GreenGrape's blog

Codeforces Round #461 (Div. 2)
By GreenGrape, 7 years ago, translation, In English

Hello, Codeforces! Haven't seen ya for a while :D

Codeforces Round #461 will take place on Wednesday, 20:15 MSK. Note that the round is scheduled a bit later than usual.

The round will be rated for all division 2 participants. Division 1 is heartly welcome aswell :)

My gratitude to Kolya (KAN) for round coordination, Grisha (vintage_Vlad_Makeev), Oleg (xen), AmirReza (Arpa) and Senya (craborac) for testing and Mike (MikeMirzayanov) for awesome Codeforces and Polygon. A special thanks goes to Dima (Dmitriy.Belichenko) and Camille (pseuda) for sharing a problem sketch with me.

The main character of this round will be Imp the playful monster. Yet the statements are guaranteed to be short :)

There will be six problems with the following scoring:
500 — 1000 — 1250 — 1500 — 2000 — 2750

GL & HF!

Note that the round has been moved 40 minutes later to avoid clashing with CSAcademy Round.

UPD. The contest is over. Well...

UPD. System testing is over! Editorial.

Congratz to the winners!

Div. 2:

  1. sorry-haghani
  2. I_hate_Shadow_Light
  3. SashaShlyapik
  4. Pojiloy_gibon
  5. Marisa

Div. 1 (unofficial):

  1. dotorya
  2. Vercingetorix
  3. dreamoon_love_AA
  4. kmjp
  5. Anadi

UPD. The complete editorial is out (D/F).

  • Vote: I like it
  • +674
  • Vote: I do not like it

Comments Comments (251)
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7 years ago, # |
← Rev. 2   Vote: I like it +44 Vote: I do not like it

Really loved your last round but hopefully this time, there will be a better difficulty distribution in the problems!

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    7 years ago, # ^ |
      Vote: I like it +74 Vote: I do not like it

    I tried to make it a bit more balanced this time :)

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      7 years ago, # ^ |
        Vote: I like it +45 Vote: I do not like it

      I am still trying to solve your C up to this day.

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

        Hey! This might help. Spoilers for problem C : Cave Painting

        Instead of checking whether the remainder left by dividing n by 1 to k are different or not, lets find the max K such that when n is divided by 1 to n, it gives different remainder all the times. For e.g. whenever n is even, K = 1

        Note that all these 1 to K will give remainders as 0,1,2,...K-1 as they are different and they don't have other choice.

        So just find K for the given n and check if it is greater than equal to k.

        Hope I am clear.

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

codeforces contests are always great and amazing , i always try to attend the contests , but my bad , may be i will not be able to participate in this contest due to some travelling stuffs , and it is much painful for me , competitive programming is my passion

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7 years ago, # |
  Vote: I like it +38 Vote: I do not like it

Ohh god CM problem setter again.

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    7 years ago, # ^ |
      Vote: I like it +48 Vote: I do not like it

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    7 years ago, # ^ |
      Vote: I like it -28 Vote: I do not like it

    How to download Test #15 of 650C Table Compression?

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      7 years ago, # ^ |
        Vote: I like it +1 Vote: I do not like it

      In input :

      1000 1000

      1002 1002 1002 1002..... (1000 times)

      In output :

      1000 1000 1000 1000..... (1000 times)

      Coz you were asking so many times.

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

        the other numbers must be n*m(1000*1000). And also if there is only 1002(1000*1000 times) the answer must be 1(1000*1000 times).

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

          oh but downloading the full test cases is not possible

          Moreover it is highly unlikely that you will be able to debug your code with such a large input /output file

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7 years ago, # |
  Vote: I like it -55 Vote: I do not like it

Is it rated?

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7 years ago, # |
  Vote: I like it +46 Vote: I do not like it

Obligatory comment: "Hope the problem statements are as short as the announcement"

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

Have seen you being quite responsive in the last contest. Keep up the good work, and well, wish a quality and balanced contest ;)

P/s: Yup, I still can't figure out the most optimal income when using Perun's ult xD

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7 years ago, # |
  Vote: I like it +13 Vote: I do not like it

Oh... This contest is even later than others T_T

Too bad I can't enter it because of... time zone problem...

