Let the denominations be D[] = {1000,500,100},
and amount be N.
Let C[] , denotes the count of each denomination.
for ( i=0 ; i < 2 ; i++) {
C[i] = (N-1)/D[i] ;
N = N - D[i]*C[i] ;
}
C[2] = N/D[2] ;
For N=4800, C[] = {4, 1, 8}
For N= 2000, C[] = {1, 1, 5}, as required.
Nice observation :) .
PS: Its the Newton who appreciated the falling apple. There aren't many who
really appreciate the happenings from our normal life. [?]
On Sat, Sep 19, 2009 at 11:50 PM, eSKay <[email protected]> wrote:
>
> for example: if I draw 2000, what I get is
> 1000+500+100+100+100+100+100.
>
> What algorithm can be used to decide how to break up the entered
> amount?
>
> >
>
--
Shishir Mittal
Ph: +91 9936 180 121
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