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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