a question
4 the three first crack as they are snapers
1 OFF
2 OFF
3 ON?

2010/5/10 Greg D <[email protected]>

> I did something simpler than they proposed, but it solved the large
> problem in seconds, so I score it as "good enough". :-)
>
> Since the groups have to board in order, for each group you can
> calculate how many people will ride, and which group will be first the
> next time.  So I built an N x 2 array, and filled in those two values
> for each group.  After that it's a simple process:
>
> 1: Pull up group 0's info.
> 2: Add # riders you get when group 0 is front of the line.
> 3: Find out which group will be at the beginning of the line for the
> next ride.
> 4: If there's another ride to be done, go to 1, using the new group's
> info.
>
> Greg
>
> On May 9, 8:33 am, Pakku <[email protected]> wrote:
> > Guys,
> >
> > I am wondering what all optimization we can really do.
> > Though, in the qualification, for the theme park problem, i could not
> > solve the large input problem in the given time frame, i optimized it
> > latter in the practice mode.
> >
> > These are the optimization that I did.
> >
> > for each test case,
> >
> > 1. if capacity of roller coaster >= total strength [ sum of all groups
> > ] then simply total earnings = total strength * number of rides.
> > 2. detect if can get a rep if yes, then
> > TotalEarnings =  ( EarningsForRep * (RidesPerDay / TripsNeededToGetARep )
> );
> > and then TotalEarnings += earnings for the TripsStillLeftInADay ;
> >
> > Apart from these 2 optimization is there any other place we can
> > optimize. what is the best possible optimized solution.
> > anyone willing to share the best solution.
> >
> > -Prakash
> >
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-- 
Darwin Plazas Ortega
Programador
----------------
Si buscas resultados distintos, no hagas siempre lo mismo.  (Albert
Einstein)

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