mohammedehab2002's blog

Codeforces Round #628 (Div.2)
By mohammedehab2002, history, 5 years ago, In English

Hi!

Codeforces round #628 (vintage Codeforces round #2) will take place on 14.03.2020 17:35 (Московское время). It's rated for the second division, but, as usual, first division participants can take part out of competition.

The problems were created by me. I'd like to thank, and orz, antontrygubO_o for coordinating the round; ramchandra, pajenegod, aryanc403, taran_1407, Kuroni, mcdx9524, dorijanlendvaj, Andreasyan, 300iq, zoooma13, Osama_Alkhodairy, Mohammad_Yasser, and TripleM5da (special shout-out) for testing the round; and of course MikeMirzayanov for the great codeforces and polygon platforms.

You'll be given 6 problems and 2 hours to solve them.

UPD: the scoring distribution will be 500-750-1250-1750-2500-2750.

UPD: the editorial is out.

UPD: congratulations to the winners!

Div.1:-

  1. tmwilliamlin168
  2. jqdai0815
  3. Um_nik
  4. imeimi
  5. HIR180

Div.2:-

  1. davooddkareshki
  2. rainboy
  3. PouyaNavid
  4. _Lucien
  5. socho

Good luck & Have fun!

  • Vote: I like it
  • +671
  • Vote: I do not like it

Comments Comments (159)
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5 years ago, # |
  Vote: I like it +320 Vote: I do not like it

Good to see Round pi*200 on pi day.

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5 years ago, # |
← Rev. 2   Vote: I like it +6 Vote: I do not like it

Are you sure you're going to be available on that day?

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

    Depends on the timezone you consider ;-;

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5 years ago, # |
  Vote: I like it +55 Vote: I do not like it

I'd like to thank, and orz, antontrygubO_o

inb4 anton quits coordinating rounds like he quit AC

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

tau is superior

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5 years ago, # |
← Rev. 2   Vote: I like it +55 Vote: I do not like it

Another contest by one of my favourite problem setters — mohammedehab2002. I'm going to revise XOR problems to prepare !

UPDATE — I will lose ratings but I enjoyed the hell out of it ! Thanks mohammedehab2002 for preparing an excellent contest and I hope to see many more from you in the future !

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

    yeah...looking forward for "Ehab and..." problem :3

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

make the problem description short like the announcement ^_^

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

Most of the Problem title would look like 'Ehab and bla bla bla ....'

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

orzing is forbidden

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

what do you mean by (vintage Codeforces round #2) ?

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5 years ago, # |
← Rev. 2   Vote: I like it +21 Vote: I do not like it

*le xor problems
here we go again.

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

If i registers for contest. But will not give this contest. Will it affects my rating?

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

    No, as long as there is no submission from your account

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

    No, but if you do at least one submission during the contest you will be affected.

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

      if submission doesn't pass first test_case, then it doesn't affect rating.

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

        No. If you do at least one submission during the contest, it will affect your rating (even if it fails the first test case). But of course, it doesn't add to your penalties (-50 points for the classic Div. 1/2 or +10 minutes for Educational or Div. 3).

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

I think you missed this:

مرحبا كودفورسز (hello codeforces)

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5 years ago, # |
  Vote: I like it +70 Vote: I do not like it

xor problems, here we go again.

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

Is it rated till 1900 or 2100 ?

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

    As mentioned here.

    The 1900+ rating means that you're part of the first division

    So, 1900 i believe.

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

      No, when a div1 round and a div2 round both are going on simultaneously then the div2 round is rated for participants upto 1900 rating. If only a div2 round is being conducted, its rated for participants having a rating upto 2100

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        5 years ago, # ^ |
        ← Rev. 3   Vote: I like it +21 Vote: I do not like it

        Well I know this fact. And I'm sure even you would know that each and every time, any Div 2 contest is hosted alone, it is always mentioned in the contest announcement post 'rated till 2100'. That was why I preferred confirming once.

