Pa_sha's blog

Codeforces Round 891 (Div. 3)
By Pa_sha, 3 years ago, In English

Hello Codeforces!

We are pleased to invite you all to participate in Codeforces Round 891 (Div. 3), which will start on Aug/07/2023 17:35 (Moscow time).

The format of the event will be like any Div. 3 rounds:

  • 6-8 tasks;

  • ICPC rules with a penalty of 10 minutes for an incorrect submission;

  • 12-hour phase of open hacks after the end of the round (hacks do not give additional points)

  • after the end of the open hacking phase, all solutions will be tested on the updated set of tests, and the ratings recalculated

  • by default, only "trusted" participants are shown in the results table.

We encourage participants with a rating of 1600+ not to create new accounts but to participate unofficially.

Only trusted participants of the third division will be included in the official standings table. This is a forced measure for combating unsporting behavior. To qualify as a trusted participant of the third division, you must:

  • take part in at least five rated rounds (and solve at least one problem in each of them),
  • do not have a point of 1900 or higher in the rating.

Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600 (or you are a newcomer/unrated), then the round will be rated for you.

Problems have been created and written by our team: FBI, Esestree, JuicyGrape and Pa_sha.

We would like to thank

Good luck!

UPD: There was an error on problem F. We fixed the tests and rejudged solutions . We apologize for that. It only affected a few people whose solutions passes all tests now. Also some hacks were already added to the tests and broke some of solutions.

UPD2: Editorial

  • Vote: I like it
  • +234
  • Vote: I do not like it

Comments Comments (253)
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3 years ago, hide # |
 
Vote: I like it +78 Vote: I do not like it

As a developer who happened to be the leader of the team, I would say you just these two words. Good luck!

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

I hope the contest will be good, in advance I congratulate you on becoming the authors of the contest on cf

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

Keep Calm & Create Interesting Competitions For All

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

As a coordinator the guys have prepared an interesting competition for you!

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

As a tester, I hope div3 users will enjoy this contest from well-prepared problems.

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

As a first-time tester, this is the best contest I’ve tested :) I hope you’ll like the problems!!

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

We know hacks do not give additional points, they just take away points .,.

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

Hoping it will be a great round of Div.3. Best of luck to all contestants!

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

As a coach of these guys, I am delighted to see their first performance as problem setters at Codeforces, as well as to see positive feedback from the testers. I do believe that problems are well-prepared and fascinating. Also, I hope that everyone (including Grothendieck, the mathematician after whom we called our team) will enjoy the contest.

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3 years ago, hide # |
← Rev. 2  
Vote: I like it +37 Vote: I do not like it

As a tester, I can assure you that the problems are of high quality, and you will definitely enjoy solving them. Please, read all the statements. Even if you find yourself stuck on a problem — try to solve the next one.

Good luck! May the skill be with you.

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

As a first-time tester, all the problems are fun and well-prepared. I hope you will enjoy the contest and earn a positive delta! Good luck!

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

First Unrated Div3 :)

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

JuicyGrape orz

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

Tomorrow : Problem A : dp on a tree ):

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3 years ago, hide # |
 
Vote: I like it +14 Vote: I do not like it

Ukranian Contest!

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

Hoping to become cyan in this round.

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

Good Luck everyone!!!

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

Wish everyone can enjoy the Div.3 round, let's gooooooooo!

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

Ukrain rocks!!!

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

I will solve at least five problems.

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

hope a good round, although i will be unrated.

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

Hope that I can get back to expert.

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

I wish good luck to the participants and no problems during the round to the developers! I am also very glad that now more and more different teams are preparing div-3 rounds! ✨

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

hope everyone rating increse after div 3 , good luck everyone !

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

Currently, There's a variety in teams who prepare DIV 3. Interesting UwU

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

Hope to become expert

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

Hope to be Specialist this round.

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

I really like it when I see the problem setters of the contest have solved thousands of problems, when their rating graphs has lots of ups and down, because they know the struggles of people who have practiced a lot and lot but still face problems during contest. And they set problem keeping these things in mind.

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

Hope to become expert!

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

A div 3 prepared by Ukrainian and Russian, I love this.

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

Hope to become at least Specialist in this round . ;(

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

안녕, i am a (beta) tester

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

Hopefully i will be pupil again. Good Luck guys!

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

Wow, so the writers are all from Ukraine!

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

Good Luck!

