To piss or not to piss

When a guy goes into the bathroom, which urinal does he pick? Here we
consider the following protocol: Each guy always chooses the urinal that
puts him furthest from anyone else peeing. In case of a tie, he chooses
the urinal closest to the entrance door (assume it to the left of the
bathroom). Additionally, at least one empty urinal is required between
any two guys. For instance, if there are five urinals, they fill up like
this:

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The first guy takes the left urinal, the second guy takes the right
urinal, and the third guy takes the middle one. At this point, the
urinals are jammed: no further guys can pee without awkwardness. But
it’s pretty efficient; 60% of the urinals are used. By contrast, if
there are seven urinals, they do not fill up so efficiently; less than
43% of the urinals are used:

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There should be room for four guys to pee without awkwardness, but
because the third guy followed the protocol, there are no options left
for the fourth guy. In order to increase efficiency, some walls can be
placed between two urinals, and then independently apply the protocol on
each one of the parts, from left to right. If we are allowed to place
one wall in the last example, efficiency would increase to more than
57%:

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Placing more walls the efficiency would increase further, reaching 100%
when using 6 walls. Your task in this problem is to compute the minimum
number of walls necessary to reach a desired efficiency, given the
number of urinals.

Input

Input consists of several cases. Every case has the number of urinals
(between 1 and 500), and the desired percentage of efficiency (a natural
number between 0 and 100).

Output

For every case, print the minimum number of walls needed to reach the
required efficiency.

Problem information

Author: Albert Graells

Generation: 2026-01-25T11:41:31.053Z

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