@Nishant :
Don's algo :
~~~~~~~~
first he is counting #of bits of all numbers from 0 to 65536 and
maintaining in an array bitCount.
Converted the input array into of type short. short range is 0 to 65536.
for the given input, he is getting the number of bits set using bitCount[]
.. its like memorization process...
On Sun, May 26, 2013 at 9:18 PM, Nishant Pandey <
nishant.bits.me...@gmail.com> wrote:
> @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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