Given a number
Examples:
Input : N = 6 Output : 1 Explanation: The only pair is (3, 6) which satisfies all of the given condition, 3<6 and 18%9=0. Input : N = 15 Output : 4
The basic approach is to iterate using two loops carefully maintaining the given condition 1 <= x < y < N and generate all possible valid pairs and count such pairs for which the product of their values is divisible by sum.
Below is the implementation of the above approach:
C++
// C++ program to count pairs of numbers // from 1 to N with Product divisible // by their Sum #include <bits/stdc++.h> using namespace std;
// Function to count pairs int countPairs( int n)
{ // variable to store count
int count = 0;
// Generate all possible pairs such that
// 1 <= x < y < n
for ( int x = 1; x < n; x++) {
for ( int y = x + 1; y <= n; y++) {
if ((y * x) % (y + x) == 0)
count++;
}
}
return count;
} // Driver code int main()
{ int n = 15;
cout << countPairs(n);
return 0;
} |
Java
// Java program to count pairs of numbers // from 1 to N with Product divisible // by their Sum import java.io.*;
class GFG {
// Function to count pairs static int countPairs( int n)
{ // variable to store count
int count = 0 ;
// Generate all possible pairs such that
// 1 <= x < y < n
for ( int x = 1 ; x < n; x++) {
for ( int y = x + 1 ; y <= n; y++) {
if ((y * x) % (y + x) == 0 )
count++;
}
}
return count;
} // Driver code public static void main (String[] args) {
int n = 15 ;
System.out.println(countPairs(n));
}
} // This code is contributed by anuj_67.. |
Python3
# Python 3 program to count pairs of numbers # from 1 to N with Product divisible # by their Sum # Function to count pairs def countPairs(n):
# variable to store count
count = 0
# Generate all possible pairs such that
# 1 <= x < y < n
for x in range ( 1 , n):
for y in range (x + 1 , n + 1 ):
if ((y * x) % (y + x) = = 0 ):
count + = 1
return count
# Driver code n = 15
print (countPairs(n))
# This code is contributed # by PrinciRaj1992 |
C#
// C# program to count pairs of numbers // from 1 to N with Product divisible // by their Sum using System;
class GFG
{ // Function to count pairs static int countPairs( int n)
{ // variable to store count
int count = 0;
// Generate all possible pairs
// such that 1 <= x < y < n
for ( int x = 1; x < n; x++)
{
for ( int y = x + 1; y <= n; y++)
{
if ((y * x) % (y + x) == 0)
count++;
}
}
return count;
} // Driver code public static void Main ()
{ int n = 15;
Console.WriteLine(countPairs(n));
} } // This code is contributed by anuj_67 |
PHP
<?php // PHP program to count pairs of // numbers from 1 to N with Product // divisible by their Sum // Function to count pairs function countPairs( $n )
{ // variable to store count
$count = 0;
// Generate all possible pairs
// such that 1 <= x < y < n
for ( $x = 1; $x < $n ; $x ++)
{
for ( $y = $x + 1; $y <= $n ; $y ++)
{
if (( $y * $x ) % ( $y + $x ) == 0)
$count ++;
}
}
return $count ;
} // Driver code $n = 15;
echo countPairs( $n );
// This code is contributed by ajit ?> |
Javascript
<script> // Javascript program to count pairs of numbers // from 1 to N with Product divisible // by their Sum // Function to count pairs function countPairs(n)
{ // variable to store count
let count = 0;
// Generate all possible pairs such that
// 1 <= x < y < n
for (let x = 1; x < n; x++) {
for (let y = x + 1; y <= n; y++) {
if ((y * x) % (y + x) == 0)
count++;
}
}
return count;
} // Driver code let n = 15; document.write(countPairs(n)); // This code is contributed by souravmahato348. </script> |
Output:
4
Time Complexity : O(N2)
Auxiliary Space: O(1)