By BledDest, 3 years ago, In English

Hello Codeforces!

On Oct/10/2021 12:05 (Moscow time) Educational Codeforces Round 115 (Rated for Div. 2) will start. Note that the start time is unusual.

Series of Educational Rounds continue being held as Harbour.Space University initiative! You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.

This round will be rated for the participants with rating lower than 2100. It will be held on extended ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.

You will be given 6 or 7 problems and 2 hours to solve them.

The problems were invented and prepared by Alex fcspartakm Frolov, Mikhail awoo Piklyaev, Max Neon Mescheryakov and me. Also huge thanks to Mike MikeMirzayanov Mirzayanov for great systems Polygon and Codeforces.

Good luck to all the participants!

Our friends at Harbour.Space also have a message for you:

Hello once again, Codeforces!

We are thrilled to congratulate our faculty member Nikolay KAN Kalinin on his first place at the ICPC World Finals, celebrated in Moscow, Russia. Years of training by Nikolai and his team from Nizhny Novgorod State University lead them to the top of the scoreboard, defeating teams from 116 other universities and becoming world champions.

We would also like to congratulate our future student Egor 244mhq Dubovik, who won the silver medal with the Belarusian State University. Egor will join our Masters in Computer Science in the coming weeks.

We are looking forward to seeing Nikolay again next January when he teaches his course on Advanced Algorithms and Data Structures alongside Mike Mirzayanov. In this course, students focus on key and in-depth algorithms and data structures that form a modern computer specialist’s toolkit.

We are always excited to see Codeforces participants as our students here at Harbour.Space, so once again we’ve given a special discount (up to 70%) for the single course participation in Barcelona, Spain (travel cost and accommodation are not included).

Reserve your spot →

Codeforces and Harbour.Space

Good luck on your round, and see you next time!

Harbour.Space University

UPD: The editorial can be found here.

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3 years ago, # |
  Vote: I like it +40 Vote: I do not like it

Note that the start time is unusual

is becoming quite usual now.

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3 years ago, # |
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Great time for Chinese players!

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3 years ago, # |
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Thank you for the contest. I hope that my rating will be increase after doing this contest

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3 years ago, # |
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Why is it so early dudeee, some people have school

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    3 years ago, # ^ |
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    Yes of course "School in Sunday"

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      3 years ago, # ^ |
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      what's wrong with school on sunday? some schools have weekends on Friday and Saturday instead of Saturday and sunday

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    3 years ago, # ^ |
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    Some people also have school during the usual contest time.

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    3 years ago, # ^ |
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    For Chinese and Japanese,this tmie is better than usual time, as the usual time is 10:35~01:05 in Peking, China.

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3 years ago, # |
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Too sad we are not seeing vovuh's handle in writers part of an educational round...

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3 years ago, # |
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[deleted]

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3 years ago, # |
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All the best guys!

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3 years ago, # |
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Why the participation has reduced a little bit from previous 2-3 contests?

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    3 years ago, # ^ |
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    Due to universities mid exams I guess !!

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3 years ago, # |
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RIP sleep schedule. Gotta get up at 5am for this contest.

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    3 years ago, # ^ |
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    In China we have to stay up until 2am to participate in contests with usual times :(

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3 years ago, # |
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Hoping for a great contest OvO

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3 years ago, # |
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The start time of this contest is very friendly to Chinese contestants! Hope I'll perform well in this contest.

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3 years ago, # |
  Vote: I like it -32 Vote: I do not like it

In China, you can take a nap, wake up and start playing games

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    3 years ago, # ^ |
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    The game I said refers to playing codeforces. Please don't get me wrong

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3 years ago, # |
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I hope I will be specialist after this round.

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3 years ago, # |
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I enjoyed this round There is really very interesting problems

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3 years ago, # |
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How to solve D?

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    3 years ago, # ^ |
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    I tried to use inclusion-exclusion principle. The answer is (different a) + (different b) — (different a and b). I have figure out how to calculate "different a" and "different b", but didn't figure out how to calculate "different a and b".

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      3 years ago, # ^ |
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      Lol, I did the same. Couldn't pass it during the contest. Later debugged and got accepted. Definitely it was an overkill..

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    3 years ago, # ^ |
    Rev. 2   Vote: I like it +33 Vote: I do not like it

    I calculated the total possible solutions which will be nC3 and then subtracted the number of bad combinations(observe that in a bad combination 2 problems will be having the same topic type and 2 problems will be having the same difficulty).

