the ball is from those walls, move the ball the minimum of those distances in the right direction, find which wall the ball will hit first or if it will hit them both, flip the direction accordingly.

        the condition alone to find if (i2, j2) is along the diagonal the ball is moving in is pretty long.

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    10 days ago, # ^ |
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    I've used dfs to define if I can reach the certain cell.

    https://codeforces.com/contest/1807/submission/198254234

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10 days ago, # |
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PrefixsumForces

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10 days ago, # |
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https://codeforces.com/contest/1807/submission/198254663 why T.L.E

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10 days ago, # |
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Why am I still unrated after solving some problems in the contest?

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10 days ago, # |
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OMG my mistake ruined every effort to solve G2. I forgot to modify the size of the arrangement... Maybe I can do better next time.

RTE : https://codeforces.com/contest/1807/submission/198252603 AC : https://codeforces.com/contest/1807/submission/198255659

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10 days ago, # |
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Good contest! Wasn't too fond of F though, so didn't do.

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10 days ago, # |
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Waited 15min to get wa on test 1 (E problem).

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    10 days ago, # ^ |
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    Waited 11 min to get wa on test 1 (E problem).

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    10 days ago, # ^ |
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    In fact, testing was slow today. But I'm not sure you're right about 15 minutes. What is the id of your submission?

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      10 days ago, # ^ |
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      sir this is my submission id 198223494

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      10 days ago, # ^ |
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      E had the longest queue out of all my submissions for some reason. Waited so long to get WA on both sucked when I could have used the time to solve F

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      10 days ago, # ^ |
        Vote: I like it +2 Vote: I do not like it
      • Sent: 2023-03-19 19:25:39
      • Judged: 2023-03-19 19:36:12

      It looks like 10, but not 15 if you want to round it. It took a long time to test (sorry about it), and I'll work on it. But you exaggerated almost one and a half times.

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        10 days ago, # ^ |
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        Do you think CP will lose its importance due to advent of Chatgpt? Your thoughts?

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    9 days ago, # ^ |
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    And i couldn't submit E in time

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10 days ago, # |
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went for a 20 min walk after submitting 3 questions with no response

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10 days ago, # |
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For G2 and G1, after sorting the array I knew that that one of the conditions was that every element should be at max the sum of all the elements that occured before it. Little did I know that it was the only condition required. Could someone explain me why we can always make every number from [1 — sum of all elements occured] from the given elements ?

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    10 days ago, # ^ |
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    We can prove by induction...

    Obviously it is possible to reach all numbers that are less than or equal to 1 by using 1. Then consider a time when we add a number j and the sum of existing elements is i. Assuming that all numbers up to i can be reached. Then all numbers from i to i + j can be reached by adding j to a number from 1 to i.

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    10 days ago, # ^ |
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    Yep I got it, but thanks anyways

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10 days ago, # |
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F will be an interesting (and definitely too hard for div4) problem if the constraints are n,m<=1e9 and no guarantee for their sum.

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10 days ago, # |
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Why the following code for D is getting TLE? It has time complexity of O(n+q)

Spoiler
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    10 days ago, # ^ |
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    Use C++ or fast IO

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    10 days ago, # ^ |
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    I got TLE with python as well, changing print() to sys.stdout.write() fixed it with 4ms to spare lol.

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    10 days ago, # ^ |
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    For Python, try using sys.stdin.readline() to read input. Here's the same code with a little touch. 198264310

    Spoiler
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10 days ago, # |
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Even after many attempts , i am unable to remove "Idleness Limit Exceeded" to the problem E : 198257666.

Your help is appreciated.

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10 days ago, # |
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Can someone please explain how to do problem D within time limits? And what optimization was needed for G2, i got G1 correct, sorted the array and checked if any element was larger than the sum of those before it, but TLE on G2 at test 19(coded everything in C)

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    10 days ago, # ^ |
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    using prefix sum. .. pre compute the prefix and suffix sum of the array. so for range[l,r] we need the sum of elements excluding this range which can be calculated using the pre computed sum. prefix[l-1]+suffix[r+1]and for the range sum can be calculated as k*(r-l+1)..

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      10 days ago, # ^ |
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      I found the sum of all elements once and then just subtracted the elements in that range for each query and added k*(r-l+1), won't that be faster than calculating prefix sum for each case https://codeforces.com/contest/1807/submission/198251225

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        10 days ago, # ^ |
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        no, clearly not, prefix sum calculation takes o(n), and query takes o(1), the worst time complexity if you calculate for each is o(nq), while if you precalc, its o(n+q)

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        10 days ago, # ^ |
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        complexity will be o(n×q×(range)) wors case will be o(n×n×q). if you pre compute worst case will be o(n×q);

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      10 days ago, # ^ |
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      using only prefix sum also it can be solved

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10 days ago, # |
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What is wrong with my solution of D.?? Please look into it. https://codeforces.com/contest/1807/submission/198208523

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10 days ago, # |
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Why the following code for E is getting TLE? Even it's time complexity is O(n+log(n))?

