Contar pares de números del 1 al N con Producto divisible por su Suma

Dado un número  . La tarea es contar pares (x, y) de modo que x*y sea divisible por (x+y) y la condición 1 <= x < y < N se cumple.
Ejemplos : 
 

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

El enfoque básico es iterar usando dos bucles manteniendo cuidadosamente la condición dada 1 <= x < y < N y generar todos los pares válidos posibles y contar esos pares para los cuales el producto de sus valores es divisible por la suma.
A continuación se muestra la implementación del enfoque anterior: 
 

// 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 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..

# 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# 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 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
?>

<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>

4

Tiempo Complejidad : O(N ² )

Espacio Auxiliar: O(1)