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    3 years ago, # ^ |
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    Try to calculate the number of bad combinations and then it's trivial.

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3 years ago, # |
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How to solve E?

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    3 years ago, # ^ |
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    with dp. the main point is for each flip you just have to update O(n) dp values which makes it O(nq) that fits in time limit

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      3 years ago, # ^ |
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      I was wondering the whole time why's $$$q \leq 10^{4}$$$ and not $$$10^{5}$$$, I knew that solution was somewhat related to it. Unfortunalely, I wasn't able to figure it out in time.

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      3 years ago, # ^ |
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      Can you please explain further, my brain is fried up right now and I am unable to see the dp states..:P

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        3 years ago, # ^ |
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        yes sure let dp[x][y] be the number of stair cases that start from point x,y also dp1[x][y] : number of stair cases from x,y that second move is right and dp2[x][y] : number of stair cases from x , y that second move is down .

        now we can easily update all these dp values (you can see my submission) sorry for bad english

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          3 years ago, # ^ |
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          No no the english is quite clear, and thanks. I think I will see ur submission now.

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        3 years ago, # ^ |
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        and also for each query update you just have to update these points : (x,y) (x-1,y) (x,y-1)

        (x-1,y-1) (x-2,y-1) (x-1,y-2)

        ...

        which is in total at most 3*min(n,m)

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          3 years ago, # ^ |
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          Ahh yes I see.. BTW do u think this was solvable by Segment Tree or some similar data structure? I would think so considering the type of problem.

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          3 years ago, # ^ |
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          Hey, can you please explain why only those points have to be updated? I think points like (x, y-2) also need to be updated

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    3 years ago, # ^ |
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    There is a technique know in Brazil as Color Update, it consists in store in a set intervals that we assign a given color, each position can have at most one color. You can use this idea to color an interval, ask the maximal interval of the same color that contains a given point, etc. Using this idea just build maximal staircases, for each position store in which maximal staircases it is, and in which position. Then using Color Update technique the solution is straightforward.

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3 years ago, # |
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How to solve D ? I found it much harder than E or am i missing something ?

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    3 years ago, # ^ |
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    just find cases where the given condition fails and remove it from total cases

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    3 years ago, # ^ |
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    Inclusion and Exclusion Principle, the total number we can get without any constrains is (N choose 3), then we subtract the invalid ones.

    the invalid ones must have 2 similar elements in both sides so we just count that.

    https://codeforces.net/contest/1598/submission/131438774

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      3 years ago, # ^ |
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      could you elaborate how you calculated invalid ones

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        3 years ago, # ^ |
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        Invalid ones should have at least two similar elements in both arrays let's say we picked the current element.

        A is the topic
        B is the difficulty
        so
        A should appear at least one more time.
        B also should appear at least one more time.

        so we count the number of other A's multiply by the number of other B's and keep subtracting this from the answer, and this will work nicely with the constrains we have as (A,B) can't repeat again together.
        sorry it's hard to explain, but you can go through my code for better understanding.

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        3 years ago, # ^ |
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        Use Multiplication Principle.

        Preprocess each element's number of occurrences , and mark them as cnta[] and cntb[].Enumerate i from 1 to n , and the invalid ones are sigema((cnta[a[i]] — 1) * (cntb[b[i]] — 1)).Hope you can understand it.(My English isn't good).

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          3 years ago, # ^ |
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          thanks man.

          understood.

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            3 years ago, # ^ |
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            What's more.Because of "It is guaranteed that there are no two problems that have the same topic and difficulty at the same time".

            Invalid ones are a group with two identical "elements A" and two identical "elements B". So we just need to select the first pair first, and then select the other two pairs which have one "element A" and one "element B" which are the same as the first pair. The formula is just written in my last comment.

            I think this comment is more clear than the last one.By the way , wish you good luck. :)

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              3 years ago, # ^ |
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              yup , just got the code accepted.

              this comment really summed it up for me ^_^

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    3 years ago, # ^ |
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    Definitely missing something. $$$D$$$ was a nice exercise of complement counting. I solved it by adding edges $$$a[i], b[i]$$$ to empty bipartite graph with $$$2$$$ partite sets of size $$$n$$$ and then the problem becomes equivalent to counting the number of paths with $$$3$$$ edges in this graph.(at this point the problem is quite easy by storing frequency array of both $$$a$$$ and $$$b$$$) Of course this is only for better visualization. In reality you never actually do a graph algorithm to solve the problem. But I personally found this as a nice way to see the problem.

