COCI

Will someone participate in the COCI competition tomorrow. Can someone give me more information about the competition ( number of tasks, duration , etc. ) ? :)

I will. Duration is 3 hours, number of tasks is 6. Theis difficulties are very wide, from very easy first task to practically unsolvable (for me, at least) sixth, so point distribution is very real. I mean, participation in COCI will give you accurate picture of your skills.

Grading is like on normal Z competitions: you submit your source and it gets tested on those cases given in your problem description. If it passes those tests, then it's being put for grading after the contest ends. Number of test cases and points per case are different for every task.

That's about it... If you need to know anything else, just ask. Maybe this link can help too.

http://www.hsin.hr/coci/announcement.html

Good luck to you and all other contestants tomorow! ;)

Okay, thanks for the explanation. Good luck to you, too and all other competitors. ;-)

I will participate too. Sortof. We'll be having HONI ( native version of COCI ) and I'll upload same codes on both.

Good luck all ;)

I'm going to do it too. See you there =)

How many problems have you solved? I've solved 1. and 2. task and i have the idea for solving the 3. and 4. task. :-)

Only 176 points :((

:((( Terrible! Only 120...

I can't believe it, my code for the first task isn't true. :( Only 48 points...

The first task is bubble sort. I have just 115

I solved first 3. 100%,and i had no idea how to solve the rest(that would work 100%). i tried something but all i got was 0.

129

I know. But I wasn't careful. I've double printed some steps of the sort. :(

Are we allowed to upload the tasks from COCI on z-trening?

I received full credit for the first 3 and partial credit for 4 and 5. Total = 280

What was your idea for 4. i 5.?

Can someone add COCI's problems? I couldn't participate today, but I know that their problems are always nice.

280 here same :) 1., 2., 3. full
4. just last test case TLE
5. 4 cases passed

@msantl: Nothing smart, I just mostly used brute force.

#4: I recursed through all possible ways of coloring one black and one white our of 4. It very inefficient but I was lucky that the cases weren't that bad. It got 70 points.

#5: I traversed through the lattice points and check to see what type of bank they were. I noticed that if the gcd(dx, dy) was 1 then the point was visible directly. (dx and dy are the changes in X and Y between the guard and bank). This was enough for 60 points. I'm curious to see what kind of optimization tricks were used to get full points.

@frank44:
#5: If you want to find number of points not visible by guard (at postion (A, 0)) for some fixed y = YY you can do this:
sum = 0
Take all prime divisors of YY - A
for each combination comb of divisors
sum += (-1) ^ comb.size * L / multiple_of_divisors_in_comb

sum contains wanted number

In similar way you solve each sub-problem. Idea is to use inclusion-exclusion method.

How did you get this idea during the competiton? Great work, I admire you... =S

@boba5551
wow, I'm with iggy91, that is really nice solution. I would never have thought of that about using PIE.

What is PIE?

3.141592... just kidding. It's the Principle of Inclusion-Exclusion (PIE for short).

xD Ok, tnx.