interlude's blog

By interlude, history, 3 weeks ago, In English

Hello Codeforces!

CQXYM and I are glad to invite you to Codeforces Round #745 (Div. 1) and Codeforces Round #745 (Div. 2), which will be held on Sep/30/2021 13:15 (Moscow time). Note the unusual time of the round.

Each division will have 6 problems and 2 hours to solve them. All problems were written and prepared by CQXYM and me. The round will be rated for both divisions.

We would like to thank:

This is our first round, and great efforts have been put into preparing this round. Were you to kindly participate in this round, we would be very grateful and hope you will enjoy it.

Good luck!

UPD: Here are the scoring distributions:

Div. 1: $$$500$$$ — $$$1000$$$ — $$$1750$$$ — $$$2000$$$ — $$$3500$$$ — $$$3500$$$

Div. 2: $$$500$$$ — $$$1000$$$ — $$$1500$$$ — $$$1750$$$ — $$$2500$$$ — $$$2750$$$

And we would also like to thank Um_nik for testing the round.

UPD2: Editorial is out!

UPD3:

Winners

Congratulations to the winners:

Div1:

  1. medium_waxberry

  2. maroonrk

  3. Radewoosh

  4. Karry5307_AK_NOI2021

  5. hos.lyric

Div2:

  1. _ywy_c_asm_

  2. lpf666

  3. Fuzen

  4. leap_frog

  5. OguriCap

 
 
 
 
  • Vote: I like it
  • -1219
  • Vote: I do not like it

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3 weeks ago, # |
Rev. 2   Vote: I like it -155 Vote: I do not like it

As a tester I would encourage you to participate in this round and assure you that the problems are interesting and good .

Update :- My review to testers

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3 weeks ago, # |
  Vote: I like it +27 Vote: I do not like it

Here it is!

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3 weeks ago, # |
  Vote: I like it -57 Vote: I do not like it

Mathforces?

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3 weeks ago, # |
Rev. 2   Vote: I like it +40 Vote: I do not like it

Another Chinese round and friendly time for Chinese! Looking forward to great problems and having fun!

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3 weeks ago, # |
  Vote: I like it -61 Vote: I do not like it

Sad, why isn't 1-gon a VIP tester :(

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    »
    3 weeks ago, # ^ |
      Vote: I like it +170 Vote: I do not like it

    I'm only moderately important this time.

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    3 weeks ago, # ^ |
      Vote: I like it -81 Vote: I do not like it

    What? 1-gon's comment didn't hit 100 upvotes in a day?? Impossible!)

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

This time is good for me.

Orz CQXYM!

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3 weeks ago, # |
Rev. 2   Vote: I like it -58 Vote: I do not like it

Second Shortest announcement after Tourist's Round announcement Since I started using CF.

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3 weeks ago, # |
  Vote: I like it -50 Vote: I do not like it

Bit manipulation problems?

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

I hope.

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3 weeks ago, # |
  Vote: I like it +65 Vote: I do not like it

I don't have anything against Chinese rounds, but... really, morning on a weekday?

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    »
    3 weeks ago, # ^ |
      Vote: I like it +154 Vote: I do not like it

    Probably people in different time zones exist.

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    »
    3 weeks ago, # ^ |
      Vote: I like it +30 Vote: I do not like it

    In China it is 6:05p.m. on the day before the National Day holiday. You can imagine the experience of participating a CF round after school :)

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

      But (maybe?) most middle school students have to stay at school and study at night.

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

        It often happens on weekdays. But if tomorrow is holiday, students can go back home in the afternoon.

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    »
    2 weeks ago, # ^ |
      Vote: I like it +23 Vote: I do not like it

    I think that it's because the National Day holiday in China is coming.

    However authors don't know 6 p.m. is even worse than 10 p.m. for many Chinese because most of schools end at 9 :(

    • »
      »
      »
      2 weeks ago, # ^ |
        Vote: I like it -34 Vote: I do not like it

      However we must take the fact that ICPC World Final will be held today into consideration. Thus we choose 6 p.m. over 9 p.m.

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3 weeks ago, # |
Rev. 2   Vote: I like it +7 Vote: I do not like it

Rare time....but maybe just right for Chinese contestants like me. I'm a little EXCITED. Hope all of us can enjoy it!

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

I finnish school 5 minutes before the round, can't join, but I wish those who read the greatest of luck.

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    »
    3 weeks ago, # ^ |
      Vote: I like it +71 Vote: I do not like it

    I managed to convince the teacher to let me compete at school.

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

Big thanks for the time of the round from Asia! :)

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

Great, 3:05 AM.

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

    i guess that's an i for an i for us having to participate from 10:30pm to almost 1am before...

