Finding square roots using long division.

[jahuja73]

If you want to continue beyond this you must add new groups of zeroes to "a", one group for the number of decimals you want

Eg. the square root of 3:
With your method you get 1.
Next step, decide the number of decimals, say 3.
Extend "a" to [03, 00, 00, 00]
and count with the same method

Thanks, seems easy for integers, but still clueless for floats as 1235.6789.
The coding is too difficult as have jobs:
1. In the driver code, need know length of float.
Need break in units of two based on that..

If make modifications in the Driver code as follows:

#Driver code 
x = 1235.789
# Find length of integer
y=x
length =0
"""
while (y > 0): 
    y //=100
    length+= 1
"""
while (y > 0): 
    if (y>=1):
        print 'integer part'
        y = y //100
        print 'y :', y
        length += 1
    elif (y <1) :
        print 'decimal part'
        y = y*100 
        y /= 100
        print 'y :', y
        length += 1
 

Then, get:

Output:

integer part
y : 12.0
integer part
y : 0.0
length : 2

Unable to find a way out.