majk's blog

Codeforces Global Round 4
By majk, 5 years ago, In English

Hi!

On Jul/20/2019 18:35 (Moscow time) we will host Codeforces Global Round 4.

It is the fourth round of a new series of Codeforces Global Rounds supported by XTX Markets. The rounds are open for everybody, the rating will be updated for everybody.

You will be given 8 problems and 150 minutes to solve them. The scoring distribution will be 500-750-1250-1750-2000-(1500+1500)+3250+4000.

The prizes for this round:

  • 30 best participants get a t-shirt.
  • 20 t-shirts are randomly distributed among those with ranks between 31 and 500, inclusive.

The prizes for the 6-round series in 2019:

  • In each round top-100 participants get points according to the table.
  • The final result for each participant is equal to the sum of points he gets in the four rounds he placed the highest.
  • The best 20 participants over the whole series get sweatshirts and place certificates.

The problems of this round were developed by me. I will be available on Discord to discuss the problems after the contest.

I would like to thank:

UPD: The round is finished. I hope you enjoyed the problems! I am very sorry about the issue on E.

UPD2: Editorial

UPD3: Congratulations to:

  1. ainta
  2. Radewoosh
  3. tzuyu_chou
  4. LHiC
  5. Benq

UPD4: The results after four rounds.

Announcement of Codeforces Global Round 4
  • Vote: I like it
  • +875
  • Vote: I do not like it

Comments Comments (240)
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5 years ago, # |
  Vote: I like it +293 Vote: I do not like it

What about ISeeDS? He has feelings too

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5 years ago, # |
  Vote: I like it +71 Vote: I do not like it

And, of course, Accepted_ as a welcome guest of any contest

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5 years ago, # |
  Vote: I like it +41 Vote: I do not like it

And, wish that wrong_answer stays away

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

Sounds great!But hope for no reading comprehension problems....

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5 years ago, # |
  Vote: I like it +103 Vote: I do not like it

An old comment from the setter of the contest :)

For those who don't want to click
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    5 years ago, # ^ |
      Vote: I like it +42 Vote: I do not like it

    The one "for those who don't want to click" is good for me. I'm so lazy.

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5 years ago, # |
  Vote: I like it +20 Vote: I do not like it

Tourist is coming soon

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5 years ago, # |
  Vote: I like it -25 Vote: I do not like it

Global Round = div1 + div2 + div3 = div6

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    5 years ago, # ^ |
      Vote: I like it +152 Vote: I do not like it

    [Error] no match for 'operator+' (operand types are 'div.' and 'div.')

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

    Global round = max(div1, div2, div3) = div1; for divisions we use max;

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    5 years ago, # ^ |
      Vote: I like it +69 Vote: I do not like it

    I think that div1 > div2 > div3. My estimation is that div1 + div2 + div3 = div0.9

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    5 years ago, # ^ |
      Vote: I like it -191 Vote: I do not like it

    Let's downvote Errichto to make his contribution lower than Radewoosh

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

      Let's downvote him to make his contribution lower than r_pi.

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

      Why would you want to make contribution of Errichto less? I think his over all effort in helping the community is much higher than anyone else here.

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

8 problems and just 150 minutes is it enough????

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

    Generally, educational rounds have 7 problems and 120 minutes of time. So comparatively it is enough.

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

      Are you compare the educational with global round ?? educational is for div2 and global round for all division

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

        For div3 also... (you can check contest history)

        and for div1 users if you add 2-3 easy problems in contest, it doesn't change anything for them.

        Required time for contest depends more on difficulty level than number of questions.

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

      And also this round needs to test the abilities of people range from beginners to masters so it definitely needs to have enough time for all the people to solve the problems.

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    5 years ago, # ^ |
      Vote: I like it +25 Vote: I do not like it

    No, of course not. We need 300 minutes.

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

Btw, did anyone notice that majk is setting a global round and a local round these days?

