Maximum count of pairwise co-prime and common divisors of two given numbers

Given an array of pairs arr[] of two numbers {N, M}, the task is to find the maximum count of common divisors for each pair N and M such that every pairs between the common divisor is co-prime.

A number x is a common divisor of N and M if, N%x = 0 and M%x = 0
Two numbers are co-prime if their Greatest Common Divisor is 1. 
 

Examples:

Input: arr[][] = {{12, 18}, {420, 660}} 
Output: 3 4 
Explanation: 
For pair (12, 18): 
{1, 2, 3} are common divisors of both 12 and 18, and are pairwise co-prime. 
For pair (420, 660): 
{1, 2, 3, 5} are common divisors of both 12 and 18, and are pairwise co-prime.
Input: arr[][] = {{8, 18}, {20, 66}} 
Output: 2 2 
 

Approach: The maximum count of common divisors of N and M such that the GCD of all the pairs between them is always 1 is 1 and all the common prime divisors of N and M. To count all the common prime divisors the idea is to find the GCD(say G) of the given two numbers and then count the number of prime divisors of the number G.



Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the gcd of
// two numbers
int gcd(int x, int y)
{
    if (x % y == 0)
        return y;
    else
        return gcd(y, x % y);
}
 
// Function to of pairwise co-prime
// and common divisors of two numbers
int countPairwiseCoprime(int N, int M)
{
    // Initialize answer with 1,
    // to include 1 in the count
    int answer = 1;
 
    // Count of primes of gcd(N, M)
    int g = gcd(N, M);
    int temp = g;
 
    // Finding prime factors of gcd
    for (int i = 2; i * i <= g; i++) {
 
        // Increment count if it is
        // divisible by i
        if (temp % i == 0) {
            answer++;
 
            while (temp % i == 0)
                temp /= i;
        }
    }
    if (temp != 1)
        answer++;
 
    // Return the total count
    return answer;
}
 
void countCoprimePair(int arr[][2], int N)
{
 
    // Function Call for each pair
    // to calculate the count of
    // pairwise co-prime divisors
    for (int i = 0; i < N; i++) {
        cout << countPairwiseCoprime(arr[i][0],
                                     arr[i][1])
             << ' ';
    }
}
 
// Driver Code
int main()
{
    // Given array of pairs
    int arr[][2] = { { 12, 18 }, { 420, 660 } };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    countCoprimePair(arr, N);
    return 0;
}

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Java

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// Java program for the above approach
class GFG{
 
// Function to find the gcd of
// two numbers
static int gcd(int x, int y)
{
    if (x % y == 0)
        return y;
    else
        return gcd(y, x % y);
}
 
// Function to of pairwise co-prime
// and common divisors of two numbers
static int countPairwiseCoprime(int N, int M)
{
    // Initialize answer with 1,
    // to include 1 in the count
    int answer = 1;
 
    // Count of primes of gcd(N, M)
    int g = gcd(N, M);
    int temp = g;
 
    // Finding prime factors of gcd
    for (int i = 2; i * i <= g; i++)
    {
 
        // Increment count if it is
        // divisible by i
        if (temp % i == 0)
        {
            answer++;
 
            while (temp % i == 0)
                temp /= i;
        }
    }
    if (temp != 1)
        answer++;
 
    // Return the total count
    return answer;
}
 
static void countCoprimePair(int arr[][], int N)
{
 
    // Function Call for each pair
    // to calculate the count of
    // pairwise co-prime divisors
    for (int i = 0; i < N; i++)
    {
        System.out.print(countPairwiseCoprime(arr[i][0],
                                               arr[i][1]) + " ");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    // Given array of pairs
    int arr[][] = { { 12, 18 }, { 420, 660 } };
    int N = arr.length;
 
    // Function Call
    countCoprimePair(arr, N);
}
}
 
// This code is contributed by Rajput-Ji

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Python3

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# Python3 program for the above approach
 
# Function to find the gcd of
# two numbers
def gcd(x, y):
    if (x % y == 0):
        return y
    else:
        return gcd(y, x % y)
 
# Function to of pairwise co-prime
# and common divisors of two numbers
def countPairwiseCoprime(N, M):
 
    # Initialize answer with 1,
    # to include 1 in the count
    answer = 1
 
    # Count of primes of gcd(N, M)
    g = gcd(N, M)
    temp = g
 
    # Finding prime factors of gcd
    for i in range(2, g + 1):
 
        if i * i > g:
            break
 
        # Increment count if it is
        # divisible by i
        if (temp % i == 0) :
            answer += 1
 
            while (temp % i == 0):
                temp //= i
 
    if (temp != 1):
        answer += 1
 
    # Return the total count
    return answer
 
def countCoprimePair(arr, N):
 
    # Function Call for each pair
    # to calculate the count of
    # pairwise co-prime divisors
    for i in range(N):
        print(countPairwiseCoprime(arr[i][0],
                                   arr[i][1]),
                                     end = " ")
 
# Driver Code
if __name__ == '__main__':
   
    # Given array of pairs
    arr= [ [ 12, 18 ], [ 420, 660 ] ]
    N = len(arr)
 
    # Function Call
    countCoprimePair(arr, N)
 
# This code is contributed by Mohit Kumar

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C#

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// C# program for the above approach
using System;
class GFG{
 
// Function to find the gcd of
// two numbers
static int gcd(int x, int y)
{
    if (x % y == 0)
        return y;
    else
        return gcd(y, x % y);
}
 
// Function to of pairwise co-prime
// and common divisors of two numbers
static int countPairwiseCoprime(int N, int M)
{
    // Initialize answer with 1,
    // to include 1 in the count
    int answer = 1;
 
    // Count of primes of gcd(N, M)
    int g = gcd(N, M);
    int temp = g;
 
    // Finding prime factors of gcd
    for (int i = 2; i * i <= g; i++)
    {
 
        // Increment count if it is
        // divisible by i
        if (temp % i == 0)
        {
            answer++;
 
            while (temp % i == 0)
                temp /= i;
        }
    }
    if (temp != 1)
        answer++;
 
    // Return the total count
    return answer;
}
 
static void countCoprimePair(int [,]arr, int N)
{
 
    // Function Call for each pair
    // to calculate the count of
    // pairwise co-prime divisors
    for (int i = 0; i < N; i++)
    {
        Console.Write(countPairwiseCoprime(arr[i, 0],
                                           arr[i, 1]) + " ");
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    // Given array of pairs
    int [,]arr = { { 12, 18 }, { 420, 660 } };
    int N = arr.GetLength(0);
 
    // Function Call
    countCoprimePair(arr, N);
}
}
 
// This code is contributed by Rajput-Ji

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Output: 

3 4





 

Time Complexity: O(X*(sqrt(N) + sqrt(M))), where X is the number of pairs and N & M are two pairs in arr[].
Auxiliary Space: O(1)

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Improved By : mohit kumar 29, Rajput-Ji