Hi Dave,
thanks that made perfect sense.
-sarvesh
On Sun, Aug 21, 2011 at 11:28 PM, Dave <dave_and_da...@juno.com> wrote:
> @Sarvesh: This can be coded like this:
>
> int i, ones, bits[8] = {0, 1, 1, 2, 1, 2, 2, 3}; // no of 1-bits in 0
> to 7
> for( ones = 0, i = 1 ; i < N ; ++i )
> ones += bits[A[i]];
>
> Dave
>
> On Aug 21, 5:55 am, sarvesh saran <aquarian.thun...@gmail.com> wrote:
> > Hi Prakash,
> >
> > I'll paste the exact description of the problem:
> >
> > A non-empty array A of N elements contains octal representation of a
> > non-negative integer K, i.e. each element of A belongs to the interval
> [0;
> > 7] (both ends included).
> >
> > Write a function that returns the number of bits set to 1 in the binary
> > representation of K.
> >
> > thanks,
> >
> > Sarvesh
> >
> > i.e take any decimal number, convert to base 8 and then store each digit
> of
> > base 8 representation in an array.
> >
> > So the question is, given such an array get back the original number.
> >
> > thanks,
> > Sarvesh
> >
> >
> >
> > On Sun, Aug 21, 2011 at 4:13 PM, Prakash D <cegprak...@gmail.com> wrote:
> > > A[i]<<3*i
> >
> > > why is it needed to convert from base 8 to base 10??
> >
> > > On Sun, Aug 21, 2011 at 4:07 PM, Sanjay Rajpal <srn...@gmail.com>
> wrote:
> >
> > >> Hi your intention was logical OR or BITWISE OR ?
> >
> > >> u did Logical.
> > >> Sanju
> > >> :)
> >
> > >> On Sun, Aug 21, 2011 at 3:30 AM, sarvesh saran <
> > >> aquarian.thun...@gmail.com> wrote:
> >
> > >>> Hi Nitin,
> >
> > >>> thanks that makes sense. I will try that out.
> >
> > >>> I have another question. Is there a really fast way of converting a
> > >>> hexadecimal string say "02F9A" to its decimal representation in C++?
> >
> > >>> thanks,
> > >>> Sarvesh
> >
> > >>> thanks,
> > >>> Sarvesh
> >
> > >>> On Sun, Aug 21, 2011 at 3:41 PM, Nitin Nizhawan <
> > >>> nitin.nizha...@gmail.com> wrote:
> >
> > >>>> int num = 0;
> > >>>> for(int i=0;i<A.size();i++){
> > >>>> num=num||(A[i]<<3*i);
> > >>>> }
> > >>>> printf("%d",num);
> >
> > >>>> I think this will do. Given the number is with in the range of
> integer.
> >
> > >>>> On Sun, Aug 21, 2011 at 3:40 PM, Nitin Nizhawan <
> > >>>> nitin.nizha...@gmail.com> wrote:
> >
> > >>>>> int num = 0;
> > >>>>> for(int i=0;i<A.size();i++){
> > >>>>> num=num||(A[i]<3*i);
> > >>>>> }
> > >>>>> printf("%d",num);
> >
> > >>>>> I think this will do.
> >
> > >>>>> On Sun, Aug 21, 2011 at 2:25 PM, sarvesh saran <
> > >>>>> aquarian.thun...@gmail.com> wrote:
> >
> > >>>>>> Hi,
> >
> > >>>>>> I have a vector<int> A or an array (for C guys) that contains the
> > >>>>>> octal representation of a number.
> >
> > >>>>>> So the array can be something like: [1,5,7] or [7,7,5,6,3,4,2] etc
> >
> > >>>>>> i.e no number in the array can be >= 8.
> >
> > >>>>>> Now given this array, I need to convert it its decimal
> representation.
> >
> > >>>>>> The naive way to do it would be to scan array from left to right,
> take
> > >>>>>> each digit, multiply by 8 pow (x) where x is from 0 to ...n and
> compute sum.
> >
> > >>>>>> i.e something like:
> >
> > >>>>>> int oct = 1;
> > >>>>>> int num = 0;
> >
> > >>>>>> for(<array length>){
> > >>>>>> num+= oct * A[i];
> > >>>>>> oct = oct * 8;
> > >>>>>> }
> >
> > >>>>>> is there a faster way to do this? maybe using some STL container
> or algorithm. ?
> >
> > >>>>>> thanks,
> > >>>>>> sarvesh
> >
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