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

Hoping it will be a great round of Div.3. Best of luck to all contestants!

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

I hope to get close to 1600 or above.

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

John Cena wishes you luck

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

Hope that the statement is clear and short

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

Speedforces!!

Speedcoding is a different kind of fun...

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

idk but i feel like this contest might have been a little bit hard :)

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

I see a lot of negative comments, and I don't know why, as the statements were pretty clear and problems seemed fine, just like in the last 2 contests.

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

G is absolutely beautiful!

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

    Didn't solve but seems pretty elegant to me as well!

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

    Is it 2^(number of pairs u-v where dist(u,v) < S) ??? I couldn't think of a concrete solution

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

      No, the main idea is based on Kruskal's algorithm. If you know Kruskal's then you'll recall that it adds edges in increasing order of weight, but if an edge creates a cycle, then it skips it. Now if you apply Kruskal's to the given tree (add edges in order of weight), you'll be able to deduce some of the edges that need to have larger weights than the current edge. (bad explanation, ik, but it should get the main idea across.)

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

      I believe that if any of the edge has a weight s initially, then it's 0.

      Otherwise, it's a combinatorics problem — You can try to insert every possible edge with every possible combination with every weight in range of [max weight in the given tree+1, S].

      P.S: 0 if w_i==s is wrong. An edge can be added without changing the MST.

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

      80 is a possible output as shown in testcases; so unlikely

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

    kind of standard actually, ive seen this exact thing before (counting how many pairs of nodes each edge is maximum of)

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

How to solve F?

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

    Hint: for each pair of x and y, there is only one set of numbers which satisfies a + b = x and a * b = y

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

    If ai is part of a pair then ai+(y/ai)=x, which gives a quadratic equation. Solve it and you get the values of ai and aj. The rest is simple combinatorics. Be careful of the case when the quadratic equation has zero or one solutions.

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

    For every $$$(x,y)$$$ the is at most one unordered pair of integers $$$(a,b)$$$ such that $$$a+b = x$$$ and $$$ab = y$$$. You can find $$$a$$$ and $$$b$$$ via quadratic quation.

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

    You can store the frequency distribution of a_i in a map and then just try to solve the quadratic. If roots are equal, answer is n choose 2 where n is frequency of the root otherwise, it is n1*n2

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

    Solved it 5 minutes after contest sadge. The answer is pretty simple, just notice that you can rewrite your two equations as two quadratic equations and then apply the quadratic formula.

    a_i + a_j = x means that a_i = x — a_j. And if you insert this in a_i * a_j = y you get a_j*(x-a_j) = y which can be written as a_j ² — a_j x + y. And the same for a_i.

    Then quadratic equation gives you two possible solutions x — sqrt(x*x — 4*y) and x + sqrt(x*x — 4*y). So you just see how many of these elements are in your array to make an answer out of and you get your answer :)

    https://mirror.codeforces.com/contest/1857/submission/217739412

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

      Bro i saw your code what is (x+thing)%2 in that can you please explain

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

        Sorry for the non-descriptive name, but thing is just the square root of the discriminant. You can see that in the quadratic formula the answer is (-b +/- thing)/2a. Which is (x +/- thing)/2 (which you see in the code)

        The reason we have (x + thing)%2 in our code, is just because we are working with integers and not floating point numbers. If (x+thing) was an odd number, then the quadratic formula would've given us a non-integer as an answer, which we know we can't have, so we exclude those cases.

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

How to solve G ?

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    3 years ago, hide # ^ |
    ← Rev. 4  
    Vote: I like it +6 Vote: I do not like it

    Consider Kruskal algorithm performed on input tree. On each step we're connecting 2 components via edge with weight $$$w$$$. The other edges connecting these two components must either have higher weight or just not exist, so we multiply our answer by $$$(S-w+1)^{S_1S_2-1}$$$, where $$$S_1$$$ means size of first component and $$$S_2$$$ size of the second one.

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

B was a little unclear for me. also hard.

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

Thanks for the round. I solved until problem E. Problems up to E were fine for me, except for problem B.

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

Missed F by 30 seconds. F

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

I feel like Problem B was too hard for D3B.

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

What's the test 24 on pG:(

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

very easy with problem F, but i wasted time for B, and D, E , F is more easy than D or E.

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

B = C >> F for me, solved in in 7 minutes but for the next hour couldn't solve and implement B and C correctly. Feels so bad ;(

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

Problem statement of B is the ultimate example of "How not to write a problem statement".