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5 years ago, # |
  Vote: I like it +21 Vote: I do not like it

Although we are Chinese,we still try to join in every Global Round.

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    5 years ago, # ^ |
      Vote: I like it -18 Vote: I do not like it

    But there isn't a "Codeforces Global Round 1" in your contest history.

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      5 years ago, # ^ |
        Vote: I like it +47 Vote: I do not like it

      he didn’t say he did join, he just said he tries to join

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    5 years ago, # ^ |
      Vote: I like it +23 Vote: I do not like it

    Most rounds is a bit late for Chinese,hope the contest can be earlier,and we won't stay up late and feel sleepy the next day.

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

      I agree with your suggestion. I hope that in the future contests in Codeforces could start earlier, maybe 20:35 or 21:05 UTC+8 will work, so that we don't have to stay up very late and feel sleepy the next day.

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      5 years ago, # ^ |
        Vote: I like it -16 Vote: I do not like it

      Most rounds are a bit early for people like me on the US west coast.

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

Pretest?

TLE?

Editorial?

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

Global round 3 is good go for global round 4

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

Wish you a job promotion :-)

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

Thanks MikeMirzayanov also for updating and adding amazing features in CODEFORCES regularly.

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

No one is talking about Memory_Limit_Exceeded. That's sad

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

i hope that will be funny:)

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

I hope that the meaning of all the problems are easy to understand.

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5 years ago, # |
  Vote: I like it +130 Vote: I do not like it

It could be the last contest which is rated for Div.1, and the last big global round before IOI 2019 Azerbaijan.
I hope everyone's good luck, as a participant of IOI 2019.

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

Hope this will not happen :3

kkkkk

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5 years ago, # |
← Rev. 5   Vote: I like it 0 Vote: I do not like it

have a good contes :)

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5 years ago, # |
  Vote: I like it -23 Vote: I do not like it

i hope i become yellow after this

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

What about UKE,OLE,_Wrong_Answer_

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5 years ago, # |
  Vote: I like it +21 Vote: I do not like it

hope i InvincibleMario one day

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

Did anyone notice 'alice' and 'bob' in the blog tags? Be ready for some good game theory problems.

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    5 years ago, # ^ |
    ← Rev. 2   Vote: I like it +64 Vote: I do not like it

    :3 Meanwhile, the problem will be like:

    ***
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      5 years ago, # ^ |
        Vote: I like it +46 Vote: I do not like it

      They start the game by drawing convex polygon in which the area of it doesn't exceed X squared units and is at least Y squared units, if a player can't draw a convex polygon without crossing an already drawn convex he loses, determine the winning player if both played optimally.

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        5 years ago, # ^ |
        ← Rev. 3   Vote: I like it +65 Vote: I do not like it

        This problem is really easy. Solution:

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

          Interesting, someone could make various problems by tweaking its details.

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

majk come back with a contest after Good Bye 2018

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

Hope I will be either retired.coder.1 or WHITE2302 . I don't want to be grey anymore.

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

Why no one has made an account with a nickname "Idleness_limit_exceeded"? :c

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

How Global Round's rating distribution executes? Do Div1 and Div2 contestants get seperate rating? How likely Div2 contestants increase rating? Thanks!

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5 years ago, # |
← Rev. 4   Vote: I like it +18 Vote: I do not like it

TimeLimitExceed and TimeLimitExceeded will be sad if they shrink to TLE

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5 years ago, # |
  Vote: I like it +52 Vote: I do not like it

I think TLE is abbreviation for --> tourist limit exceed please don't gimme negative feedback I am just joking

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

an AK-king jqdai0815

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5 years ago, # |
  Vote: I like it -51 Vote: I do not like it

Why arsijo again?

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

    why not ? because of just one mistake he did before ? i'm afraid for you ;

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

Should I do it? I am new in programming.

