Supposedly Trivial math question.

John Shaver bobjohnbob at gmail.com
Thu Oct 3 11:52:00 MDT 2013


Hmm set the 'arbitrary' base as base 2 and just use the bits :)

http://ideone.com/lxkd6c

Not very useful for converting to anything other than base 2 though.

-John Shaver




On Tue, Oct 1, 2013 at 2:48 PM, S. Dale Morrey <sdalemorrey at gmail.com>wrote:

> static count x=1 was supposed to be static count =1;  Then a count++ there
> somewhere before the loop
>
>
> On Tue, Oct 1, 2013 at 2:45 PM, S. Dale Morrey <sdalemorrey at gmail.com
> >wrote:
>
> >
> > The base is arbitrary, meaning if the customer is giving base 10 money
> > then all change must be base 10, i.e. 1's 10s, 100's etc.  As stated
> > earlier each bill is a power of the previous bill, there are no 0.25's or
> > 20's or other Americanisms to deal with.  What we would call 25 cents,
> > would optimally be handled (in base 10) with 2 tens and 5 ones, there
> would
> > not be a 25 coin bill.
> > Near the end he allowed me to just pick a base and lock my answer to
> that,
> > i.e. I no longer had to account for a different base.  So I included that
> > in the instructions upfront.
> >
> > My solution (which I've only recently worked out), looks like this.
> >
> > function makeChange(int base, int amountDue, int amountGiven){
> >
> >           changeDue = amountGiven - amountDue;
> >          return breakBills(changeDue, base);
> > }
> >
> > function breakBills(int amount,int base ){
> >             static count x = 1;
> >             dividend = floor(amount / base);
> >             remainder = amount mod base;
> >              if(remainder == 0){
> >                  return;
> >              }else{
> >                   return dividend+"-" + (count * base) + "'s, " +
> > breakBills(remainder,base);
> >              }
> >
> > }
> >
> > Theoretically this might be javascript, frankly it's out of thing air, I
> > haven't tested it yet, but it looks like it should work right?
> >
> > }
> >
> >
> >
> >
> >
> > On Tue, Oct 1, 2013 at 11:18 AM, Levi Pearson <levipearson at gmail.com
> >wrote:
> >
> >> On Tue, Oct 1, 2013 at 11:10 AM, Richard Holden <rholdeniii at gmail.com>
> >> wrote:
> >> > Unless I'm reading it wrong it doesn't actually seem to require all of
> >> the
> >> > division. They say that the smallest fractional unit can be taken to
> be
> >> a
> >> > whole unit, so in the values stated the 123.45 and 1300.00 get changed
> >> to
> >> > whole units as 12345 integer and 130000 integer. That gives us the
> >> > calculation for the integer change as
> >> >
> >> >  130000
> >> > - 12345
> >> > -------
> >> >  117655
> >> >
> >> > To figure out the change we just convert the integer into a string
> >> > (assuming that the language of choice correctly converts) which should
> >> give
> >> > us "117655". If we reverse the string ("556711") and iterate over the
> >> > characters then the index becomes the power to which we raise the
> base.
> >> > i.e. pulling out the second 5 (index value 1) we could output
> 5-(10^1)s
> >> or
> >> > 5-10s, etc. Giving "5-1s 5-10s 6-100s 7-1000s 1-10000s 1-100000s"
> >> >
> >> > To do this in base 2 it just becomes
> >> > 5-(2^0)s 5-(2^1)s 6-(2^2)s 7-(2^3)s 1-(2^4)s 1-(2^5)s
> >> >
> >>
> >> This is true, and is usable if you get the input in the form of a
> >> string in the correct base.  If you get it as a number, either you or
> >> a library function would have to do the divisions in order to convert
> >> the number to a string in the correct base in the first place.  If
> >> that's the case, there's no point in using a string; you might as well
> >> use an array of digit values with the index as the power to which
> >> you'll raise the base before multiplying.  The problem definition is
> >> underspecified with respect to input/output, so it's not clear whether
> >> this approach is valid or not, but it's certainly sound if it is
> >> applicable.
> >>
> >>         --Levi
> >>
> >> /*
> >> PLUG: http://plug.org, #utah on irc.freenode.net
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> >>
> >
> >
>
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