Answer / raghuram

#include<iostream.h>
#include<conio.h>
/*no. of 1's in no. of 1's steps*/
int count(unsigned long int n)
{
int count=0;
while(n)
{
count++;
n=n&n-1;
}
return count ;
}

Answer / vijay

#include<stdio.h>
main()
{
int x=123,i=0;
while(x)
{
x=x&(x-1);
i++;
}
printf("Number of set bits are %d \n",i);
}

Answer / xyz

main()
{
int i=1177;
int j=0;
while(i>0)
{
if((i%2)!=0)
j++;
i=i/2;
}
printf("The number of one is %d\n", j);
}

Answer / lawrence

@Turk
your solution is O(n) not O(logn). your algorithm still
counts each zeros and ones.

Answer / sunny arora

/* This is a table which will contain number of ones
corrosponding to number value */
static int bits_in_char [256] ;

int bitcount (unsigned int n)
{
// works only for 32-bit ints

return bits_in_char [n & 0xffu]
+ bits_in_char [(n >> 8) & 0xffu]
+ bits_in_char [(n >> 16) & 0xffu]
+ bits_in_char [(n >> 24) & 0xffu] ;
}

Answer / turk

Here is an O(logn) solution

#include<stdio.h>

int numberOfOnesByte(unsigned char c, int length){
if (c==0)
return 0;
if (c==1)
return 1;
unsigned char left, right;
int lengthP = length/2;
left = c>>lengthP;
right = c-(left<<lengthP);
return numberOfOnesByte(left,length-lengthP)+numberOfOnesByte(right,lengthP);
}
int numberOfOnes(unsigned char *array, int start, int end, int length){
if(length==1){
return numberOfOnesByte(array[start],8);
} else {
int lengthP = length/2;
return numberOfOnes(array,start,start+lengthP-1,lengthP)+
numberOfOnes(array,start+lengthP,end,length-lengthP);
}
}
int main(){
unsigned char array[8] = {0xFF,0XAA,0xFF,0XAA,0xFF,0XAA,0xFF,0XAA};
printf("number of bits %d\n",numberOfOnes(array,0,7,8));
return 0;
}

Answer / pgmrsf

static void NumOfOnes(uint n)
{
int c = 0;
for (int i=0; i<10; i++)
{
if ((n & (int)Math.Pow(2, i)) == (int)Math.Pow(2,
i))
{
c++;
}
}
Console.WriteLine("{0}", c);
}

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