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    5 years ago, # ^ |
    ← Rev. 3   Vote: I like it -20 Vote: I do not like it

    [DELETED]

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

    you can do some contests generally -- pratice more. There're so many basic problems for newbies, such as Div2A/B

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

Everything is gonna be Okay

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5 years ago, # |
← Rev. 3   Vote: I like it +31 Vote: I do not like it

madhav_1999 will become master today! — Prokhar

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

Could you please postpone the contest 15 minute later ?

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

And hopefully the round won't become Unrated.

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5 years ago, # |
  Vote: I like it +22 Vote: I do not like it

Is there a page with current standings?

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

I hope that the contest will be good,funny and interesting like the announcement .

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

gl hf

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

i have to stay up late

i look like the panda in my photo

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

When this "prior to round" happen????? Only 15 minutes left......

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

speedforces strikes again again

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    5 years ago, # ^ |
      Vote: I like it +110 Vote: I do not like it

    Now that the round is over, I feel I should comment on that. We considered both D and E to be tricky problems, even some very good testers had problems with them. Perhaps C and D turned out to be a bit easier that anticipaded.

    F on the other hand, was a different beast. I originally proposed a completely different problem — with in my opinion a cute observation. However, it was geometry and testers almost went on a riot because of the implementation difficulty.

    About $$$10$$$ days before the contest I came up with the problem that is now $$$F1$$$, but the testers saw it as too easy, even easier than D or E — and the gap between $$$F$$$ and $$$G$$$ was afwul.

    So I realised then that we can increase $$$m$$$ and it adds to the problem. Suddenly, it was too difficult and it was the gap between E and F that was brutal. With the contest fast approaching and the tester's ranks dwindling, we realised that the safest way to mitigate was to split F.

    Some small gap still remained though, but I'd say that the ratio of ACs on adjacent problems around $$$4$$$ is fine (after all, it has to reduce from couple thousand to only a few in around $$$6$$$ problems). Please try to understand that it is sometimes quite difficult to judge the difficulty based on a small sample of testers. Additionally, if you find out two days before contest it is going to be speedforces, the best solution is typically to keep it that way rather than to perform some ninja fixes.

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

What happened just now? My rank changed from 900+ to 400+ and to 900+ again...

[Edit] Got it.

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5 years ago, # |
  Vote: I like it +35 Vote: I do not like it

i am suffering subsequence string phobia.

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

" The round is still rated. " ?????? What the f**k ??????

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

    My friend locked his problem E but he can't change his code after rejudged and already lost his rating

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      5 years ago, # ^ |
        Vote: I like it +42 Vote: I do not like it

      You seem to care a lot about your friend! I think HE can ask organizers to make it unrated just for him.

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

    Why unrated after rejudged? I can't understand.

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    5 years ago, # ^ |
      Vote: I like it -23 Vote: I do not like it

    The round is always UNRATED or at the edge of UNRATED if arsijo is the coordinator.

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

How to solve G?

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    5 years ago, # ^ |
    ← Rev. 3   Vote: I like it +18 Vote: I do not like it

    Actually, I didn't get pretest passed in this problem, but I have an idea.

    Solution 1
    You can replace the problem to a new problem as follows by using euler tour method:

    • Initially, there is an array $$$a = $$${$$$a_1, a_2, ..., a_N$$$} and $$$b = $$${$$$b_1, b_2, ..., b_N$$$}.
    • Query type 1: Add $$$a_l$$$, $$$a_{l+1}$$$, $$$a_{l+2}$$$, ..., $$$a{r}$$$ by $$$x$$$.
    • Query type 2: Get the maximum value among {$$$|a_l| * |b_l|, |a_{l+1} * b_{l+1}|, ..., |a_r * b_r|$$$}.

    To solve this new problem, you can use sqrt decomposition by query.
    Let's think about to solve from query $$$1$$$ to $$$B$$$ with $$$O(N + B^2 log B)$$$ complexity.

