@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 <[email protected]> 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 <[email protected]> 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 <[email protected]> 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 <
> >> [email protected]> 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 <
> >>> [email protected]> 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 <
> >>>> [email protected]> 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 <
> >>>>> [email protected]> 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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