Sammarize's blog

Codeforces Round 313
By Sammarize, 9 years ago, translation, In English

Good day for all!

I invite you to participiate in the Codeforces round 313, which is prepeared by me and tunyash. Each of us is prepared four rounds then it is our fifth of ninth round to your notice. I figured out almost all problems (except D div.1), wrote the statements and analysis of all the problems, and tunyash has developed all problems.

Gerald is not coordinator yet and you probobly missed him. In this round you will meet him again and help him in his ordinary life problems.

I want to thank Zlobober, our translator Maria Belova (Delinur) and MikeMirzayanov and all Codeforces team for this platform.

This round will be held in unusual time — 17:00 Moscow Time.

Contest finished! Welcome to editoral:
Short editoral.
Extended editoral.

Div.1 scoring distribution:

500 — 1000 — 1500 — 2250 — 2250

Div.2 scoring distribution:

500 — 1000 — 1500 — 2000 — 2500

I wish you to enjoy solving problems!

Congratulations to winners!

Div. 1:
1. jqdai0815
2. qwerty787788
3. SirShokoladina
4. ainu7
5. Endagorion

Div. 2:
1. goons_will_rule
2. lbn187
3. crawling
4. loveannie
5. Jagabee

  • Vote: I like it
  • +330
  • Vote: I do not like it

Comments Comments (217)
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9 years ago, # |
  Vote: I like it +131 Vote: I do not like it

This is always my first thought when I see 313 anywhere:

:)

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

Wow!I'm looking forward to the coming contest!

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9 years ago, # |
  Vote: I like it +78 Vote: I do not like it

>mfw there's finally a round after a long time

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    9 years ago, # ^ |
      Vote: I like it +39 Vote: I do not like it

    I think Zlobober is dead. It's a long time he doesn't reply anyone.

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

      Don't you think it's a bad idea to joke about someone's death?

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        9 years ago, # ^ |
          Vote: I like it +72 Vote: I do not like it

        Nope

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

        He's not joking. he is so serious.

        but why so serious ???

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

        I think he said that nick (Zlobober) wasn't online for a long time :)

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

Another unusual time! I'm happy because I'll be able to sleep before 02:00... though it might collide with others' schedules.

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9 years ago, # |
  Vote: I like it -38 Vote: I do not like it

it is so funny

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

    Really Don Rosa drew it like this :o? It's like a profanation made by a saint ;__;

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

17:00 ? Srsly? Every normal people work in this time (in Russia) -_-

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    9 years ago, # ^ |
      Vote: I like it -37 Vote: I do not like it

    Normal people earn money from home :P

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    9 years ago, # ^ |
      Vote: I like it +156 Vote: I do not like it

    Are programmers normal people? :D

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

      :D

      you are funny always :D

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

    It is only people in our time zone, who works without flexible schedule. It is very few people. Every contest time not comfortable for someone.

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      9 years ago, # ^ |
        Vote: I like it -45 Vote: I do not like it

      "Every contest time not comfortable for someone." — of course, but the time 19:00 — 20:00 is more convenient for more people IMHO.

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        9 years ago, # ^ |
          Vote: I like it +75 Vote: I do not like it

        And hence we must schedule all contest in this time, ignoring all other people?

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

        It's not for me :D. Currently it will be 7-9 AM in my timezone, so I can just go later to work, but 9-11 AM will be too late :P

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

    We take 'usual time' CF rounds at 01:30, and it takes two hours, so if you are a student or have a schedule and you REALLY want to participate, you should spare your sleep.

    I'm not saying that the time CF rounds start should iterate over and over. I just want you to know that there are many people around the globe who cannot(or wouldn't) participate because of the time zone. Give us some chance, please.

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

Thanks guys for taking the time doing all this work!

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

313 is a holy number for Muslims :)

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

i think it's a good time for china coders

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

which rounds did you prepare?

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

3.13 is my birthday! Hope there be a rating harvest!