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      3 years ago, # ^ |
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      Can you please elaborate your visualization?

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        3 years ago, # ^ |
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        Uhh just add vertices in the graph and do simple case work for counting the complement. You will find every valid element in the complement corresponds to a path of length $$$3$$$ in the graph u made...

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    3 years ago, # ^ |
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    You are not alone.

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3 years ago, # |
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Can E be solved with link cut tree ?

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3 years ago, # |
  Vote: I like it -70 Vote: I do not like it

Definitely moving up from Pupil today!! How was the contest for y`all guys?

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    3 years ago, # ^ |
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    There are so many comments in your code :-\

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      3 years ago, # ^ |
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      Removing the comments in your submission of problem d, its remarkably similar to the solution in this image

      Spoiler

      I don't know if you will be caught by plag check but, What's more shocking is the audacity of your comment " Definitely moving up from Pupil today!! How was the contest for y`all guys? "

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    3 years ago, # ^ |
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    3 years ago, # ^ |
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    You seem to be a Cheater

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      3 years ago, # ^ |
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      Ironical to spam comments just to avoid plag check. Where is your dignity lol

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    3 years ago, # ^ |
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    you are not going to see any rating change.

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    3 years ago, # ^ |
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    You piece of shit. Look into the mirror and ask yourself what have you gained from cheating. You are a disgrace. Your parents will be ashamed of you.

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    3 years ago, # ^ |
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    Someone, Please hack this cheater's solution The same copied codes by others are hacked by many people

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3 years ago, # |
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solved D but tough Java constraints :( https://codeforces.net/contest/1598/submission/131447806

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

A round a day keeps my rating away.

It seems that I will get about -50 rating after this round. I hope I can remain an Expert.

By the way , Problem D is such an excellent problem although I didn't solve it in this round. It really gives me a lesson.

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3 years ago, # |
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Talk about the shortcomings of this game (this does not mean that I do not support the author, I support every author of CF), the data range of the C question is very intriguing (why is there a data range that cannot be covered by int64?), I am useless Over int128, so I spent nearly 20 minutes on compilation errors. Of course, the other topics are very good, thank you very much to the authors!

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    3 years ago, # ^ |
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    I guess you checked for $$$sum \cdot (n-2)=(sum-a_1-a_2)\cdot n$$$?

    Notice, that you can shorten this check to $$$2 \cdot sum = a_1 n + a_2 n$$$ for which both sides do fit in a long long.

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    3 years ago, # ^ |
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    int64 works fine for C. You just need to reduce the equation with algebra.

    Spoiler
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How To solve D ?

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    3 years ago, # ^ |
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    Let be ai = task, notice, that for a we can take (n * (n — 1) / 2) other tasks and lets just remove pairs of tasks that we can not use with a. This is the amount of the same difficulty * the amount of the same topic

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    3 years ago, # ^ |
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    If we consider pairs as coordinates on 2D plane and since all points are distinct, only way three points selected at random dis-satisfy the given conditions is when they form the right angle triangle. Hence subtract number of right angle triangles from NC3.

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3 years ago, # |
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what is the meaning of open hacking phase is running, can i get points if i solve problems or what it is ?

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    3 years ago, # ^ |
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    You can hack other's submissions. Hacking doesn't earn or loss points in this contest. Feel free to hack if you find any bugs in other's code.

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I misunderstood D and thought that both the given conditions should be satisfied. Later obtain the actual answer easily from the answer I calculated. Anyone did the same?

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    3 years ago, # ^ |
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    D was combinations problem though looks like graph

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      3 years ago, # ^ |
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      break into smaller subproblems simplify using pnc and generalize the formula to compute in O(nlogn)

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    3 years ago, # ^ |
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    why u didnt check examples?

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      3 years ago, # ^ |
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      I'm lazy and I wanted to solve the problem faster.

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3 years ago, # |
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In C why (sum*(n-2))==(sum-a1-s2))*n is wrong?