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3 weeks ago, # |
  Vote: I like it +27 Vote: I do not like it

interlude I think we need to update this blog to make this appear on top of the div.3 contest. Many people might not see it. Especially the unusual start time of the round. I didn't see this blog existed. Because the first post from the top was round 744. thanks for considering. :) good luck to the people who are participating.

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3 weeks ago, # |
  Vote: I like it +72 Vote: I do not like it
lighthearted meme
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    »
    3 weeks ago, # ^ |
      Vote: I like it -15 Vote: I do not like it

    For me College classes <<<< Codeforces round.Ha agar test ya lab wagira ho tab bat alag hai

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

      You High on Weed . :) __ | | | | | | | | ( | | | | )

      Hope you Understand my feeling.

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

      That's not a good idea to play jokes like this on the Internet. We can sure that there was something wrong with the education you get from your parents.

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        2 days ago, # ^ |
          Vote: I like it -15 Vote: I do not like it

        заткнись телочка.еще что-то скажешь про моих родителей, зубы выбью тебе, тварь.

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    »
    3 weeks ago, # ^ |
      Vote: I like it +66 Vote: I do not like it
    Meanwhile me
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    2 weeks ago, # ^ |
      Vote: I like it +31 Vote: I do not like it
    Me:
    • »
      »
      »
      2 weeks ago, # ^ |
        Vote: I like it +1 Vote: I do not like it

      Now that the contest is over, are you happy with your decision?

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

        i guess yes as he got a very good rank in this one

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

    I have a test in my class today that completely overlaps with the contest timings. Cant skip them even if I want to

    :sobb:

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

Another great round!! Hope I can do better than before in it.

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

Hi there. Could the contest be at 18.35?

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

    The round authors are Chinese; I assume starting at 22:35 is not comfortable to them.

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

Hoping to become Pupil after this round...Best of Luck everyone!!

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

As a Chinese,I usually leave school at 6:30 PM,so I cannot enjoy this round :(

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

    Leave school, but don't leave a contest. (Don't take it seriously.)

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

    And now I think I am lucky :D.

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3 weeks ago, # |
  Vote: I like it -26 Vote: I do not like it

i do remember that one of the past contest nearly at the same time as of this one and it went bad ...........

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

Can you somehow postpone the competition for 2 — 3 hours? I think it would be more convenient to write everything in the evening than in the afternoon. usually the competitions are held at 17:30 Moscow time and this is very convenient.

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

Is 1900-2100 no longer in Div 2?

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

    People rated <2100 can participate in div2, when it is div2Only round(so no parallel div1), but when there is a parallel div1 round, 1900+ have to participate there. It has been like this for a long time

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2 weeks ago, # |
  Vote: I like it -51 Vote: I do not like it

A way to increase contribution from 0 to 100:

0-20: leave stupid comments

20-40: write a blog on how to increase contribution from 0 to 20

40-60: write a blog on how to increase contribution from 20 to 40

60-80: write a blog on how to increase contribution from 40 to 60

80-100: write a blog on how to increase contribution from 60 to 80

  • »
    »
    2 weeks ago, # ^ |
      Vote: I like it -9 Vote: I do not like it

    A way to increase contribution from 0 to 126:

    0-126: leave stupid comments

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2 weeks ago, # |
  Vote: I like it -9 Vote: I do not like it

Big thanks for the friendly time of the round:)

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

Starting My Codeforces Journey From your round...

Hope to have my first giving round and your first making round great!

Thank you

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

My first div2 round here. Hopefully it doesnt go downhill. fingers crossed. I like this time slot :)

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

All the best to all the participants. :))

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

From contest authors and testers, I'm waiting really good contest!

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2 weeks ago, # |
  Vote: I like it -118 Vote: I do not like it
Greeting all chinese people!
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    »
    2 weeks ago, # ^ |
    Rev. 3   Vote: I like it -27 Vote: I do not like it

    Idk what's up with everyone, so retracting it for the better!

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    »
    2 weeks ago, # ^ |
      Vote: I like it +38 Vote: I do not like it

    I think the content you posted is not appropriate, because we all know that the Great Firewall in Mainland China blocks almost all "user-generated content" websites, and codeforces as a "user-generated content website" is not blocked by it because There is very little political content on codeforces. If you post more content in this area, the result is likely to be that codeforces is blocked by the Great Firewall of China. As a result, it will cause great inconvenience to users in mainland China. They need to cross the Great Firewall to access codeforces.

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      »
      2 weeks ago, # ^ |
        Vote: I like it -14 Vote: I do not like it

      Good.

