By scott_wu, 22 months ago, In English,

Hello everyone! The first round of the 8VC Venture Cup will be held on Saturday, February 13th at 12:35PM EST. ecnerwal and I are the problem setters. We want to thank GlebsHP for his help in preparing the contest, Delinur for translating the problems, and MikeMirzayanov for creating the Codeforces platform

The contest is for competitors in both divisions and contains seven problems. The scoring distribution is as follows:

500 — 750 — 1000 — 1500 — 2000 — 2500 — 3000

The contest will be slightly longer than usual — two and a half hours. The top 200 contestants will advance to the final round, and the top 20 local finishers will be invited to Woodside, CA to compete onsite. Good luck!

UPD: System testing is now over. Congratulations to the top contestants:

  1. Petr
  2. TooDifficuIt
  3. ilyakor
  4. bmerry
  5. Errichto

The top 200 contestants will advance to the final round in two weeks. Congratulations!

The editorial can be found here.

 
 
 
 
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22 months ago, # |
Rev. 2   Vote: I like it -64 Vote: I do not like it

I hope This contest will be very interesting and there are many hacks. Good luck every one !!!

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    22 months ago, # ^ |
      Vote: I like it +42 Vote: I do not like it

    How do you know?

    And by the way being among the top 100 while having solved only 3 problems (because of hacks) is not a good thing.

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      22 months ago, # ^ |
      Rev. 3   Vote: I like it +41 Vote: I do not like it

      being among the top 100 while having solved only 3 problems (because of hacks) is not a good thing. but being among the top 20 while having solved only 3 problems (because of hacks) is a very good thing. :D

      I think nothing is not good in being in the top 100, if you solve problems or make successful hacks both are allowed in CF rounds. If you did not cheat then you deserve this place.

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

        I was about to say among the top 20 but then changed it to 100 when I remembered that there're gonna be Div. 1 participants :D

        And by the way I am proud of my rank in that contest, but I'm not proud of my performance. A participant who solved the forth problem would have deserved my rank much better.

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        22 months ago, # ^ |
          Vote: I like it +11 Vote: I do not like it

        Hacks are good when you open the source code of another person and try to find a bug (which is really challenging and who makes it successfully deserves a higher rank), not when you find a test case that's not included in the pretests and keep refreshing the room for new participants to hack with that same test.

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22 months ago, # |
  Vote: I like it +19 Vote: I do not like it

The score gaps of the first three problems is close => Must solve fast.

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22 months ago, # |
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tourist is joining, how high will his rating go?

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22 months ago, # |
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Hope that all legendary masters will compete. Very big prizes!!!

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    22 months ago, # ^ |
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    I am not sure, but scott_wu would probably win ;)

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      22 months ago, # ^ |
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      I notice that codeforces authors seldom participate in their own contests, do they?

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22 months ago, # |
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Is it rated?

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22 months ago, # |
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Cant wait for it.

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22 months ago, # |
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Good luck everybody.

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    22 months ago, # ^ |
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    How do you know that letters will be ABCDEFG? Are you a cheater?

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      22 months ago, # ^ |
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      Yes i am a cheater.

      First problem A.

      Distrubition 500.

      A: binary search, segment trees

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22 months ago, # |
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What does top 20 local finishers mean? Is there also an onsite version of this round?

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    22 months ago, # ^ |
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    Local means people who can get to San-Francisco by their own (organizers do not cover transportation expenses)

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    22 months ago, # ^ |
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    People who marked "I would like to compete onsite in finals" during a registration. Organizers don't pay for travel so likely only people living close to Woodside want to attend onsite finals. Thus, "local finishers".

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22 months ago, # |
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Перевод на русский будет?

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22 months ago, # |
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It feels nice, inspiring and much exciting to compete with the Div 1 participants.

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22 months ago, # |
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So lately!

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22 months ago, # |
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Oh money money.

Wwonder who will be the first , and takes the money.

1st place 2500$

tourist will still rich ))

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22 months ago, # |
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I have a few problems solved, but I didn't succesfully register :-(

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22 months ago, # |
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Sorry I'm asking this here.If someone registers for a contest and sees the problems but makes no submission will their rating be changed ?

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    22 months ago, # ^ |
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    No, rating change only happens after you make a submission.

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      22 months ago, # ^ |
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      Yayyyy :)))

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      22 months ago, # ^ |
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      I personally think the current system should be changed — anyone who sees problems should be automatically counted. There seem to be a lot of people who register but decide not to submit anything after they find out they can't solve enough problems to boost their rating. About 5500 registered, but only about 3000 are in standings. There are probably some people who couldn't just make it to the contest, but there are also probably a lot who just look at the problems and go away after finding them too difficult.

