水平线和垂直线为H时形成的三角形总数

给定一个三角形ABC 。 H卧式线从侧AB到AC(如图1中所示。)和V的垂直线从顶点A到BC侧被吸引，任务是找到没有总。形成的三角形。
例子：

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 
 

 

方法：如下面的图片所示，我们可以为上述问题导出一个通用公式：

1. 如果只有h条水平线，则三角形总数为(h +1) 。
2. 如果只有v条垂直线，则三角形总数为(v + 1)*(v + 2)/ 2 。

1. 因此，总三角形是由水平线形成的三角形*由垂直线形成的三角形，即(h + 1)*((v + 1)*(v + 2)/ 2) 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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)))

Javascript

 
// 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
 

输出：

18

时间复杂度： O(1)
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