Encuentre los números a y b que satisfagan las condiciones dadas

Dado un entero n , la tarea es encontrar dos enteros a y b que satisfagan las siguientes condiciones: 
 

1. un % b = 0
2. a * b > n
3. un / segundo < norte

Si ningún par satisface las condiciones anteriores, imprima -1 . 
Nota: Puede haber varios pares (a, b) que satisfagan las condiciones anteriores para n .
Ejemplos: 
 

Input: n = 10
Output: a = 90, b = 10
90 % 10 = 0
90 * 10 = 900 > 10
90 / 10 = 9 < 10
All three conditions are satisfied.

Input: n = 1
Output: -1

Enfoque: supongamos que b = n , al tomar esta suposición , se puede encontrar a en función de las condiciones dadas: 
 

• (a % b = 0) => a debe ser múltiplo de b .
• (a / b < n) => a / b = n – 1 que es < n .
• (un * segundo > norte) => un = norte .

A continuación se muestra la implementación del enfoque anterior:
 

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the required numbers
void find(int n)
{
    // Suppose b = n and we want a % b = 0 and also
    // (a / b) < n so a = b * (n - 1)
    int b = n;
    int a = b * (n - 1);
 
    // Special case if n = 1
    // we get a = 0 so (a * b) < n
    if (a * b > n && a / b < n) {
        cout << "a = " << a << ", b = " << b;
    }
 
    // If no pair satisfies the conditions
    else
        cout << -1 << endl;
}
 
// Driver code
int main()
{
    int n = 10;
    find(n);
    return 0;
}

// Java implementation of the above approach
 
public class GFG{
 
    // Function to print the required numbers
    static void find(int n)
    {
        // Suppose b = n and we want a % b = 0 and also
        // (a / b) < n so a = b * (n - 1)
        int b = n;
        int a = b * (n - 1);
     
        // Special case if n = 1
        // we get a = 0 so (a * b) < n
        if (a * b > n && a / b < n) {
            System.out.print("a = " + a + ", b = " + b);
        }
     
        // If no pair satisfies the conditions
        else
            System.out.println(-1);
    }
     
    // Driver code
    public static void main(String []args)
    {
        int n = 10;
        find(n);
    }
    // This code is contributed by Ryuga
}

# Python3 implementation of the above approach
 
# Function to print the required numbers
def find(n):
 
    # Suppose b = n and we want a % b = 0
    # and also (a / b) < n so a = b * (n - 1)
    b = n
    a = b * (n - 1)
 
    # Special case if n = 1
    # we get a = 0 so (a * b) < n
    if a * b > n and a // b < n:
        print("a = {}, b = {}" . format(a, b))
     
    # If no pair satisfies the conditions
    else:
        print(-1)
 
# Driver Code
if __name__ == "__main__":
 
    n = 10
    find(n)
 
# This code is contributed by Rituraj Jain

// C# implementation of the above approach
using System;
 
class GFG
{
 
// Function to print the required numbers
static void find(int n)
{
    // Suppose b = n and we want a % b = 0
    // and also (a / b) < n so a = b * (n - 1)
    int b = n;
    int a = b * (n - 1);
 
    // Special case if n = 1
    // we get a = 0 so (a * b) < n
    if (a * b > n && a / b < n)
    {
        Console.Write("a = " + a + ", b = " + b);
    }
 
    // If no pair satisfies the conditions
    else
        Console.WriteLine(-1);
}
 
// Driver code
public static void Main()
{
    int n = 10;
    find(n);
}
}
 
// This code is contributed
// by Akanksha Rai

<?php
// PHP implementation of the above approach
 
// Function to print the required numbers
function find($n)
{
    // Suppose b = n and we want a % b = 0 and also
    // (a / b) < n so a = b * (n - 1)
    $b = $n;
    $a = $b * ($n - 1);
 
    // Special case if n = 1
    // we get a = 0 so (a * b) < n
    if ($a * $b > $n && $a / $b <$n) {
        echo "a = " , $a , ", b = " , $b;
    }
 
    // If no pair satisfies the conditions
    else
        echo -1 ;
}
 
// Driver code
 
    $n = 10;
    find($n);
// This code is contributed
// by inder_verma..
?>

<script>
    // Javascript implementation of the above approach
     
    // Function to print the required numbers
    function find(n)
    {
     
        // Suppose b = n and we want a % b = 0
        // and also (a / b) < n so a = b * (n - 1)
        let b = n;
        let a = b * (n - 1);
 
        // Special case if n = 1
        // we get a = 0 so (a * b) < n
        if (a * b > n && a / b < n)
        {
            document.write("a = " + a + ", b = " + b);
        }
 
        // If no pair satisfies the conditions
        else
            document.write(-1);
    }
     
    let n = 10;
    find(n);
     
    // This code is contributed by surehs07.
</script>

a = 90, b = 10

Complejidad Temporal: O(1), ya que no hay bucle ni recursividad.
Espacio Auxiliar: O(1), ya que no se ha ocupado ningún espacio extra.