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By Endagorion, history, 5 months ago, translation, In English,
Tutorial is loading...
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Tutorial is loading...
Tutorial is loading...
Tutorial is loading...
Tutorial is loading...
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5 months ago, # |
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Auto comment: topic has been translated by Endagorion (original revision, translated revision, compare)

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5 months ago, # |
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Why it shows Tutorial is not available ?

UPD: It's available now :)

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5 months ago, # |
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Can anyone explain how Div2 B can be solved using Ternary Search

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    5 months ago, # ^ |
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    I have done this.

    We analyze total time in relation to position, I will call it p selected. Time of each point x with speed v is abs(x - p) / v. If we make p vary, as v and x are fixed we'll see an abs function

    Which is the total time? The worst of all those times. formally max(v1, v2, ..vn). If we make drawings we can see that if we plot the total time in function to the position selected, that graphic has no local maximum, and the left and right sides tend to positive infinite. So we observe that as the function is continuous we need to have a unique local minimum somewhere in the center of the graphic (otherwise the sides can't tend both to infinity). There can't be more than one minimum, because otherwise we would need a maximum to exist and we accepted it didn't exist. The left side of the minimum must be strictly decreasing (as it comes from positive infinity), and the right side strictly increasing (as it goes to positive infinitive), and so, the graphic is something like this

     Note that the left derivative is negative and the right one is positive. Which means we can make a ternary search on the point when it changes, from negative, to positive, that will be a local minimum, as the only one, the global one, hence the solution of the problem.

    Evaluating the function will be O(n), so total complexity the product of O(n) and your number of ternary search iterations (of logarithmic behavior)

    25253744

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      4 months ago, # ^ |
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      Great explanation, I have seen this problem on virtual contest and tried to solve with similar approach. Intuitively I felt that there should be only one point where the required time is minimum. But couldn't solve during the contest, after reading this comment everything clicked and got AC.

      By the way, I have checked your solution and why use mabs? cpp has its own fabs which does exactly same

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        4 months ago, # ^ |
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        By ignorance :) Sometimes, during a contest it is faster to write these simple functions rather than open google haha. Also sometimes I decide to avoid to use some c++ functions to reduce the risk of generate unexpected algorithm complexity, however this is not the case :)

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5 months ago, # |
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for Div.2 D (Div.1 B)
somebody says try all ai, if there is the conflict then try all bi. if these two cases both have conflict then the answer is NO.
like submission 25322923 and try the input:

  4 
ABC DDD
ABC EEE
QWE FFF
QWR FFF
get the output: NO but isn't
 ABD, ABE, QWE, QWR 
a solution?
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    5 months ago, # ^ |
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    Here we don't try all b_i when there is some conflicts for a_i. But only those b_i whose corresponding a_i conflict with each other.

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      5 months ago, # ^ |
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      I know but the somebody's code doesn't seems like that.

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5 months ago, # |
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Can someone please help me understand the editorial of problem 782B — The Meeting Place Cannot Be Changed. how to iterate for all t's and how to apply binary search.

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5 months ago, # |
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782B can be solved without binary search. Working time is much better comparing to binary search solutions. heuristic with enumeration all pairs & O(n*log(n)) sorting + O(n) scan

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5 months ago, # |
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in div2 E, why is it guaranteed that the tour exists? and even if it doesn't which, why is writing the nodes in that way is guaranteed to generate the answer?

edit: i understood the solution, but why is it called euler tour?

and is it me, or mentioning that traversing an edge twice is fine had to be mentioned in the statement?

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    5 months ago, # ^ |
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    would someone tell me if we can visit the same edge twice? and if so, why is it called "euler tour" in the editorial?

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      5 months ago, # ^ |
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      Yeah, we can visit the same edge twice. Suppose in the dfs ordering of nodes, we backtrack all edges (in the dfs tree) once.

      -> Total edges traversed = 2*(n-1) -> If we divide into k parts, we get [2*(n-1)/k] <= [2n/k] -> We can always assign dfs (preorder) accordingly.

      I have no idea why it's called Euler tour, since that means we should be visiting each edge once, and end up on the beginning node.

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        5 months ago, # ^ |
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        You should code simple recursive algorithm for Euler tour. But write vertice twice first time on visiting and another time after recursive return (on each return to the vertice). This will produce required sequence.

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In div1D, what is meant by "optimizing boolean multiplication with bitsets" and how do we achieve this optimization?

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Can you figure out what is wrong with my code in problem 781B (wrong in test 16)? 25333425

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what meaning "The route is a closed polygon line in the place, with all segments parallel to one of the axes" in Intranet of Buses? help on Intranet of Buses!

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5 months ago, # |
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I'm getting 'Tutorial is not available'.

And tutorials for some other rounds are also not available currently see this

KAN MikeMirzayanov please fix this.

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In Div1 B (football league problem) what is the mistake in my algorithm? I really don't see the bug. http://codeforces.com/contest/780/submission/25485848

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can someone please find why am I getting runtime error in python2.7 in problem C div 2?

Here is my code, a big thanks in advance

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in task d(div 2) test 5 aaa b aaa c aaq z aab z aab o craches lots of solutions. The result is probably YES aab,aac,aaq,aaz,aao but it isn't)

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4 months ago, # |
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What's the idea behind giving 5 seconds for div 2 B if the intended solution is 46 ms 26456429 . Is it to allow for N^2 solution or to trick people into attempting N^2 solution?

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    4 months ago, # ^ |
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    I believe it was done to allow solutions using slower programming languages.

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      4 months ago, # ^ |
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      interesting. There is programming language 100x slower than c++? If so, what language is that?

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        4 months ago, # ^ |
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        I'll give you a hint: the name of this language starts with letter p.

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4 months ago, # |
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Can anyone please explain me why the complexity in DIV 2 B is O(n * log (eta inverse) ) and not O(n * log ( max (eta inverse,(h-l) ) ), here h=upper bound and l = lower bound of the binary search respectively. Thank you in Advance :)