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9 years ago, # |
← Rev. 3   Vote: I like it +19 Vote: I do not like it

tagged "codefores" :p if its a mistake, surprisingly it was done before :p

UPD: tag was removed :p

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

" Scoring distribution will be known closer to the beginning of round ," this time there is new sentence ;) !!!!!!!!!!!!!!!!!!!!!!!

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9 years ago, # |
  Vote: I like it -43 Vote: I do not like it

tnx for usual time

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

Anybody know at what time the practice problems will be unavailable for judging because of this round! It's very important for me to know... thanks!

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9 years ago, # |
  Vote: I like it +78 Vote: I do not like it

My last contest before IOI!

I hope I can practice with this round.

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

Good luck to IOIers. (Not everyone, hence it is possible :P)

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

Good luck for your round Sammarize :D

I'm very excited to take part in it!

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9 years ago, # |
← Rev. 2   Vote: I like it -11 Vote: I do not like it

My previous dorm room no. was 313 . Loved that room :P hope this contest is some kind of good luck charm for me!

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

wow...scoring distribution occurred so soon :D, really unusual round

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

it seems that 313 has lots of significance :/

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

Can't hack... Codes don't open...

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

I didn't have brave to submit something, but problem set is very nice ! I like first four problem ( in div 2).

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9 years ago, # |
  Vote: I like it +29 Vote: I do not like it

UNSUCCESSFUL HACKING ATTEMPT AT 19:00:00

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

geometry geometry and geometry :P

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

Div1 B has already been given for training to the Junior National Team in our country and the teacher who gave us the problem said that she took it from some Russian site... And announcing the thing for odd length wasn't fair, I needed to resubmit. And it wasn't even added to the statement...

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    9 years ago, # ^ |
    ← Rev. 2   Vote: I like it +51 Vote: I do not like it

    The statement clearly said "If we split string a into two halves of the same size". You can't do that if it has odd length. If anything, it's unfair that the announcement caters to people who can't read.

    EDIT: Ok, I reread the statement more carefully, and I see why the clarification was necessary. "If we split a into two equal halves, then some things are equal" can also be understood as "For every possible way to split a into two equal halves, some things are equal". In this case, if a has even length, there is of course exactly 1 way to split it. But if it has odd length, there are 0 ways, so indeed the things are equal for every way. However, this would imply that the answer is always "YES", which is inconsistent with the sample tests, so you should notice that something is wrong before submitting.

    So, clarification is indeed necessary, I retract my slightly rude comment above, but I stand by my claim that it's impossible to read it in a wrong but consistent way.

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

    "If we split string a into two halves of the same size..."

    If size is odd you can't divide it into two halves of the same size, do you ?

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

    It's a formal consequence of the statement: string of odd length cannot be divided into two halves of equal lengths, therefore strings of odd length are equivalent iff they are equal

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

      Okay, my bad, however it would be better if there was such sample. And still, the problem is known and I guess a lot of people just coded it with no thinking :)

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

Can someone please help me catch the mistake in this code? It never gives correct answer but I think the reasoning behind it is corect.12182932

Essentially I am trying to sort the strings(swapping the halves of the blocks), first in blocks of length 2, then blocks of length 8 and so on. sorting for all blocks of powers of 2 that divide the length of the strings.

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

    Blocks can have odd length (e.g. string of 6 letters has 2 blocks of length 3).

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

      Oh yeah, so I had to make d the greatest odd factor of the string length and then the blocks should be of length 2d, 4d, 8d... right? thank you.

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

Problem C (div 1) is an old problem.

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

Does solution for DIV 1-B is just simply write a recursion function that check what the problem statement said?!?

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

    I think NO. It will be TL. I've used something like merge-sort in O(NlogN).

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

      Well , I got AC , with the recursion function , the problem is only O(lgn)+n , not even O(nlgn) .12177343

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

        T(n) = 4 * T(n/2) + O(1) So it's actually n^2.

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

          Isn't T(n) = 4*T(n/2) + O(n)? O(n) as we need to compare strings if they are of odd length. The complexity is n^2 even for this, but still.

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

        Your complexity is much more actually.

