Single bit shift...
int divide(int n)
{
n-=1;
n>>=1;
return n;
}
On 7/30/11, tech rascal <techrascal...@gmail.com> wrote:
> hw will u get the ans on repeated subtraction from the sum of the digits??
> I mean ....if I hv 2 divide 27 by 3 thn first I'll find sum of the digits
> i.e, 2+7=9
> then I'll apply repeated subtraction on 9, so hw will i reach to the ans??
>
>
> On Sat, Jul 30, 2011 at 10:05 AM, nivedita arora <
> vivaciousnived...@gmail.com> wrote:
>
>>
>> @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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