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        22 months ago, # ^ |
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        Unfortunately in your case there will be a lot of twice account guy in CF, every cheater can register another account to see whether problems are difficult or not. As consequence we will have inflation of rating(all fake account will fall in rating). Anyway, cheaters gonna cheat.

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          22 months ago, # ^ |
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          That is true — there is no perfect way to block cheating. However, a lot more people will hesitate when they realize they are doing something that explicitly breaks the rules (IIRC registering and not submitting due to problem difficulty is not breaking the rules).

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        22 months ago, # ^ |
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        I've found out that even if someone who submitted their solution(s) thinks that he'll get less points than he wants to, he starts decreasing his current points by bad hacks until it turns negative, which will not effect his rating.

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22 months ago, # |
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"Without looking, they hand their balls to Harry"

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22 months ago, # |
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Okay, for E, am I right that I only need 1,2,3 or 4 elements? I couldn't handle the 4-case :(

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    22 months ago, # ^ |
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    If I'm right, you never need an even number of elements, but you might need an odd number of elements bigger than 4. Example:

    5 0 0 0 13 13

    Optimal solution is to take all 5 elements.

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      22 months ago, # ^ |
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      Oh, thanks! :)

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      22 months ago, # ^ |
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      Why odd number of elements is enough? The problem kind of pushes towards this conclusion but I could not see why is it true.

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        22 months ago, # ^ |
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        I'm not going to post the whole algebra here, but basically, suppose you have an optimal subset with an even number of elements, the middle ones being A1 and A2. Now consider the same subset without A2, and you'll see the simple skewedness can only increase by removing that element, which means the original subset wasn't actually optimal.

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    22 months ago, # ^ |
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    NO 6 2 2 2 3 12 12

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22 months ago, # |
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How to hack C? QAQ

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    22 months ago, # ^ |
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    I used 9 4

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    22 months ago, # ^ |
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    6 3

    I think I got hacked this way. I used greedy then after getting hacked I switched to binary search.

    P.S. I thank the guy who hacked me so that I at least have 400+ points instead of 0 by failing System Tests.

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      22 months ago, # ^ |
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      Is 14 the right answer? If it is, then greedy probably works.

      Here is the link to my greedy solution: link.

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      22 months ago, # ^ |
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      Greedy? Binary search? Take a look at my submission. It passed all hacks described here, and I hope, it will pass sys. tests: 16001257 UPD: Yes, it passed. But maybe it is greedy too.

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    22 months ago, # ^ |
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    7 6 -> ans 20

    6 4 -> ans 15

    I used these two.

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    22 months ago, # ^ |
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    I used just 2 random big numbers.

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22 months ago, # |
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difficulty : A<B<C<D<<<<<<<<<<<<<<<<E,F<<<<<<<<<G

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22 months ago, # |
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How to solve F?

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    22 months ago, # ^ |
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    Sort by value and then dp[open_intervals][cost_so_far]. When we move from i to i+1 then cost_so_far changes to cost_so_far + open_intervals * (t[i+1]-t[i]).

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      22 months ago, # ^ |
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      Can you please define Open Intervals ?

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        22 months ago, # ^ |
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        The number of groups (intervals) for which we have already processed at least one element but there is at least one non-processed element. In other words, the first element of a group (interval) is not greater than i, and the last element is greater than i.

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      22 months ago, # ^ |
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      Another possible solution (with much worse complexity, yet fast enough to pass) :

      We are interested in having difference between sum of all "left endpoints" and all "right endpoints" at most k. Let's calculate dp[open_intervals][current_difference]. When considering new element, it can either start new interval (and increase current sum of "left endpoints") or join some existing interval; in both cases it can either close interval (and increase sum of "right endpoints") or keep it open.

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        22 months ago, # ^ |
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        I did this and got TLE :(

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        22 months ago, # ^ |
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        It sounds like you're describing the same solution. The "current difference" can be big, but that can be solved by indexing using "current_difference-open_intervals*current_number"; if you write down how it changes the rules for opening/closing, you arrive at errichto's formula.

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          22 months ago, # ^ |
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          It's easy to see that it isn't the same solution because the complexities are different. The solution which I_love_Tanya_Romanova described with your optimization (which is what I implemented) has (n * max(ai) * k) states, while errichto's doesn't depend on the actual values of the integers in the array. Of course, as max(ai) = 500 in this problem, it's clearly good enough.