Wish you guys good luck and high ratings anyway ^_^

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I hope my rating will be more than 1900. Then this contest is unrated for me. I don't like rated contest. My rating will down if I don't do well in this contest. CONTEST, CONTEST and CONTEST ... RATING, RATING and RATING ... TRAINING, TRAINING and TRAINING ... ......

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    7 years ago, # ^ |
      Vote: I like it -22 Vote: I do not like it

    if you cheat in this contest, this contest is unrated for you too, and if nobody discover it, you can have a high rating:)

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7 years ago, # |
← Rev. 2   Vote: I like it -22 Vote: I do not like it

Good luck to all

let's see, can i be a Candidate master at the end of this round or not ...

wish me luck... :)

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Good Luck!) Hope you will be a Candidate

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    7 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    why people doesnt like this post whats wrong about it

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      He didn't get cm :\

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        7 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        sorry,what is cm?

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        7 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        you're right, i was far away from my ideal performance last night ...

        too upset for it... :(

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

        well ...

        now i'm :)

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7 years ago, # |
  Vote: I like it -6 Vote: I do not like it

So sad that magic is over, can't troll with different colored grapes anymore :(

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I wondered what is different from participant division 2 and division1

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    7 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it
    -The Blue Color and all the colors below it are for Div2 contestants.
    -the purple color and all the colors above are for Div1 contestants.
    -Div1 guys can participate in Div2 Rounds unofficially.
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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

really awesome, wait for some blossom

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7 years ago, # |
  Vote: I like it +33 Vote: I do not like it

A prime number contest!

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7 years ago, # |
  Vote: I like it -19 Vote: I do not like it

not u again

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7 years ago, # |
  Vote: I like it +13 Vote: I do not like it

"Yet the statements are guaranteed to be short :)" this is the best announcement so far.

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7 years ago, # |
  Vote: I like it -8 Vote: I do not like it

Yet the statements are guaranteed to be short :)

Thank you very much:)

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7 years ago, # |
  Vote: I like it +11 Vote: I do not like it

Unusual early scoring!

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7 years ago, # |
← Rev. 2   Vote: I like it +23 Vote: I do not like it

"Codeforces и Polygon"

Learning some russian.

Learning everywhere. :)

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    7 years ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    Learning so much that you start to understand Croatian. (Source: COCI 2016/17 Round #6 solution, p. 4, line 8)

    Let Pa < Pb i k * Pa ≤ Pb < (k + 1) * Pa.

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

Hope another great problem set from this amazing head. Big chance to rise up. CF is too good to stay away from. :)

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7 years ago, # |
  Vote: I like it -7 Vote: I do not like it

Can anyone help me with one problem? I have good answer on test 2 on my computer, but on codeforces I get WA no matter what I do. Obviously compiler settings or something isnt same. Just visit my problem and check last submissions, problem C from few rounds before. 34983468

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Test ur program on custom invocation and u will undrestand why got WA.

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7 years ago, # |
← Rev. 2   Vote: I like it +14 Vote: I do not like it

It's quite strange no one noted that this round coincides with CSA round #68 tomorrow! Can't you just delay it 30 more mins? GreenGrape

UPD: the contest is delay 40 mins!

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    7 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    More like, CSA coincides with Codeforces Round 461.

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

      linguistically, is there a difference?!

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

      I mean, CSA consistently hosts contests only in that time slot.

      I guess it went under the radar since they start 30mins earlier until recently.

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    7 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    CF round moved by 40 minutes.

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

Glad to hear this: Yet the statements are guaranteed to be short :) Looking forward to it !!!

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7 years ago, # |
  Vote: I like it -11 Vote: I do not like it

How to download Test #15 of 650C Table Compression?

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    7 years ago, # ^ |
      Vote: I like it +16 Vote: I do not like it

    Okay, this should be the last time I see this anywhere.

