http://ideone.com/4wj5t

On Sep 3, 10:52 pm, Discover <maniksinghal.n...@gmail.com> wrote:
> But how the number(in decimal form) will be displayed....if ques
> demands so.
>
> On Sep 2, 1:49 pm, saurabh singh <saurabh.n...@gmail.com> wrote:
>
>
>
> > Suppose the number of shifts be x.
> > Also let the integer be represented by 16 bits on that machine.
> > Now take int n= (int)(x/16 + 0.5), to take the upper cap on result :) .
> > SO having 2^x will be same as doing 2<<(x-1) so essentially if we represent
> > the resultant number in a linked list of nodes, where each node can store an
> > integer, then the bit position at 2+x-1=x+1 will be set as 1 and this
> > (x+1)th nit will fall in (x+1)/16th node in linked list also in that node
> > (x+1)%16 will give the position of bit to be set.
>
> > On Thu, Sep 2, 2010 at 1:32 PM, ashish agarwal <ashish.cooldude...@gmail.com
>
> > > wrote:
> > > I think it will be 1<<x
>
> > > On Wed, Sep 1, 2010 at 10:53 PM, Yan Wang 
> > > <wangyanadam1...@gmail.com>wrote:
>
> > >> Maybe you misunderstand the question.
> > >> The question is how to compute 2^X where 00000 <= X <= 99999?
> > >> How?
>
> > >> On Wed, Sep 1, 2010 at 10:48 PM, Ruturaj <rutura...@gmail.com> wrote:
> > >> > a 5 digit number is of order 10^5 which is << 10^16 of which int in C
> > >> > is of size.
> > >> > Just multiply both numbers.
>
> > >> > On Sep 2, 10:39 am, prasad rao <prasadg...@gmail.com> wrote:
> > >> >> Program that takes a 5 digit number and calculates 2 power that number
> > >> and
> > >> >> prints it and should not use the Big-integer and Exponential
> > >> Function's.
>
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> > Thanks & Regards,
> > Saurabh

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