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    3 years ago, # ^ |
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    overflow.

    sum can be at max 2e5 * 1e9 == 2e14

    And then sum*n==2e14*2e5 == 4e19 which is more than the limit of the long data type

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    3 years ago, # ^ |
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    I don't think this is wrong, maybe its an overflow issue

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    3 years ago, # ^ |
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    bro power will reach to something 10^20 and long long has range 10^18

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      3 years ago, # ^ |
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      Can't we find mean using this method ?

      double k = 0; //k is the mean
      int t = 1;
      for (double x : arr) {
          k += (x - k) / t;
          ++t;
      }
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        3 years ago, # ^ |
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        avoid doing such operations bro.

        it won't give you the correct ans.

        there are always chances of floating pint error,

        performing such operation is not advisable

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3 years ago, # |
Rev. 5   Vote: I like it -67 Vote: I do not like it
  • I love Russian Contests.
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    3 years ago, # ^ |
      Vote: I like it +36 Vote: I do not like it

    The fact is that there are many Chinese top coders like Miracle03 and QAQAutoMaton are busy in teaching some new coders.So they don't have time to write more contests.The rounds you see are written by some students who love coding , and I think they're good enough for these younger coders.I think they need more encouragement but not prejudice.

    I do hope you could post comments with less discrimination and prejudice.And I think Chinese writers can get progress very soon.Best wish.

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3 years ago, # |
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Too good questions (:)

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3 years ago, # |
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Where is the solution of this contest ? I think I need it!

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3 years ago, # |
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there is a solution for problem E

notice that each staircase's length at most 2000

and each point have 4 staircases at most

so U can use brute force to solve it

the complexity will be $$$O(Q(N+M))$$$

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    3 years ago, # ^ |
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    Oh, i mean that you can start only 4 staircases at each free cell

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when will publish editorial?

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3 years ago, # |
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Awesome round. Love problem F

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    3 years ago, # ^ |
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    F has too many details :( Is it only for me?

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solved E just several seconds after the round ends, what a pity! :(

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3 years ago, # |
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Awesome round with much less extra hackings (comparing with other Edu rounds) ,excepting problem D.

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Can anyone help me with the approach to solve B?

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    3 years ago, # ^ |
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    There are only 5 days. We can enumerate all pairs of days and find if it is possible to seperate students into two groups. Let's say we choose day A and day B. If there are some students (possibly one) who can't attend day A and day B, then {A, B} isn't the answer. We count the students who can only attend at day A, students who can only attend at day B, and students who can attend at both day A and B.

    Code
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3 years ago, # |
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Nice round. Totally enjoyed it!!

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3 years ago, # |
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Please explain how to solve D. >_<

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    3 years ago, # ^ |
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    Substract the illegal ways from all the ways.

    There are in total $$$\dbinom n 3$$$ ways to choose $$$3$$$ elements among $$$n$$$ of them. Now considering how many of them are illegal, that is, none of the two conditions in the statement is satisfied.

    To calculate it, we first define cnta[x] as how many times topic x has occured and cntb[x] as how many times difficulties y has occured.

    Then assume that we choose the $$$i$$$ th problem, since no two problems share the same $$$a_i$$$ and $$$b_i$$$, we can assume the second problem we choose shares the same $$$a$$$ with $$$a_i$$$ and the third problem shares the same $$$b$$$ with $$$b_i$$$. So we just need to calculate 1ll * (cnta[a[i]] - 1) * (cntb[b[i] - 1]).

    You can view my code 131440598 for more details.

    Sorry for my poor English(

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      3 years ago, # ^ |
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      Can you provide a more rigorous proof?

      It seems that a lot of the solutions for D are getting hacked (mine included) and they all count the answer the same way you are describing. I don't know why though.

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        3 years ago, # ^ |
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        They are hack just because they didn't use long long

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        3 years ago, # ^ |
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        In your code, when it comes to nA += (ca[a[i]]-1) * (cb[b[i]]-1);, I suppose that it overflows the int limitation

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      3 years ago, # ^ |
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      thank you

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3 years ago, # |
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The penalty for each incorrect submission until the submission with a full solution is 10 minutes What is the meaning of this?

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    3 years ago, # ^ |
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    Penalties are for the leader board. More time used means lower ranking.

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      3 years ago, # ^ |
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      Oh okay. I am a little confused, will I get penalities for multiple wrong answer submissions for a problem even If i don't solve it?

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        3 years ago, # ^ |
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        No, penalty will be added only for those problems which are accepted.

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Any hack tc for D ?

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    3 years ago, # ^ |
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    I think the hacks exploit this.

    I believe you can have all numbers from $$$1$$$ to $$$100000$$$ and create the pairs $$$(1, 100000)\ (2, 100000)\ \dots\ (100000, 100000)\ (1, 1)\ (1, 2)\ \dots\ (1, 99999)$$$.