    You cannot do that. At all.

    (Well, except if the problem setter provide the full package publicly).

    For convenience, all test cases are compressed in case they are too big, so large input/output files are near to complete omittance.

    Usually, if you get stuck with a big tests, it can only mean a few things:

    • You ran out of memory (MLE verdict).

    • You ran out of time / your algorithm was too complex (TLE verdict).

    • You handled the undefined behaviors poorly (WA/RE verdict).

    These tests are mostly used to tests how a code works against large input streams, and not to be corner cases, so please check your own source code for places that can be optimized, not just spamming the blogs all the time.

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7 years ago, # |
  Vote: I like it +34 Vote: I do not like it

Normal guy's Imp.

Codeforces member's Imp.

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7 years ago, # |
  Vote: I like it +117 Vote: I do not like it

Is it legal?

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    7 years ago, # ^ |
      Vote: I like it +34 Vote: I do not like it

    I registered him manually, just a prank ^^

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      7 years ago, # ^ |
        Vote: I like it +54 Vote: I do not like it

      How is registering someone in a contest a prank?

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

        It’s just weird so a little funny. Pseudo will be like “wtf, dude” .. hence.

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

      Seems someone also wanted to prank you and registered you manually :P See you on leaderboard :D

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        7 years ago, # ^ |
          Vote: I like it +5 Vote: I do not like it

        This will be the first time someone solve all problems in 1 minute :D

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7 years ago, # |
  Vote: I like it +11 Vote: I do not like it

I am gonna solve A, B and C today !

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7 years ago, # |
← Rev. 7   Vote: I like it +55 Vote: I do not like it

Wow :O Rounds

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    7 years ago, # ^ |
      Vote: I like it +15 Vote: I do not like it

    And I thought heaven didn't exist.

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

Two line summary of the announcement :
1. The statements are guaranteed to be short :)
2. There will be six problems with the following scoring: 500 — 1000 — 1250 — 1500 — 2000 — 2750

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

The time is a little late for me,but I am still glad to take part in the round.

It's really a good chance to train myself. :)

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I just want to say it's too late for Asian contestants . Hope I will stay alive for being such late......

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    One hour earlier, but yeah, I feel you, dude...

    12:15 AM in UTC+7 (Hanoi, Vietnam). Uh-oh...

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Hope I can become Expert!

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7 years ago, # |
  Vote: I like it -8 Vote: I do not like it

Sanwli saloni, adaayein manmohni Tere jaisi beauty kisi ki vi ni honi Thande ki botal main tera opener Tujhe ghatt ghatt main pee lon Coca Cola tu ROUND GONNA BE AMAZING !

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

    You Should Behave On website like codeforces , though fun is necessary , but people seach for good comments over here . I hope you understand it ! Good Luck for contest .

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7 years ago, # |
  Vote: I like it +1 Vote: I do not like it

i think i will be green this time,thank you GreenGrape.

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7 years ago, # |
  Vote: I like it -24 Vote: I do not like it

is it rated??

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

can't wait.....

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7 years ago, # |
  Vote: I like it +11 Vote: I do not like it

Again a round with near 7000 participation :)

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

I hope the long queue is fixed before the contest starts

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

    I hope the int vector is fixed setprecision(10) after the contest end ;!

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7 years ago, # |
  Vote: I like it +1 Vote: I do not like it
How's this possible?!
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    7 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    What's wrong? I thought you can lock the problem as long as you have the "Pretest passed" verdict, irrespective of whether you get hacked or not. Correct me if I am wrong.

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

      Nope, you can't lock after you got hacked. But I think it happened because times are very close.

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        7 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Oh okay, that's interesting. This probably happened because of the long hack queues today.

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7 years ago, # |
  Vote: I like it +8 Vote: I do not like it

very long queue in hacks

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7 years ago, # |
  Vote: I like it +6 Vote: I do not like it

HackerRank will be proud of this contest

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7 years ago, # |
  Vote: I like it +27 Vote: I do not like it

I think this contest are more "Successful hacking attempt" messages than "Accepted"

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

A round full of hacks. Even one of my friends hacked me, nice contest

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

Hackforces in a nutshell...

