From: "Jesús Gabriel y Galán" <jgabrielygalan@...>
Date: 2012-09-26T18:44:24+09:00
Subject: Re: inject problem

On Wed, Sep 26, 2012 at 9:39 AM, Roelof Wobben <rwobben@hotmail.com> wrote:
> Now refractor the if then 's and I hope I can decline it to less then 5 if
> then's.
> But that will be difficult.

I think you should not structure the code this way. I think you should
loop through the counts, finding and removing sets. Something like
(pseudocode):

while there are sets of 3 ones:
  remove the set from the hash (h[1] -= 3)
  sum 1000 to the score
end

while there are sets of 3 any other pip count:
  remove the set from the hash (h[x] -= 3)
  sum 100*x to the score
end

sum 100 * (remaining number of ones)
sum 50 * (remaining number of fives)

This way, your code is more simple and robust. For example this will
also work if you throw 9 dice, while your current algorithm does not.

Jesus.