    In that way there are $$$2\cdot 10^5 - 1$$$ pairs (possible because $$$n\leq2\cdot 10^5$$$) but the product of counts will overflow if they are initialized as int. (for example considering the first pair the counts will both be $$$99999$$$)

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      3 years ago, # ^ |
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      I used size_t( vector.size() ) in my code, but it still be hacked.Do you know why?

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

        You've been overflown. Thought your ans is 'long long', you are not yet familiar how expression is evaluated. It's in this line:

        ans += (cntb[d].size() - 1) * (x.se.size() - 1);

        Try to output: long long ans = 100'000 * 100'000;

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

          Thanks for your reply!
          Could give me one testcase? I test my code with input:
          n = 200000
          a = [1, ..., 1, 100001, 100002, ..., 200000]
          b = [1, 2, ..., 100000, 1, ..., 1 ]
          When x.fi = 1, d = 1, that line is:
          ans += (100000 — 1) * (100001 — 1). It seems that my code is correct on this testcase with the ouput of 1333303333500000.

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            3 years ago, # ^ |
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            Try for this one. But with very large sequence, (1, 1), (1, 2), (1, 3), ... , (2, 1), (3, 1), .....

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3 years ago, # |
Rev. 4   Vote: I like it -10 Vote: I do not like it

Nice problems :)))

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3 years ago, # |
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Nothing to say!XD

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

Why O(Q*(N+M) + N*M) is failing for problem E (TLE on testcase 41 (not included in pretests))?
My Code : https://codeforces.net/contest/1598/submission/131480763

Edit : Finally my code was accepted. Replaced vector with pair which brought a major change in execution time (approx 7 times faster). Always try not to use small vectors (if possible).
AC Code : https://codeforces.net/contest/1598/submission/131498204

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    3 years ago, # ^ |
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    Assuming that you are doing a bfs for each query : giving the fact the number of situations is at most (nb of x * nb of y * nb dir = N*M*2), the total complexity of your algorithm is O(N*M*2*Q + N*M) = O(N*M*Q), which is too slow. Sorry for my bad English.

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      3 years ago, # ^ |
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      No, the time complexity is fine actually because for changing a particular cell you just need to update max O(n+m) cells, not the whole dp table. I had run test case 41 on codeforces and it's taking 3.3 seconds to give the output. The runtime is slow. Can anyone please optimize this code?

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    3 years ago, # ^ |
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    To make your code cleaner, you can use array<ll, 4> instead of those pairs, here is the AC code with that modification 131514905

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3 years ago, # |
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I just came to know that Um_nik was in the 1st place.. I thought that he finished University and I was watching the closing ceremony and thought that this guy behind the mask has a familiar hairstyle and then after googling I got to know that he was our Um_nik. I tagged you because I wanted to congratulate you publicly but without a blog.. Congratulations and wow you are a hardcore person from specialist to red and then ICPC Champion... way to go man I couldn't be a great coder winning contests like you do till now but it's ok coz everyone's life is different.. but it feels good to see an inspiration of hardwork and straight forward man(this i am saying by your blogs and comments) like you at the top.

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3 years ago, # |
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Can somebody give a testcase where this is going wrong

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    3 years ago, # ^ |
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    try this

    1
    6
    1 0 0 1 0
    0 0 0 1 0
    0 1 0 0 1
    0 1 0 1 1
    0 1 0 1 1
    0 1 0 1 1
    
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3 years ago, # |
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For problem C I used double still I think it is overflowing somewhere How can double overflow here.Please help My submission

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    3 years ago, # ^ |
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    just use (long long) instead of (int) here: cout<<int(s)<<'\n'; int has been overflowed, not double

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3 years ago, # |
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When will the solution be available

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3 years ago, # |
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Does anyone know why the unordered_map solutions to C are resulting in TLE? Is it because of some crafty input which is causing everything in the hash table to be put into single bucket?

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    3 years ago, # ^ |
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    yeah its because of anti hash test case which is causing the unordered map to have a bad hashing

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      3 years ago, # ^ |
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      That's really interesting. Do you know where I could read more on this or how I could generate such cases?

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        3 years ago, # ^ |
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        yeah you can read about it here, I have also used these techniques to hack 20+ submissions of C

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      3 years ago, # ^ |
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      my question is that is it valid to hack solution on the basis of short coming of unordered hash , should this be concern of the problem . I think its a useless ground to discard out solution specially in problem C

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      3 years ago, # ^ |
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      What about hash map implementations in other languages (say Python dict)? Can they be hacked like this?