        I don't understand why people here just start downvoting the comments. If the query was so stupid, please accept my apology and don't mind :)

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          5 years ago, # ^ |
          ← Rev. 3   Vote: I like it +26 Vote: I do not like it

          Yes, I agree, people often downvote genuine comments as well without a proper reason, maybe just because it has already accumulated a few downvotes.

          Maybe this comment of mine will be downvoted for no reason.

          But one more thing is there, maybe people thought your question was a little stupid (the part you wrote about the 2100 rating always being mentioned wasn't written)

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

      2100

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5 years ago, # |
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I hope the the expected XOR Problem is replaced by a constructive or number theory problem. orz

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5 years ago, # |
← Rev. 3   Vote: I like it -67 Vote: I do not like it

1)

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

Is it rated?

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5 years ago, # |
  Vote: I like it -84 Vote: I do not like it

Hello. Can you please reschedule the round for 12 hours (Moscow time), in connection with the holding of another Olympiad at 15:00 (Moscow time). Thanks in advance

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5 years ago, # |
  Vote: I like it -14 Vote: I do not like it

Hello. Can you please reschedule the round for 12 hours (Moscow time), in connection with the holding of another Olympiad at 15:00 (Moscow time). Thanks in advance

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

    You can give virtual ,as it will be unrated for you anyhow.

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    5 years ago, # ^ |
      Vote: I like it +101 Vote: I do not like it

    I'll ask my coordinator.

    Hello. Can you please reschedule the round for 12 hours (Moscow time), in connection with the holding of another Olympiad at 15:00 (Moscow time). Thanks in advance

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

Get ready for more XOR problems !

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5 years ago, # |
  Vote: I like it -27 Vote: I do not like it

Can the contest be delayed 1 hour please because I wouldn't be available at this time. mohammedehab2002

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

    Hello. Can you please reschedule the round for 12 hours (Moscow time), in connection with the holding of another Olympiad at 15:00 (Moscow time). Thanks in advance

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    5 years ago, # ^ |
      Vote: I like it +55 Vote: I do not like it

    Hello, aihay.

    Ah, i see, Mike didn't discuss contest timing with you this time. Forgive him, he's a bit busy preventing corona from spreading through CF, you know.

    I'll make sure this will not happen again! My apologies for this awful inconvenience. CF stuff will be punished, I promise!

    Hope you like the new timing! If you have any other problems with CF feel free to DM me!

    Best regards, kostia244

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

Will this contest effect my rating ....if my current rating is 1255,._???

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

Xor here u are

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

Waiting for interesting problems as expected from you

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

Hope that I can solve ABC <3

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5 years ago, # |
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Seeing the score distribution: 500-750-1250, I think the difficulty gap between A and B will not be as big as other contests.

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    5 years ago, # ^ |
      Vote: I like it -62 Vote: I do not like it

    Oh, you are so smart!

    Why you are still newbie, phps?

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

It's exciting that Codeforces is holding more and more vintage rounds (ʃƪ ˘ ³˘)

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

Why a lot of comments are about XOR-problem. Could you explain me this joke

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

Lets starts xorforces...

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

owee great new XOR problem is coming ... :)

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

Maybe today I will reach master before CoronaVirus reaches me :D

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

tedesjo fire ||| _ __ _ ## (_ " + __ _ )) )) + || | | _+ _

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

15600+ registration! Get ready for long queue! O_o

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5 years ago, # |
← Rev. 2   Vote: I like it -51 Vote: I do not like it

Deleted

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

timelimit on c with O(n) with python :(

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

MathForces > ArrayForces

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

I did A and B in 30 minutes, still my rating is going down, C is too much for me, oh well.

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

As expected , Another xor problem D.

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5 years ago, # |
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When I submit, the output that is on your testing is different from the output when I run the program on my computer. How do I fix this?

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

How to solve C?

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

    set smallest numbers to edges to leaf, and others randomly)

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

      You would have to take care that the parent on the leafs where you place the edges 0 and 1 is not the same vertex. Because if it is, that MEX will be 2 instead of 1.