Anyway, to be serious, maybe it was just me, but I think the scoring is still a little bit underrated. This should be 500-1000-1500-1750-2250-3000. C is sufficient to be called a Div2C, D is somehow quite twisted (got 4 WAs verdict on the 4th problems, all on pretest 4).

Not really sure about my opinion about E and F, coz' I have just read it for a short while.

After all, it was fun actually. And these Imps are just cute ^^

UPD: I took my words about D back. Somehow I wish I could think sharper...

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

Can someone please provide explanation for third testcase of problem E.

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    7 years ago, # ^ |
      Vote: I like it +18 Vote: I do not like it

    I think you are confused just like I was — summoning a bird doesn't increase your mana, it increases you maximum mana.

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

    Take 1 bird from the first nest (you don't have enough mana to take two birds), take 10 birds from second nest (your mana capacity is 17 and you have 15 mana so you can take them all).

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Problem D is solvable with radix sort?

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

    I just used normal sorting with a comparing function that returns numS[a] * numH[b] > numS[b] * numH[a]

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Hell man! :'( I tried to do it with normal comparison too but that condition didn't pop up to my mind! How can you prove this works ?

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

        Given a string s, denote f(s) to be the number of subsequences 'sh'.

        Now given strings s1, s2, ... sn, it is easy to prove that if we denote xi, yi as the number of 's' and 'h' in string si, that .

        To maximize our desired value, all we need to look is the value .

        For n = 2, it's trivial that the comparing function wewark described works.

        We will induct on n — say the comparing function works for n and we prove it works for n + 1.

        Clearly, , as after deciding which sn + 1 to choose, we are left with optimizing the value with n strings. (induction hypothesis is used here)

        Now, all we need to do is prove that we need to put sn + 1 as the string with the highest value. This is simply greedy, easy to prove.

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

The 1st problem made for the hack..

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7 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Idea behind C ?

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    For small Ks (probably <= 1000) you will find two exact values, so just implement

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    just do it with brute force :)

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Ok. Limits are very large so how do you guarantee fast running time ?

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        7 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        it will find a pair soon

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

          Thats what my questions is. How to prove that ?

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            7 years ago, # ^ |
            ← Rev. 2   Vote: I like it +7 Vote: I do not like it

            It is known that (a very important consequence/equivalence of prime number theorem) .

            For n, k to be good, (n + 1) must be a multiple of . So n must be like ek.

            This should be good enough, as running time is .

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              7 years ago, # ^ |
                Vote: I like it +13 Vote: I do not like it

              Or, say primes >10 will multiply, so it be over 10^18 fast...

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

              Bruteforce works. :)

              This was my approach.
              gcd(1,2,...,k) = 1

              Therefore, lcm(1,2,...,k) = 1*2*3*..*k

              Also, since the remainders have to be distinct, one observation is that for any i between 1 to k, n%i should be equal to i-1.

              This means: n = i*C + i-1 ==> (n+1)%i = 0.

              Therefore n+1 should divide all the numbers from 1 to k. This also means that n+1 should be equal to the least common multiple of 1,2,..,k (in this case its just the product.) or any multiple of this lcm.

              Looping from 1 to k and stopping the loop whenever product crosses n+1. Product crosses n+1 easily by the time we reach i=50 in the loop. We return "No" in this case.

              If k! is equal to n+1 then we return "Yes"

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

        try this

        bool f;

        int top=0;

        for(int n=1;n<=100000;n++){

        for(int k=1;k<n;k++){

        f=true;

        for(int i=1;i<k&&f;i++){

        for(int j=i+1;j<=k&&f;j++){

        if(n%i==n%j)f=false;

        }}

        if(!f){

        top=max(top,k);

        break; }}}

        printf("%d\n",top);

        just look at the output and you will find out why it is possible to do it in brute force

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

    People have this idea:

    Since you need to have all unique values till k. They should be obtained serially from 1,2,3,..k like this: 0,1,2,3,4,5...k-1

    so, n should be of the form: 2x+1, 3x+2, 4x+3, 5x+4, 6x+5, 7x+6,... all at the same time. But if something is of the form 5x+4 .. it cannot be of the form 2x+1 .. (coz one is odd and other is even, for even x, similarly chose a pair for odd x)

    Edit:

    The conclusion seems flawed. Maybe we can continue the logic like this:

    Number has to be of form: 2x+1, so 3,5,7,9.. 3x+2, so 5,8,11,14.. ... ....

