Pinely Round 5 (Div.1 + Div.2)
By pinely, 6 months ago, translation, In English

header Hey, Codeforces!** (*^‿^*)

Pinely is here! We are glad to invite you to participate in Pinely Round 5 (Div. 1 + Div. 2) which will start on which will start on Oct/30/2025 19:35 (Moscow time) Note the unusual time of the round.

  • You will be given 8 problems and 3 hours to solve them. The round will be rated for everyone.

The problems were authored and prepared by pinely Endagorion ch_egor aleex AndreySergunin budalnik cdkrot gepardo gleb.shnshn Ormlis VArtem along with AmShZ and our special friend Golovanov399

Thanks for all your ideas, remarks, improvements and attention for details (*ˊᗜˋ*)/ᵗᑋᵃᐢᵏ ᵞᵒᵘ*

We would like to express our special thanks:

And, of course, you for participating (ᵔ◡ᵔ)

Prizes: The top 15 contestants will receive branded Pinely zip hoodies. Each hoodie can be personalized with custom name patches, making every prize truly unique.

To learn more about pinely you can visit our site or you can find our colleagues on CF in pinely organization.

Current openings are HERE

UPD: Score Distribution

500 — 1000 — 1750 — 2250 — 3000 — 3250 — 3750 — 5000

☆ﾐ(o*･ω･)ﾉ Editorial is HERE

Congrats to winners! And everyone for participating ♥♥♥

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

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6 months ago, hide # |
← Rev. 3  
Vote: I like it -29 Vote: I do not like it

good luck may you get positive delta

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6 months ago, hide # |
← Rev. 3  
Vote: I like it -66 Vote: I do not like it

good luck everyone

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6 months ago, hide # |
 
Vote: I like it -36 Vote: I do not like it

Pinely Round is back after 15 months!

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6 months ago, hide # |
 
Vote: I like it -23 Vote: I do not like it

let's goooo, pinely personalized hoodies! Thank you for everything you do for the CP community :)

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6 months ago, hide # |
← Rev. 2  
Vote: I like it -19 Vote: I do not like it

:(

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

Good luck everyone :)

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6 months ago, hide # |
 
Vote: I like it -61 Vote: I do not like it

Hot round

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

Golovanov399 you special fried chicken

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

As a tester, I wish everyone a great time enjoying these interesting problems.

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

Note the unusual start time?

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

vary nice

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

I wish that there is atleast 1 interactive problem!

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

Nice photo. It reminds me a little of those photos where they find narcotraffickers and the police turn around to avoid exposing their faces. No offense.

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

All the best everyone !!

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

Score distribution?

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

Whoever read this comment will be increase rating

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

good luck everyone (*^‿^*)

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

Goodluck everyone.

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

Tips to become permanent speaclist...not just crossing the 1400 ? Please enlighten me.

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

I want a hoodie too :(

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

Well, I expected to join this contest as my first one, but the time isn't so friendly for Chinese competitors(I think so).

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6 months ago, hide # |
← Rev. 2  
Vote: I like it -8 Vote: I do not like it

worst contest of my life.

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

Will this contest be too hard for newbies?

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

Score distribution?

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

Any update on score distribution and the presence of an Interactive problem?

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

There're 2 which will start on on the paragraph 2.

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

Only 5 hrs remaining,where is the score distribution?

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6 months ago, hide # |
← Rev. 2  
Vote: I like it +14 Vote: I do not like it

Where are the score distribution?

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

Score distribution is on the way (￣^￣)ゞ

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

This was my first time testing. As a tester, GLHF!

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

wish it wasn't this late, but hope It's a great round for everyone :)

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

user:SirTechnical

isn't rendered correctly ig .

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6 months ago, hide # |
 
Vote: I like it +53 Vote: I do not like it

It is two hours before contest but we still don't know score distribution.

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

I took a hiatus from competitive programming and decided to restart this week. I took a look into my contest history and what a surprise — the last contest I participated in was Pinely Round 1! What a coincidence. It must mean positive delta, so let's go.

