Hi! On 2014-05-04 17:48, LeppyR64 wrote: > I didn't participate in this round since I qualified through Round > 1A. > > In analysis I noticed that a large amount of people near the bubble > for the top 1000 who got a wrong answer for A Large. I'm a little > confused that I fail to see the pitfall after successfully solving > the small input.
I'm one of those unlucky, because I was too confident about this "simple" problem (and focused later on the other two problems). For two strings, the number of moves required to bring a particular "position" to the same length is the difference between the counts in the two strings, i. e., for aaaabb aabb two (4 - 2) moves are required to bring aaaa and aa to the same length. I rushed way too much with the large input and used "maximum number - minimum number", which is of course wrong. Instead, one has to find a common length to bring all pieces to, and then count the difference of *each* piece to this common length. In order to minimise the L1-differences, the median should be chosen as the common length. This would have been an almost trivial change to make, but alas. If I wasn't clear here, let me know and I can try to elaborate in more detail on something. Yours, Daniel -- http://www.domob.eu/ OpenPGP: 901C 5216 0537 1D2A F071 5A0E 4D94 6EED 04F7 CF52 Namecoin: id/domob -> https://nameid.org/?name=domob -- Done: Arc-Bar-Cav-Hea-Kni-Ran-Rog-Sam-Tou-Val-Wiz To go: Mon-Pri
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