Total number of triangles formed when there are H horizontal and V vertical lines

Given a triangle ABC. H horizontal lines from side AB to AC (as shown in fig.) and V vertical lines from vertex A to side BC are drawn, the task is to find the total no. of triangles formed.
Examples:

Input: H = 2, V = 2 
Output: 18 

As we see in the image above, total triangles formed are 18.
Input: H = 3, V = 4 
Output: 60 

Approach: As we see in the images below, we can derive a general formula for above problem: 

1. If there are only h horizontal lines then total triangles are (h + 1).
2. If there are only v vertical lines then total triangles are (v + 1) * (v + 2) / 2.. 
   

1. So, total triangles are Triangles formed by horizontal lines * Triangles formed by vertical lines i.e. (h + 1) * (( v + 1) * (v + 2) / 2).

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define LLI long long int

// Function to return total triangles
LLI totalTriangles(LLI h, LLI v)
{
    // Only possible triangle is
    // the given triangle
    if (h == 0 && v == 0)
        return 1;

    // If only vertical lines are present
    if (h == 0)
        return ((v + 1) * (v + 2) / 2);

    // If only horizontal lines are present
    if (v == 0)
        return (h + 1);

    // Return total triangles
    LLI Total = (h + 1) * ((v + 1) * (v + 2) / 2);

    return Total;
}

// Driver code
int main()
{
    int h = 2, v = 2;
    cout << totalTriangles(h, v);

    return 0;
}

Java

// Java implementation of the approach
class GFG {

    // Function to return total triangles
    public static int totalTriangles(int h, int v)
    {
        // Only possible triangle is
        // the given triangle
        if (h == 0 && v == 0)
            return 1;

        // If only vertical lines are present
        if (h == 0)
            return ((v + 1) * (v + 2) / 2);

        // If only horizontal lines are present
        if (v == 0)
            return (h + 1);

        // Return total triangles
        int total = (h + 1) * ((v + 1) * (v + 2) / 2);

        return total;
    }

    // Driver code
    public static void main(String[] args)
    {
        int h = 2, v = 2;
        System.out.print(totalTriangles(h, v));
    }
}

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to return total triangles
    public static int totalTriangles(int h, int v)
    {
        // Only possible triangle is
        // the given triangle
        if (h == 0 && v == 0)
            return 1;

        // If only vertical lines are present
        if (h == 0)
            return ((v + 1) * (v + 2) / 2);

        // If only horizontal lines are present
        if (v == 0)
            return (h + 1);

        // Return total triangles
        int total = (h + 1) * ((v + 1) * (v + 2) / 2);

        return total;
    }

    // Driver code
    public static void Main()
    {
        int h = 2, v = 2;
        Console.Write(totalTriangles(h, v));
    }
}

// This code is contributed by Ryuga

Python3

# Python3 implementation of the approach

# Function to return total triangles
def totalTriangles(h, v):

    # Only possible triangle is
    # the given triangle
    if (h == 0 and v == 0):
        return 1

    # If only vertical lines are present
    if (h == 0):
        return ((v + 1) * (v + 2) / 2)

    # If only horizontal lines are present
    if (v == 0):
        return (h + 1)

    # Return total triangles
    total = (h + 1) * ((v + 1) * (v + 2) / 2)

    return total

# Driver code
h = 2
v = 2
print(int(totalTriangles(h, v)))

PHP

<?php
// PHP implementation of the above approach

// Function to return total triangles
function totalTriangles($h, $v)
{
    // Only possible triangle is
    // the given triangle
    if ($h == 0 && $v == 0)
        return 1;

    // If only vertical lines are present
    if ($h == 0)
        return (($v + 1) * ($v + 2) / 2);

    // If only horizontal lines are present
    if ($v == 0)
        return ($h + 1);

    // Return total triangles
    $Total = ($h + 1) * (($v + 1) *
                         ($v + 2) / 2);

    return $Total;
}

// Driver code
$h = 2;
$v = 2;
echo totalTriangles($h, $v);

// This code is contributed by Arnab Kundu
?>

Javascript

<script>

// javascript implementation of the approach
// Function to return total triangles

function totalTriangles(h , v)
{
    // Only possible triangle is
    // the given triangle
    if (h == 0 && v == 0)
        return 1;

    // If only vertical lines are present
    if (h == 0)
        return ((v + 1) * (v + 2) / 2);

    // If only horizontal lines are present
    if (v == 0)
        return (h + 1);

    // Return total triangles
    var total = (h + 1) * ((v + 1) * (v + 2) / 2);

    return total;
}

// Driver code
var h = 2, v = 2;
document.write(totalTriangles(h, v));

// This code contributed by shikhasingrajput

</script>

Output:

18

Time Complexity: O(1) 
Auxiliary Space: O(1)