Hi all
International Olympiad in Informatics 2014 is coming close and we decided to host 2 practice contest for our hackers.
There will be two contests: Practice Contest 1 and 2. Both contests consists of 5 interesting challenges based on IOI syllabus and will run for 3 hours. IOI 2014 Practice Contest 1 will commence on 5 July 2014 at 15:30 UTC and IOI 2014 Practice Contest 2 wil start at 6 July 2014 at 15:30 UTC.
IOI is conducted for High School students but we would like everyone to join the spirit.
Contributers: kevinsogo, Peng Cao, Guissmo Asuncion.
Languages Allowed: C, C++, Pascal, Java
Scoring: ACM ICPC scoring will be followed.
Happy Coding!
Too bad the first one is at the same time as TCO Round 2C. Although it makes sense a bit, since most of the IOI participants are not eligible for TCO yet :3
>_>
First, there are probably plenty of IOI participants eligible for TCO, it just depends on the country. I myself went to IOI when I was 18 and 19 years old.
Second, this is either forcing people to choose one contest, or just negligence. There's just 1 major contest going on tomorrow (TCO), and it's not even very long, so why the hell does there have to be a collision again (since this happens rather frequently)?
Well, yeah. Checking with http://clist.by before setting up a contest time shouldn't be that hard :|
Can you open clist.by now. I can not.
A gentle reminder. Contest starts within an hour. :)
The time links don't work, probably because of annoying characters in them. Here's a way to fix that.
Okay, this contest is awkward and slacked off in many ways, but I won't flame too much and ask just 1 question:
WHY IS AN IOI PRACTICE CONTEST USING ACM ICPC RULES??!!!!1!1
How to solve last 2 problems from the second contest?
The last one: here
The last but one: "Divide and conquer optimization" from here.
How do I solve the second problem efficiently? I got time out on a couple cases solving it the best way I could think of.
I had a vector for every possible factor. Then I used trial division to factorize the integers and added the original number to every vector corresponding its factors. The query could be then done simply using using lower_bound and upper_bound to vector corresponding the given factor and calculating the distance between lower_bound and upper_bound.