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22 months ago, # |
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For Problem F, can someone explain me why the answer at the second test is 13 and not 15 ? :/

EDIT : ok I got it.

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    22 months ago, # ^ |
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    hey how to get 13 not 15?

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      22 months ago, # ^ |
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      The total imbalance must me lower than k, and not the imbalance of each group.

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22 months ago, # |
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I guess, the statement of problem C is not describe clearly.

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22 months ago, # |
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C had weak pretests / made for hacks :D I wish i read "alphabetically" earlier in B :( Cost me 3 WA

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22 months ago, # |
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So, can anyone please explain why the result of 2nd sample case in task F is 13?

I've manually calced it on paper like 10 times in different ways and keep getting 15.

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    22 months ago, # ^ |
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    All(15) — [7,10][8,9] — [7,9][8,10]

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    22 months ago, # ^ |
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    You need to keep TOTAL imbalance at most k (meaning sum of imbalances of sets)

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      22 months ago, # ^ |
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      Ah I see. I've somehow missed it :(

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22 months ago, # |
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I'm sure that there are many people confusing when they read "so no two students’ towers may contain the same number of blocks" in problem C. While it should be no two towers have the same height.

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    22 months ago, # ^ |
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    I also got confused when the problem said that the students were stacking the blocks lengthwise.

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    22 months ago, # ^ |
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    There is a clear distinction between pieces and blocks in the statement. IMHO no confusion.

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22 months ago, # |
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I'm really upset. I can't pass the Examples while I'm working on F. Finally I found a bug but there are only 16 seconds left. I was too exciting to submit my code properly. (I had a mistake when copying my program and it got CE.)

upd:Now I feel much better because I got TLE on test 11. (⊙o⊙)

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    22 months ago, # ^ |
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    Unluckily. But I didn't know how to solve F, even now QwQ

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22 months ago, # |
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Could anyone explain test 8 for E?

I have completely no idea :(

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22 months ago, # |
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In E, I quickly found out that we need 2 intervals (in the array sorted left to right), where the middle element/s have to be in the first one and the second one touches the right end. I couldn't move from that point for over an hour...

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    22 months ago, # ^ |
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    I saw several passed solutions that looks like this: make ternary search in one part, in another, make binary search in another place, and check another 100 places. I don't think that it is solution supposed by authors.

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    22 months ago, # ^ |
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    Let's assume that the median is fixed. We need to find optimal k (that is, the length of the suffix). Key observation: if we consider "skewness" as a function of k, it's convex and ternary search is applicable to it(can be verified by writing down all sums and seeing what happens if we increase k by one).

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      22 months ago, # ^ |
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      I see. I briefly considered ternary search as one of my guesses, but didn't try to prove con[vex/cave]ity.

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      22 months ago, # ^ |
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      All I could guess from looking at examples, was that it was a monotonous function, so I coded binary search and it failed pretest 8, guess I looked at poor examples..

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      22 months ago, # ^ |
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      I got WA7 with that solution. Is there any problems with floating values?

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      22 months ago, # ^ |
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      How do you handle equal values of mean with ternary search?

      e.g. 0 0 0 0 1 2 2 2 2

      If you fix median = 1, then there're lots of equal value of mean for different values of k

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        22 months ago, # ^ |
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        If the mean doesn't change, it doesn't matter which one you pick :D

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          22 months ago, # ^ |
          Rev. 2   Vote: I like it +8 Vote: I do not like it

          That's a small example. It could be something like:

          0 1 1 1 1 10 99 100 100 100 100

          If you fix median = 10, the mean changes at some point, but your ternary search doesn't find it.

          Zlobober's comment answered my question though.

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            22 months ago, # ^ |
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            change if (f2(mid, k) < f2(mid+1, k)) { r = mid; } else l = mid; to if (f2(mid, k) < f2(mid+1, k)) { r = mid; } else l = mid + 1; you will get ac :)

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            22 months ago, # ^ |
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            My ternary search doesn't find anything because I don't have any ternary search :D

            In fact, I said the same thing as Zlobober. It doesn't matter which value you pick out of an interval of identical ones. Look at ternary search as binary search for a point where the difference between consecutive values of the given function (a prerequisite of ternary search is that it's monotonous) changes sign from  > 0 to  ≤ 0; it's more obvious that way.

