Given two numbers A and B, the task is to Divide the two numbers A and B by their common divisors. The numbers A and B is less than 10^8.
Examples:
Input: A = 10, B = 15 Output: A = 2, B = 3 The common factors are 1, 5 Input: A = 100, B = 150 Output: A = 2, B = 3
Naive Approach: Iterate from i = 1 to minimum of A and B and check whether i is a factor of both A and B. If i is a factor of A and B then Divide the two numbers A and B by i.
Below is the implementation of the above approach:
// C++ implementation of above approach #include <bits/stdc++.h> using namespace std;
// print the numbers after dividing // them by their common factors void divide( int a, int b)
{ // iterate from 1 to minimum of a and b
for ( int i = 2; i <= min(a, b); i++) {
// if i is the common factor
// of both the numbers
while (a % i == 0 && b % i == 0) {
a = a / i;
b = b / i;
}
}
cout << "A = " << a << ", B = " << b << endl;
} // Driver code int main()
{ int A = 10, B = 15;
// divide A and B by their common factors
divide(A, B);
return 0;
} |
// C implementation of above approach #include <stdio.h> // print the numbers after dividing // them by their common factors void divide( int a, int b)
{ // iterate from 1 to minimum of a and b
int min;
min = a;
if (min > b)
min = b;
for ( int i = 2; i <= min; i++) {
// if i is the common factor
// of both the numbers
while (a % i == 0 && b % i == 0) {
a = a / i;
b = b / i;
}
}
printf ( "A = %d, B = %d\n" , a, b);
} // Driver code int main()
{ int A = 10, B = 15;
// divide A and B by their common factors
divide(A, B);
return 0;
} // This code is contributed by kothvvsakash |
// Java implementation of above approach import java.util.*;
class solution
{ // print the numbers after dividing // them by their common factors static void divide( int a, int b)
{ // iterate from 1 to minimum of a and b
for ( int i = 2 ; i <= Math.min(a, b); i++) {
// if i is the common factor
// of both the numbers
while (a % i == 0 && b % i == 0 ) {
a = a / i;
b = b / i;
}
}
System.out.println( "A = " +a+ ", B = " +b);
} // Driver code public static void main(String args[])
{ int A = 10 , B = 15 ;
// divide A and B by their common factors
divide(A, B);
} } // This code is contributed by // Surendra_Gangwar |
# Python3 implementation of above approach # print the numbers after dividing # them by their common factors def divide(a, b) :
# iterate from 1 to minimum of a and b
for i in range ( 2 , min (a, b) + 1 ) :
# if i is the common factor
# of both the numbers
while (a % i = = 0 and b % i = = 0 ) :
a = a / / i
b = b / / i
print ( "A =" , a, ", B =" , b)
# Driver code if __name__ = = "__main__" :
A, B = 10 , 15
# divide A and B by their
# common factors
divide(A, B)
# This code is contributed by Ryuga |
// C# implementation of above approach using System;
class GFG
{ // print the numbers after dividing // them by their common factors static void divide( int a, int b)
{ // iterate from 1 to minimum of a and b
for ( int i = 2; i <= Math.Min(a, b); i++)
{
// if i is the common factor
// of both the numbers
while (a % i == 0 && b % i == 0)
{
a = a / i;
b = b / i;
}
}
Console.WriteLine( "A = " +a+ ", B = " +b);
} // Driver code static public void Main ()
{ int A = 10, B = 15;
// divide A and B by their common factors
divide(A, B);
} } // This code is contributed by ajit. |
<?php // PHP implementation of above approach // print the numbers after dividing // them by their common factors function divide( $a , $b )
{ // iterate from 1 to minimum of a and b
for ( $i = 2; $i <= min( $a , $b ); $i ++)
{
// if i is the common factor
// of both the numbers
while ( $a % $i == 0 &&
$b % $i == 0)
{
$a = $a / $i ;
$b = $b / $i ;
}
}
echo "A = " , $a , ", B = " , $b , "\n" ;
} // Driver code $A = 10;
$B = 15;
// divide A and B by their common factors divide( $A , $B );
// This code is contributed by Sach_Code ?> |
<script> // Javascript implementation of above approach // print the numbers after dividing // them by their common factors function divide(a, b)
{ // iterate from 1 to minimum of a and b
for (let i = 2; i <= Math.min(a, b); i++) {
// if i is the common factor
// of both the numbers
while (a % i == 0 && b % i == 0) {
a = a / i;
b = b / i;
}
}
document.write( "A = " + a + ", B = " + b + "<br>" );
} // Driver code let A = 10, B = 15;
// divide A and B by their common factors
divide(A, B);
// This code is contributed by Mayank Tyagi </script> |
A = 2, B = 3
Time Complexity: O(min(a,b))
Auxiliary Space: O(1)
An efficient approach is to use the same concept used in Common divisors of two numbers. Calculate the greatest common divisor (gcd) of given two numbers, and then divide the numbers by their gcd.