    • You can divide at most $$$2 * B + 2$$$ subsegments, that the value of $$$a_i$$$ that increased in add query is the same in subsegment.
    • Add query: Just add $$$x$$$ in proper subsegments. Complexity $$$O(B)$$$ for each query.
    • Get-maximum query: Use Convex-Hull Trick and binary search. The convex-hull can calculate with precount. Complexity: $$$O(N)$$$ for precount, $$$O(B log B)$$$ for each query.

    Since you should process $$$B$$$ queries, The total complexity that takes to calculate all answer in query [$1, B$] is $$$O(N + B^{2}logB)$$$.

    So, how to solve in not only query $$$[1, B]$$$, but also $$$[B + 1, 2B]$$$, $$$[2B + 1, 3B]$$$, $$$[3B + 1, 4B]$$$, and so on?
    You can find the value of $$$a_i$$$ when query $$$B, 2B, 3B, 4B, ..., $$$ has just finished, by using cumulative sum method. Since the value of $$$b_i$$$ does not change, you can also solve in query $$$[B + 1, 2B]$$$ and so on, as same as query $$$[1, B]$$$.

    If the backet size $$$B$$$ is nearly $$$\frac{B^{0.5}}{log_{2}B}$$$, the total complexity will be $$$O(B^{1.5} * sqrt(log_{2}B))$$$.

    Solution 2
    Since the constraints is $$$n \leq 200,000$$$, $$$q \leq 100,000$$$ and the time limit is 5sec, since the process is very light, I think that even straight-forward brute force $$$O(NQ)$$$ which is very very very very fast may pass.
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5 years ago, # |
  Vote: I like it 0 Vote: I do not like it

How to solve E?

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    5 years ago, # ^ |
      Vote: I like it +25 Vote: I do not like it

    If you consider two first and two last characters of the string, you are able to find one in first two and one in last two that match. Add them to your palindrome.

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

    Hint: Since no two characters are adjacent, and there are only 3 characters in the alphabet, there exists a character in the first two characters of the string that equals a character in the last two characters of the string.

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

    Since no consecutive letters are the same, you can make any even-length palindrome into a longer odd-length palindrome. So let's fix the middle of the palindrome, how about s[n/2]. Then you can show that from the midpoint, you never have to go more than two spaces to the left, or two spaces to the right to find two letters that are the same.

    So at worst, it will take 4 spaces to add 2 letters to your palindrome.

    Be careful of n=4 case (may need to try other midpoint).

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

How to solve E

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5 years ago, # |
← Rev. 4   Vote: I like it +65 Vote: I do not like it

E's solution is cute, although the problem is a little contrived.

EDIT: The key is pigeonhole principle. Take the last two and first two letters, there must be two letters that are equal (and not adjacent, since no two adjacent are the same). Take these two and recurse. If there are fewer than four characters, just take any one.

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

    Can you share your cute solution :)

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

    really kyot solution

    i overcomplicated the problem :\

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    5 years ago, # ^ |
      Vote: I like it +31 Vote: I do not like it

    Am I the only one to try solving problem without that "no repeated letters" thing for 15-20 minutes before realizing that problem is totally different?

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

      Just scroll down, I solved it without knowing that.

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

        How? And can you prove your solution?

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

          I can't prove it, but it makes sense.

          My solution optimizes the $$$O(N^2)$$$ dp solution using next and before arrays and limiting the search on the first point the answer reaches the needed length.

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

Good tasks, thanks)

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

My girl friend is beautiful, smart, kind, rich, humourous and never blame anything on me!

I consider her a perfect girl friend.

Unfortunately, I discover that SHE is actually a man.

That's my point on the checker of the problem E.

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5 years ago, # |
  Vote: I like it +23 Vote: I do not like it

Not so many Pretest :(

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5 years ago, # |
← Rev. 3   Vote: I like it +82 Vote: I do not like it

We had an issue with the problem E. The rejudge changed ~10 AC submission to WA in pretests. Since only ~10 people were involved in this issue, we do not see a reason to make this round unrated. If you have another opinion, you are welcome to share it.