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        22 months ago, # ^ |
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        Actually, as you take elements from right to left from both sides, there may be no more than one segment of equal values for the function we are willing to maximize. And this segment corresponds to the maximum value of a function as it is concave.

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      22 months ago, # ^ |
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      Could you prove why it is convex? I spent a while trying to prove this during contest, but my thought were too messy, so I took a risk and submitted it without a proof and as it turned out, that was a good decision, however I am still not convinced why this works.

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        22 months ago, # ^ |
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        Not sure how to prove it's convex, but it's somewhat easy to see it's bitonic, which is good enough.

        What happens when we increase size by two? We move from S/n to (S+a+b)/(n+2). As we know, the latter lies between S/n and (a+b)/2. (a+b)/2 continuously decreases (since we move to the left). So while (a+b)/2 is greater than the current mean, we will get an increase, and when it starts being less, we will always get a decrease.

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          22 months ago, # ^ |
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          And it seems to me that it's not actually convex. If we have something like

          aaa...aaabbb...bbbmxxx...xxxyyy...yyy

          (number of bs=number of ys<<number of as=number of xs)

          then we'll have two almost flat segments.

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            22 months ago, # ^ |
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            Can you show an actual example? I'm quite satisfied with an explanation I provided below.

            UPD: got it, my explanation below is an explanation of bitonicity. Of course the slope of a line joining the origin and the point on the convex polygon doesn't have to be convex itself.

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              22 months ago, # ^ |
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              import matplotlib.pyplot as plt
              %matplotlib inline
              n1 = 100
              n2 = 10
              vals = [0.0] * n1 + [1.0] * n2 + [2.0] + [3.0] * n1 + [4.0] * n2
              def Mean(count):
                  all = vals[n1 + n2 - count : n1 + n2 + 1] + vals[len(vals) - count:]
                  return sum(all) / len(all)
              plt.plot(range(n1 + n2 + 1), [Mean(x) for x in range(n1 + n2 + 1)])
              
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          22 months ago, # ^ |
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          Ahhh, right, thanks :). I had similar thoughts during contest, but couldn't join them to get that proof, however if I'm not mistaken there's still a flaw in your proof (or something that doesn't sound as you wanted to) -> "current mean always increases" (which is obviously just a silly mistake since you wanted to show it's bitonic :). That mean increases up to point where that (a+b)/2 becomes smaller than mean. After it, mean also decreases, but slower than (a+b)/2 (it will be larger than previous value of (a+b)/2), so (a+b)/2 will stay smaller than mean, so after that point mean is decreasing.

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            22 months ago, # ^ |
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            Yeah, the first version was incorrect, but I've updated my comment as if it never happened :)

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        22 months ago, # ^ |
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        It can be seen as a slope of a line joining the origin and the point whose x-coordinate is number of taken integers and y-coordinate is their sum. As you increase the number of taken integers, the differences between consecutive y's are non-increasing (since the numbers that we add are non-increasing), so it looks like a concave polygon in positive quadrant that we draw a tangent to from an origin.

        UPD: this is not an explanation why the function we are willing to maximize is convex, but it is a proof that it is bitonic (i. e. first increases, then decreases).

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22 months ago, # |
Rev. 2   Vote: I like it +24 Vote: I do not like it

My B passed pretests during the contest. Why did it fail test 1 during system tests?

Edit: I found a typo that makes it fail pretest 1. Yet it somehow passed all pretests the first go around. It was merely a typo and I fixed it here. I would have quickly fixed it during the contests if it told me that pretest 1 failed.

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22 months ago, # |
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207 T_T

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22 months ago, # |
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Is me only one who thinks that problem C should be more elaborated :/. I read PS thrice before i figured out what it meant :(

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    22 months ago, # ^ |
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    I agree. Russian statement is about impossible to understand (at last I switched the language).

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      22 months ago, # ^ |
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      Even in English it took some time to understand.

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    22 months ago, # ^ |
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    No. It also has the first explanation wrong. It should be written '2' instead of '4'...

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      22 months ago, # ^ |
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      It isn't wrong. 2, 3, 6, 9 and 3,4,6,9 both are acceptable i think

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      22 months ago, # ^ |
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      4 is also acceptable so the explanation is right.

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        22 months ago, # ^ |
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        Oh, so maybe I didn't understand it even after the end of the contest :) It happens.

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      22 months ago, # ^ |
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      Both 2 and 4 is correct.

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      22 months ago, # ^ |
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      how 4 came?

