@Shishir
cool! this seems to work!!
C[i] = (N-1)/D[i] ; //this is super cool
On Sep 21, 10:29 am, Shishir Mittal <1987.shis...@gmail.com> wrote:
> 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 <catchyouraak...@gmail.com> 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
>
> 329.gif
> < 1KViewDownload
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