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 pair between the common divisor are 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++ 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;
} |
// 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 |
# 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 |
// 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 |
<script> // JavaScript program for the above approach // Function to find the gcd of // two numbers function 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 function countPairwiseCoprime(N, M)
{ // Initialize answer with 1,
// to include 1 in the count
let answer = 1;
// Count of primes of gcd(N, M)
let g = gcd(N, M);
let temp = g;
// Finding prime factors of gcd
for (let 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;
} function countCoprimePair(arr, N)
{ // Function Call for each pair
// to calculate the count of
// pairwise co-prime divisors
for (let i = 0; i < N; i++)
{
document.write(countPairwiseCoprime(arr[i][0],
arr[i][1]) + " " );
}
} // Driver Code // Given array of pairs
let arr = [[ 12, 18 ], [ 420, 660 ]];
let N = arr.length;
// Function Call
countCoprimePair(arr, N);
</script> |
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)