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      »
      »
      2 weeks ago, # ^ |
      Rev. 2   Vote: I like it -87 Vote: I do not like it

      However, according to some netizens in the QQ group Universal OJ Users, almost everyone in China mainland who use Codeforces can across the Great Firewall

      But anyway, I don't think any political content posted on codeforces is appropirate because codeforces is a website for people who love programming. I think whether it will make codeforces blocked or not it's not good to post content which is not about algorithms on codeforces

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      »
      »
      2 weeks ago, # ^ |
      Rev. 2   Vote: I like it -121 Vote: I do not like it

      Although I heard some netizens say, and I also think whether Codeforces is blocked or not doesn't matter to many codeforces users, I think after all using websites not blocked is more convenient.

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    »
    2 weeks ago, # ^ |
      Vote: I like it +72 Vote: I do not like it

    Lol, this is pretty good, don't know why it's being downvoted

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

can anyone tell me what is the process of penalty in DIV 2 round

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    »
    2 weeks ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    Technically, there are no 'penalties' on time count. But yes, you get less points if you solve a problem after incorrect attempts.

    Each problem is assigned points (e.g., Problem A in this round is 500), the faster you solve a problem, the more points you get. The amount of points you can achieve gradually decreases over time.

    Problem A points deteriorates at a rate of 2 points per minute (you get 498 points if you solve after 1 minute, 496 points if you solve after 2 minutes, 494 points if you solve after 3 minutes and so on)

    Plus, every incorrect attempt costs you 50 points.

    Now let's see, if you solve that problem A after 10 minutes on your 2nd attempt, you get (500) - (10*2) for 10 minutes to solve - (50) for one incorrect attempt = 430.

    If you solve that problem A after 15 minutes on your 3rd attempt, you get (500) - (15*2) for 15 minutes to solve - (50*2) for two incorrect attempt = 370.

    Additionally, the minimum points you get on any problem is lower bounded by 30% of the max points on that particular problem. e.g., irrelevant of how many incorrect attempts or how much time, you will get a minimum of 500 * 30% = 150 for that problem A.

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

Round delayed by 10 minutes

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

Thanks for wishing us good luck

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

Time added 10 minutes!

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

contest extended by 10 min!

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

Why this sudden addition of ten minutes in starting the round??

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2 weeks ago, # |
  Vote: I like it +69 Vote: I do not like it

Hopefully the delay doesn't signify an unrated round.

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2 weeks ago, # |
  Vote: I like it +87 Vote: I do not like it

now what to do with 10 min of my life??

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

how many point i need to join to div 1 ???

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

I hope it'll be the only problem of the contest.
Wish all the best!!!

P.S. The "problem" here doesn't mean the problems that we're going to solve in the contest.

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2 weeks ago, # |
  Vote: I like it +90 Vote: I do not like it

This doesn't really matter, but why would you use "obsidian" instead of "black"?

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2 weeks ago, # |
Rev. 4   Vote: I like it +190 Vote: I do not like it

If you can't write a nice problemset for cf, just move the problemset to other websites.

Terrible contest.

edit: Chinese slang

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    2 weeks ago, # ^ |
      Vote: I like it -15 Vote: I do not like it

    ?______?

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

      It means he hates the idea provider due to hard problems, but it may be Chinese slang that is very disruptive and not appropriate to show in public.

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

First time not being able to solve (or not wanting to implement such an ugly solution for) even the first problem. Might just be me, but I think this contest needed a lot more tuning.

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    2 weeks ago, # ^ |
    Rev. 2   Vote: I like it -49 Vote: I do not like it

    Thought it was one of best contests I competed in, besides B and C out of order. Finally some actual implementation and more ideas than just greedy/math with one for loop required...

    If you can't solve a 30 line solution due to implementation I'm not sure you should be in div1. I've long said there is severe lack of implementation skill for most low div1 users and below.

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

Hardforces (: (at least for me )

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

Coding is very hard

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

div.1 too hard for me :(

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2 weeks ago, # |
  Vote: I like it +154 Vote: I do not like it

Why would you write statement like this:

"and the diameter of the graph must be less than $$$k-1$$$?"

Couldn't you write "diameter of the graph must be less than $$$k$$$"?

Or even better, "diameter of the graph must be less than or equal to $$$k$$$?

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

    For real, my code was so messed up I had to rewrite because I got confused in the middle.

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2 weeks ago, # |
Rev. 5   Vote: I like it +27 Vote: I do not like it

I'm sorry but i am happy because today it's not only me who can't solve a single problem.

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2 weeks ago, # |
  Vote: I like it +90 Vote: I do not like it

How to unregister?

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

Not my cup of tea...

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

Damn bro you didn't have to kill us with those problem statements!