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    5 years ago, # ^ |
      Vote: I like it +216 Vote: I do not like it

    You can make the contest unrated for these 10 participants.

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    5 years ago, # ^ |
    ← Rev. 2   Vote: I like it +13 Vote: I do not like it

    Well, After 1.5 hour I noticed that my ac submission has wa. It's kinda unfair.

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5 years ago, # |
  Vote: I like it +28 Vote: I do not like it

Thanks to Pretest XD

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

    I saw you hacked A. Can you tell me please what was the hack for it ?

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      5 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it 
      5
5 3 3 3 1

      That would have hacked me because I misread the problem and thought that you can capture the 1 to have a total coalition size of 6, which would enable you to capture the 3s.

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

first four problems were not THAT hard as in previous contests

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

pretests too strong barely 10 hacks total

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5 years ago, # |
  Vote: I like it +175 Vote: I do not like it

Oh, nice problem G, I've seen very similar things only a million times and it always ensured me at least one hour of fun coding and debugging, so happy to be given another chance.

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

    Isn't problem F nice as well? Or I guess all the fun with it was too short for you to feel the excitement?

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5 years ago, # |
← Rev. 2   Vote: I like it +13 Vote: I do not like it

I would like to thank:

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

In Problem A, for testcase: 2 6 2

shouldn't the answer be 1 1 as the question mentions "If Alice's party has enough people to create a coalition on her own, she can invite no parties." i got an "unsuccessful hacking verdict" defender's output was: 2 1 2

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

    2 1 2 is also correct. You don't have to minimize number of parties.

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

      Yes but as i said the question mentions "If Alice's party has enough people to create a coalition on her own, she can invite no parties."

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

        "She can invite no parties", not "she cannot invite parties".

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

          To be fair, I understand the confusion now. English is ambiguous sometimes.

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

          oh damn. need to take english lessons now...

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

    I feel that mate. I got that twice

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

    Both outputs are correct. Multiple solutions possible.

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

    That's a good point, my solution also gives "2 1 2" but it passed system test... Maybe the intended meaning of "she can invite no parties" is "it is allowed that she doesn't invite any parties".

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5 years ago, # |
← Rev. 2   Vote: I like it +8 Vote: I do not like it

no two consecutive characters are the same

FML, I solved E for general strings

UPD: got TLE

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

Just little FYI from a noob coder if anyone doesn't know already..."str = str + xyz" takes waayyyyyy too much time compared to "str.push_back(xyz)"... got to learn that the hard way today :(

Well at least learned something :D

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    5 years ago, # ^ |
      Vote: I like it +22 Vote: I do not like it

    You can also do the following: str += xyz

    str += something in strings is much faster than str = str + something for C++

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

      Well, += just adds to the end of the string (I assume that it just push_backs), while str = str + xyz has to copy the whole string, then append, and then save that value in the original variable.

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

      Thanks! Noted.

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

    Another noob coder here... took me three TLEs to realize the thing.

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

Imo score distribution wasn't good enough (maybe it's just for me). I spent 1:30 for B, 0:15 for C and 0:30 for D.

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5 years ago, # |
  Vote: I like it +30 Vote: I do not like it

Just look at this submission: 57419391

Brute force wins again!

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

Every time I finished the code dugging, the game had just ended. What a pity！

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

    I hope that I will find the bug more quickly next time.

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      5 years ago, # ^ |
        Vote: I like it +28 Vote: I do not like it

      You can optimize your wish by hoping to not make a bug next time

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

I only now noticed abs in G. Can't even read :(

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

What is the reason for the low constraint at problem D? It could be solved even with $$$10^6$$$ limit

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    5 years ago, # ^ |
      Vote: I like it +40 Vote: I do not like it

    That would be a huge spoiler to the solution (what number of edges can be linear to the number of vertexes).

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

      Exactly. Also, the time complexity wasn't supposed to be critical in this problem.

