Also the two neighbours submissions have pretty different amount of includes...

You also can sort all of the cubes and use two iterators for the current upper cube and the corresponding cube it lays on.

Your friends are lying to you to collect some thumbs up!

Что должно выдавать в D(40) на тесте N = 20, K = 1, A[i] = 50 (для всех i = {0, ..., 19})? А еще спешу огорчить тех, у кого в B(20), на тесте 3 7 7, выдает 9:(

If we do what you say by the R = 2-eps, everythin will be OK:) Because it'll be uncovered point in a center of rectangle.

UPD: And if we don't reduce the sides of rectangle (but increase the radius of all circles by 2), it would means that the point with coordinate (4; 1.9) is a center of result circle. But it's false. Because it intersects with bottom side of rectangle.

Yes, but not "_the above mentioned rectangle_", because you have to decrease the sides of it.

We need to be able to check: is it possible to place a circle of radius R. This problem is equivalent to search uncovered point inside the rectangle, after increasing the radius of the circles on R. Also you have to remember to reduce the sides of the rectangle on R.

You have to use binary search by the answer. So you need to increase the radius of each circle at the checked value and determine "Whether there is any point which is located out of all circles?". It can be solved by using the divide and conquer algorithm. If all the circles don't intersect current rectangle, then we've already found uncovered points. If some circles intersect our rectangle, then we need to devide our rectangle into 4 smaller and return to the first part of algorithm. Otherwise, if the current rectangle is entirely inside any circle, then we exit from current branch of recursion. It works O(N * log(MaxAns/eps)).. But I'm not sure in this asymptotic.

Секунд 15 вчитывался в сообщение.. вот уж грамотей..

А что такого в D на 12ом тесте?

I think, you cannot make the conclusions becouse of first and second tasks.

Just a bug)