Supposedly Trivial math question.

S. Dale Morrey sdalemorrey at gmail.com
Tue Oct 1 14:48:43 MDT 2013


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
>>
>> /*
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>>
>
>


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