// C++ implementation of above approach #include <bits/stdc++.h> using namespace std;
// Function to calculate gcd of two numbers int gcd( int a, int b)
{ if (a == 0)
return b;
return gcd(b % a, a);
} // Function to calculate all common divisors // of two given numbers // a, b --> input integer numbers void commDiv( int a, int b)
{ // find gcd of a, b
int n = gcd(a, b);
a = a / n;
b = b / n;
cout << "A = " << a << ", B = " << b << endl;
} // Driver code int main()
{ int a = 10, b = 15;
commDiv(a, b);
return 0;
} |
// C implementation of above approach #include <stdio.h> // Function to calculate gcd of two numbers int gcd( int a, int b)
{ if (a == 0)
return b;
return gcd(b % a, a);
} // Function to calculate all common divisors // of two given numbers // a, b --> input integer numbers void commDiv( int a, int b)
{ // find gcd of a, b
int n = gcd(a, b);
a = a / n;
b = b / n;
printf ( "A = %d, B = %d\n" ,a,b);
} // Driver code int main()
{ int a = 10, b = 15;
commDiv(a, b);
return 0;
} // This code is contributed by kothvvsaakash |
// Java implementation of above approach class GFG
{ // Function to calculate gcd // of two numbers static int gcd( int a, int b)
{ if (a == 0 )
return b;
return gcd(b % a, a);
} // Function to calculate all common // divisors of two given numbers // a, b --> input integer numbers static void commDiv( int a, int b)
{ // find gcd of a, b
int n = gcd(a, b);
a = a / n;
b = b / n;
System.out.println( "A = " + a +
", B = " + b);
} // Driver code public static void main(String[] args)
{ int a = 10 , b = 15 ;
commDiv(a, b);
} } // This code is contributed // by Code_Mech |
# Python3 implementation of above approach # Function to calculate gcd of two numbers def gcd(a, b):
if (a = = 0 ):
return b
return gcd(b % a, a)
# Function to calculate all common # divisors of two given numbers # a, b --> input eger numbers def commDiv(a, b):
# find gcd of a, b
n = gcd(a, b)
a = a / / n
b = b / / n
print ( "A =" , a, ", B =" , b)
# Driver code a, b = 10 , 15
commDiv(a, b) # This code is contributed # by mohit kumar |
// C# implementation of above approach using System;
class GFG
{ // Function to calculate gcd // of two numbers static int gcd( int a, int b)
{ if (a == 0)
return b;
return gcd(b % a, a);
} // Function to calculate all common // divisors of two given numbers // a, b --> input integer numbers static void commDiv( int a, int b)
{ // find gcd of a, b
int n = gcd(a, b);
a = a / n;
b = b / n;
Console.WriteLine( "A = " + a +
", B = " + b);
} // Driver code public static void Main()
{ int a = 10, b = 15;
commDiv(a, b);
} } // This code is contributed // by Code_Mech |
<?php // PHP implementation of above approach // Function to calculate gcd of // two numbers function gcd( $a , $b )
{ if ( $a == 0)
return $b ;
return gcd( $b % $a , $a );
} // Function to calculate all common // divisors of two given numbers // a, b --> input integer numbers function commDiv( $a , $b )
{ // find gcd of a, b
$n = gcd( $a , $b );
$a = (int)( $a / $n );
$b = (int)( $b / $n );
echo "A = " . $a .
", B = " . $b . "\n" ;
} // Driver code $a = 10;
$b = 15;
commDiv( $a , $b );
// This code is contributed by mits ?> |
<script> // Javascript implementation of above approach
// Function to calculate gcd
// of two numbers
function gcd(a, b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Function to calculate all common
// divisors of two given numbers
// a, b --> input integer numbers
function commDiv(a, b)
{
// find gcd of a, b
let n = gcd(a, b);
a = parseInt(a / n, 10);
b = parseInt(b / n, 10);
document.write( "A = " + a + ", B = " + b);
}
let a = 10, b = 15;
commDiv(a, b);
</script> |
A = 2, B = 3
Time complexity : O(log(min(a, b)))
Auxiliary Space : O(log(min(a, b))), for recursive stack space while calculating GCD.