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

    I guess the $$$10^3$$$ is for people to easily test if their solution is correct in $$$O(n^2)$$$ time, coz it's not easy to prove that it works.

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    5 years ago, # ^ |
      Vote: I like it +87 Vote: I do not like it

    How can I output my graph with $$$10^{12}$$$ edges in that case?

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

      Nohow? You'll get TLE in that case. And it's your problem, not autors

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      5 years ago, # ^ |
        Vote: I like it +17 Vote: I do not like it

      fast io

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      5 years ago, # ^ |
        Vote: I like it +66 Vote: I do not like it

      You should try bitsets, I heard they're pretty fast

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

    So that users like me can go wrong , and find primes( degrees) using Goldbach conj. and stuck there.

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

In problem E, will there ever be a case where answer is IMPOSSIBLE?

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

    No.

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    5 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it
    Spoiler
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5 years ago, # |
  Vote: I like it +19 Vote: I do not like it

I think some people just guested problem C after seing samples

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

    Right. No idea why it's true xd.

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

      Call the vertical segment white if it's white from the left/black from the right. Note that if segment is white then all others in the same row also white. Same for horizontal

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

      4 choices for top-left square. Two choices for the rest along the top and the left. Only 1 choice for the remaining squares. i.e. if you choose the top row and the left column every other square is forced to be a single orientation.

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5 years ago, # |
  Vote: I like it +28 Vote: I do not like it

Problem C has best samples ever

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    5 years ago, # ^ |
      Vote: I like it +30 Vote: I do not like it

    It would have been better if only one test case w=2 and h=2 was given. Then most of them who solved it just by test case would have to think of proof(including me lol).

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

E has got three greedy tags and two string tags...

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5 years ago, # |
  Vote: I like it +82 Vote: I do not like it

It it just me, or prob.A's statement is TOO HARD to read? (you can check my submissions)

I'm not familiar with the Parliamentary System in some western countries :(

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

ovvo

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5 years ago, # |
← Rev. 2   Vote: I like it +19 Vote: I do not like it

In D, If n is prime, then the answer is a circular graph. Otherwise, create a circular graph at first and then create extra edges by connecting nodes with degree 2 to make the number of edges a prime number.

Is this the only solution?

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

Problem G can be solved using sqrt decomposition. In the contest I chose a bad block size and didn't get AC. I'm so done right now.

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

    Using dfs order we can turn G in to the following range query problems:

    Given 2n number a[1], a[2], a[3], ..., a[n] and b[1], b[2], b[3], ..., b[n]. You have to query max(abs(a[i])*b[i]) for some l<=i<=r and add a positive number x to a[l], a[l+1], ..., a[r]. Note that b[i] is set from the begining.

    Now for each i from 1 to n, let w[i] be the sum of x added to a[i], then you have a graph of at most 2 lines representing the value of abs(a[i]+w[i])*b[i]. To get the answer for some query, just get max of all (a[i]+w[i])*b[i].

    To update and query fast, split the array in to blocks of size S. For each block you can represent the lines using a Lichao tree or using CHT. I "upsolved" using CHT, but maybe Lichao tree will work as well. For each update query, simply update all the elements from the left most and right most block, and rebuild the two blocks, for all the blocks in between, keep an array w[] to know the elements in the blocks have been added by how much, this take O(Slog(S)+N/S). For each get query, you can just do the same, get the elements from left most and right most block, query each of the inbetween block in O(log(S)), so this take O(S+N/S*log(S)). Also this approach can be used even if you want to add negative number to a[].

    Honestly I didn't think this would pass, and I hoped that it gets TLE. But it works, and I chose the wrong S, so I got TLE in the contest. This disappointed me very much on top of the also very dumb mistake I got in F2. Still the problems are very good. Codeforces rules are very punishing, so something like this is to be expected.

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

      For anyone interested in implementing this approach, S=100, 200, 300, even 400 will get AC. Just don't be bad like me and use 500 or something.