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22 months ago, # |
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Btw, it is me, or today codeforces was running like 5 times faster than usual? Literally no page load longer than 1 second, and pretests was really fast too.

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    22 months ago, # ^ |
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    Me too, No lag during hacks too. And its combined round

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    22 months ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    System tests was extremely fast too.

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    22 months ago, # ^ |
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    It was perfectly prepared round at all. And i have not any problems with translation. Thanks a lot for everyone!

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

Conclusion : US Round = Math Round

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

It's 4:23 a.m. in China, And I think I must go to bed to have a rest. So Tired QwQ

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

Can any one please tell me why this submission to D fails in 6th test 16007337

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    22 months ago, # ^ |
      Vote: I like it -10 Vote: I do not like it

    (2000*1999)^3 > 2^64

    You have to divide by 8 somewhere in between.

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

    Overflow at B = N * N * N * (N - 1) * (N - 1) * (N - 1).
    Before division this term overflows. Your accepted code with the changes 16008548

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22 months ago, # |
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Failed due to integer overflow on problem D :(. Was sitting for so long not understanding why my code was failing :(

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22 months ago, # |
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Auto comment: topic has been updated by scott_wu (previous revision, new revision, compare).

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

Who thinks B was harder than C ?

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    22 months ago, # ^ |
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    I didn't even understand C.

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    22 months ago, # ^ |
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    B is easier as algorithm, but harder in terms of implementation accuracy (a lot of possible cases).

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      22 months ago, # ^ |
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      No, you are wrong. You can take a look to my solution — there is not a lot of different cases. And algorithm is even easier as yours :) 16007975 The only thing because of I have not solved it is 200 instead of 201( I had lost a lot of time and was very busy, so, done really stupid mistake.

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        22 months ago, # ^ |
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        So, you are saying that 10 IF statements is not a lot of cases? :)

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          22 months ago, # ^ |
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          Okay... but still not 14 "if" and 6 "else" ;)

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

            Now just multiply all your IFs on recursion depth and compare :)

            "Easier" is not a synonym for "shorter code".

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      22 months ago, # ^ |
      Rev. 7   Vote: I like it +5 Vote: I do not like it

      I've found a short and easy solution: 15992405 without many ifs and elses

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

The top 200 contestants will advance to the final round in two weeks.

Currently, there are 2 people in 200th place (me and Nikitosh), so what will it be? XD

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22 months ago, # |
Rev. 3   Vote: I like it +34 Vote: I do not like it

Memory limit exceeded in F, though the array is int[][][] dp = new int[n + 1][n + 2][K + 1], which is 201*202*1001*4 ~ 160Mb, while memory limit is 256Mb. 16003432

Running max test (n=200, k=1000) in "Custom Invocation" shows 309828 KB of memory usage.

Running test with (n=184, k=1000) shows only 128488 KB.

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

Thanks for the contest. I'm interested in editorial

Reminds me that I must work more on proper technical execution, I spent too much time finding "i" instead of "1" bugs. Also I knew that numbers in D must be sorted, I thought I sorted them and did not understand why pretest fail.. Only a few minutes after the end of contest I realized how close to success I was.

A. Is a nice example of "read constrains" problem. Surely naive solution is not the fastest, but it is fast enough.

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22 months ago, # |
  Vote: I like it +39 Vote: I do not like it

Identical participant output results in different checker's comment:

http://codeforces.com/contest/626/submission/16002527

participants output: 3 0 475712 999999

checker output: wrong answer jury found better answer: expected simple skewness 262143/3, found simple skewness 48575/48575

http://codeforces.com/contest/626/submission/16008372

participants output: 3 0 475712 999999

checker output: wrong answer jury found better answer: expected simple skewness 262143/3, found simple skewness 48575/3

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

    Also seems to give Denial of Judgement when I try to print some stuff to debug: [submission:http://codeforces.com/contest/626/submission/16009256]

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      22 months ago, # ^ |
        Vote: I like it +18 Vote: I do not like it

      Yes, checker crashed if you output some large number first. That happens. No one was affected during the contest anyway.

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      22 months ago, # ^ |
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      Checker is OK now, submission was rejudged.

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    22 months ago, # ^ |
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    We noticed bug in checker's comment during the contest and fixed it. So the first submission was made before the bug was fixed.

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22 months ago, # |
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How to solve D?

