Program to find GCD or HCF of two numbers using Middle School Procedure

Given two positive integers M and N, the task is to find the greatest common divisor (GCD) using Middle School Procedure.
Note: GCD of two integers is the largest positive integer that divides both of the integers.

Examples:

Input: m = 12, n = 14
Output: 2
Prime factor of 12 =  1*2*2*3
Prime factor of 14 = 1*2*7
GCD(12, 14) = 2

Input: m = 5, n = 10
Output: 5
Prime factor of 10 = 1*2*5
Prime factor of 5 = 1*5
GCD(5, 10) = 5


The algorithm to find GCD using Middle School procedure GCD(m, n):

  1. Find the prime factorization of m.
  2. Find the prime factorization of n.
  3. Find all the common prime factors.
  4. Compute the product of all the common prime factors and return it as gcd(m, n).

Below is the implementation of above algorithm:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of above algorithm
#include <bits/stdc++.h>
#define MAXFACTORS 1024
using namespace std;
  
// struct to store factorization of m and n
typedef struct
{
  
    int size;
    int factor[MAXFACTORS + 1];
    int exponent[MAXFACTORS + 1];
  
} FACTORIZATION;
  
// Function to find the factorization of M and N
void FindFactorization(int x, FACTORIZATION* factorization)
{
    int i, j = 1;
    int n = x, c = 0;
    int k = 1;
    factorization->factor[0] = 1;
    factorization->exponent[0] = 1;
  
    for (i = 2; i <= n; i++) {
        c = 0;
  
        while (n % i == 0) {
            c++;
  
            // factorization->factor[j]=i;
            n = n / i;
            // j++;
        }
  
        if (c > 0) {
            factorization->exponent[k] = c;
            factorization->factor[k] = i;
            k++;
        }
    }
  
    factorization->size = k - 1;
}
  
// Function to print the factors
void DisplayFactorization(int x, FACTORIZATION factorization)
{
  
    int i;
    cout << "Prime factor of << x << = ";
  
    for (i = 0; i <= factorization.size; i++) {
  
        cout << factorization.factor[i];
  
        if (factorization.exponent[i] > 1)
            cout << "^" << factorization.exponent[i];
  
        if (i < factorization.size)
            cout << "*";
  
        else
            cout << "\n";
    }
}
  
// function to find the gcd using Middle School procedure
int gcdMiddleSchoolProcedure(int m, int n)
{
  
    FACTORIZATION mFactorization, nFactorization;
  
    int r, mi, ni, i, k, x = 1, j;
  
    // Step 1.
    FindFactorization(m, &mFactorization);
    DisplayFactorization(m, mFactorization);
  
    // Step 2.
    FindFactorization(n, &nFactorization);
    DisplayFactorization(n, nFactorization);
  
    // Steps 3 and 4.
    // Procedure algorithm for computing the
    // greatest common divisor.
    int min;
    i = 1;
    j = 1;
    while (i <= mFactorization.size && j <= nFactorization.size) {
        if (mFactorization.factor[i] < nFactorization.factor[j])
            i++;
  
        else if (nFactorization.factor[j] < mFactorization.factor[i])
            j++;
  
        else /* if arr1[i] == arr2[j] */
        {
            min = mFactorization.exponent[i] > nFactorization.exponent[j]
                      ? nFactorization.exponent[j]
                      : mFactorization.exponent[i];
  
            x = x * mFactorization.factor[i] * min;
            i++;
            j++;
        }
    }
  
    return x;
}
  
// Driver code
int main()
  
{
  
    int m = 10, n = 15;
    cout << "GCD(" << m << ", " << n << ") = "
         << gcdMiddleSchoolProcedure(m, n);
  
    return (0);
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of above algorithm 
class GFG
{
static final int MAXFACTORS = 1024 ;
  
// class to store factorization 
// of m and n 
static class FACTORIZATION
    int size; 
    int factor[] = new int[MAXFACTORS + 1]; 
    int exponent[] = new int[MAXFACTORS + 1]; 
  
  
// Function to find the 
// factorization of M and N 
static void FindFactorization(int x, FACTORIZATION 
                                     factorization) 
    int i, j = 1
    int n = x, c = 0
    int k = 1
    factorization.factor[0] = 1
    factorization.exponent[0] = 1
  
    for (i = 2; i <= n; i++) 
    
        c = 0
  
        while (n % i == 0
        
            c++; 
  
            // factorization.factor[j]=i; 
            n = n / i; 
            // j++; 
        
  
        if (c > 0
        
            factorization.exponent[k] = c; 
            factorization.factor[k] = i; 
            k++; 
        
    
  
    factorization.size = k - 1
  
// Function to print the factors 
static void DisplayFactorization(int x, FACTORIZATION 
                                        factorization) 
    int i; 
    System.out.print("Prime factor of " + x + " = "); 
  
    for (i = 0;
         i <= factorization.size; i++)
    
  
        System.out.print(factorization.factor[i]); 
  
        if (factorization.exponent[i] > 1
            System.out.print( "^" +
                       factorization.exponent[i]); 
  
        if (i < factorization.size) 
            System.out.print("*"); 
  
        else
            System.out.println( ); 
    
  
// function to find the gcd 
// using Middle School procedure 
static int gcdMiddleSchoolProcedure(int m, int n) 
  
    FACTORIZATION mFactorization = new FACTORIZATION();
    FACTORIZATION nFactorization = new FACTORIZATION(); 
  
    int r, mi, ni, i, k, x = 1, j; 
  
    // Step 1. 
    FindFactorization(m, mFactorization); 
    DisplayFactorization(m, mFactorization); 
  
    // Step 2. 
    FindFactorization(n, nFactorization); 
    DisplayFactorization(n, nFactorization); 
  
    // Steps 3 and 4. 
    // Procedure algorithm for computing the 
    // greatest common divisor. 
    int min; 
    i = 1
    j = 1
    while (i <= mFactorization.size && 
           j <= nFactorization.size) 
    
        if (mFactorization.factor[i] < 
            nFactorization.factor[j]) 
            i++; 
  
        else if (nFactorization.factor[j] < 
                 mFactorization.factor[i]) 
            j++; 
  
        else /* if arr1[i] == arr2[j] */
        
            min = mFactorization.exponent[i] > 
                  nFactorization.exponent[j] ?
                  nFactorization.exponent[j] :
                  mFactorization.exponent[i]; 
  
            x = x * mFactorization.factor[i] * min; 
            i++; 
            j++; 
        
    
  
    return x; 
  
// Driver code 
public static void main(String args[])
    int m = 10, n = 15
    System.out.print("GCD(" + m + ", " + n + ") = "
                     gcdMiddleSchoolProcedure(m, n)); 
}
  
// This code is contributed by Arnab Kundu

chevron_right


Output:

Prime factor of 10 = 1*2*5
Prime factor of 15 = 1*3*5
GCD(10, 15) = 5


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : andrew1234