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

I need the score distribution, SIRS !!!

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

Ta Da!

Score Distribution here 500 — 1000 — 1750 — 2250 — 3000 — 3250 — 3750 — 5000

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6 months ago, hide # |
← Rev. 2  
Vote: I like it -14 Vote: I do not like it

CDN

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

My friend Meeple got banned, but i am sure he wasn't cheating. Can he get the reason for being banned?

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6 months ago, hide # |
← Rev. 2  
Vote: I like it +54 Vote: I do not like it

Wrong answer on pretest 2

Wrong answer on pretest 2

Wrong answer on pretest 2

Wrong answer on pretest 2

Wrong answer on pretest 2

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6 months ago, hide # |
← Rev. 2  
Vote: I like it -13 Vote: I do not like it

Too many alts in the scoreboard

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

i dont know that task c is easier than b i cant do b(i just spend 2 hour to task b) =((((

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6 months ago, hide # |
← Rev. 2  
Vote: I like it +43 Vote: I do not like it

my image

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

Implementing B is such a pain in the ass XD

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

    Depends on your idea. Just checking 4 types of paths for every (i, j) works. It's just a recursion

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      6 months ago, hide # ^ |
       
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      Shit!!!!! should have seen that :)

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      6 months ago, hide # ^ |
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      Staircase can be either diagonal , or anti-diagonal. Lets solve for diagonal first...

      Part 1 : There is even simpler approach. if grid[i][j] == '#' then for any Diagonal parallel staircase pattern, (i-j) can have at most 2 distinct value. (handle case of 2*2 square ).

      Part 2: Now, how to handle anti-diagonal staircase ? (simply reverse all the strings in the grid,,, anti-diagonal will become diagonal ). Make a call to Part1 function.

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

      I have also seen that, but I missed the fact that is not enough to only check the first "#" :(, because you can hvve something like:

      ...##...

      ..#....

      I lost 2 hours finding out that break... Problem C was much, much easier. I assume problem D was also easier than B given the same amount of time for it.

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6 months ago, hide # |
← Rev. 2  
Vote: I like it +16 Vote: I do not like it
Idea for B
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6 months ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Here come the downvotes lol. What even was B, C was so much easier. Can someone drop some hints for B?

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

Problem B is harder than usual. For me, it's harder than problem D.

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

What are the ideas for D?

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

worst contest of my life.

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

Problem B >>>> Problem C

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

I don't know if I got lucky with $$$B$$$ or if I've really improved at solving constructives.

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

Any hints for c?

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

    I guessed that prefer maximum value when points can be earned else just choose minimum value,

    also more comfortable in guessing because pretest = system test

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

    you can make max floor(sum(ai)/X) of products that gives you bonus points. Let this be K. Sort all products, take K most valued.

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

    Think of how you can increase bonus by using largest prices

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

    We can solve the problem by Greedy Algorithm and Double Pointers.

    Consider currently the total cost is $$$S = qX + r$$$, where $$$0 \leq r \lt X$$$, then now we have loyalty score of $$$q$$$.

    Getting $$$a_i$$$ bonus points is equivalent to $$$r + a_i \geq X$$$, since our new cost became $$$S = qX + r + a_i = (q+1)X + r'$$$, where $$$0 \leq r' \lt X$$$, $$$r' = r + a_i - X$$$, and that means our loyalty score increases from $$$q$$$ to $$$q+1$$$.

    It's better for us to get bonus points when purchasing expensive items. Then we sort the original array at first, initialize $$$l = 1$$$ and $$$r = n$$$. We need to record our current cost $$$cur$$$ in modular $$$X$$$, so at first $$$cur = 0$$$. Every time we do the following process:

    • If $$$cur + a[r] \geq X$$$, we immediately get $$$a[r]$$$ bonus points, and $$$cur$$$ becomes $$$(cur + a[r]) \mod X$$$, and then decrease $$$r$$$ by $$$1$$$.
    • Otherwise, we buy $$$a[l]$$$ to make least waste, $$$cur$$$ becomes $$$cur + a[l]$$$, and then increase $$$l$$$ by $$$1$$$.