      Edit: It turns out I am completly wrong with this. There is no need to take care for the parent. The only thing to take care of is not placing the edge 2 between the edges 0 and 1. For that it is perfectly fine to simply place the numbers 0,1,2 in leaf edges.

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

    If it's a chain, then fill all the edges randomly.

    Otherwise, choose a node of which degree is 3 or above, fill three of the edges using 0, 1, 2; others random.

    It can be proved to be one of the best answers.

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

    Here's what I did : Sort all vertices by their degree in descending order. Now, starting from the vertice with the maximum degree, try and fill all its unfilled edges with numbers in a serial manner( Maintain a count variable. Fill edges with the count variable and increment it everytime you fill an edge )

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5 years ago, # |
  Vote: I like it +239 Vote: I do not like it

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

How to solve problem D?

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

    I tried to construct bit by bit, starting at lowest bit and the xor input. Then add bits to get the sum, by adding the one lower bit twice to two numbers in the array. And so on...

    Did not work for some reason.

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

    max length of the answer is 3

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

    The answer is at most 3 or it is impossible.

    If $$$u$$$ and $$$v$$$ don't have same parity or $$$u>v$$$, it is impossible.

    If $$$u=v$$$ the answer is $$$0$$$ if $$$u=0$$$, $$$1$$$ otherwise.

    Otherwise, try to build a pair which xor gives $$$u$$$ and sum gives $$$v$$$. Let $$$t=(v-u)/2$$$. If $$$u$$$ and $$$t$$$ don't have any bits in common, a pair of solution is $$$u+t$$$ and $$$t$$$.

    Otherwise, $$$t$$$, $$$t$$$, $$$u$$$ is a solution.

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

    Look at the bitwise representation of $$$u$$$. Every bit that equals 1 means that there is an odd # of elements with that set bit. Every bit that equals 0 means there is an even # of elements with that set bit (because 1^1^... odd number of times = 1, 1^1^... even number of times = 0). So one of the array elements is at least equal to u. Suppose first element of array = $$$u$$$. Then we have to add two elements such that we don't change the parity of set bits; i.e., we must add two of the same element, call it $$$y$$$. This is 3 total. But then we can consolidate one of the $$$y$$$ into $$$u$$$, iff $$$u$$$ AND $$$y$$$ equals 0.

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      5 years ago, # ^ |
      ← Rev. 2   Vote: I like it +6 Vote: I do not like it

      I had this idea, too. if u-v is even, add (u-v)/2 twice. But how does it work if u-v is odd?

      We cannot add both numbers, because one bit will be set wrong then.

      Edit: Fuck it, the parity of u and v is same! I had this before, checked and output -1. So, if the parity is same, the difference is allways even. The odd case does not exist.

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

    You can make array of one element if the XOR value equals to SUM,

    the one's bits in XOR must appear in odd numbers in array.

    so we can make array of two elements if the following conditions comes true,

    one of the two numbers has the one's bits in XOR

    and the remainder rem = (XOR — SUM),

    can be distributed among two numbers (must be half in the first and half in the second to get XOR ZERO)

    and the XOR of those two number must be the same of XOR in the input

    and you can make array of three elements if the following conditions comes true,

    one of the three numbers has the one's bits in XOR

    and the remainder rem = (XOR — SUM),

    distributed among the other two numbers equally to get (XOR zero) when XOR them,

    so when we XOR the input with zero we got XOR.

    otherwise there is no answer.

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5 years ago, # |
  Vote: I like it +107 Vote: I do not like it

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5 years ago, # |
  Vote: I like it +32 Vote: I do not like it

Me after solving ABCD:

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

what is test case 18 in problem D

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

    I got wrong answer in testcase 18 because of not handling the answer for size 3 properly.

    Basically, for size 3 you have to print u, (v — u)/ 2, (v — u) / 2

    Instead, I printed u, v / 2, v / 2

    I think you have missed this point too.