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

I solved C using this. Can it be optimized further? 217736885

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

What does "Unexpected verdict" mean in hacking?

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

guys when does results will come

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

Appreciate it :) I really enjoyed the round.
I wish I could notice earlier the fact that
the description of problem B was actually describing normal round operation, but still I had lessons learned.

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

B was an unclear statement take me a lot of time to understand!

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

Good contest. Enjoyed. But F, it was just like: "Hey! do you know quadratic equations?"

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

rename this contest to Div.2 ):

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

What the hell was B? very bad problem statement!

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

solved E, F but unable to solve D :v. Wasting time for thinking about graph and reverse graph :v

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    3 years ago, hide # ^ |
    ← Rev. 2  
    Vote: I like it +11 Vote: I do not like it

    I was lucky enough to consider transpose the equation au−av ≥ bu−bv into au−bu ≥ av−bv.
    Then the problem turns into finding maximum values in the new array c = a - b

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

Why does this code give WA in problem G? Can you help me please to find a mistake in the code? https://mirror.codeforces.com/contest/1857/submission/217742367

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

Solutions for problems A, B, C, D and E were on youtube during the contest again. Can't codeforces contact Youtuble to delay codeforces related videos when there is a contest? + Why do people even cheat? Why are they getting from cheating?

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

FBI Why where is no such test in F that $$$x^2 - 4y$$$ is not a full square?

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

I think the problem statement in C is kind of easy to be misunderstood.

It's quite sad that I couldn't solve C despite solving (preliminarily) D E F.

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

Mathforces

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

How can we solve problem C? what's the intuition behind and are there any similar problems?

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

    Math.
    Sort array b, then the minimum element shows up (n-1) times, next smaller value shows up (n-2) times, ... so on and the last value(largest one) can't be determined.
    So we can simply choose the maximum value for the last one, which is 10^9 in this problem.

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

      Ah thanks, i just realized b was result of all possible pairs not subarrays, should have re read the statement. -_-

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

I really enjoyed the contest, and in my opinion B wasn't that bad like other comments say, but the implementation was tricky

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

Can somebody look into my solution for C: https://mirror.codeforces.com/contest/1857/submission/217718886 I am not able to find the mistake. Thanks.

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

Guys please give some hints to D, I's stuck there

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

None of the CF extensions are working. Not even CF Get Rating or CF Analytics. It shows codeforces API error. I just wanted to know my predicted delta. Anybody else with the same error?

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

I dont know what this line mean after the TC "It is guaranteed that the sum of n over all tests does not exceed 10^3"

Most of the prb I notice but not able to understand what they say I need to sum n values of every TC, or the array sum of n not exceed 10^3 or what they mean

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

    Each test has some number of test CASES. I believe you're referring to problem C here.

    There are some (unknown) number of tests, but that doesn't matter since your code is run once for each test. The time limit is also per test.

    For this problem, there can be up to 200 test cases per test. Each test case can have n up to 1000, but the sum of all test cases in one test cannot exceed 1000.

    So if there is a test case with n = 1000, there is only one test case. Or there could be 10 test cases with n = 100 each. Or 2 test cases with n = 700 and n = 300.

    Usually, the worst-case scenario test is one TEST CASE with the maximum n and this is the important takeaway.

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

I my opinion hardness of Problem G sharply increased.

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3 years ago, hide # |
← Rev. 3  
Vote: I like it +61 Vote: I do not like it

Good contest overall problems was fun! However I noticed there is a bad test case on problem F. In the constraints, it stated that x and y had to have at least an absolute value of 1 or greater. However, test case 7 had queries where x = 0, causing my code to fail. :(((((((( I've noticed a lot of other people failed this case too, so ig i have the power to make this unrated :p

Submission: https://mirror.codeforces.com/contest/1857/submission/217663720

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

problem D hadn't pretest like: 

1 
2
-1000000000 -1000000000
1000000000 1000000000

is it good?

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3 years ago, hide # |
← Rev. 2  
Vote: I like it +38 Vote: I do not like it

Problem statement of F mentions $$$1 \lt =|x|$$$ but testcase 7 has $$$ x = 0$$$ and $$$ y = -9$$$ as one of the queries.

This should not fail in my opinion if $$$x \neq 0$$$ was ensured.