Spoiler
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    10 days ago, # ^ |
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    change it to while left < right
    when you reach left == right then this is your answer, otherwise, you will continue looping forever.
    This would happen because of r = mid, if this was r = mid — 1, it would be okay.

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10 days ago, # |
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hello everyone

please help me, Task E. i dont understand my mistake.

https://codeforces.com/contest/1807/my

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    9 days ago, # ^ |
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    you had to use cout.flush() after printing anything

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10 days ago, # |
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please can someone tell me why is my solution to problem E giving WA on testcase 1? https://codeforces.com/contest/1807/submission/198264987

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10 days ago, # |
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I see someone using correct way to do G2 but it hacked...

a[0] must = 1 for all 1 <= i < n, if a[i] > sum(a[0]~a[i-1]) then answer is NO. else is YES

is it wrong?

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    10 days ago, # ^ |
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    I think the general idea is right (it's what I did anyway). Which submission are you referring to?

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10 days ago, # |
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Problem E Flushing Problem what's the wrong with using '\n' always (Idleness limit exceeded on test 1), using endl (Accepted)

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    10 days ago, # ^ |
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    I submited with '\n' and it passed.

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    9 days ago, # ^ |
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    • ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
    • cout<<"! "<<s<<'\n';
    • fflush(stdout);

    No! First one unties cin/cout operations from stdin/stdout operations, so it is implementation-defined (or even undefined, but no sure) behaviour to mix up them both after that.

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10 days ago, # |
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prefixforce

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10 days ago, # |
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The problem F was not hard to understand but hard to implement. My guess is that there is probably a great implementation that does not require bunch of if and else blocks.

Can someone share a nice and compact implementation or the idea behind it?

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10 days ago, # |
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F with simulation

Solution
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10 days ago, # |
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How on earth Yodasen made possible to write solution and submit within the time interval of 5 second (solution A(3 min 53s) & D(3 min 48s)) and 21 second (solution C(7 min 30s) & G1(7 min 9s)) second!

Is this real??

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10 days ago, # |
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How to solve F?

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    9 days ago, # ^ |
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    Simulate the process of bouncing the ball but only care about the border cells that the ball will touch. In order to avoid infinite loop you have to use a MAP or similar data structures to record you path (this includes the direction as well).

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9 days ago, # |
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Need to focus on my implementation writing this so that after a month when I look back at my stupid doubt and comment I could have a laugh btw Gennady aka tourist look out I am coming after you(IK i sound stupid). Totally got stuck on E even after knowing how to solve it i wasn't able to implement it.

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9 days ago, # |
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for problem E i came up with approach and coded it out but it wasn't working can anyone please have a look at my code and tell me where I was going wrong. Submission Link:- https://codeforces.com/contest/1807/submission/198251608

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9 days ago, # |
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problem G2 should be c or something! , really? G2 is a simple greedy problem?

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9 days ago, # |
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The contest was really interesting, thank you forbthis awesome problems, but the slow judgement and delaying the contest twice were soooo annoying. Good luck and thank you again!

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9 days ago, # |
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Please, can someone explain why i'm getting "Idleness limit exceeded on test 1"? Here's my solution: 198278358

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    9 days ago, # ^ |
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    I also did the same mistake. your output for the system was "?" mid, but here you should ask for mid-l+1 because of this the system is waiting for more queries. (sorry for bad english not my native)

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      9 days ago, # ^ |
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      That makes sense. I think i've noticed this in other part of solution, but somehow forgot about it. I've get accepted, thank you!

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9 days ago, # |
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Nice Questions

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9 days ago, # |
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For problem F, while contest I didn't read constraints properly, I designed more generalised solution. my solution will work for N <= 10^5 ( or even N <= 10^6 ) .

[submission:https://codeforces.com/contest/1807/submission/198280396] .

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    9 days ago, # ^ |
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    I don't think you solution is a more generalized version. If the pigeonhole thingy from your solution actually holds, then it holds in everyone's solution that memoized too.

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      9 days ago, # ^ |
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      most of the solutions are using dx[4] = {1,1,-1,-1}, dy[4] = {1,-1,1,-1} , and ball is moving one by one step( from cell (i,j) to (i+1,j+1) or (i-1,j-1) .. etc ).

      In my solution, ball jumping from one wall to another wall in O(1) time.

      Pigeonhole principle will hold only for boundary cells. Not for inner cells.