    Though, thankfully i spotted my mistake quickly during the contest time. :3

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      5 years ago, # ^ |
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      i got my mistake something like your's..thanx for answer

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5 years ago, # |
  Vote: I like it -14 Vote: I do not like it

Is: (a + b + c) = (a XOR b XOR c) + 2 * (a AND b AND c) ?

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

is E shortest cycle detection after reducing prime divisors to a graph?

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

How to solve C & D?

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

How to solve the case in E, when you must find the shortest cycle in the graph? :c

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

    Disclaimer: I don't have an accepted version yet.

    Spoiler
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5 years ago, # |
  Vote: I like it -13 Vote: I do not like it

It is rated?

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

How to Solve C?? Is there any Pattern ?

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

    Label any 3 edges, which connects a leaf vertice, with 0,1,2

    You can label other edges with any numbers avaliable.

    If the tree is a chain, all of the labeling way are acceptable.

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

    if you find one node like x that has 3 neighbors then use 0,1,2 for labels of edges between x and neighbors and you can chose labels of other edges in any order and you will get MEX(u,v) = 2

    if you can't find node like x then chose all labels of edges in any order and you will get MEX(u,v) = n — 1

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

How to solve D?

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5 years ago, # |
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Was problem F related to finding if the graph is triangler graph?

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

For problem C - I took number of leaf nodes of the tree count and assigned count to n-2-count to those edges and remaining values to the nodes with children. Was I wrong?

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

    For second sample, count=3, you assigned 3 to (6-2-3)=1. Even 0 must be used for those edges.

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

So, how do you solve E?

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5 years ago, # |
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Please help me in solving question D. Thank You.

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5 years ago, # |
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Does anybody know how to solve F?? i thought the articulation point.. but i can't solve this

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    5 years ago, # ^ |
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    Please help me in solving D. Thank You.

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

      you think that first value is v and rest value is adding number to u to satisfy xor

      if bit of difference of v-u is 1, you can make two number (1<<(bit-1) and (1<<(bit-1) because (1<<(bit-1))^(1<<(bit-1)) = 0 and (1<<(bit-1)) + (1<<(bit-1)) = (1<<bit)

      so you can make three values in array

      but if (1<<(bit-1))^v == 1, you just add the number v to (1<<(bit-1)) so you probably make the two number in array

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

The editorial is out. It's the fastest editorial I've ever seen. Orz

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

E:

We can divide every number by square of a prime until it is a product of primes. If there is an $$$1$$$ or 2 indices with the same number, we already found an answer.

Also observed that in each number there will be at most one prime larger than 1000.

What can be done next to get the solution?

UPD: seems I miss that its "divisors" not "prime divisors", thanks editorial.

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5 years ago, # |
  Vote: I like it +44 Vote: I do not like it

Screencast

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5 years ago, # |
  Vote: I like it +43 Vote: I do not like it

Screencast with commentary

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

Hey, can anyone please tell me the total hacking time for this particular contest?[contest:628][contest:628(Div2)]

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

    There is no hacking time, when it is finished, it is finished.

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

    Usually, it is the Div 3 and the Educational Rounds which have hacking phases after the coding phase of the contest. In Div 2 and Div 1 contests, there is only the coding phase but not hacking phase.

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

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5 years ago, # |
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Strangely, I started to read coronavirus instead of codeforces :o

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

Other Solution for C:

count = no of leaf node(node with degree == 1)

if(count >= 3):set any three leaf node to 0, 1, 2 that will set the upper boundary for MEX to 2

because any path in tree can have maximum 2 leaf nodes

if(count < 3) arbitary

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

    I did this too! The editorial solution made me realize that there exist at least 3 leaves iff there exists a vertex of degree 3. It is a special case of this theorem

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

Can anyone help with what is problem with test 143 in E? Its too large to debug by hand, and I can't find small counter to my code...

My code: https://mirror.codeforces.com/contest/1325/submission/73272537

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

Lol. Didn't solve D because I didn't consider just 1 case: if(u == v) {cout<<1<<'\n'<<u; return 0;}

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

My Java solution for E passed in 2979 ms. Talk about scraping the time limit.

