Derek Teixeira wrote:
> okay, cool. i found the answer to the question on a forum online ...

>   m_mod = input%1000

Puts the remainder of division by 1000 (i.e. the three last digits of the
number) in m_mod.

>   d_mod = input%500

Remainder of division by 500 (i.e. the three last digits or if that number
would be higher than 500, the three last digits minus 500)

>   c_mod = input%100

2 last digits.

>   l_mod = input%50

2 last digits minus 50 if necessary.

>   x_mod = input%10

Last digit.

>   v_mod = input%5

Last digit minus 5 if necessary.

>   m_div = input/1000

Divides the number by 1000, cutting of everything after the point (because
this is integer division). So basically this returns everything but the
three last digits.

>   d_div = m_mod/500

This is one if the third to last digit (i.e. the one who specifies the
hundreds) is 5 or higher and zero otherwise.

>   c_div = d_mod/100

The third to last digit minus 5 if it's 5 or higher.

>   l_div = c_mod/50
>   x_div = l_mod/10

>   v_div = x_mod/5
>   i_div = v_mod/1

The same thing for the second to last and last digit accordingly.


> do you think you explain to me why the % and / works here?

I hope this is what I just did and you understand it now.

HTH,
Sebastian Hungerecker
-- 
Ist so, weil ist so
Bleibt so, weil war so