将(2 / N)表示为形式(1 / m)的三个不同的正整数之和

给定一个正整数N ，任务是将分数2 / N表示为形式为/ m的三个不同的正整数之和，即(2 / N)=(1 / x)+(1 / y)+( 1 / z)并打印x ， y和z 。
例子：

Input: N = 3 
Output: 3 4 12 
(1 / 3) + (1 / 4) + (1 / 12) = ((4 + 3 + 1) / 12) 
= (8 / 12) = (2 / 3) i.e. 2 / N
Input: N = 28 
Output: 28 29 812 
 

方法：很容易推断出，对于N = 1 ，将没有解决方案。对于N> 1 ， (2 / N)可以表示为(1 / N)+(1 / N) ，问题可以简化为用两个分数之和表示。现在，找到(1 / N)与1 /(N + 1)之间的差，并得到分数1 /(N *(N + 1)) 。因此，解为(2 / N)=(1 / N)+(1 /(N + 1))+(1 /(N *(N + 1)))其中x = N ， y = N + 1并且z = N *(N +1) 。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to find the required fractions
void find_numbers(int N)
{
    // Base condition
    if (N == 1) {
        cout << -1;
    }
 
    // For N > 1
    else {
        cout << N << " " << N + 1 << " "
             << N * (N + 1);
    }
}
 
// Driver code
int main()
{
    int N = 5;
 
    find_numbers(N);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
// Function to find the required fractions
static void find_numbers(int N)
{
    // Base condition
    if (N == 1)
    {
        System.out.print(-1);
    }
 
    // For N > 1
    else
    {
        System.out.print(N + " " + (N + 1) +
                             " " + (N * (N + 1)));
    }
}
 
// Driver code
public static void main(String []args)
{
    int N = 5;
 
    find_numbers(N);
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
 
# Function to find the required fractions
def find_numbers(N) :
 
    # Base condition
    if (N == 1) :
        print(-1, end = "");
 
    # For N > 1
    else :
        print(N, N + 1 , N * (N + 1));
 
# Driver code
if __name__ == "__main__" :
 
    N = 5;
 
    find_numbers(N);
     
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to find the required fractions
static void find_numbers(int N)
{
    // Base condition
    if (N == 1)
    {
        Console.Write(-1);
    }
 
    // For N > 1
    else
    {
        Console.Write(N + " " + (N + 1) +
                          " " + (N * (N + 1)));
    }
}
 
// Driver code
public static void Main(String []args)
{
    int N = 5;
 
    find_numbers(N);
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

// javascript implementation of the approach   
// Function to find the required fractions
    function find_numbers(N)
    {
     
        // Base condition
        if (N == 1) {
            document.write(-1);
        }
 
        // For N > 1
        else {
            document.write(N + " " + (N + 1) + " " + (N * (N + 1)));
        }
    }
 
    // Driver code
        var N = 5;
        find_numbers(N);
 
// This code is contributed by gauravrajput1

输出：

5 6 30