    It kind of creating a sieve of eratosthenes, which generates a prime number.

    I don't what that gets us :)

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7 years ago, # |
  Vote: I like it +9 Vote: I do not like it

Locking after being hacked

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7 years ago, # |
  Vote: I like it +6 Vote: I do not like it

how to solve D?

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    sorting : )

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      merge or bubble

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

        consider to string a and b

        let the number of 's' in a be a.S and the number of 'h' be a.H

        if you arrange a and b be ab

        the you will get a.S*b.H pairs

        so we know that if a.S*b.H>b.S*a.H then a should be in front of b

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

What was the hack for A?

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    7 years ago, # ^ |
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    if you donnot have a copy,you cannot get more copy

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    7 years ago, # ^ |
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    Most common one: x 1 (while x is a positive even number).

    Expected answer: "NO" (as there are no initial copies to generate new ones)

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

    0 0

    0 1

    1 0

    3 0

    2 1

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    7 years ago, # ^ |
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    2 1

    output

    NO

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    7 years ago, # ^ |
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    something!=0, y=0 and something!=0, y=1

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7 years ago, # |
  Vote: I like it +5 Vote: I do not like it

Problem A just burn my room

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

What is wrong this this solution for E?

Let dp[i][j] be the max mana we have at tree i with j summoned in the trees before us. The max capacity is always w + j·b for a fixed j so we don't have to store that. I used long long int here and for dp table so no overflow. The answer is the largest i such that dp[n][i] is valid. I kept getting WA on pretest 5.

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    7 years ago, # ^ |
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    it's correct

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Did you use sliding window maximum?

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

    problem is that max summoned birds numbercan be ci * n = 103 * 104 = 107, so your array will have size dp[103][107].

    EDIT: I didn't read task carefully :{

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7 years ago, # |
  Vote: I like it +1 Vote: I do not like it

How to solve B? Seems easy cause lot of people solved it, but I only solved C and D.

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    7 years ago, # ^ |
      Vote: I like it +7 Vote: I do not like it 
    ans = 0
for a = 1 .. n
  for b = a .. n
    c = a ^ b
    if b <= c <= n and a + b > c
      ans++
print ans
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      7 years ago, # ^ |
        Vote: I like it +3 Vote: I do not like it

      Wow.. I'm a fool :) Thanks for your reply! (And all the others too)

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    7 years ago, # ^ |
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    can you please tell me how to solve C?

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      7 years ago, # ^ |
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      Try 1~50

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      7 years ago, # ^ |
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      n mod 1 , ... , n mod k has to be different. but you can easily notice that n mod 1 is always 0. So, n mod 2 must be equal to 1, because n mod 2 is 0 or 1, but it cannot be 0. ... n mod i must be equal to i-1, which means that n+1 is divisible by 1, 2, ..., k.

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

    For a <= n, b <= n, let c = a ^ b, if all other conditions hold, then increment answer

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

    a^b^c = 0 implies a^b = c. So bruteforce for a, b, and check if the 3 numbers can form sides of a triangle.

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    You can iterate through all possible cases of a and b. c can be calculated by bitwise xor:

    As a xor b xor c = 0, we can agree that c = a xor b.

    If c is not lower than a or b, and not greater than n, and a,b,c forms a non-degenerate triangle (simply check that if all 3 inequalities of triangles are proved right), then you can add 1 to the answer.

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

    Simply iterate over all a, b values . Check a + b > c, a + c > b and b + c > a

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    7 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    I actually passed it with brute forcing which is O(N^3). It gave me 780ms even though it doesn't seem to be the intended solution.

