@DOn can u explain ur first algo, it would be helpful.
On Wed, May 22, 2013 at 7:28 PM, Don <dondod...@gmail.com> wrote:
> My program works with any numbers.
> Don
>
> On May 22, 3:45 am, Pramida Tumma <pramida.tu...@gmail.com> wrote:
> > This above program works only if the array contains consecutive numbers
> > starting from 1 to n. What to do if the array contains random numbers?
> >
> >
> >
> >
> >
> >
> >
> > On Fri, May 17, 2013 at 6:55 PM, Don <dondod...@gmail.com> wrote:
> > > Counting the set bits in one integer is not the problem which was
> > > asked.
> > > However, I think that something like this is both faster and more easy
> > > to understand than what you have written:
> >
> > > int bitCount(unsigned int x)
> > > {
> > > int result = 0;
> > > while(x)
> > > {
> > > if (x & 1) ++result;
> > > x >>= 1;
> > > }
> > > return result;
> > > }
> >
> > > On May 17, 8:32 am, bhargav <ybbkris...@gmail.com> wrote:
> > > > as bitwise operators are fast can count by following logic, works
> oly fr
> > > > +ve, just a tweak will make it to work with -ves also ..
> >
> > > > #include <stdio.h>
> > > > main() {
> > > > unsigned int x=12312,a;
> > > > a=x<<1;
> > > > //printf("%u",a);
> > > > int count=0;
> > > > while(x>0) {
> > > > a = x<<1;
> > > > //printf("%u \n",a);
> > > > if(a<x)
> > > > count++;
> > > > x=a;}
> >
> > > > printf("%d\n",count );
> > > > getch();
> >
> > > > }
> >
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