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

Man!! explain me the problems. Cant understand anything from the statements

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

its my 2nd chinese round and i find it harder the usual div2 coz i can solve avg 2 problem but in this round i can't do even A. Chinese loves maths too much ≧ ﹏ ≦

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

only 3.7k submissions to problem A, btw WAonpretest2forces :(

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2 weeks ago, # |
  Vote: I like it +72 Vote: I do not like it
At least I'm not alone, here we are :')
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    »
    2 weeks ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Getting demoralized when not able to solve any problem.

    then coming to comments-->turns out to be a natural therapy

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

Not even able to solve one quest, very tough round.

but atleast doing no submission won't affect my rating ^-^

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

    I thought if you presented, no matter have you submit or no, It will affect your rating?

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

The wording of some of the problems are hard to understand :(

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2 weeks ago, # |
Rev. 3   Vote: I like it -8 Vote: I do not like it

Bruh, I don't even feel like trying C cause pretty sure it's beyond my capabilities.

Spoiler
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2 weeks ago, # |
Rev. 2   Vote: I like it +8 Vote: I do not like it

In the last 12 rounds held over the past 3-4 months, this is the only round where I had to spend ~10 minutes on both A and B just to be able to interpret properly what the question wants to say, never mind attempt to solve them. Statements were weirdly put, and maybe difficulty wasn't properly calibrated going by the meagre amount of submissions ~4k for div2 A even after an hour.

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

statement of div2 c is beyond my scope :')

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

one silly mistake costed 4 penalties on problem B :(

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2 weeks ago, # |
  Vote: I like it +398 Vote: I do not like it


Thanks a lot for clarifying this, before the round I was worried that my solution might also be tested on a negative number of test cases.

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

I am even not able to solve B this time, my brain exploded thinking of it :(

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

    I am not even able to solve A :/

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2 weeks ago, # |
  Vote: I like it +155 Vote: I do not like it

A flaw in the rating system exposed brutally by this round: in div 1, those who do manage to solve A, but do so late and/or with penalty attempts, will see their rating fall significantly, while those who bail out completely on solving A will see no change to their rating.

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    2 weeks ago, # ^ |
      Vote: I like it +73 Vote: I do not like it

    that's why A should always be comparatively easy in both divisions, can you imagine less than 6k participation in div2 :(

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2 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it
Me on this contest be like:
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2 weeks ago, # |
  Vote: I like it +27 Vote: I do not like it

Damn, That's the hardest A and C I've ever seen

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

HardForces :(

UPD: Sad, Got FST on Div.2 B :(

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

Today I will get -101 rating change :(

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

It`s really good, but is so hard:(

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

20000 registers 6000 contestants hmmm makes sense

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2 weeks ago, # |
Rev. 2   Vote: I like it +15 Vote: I do not like it

Didn't participate in a Div. 0 before, but today i did.

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

I Just guessed the solution for A after 1hr 20 mins of waste analysis and it worked :/ and then solved B under 30 mins , strange round

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

Damn, couldn't even solve A.This contest really was amazing!

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

Is O(N^3) the intended solution for Div1 A or are you expecting O(N^2)?

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

    $$$O(N^3)$$$.

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

      Amazing. I had an O(N^3) time with O(N^3) space initially which got MLE.

      Optimized it to O(N^3) time with O(N^2) memory and still got TLE.

      Tried optimizing it further and got WA.

      What a waste of 2 hours for me.

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

        I solved it O(N^3 * 16) it gave me TLE but i break when i find solution with value greater than 16 and passed the pretests

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        2 weeks ago, # ^ |
          Vote: I like it +25 Vote: I do not like it

        Atleast you were writing codes. I was just crying for 2 hrs T_T

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

    Yes, my O(n^3) passed

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

    I think O(n^3)

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

Are the CF div2 rounds are getting harder or me getting dumber?

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    2 weeks ago, # ^ |
      Vote: I like it +19 Vote: I do not like it

    You just happened to be in the wrong contest at the wrong time

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

    I am trying again on CF after spending a long time on Leetcode and DSA style problems. I am thinking to quit. It feels like a waste of time now but I am interested too lol. I don't know. I can spend this time on learning other skills like more backend. Will maybe just try to solve some harder problems for my range in the next few days.

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

A contest with really high difficulty...

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

So is there anyone tell me how to solve Div.1A/Div.2C. I got WA on pretest 18 :(

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

    You can discover the property that the answer will not be more than 16. and just do it in brute force. it will be O(n * m * 16 * 16). Hint : try to use the number larger than 16, maybe 17.

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

      I found that but failed to have further a further thought. Could you please tell me what the (16 * 16) is?

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

    You are just checking min answer for the 5x4 matrix. But there may be test cases like 8x8 size 01111110 10000001 10000001 10000001 10000001 10000001 10000001 11111110 In this case, the answer is 0. But according to your approach, the answer will come to more than zero.