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

        The only difference from your code to my code (12180789) is that I used strncmp and I got TLE. I replaced strncmp by for and... AC -.-".

        strncmp nevermore.

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

    Let's define the following operation on string S: We can split it into two halves and if we want we can swap them. Then we can recursively apply the operation to both of its halves.

    Now we need to make the observation that if after applying the operation for some string X, we can obtain Y, then after applying the operation for Y we can obtain X.

    And for the given two strings, we can recursively find the least lexicographically string that can be obtained from them. Let's say that those strings are A' and B'. If A'==B' then the answer is YES, otherwise it's NO.

    But it seems like almost nobody made those observations since the problem is known and most of the Russian-speaking contestants may have seen it before :)

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

    By the master theorem, your solution would have O(n^2) complexity.

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

    I wrote this recursion and it didn't got TL. I think that it is because I used random to choose the order in which I call recursive function.

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

      No. I removed this randomization, but it still works.

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

      Solution with random (265 ms): http://codeforces.me/contest/559/submission/12169720

      Solution without random (1435 ms): http://codeforces.me/contest/559/submission/12183383

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      9 years ago, # ^ |
        Vote: I like it +24 Vote: I do not like it

      I wrote a naive recursive implementation and got AC 124ms here: 12177962

      I still don't get how my solution ran so fast.

      No random, no optimizing, no anything.

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

        It is probably the order of the recursion call. Check 12187196 and 12187385.

        There is something called luck related to this problem. And seriously, letting us pass O(n^2) solution is not good at all.

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

          It has nothing to do with luck. Look at this C++ code:

          int main()
{
    int x, y;
    cin >> x >> y;
    cout << ((y == 1) && ((1 / x) == 5)) << endl;
    return 0;
}

          What will be the output when you input x = 0 and y != 1?

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

            Yep, but the thing is that you already know this test case and you are sure that first of these equalities will be true. Imagine that y != 1 in all of the test cases and pretests wouldn't contain case x = 0, but final tests would. That's where you are more lucky than someone, who sent cout << (((1 / x) == 5) && (y == 1)) << endl; and got RE (if I'm not mistaken) after final testing.

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              9 years ago, # ^ |
              ← Rev. 3   Vote: I like it +12 Vote: I do not like it

              I agree with you, order matters in this example. But my point was that C++ optimizes bool operations. If first operand of && evaluated to false, then other operand does not compute. Applying this reasoning to solutions with 4 recursive calls in a (A&&B)||(C&&D) manner, we can conclude that there are maximum of 3 calls actually made. Let's look:

              -If A == true and B == true, then neither C nor D are computed, because || already evaluates to true

              -If A == false, then B does not compute, because A&&B is false for sure

              -If C == false, then D does not compute

              -At last, if A == false, C == true, D == true, we get 3 recursive calls, which is maximum.

              So, basically, solutions written in C++ in this manner are not O(n2), but rather .

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

                Maybe i'm missing something, but what about A=true, B=false, C=true, D=true? Then we will get 4 calls. Why isn't it possible?

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

                  Let's use  =  sign for equivalence. Then we have:

                  a1 = b1

                  a2 ≠ b2

                  a2 = b1

                  a1 = b2

                  We know that equivalence relation is transitive, so:

                  a2 = b1 = a1 = b2

                  But we suggested that a2 ≠ b2. It means that such situation is impossible.

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

                Looks like A=true, B=false, C=true and D=false is still possible, for example, when s=XX and t=XY, and that's 4 computations.

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

                  If we take the meaning of A, B, C and D from the statement, then:
                  A = 'X is equal to X' = true
                  B = 'X is equal to Y' = false
                  C = 'X is equal to Y' = true
                  That's contradiction.

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

                  Right you are, my bad!

                  But: another example, checked by writing a program this time.

                  s=XY, t=XX

                  A = (X==X) = true
B = (Y==X) = false
C = (X==X) = true
D = (Y==X) = false

                  A byproduct of checking is that this example is unique (if we don't consider symmetrical cases).

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

                  That is brilliant! My apologies for claiming that 3 calls are sufficient.

