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Program to find GCD of floating point numbers

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The greatest common divisor (GCD) of two or more numbers, which are not all zero, is the largest positive number that divides each of the numbers. 
Example: 
 

Input  : 0.3, 0.9
Output : 0.3

Input  : 0.48, 0.108
Output : 0.012


 


The simplest approach to solve this problem is :
a=1.20 
b=22.5 
Expressing each of the numbers without decimals as the product of primes we get:
120=2^3*3*5
2250=2*3^2*5^3
Now, H.C.F. of 120 and 2250 = 2*3*5=30 
Therefore,the H.C.F. of 1.20 and 22.5=0.30 
(taking 2 decimal places)
We can do this using the Euclidean algorithm. This algorithm indicates that if the smaller number is subtracted from a bigger number, GCD of two numbers doesn’t change. 
 

C++

// CPP code for finding the GCD of two floating
// numbers.
#include <bits/stdc++.h>
using namespace std;
 
// Recursive function to return gcd of a and b
double gcd(double a, double b)
{
    if (a < b)
        return gcd(b, a);
 
    // base case
    if (fabs(b) < 0.001)
        return a;
 
    else
        return (gcd(b, a - floor(a / b) * b));
}
 
// Driver Function.
int main()
{
    double a = 1.20, b = 22.5;
    cout << gcd(a, b);
    return 0;
}

                    

Java

// JAVA code for finding the GCD of
// two floating numbers.
import java.io.*;
 
class GFG {
     
    // Recursive function to return gcd
    // of a and b
    static double gcd(double a, double b)
    {
        if (a < b)
            return gcd(b, a);
      
        // base case
        if (Math.abs(b) < 0.001)
            return a;
      
        else
            return (gcd(b, a -
                   Math.floor(a / b) * b));
    }
      
    // Driver Function.
    public static void main(String args[])
    {
        double a = 1.20, b = 22.5;
        System.out.printf("%.1f" ,gcd(a, b));
    }
}
 
/*This code is contributed by Nikita Tiwari.*/

                    

Python

# Python code for finding the GCD of
# two floating numbers.
 
import math
 
# Recursive function to return gcd
# of a and b
def gcd(a,b) :
    if (a < b) :
        return gcd(b, a)
     
    # base case
    if (abs(b) < 0.001) :
        return a
    else :
        return (gcd(b, a - math.floor(a / b) * b))
     
      
# Driver Function.
a = 1.20
b = 22.5
print('{0:.1f}'.format(gcd(a, b)))
 
# This code is contributed by Nikita Tiwari.

                    

C#

// C# code for finding the GCD of
// two floating numbers.
using System;
 
class GFG {
     
    // Recursive function to return gcd
    // of a and b
    static float  gcd(double a, double b)
    {
        if (a < b)
            return gcd(b, a);
     
        // base case
        if (Math.Abs(b) < 0.001)
            return (float)a;
     
        else
            return (float)(gcd(b, a -
                Math.Floor(a / b) * b));
    }
     
    // Driver Function.
    public static void Main()
    {
        double a = 1.20, b = 22.5;
 
        Console.WriteLine(gcd(a, b));
    }
}
 
// This code is contributed by vt_m.

                    

PHP

<?php
// PHP code for finding the GCD
// of two floating numbers.
 
// Recursive function to
// return gcd of a and b
function gcd($a, $b)
{
    if ($a < $b)
        return gcd($b, $a);
 
    // base case
    if (abs($b) < 0.001)
        return $a;
 
    else
        return (gcd($b, $a -
              floor($a / $b) * $b));
}
 
// Driver Code
$a = 1.20;
$b = 22.5;
echo gcd($a, $b);
 
// This code is contributed
// by aj_36
?>

                    

Javascript

<script>
// javascript code for finding the GCD of
// two floating numbers.
 
    // Recursive function to return gcd
    // of a and b
    function gcd(a , b)
    {
        if (a < b)
            return gcd(b, a);
 
        // base case
        if (Math.abs(b) < 0.001)
            return a;
        else
            return (gcd(b, a - Math.floor(a / b) * b));
    }
 
    // Driver Function.   
    var a = 1.20, b = 22.5;
    document.write( gcd(a, b).toFixed(1));
 
// This code is contributed by aashish1995
</script>

                    

Output: 
 

0.3

Time Complexity: O(log n) 
Auxiliary Space: O(log n)




 



Last Updated : 23 Jun, 2022
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