Edit: incorrect link

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3 years ago, hide # |
← Rev. 3  
Vote: I like it +4 Vote: I do not like it

F's pretests were really weak i forgot to check the most important condition whether the (-b + sqrt(D)) is divisibly by 2 or not :O. A lot of hacks incoming

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

    I don't think it's possible for (x + sqrt(x^2 - 4y))to be odd. 4y is always even. If x is odd, x^2 must be odd. Then odd - even is odd. So sqrt must be odd. So, x + D is even.

    Similarly you can argue for the case where x is even.

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

Just wanted to get some clarification, for the problem D of today's contest. I got this weird error in the following submissions (using java):

When I submitted equivalent code using long, it worked (using java): - https://mirror.codeforces.com/contest/1857/submission/217686813

Using int, cpp code worked though: - https://mirror.codeforces.com/contest/1857/submission/217741962

Anyone has any idea what might have happened? I would really appreciate the help. Thanks in advance.

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

Task C is complete constructive shit. I liked all the other tasks very much.

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

Can anyone suggest me some better approach in E? MY submission https://mirror.codeforces.com/contest/1857/submission/217726880

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

    i think, the only optimization here can be not finding lower_bound(vb.begin(), vb.end(), va[i]), but using just i instead, because anyway you get 1 from segments like [a, a] and lower_bound adds logN to your complexity. The idea seems ok.

    Edit. I'm sorry for not noticing that, but this way you should use vb for iterating through the indices, not va. So, you don't actually need va except for reading the input :)

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3 years ago, hide # |
← Rev. 2  
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well i got RE for 7 times in D 217669362 (exit with 11:java.lang.IllegalArgumentException: Comparison method violates its general contract!) and solved in this version 217692889 ...never got exception like this before,can anyone tell me why?

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

Why were all problems except G math-y?

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

look at this guy's submission Stark663

I think he is a cheater

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

    no,but he may be banned because he violated the rules that you are agreeing to before registering to any CF contest)

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

Video Editorial For problems A to F.

https://youtu.be/6s1vEnzK7vI

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3 years ago, hide # |
 
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Why am I getting MLE at test case 4 in F? MY submission https://mirror.codeforces.com/contest/1857/submission/217743843 How can I optimize it?

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    3 years ago, hide # ^ |
     
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    Even if it didn't MLE, it would have probably been hacked (TLE) because you used unordered_map, see this blog for more details

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    3 years ago, hide # ^ |
     
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    Not necessarily related to MLE, but I wonder what's C++ sqrt()'s behavior on negative input because x * x — 4 * y could be negative.

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3 years ago, hide # |
 
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made video explaining A&B&C&D:

https://youtu.be/elQ4jmUfYqM

thought would be usefull for community

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3 years ago, hide # |
 
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It was contest with the most creative (also good) problems I have ever participated.

Although B may be difficult for many people, it can be in best problems for last contests. Also E and G were pretty good. But I think F was too mathematical, perhaps you could add another problem which more greedy or more algorithmic.

Thanks for the great contest, I love it :D

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3 years ago, hide # |
 
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how this code got hacked https://mirror.codeforces.com/contest/1857/submission/217662800

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TY for the contest <3

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3 years ago, hide # |
 
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Please check this blog , Problem F constrains has (1 $$$\le$$$ $$$|x|$$$ $$$\le$$$ 10^9).

Meanwhile in test 7 many test cases have x = 0.

I don't know what should happen in this case, but please do something about it.

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3 years ago, hide # |
← Rev. 2  
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I get it nvm

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3 years ago, hide # |
 
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open upsolving please

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3 years ago, hide # |
 
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If I hacked someone, but it was unsuccessful. Do I lose points?

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3 years ago, hide # |
 
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I don't know why my solution is failing for F: https://mirror.codeforces.com/contest/1857/submission/217724931 Can somebody help me?

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    3 years ago, hide # ^ |
     
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    you need to check again if temp*(x-temp)!=y

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      3 years ago, hide # ^ |
       
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      thanks, it got AC

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      3 years ago, hide # ^ |
       
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      Thanks, I got AC by adding this. But I'm trouble in why is it necessary? I try to check sqrt(x * x - 4 * y) * sqrt(x * x - 4 * y) != x * x - 4 * y but got WA too. Thanks for your patience.

