在给定的等边三角形内可以形成的等边三角形的最大数量

给定两个整数N和K ，其中 N 表示更大的等边三角形的单位大小，任务是找到大小为 K 的等边三角形在 N 边的更大三角形中的数量。

例子：

Input: N = 4, K = 3

Output: 3
Explanation:
There are 3 equilateral triangles of 3 unit size which are present in the Bigger equilateral triangle of size 4 units.

Input: N = 4, K = 2
Output: 7
Explanation:
There are 7 equilateral triangles of 2 unit size which are present in the Bigger equilateral triangle of size 4 units. 

朴素的方法：这个想法是迭代更大的等边三角形的所有可能大小，以检查具有所需大小K的三角形数量并打印三角形的总数。

时间复杂度： O(N)
辅助空间： O(1)

高效的方法：要优化上述方法，请注意以下几点：

• 在大小N 中存在的大小为K的向上方向具有峰值的三角形的数量等于((N – K +1 ) * (N – K + 2))/2 。
• 在大小N 中存在的大小为K的向下方向具有峰值的倒三角形的数量等于((N – 2K + 1) * (N – 2K + 2))/2 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the number of
// equilateral triangle formed
// within another triangle
int No_of_Triangle(int N, int K)
{
    // Check for the valid condition
    if (N < K)
        return -1;
 
    else {
 
        int Tri_up = 0;
 
        // Number of triangles having
        // upward peak
        Tri_up = ((N - K + 1)
                  * (N - K + 2))
                 / 2;
 
        int Tri_down = 0;
 
        // Number of inverted triangles
        Tri_down = ((N - 2 * K + 1)
                    * (N - 2 * K + 2))
                   / 2;
 
        // Total no. of K sized triangle
        return Tri_up + Tri_down;
    }
}
 
// Driver Code
int main()
{
    // Given N and K
    int N = 4, K = 2;
 
    // Function Call
    cout << No_of_Triangle(N, K);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
 
// Function to find the number of
// equilateral triangle formed
// within another triangle
static int No_of_Triangle(int N, int K)
{
    // Check for the valid condition
    if (N < K)
        return -1;
 
    else
    {
        int Tri_up = 0;
 
        // Number of triangles having
        // upward peak
        Tri_up = ((N - K + 1) * (N - K + 2)) / 2;
 
        int Tri_down = 0;
 
        // Number of inverted triangles
        Tri_down = ((N - 2 * K + 1) *
                    (N - 2 * K + 2)) / 2;
 
        // Total no. of K sized triangle
        return Tri_up + Tri_down;
    }
}
 
// Driver Code
public static void main(String[] args)
{
    // Given N and K
    int N = 4, K = 2;
 
    // Function Call
    System.out.print(No_of_Triangle(N, K));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 program for the above approach
 
# Function to find the number of
# equilateral triangle formed
# within another triangle
def No_of_Triangle(N, K):
   
    # Check for the valid condition
    if (N < K):
        return -1;
 
    else:
        Tri_up = 0;
 
        # Number of triangles having
        # upward peak
        Tri_up = ((N - K + 1) *
                  (N - K + 2)) // 2;
 
        Tri_down = 0;
 
        # Number of inverted triangles
        Tri_down = ((N - 2 * K + 1) *
                    (N - 2 * K + 2)) // 2;
 
        # Total no. of K sized triangle
        return Tri_up + Tri_down;
     
# Driver Code
if __name__ == '__main__':
    # Given N and K
    N = 4; K = 2;
 
    # Function Call
    print(No_of_Triangle(N, K));
 
# This code is contributed by sapnasingh4991

C#

// C# program for the above approach
using System;
class GFG{
 
// Function to find the number of
// equilateral triangle formed
// within another triangle
static int No_of_Triangle(int N, int K)
{
    // Check for the valid condition
    if (N < K)
        return -1;
 
    else
    {
        int Tri_up = 0;
 
        // Number of triangles having
        // upward peak
        Tri_up = ((N - K + 1) * (N - K + 2)) / 2;
 
        int Tri_down = 0;
 
        // Number of inverted triangles
        Tri_down = ((N - 2 * K + 1) *
                    (N - 2 * K + 2)) / 2;
 
        // Total no. of K sized triangle
        return Tri_up + Tri_down;
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    // Given N and K
    int N = 4, K = 2;
 
    // Function Call
    Console.Write(No_of_Triangle(N, K));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// JavaScript program for the above approach
 
// Function to find the number of
// equilateral triangle formed
// within another triangle
function No_of_Triangle(N, K)
{
    // Check for the valid condition
    if (N < K)
        return -1;
 
    else {
 
        let Tri_up = 0;
 
        // Number of triangles having
        // upward peak
        Tri_up = Math.floor(((N - K + 1)
                * (N - K + 2))
                / 2);
 
        let Tri_down = 0;
 
        // Number of inverted triangles
        Tri_down = Math.floor(((N - 2 * K + 1)
                    * (N - 2 * K + 2))
                / 2);
 
        // Total no. of K sized triangle
        return Tri_up + Tri_down;
    }
}
 
// Driver Code
 
    // Given N and K
    let N = 4, K = 2;
 
    // Function Call
    document.write(No_of_Triangle(N, K));
 
 
// This code is contributed by Surbhi Tyagi.
 

输出：

7

时间复杂度： O(1)
辅助空间： O(1)

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