I think the question is in the written round!!!

On Tue, Jul 24, 2012 at 9:11 PM, algo bard <algo.b...@gmail.com> wrote:

> #include<stdio.h>
> #define RANGE_START 0
> #define RANGE_END 100
>
> int main()
> {
>     int i,n,ctr=0;
>
>     for( i=RANGE_START ; i<=RANGE_END ; i++)
>     {
>         n = i;
>         while(n)  // Using Brian Kernighan's Algorithm to count the number
> of set bits in a number.
>         {
>             n= n&n-1;
>             ctr++;
>         }
>
>     }
>
>     printf("%d ",ctr);
> }
>
> TC = No. of set bits in the given range of numbers.
>
>
> On Tue, Jul 24, 2012 at 7:56 PM, Lomash Goyal <lomesh.go...@gmail.com>wrote:
>
>> // count number of 1's upto n.cpp : Defines the entry point for the
>> console application.
>> //
>>
>> #include "stdafx.h"
>> #include<math.h>
>> #include<conio.h>
>>  //the following functions will count number of bits in the number
>> int countbits(int n)
>> {
>>    int count=0;
>>    while(n)
>>   {
>>     n/=2;
>>     count++;
>>   }
>> return count;
>> }
>>
>>
>> int countnumberof1(int number)
>> {
>>     if(number==0)
>>  return 0;
>> if(number==1)
>> return 1;
>>  if(number==2)
>> return 2;
>> if(number==3)
>>  return 4;
>>  if(number>3)
>>  {
>> int bits=countbits(number);
>> if(number==pow(2.0,bits)-1)
>>  {
>> return pow(2.0,bits-1)+2*countnumberof1(pow(2.0,bits-1)-1);
>> }
>>  else return
>> pow(2.0,bits-2)+2*countnumberof1(pow(2.0,bits-2)-1)+countnumberof1(number-(pow(2.0,bits-1)))+number-(pow(2.0,bits-1))+1;
>> }
>> }
>> int _tmain(int argc, _TCHAR* argv[])
>> {
>> printf("%d",countnumberof1(10));
>> getch();
>>  return 0;
>> }
>>
>>
>> On Tue, Jul 24, 2012 at 3:09 PM, ruru <soupti...@gmail.com> wrote:
>>
>>> find no. of 1's in binary format of numbers from 1 to 100. like for
>>> 1 to 10 answer is 17
>>>
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>>
>>
>> --
>> Regards
>>
>> Lomash Goyal
>>
>> *
>> *
>>
>>
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>
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-- 
With regards,
Manish kumar untwal
Indian Institute of Information Technology
Allahabad (2009-2013 batch)

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