➤ Hello World! in C++
➤ Print Number in C++
➤ Add 2 Numbers C++
➤ Arithmetic Operation
➤ Sum Avg of 3 Number
➤ Area Program in C++
➤ Simple Interest in C++
➤ Find ASCII value in C++
➤ Swap 2 Number in C++
Flow Control Programs
➤ Even-Odd in C++
➤ +ve, -ve, 0 in C++
➤ Vowel-Consonant
➤ Greatest of 3 no.
➤ Check Leap Year
➤ Calculator Program
➤ Reverse a Number
➤ Sum of Natural Number
➤ GCD of 2 Number
➤ LCM of 2 Number
➤ Find Power in C++
➤ Fibonacci Series in C++
➤ Palindrome Number
➤ Find Factorial in C++
➤ Factorial Using Recursion
➤ Prime Number in C++
➤ Prime Number b/w 1-N
Array
➤ Linear Search in C++
➤ Binary Search in C++
Others
➤ Introduction to C++
➤ Data Types in C++
➤ Range of Data Types
➤ Void main, main vs int main
In this post, we will develop a program to find GCD or HCF of two numbers in C++. We will also develop program using recursion and function.
The Greatest Common Divisor (GCD) of two or more numbers is the greatest number which divides each of them exactly. Greatest Common Measure(GCM) and highest common factor (HCF) are the other terms used to refer to HCF.
Example: HCF of 60 and 75 = 15 because 15 is the highest number which divides both 60 and 75 exactly.
C++ program to find GCD (Greatest common divisior) through subtracting numbers.
#include<iostream>
using namespace std;
int main()
{
// declare two variables
int num1, num2;
// take inputs
cout << "Enter Two Integers: ";
cin >> num1 >> num2;
// find GCD
while( num1 != num2 )
{
if(num1 > num2)
num1 -= num2 ;
else
num2 -= num1 ;
}
// display result
cout << "GCD = "<< num1 << endl;
return 0;
}
Output:-
Enter Two Integers: 12 15
GCD = 3
The num1 and num2 are two variables which stores value of two integers. Inside while loop, if num1 will be greater than num2 then num1 = num1 – num2 will evaluated otherwise num2 = num2 – num1. This process continuous until num1 and num2 becomes equal.
if(num1 > num2)
num1 -= num2 ;
else
num2 -= num1 ;
These line of code can be replaced by,
( num1 > num2 ) ? (num1 -= num2) : (num2 -= num1);
Another way
#include<iostream>
using namespace std;
int main()
{
// declare variables
int num1, num2, gcd;
// take inputs
cout << "Enter Two Integers: ";
cin >> num1 >> num2;
// swap num1 and num2
// if num2 is greater than num1
if(num2 > num1)
{
// swap using bitwise XOR operator
num1^=num2^=num1^=num2;
}
for(int i=1; i<=num1; i++ )
{
if(num1%i==0 && num2%i==0)
gcd=i;
}
cout << "GCD = " << gcd << endl;
return 0;
}
C++ Program to Find GCD of Two Numbers Using Function
In this program we can use any of the above technique to find the GCD of two numbers in C++.
#include<iostream>
using namespace std;
int findGCD(int, int);
int main()
{
// declare variables
int num1, num2, gcd;
// take inputs
cout << "Enter Two Integers: ";
cin >> num1 >> num2;
// find GCD
gcd = findGCD(num1, num2);
cout << "GCD = " << gcd << endl;
return 0;
}
// function to find GCD
int findGCD(int m, int n)
{
while( m != n )
( m > n ) ? (m -= n) : (n -= m);
return m;
}
GCD of Two Numbers in C++ Using LCM
The product of two numbers a and b is equal to the product of GCD(a,b) and LCM(a,b).
a*b = GCD(a,b) * LCM(a,b)
In the below GCD program in java, we first find the LCM of numbers then the using the given formula GCD of numbers will be calculated. The Formula used for this purpose is:-
GCD(a,b) = (a*b) / LCM(a,b)
#include<iostream>
using namespace std;
long findGCD(int, int);
long findLCM(int, int);
int main()
{
// declare variables
int num1, num2;
long gcd;
// take inputs
cout << "Enter Two Integers: ";
cin >> num1 >> num2;
// find GCD
gcd = findGCD(num1, num2);
cout << "GCD = " << gcd << endl;
return 0;
}
// function to find GCD of 2 numbers
// GCD = ( n1*n2 ) / LCM(n1,n2)
long findGCD(int m, int n)
{
return (m*n) / findLCM(m, n);
}
// function to find LCM of 2 numbers
long findLCM(int n1, int n2)
{
// declare temporary variable to find lcm
long minMultiple = 0;
// find smallest and largest number
int smallest = (n1 < n2) ? n1 : n2;
int largest = (n1 > n2) ? n1 : n2;
// assign smallest number to minMultiple
minMultiple = smallest;
// loop
while(true) {
if(minMultiple % largest == 0)
return minMultiple;
minMultiple = minMultiple + smallest ;
}
}
GCD Using Recursion C++
We can also use the recursion technique to find the GCD of two numbers. A technique of defining the method/function that contains a call to itself is called the recursion. The recursive function/method allows us to divide the complex problem into identical single simple cases that can handle easily. This is also a well-known computer programming technique: divide and conquer.
#include<iostream>
using namespace std;
int findGCD(int, int);
int main()
{
// declare variables
int num1, num2, gcd;
// take inputs
cout << "Enter Two Integers: ";
cin >> num1 >> num2;
// find GCD
gcd = findGCD(num1, num2);
cout << "GCD = " << gcd << endl;
return 0;
}
// recursive function to find GCD of 2 numbers
int findGCD(int num1, int num2)
{
if(num2 == 0) // base case
return num1;
else // general case
return findGCD(num2, num1%num2);
}
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