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

    Wow, it passed the final test. I'm surprised...

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

What the the possible end configurations in C? I got a point, an equilateral triangle, a trapezoid, a rhombus. What did I miss?

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

Can anybody please see my Div2 E code, and tell me if it is TLE because of python, or my algorithm needs optimizing? :( Wasted 4 submissions on this.

http://paste.ubuntu.com/11920717/

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

    No, a O(H*W*log(N) ) algorithm wont pass with H,W=10^5 and N=1000. The proper solution is O(max(H,W)*N). I couldnt get it correct in time, though...

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      9 years ago, # ^ |
        Vote: I like it +29 Vote: I do not like it

      No the correct algorithm is O((W+H) + N^2). The W+H comes from precomputing binomial coefficients.

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9 years ago, # |
← Rev. 2   Vote: I like it +24 Vote: I do not like it

I feel like problem C(Div1) is famous. I couldn't solve it but I'm pretty sure I've seen it around more than once.

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

How to hack Div1B, those who just recursively check equivalence?

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

    It passed the systest, so the correct question is, how to prove its runtime is better than O(N^2).

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      9 years ago, # ^ |
      ← Rev. 8   Vote: I like it +35 Vote: I do not like it

      I coded my B as a recursive function satisfying T(n) = 3T(n / 2) + O(n), which is n1.5 approximately. This is much less likely to fail than a O(n2). You can check my code for details on how you do the reduction on number of checks needed.

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

        Clever. But that doesn't explain why some people passed without using this trick, for example 12178491

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

          I don't think there is a test case for this solution which will run N^2. ANDs would eventually cancel two extra recursion calls.

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

        Really good observation

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

        Didn't actually check your code, but I guess you wrote A?B:(C&&D) instead of (A&&B)||(C&&D) while B, C, D is compute only when needed.

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

        Can you explain why you can go directly to the return statement in your if without checking the else?

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

          If you're asking about my solution, what it does is:

          Use ~ to denote equivalence. Note that it is transitive.

          Say A splits in a_1, a_2. B splits into b_1, b_2. If a_1 ~ b_1, it suffices to check that a_2 ~ b_2. Indeed, if the other case holds (a_1 ~ b_2 and a_2 ~ b_1), then b_2 ~ a_1 ~ b_1 ~ a_2, so we would get that a_2 ~ b_2 anyways.

          Otherwise, a_1 is not equivalent to b_1, so just check a_1 ~ b_2 and a_2 ~ b_1.

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

      Why did mine fail then? High constant?:/

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

        I think so. Copying substring is way too slow to pass this problem. A char* solution can pass it easily and unexpectedly. 12182556

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

      I know recursion enough good, because I couldn't invent something smarter for half hour and some guys solve it in 5 minutes :)

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

Anyone got accepted problem C(Div.1) using O(n^2*log(10^9)) time ? :\

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

    No... But I got accepted in O(N^3)...

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

      O(N^3) how? I did it in O(N^2).

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

        Well my solution is pretty stupid, it calculated useless values. My dynamic programming is F[A][B] = number of paths from black cell A to black cell B without passing any other black cells along the path. Computing every state takes linear time so O(N^3) in total. The question is — why doesn't it time limit? I'd bet on weak tests.

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

          Answer for every question, Codeforces is very fast :)

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

          2000^3 is OK for CF servers :D

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

            But not for multiplying! Even 10^8 multiplications gonna catch TL!

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

          You have O(N^2), actually. Your solution computes F[A][B] only for states, that have the same A.

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

            Oh.. well that's lucky then :D

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

    I promise to always precalculate modulo inverse from now! Rank 64 to Rank 213 :(

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

When can we use the training problems again? Now it seems blocked...

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

Ok, so problem B was in an ACM training around a year ago, and C was in a Slovak OI nationals around 6 years ago. Naturally, I knew the solutions of both by heart.

D: Pick's theorem was obvious, but I couldn't figure out how to cut down the boundary part to better than in contest. Oh well, time to keep going with those 480.6 pushups.

