Wednesday, March 19, 2014

Divide a number by 3 without using any of operators (%,/,*)


Problem
 Divide a number by 3 without using any of operators (%,/,*)

Solution

Method 1

int divideby3 (int num)
{
  int sum = 0;
  while (num > 3) {
        sum += num >> 2;
                num = (num >> 2) + (num & 3);
  }
  if (num == 3) ++sum;
  return sum; 
}

Dry running the above method : 
For example if your number is 10 then convert to binary
10=>00001010
if 10 > 3 then shift the binary number 2 bits
Now num will be 00000010 ie, 2
Now sum will be 2
num=(right shift the number by 2 bits)+(number BITWISE AND 3)
num= 2+2

Now the number will be 4 then, 4>3 => true so loop will be repeated
4=>00000100 then shift the binary number 2 bits
Now sum=2+1 ie, sum=3
num=(shift the number(00000100) by 2 bits)+(number BITWISE AND 3)
num=1+0
ie remainder=1

Thanks.

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