    The process terminates when $$$l \gt r$$$, that means we have selected all the items. We have calculated the final bonus points in the process, and we can use a vector to record the order we purchase items, finally just print them out is OK.

    For each testcase the time complexity is $$$O(n\log n)$$$, which is when we are sorting the array.

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

Spent ~45min on a wrong-understood problem

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

B — you can only make staircase like pattern, bit more code than I expect for B — i am hoping authors have a simpler solution in mind for choosing the problem as B

C — greedy guessforces

D — can't solve, please give ideas !!!

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

I don't even want to upsolve B. Why do you people keep such problems which don't happen even after staring at the screen for hours and hours? My friends advised me against participating in pinely rounds telling that they are trash contests which just test constructives and are good for dropping 200 points. Yeah, I am the fool here.

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

    If you're weak in any topic, you should solve more problems of that topic and get good at it. Not complain aobut it.

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

      after reading the solution of b i feel like a stupid shit because in my mind i was stuck proving things which were untrue. Skill issue from my side.

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

        What were you proving? If you want the intuition for B's solution, then try drawing a grid and coloring the cells. If you start at a cell, and keep coloring the orthogonal cells while following the 3rd condition, you will realize that either you can form a 2x2 square, or a staircase like pattern.

        My solution was just to check for these cases. If there are total 4 black cells then check for a 2x2 square. Also, check for all the different staircase patterns possible.

        Implementing the staircase check can be tricky, but you can just repeat code if all else fails.

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

Image

For problem B, check all black cells belong to 1 of 4 these paths

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

Problem B was B div 2 type but much harder than c, I think problem b worth more score

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6 months ago, hide # |
 
Vote: I like it -14 Vote: I do not like it

Big thanks to the organisers for the worst round I’ve ever taken part in.

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

why my code is skipped? could someone please explain? Does that mean the system thinks I cheated? That's not right... 346728132

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    6 months ago, hide # ^ |
     
    Vote: I like it +19 Vote: I do not like it

    If you have multiple submissions for the same problem which passed pretests, only the last one counts, others are skipped.

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

      oh my, thank you. I just wanted to check whether another approach is correct. It's lucky that the time span is not so long, so I won't lose so much score. Definitely won't do that again in the future.

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

    coz you made one more submission later.. only last one counts.

    I think in this case cheating is not the reason

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

    Do not resubmit. Especially we don't have hacks for A-D now.

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

B>>>>C

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

This does happen when the number of testers are less than that of authors.

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

    I was surprised only 12 testers test this round. In my round we have nearly 40 testers with at least two testers for each level from grey to LGM. If they have more expert testers or below they will not make B so hard and C so easy, at least they will swap B and C.

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6 months ago, hide # |
 
Vote: I like it +45 Vote: I do not like it

I feel some good things and bad things in this contest, so please let me share:

  1. Revealing PT=ST, and Disabling hack is a good decision in nowadays CP. But why only A-D? In my room, no one get AC for >=E except me, so there is no chance to hack for me, but in some other rooms, someone may have chance.
  2. Though the easiest solution is very light (and it's not so hard to reach it for Div1 participants imo), B is too tough for it's score and position. Maybe adding few <=blue testers resolve this issue?
  3. C is very cute as first-half problem, I love it!

overall, the contest is interesting for me, but there is a room for improvement. Thanks for the round!

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6 months ago, hide # |
← Rev. 2  
Vote: I like it +13 Vote: I do not like it

What a round!! I'm never participating in a Div 1+2 again , it feels like an unbreakable curse.

You made me lose delta in under an hour. The curse is now officially proven.

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6 months ago, hide # |
← Rev. 4  
Vote: I like it +11 Vote: I do not like it

I think the testers team should have had some people with rating < 1900. That would have given you a better indication of the problems difficulty.

Thanks for the time you put into preparing the problems though!