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

Any O(nlogn) solution for Div1C/Div2E? I only came up with sqrt decomposition XD

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    2 weeks ago, # ^ |
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    wasn't time limit a little too tight for sqrt decomp? :(

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

      O(nsqrt(nlogn)) should work for n=100k.

      Example: 1540D - Inverse Inversions

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

        It has 5 sec limit. 1 sec limit kept me under the impression that nsqrt(n) is being punished here. anyways my fault :(

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

        Not in this task, not with 1 second time limit, IMO. 1.5 seconds would probably be enough.

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

      I passed system tests with $$$O(n\cdot \sqrt{n}\cdot \log{n})$$$.

      I used a fenwick tree for big values of $$$x[i]+y[i]$$$ instead of just maintaining when to add or subtract something.

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

        Weak systests :/ But I see that this solution was already hacked after contest. My solution with FT was a bit too slow too.

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    2 weeks ago, # ^ |
      Vote: I like it -84 Vote: I do not like it

    I can only come up with $$$O(n\sqrt{n})$$$ solution too.

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    2 weeks ago, # ^ |
      Vote: I like it +26 Vote: I do not like it

    The TL may be a bit tight for O(N sqrt(N))

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      2 weeks ago, # ^ |
        Vote: I like it -126 Vote: I do not like it

      In fact the constant of the solution is very small. I think even $$$O(n\sqrt{n}log_n)$$$ solution can also pass.

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

        And O(n^5) can pass B with some optimization...

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          2 weeks ago, # ^ |
            Vote: I like it -130 Vote: I do not like it

          $$$O(n^5)$$$ is the expected complexity.

          In fact you will find the constant is extreme small. A lot of state is useless.

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            2 weeks ago, # ^ |
              Vote: I like it +36 Vote: I do not like it

            That's really unexpected.. I thought author may have come up with something smarter to get $$$O(n^4)$$$ solution and set $$$n=100$$$ to cut $$$O(n^5)$$$ one..

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

              In fact there do have a solution works in $$$O(n^3log^2n)$$$, but it's too difficult for D1B.

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            2 weeks ago, # ^ |
              Vote: I like it +146 Vote: I do not like it

            Well, that's surprising. Why didn't you set $$$n \leq 50$$$? It would be much easier to understand that the solution is fast enough and it's worth coding. Was it that important to fail $$$O(n^6)$$$ solution?

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

              Um_nik played these games.

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              2 weeks ago, # ^ |
                Vote: I like it -65 Vote: I do not like it

              Sorry but I didn't come up with any $$$O(n^6)$$$ solution.

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                2 weeks ago, # ^ |
                  Vote: I like it +107 Vote: I do not like it

                Then my question remains: why $$$n \leq 100$$$? Only to make some people not implement a correct solution?

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

                  "In fact the constant of the solution is very small." -interlude

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

                  I understand that the solution is fast enough. It would still be with $$$n \leq 50$$$. And this fact would be clear before implementing.

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

                  Just because the $$$O(n^5)$$$ solution can pass the task and it runs very fast. In fact you will find that the answer will be 0 if $$$m,k$$$ is larger than $$$\frac{n}{2}$$$.

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                  2 weeks ago, # ^ |
                    Vote: I like it +45 Vote: I do not like it

                  Not always, my $$$O(n^5)$$$ solution got TL.

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            2 weeks ago, # ^ |
              Vote: I like it +38 Vote: I do not like it

            Are u serious? Only because of this, I didn't code the solution and thought for an O(n^4) solution for almost 1.5 hrs. And the constant isn't that small, recursion won't pass easily without optimizations (returning wherever you can).

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

        However my O(N sqrt(N)) solution didn't pass.

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

      I failed on both BC for TL

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

        I think your solution for C is slow because of "cin and cout" .

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

          I'm sorry but as a matter of fact,coder's submission has got another TL after replace cin/cout with faster IO streams. Isn't it obvious that this failure it due to implementing or Time Limit but not IO ways?

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          2 weeks ago, # ^ |
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          You are right..my bad

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    2 weeks ago, # ^ |
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    O(nsqrt(n)), you can operate trains whose x+y>=sqrt(n) and whose x+y<sqrt(n) in different way.

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2 weeks ago, # |
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Perfect! Problems are challengable and of benifits. especially C, it`s great on both Thinking and Realization. All in all, great round.

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2 weeks ago, # |
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Ooh man! one of the toughest A in div 2!!!

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

    maybe you can just look at sample test cases to find the pattern without thinking much or proving anything . At least that's what I did XD

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

      I figured it out with PnC but still I would say it shouldn't have had been an A problem, maybe B or B1 if possible!