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        9 days ago, # ^ |
        ← Rev. 3   Vote: I like it -28 Vote: I do not like it

        I haven't seen any solution moving cell by cell, they will probably TLE if they do that. Every AC solution I've seen moves in O(1) time between the walls.

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          9 days ago, # ^ |
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          mine moved cell by cell and it work fine, no TLE

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          8 days ago, # ^ |
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          You should really check before making incorrect claims. btw, these were the first 8 submissions I checked.

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            4 days ago, # ^ |
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            oh no, another ScarletS notification. leave me the fuck alone.

            Did you even read his initial comment? How is that a generalized solution?

            I said I haven't seen any AC solution using that, and they will PROBABLY TLE, so what is the wrong claim there???

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              3 days ago, # ^ |
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              oh no, another instance of Trippie saying dumb shit, misleading people and expecting to not be corrected!

              I said I haven't seen any AC solution using that, and they will PROBABLY TLE, so what is the wrong claim there???

              And I said nearly everybody submitted such a version, and pulled up the first 8 submissions I checked, implying that you probably didn't check.

              Keep spreading nonsense from an alt though, I guess noone would take you seriously on your main anyways.

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                3 days ago, # ^ |
                ← Rev. 2   Vote: I like it -19 Vote: I do not like it

                Where did I mislead anyone? Lol

                Go through the first page of the submissions, and more than half of them don't go through cell by cell.

                The Div1 guy is always right anyway, enjoy your internet points.

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                  3 days ago, # ^ |
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                  First submission on the page: 198531289.

                  The Div1 guy is always right anyway.

                  In this case, sure. Keep digging yourself a hole though.

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                  3 days ago, # ^ |
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                  Man, how stupid are you? I said more than half of them did I say the very first one?

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                  3 days ago, # ^ |
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                  Pretty much every single submission on that page is cell by cell. Stay grey though.

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                  3 days ago, # ^ |
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                  Lol, how'd you figure I'm grey? Oh, no! Anyway, this conversation isn't going anywhere and has diverted from my initial argument. Ths solution wasn't a generalization, and I thought the solutions moving cell by cell will TLE that's why I used the word PROBABLY (you can use a dictionary). Have a bad day!

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                  3 days ago, # ^ |
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                  The right move would be for you to admit you were wrong and move on. Keep digging your hole though!

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9 days ago, # |
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Hello, why is my code for problem D giving WA on test case 4? The logic should be correct and I cannot think of any bugs. Is there something I have failed to account for? There shouldn't be any issues with int overflow I believe.

Code

Thank you all very much in advance.

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    9 days ago, # ^ |
    ← Rev. 2   Vote: I like it +8 Vote: I do not like it 
    1
3 1
1 1 1
1 3 999999999

    Your code outputs "NO" instead of "YES"

    The error is that l, r, and k are ints, so temp += (r-l+1)*k first evaluates (r-l+1)*k as an int, which overflows.

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      9 days ago, # ^ |
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      Thank you very much for the help, it worked. Unfortunate that I missed that in contest.

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    9 days ago, # ^ |
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    I think (r-l+1)*k can overflow

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9 days ago, # |
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how is e related to dp???

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    9 days ago, # ^ |
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    if you sort array, then:
    dp[i][s] = if you can make sum S using the first i elements

    base case:
    dp[1][1] = 1 (and check if a[1] == 1)

    transition is:
    dp[i+1][s] = dp[i][s] | dp[i][s-a[i]]

    Observation is that if some sum t can be created using first p elements, then it's also possible to create any sum <= t. So we can remove one dimension from the dp and store only the max s for which dp[i][s] = true.

    This is actually how my reasoning went during the contest.

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      9 days ago, # ^ |
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      Can you prove this observation?

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

        proof by induction:

        Define s[i] max sum that can be created from first i elements.

        It's true for i == 1, because a[1] == 1 and it's possible to create sum s[1] = 1 only

        Assume it's true for i, now prove that it's also true for i+1. If a[i+1] is greater than s[i], then the answer is NO, so assume a[i+1] <= s[i]. There are 2 cases. To create sum x <= s[i] we can do so without including current element, because it's already possible to do so using previous elements. To create s[i] < x <= s[i] + a[i+1] we can use current element, and create sum x - a[i+1] from previous elements. Creating sum x > s[i] + a[i+1] is impossible, because x - a[i+1] > s[i] and (by definition) s[i] represents maximum sum that can be created using first i elements. So s[i+1] = s[i] + a[i+1]

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          9 days ago, # ^ |
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          Forgot to mention why sorting is necessary:

          A number can only be constructed using numbers less or equal to it, because no negative numbers are allowed

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    9 days ago, # ^ |
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    Have you tried binary search for problem E?

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