I have tried to make the same plan work, but unfortunately I could not solve F in time — my solution was written at the end of the contest, but it had two issues:

- It did not work on the samples
- Its complexity was O(
*n*^{2}*log(*n*)) in the worst case (even though with a 7-second time limit this was not obviously bad)

I could make it work on the samples in just a couple of minutes after the end of the contest, and the solution passed the system tests — only to be hacked as the complexity was in fact too big. I would've probably reached the second place if I had those couple of minutes, and I'm still on the fence whether the fact that I'd have squeezed in an incorrect solution makes me regret this more or less :)

I'd like to highlight another problem though, for which I had completely no working ideas during the contest: problem D. You are given

*n*<=2000 points on the plane such that no three points lie on the same line. Each point has an integer label between 1 and*n*, and all labels are distinct. Your goal is to rearrange the labels in such a way that the*i*-th point has label*i*. You are allowed to take any two points and swap their labels, but when you do that you must draw a straight line segment connecting those two points. The segments you draw must not intersect except at their endpoints (in particular, you are not allowed to draw the same segment twice). Can you see a way to achieve the goal?In my previous summary, I have mentioned a Code Jam problem: you are given up to 10

^{15}cards, each with a prime number between 2 and 499. You need to split the cards into two piles, such that the sum of the numbers in the first pile is equal to the product of the numbers in the second pile, or tell that it's impossible.The key idea here is that the product grows very quickly, so the second pile will always be very small. Therefore the sum of the numbers in the second pile will be small. But the sum of the numbers in the first pile is the sum of all given numbers minus the sum of the numbers in the second pile. If the sum of all given numbers is

*S*, we just need to check all numbers in the segment [*S*-*C*,*S*] as the candidates for the sum of the numbers in the first pile (which is equal to the product of the numbers in the second pile) for some relatively small value of*C*.How do we check a particular candidate? Here the fact that all numbers are prime comes into play: since we know that the candidate is a product of some subset of the given numbers, and all of them are prime, there is in fact a unique decomposition for it as a product of primes. Factorizing a number this big can still be problematic, but since we're only interested in prime factors up to 499, it is fast enough.

You can check the official analysis for more details, such as what is a reasonable value for

*C*. Thanks for reading, and check back next week!
Why is the blog's title "yuri"?

ReplyDeleteI had two things in mind:

Delete1) The featured Codeforces round used the characters from "Yagate Kimi ni Naru" manga, which according to Wikpedia is a yuri manga series.

2) The 60th anniversary of Yuri Gagarin's first spaceflight was last week :)

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ReplyDelete