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3 years ago, # |
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Can someone please let me know the problem with my Problem-C solution. It is giving TLE on Test-17. I have used a standard unordered_map based solution, with O(n) complexity as per my understanding.

Any help will be appreciated. Thank you !

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    3 years ago, # ^ |
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    Unordered map does not work as efficient as hashing. Technique

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      3 years ago, # ^ |
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      Is not unordered map == Hashing ?

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        3 years ago, # ^ |
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        unordered map worst time complexity is O(n) , so anti hash test cases has been used to hack unordered map without custom hashing

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    3 years ago, # ^ |
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    Notice that unordered_map is a hash table and the complexity is average O(n). I believe that is some input that is causing it to get the worst case complexity which is O(n^2)..

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      3 years ago, # ^ |
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      So should I prefer ordered_map in future ?

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        3 years ago, # ^ |
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        I don't know to be honest. I myself used unordered_map in the contest.. I think that if the log n operations are acceptable we should prefer map since.. Though I think someone more experienced would be able to answer better.

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        3 years ago, # ^ |
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        you can still use an unordered map but with custom hashing ,you can copy paste the template from this blog,because in edu round people have enough time to make anti hash test cases.

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

          Thanks KingRayuga, will keep in mind.

          I think we can also go with ordered_map as it seems that the time limits are flexible enough to accommodate O(n*logn) solution even when best one is O(n). Please, correct me if I am wrong.

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    3 years ago, # ^ |
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    Just read someone else's comment here and tried with ordered map. Solution Accepted !

    So what should I learn/conclude from here regarding the use of ordered/unordered_map. Like when to prefer ordered_map over unordered (because here I thought using unordered should work, but id didn't eventually).

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3 years ago, # |
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The test cases used for successful hacks will be included in the main tests?

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3 years ago, # |
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Probably not relevant to the blog but I have a solution which I think would work for non-distinct pairs for D. Would anyone like to test it out or perhaps reply with a solution of their own which also works for non-distinct pairs.

Here's my submission

https://codeforces.net/contest/1598/submission/131482336

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3 years ago, # |
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same solution using unoreded_map is tle (after hack) but using map is accepted.

I used unordered_map in contest

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3 years ago, # |
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When will the ratings change?

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3 years ago, # |
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How to solve problem F? I have been thinking for a long time but still have no ideas. Can anyone tell me the idea of this problem? Thx.

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    3 years ago, # ^ |
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    Hint
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3 years ago, # |
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Why ratings change taking so long?

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    3 years ago, # ^ |
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    I guess because the time of the contest, when the hacking phase finished the timing was too late for Mike (Just a guess) :D

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    3 years ago, # ^ |
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    yes editorial was also late

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3 years ago, # |
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Rating change is taking longer than I thought!!!

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    3 years ago, # ^ |
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    I think it's because the unusual time of this round, but also, I have to say that the rating change of this round is the longest I have waited...

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3 years ago, # |
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when does system testing starts?

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3 years ago, # |
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MikeMirzayanov please remove the cheaters and update the ratings thanks !

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    3 years ago, # ^ |
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    why the systetm testing is caried out multiple times. is this for rejudging with additional testcases.?

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3 years ago, # |
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3 years ago, # |
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why system testing is done so many times ?

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3 years ago, # |
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unordered_map's TLE is about hash, but how about multiset?

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3 years ago, # |
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I will become pupil but why the delay in giving ratings.

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3 years ago, # |
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I find 2 solutions are copied. But even why were they not skipped

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3 years ago, # |
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Please upload editorial as soon as possible. Thank you for the contest.

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-1

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    3 years ago, # ^ |
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    I agree w you. I found many cheaters in Educational round 115 but their rating still increase

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3 years ago, # |
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Hello everyone, I have written a blog on improving codeforces rating that I used to follow and it is a really good routine. Do read the blog here.

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3 years ago, # |
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when will cheaters be remove. I found many cheaters but they were not removed

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Can anyone tell me why my rating was decreased by 29 even when my verdict was OK on a single submission. No other submissions were done .Just one and that too with OK verdict

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    3 years ago, # ^ |
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    Rating change depends on your rank

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    3 years ago, # ^ |
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    If your rating had been increasing whenever you were making a single correct submission in a round you would have been a legendary grandmaster by now.

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3 years ago, # |
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-1

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3 years ago, # |
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Have the cheaters been removed? As there are many remaining..