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6 months ago, hide # |
← Rev. 2  
Vote: I like it +16 Vote: I do not like it

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

why my problem A hasn't get into Final test? 346670098

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

    In this round, hacks for problems A-D are disabled. In problems A-D, system tests are equal to pretests. In problems E-H, system tests are pretests plus hacks.

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

      But my score of this problem gone anf the system test is over and it showed me -1 on this question but I truly passed it in the exam you can see my submmisions

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

Rage Bait contest !!!

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

Why is hacking of question C prohibited? I passed the test 346709015, but I later discovered that there were issues with my code, such as 1 4 4 3 3 3 10 The answer might be 19, but my output was not that.

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

    I don't think so, all permutations of 3 3 3 10 gives an answer of 16

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

    Sorry, I overlooked the restrictions of the question. Nevertheless, I still feel that my code is not quite correct.

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

      Why do you think it's incorrect, even though it passed system tests and you are not able to find a counter test case? FWIW, I stress tested your solution for $$$1 \le n \le 100, 1 \le X \le 1000$$$, with randomly generated arrays and your answer matches that of my own solution.

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

        Thank you for testing my code. In fact, the test cases I provided earlier were actually my own counterexamples, but they were illegal data. My code was also developed without the constraint that $$$a_i \lt = X$$$. I made a mistake in interpreting the constraints of the problem.

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

great contest! thank you very much. i would say that B is too hard for B. D is such a cool problem! and F is very nice.

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

    I think the contest setters underestimated how hard it is to make the necessary observation.

    I didn't see it until I looked at the editorial, and all the while I thought it was DFS(I have a habit of solving problems with DFS when they ain't DFS)

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6 months ago, hide # |
← Rev. 2  
Vote: I like it +10 Vote: I do not like it

I disliked the round. I think it was poorly prepared, which is hinted by the late anouncement of the score distribution and by the lack of the list of testers.

  • B: Bad problem for a B problem. Too observation heavy, too implementation heavy if done in a suboptimal way.
  • C: I was too tired after B. After the contest I checked the solve count and the comments about it and found out that it was just a guess problem.

I had, I believe, my worst performance this round. But I also think that the round was one of the worst I've seen so far in terms of the problem set balance.

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

Was the solution of E using this technique planned? Or were the constraints just randomly chosen to be small?

My solution is a bit of an overkill, it runs in $$$O(n \sqrt n)$$$ (or maybe less if the queries are actually in a good order which I didn't manage to see) and calculates prefix sum of $$$C_n^k$$$

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    6 months ago, hide # ^ |
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    Short answer: well, yes (at least my solution works like this)

    You actually can save the current value of $$$\Sigma_{k=0}^yC_x^k$$$ where $$$x$$$ is the number of free cells and $$$y$$$ is the max number of ones we can place there. Now let's see that both $$$x$$$, $$$y$$$ change only by one each time (if moving from right to left), thus it's easy to recalc. If $$$A$$$ is the current value — you can increment $$$x$$$ making $$$A := 2 \cdot A - C_x^y$$$ and easily increment / decrement $$$y$$$ when necessary.

    Time complexity is $$$O(N)$$$

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

lol. why are there soooo many submissions for H??

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

oh no. I thought I coud get positive delta, maybe due to the starting and ending time there are many registered participants didn't participate so my rank is actually a lot lower than it seems to be.

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

I have wasted 2.5h on problem D, any hints please?

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

Good contest! Thanks

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

orz 353cerega

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

Thankyou for the great round, the questions were great. B was actually easy once you figured out the 3 cases. Overall missed specialist by not being able to solve B quick enough and also not being able to solve D, else could’ve got a delta of ~150

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

Thanks for the great round — I particularly enjoyed problems D and E. For D, I’ve seen a large number of different solutions (and mine is different from all others I’ve seen :)), which really adds to the beauty of the problem.

