How To Find Gcd Of A Number
GCD (Greatest Mutual Divisor) or HCF (Highest Common Factor) of two numbers is the largest number that divides both of them.
For example GCD of 20 and 28 is 4 and GCD of 98 and 56 is 14. For solution suppose a=98 & b=56
a>b so put a= a-b and b is remain same and so a=98-56=42 & b= 56 . Now b>a so b=b-a and a is same
b= 56-42 = xiv & a= 42 . 42 is three times of 14 so HCF is 14 . besides a=36 & b=60 ,here b>a so b = 24 & a= 36 at present a>b so a= 12 & b= 24 . 12 is HCF of 36 and threescore . This concept is always satisfying.
A simple solution is to observe all prime factors of both numbers, so observe intersection of all factors present in both numbers. Finally return product of elements in the intersection.
An efficient solution is to use Euclidean algorithm which is the main algorithm used for this purpose. The thought is, GCD of two numbers doesn't change if smaller number is subtracted from a bigger number.
C++
#include <iostream>
using namespace std;
int gcd( int a, int b)
{
if (a == 0)
return b;
if (b == 0)
render a;
if (a == b)
return a;
if (a > b)
return gcd(a-b, b);
return gcd(a, b-a);
}
int main()
{
int a = 98, b = 56;
cout<< "GCD of " <<a<< " and " <<b<< " is " <<gcd(a, b);
return 0;
}
C
#include <stdio.h>
int gcd( int a, int b)
{
if (a == 0)
return b;
if (b == 0)
render a;
if (a == b)
return a;
if (a > b)
return gcd(a-b, b);
render gcd(a, b-a);
}
int main()
{
int a = 98, b = 56;
printf ( "GCD of %d and %d is %d " , a, b, gcd(a, b));
return 0;
}
Java
class Test
{
static int gcd( int a, int b)
{
if (a == 0 )
return b;
if (b == 0 )
return a;
if (a == b)
return a;
if (a > b)
return gcd(a-b, b);
return gcd(a, b-a);
}
public static void principal(String[] args)
{
int a = 98 , b = 56 ;
System.out.println( "GCD of " + a + " and " + b + " is " + gcd(a, b));
}
}
Python3
def gcd(a,b):
if (a = = 0 ):
return b
if (b = = 0 ):
return a
if (a = = b):
return a
if (a > b):
return gcd(a - b, b)
render gcd(a, b - a)
a = 98
b = 56
if (gcd(a, b)):
print ( 'GCD of' , a, 'and' , b, 'is' , gcd(a, b))
else :
print ( 'not institute' )
C#
using System;
class GFG {
static int gcd( int a, int b)
{
if (a == 0)
return b;
if (b == 0)
render a;
if (a == b)
return a;
if (a > b)
return gcd(a - b, b);
return gcd(a, b - a);
}
public static void Main()
{
int a = 98, b = 56;
Console.WriteLine( "GCD of "
+ a + " and " + b + " is "
+ gcd(a, b));
}
}
PHP
<?php
function gcd( $a , $b )
{
if ( $a == 0)
return $b ;
if ( $b == 0)
return $a ;
if ( $a == $b )
return $a ;
if ( $a > $b )
return gcd( $a - $b , $b ) ;
return gcd( $a , $b - $a ) ;
}
$a = 98 ;
$b = 56 ;
repeat "GCD of $a and $b is " , gcd( $a , $b ) ;
?>
Javascript
<script>
function gcd(a, b)
{
if (a == 0)
render b;
if (b == 0)
render a;
if (a == b)
return a;
if (a > b)
render gcd(a-b, b);
render gcd(a, b-a);
}
permit a = 98, b = 56;
document.write( "GCD of " + a + " and " + b + " is " + gcd(a, b));
</script>
Output:
GCD of 98 and 56 is fourteen
Time Complication: O(max(a,b))
Auxiliary Space: O(max(a,b))
Dynamic Programming Approach (Meridian Downwardly Using Memoization) :
C++
#include <bits/stdc++.h>
using namespace std;
int static dp[1001][1001];
int gcd( int a, int b)
{
if (a == 0)
return b;
if (b == 0)
return a;
if (a == b)
return a;
if (dp[a][b] != -1)
return dp[a][b];
if (a > b)
dp[a][b] = gcd(a-b, b);
else
dp[a][b] = gcd(a, b-a);
render dp[a][b];
}
int main()
{
int a = 98, b = 56;
memset (dp, -1, sizeof (dp));
cout<< "GCD of " <<a<< " and " <<b<< " is " <<gcd(a, b);
render 0;
}
Java
import coffee.util.*;
public class GFG
{
static int [][]dp = new int [ 1001 ][ 1001 ];
static int gcd( int a, int b)
{
if (a == 0 )
return b;
if (b == 0 )
return a;
if (a == b)
return a;
if (dp[a][b] != - one )
return dp[a][b];
if (a > b)
dp[a][b] = gcd(a-b, b);
else
dp[a][b] = gcd(a, b-a);
return dp[a][b];
}
public static void master(Cord[] args)
{
for ( int i = 0 ; i < 1001 ; i++) {
for ( int j = 0 ; j < 1001 ; j++) {
dp[i][j] = - 1 ;
}
}
int a = 98 , b = 56 ;
System.out.println( "GCD of " + a + " and " + b + " is " + gcd(a, b));
}
}
Python3
dp = [[ - 1 for i in range ( 1001 )] for j in range ( 1001 )]
def gcd(a,b):
if (a = = 0 ):
return b
if (b = = 0 ):
render a
if (a = = b):
return a
if (dp[a][b] ! = - 1 ):
return dp[a][b]
if (a > b):