UPD: Just did E, it was much easier for me than D. I've got to be more daring...

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

Div.1 pC let me recall this problem : ICPC 2009 Asia Phuket Regional — Your ways

When I practiced it, I misread it as problem of today.

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

Is there anyone else who saw "Judge protocol is inaccessible" error during challenge process? I saw that error and submitted the same wrong hack submission again. Several minutes later, all the submissions which gave that error were rejudged and considered "unsuccessful hacking attempt", so I got plenty of minus scores :p

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

I particularly didn't like this round considering problems B and C were old problems and problem A was just simply googlable.

Plus, then you read comments about how simple brute force passed in B and O(N^3) passed in C.

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

    A was googlable. ? oh damn. I could have been purple .

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

    "A+B problem" is also googlable xd. Funny how people can think of googling such adhoc and simple things.

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

      It all depends on how you are approaching the problem. Most people, I think, thought of finding the area of the special polygon.

      Its a formula which you can a) derive for hexagons in the contest, or b) google and get answer in a minute.

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

      Come on, don't make so categorical judgements :)

      Open standings of div2 and you'll see that 70% of contestants there wasn't able to solve this problem. One which is simple for you — isn't that simple for someone else.

      Moreover, it took me 3 attempts to solve this "adhoc and simple" one :) At least I didn't google it...

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

        Maybe I didn't express it clearly, but mainly I wondered who even bothers of writing such things in Internet out so that they can be easily googlable :P.

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

Did anyone get Accepted in Div1-B using hashing? My solution 12181985 failed and not sure why.

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

    Yes 12179329, Time Complexity O(nlog^2n)

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

    Equality check using only the hash may give you a false positive. You should do the actual string comparison in case of matching hashes, to be sure. But then you risk getting a TLE on something like test 89, like I did: 12185730 :)

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

      Yep, also tried using hashing and got TLE on 89 12171361

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

      it turned out that the reason of the TL isn't the actual string comparison of matching hashes but the fact that the branching factor for each recursive step is 4 then, in the worst case, the overall complexity is O(n2).

      If you try to choose randomly the order of recursive calls you make for each recursive step I guess it is hard to create a test case that requires 4 calls for each recursive step.

      This is the AC solution using hashing (and just using the equality check of hashes to determine equivalence) with random order of recursive calls: 12187657

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

    Hashed successfully, using several primes: 12176236

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

    I got it accepted using hashing after changing the order of recursive function calls. (http://codeforces.me/contest/560/submission/12188786)

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9 years ago, # |
← Rev. 3   Vote: I like it +14 Vote: I do not like it

Here is an O(nlgn) solution of div1 B: Let the input strings be S and T, find the lexicographically smallest strings which are equivalent to S and T respectively, and check if they are equal. code : 12173245

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

Why does 12166408 fail? Precision problems? It seems to do the same thing as everyone else.

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

    (int)(ans + 0.5)?

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      9 years ago, # ^ |
      ← Rev. 2   Vote: I like it -12 Vote: I do not like it

      I think this is to print closest integer to ans. So, if ans is something like 0.9999 due to precision errors, we want to print 1.

      Edit: I didn't see the solution actually and thought you were asking why is that being done. Of course, you are right, that is the mistake.

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        9 years ago, # ^ |
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        It should say (int)(ans+0.5) but it says (int)ans+0.5.

        Operator precedence strikes once again.

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      9 years ago, # ^ |
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      Oh wow, he's gotta be mad.

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

    Yeah, (int) ans + 0.5 should be (int) (ans + 0.5) (12185888).

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

      Is there any reason behind writing "(int) n", but not "int(n)"? Is latter variant considered something like bad style? It always seemed ok and even more logical for me, but you, W4yneb0t and ainu7 wrote it another way.

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

        As far as I understand it, (int)n is called a typecast, while int(n) is called a constructor, with the difference being that one of them causes stupid bugs like this and the other doesn't. I write int(n) in my own code.

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

          Multiple sources (cplusplus.com, wikibooks.org) state that both are called type casts. Perhaps it doesn't hurt to think of the latter one as a constructor though.