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

      Your condition ( c < a + b ) in the third loop in your code makes it about O(n^2) i belive

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7 years ago, # |
  Vote: I like it +1 Vote: I do not like it

Test for hack C: 1 2

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    7 years ago, # ^ |
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    Isn't it "yes" ?

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

      Yes. Many people (me too) have writen: if( n < k ) cout << "No";

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        7 years ago, # ^ |
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        Bad luck, man

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        7 years ago, # ^ |
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        fortunately I had written n<=k so it failed pretest and I was able to correct it. and now i'm purple.

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    7 years ago, # ^ |
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    pretty sure this is the case that got me

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7 years ago, # |
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How to solve F? I tried taking all N numbers initially and consecutively removing the one which will decrease f(I) the most but such that f(I) ≥ k will remain true. Just plain intuition, didn't work.

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I bruteforced n<=17, implemented it for larger n, it worked well for all local inputs. Still kept getting TLE on the 7th pretest. I just wonder whether it was my complexity, I/O, special countertest designed to make this kind of solution fail...

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Explanation — Problem F

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Man, them noise and echo are simply unbearable which makes your speech unrecognizable.

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7 years ago, # |
  Vote: I like it +19 Vote: I do not like it

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    7 years ago, # ^ |
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    Is it so interesting to post one picture several times?

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7 years ago, # |
  Vote: I like it +1 Vote: I do not like it

In question D, what is the pretest 4?

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

I think there should be a limit on the number of hacks for each problem... Yeah hacks are a part of CF, but when a hacker with ABC has more points than a non-hacker with ABCD, I think there is a problem.

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can anyone give me some idea for 922F - Divisibility

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    7 years ago, # ^ |
      Vote: I like it +6 Vote: I do not like it 
    for i = 1 .. n
  for j = i .. n, j += i
    d[j]++
if sum(d) < k
  print no
otherwise, kick out some numbers > n/2 - since they do not have multiples - so  that number of pairs will decrease to k
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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

can someone help me find why i was hacked in problem C? 35018691

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7 years ago, # |
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Short statements not means a lot of hacks.

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7 years ago, # |
  Vote: I like it +11 Vote: I do not like it

This was the biggest slaughter since I started to participate at contests

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

When will the system testing start ?

Update: System testing started :)

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7 years ago, # |
  Vote: I like it +37 Vote: I do not like it

When someone asks how life is going on?

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7 years ago, # |
  Vote: I like it +13 Vote: I do not like it

While everybody is discussing today's contest, I just found my love <3

Thanks Mike :)

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7 years ago, # |
← Rev. 2   Vote: I like it +75 Vote: I do not like it

Isn't it obvious that current hacking system sucks? It's a brilliant idea, but really bad realization. Hacking should give you points (maybe even more than 100) if you find hidden not obvious bug in someone's code. But if you just take n = 0 corner case in div2A and it gives you 1500 points that's so unfair: you get that much points not because you are smart or your debugging skills are better than average, but because you have gray guys in your room who always do this mistake.

That's why I think that hacking points must be not a constant, but a function, which depends on how much hacks there are on the problem.

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

    if problem difficulty is 500 points so on one hack your receive 100 is 20% but if problem difficulty is 2500 points so on one hack your receive 100 is only 4%

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

    til i am a grey

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Agreed. I hacked 18 solutions (Problem A) using only two test cases And guess what? My own solution missed one corner case and I got WA in system test. But 1800 points gave me an unfair advantage.

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

Many among you would be disappointed because of all the hacks, but I feel that the problems were super exciting and I enjoyed all the brainstorming while getting hacked.

+1 to GreenGrape

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I don't know why, but my submission got stuck at test 39 for last 10 min.s. I am on page 6 of filtered list

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7 years ago, # |
  Vote: I like it +57 Vote: I do not like it

When you find one of the testcases for A hackerman.jpg

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

I think that there is a problem with D testcases. This is my AC submission.