Just a small proposal (though there might be downsides I haven’t fully considered): would it make sense to somehow group Div. 1 participants in the same rooms for contests where hacks are disabled on the easier problems? I really enjoy the hacking aspect of the contest, and I think most of the people solving any of the harder problems were among the only ones (if not the only ones) to solve anything >= E in their room during the round.

Thanks again!

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    6 months ago, hide # ^ |
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    isn't D dp + binary search. What other ideas are there can you please tell me?

    For E, I noticed that once we switch 1->0 or 0->1 we need sum 0 for next k-1 always. But could not code in time.

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    6 months ago, hide # ^ |
     
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    group Div. 1 participants in the same rooms for contests where hacks are disabled on the easier problems?

    There are similar plans.

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

      I really hope Codeforces doesn't go in this direction with regards to hacking... In the recent years because of the strong tests, hacking can be mostly ignored, which is a great thing.

      My personal feeling on hacking is that it incentivizes bad behavior, such as making your code harder to read or putting in something which looks like a bug in your code which actually isn't. As a contestant, I don't want to have think about these concerns. It's great that everyone can choose to solve the problems in their own style, programming language, etc without having to worry about what others are doing. (Not to mention hacks adding unneeded randomness/unfairness to the results.)

      I enjoy CF for the well-prepared and interesting problemsets, and trying to solve those problems quickly. Going back to the days of hacking would cheapen the contest experience which in my opinion is great as it is.

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        6 months ago, hide # ^ |
         
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        My personal feeling on hacking is that it incentivizes bad behavior, such as making your code harder to read or putting in something which looks like a bug in your code which actually isn't.

        Obfuscation is technically against the rules on Codeforces. The extent of what is considered obfuscation would be subjective, but the rules do state that it is prohibited.

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        6 months ago, hide # ^ |
         
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        I can't say I disagree with more-or-less any of the points you are making, but my take on this would be a situation (which happens pretty frequently atleast around the rating I am):

        • You've just solved a problem (let's say E), and there is ~20 minutes before the end of the round. The top contestants took ~20 minutes to solve F (the next one), so there is no real chance of that problem being solved by someone mediocre. In that case, spending the ending of the contest taking a look ot others' codes (potentially learning something from them) + being able to climb up a few places if successful is a very fun thing to do, at least in my opinion.

        I fully agree that it adds some randomness and that it is imperfect, but I still think that there might be some pros of having hacking included.

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6 months ago, hide # |
 
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Legend tourist is back!

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6 months ago, hide # |
 
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Whoever got a negative delta downvote this announcement blog

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i think B harder than C

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Problem B was the worst I've ever seen.

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6 months ago, hide # |
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In problem B 3rd test case

3

.##

.##

...

How does it satisfy: condition 2 & 3

if I make (3,1) to black 3

.##

.##

#..

Then condition 2 :it should be possible to go from any black cell to any other by crossing several vertical or horizontal cell borders while visiting only black cells. You can't go directly through the corner of a cell. Then (2,2) to (3,1) we have to go through corner.

If we make another white cell to black we will violate condition three.

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    6 months ago, hide # ^ |
     
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    and why would you do that? it says we want to connect all black cells that are present "initially". In this case there is just one component so its fine already. This was a cool pattern observation problem though.

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      6 months ago, hide # ^ |
       
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      In question it is not mentioned initially and also they mentioned to achieve the following conditions we can change . to # .

      Thank you for replying I'm thinking, I was missing something

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        6 months ago, hide # ^ |
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        You can change '.' to '#', but you are not forced to do that. You can also just leave everything as is. In that one example there is no need to change anything, because all 3 conditions are already fulfilled.

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          If I change . to # it violates condition it violates 2 or if I make another change it will violate 3 as well. Then according to output mentioned

          For each test case, print "YES" if it is possible to paint some white cells black to satisfy all the conditions, and "NO" otherwise. __

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            6 months ago, hide # ^ |
             
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            "some white cells" can also be "no white cells". You can paint no white cells and the conditions are satisfied.

