ENG  RUSTimus Online Judge
Online Judge
Problems
Authors
Online contests
About Online Judge
Frequently asked questions
Site news
Webboard
Links
Problem set
Submit solution
Judge status
Guide
Register
Update your info
Authors ranklist
Current contest
Scheduled contests
Past contests
Rules
back to board

Discussion of Problem 1572. Yekaterinozavodsk Great Well

Nothing Controversial
Posted by jagatsastry 2 Dec 2007 23:27
Well people, I've made corrections to my post. I'd made a silly mistake.Nothing controversial here.

The solution:
 Take the first sample test case:
2 10
3
3 20
1 5
2 11

The max linear size of the hole is 10*root(2).
3 20: Min linear size of the cover is 20*root(3)/2. (doesnt pass through)
This is because the triangle can be tilted such that it passes with its projection being 20/root(2)=10*root(2).
1 5: min linear size=10 i.e. the diameter(passes through)
2 11: Min linear size=11 i.e. side of the square.(passes through)

Hence the answer is 2.


Edited by author 03.12.2007 16:48
Re: Controversial
Posted by KIRILL(ArcSTUpid coder:) 3 Dec 2007 00:14
for triangle a*sqrt(3)/2
Re: Controversial
Posted by jagatsastry 3 Dec 2007 16:49
I've made changes to the post. Thanks for the clarification.