Thanks But it is used bigint but find a value without using bigint. On 4 September 2010 11:54, Shravan <shravann1...@gmail.com> wrote:
> 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. > > > > > >> > -- > > > >> > You received this message because you are subscribed to the Google > > > >> Groups "Algorithm Geeks" group. > > > >> > To post to this group, send email to algoge...@googlegroups.com. > > > >> > To unsubscribe from this group, send email to > > > >> algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > <algogeeks%2bunsubscr...@googlegroups .com> > > > >> . > > > >> > For more options, visit this group at > > > >>http://groups.google.com/group/algogeeks?hl=en. > > > > > >> -- > > > >> You received this message because you are subscribed to the Google > Groups > > > >> "Algorithm Geeks" group. > > > >> To post to this group, send email to algoge...@googlegroups.com. > > > >> To unsubscribe from this group, send email to > > > >> algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > <algogeeks%2bunsubscr...@googlegroups .com> > > > >> . > > > >> For more options, visit this group at > > > >>http://groups.google.com/group/algogeeks?hl=en. > > > > > > -- > > > > You received this message because you are subscribed to the Google > Groups > > > > "Algorithm Geeks" group. > > > > To post to this group, send email to algoge...@googlegroups.com. > > > > To unsubscribe from this group, send email to > > > > algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > <algogeeks%2bunsubscr...@googlegroups .com> > > > > . > > > > For more options, visit this group at > > > >http://groups.google.com/group/algogeeks?hl=en. > > > > > -- > > > Thanks & Regards, > > > Saurabh > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algoge...@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