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

    The hardest ever

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

So difficult.

UPD : also fstforces.

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2 weeks ago, # |
  Vote: I like it +48 Vote: I do not like it

Why is the diameter of the graph given as <=k-1, why not <=k ?

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    2 weeks ago, # ^ |
      Vote: I like it +32 Vote: I do not like it

    It's < k-1, My guess is just to add more confusion.

    it's like let's make the contest very hard and confusing for the people who decided to participate at 3 AM as they already have a brain fog at this time :D

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

Am I the only one who has a bad feeling about system tests?

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2 weeks ago, # |
  Vote: I like it +40 Vote: I do not like it

Well at least the judge queue will be short lol

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

Please, if you will write a round again in the future think about the quality of statements twice before. Some things were not very clear, like in div2 B where you should have specified strictly less, as you did while the round was going. Also, talking about B, it wouldn't be a bad idea to also write in bold important words, not just numbers and variables(here I'm talking about the word CONNECTED); you can see in other past problems that this is sometimes done, especially for important conditions like the connectivity one in B(for example for saying that all the numbers in an array need to be different). One last thing about this problem would be the limits for k, like were those special cases of k <= 1 really that important to be placed in the tests, ore are they even in the tests?

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

    I am really confused why the diameter of the graph must be strictly less than k−1. other than k, it causes unnecessary confusion.

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2 weeks ago, # |
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hardforce

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    2 weeks ago, # ^ |
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    don't worry, system_fail_forces also coming soon.

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2 weeks ago, # |
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Couldnt solve anything. PS: Writing bruteforce and searching on OEIS is not "solving".

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

I feel like today's contest is suitable for Div 1.5 and Div 0.5

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2 weeks ago, # |
  Vote: I like it +25 Vote: I do not like it

The round had nice problems (I read only div1A-C tho), though ridiculously hard. Could've easily been split into two separate div1 rounds with some easier, filler problems. I want to express my deepest sympathy to all div2 participants, who had those super tough problems given as C-E. Div1A should've been div2E IMO. To the authors, you did very good job with inventing interesting problems. Only that you underestimated their hardness and the overall contest turned out too hard.

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

    That is the job of coordinators and testers but I don't know how can they all agree to serve all these problems together.

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

Some solutions for Div2 B will fail for this test: n = 1, m = 0, k = 1

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    2 weeks ago, # ^ |
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    is it NO ?

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      2 weeks ago, # ^ |
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      Yeah, some solutions didn't consider that the given diameter could be negative as well.

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        2 weeks ago, # ^ |
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        In first sample he told u diameter of the graph is zero, its obv. not strictly less than 0

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

        Although I honestly don't see why the authors would want to make k start from 0 and not 1.

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        2 weeks ago, # ^ |
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        I felt,The definition of diameter is valid only when there are atleast 2 vertices. Is this assumption wrong?

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      2 weeks ago, # ^ |
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      Yes. The Answer will be NO

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

    YES

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

Honestly how to iterate on this map in c++ for Div1 C ? map[(xi + yi)][day % (xi + yi)][xi].push_back({st_j, end_j}) Plus is it really the solution ?

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

    Just use map.insert.

    The solution is about O(n sqrt m)

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

Maybe the pretest is too weak?

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

system testing is super fast today!

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

Despite so many testers spanning across different ratings, a pretty unbalanced round.

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

Saw a pattern in the ans of A for n = 2 , half of the total-permutations and it actually worked for larger n

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2 weeks ago, # |
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You could bold "connected" in B :/

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2 weeks ago, # |
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k = 0 on problem B. Damn you guys got me!!!

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2 weeks ago, # |
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Awesome difficulty Forces+FST Forces = Codeforces Round #745

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2 weeks ago, # |
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system testing gotta go fast

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2 weeks ago, # |
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Did someone mistakenly swap B and C?

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    2 weeks ago, # ^ |
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    Before the contest I thought B is easier but it turns out that I was wrong.

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

Who actually proved their solutions for A lol

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    2 weeks ago, # ^ |
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    lol haha lmao . rofl . so funny .

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    2 weeks ago, # ^ |
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    Not me lol

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    2 weeks ago, # ^ |
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    I thought of something like, if a permutation is valid, it's reverse is always invalid. So half would be valid and half invalid

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

    Proof by symmetry is valid, and very simple.