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    22 months ago, # ^ |
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    Store all the possible differences in a frequency array . such that freq[i]= number of pairs such that their difference =i . eg if array is 1 2 10 freq[1]=1(2-1) freq[8]=1(10-2) freq[9]=1(10-1). Now make a cumulative array such that A[i]= number of pairs such that their difference is greater than or equal to i. Now for value for every pair of difference value find the number of pairs which have value greater than this pair . i.e if Winning points were a in first round and b in second round you need number of pairs such that winning round is greater than a+b(which is A[a+b+1]. multiply all the case and divide by (nc2)^3 Code

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22 months ago, # |
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Worst contest for me ! I submitted the same code which got WA (first submission) second time thinking i got WA on my corrected code :'( and submitting the corrected code got AC . trying not to cry

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22 months ago, # |
  Vote: I like it +112 Vote: I do not like it

Failed systests in 2 problems because of overflow

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22 months ago, # |
  Vote: I like it +26 Vote: I do not like it
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    22 months ago, # ^ |
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    Wait for my scrincast with russian rap

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      22 months ago, # ^ |
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      Can you code and sing at the same time?

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22 months ago, # |
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Are there any other contests before the final round? Or there will be 2 weeks without any contests...

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

Only one more Legendary grandmaster till all in the top rated be Legendary grandmaster. ;)

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

B: once you have many cards of the same color, it doesn't matter if you have 4 or 100000. You get the same result. This means you could implement the most naive bruteforce you can think of — and it will be fast enough — after you truncate the input to max 4 cards of each color.

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22 months ago, # |
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can someone please tell me why this solution for problem D gives TLE 16009986

let m be max element , so my solution runs in O(m^2) and m < 5000 I think 25m operation with 2 seconds isn't too much

thanx :)

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    22 months ago, # ^ |
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    Your solution complexity is O(m^2 log m^2) as you insert elements in a map; which should run in logarithmic time with a large constant factor — as you're inserting ~25M elements there. Try reducing this by removing the map and using arrays directly for storage and indexing.

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      22 months ago, # ^ |
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      thanks for correcting me ,yes it is O(m^2 log m^2) .

      fixed and got AC .

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22 months ago, # |
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No editorials?

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22 months ago, # |
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Can somebody tell me what principle/algorithm is used in this solution for 626C - Башни из кубиков 15992435, when taking a large segment from 0 to INF and then narrowing down to the correct answer? Looks like a binary search, but why does this work here?

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    22 months ago, # ^ |
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    It is binary search! :D

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    22 months ago, # ^ |
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    We are using binary search on possible answer. We know that if x is possible that all values greater than x are possible. Similarly, if x is not possible, all values less than x are not possible. So take min=0 and max=INF and calculate mid. If mid is possible then our answer will be greater than or equal to mid else less than mid. Iterate till only one possible answer is left which will be the answer.

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22 months ago, # |
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 Why I've passed pretest ? The pretest just have one testcase? :))

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    22 months ago, # ^ |
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    Why I get TLE ? on 2 RG ?

    for submission ? 16006497

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    22 months ago, # ^ |
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    Same thing happened to me. Except for test case 1. So I passed pretests but managed to fail test 1, which was just the sample test case. I'm pretty salty that the first sample case wasn't in the pretests.

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    22 months ago, # ^ |
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    I think they made a clarification that test 2 is not equal to pretest 2, and gave the data for test 2.

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      22 months ago, # ^ |
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      I will clarify my question: Why my solution gets TLE on codeforces while working fine on my machine.

      E.g.: Time: 2000 ms, memory: 7920 KB Verdict: TIME_LIMIT_EXCEEDED Input

      2 RB

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

I really need some advice about programming. Yesterday I did problem C, and I was quite sure about the solution, only to find the failed system test. When I checked, what I did wrong was missing a small, simple but crucial character "=". How do good coders avoid such silly mistake?

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    22 months ago, # ^ |
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    Practice, practice, practice,

    test your code before you submit (try and make up some test cases so you make sure all of your code in all cases have been run, except if you're very confident),

    double check your code before you submit.

    Those 3 things I guess. I also still struggle with quite a few things like that as I'm new as well. I failed D because I hadn't used long long but the rest was just fine :(. It's sad, I know, I think we all know that feeling being programmers haha.

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

I wonder how much rating gain Bus will get if he did not purposely lowered his point o.O

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

Could somebody tell me why my solution to D fails on test 2? On my machine it works perfectly, but on custom invocation it fails, even if I manually set br = 2, total = 27? My solution: http://codeforces.com/contest/626/submission/16013981

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22 months ago, # |
  Vote: I like it +43 Vote: I do not like it

Anything known about top 20 local finishers?