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        3 years ago, hide # ^ |
         
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        According to Quadratic Formula,you need to assure that ai,aj are all Integers, so sqrt(x*x -4*y) should be an Integer too, or else you may get wrong answer,so having a check is nice

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3 years ago, hide # |
 
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I managed to solve up to E during the contest and F afterward. The problems were nice, they were from various topics and I enjoyed solving them.

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3 years ago, hide # |
 
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What was the approach for E?

I got TLE here : https://mirror.codeforces.com/contest/1857/submission/217725866

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3 years ago, hide # |
← Rev. 2  
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Math tag for problems ×

Math tag for contest ✓

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3 years ago, hide # |
 
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The round was made unrated?

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3 years ago, hide # |
 
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G is beautiful

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3 years ago, hide # |
 
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Had sys test finished?

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3 years ago, hide # |
 
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It is writtent that tis contest is rated for everyone having rating less than 1900. But my rating haven't updated current rating :-806

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3 years ago, hide # |
 
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My F was hacked. Can someone tell what's the issue?

My Submission : 217815371

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3 years ago, hide # |
 
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Can anyone help me find out why this submission for F is getting tle? 217653085

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WHY did code give TLE in system testing (PROBLEM -F)

#include <bits/stdc++.h>
        using namespace std;
        #define int  long long 
        const int N = 200000;
        int p=1e9+7;

    void solve()
        {
            int n;
            cin>>n;
             unordered_map<int,int>mp;
             for(int i=0;i<n;i++)
             {
                int a;
                cin>>a;
                mp[a]++;
             }
             int q;
             cin>>q;
             while(q--)
             {
                  int a,b;
                  cin>>a>>b;
                  int val=sqrtl(a*a-4*b);
                  int count=0;
                  if(val*val==a*a-4*b)
                  {
                    if((a+val)%2==0)
                    {
                     int num1=(a+val)/2;
                      int num2=a-num1;
                      if(num1==num2 )
                      {

                        count+=(mp[num1]*(mp[num1]-1))/2;
                      }
                      else 
                      count+=mp[num1]*mp[num2];
                    }
                    else if((a-val)%2==0)
                    {
                         int num1=(a-val)/2;
                      int num2=a-num1;
                      if(num1==num2)
                      {
                        count+=(mp[num1]*(mp[num1]-1))/2;
                      }
                      else 
                      count+=mp[num1]*mp[num2];
                    }
                  }
                  cout<<count<<" ";

             }
             cout<<endl;


        }

        signed main()
        {
            int t=1;
            cin>>t;    
            while(t--){
                solve();
            }
        }
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3 years ago, hide # |
 
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Why F using unordered_map will TLE?

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3 years ago, hide # |
 
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what is causing TLE in problem E?

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3 years ago, hide # |
 
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I see the system testing has stopped at 100% for a while xD.

Hmm, it's 20:00 in my timezone and I haven't eaten yet T.T

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3 years ago, hide # |
 
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Rating changes???

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3 years ago, hide # |
← Rev. 7  
Vote: I like it +13 Vote: I do not like it

Why is the contest frozen for so long MikeMirzayanov FBI

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3 years ago, hide # |
 
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Never expected such good performance by me in such div 3 contest

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3 years ago, hide # |
 
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Problem C is much much easier than problem B

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    3 years ago, hide # ^ |
     
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    Honestly it depends on implementation skills, B is pretty easy to figure out but the implementation is difficult, while in C you need a bit more thinking with easier implementations

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3 years ago, hide # |
 
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Problem B was good, but the test cases weren't exhaustive during contest, it didn't help with the fact that codeforces was down so had to use m1 lite version which didn't show my submission time and solution was accepted, although i should have written efficient code in the first place, but this kind of gave false positive until solutions were hacked :(

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    3 years ago, hide # ^ |
     
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    hh, this kind of thing often occurs, but this pretest is still passable, at least most people's C, wa7 will realize the error later, 883's pretest is the weakest

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3 years ago, hide # |
 
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Nice!!!Through this round of Div3, I became Pupil for the first time!

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I never understand the pointing system... I mean although i just solved one question how could i lose points for that??

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    3 years ago, hide # ^ |
     
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    One of the main reason your rating is decreasing after cobtwst is that you underperformed (your performance is below your rating).

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I got a TLE because of umap on F QwQ

I was so close to AK :(

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3 years ago, hide # |
 
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Fun round for sure! Really enjoyed D, E and F, even though had to re-solve them after the round. Good tasks for a begginer, since they did indeed give me some new approaches!

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