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    2 weeks ago, # ^ |
    Rev. 2   Vote: I like it +8 Vote: I do not like it

    I think you can make a one-to-one mapping from valid permutations to invalid ones; let's say you want to count permutations which have exactly $$$n$$$ indices such that $$$p[i]<p[i+1]$$$; it's not hard to show that by negating all numbers and then adding $$$2n$$$ to all of them, you are going to have a permutation with exactly $$$n$$$ indices such that $$$p[i]>p[i+1]$$$ (note that we don't have any repeated numbers, so this condition is the exact opposite of the one we consider valid.) Let's assume $$$\Pi(i)$$$ is the count of all permutations that have $$$i$$$ consecutive increasing pairs, and $$$\neg\Pi(i)$$$ the same for decreasing pairs; we know a pair is either increasing or decreasing; so we have: $$$\Pi(i)=\neg\Pi(2n-1-i)$$$; so we have: $$$\neg\Pi(i)=\Pi(2n-1-i)$$$; therefore, we have $$$\sum_{i=0}^{n-1}\Pi(i)=\sum_{i=n}^{2n-i}\Pi(i)$$$; I think by now you can see why $$$\frac{n!}{2}$$$ works in this problem.

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2 weeks ago, # |
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What is test 4 for problem Div2B ?

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2 weeks ago, # |
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I came up with a (n^2)m solution for Div1 A but it gave TLE. Is there a more efficient solution?

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    2 weeks ago, # ^ |
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    I also have this exact complexity and it passed. I used only vectors and pairs, maybe you used some more complex data structures?

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      2 weeks ago, # ^ |
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      Hmm. Seems like the constraints were tight. I see many solutions giving TLE in the system test.

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2 weeks ago, # |
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But it actually feels good to see a tough round now and then.

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How to solve Problem C ?

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2 weeks ago, # |
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hardforces + clumsyforces + (1.5 + 0.5 ) forces +system test fail forces + ....

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2 weeks ago, # |
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It is so reassuring to see that the problems were hard for everyone! Nevertheless, they were also really engaging.

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2 weeks ago, # |
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important rule : don't join chinese contest LOL

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2 weeks ago, # |
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FSTForces :(

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2 weeks ago, # |
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shitty contest

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

I think the pretest in Div 2 B could have been a little better. Other than that, contest had really interesting problems, except C. I didn't like C, as the only solution that I thought of was implementation heavy, so I simply skipped it and moved ahead. E was good. Just missed the window for submission.

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

Annoucement-cum-Hardforces

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2 weeks ago, # |
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everything in this contest was bad. even the problems.

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2 weeks ago, # |
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Is it possible to solve Div1B faster than $$$O(n^5)$$$?

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    2 weeks ago, # ^ |
      Vote: I like it -67 Vote: I do not like it

    With 2D FFT we can solve it in $$$O(n^3log^2n)$$$.

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      2 weeks ago, # ^ |
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      I wondered if it's possible during the round. Nice :|

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      2 weeks ago, # ^ |
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      How? Don't you need division for that?

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        2 weeks ago, # ^ |
        Rev. 6   Vote: I like it 0 Vote: I do not like it

        let dp[ k ][ i ][ j ] denote number of sequences of length i that have exactly j elements with k maximums (e.g. as in the number of maximums in the process defined in statement, e.g. the variable m). Then dp[ k ][ i ][ j ] is the sum over all i1 , i2, j1 , j2 , such that i1 + i2 == i-1 , j1 + j2 == j of ((i-1 choose i1) * dp[k-1][i1][j1] ) * dp[k-1][i2][j2]. This thing can be done with 2d FFT. E.g. we are fixing the position of the maximum element.

        Edit: right, the modulo isnt guaranteed to be prime, not too sure.

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          2 weeks ago, # ^ |
            Vote: I like it -12 Vote: I do not like it

          No, its wrong.

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

          1) it should be i1+i2=i-1, though that's pretty easy to handle

          2) you forgot an (i-1 choose i1) factor, which requires dividing by i! before the convolution.

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            2 weeks ago, # ^ |
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            ohh right

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

            I am not too sure i understand part 2). Do you mean calculation of (i-1 choose i1) mod p needs division ? If so, we can pre calculate that with pascal's triangle equality.

            Edit : I see, makes sense.

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              2 weeks ago, # ^ |
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              No; you can divide by i! before convolution so you just do m convolutions of size nk; unless I'm missing something, directly calculating (i-1 choose i1) requires you to do mn^2 convolutions of size k, which is slower than the complexity interlude mentioned.

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2 weeks ago, # |
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How to solve A ?? Intution ??

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    2 weeks ago, # ^ |
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    just apply bruteforce and and think what will be the answer when n =3,4,5

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    2 weeks ago, # ^ |
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    Most people just deduced it from the given sample cases

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    2 weeks ago, # ^ |
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    $$$ans=\begin{cases}1&n=1\\\prod\limits_{i=3}^{2n} i&n\geq 2\end{cases}$$$
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      2 weeks ago, # ^ |
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      why ?? i need intution not just the plain answer :_:

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        2 weeks ago, # ^ |
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        You can find this rule by directly calculating the result when $$$n=3,4,5$$$

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

        It's in fact $$$\frac{(2n)!}{2}$$$. For any permutation with $$$k(k>=n)$$$ such $$$i$$$, you can reverse the permutation and get another unique one with $$$2n-1-k(2n-1-k<n)$$$ such $$$i$$$. So answer should be half of number of permutations.

