Apptica's blog

By Apptica, history, 5 months ago, ,

Hello everyone!
I would like to invite you to participate in HackerEarth HourStorm #7. It’s the seventh version of the short contest, that runs for 1 hour! The problem set consists of 3 traditional algorithmic tasks of various difficulties. The contest starts on the 11th of January, 9.30- PM IST

For traditional algorithmic tasks, you will receive points for every test case your solution passes — so you can get some points with partial solutions as well. Check contest page for more details about in-contest schedule and rules.

Great thanks to raghavkapoor1997 for preparing the tasks and isaf27 for testing them. As usual, there will be some prizes for the top three competitors:

$75 Amazon gift card$50 Amazon gift card

In addition, top 5 on the scoreboard with rating less than 1600 will win HackerEarth t-shirts. Good luck to everyone, and let's discuss the problems after the contest!

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 » 5 months ago, # |   0 Auto comment: topic has been updated by Apptica (previous revision, new revision, compare).
 » 5 months ago, # |   +16 I didn't participate in, but when I look at "Number Game": the statement definitely is wrong.
•  » » 5 months ago, # ^ |   0 And what's wrong with it?
•  » » » 5 months ago, # ^ |   +1 The statement might miss condition {Ai} must be pairwise distinct.
•  » » » » 5 months ago, # ^ |   0 What theorem does it use anyway?
•  » » » » » 5 months ago, # ^ |   +10 If a polynomial of degree  ≤ N is zero at N + 1 distinct points then it is zero polynomial.
 » 5 months ago, # |   +24 The problem statement for Orthogonal Pairs was not clear at all.The problem says to compute "the number of ordered quadruples" (P1, P2, P3, P4) such that the angle between the lines P1P2 and P3P4 is exactly . This was not actually what the problem was asking:First, P1, P2, P3, P4 do not have to be pairwise distinct; in particular, you are allowed to form a angle with just three points. This was not specified anywhere. Moreover, all of the following quadruples are considered the same as (P1, P2, P3, P4): (P2, P1, P3, P4), (P1, P2, P4, P3), (P3, P4, P1, P2), etc.In other words the quadruples aren't ordered at all. (And they aren't quadruples at all, but rather pairs of pairs.) A better formulation of the problem would have been to first generate all of the line segments formed by an (unordered) pair of the N points. Then count the number of (unordered) pairs of line segments that form perpendicular lines to each other.The sample explanation was also broken during the entirety of the contest, which made things even worse. But having seen the actual explanation, it did not clear up any of the above issues anyway. The only way I was able to eventually guess what the problem was asking was by reverse engineering the sample output.
•  » » 5 months ago, # ^ |   +18 The second line in the problem statement about the quadruples is ambiguous and we have updated the problem statement. We apologize for the same and thank you for the suggestion.