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

Is this solution (73271134) for F hackable, or is it accidentally correct? I sort of combined both of the editorial solutions without understanding either of them.

First, I attempt to find the independent set with exactly the editorial's "A solution using degrees" algorithm. In a set, keep a list of (degree[i], i) pairs, and keep taking the node with minimal degree and invalidating each node's neighbors, updating the degrees of the neighbors of each invalidated node in the set.

The if (n <= 16) brute-force part is only there to counter some of the manually generated edge cases, which tend to be small.

Then, I run DFS from random starting nodes. This DFS runs in arbitrary order, and when it hits an ancestor in the DFS tree, checks if the cycle is long enough/prints it. It runs until it either finds an answer, doesn't, or TLEs (might be $$$O(n^2)$$$ at worst).

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5 years ago, # |
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is it just me or has codeforces slowed down on the contests! The next one is after 5 days! :(

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

Hi, i was banned for copy pasting. But it's system error :) There are some reasons, why it is impossible to stole his solution: 1) Code is really simple and there were too many participants, that's why chance of two similar codes is really hight 2) I am from Russia, he is from France))) 3) Also i wrote him, and i believe, that he will agree with me, because it's really strange, i didn't know him before this :| Thank you for your attention and i hope that you will solve this problem, because it's really sad(

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

    Hello, I am the person the system claims Witless_Deer copied from. For the reasons he mentioned (especially reason #1), it is ridiculous to claim that he copied me. Look at our submissions for B to see for yourselves.

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5 years ago, # |
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Can anyone explain me Problem C. i am not able to understand it.

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

    Consider the paths between all nodes. If the edge with value 0 is not on such a path, the MEX of that path is 0.

    So, consider the paths with the 0. The MEX of these paths is 1, except if the edge with value 1 is also in this path. Then it is at least 2.

    So, is it possible that there is no path with edges 0 and 1 in the path?

    It turns out, no, that is not possible. Since by the simple fact that the path from two nodes of the edges with value 0 and 1 has both edges in it. So it allways exists.

    But now we are close to the solution. How can we place the edge with value 2 in a way that it is not on that path, too?

    Well, more or less simple, we place that edge not on the path between the edges 0 and 1. Then it is not.

    Last question to answer is, how do we implement this, how can we find suitable vertex and edges? The tutorial answers this clearly: We search an vertex with at least 3 edges, and use these 3. Then the edge with value 2 will not be on path between the edges 0 and 1.

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

Nice Contest <3 :)

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

I like how all Ehab contests are based on intense maths and number theory.

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

    your nickname is very suitable for your comment about math)))

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5 years ago, # |
← Rev. 2   Vote: I like it +17 Vote: I do not like it

Few Hour ago i got a message from System. I'm sharing it here.

**Attention!**
**Your solution 73232959 for the problem 1325B significantly coincides with solutions sazidnur/73232959, secsee/73233935. 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.**

My PC is too slow. So i use ideone to compile my code and also i feel comfort at online IDE but I've made a mistake. Today's morning (Bangladesh Standard Time), By mistake I change my code sharing option private to public and i didn't notice this before contest. So someone got my code on internet. I admit that it's my fault and also i promise that, I will never make this kind of mistake again.

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

I don't know why but my submissions from that contest are still shown as pretests passed. Moreover, all of them were evaluated as failed attempts. And my rating has dropped :/ I firstly thought maybe system evaluated my submissions like I cheated. However, I did not receive anything regarding to that. Can someone check this?

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

    It happens when your submissions are evaluated and you are found guilty under Plagiarism check. I think you should reach out to Codeforces Authorities regarding the same.

    Have a nice day!

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

Really amazing and Mathematical contest. Loved it.

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    5 years ago, # ^ |
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    Exactly.. One more thing which I liked was that problem statements were to the point rather than long paragraphs, and properly framed.

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

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