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

    the answer is (2*n)!/2, because each permutation whose the number of pi < pi+1 is no less than n corresponds to another permutation whose the number of pi < pi+1 is less than n after you reverse this permutation

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

    its simply a combinatrics problem ans : (2*n-1)!*(n) approach: for n there will be 2*n numbers and we can have no more than n positions with pi<pi+1 so we can place number n anywhere from pos (n+1) to 2*n then we can achieve the given condition

    code
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2 weeks ago, # |
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Tight ML in div2 C ;(

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2 weeks ago, # |
  Vote: I like it +57 Vote: I do not like it

Sometimes I miss the old times when this assumption at least sometimes was incorrect.

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

    The data of this task is actually difficult to make, and I can't come up with all the wrong solution too. Maybe we could communicate more about this task.

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

      maroonrk what's incorrect in your code?

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

        My solutions is like this:

        The essential part of this problem is to make in-degrees of all vertices except the root no less than 2. For each vertex $$$i$$$, we know to which vertex $$$j$$$ we can span an edge. Valid $$$j$$$'s form a set of ranges if we order vertices by their distances from the root, and the number of ranges is at most the degree of the vertex $$$i$$$. My solution consumes O(number of ranges of the queried vertex) time, so it's a $$$O(NQ)$$$ time algorithm. However, it passed somehow.

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

          Well, doesn't look like "we can't come up with all the wrong solution".

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2 weeks ago, # |
Rev. 2   Vote: I like it +18 Vote: I do not like it

I didn't enjoy this round very much,not only because I got a negative delta.

(For div2) The problems are way harder than a typical Div2 round,Problem ABC are not very suitable for their place and Problem DEF are so hard that very few people can solve them during the 2h contest.

Also the statement is kinda hard to understand(maybe it's my fault for not reading the statement that carefully but).You have to read it mutiple times to find certain words like "connected" or figure out what the problem is asking.

Contests are used to seperate contestants into two parts:Those who are very good at Problem Solving and those who are not that good.They are NOT used to find out who are better at English reading and typing.

upd: obviously the pretests are weak too but I blame myself

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

one of the worst contests with worst pretests with hard problems

at all, you won the worst contest

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

how to solve c?

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

Every Time I think I am now good enough to become CM , My Rating Starts Falling Until it reaches Specialist Range

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2 weeks ago, # |
Rev. 5   Vote: I like it +59 Vote: I do not like it

Sincerely, this round is not a good round. I have taken nearly 40 contests, and this contest is the first contest that I downvote.

1, the problem are so unbalanced, for div1, so many people are not able to solve A, and for div2, difficulty gap between B and C are too large, so as C and D.

2, div2 B is the worst problem I have ever met. It is so tricky, and description are soooooooo uncomfortable. What is "less than k-1" mean? Could you just write less than k" or even better "no greater than k". "n=1" is meaningless because we cannot pick two points from n=1, and "k=0 or k=1" is even more meaningless because, who want to discuss a graph with negative k?

3, div2 A are so unfriendly to newbies.

4, weak pretests

5, Except for the balance and description, a good contest should be educated. Even I cannot solve the problem during the contest and get negative delta, it doesn't matter. We can learn something after the contest, and improve our ways of thinking or coding skills.

But in this contest, I just want to say, what the hell of this ?? just a bunch of mind tricks or brain teaser? Can I learn something in the contest?

In all, getting up at 5:30 AM and take part in such contest is the most wasteful thing I have ever done.

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

    systest is pretty weak too? I uphacked myself with a case with obvious 32-bit integer overflow...

    130380959

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

I'm wondering if the pretest is too weak. I see many people got FST on 1C

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

In Div2 B, what is the expected answer for case: n = 1, m = 0 and any k. I guess according to definition of diameter given in the problem statement, diameter is not defined for n = 1 (we need at least 2 nodes to define diameter). I think the answer for this case is ambiguous and the problem should accept both "Yes" and "No" as the correct answers for the above-mentioned case.

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

    totally argee

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

    I think they made it quite clear the if n=1 the diameter is 0 by putting an example with n=1 and a note that says in this case the diameter is 0.

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

      I agree but this thing should be mentioned in the problem statement itself, most people don't see the explanation of the sample test case.

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

Why condition in div2 B use less than k-1 instead of less than just k? It just creates some confusions.