Doit
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In this problem, all employees start out with 100 units of labor left to finish.  Your goal
is to say "doit!" at various time points so as to make sure the total amount of time
required to complete all lighting fixtures is minimal.  

The main simplification in the problem is to realize that no matter when you choose to say
"doit!" (if at all), the n_- people who respond negatively will always finish their fixtures
last and the n_+ people who respond positively will always finish first.  Since "doit!"s only
speed up the n_+ people who respond positively, we can assume without loss of generality that
you choose to say "doit!"s in each of the first k time intervals for some k.

At this point, it suffices to simply try different values of k since the maximum value we need
to try is no greater than ceil(100/(r+2)) (after that time, only neutral employees or the
n_- negatively responding employees will be left, and it does no good to say "doit!" to them).
From this point on, the remainder of the problem consists of correctly computing the time
taken if you say "doit!" in the first k time intervals and is a fairly straightforward 
exercise in arithmetic, provided you make sure to round correctly.