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

        Yeah, that sucks. I tried hard to cut the solutions with extra log factor — we had two such solutions and they all needed around 8 seconds on some cases.

        On the other side of the road, there was Java that also needed to pass.

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

        Needs more coke spilling

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5 years ago, # |
← Rev. 2   Vote: I like it -41 Vote: I do not like it

I was able to complete 4 questions in this challenge. I usually use ideone for compilation of my code . My code was copied by newbi/57387826, sos_123/57387881 . I was removed from this contest.This is quite unfair and strict actions should be taken against these contestants

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    5 years ago, # ^ |
    ← Rev. 2   Vote: I like it +22 Vote: I do not like it

    As stated in codeforces rules about third party code unintentional leakage is also a violation so be careful next time.

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

    I checked the submissions you mentioned. Both of them submitted the code before you.

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

      I didn't register for the competition so I had a late penality of 10 minutes . I had compiled it way before submitting it . You can look in both of their profiles lots of there submissions are skipped .

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

        Yes, I have seen it. Their all solutions are skipped in all contests.

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5 years ago, # |
← Rev. 2   Vote: I like it +68 Vote: I do not like it

Could anyone explain why my submissions still at Pretest Passed? I solved 3 problems but my rating has been counted as 0 problems.

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5 years ago, # |
  Vote: I like it +80 Vote: I do not like it

majk:

Also majk:

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5 years ago, # |
  Vote: I like it +26 Vote: I do not like it

When t-shirt winners will announced?

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    5 years ago, # ^ |
      Vote: I like it +23 Vote: I do not like it

    We are still investigating the issue. We will post the winners after we get the official results of the round.

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

Thanks checker for good check.

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5 years ago, # |
← Rev. 4   Vote: I like it -7 Vote: I do not like it

what is wrong in this solution of problem E.

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

Can anyone prove this statement, for every n such that there is a prime between n and 3n/2?

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5 years ago, # |
  Vote: I like it +100 Vote: I do not like it

T-shirt winners!

We used these files to find the winners.

randgen.cpp
get_ranklist.py
List place Contest Rank Name
1 1178 1 ainta
2 1178 2 Radewoosh
3 1178 3 tzuyu_chou
4 1178 4 LHiC
5 1178 5 Benq
6 1178 6 Reyna
7 1178 7 maroonrk
8 1178 8 nhho
9 1178 9 mnbvmar
10 1178 10 yfzcsc
11 1178 11 isaf27
12 1178 12 MrDindows
13 1178 13 scott_wu
14 1178 14 DearMargaret
15 1178 15 Swistakk
16 1178 16 gop2024
17 1178 17 webmaster
18 1178 18 gepardo
19 1178 19 E869120
20 1178 20 dreamoon_love_AA
21 1178 21 Marcin_smu
22 1178 22 krijgertje
23 1178 23 Egor
24 1178 24 Elegia
25 1178 25 amethyst0
26 1178 26 kmjp
27 1178 26 G.Z.V.O.D.E.N
28 1178 28 aid
29 1178 29 dlalswp25
30 1178 30 tribute_to_Ukraine_2022
36 1178 35 sevenkplus
56 1178 56 Roundgod
76 1178 76 atoiz
110 1178 110 Ari
147 1178 147 tabasz
159 1178 159 asokol
169 1178 169 mohamad
174 1178 174 PinkRabbitAFO
184 1178 184 Anachor
189 1178 189 MarcotheFool
255 1178 255 Zayin
256 1178 256 rotavirus
284 1178 284 kektus
313 1178 313 Ferume
314 1178 314 Mikaeel
377 1178 377 3.14roca
401 1178 401 mghzs1997
403 1178 403 Kerim.K
449 1178 449 MonkeyKing
472 1178 471 ivanilos
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5 years ago, # |
  Vote: I like it 0 Vote: I do not like it

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

Problem C: https://gurmeet.net/puzzles/truchet-tilings/index.html

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