@brijesh-
itoa basically converts integer to string ..we are using the fact tht
a number is multiple of 3 if its sum is multiple of 3
. we have int as string and we can traverse it ..for each character
apply
 int sum+=*c-'0' (ankur missed the star :P)
then on sum we use repeated subtraction...i hope its clear .

we are using all this just coz we have to use itoa ..otherwise there
are more methods (check my frst post )


On Jul 30, 4:34 am, brijesh <brijeshupadhyay...@gmail.com> wrote:
> @ankur I didnt get this... could u or anyone please elaborate!
>
> On Jul 30, 12:43 am, Ankur Khurana <ankur.kkhur...@gmail.com> wrote:
>
>
>
>
>
>
>
> > when you use itoa , what you get is a string. get the sum of all the digits
> > , using c-'0' and then use repeated subtraction . . .
>
> > On Sat, Jul 30, 2011 at 1:01 AM, sukhmeet singh 
> > <sukhmeet2...@gmail.com>wrote:
>
> > > repeated subtraction !!
>
> > > On Sat, Jul 30, 2011 at 12:52 AM, nivedita arora <
> > > vivaciousnived...@gmail.com> wrote:
>
> > >> Without using /,% and * operators. write a function to divide a number
> > >> by 3. itoa() function is available.
>
> > >> all i cn thnk of is to use shift operator and addition ,  x/3=e^(logx-
> > >> log3) or repetitive subtraction
>
> > >> but none of them uses itoa() ..ne idea how its done?
> > >> thnks !
>
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> > --
> > Ankur Khurana
> > Computer Science
> > Netaji Subhas Institute Of Technology
> > Delhi.

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