          The reasons to use (int)n notation are perhaps tradition and/or C portability: this was the only notation available in C.

          Anyway, I just used whatever was in the source, trying to make the edit minimally intrusive.

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

        Personally, I prefer to write casts as (int) (n) in C++, since this notation is extendable to (long double) (n + 1) without change of form.

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

Could someone tell me why this code got TLE? DIV 2D Code

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

    Doesn't substring() makes a new array and copy the exact string? It may take a lot of time.

    Instead, save the string in global variable, and give only the index (of left and right bound for both string) to the recursive function.

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

    In your code for the return value instead of t1||t2 try t2||t1 Make "equivalen(a1,b1)&&equivalen(a2,b2))||(equivalen(a1,b2) && equivalen(a2,b1)" this "(equivalen(a1,b2) && equivalen(a2,b1)||equivalen(a1,b1)&&equivalen(a2,b2))"

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

For problem Div1 B I used divide and conquer but for the case when strings size is odd I called another function, this was the only reason I got TLE for test 91, I wrote code for this case in caller function and AC now. I checked "status" page and some of you have TLE for 91, so keep in mind my sad story.

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

Anyone else experiencing this problem for div 2 problem D?

In the question for operation 2, we have to recursively check for condition a or b and if either one if them is true. In the recursion code, if we check for condition b first then a, then I got an AC, if I check for condition a first then b I got a TLE (test 89)

TLE solution: http://codeforces.me/contest/560/submission/12186384 AC solution: http://codeforces.me/contest/560/submission/12186179

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

    If first condition is true then your function won't run for second condition. So order affects running time.

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

      But that means the test cases are weak right? If at all bruteforce recursion was to fail, then both codes should have got a TLE

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

        Test cases weren't strong but it's hard for problem setters to fail all possible brute-forces. Though here we have two naive solutions and both should get TLE.

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          9 years ago, # ^ |
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          Yes, I can understand that its hard for problem setters, it would be nice if a few more test cases are added now (maybe reverse strings from original cases).

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

            They could just set higher constraints. 200k is not enough sometimes.

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9 years ago, # |
  Vote: I like it +56 Vote: I do not like it

I'm doing a farewell party for some of my rating. They were supposed to leave around now, is the plane delayed?

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9 years ago, # |
← Rev. 5   Vote: I like it +79 Vote: I do not like it

Good contest! But... It seems that the testcase for div1 Problem B is not strong enough. My first submission failded a case that I wrote(It prints YES but the answer should be NO). So I resubmitted it after revising. And then I used this testcase to hack someone's program. Though the hacking attempt is unsuccessful(because this program can pass my testcase), it shows that the testcase is feasible. But now I find that my first submission can also pass this problem...... I am not very happy because this resubmission made my goal -400. :( By the way. How can I show this testcase that I wrote? Maybe many of you want to check your program with this. Thank you.

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

I can't understand why my solution of div2 problem D gives WA on test 8: 12186517. It tries to make the smallest string (from both given) using the operation: swap partitions if the first is greater than the second. Than checks if the two strings are equal.

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

Can someone explain why My Div2D got tle?? 12186568 latest submission 12188078 Thanks in advance

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

    http://codeforces.me/blog/entry/19331?#comment-241912

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

      Sorry,I actually do not get your point. Can you please elaborate your ideas. I see a lot of code passed regardless of there calling order. And my last submission 12187805 also gets timed out at case 6. Thanks in advance.

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

    This is my latest submission 12188078 . Still TLE on case 6. Please Help.

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

      You have O(n2) solution, because you do 4 checks. If you use this construction:

      if (rec(...) && rec(...) || rec(...) && rec(...)) {

      Then it will be at most 3 checks (because of boolean operators optimization and the equivalence relation), which will be O(n1.5).

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

Got TL in D div2 with python3, but the same code was accepted with PyPy. How to figure out how to send?

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

    exactly the same thing! %) Sometimes Python is faster, sometimes Pypy (( I should have used C++

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

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

Div1 C used in this contest is simply the copy of this problem. https://www.codechef.com/CDCRF15R/problems/CWAYS/ Author didn't even change the logic and used the same problem.

