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вернуться в форумNothing Controversial 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 for triangle a*sqrt(3)/2 Re: Controversial I've made changes to the post. Thanks for the clarification. |
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