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

• Difficulty Level : Basic
• Last Updated : 28 Sep, 2021

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++

 // C++ implementation of above algorithm#include #define MAXFACTORS 1024using namespace std; // struct to store factorization of m and ntypedef struct{     int size;    int factor[MAXFACTORS + 1];    int exponent[MAXFACTORS + 1]; } FACTORIZATION; // Function to find the factorization of M and Nvoid FindFactorization(int x, FACTORIZATION* factorization){    int i, j = 1;    int n = x, c = 0;    int k = 1;    factorization->factor = 1;    factorization->exponent = 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 factorsvoid 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 procedureint 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 codeint main() {     int m = 10, n = 15;    cout << "GCD(" << m << ", " << n << ") = "         << gcdMiddleSchoolProcedure(m, n);     return (0);}

Java

 // Java implementation of above algorithmclass GFG{static final int MAXFACTORS = 1024 ; // class to store factorization// of m and nstatic class FACTORIZATION{    int size;    int factor[] = new int[MAXFACTORS + 1];    int exponent[] = new int[MAXFACTORS + 1]; } // Function to find the// factorization of M and Nstatic void FindFactorization(int x, FACTORIZATION                                     factorization){    int i, j = 1;    int n = x, c = 0;    int k = 1;    factorization.factor = 1;    factorization.exponent = 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 factorsstatic 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 procedurestatic 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 codepublic 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

Python3

 # Python3 implementation of above algorithm MAXFACTORS = 1024 # class to store factorization# of m and nclass FACTORIZATION:    def __init__(self):        self.size = 0        self.factor = *(MAXFACTORS + 1)        self.exponent = *(MAXFACTORS + 1) # Function to find the# factorization of M and Ndef FindFactorization(x,factorization):    j = 1    n, c = x, 0    k = 1    factorization.factor = 1    factorization.exponent = 1        for i in range(2, n + 1):        c = 0        while (n % i == 0):            c+=1            # factorization.factor[j]=i;            n = n / i            # j++            if (c > 0):            factorization.exponent[k] = c            factorization.factor[k] = i            k+=1        factorization.size = k - 1 # Function to print the factorsdef DisplayFactorization(x,factorization):    print("Prime factor of", x, "= ", end="")        for i in range(factorization.size + 1):        print(factorization.factor[i], end = "")            if (factorization.exponent[i] > 1):            print( "^", factorization.exponent[i], sep = "", end = "")            if (i < factorization.size):            print("*", end = "")        else:            print()  # function to find the gcd# using Middle School proceduredef gcdMiddleSchoolProcedure(m,n):    mFactorization = FACTORIZATION()    nFactorization = FACTORIZATION()        x = 1        # 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.    i = 1    j = 1    while (i <= mFactorization.size and j <= nFactorization.size):        if (mFactorization.factor[i] < nFactorization.factor[j]):            i+=1            elif (nFactorization.factor[j] < mFactorization.factor[i]):            j+=1            else: # if arr1[i] == arr2[j]            if mFactorization.exponent[i] > nFactorization.exponent[j]:                Min = nFactorization.exponent[j]            else:                Min = mFactorization.exponent[i]               x = x * mFactorization.factor[i] * Min            i+=1            j+=1    return x # Driver codem, n = 10, 15print("GCD(", m, ", ", n, ") = ", gcdMiddleSchoolProcedure(m, n), sep = "") # This code is contributed by suresh07.

C#

 // C# implementation of above algorithmusing System;     public class GFG{static readonly int MAXFACTORS = 1024 ; // class to store factorization// of m and npublic class FACTORIZATION{    public int size;    public int []factor = new int[MAXFACTORS + 1];    public int []exponent = new int[MAXFACTORS + 1]; } // Function to find the// factorization of M and Nstatic void FindFactorization(int x, FACTORIZATION                                    factorization){    int i;    int n = x, c = 0;    int k = 1;    factorization.factor = 1;    factorization.exponent = 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 factorsstatic void DisplayFactorization(int x, FACTORIZATION                                        factorization){    int i;    Console.Write("Prime factor of " + x + " = ");     for (i = 0;        i <= factorization.size; i++)    {         Console.Write(factorization.factor[i]);         if (factorization.exponent[i] > 1)            Console.Write( "^" +                    factorization.exponent[i]);         if (i < factorization.size)            Console.Write("*");         else        Console.WriteLine();    }} // function to find the gcd// using Middle School procedurestatic int gcdMiddleSchoolProcedure(int m, int n){     FACTORIZATION mFactorization = new FACTORIZATION();    FACTORIZATION nFactorization = new FACTORIZATION();     int i, 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 codepublic static void Main(String []args){    int m = 10, n = 15;    Console.Write("GCD(" + m + ", " + n + ") = " +                    gcdMiddleSchoolProcedure(m, n));}} // This code contribut by Rajput-Ji

Javascript


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

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