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

    I'm pretty sure the authors didn't see this problem (you see there are not many participants in the contest you gave link at). Sometimes it happens — same natural ideas come to mind of different people.

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

Thanks for the problemset I really enjoyed it. But wasn't problem B Div. 2 Easier than problem A?

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

Finally became an IGM! Donald Duck and Taylor Swift gave me power :D!

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

    Me too, custom tests saved me from fails, slow and steady wins the race.

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

I think test cases for div1 b should be revised and solutions should be rejudged

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

    Why? Did correct solutions not get AC? If not then there won't be any rejudge.

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

      Yes. Depending on which case you handled first, you could TLE. See http://codeforces.me/blog/entry/19374.

      12188567 vs. 12188586. The only difference is in the commented/uncommented line.

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        9 years ago, # ^ |
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        Both solutions are wrong, all correct solutions got AC.

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

          Good. But why many incorrect solutions can get AC?

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

          So if I thought about the complexity before writing the solution I don't solve it, and if I write the solution without thinking I solve it? is that fair?

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

            It's bad that some wrong solutions get AC but it's impossible for setters and testers to prevent it for 100%. Is it fair? It's not cool that some people have more points than they should but well, we all had the same possibilities to use this situation — so it's fair for some definition of "fair".

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

              we all had the same possibilities to use this situation

              I can't imagine a yellow/red coder writing an O(n^2) solution for a problem where n<=2*10^5

              But you are right, it's impossible to prevent it 100%

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

                I can imagine a yellow/red coder writing an solution or properly optimising something to make his solution pass.

                And statistically, such lucky situations are negligible when the problems are tested properly. People who don't have the skill will just drop down again eventually.

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

                I wrote optimized O(N^2) solution for N = 2*10^5 in contest before (and it passed).

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

            You're not the first to think of this problem. If you solve it, tell us. Adding tests after the contest has its downsides.

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

          Actually,someone i know proves that the complexity is O(n^1.57)

          If A and B are equivalent, it will only recurse at most three subtasks so complexity is T(n) = 3T(n/2) that is approximately O(n^1.57)

          If A and B are not equivalent, it will recurse four subtasks only if two subtasks are equivalent so complexity is T(n) = 2T(n/2)+O(n^1.57) it's also O(n^1.57)

          EDIT1: I made a mistake to let T(n) = 3T(n/2) in the first case since the subtasks might not be equivalent ! But after further calculations i still believe the complexity should be O(n^1.57) Plus, i wrote a program to calculate the times of recurses and it's about 10^9 which Codeforces is still able to run in 1s

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

      And mine too 12188078 I have no clue why I am getting TLE.

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

can anyone explain div2 c. My approcah is dividing the hexagon into four triangles and calculate the area of the individual triangles and then summing them up gives area of hexagon and later divide the area of hexagon by area of the equilateral triangle .is it correct or am i going wrong somewhere ?? please tell the other methods to solve this :)

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

    Your idea is correct, you just have to implement it carefully

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

Div2/C can any body tell me what's wrong with my code , i used these theory :

area of a triangle abc = 0.5 * b *c * sin(A)

area of a triangle abc = sqrt((a+b-c)*(a-b+c)*(b+c-a)*(a+b+c))

this the code :

#define sin120 0,86602540378443864676372317075294
    vector<double> v;
    double air,air1,air2,air3,air4;
    int x,y;
    cin >> x >> y ;
    v.pb( sqrt(x*x+y*y+x*y) );
    air1=sin120*x*y*0.5;
    
    cin >> x >> y ;
    v.pb( sqrt(x*x+y*y+x*y) );
    air2=sin120*x*y*0.5;

    cin >> x >> y ;
    v.pb( sqrt(x*x+y*y+x*y) );
    air3=sin120*x*y*0.5;

    double a=v[0],b=v[1],c=v[2];
    air4=sqrt((a+b-c)*(a-b+c)*(b+c-a)*(a+b+c));
    
    air=air1+air2+air3+air4;

    cout << (air * 4)/sqrt(3);

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    9 years ago, # ^ |
    ← Rev. 4   Vote: I like it +12 Vote: I do not like it

    First, decimal representation of C++ is "." not ","

    You defined sin120 as 0,86602540378443864676372317075294 but you should change , to .