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6 months ago, hide # |
 
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Just a FYI, Axoryn is a cheater. One can easily see from his LeetCode submits that's he's cheating. Go to this link and click on his Q4 video: https://leetcode.com/contest/weekly-contest-473/ranking/?region=global_v2

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Problem B was wayyyyy harder than C :/

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Does anyone know what testcase 76 in 'test2' is? I'm failing only because of that specific testcase. in the question B

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As a tester, I wish everyone a great time enjoying these interesting problems.

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6 months ago, hide # |
 
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Can anyone let me know what is the rating for the problem D

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5 months ago, hide # |
 
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Hello!

I really don’t understand what’s going on — this is the third time I’ve received a notification saying my submission has duplicate content. I want to confirm that I have never copied any code from anywhere before submitting. Maybe I should stop competing on Codeforces because of this.

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Hello,

I received plagiarism warnings for problems 2161A and 2161D. I want to clarify that I solved both problems completely on my own and never shared or copied code from anyone. For 2161A, the solution was quite straightforward, so it’s natural that many participants could write very similar code. It’s also possible that I might have used an online IDE during the contest without realizing it could make my code public. I’ll make sure to use only private or offline environments in the future.

Kindly review my case if possible. Thank you for your time and understanding.

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Hello. My submission 346706198 for problem 2161D was marked as similar.

I recently read that blog: https://mirror.codeforces.com/blog/entry/105697

Here is written:

"In any bipartite graph, the number of edges in a maximum matching equals the number of vertices in a minimum vertex cover."

I solved the problem using that approach.

The key observation is that we can assume each number as a vertex, and there is an edge (i, j) when i < j and a[i] + 1 = a[j]. Because we put edges between odd and even numbers, it makes graph bipartite, partitioned by parity groups.

The task then reduces to covering all such edges with the minimum number of vertices, which is exactly the Minimum Vertex Cover problem on a bipartite graph. By König’s theorem, the size of the minimum vertex cover equals the size of a maximum matching as mentioned above. So we construct the bipartite graph, find the maximum matching, and the answer follows directly from |matching|.

I can provide handwritten notes if needed. Thank you for your time.

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I didn’t copy my code from anywhere; maybe it just happens to coincide.i write code with my own..i have proofe of my logic ..if you want to know..then you can knock me...i happily tell you about my logic

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Attention! Your solution 346728172 for the problem 2161B significantly coincides with solutions woony/346697295, nurdaee/346715021, MrWizard_574/346728172. Such a coincidence is a clear rules violation. Note that unintentional leakage is also a violation. For example, do not use ideone.com with the default settings (public access to your code). If you have conclusive evidence that a coincidence has occurred due to the use of a common source published before the competition, write a comment to post about the round with all the details. More information can be found at http://mirror.codeforces.com/blog/entry/8790. Such violation of the rules may be the reason for blocking your account or other penalties. In case of repeated violations, your account may be blocked.

I don't know why I got this message, I have not cheated nor do I know these guys. Also, they made this round unrated for me. What should I do?

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Hello, I received a plagiarism warning for submission [346728714] in problem 2161C. I did not share or copy code from anyone. I only submitted my solution on Codeforces and nowhere else. I implemented it myself after understanding the idea from the editorial. The similarity must be due to both of us following the same approach. Please review — this was my own work. Thank you.[submission:346728714][problem:2161C]

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Hello, I received a letter stating that my solution 346697295 for problem 2161B significantly coincides with the solutions of other participants and is included in a group of identical solutions: woony/346697295, Nurdaulet-018-2029/346715021, MrWizard_574/346728172. I wrote the code entirely on my own during the contest, I do not know the other participants mentioned, and I did not use any external sources. The problem was constructive and required checking "stairs" and "squares" because they are the only possible arrangements of black cells, and the implementation involves repetitive code segments due to the nature of building these stair patterns. This logically led to similar implementations among many participants, including myself — hence the similarity in code structure and logic. I did not use any public online IDEs and did not share my code with anyone, so I ask you to reconsider your decision and take into account that the coincidence was unintentional, resulting from a standard solution approach rather than a rules violation.

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it give me -65

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