The problem with my solution is that the comparison is made entirely on integers, and therefore it could overflow.

I'll try to generate the testcase, I think that:

2
sssss...ssssssh
hhhhh...hhhhhhs

is enough.

EDIT: oh yes, I think it was enough. My solution gave 50000, but when reversing gave 2499950000. I don't think I'm the only one who failed for this.

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    That's actually quite strange since tests 11 and 12 are made exactly to counter this.

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

      I would like to remember that overflow is an undefined behaviour, so anything could happen, even getting an incorrect AC (A guess for why this happens is that the compiler may do some simplifications like x*y/x = y — avoiding the calculation that gets the overflow — maybe something in that line is what happened)

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

        I tested it using custom invocation, and it gives the incorrect output there (50k instead of 2499..etc).

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      7 years ago, # ^ |
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      Oh, I think I know what's happening here. My submissions treats different the strings consisting of only 'h' or only 's', so that's why my submission passed.

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        7 years ago, # ^ |
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        lucky you didn't loose 100 points on tc 11.:)

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    7 years ago, # ^ |
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    35024915 this solution gets tle on your test, if i am not mistaken. if so, D should be reevaluated

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      7 years ago, # ^ |
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      Looks fine to me. Total characters are 200000.

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        7 years ago, # ^ |
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        yeah, i realized that i tested it wrong right now, sorry

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

long long contest! #define int long long accept my C and D :)

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

Can't believe my STUPID code for C got pretests passed even and finally failed on test 76th!!! I used k instead of i in my loop.

Failed.

vs

Accepted.

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7 years ago, # |
  Vote: I like it +13 Vote: I do not like it

Couldn't proof the solution for C during the contest. But now, I think I got a proof.Correct me if I'm wrong plz.

If the answer for n is "yes", then for all i from the set [1,k] will have distinct remainders.

We can easily see that for all i [1,k]

n mod i == i — 1,

so, we can say (n + 1) mod i == i == 0 ;

That means all i [1,k] must divide (n+1) simultaneously.

Which means (n+1) must be divisible by the lcm(1,2,3,........,k).

Now the lcm(1,2,3,.........,k) would be > 1e18 when k is > 41(or something) .

That's why the brute-force solution worked.

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

ohuenniy round

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7 years ago, # |
  Vote: I like it +5 Vote: I do not like it

who is sorry-haghani ?!

Haghani ...

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can anyone explain? :( 35036774 35011492

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7 years ago, # |
  Vote: I like it +5 Vote: I do not like it

Solving the B in O(n^3) no problem... i would say a O(n^2) solution was intended but anyway: http://codeforces.me/contest/922/submission/35039399

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7 years ago, # |
  Vote: I like it +8 Vote: I do not like it

Hello, world!

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Why is my code giving runtime error on test 5 in D : Link

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7 years ago, # |
  Vote: I like it +1 Vote: I do not like it

This was one of the best and most interesting contests on codeforces that i have given. So much of hacks and good difficulty distribution i should say ,awesome GreenGrape

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

GreenGrape has never been Green on Codeforces.

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

I don't understand A number problem. please, anyone, help me to understand this problem. problem link : http://codeforces.me/problemset/problem/922/A thanks

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    you start with o. Then you do magic copy to get additional o and c.

    Now you have : o,o,c

    Next you do magic copy on c and get additional c and c , so a total of 3 c,c,c.

    Likewise.

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

http://codeforces.me/contest/922/submission/35048494 why i am facing runtime error thanx in advance

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    7 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    try swapping these statement in your comp function if (!h1) return true; if (!h2) return false;

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      7 years ago, # ^ |
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      But what is logic behind this swap statement

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        7 years ago, # ^ |
          Vote: I like it +1 Vote: I do not like it

        If 2 equal objects are passed it should always return false. before it could return true.

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

Problem B, O(n^3) without pragma 35154154....using pargma only make it a little faster..35129118

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

In Problem B, why (1,2,3)is not a valid set?

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Because it's a degenerate triangle (1+2 = 3) the

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