    Second, you calculated variable air4 with wrong formula.

    Heron's formula is sqrt(s(s-a)(s-b)(s-c)) where s = (a+b+c)/2

    If you rewrite the formula and fix it right, code should be

    air4 = sqrt((a+b-c)*(a-b+c)*(b+c-a)*(a+b+c))/4.;

    Third, you used cout and printed double , without any setprecision, converting to integer or something. Answer can be large, and cout will use scientific notation such as 6e+006

    Your code should be fixed as

    cout << (int)( (air * 4)/sqrt(3)); (?)

    However, this might not correct.

    If you fix all the things, you might not get accepted because of real number error. To prevent this, rounding answer to nearest integer is preferred.

    cout << (int)( (air * 4)/sqrt(3) + 0.5 ); adding 0.5 can be used rounding answer to nearest integer, if answer is positive.

    and the code got accepted. 12190712

    Using real number is dangerous, so you should be noted.

    By the way, this problem can be solved with simple method, not using real number or any complicated method.

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

      Thanks a lot , i really appreciate it . btw i already solve it , i was just trying this methode :)

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9 years ago, # |
  Vote: I like it +64 Vote: I do not like it

NORTH KOREA DOES IT AGAIN!

I got strong suspicions against black_horse2014. First reason being of course that he's from North Korea :P.

"00:17:22 A Accepted [main tests] 00:30:36 B Accepted [main tests] 00:35:09 C Accepted [main tests] 01:16:57 D Accepted [main tests]"

His times for A, B, C were 17, 13 and 5 minutes which seems unnatural, but doesn't sound impossible. However in his previous contest we can find this:

"00:13:12 A Accepted [main tests] 00:39:13 A has been locked 00:49:40 B Wrong answer on pretest 5 [pretests] 00:53:46 B Accepted [main tests] 00:54:38 C Wrong answer on pretest 8 [pretests]"

That B and C looks highly suspicious.

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

    I dunno, there are a lot of things missing that lead to suspicions towards other Best Koreans.

    • a relatively similar template and coding style

    • I took 35 minutes to solve A+B+C as well (10 minutes on reading A,B,C,D and solving A; 9 on solving B, 16 on solving C), that doesn't mean I cheated; he could've tried to solve C first, had a bug or missing idea and found it after doing A+B

    • the minimum gap between submits in this round is 5 minutes, not seconds; in the previous one, there's a huge gap between solving and locking A that could've been spent on C, it looks like typical behaviour of someone who's unsure if the solution is correct even if it passes samples and getting AC on B boosts that confidence enough to submit; and there's still a minute gap; compare with 0 seconds and 14 seconds

    The templates in B and C are different, but they have something in similar at least (chkmin and chkmax). It's not as clear as RNS's conduct in Looksery Cup. Still, your suspicion isn't unreasonable and CF should look into it.

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

      Yes, some people might prefer submitting a few solutions at once.

      Also, I hope you don't use the word 'Korean'; many people mistake North Korea(Democratic People's Republic of, communist, nuclear, ...) as South Korea(Republic of, `88 olympics, Seoul, ...). There's even a joke about South Korean people scaring foreigners, saying something like 'I’m from North Korea!'.

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

In problem D, I got TLE in system test 91 due to my habit of using long long every-time. After the contest, I converted all the long long to int and same solution was accepted :(

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

Just curious. Why there's no "Congratulations to the winners!" since Round #312?

Anyway, congratulations for winning the round and good luck for your IOI jqdai0815!

UPD: it's been added.

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

For Div2 E, is there any notation as to how the points are placed? For example, is (1, 1) the top left of the board or the bottom?

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

No Euler circuits, so an easy win for jqdai0815 :P

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