dp[a][b] = gcd(a - b, b)
else :
dp[a][b] = gcd(a, b - a)
return dp[a][b]
a = 98
b = 56
if (gcd(a, b)):
print ( 'GCD of' , a, 'and' , b, 'is' , gcd(a, b))
else :
print ( 'not found' )
C#
using System;
grade GFG
{
static int [,]dp = new int [1001, 1001];
static int gcd( int a, int b)
{
if (a == 0)
return b;
if (b == 0)
return a;
if (a == b)
return a;
if (dp[a, b] != -1)
return dp[a, b];
if (a > b)
dp[a, b] = gcd(a-b, b);
else
dp[a, b] = gcd(a, b-a);
return dp[a, b];
}
public static void Chief()
{
for ( int i = 0; i < 1001; i++) {
for ( int j = 0; j < 1001; j++) {
dp[i, j] = -1;
}
}
int a = 98, b = 56;
Panel.Write( "GCD of " + a + " and " + b + " is " + gcd(a, b));
}
}
Javascript
var dp = new Array(1001);
for ( var i = 0; i < dp.length; i++) {
dp[i] = new Array(1001);
}
office gcd(a, b)
{
if (a == 0)
return b;
if (b == 0)
return a;
if (a == b)
return a;
if (dp[a][b] != -i)
return dp[a][b];
if (a > b)
dp[a][b] = gcd(a-b, b);
else
dp[a][b] = gcd(a, b-a);
return dp[a][b];
}
permit a = 98, b = 56;
for (allow i = 0; i < 1001; i++) {
for (let j = 0; j < 1001; j++) {
dp[i][j] = -one;
}
}
certificate.write( "GCD of " + a + " and " + b + " is " + gcd(a, b));
</script>
Output
GCD of 98 and 56 is 14
Fourth dimension Complexity: O(max(a,b))
Auxiliary Space: O(i)
A more than efficient solution is to utilize modulo operator in Euclidean algorithm.
C++
#include <iostream>
using namespace std;
int gcd( int a, int b)
{
return b == 0 ? a : gcd(b, a % b);
}
int master()
{
int a = 98, b = 56;
cout<< "GCD of " <<a<< " and " <<b<< " is " <<gcd(a, b);
return 0;
}
C
#include <stdio.h>
int gcd( int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
int chief()
{
int a = 98, b = 56;
printf ( "GCD of %d and %d is %d " , a, b, gcd(a, b));
return 0;
}
Java
class Test
{
static int gcd( int a, int b)
{
if (b == 0 )
return a;
return gcd(b, a % b);
}
public static void main(String[] args)
{
int a = 98 , b = 56 ;
Organization.out.println( "GCD of " + a + " and " + b + " is " + gcd(a, b));
}
}
Python3
def gcd(a,b):
if (b = = 0 ):
return a
return gcd(b, a % b)
a = 98
b = 56
if (gcd(a, b)):
print ( 'GCD of' , a, 'and' , b, 'is' , gcd(a, b))
else :
impress ( 'not establish' )
C#
using System;
class GFG {
static int gcd( int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
public static void Master()
{
int a = 98, b = 56;
Console.WriteLine( "GCD of "
+ a + " and " + b + " is "
+ gcd(a, b));
}
}
PHP
<?php
function gcd( $a , $b )
{
if ( $b ==0)
return $a ;
return gcd( $b , $a % $b ) ;
}
$a = 98 ;
$b = 56 ;
echo "GCD of $a and $b is " , gcd( $a , $b ) ;
?>
Javascript
<script>
office gcd(a, b){
if (b == 0){
return a;
}
return gcd(b, a % b);
}
let a = 98;
let b = 56;
document.write(`GCD of ${a} and ${b} is ${gcd(a, b)}`);
</script>
Output:
GCD of 98 and 56 is 14
Time Complication: O(log(max(a,b))
Auxiliary Space: O(log(max(a,b))
The time complexity for the to a higher place algorithm is O(log(max(a,b))) the derivation for this is obtained from the analysis of the worst-case scenario. What we practise is we inquire what are the 2 to the lowest degree numbers that accept 1 footstep, those would be (i,1). If we want to increase the number of steps to 2 while keeping the numbers as low as possible as nosotros tin take the numbers to be (i,2). Similarly, for 3 steps, the numbers would exist (2,three), four would be (3,5), 5 would exist (5,8). So we can notice a pattern here, for the nth step the numbers would exist (fib(n),fib(n+ane)). So the worst-case fourth dimension complexity would exist O(n) where a>= fib(n) and b>= fib(n+1).
Now Fibonacci series is an exponentially growing series where the ratio of nth/(n-1)th term approaches (sqrt(v)-1)/2 which is also chosen the golden ratio. So we can see that the time complexity of the algorithm increases linearly as the terms grow exponentially hence the time complexity would be log(max(a,b)).
Please refer GCD of more than than two (or assortment) numbers to find HCF of more than than two numbers.
Please write comments if you discover anything incorrect, or you want to share more information about the topic discussed above
Source: https://www.geeksforgeeks.org/c-program-find-gcd-hcf-two-numbers/
Posted by: sowellholed1992.blogspot.com
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