But how the number(in decimal form) will be displayed....if ques
demands so.

On Sep 2, 1:49 pm, saurabh singh <[email protected]> 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 <[email protected]
>
>
>
>
>
> > wrote:
> > I think it will be 1<<x
>
> > On Wed, Sep 1, 2010 at 10:53 PM, Yan Wang <[email protected]>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 <[email protected]> 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 <[email protected]> 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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