使用立方体形成的长度为 H 的金字塔中未着色细胞的计数

给定一个金字塔，用单位面积的立方体形成，给定高度H。然后将金字塔放在地上并从外面涂漆。任务是找出未着色的立方体的数量。

例子：

Input: H = 3
Output: 1
Explanation:

Input: H = 1
Output: 0

朴素方法：保持未着色的立方体的数量是金字塔边界中不存在的立方体的数量。因此，对于高度为h的每一层立方体，无色立方体的数量为$(h-2)^2$其中h>1 。因此，在每个级别添加无色立方体的数量以获得答案。

下面是上述方法的实现：

C++

// C++ program for the above problem
 
#include 
using namespace std;
 
// Function to return the number of
// uncoloured cubes
int uncolouredCubes(int H)
{
 
    // To store the number of uncoloured
    // cubes
    int res = 0;
 
    for (int h = 2; h <= H; h++) {
        res += (h - 2) * (h - 2);
    }
 
    return res;
}
 
// Driver Code
int main()
{
    int H = 3;
    cout << uncolouredCubes(H);
}

Java

// Java program for the above problem
 
 
class GFG{
 
  // Function to return the number of
  // uncoloured cubes
  static int uncolouredCubes(int H)
  {
 
      // To store the number of uncoloured
      // cubes
      int res = 0;
 
      for (int h = 2; h <= H; h++) {
          res += (h - 2) * (h - 2);
      }
 
      return res;
  }
 
  // Driver Code
  public static void main(String [] args)
  {
      int H = 3;
      System.out.println(uncolouredCubes(H));
  }
   
}
 
// This code is contributed by ihritik

Python3

# python program for the above problem
 
# Function to return the number of
# uncoloured cubes
def uncolouredCubes(H):
 
 
    # To store the number of uncoloured
    # cubes
    res = 0
 
    for h in range(2, H + 1):
        res = res + (h - 2) * (h - 2)
     
 
    return res
 
 
# Driver Code
 
H = 3
print(uncolouredCubes(H))
 
# This code is contributed by ihritik

C#

// C# program for the above problem
 
using System;
class GFG{
 
  // Function to return the number of
  // uncoloured cubes
  static int uncolouredCubes(int H)
  {
 
      // To store the number of uncoloured
      // cubes
      int res = 0;
 
      for (int h = 2; h <= H; h++) {
          res += (h - 2) * (h - 2);
      }
 
      return res;
  }
 
  // Driver Code
  public static void Main()
  {
      int H = 3;
      Console.WriteLine(uncolouredCubes(H));
  }
   
}
 
// This code is contributed by ihritik

Javascript

// JavaScript program for the above problem
 
// Function to return the number of
// uncoloured cubes
function uncolouredCubes(H){
 
 
    // To store the number of uncoloured
    // cubes
    var res = 0
 
    for(var h = 2 ; h < H + 1; h++)
        res = res + (h - 2) * (h - 2)
     
    return res
}
 
 
// Driver Code
 
var H = 3
document.write(uncolouredCubes(H))
 
// This code is contributed by AnkThon

输出

1

时间复杂度： O(N)
辅助空间： O(1)

有效方法：由于每层中在高度h处的无色立方体的数量为$(h-2)^2$ ，所以这个问题可以简化为找到第一个（H-2）个自然数的平方和。所以，答案是(n * (n + 1) * (2n + 1)) / 6其中n=H-2 。

C++

// C++ program for the above problem
 
#include 
using namespace std;
 
// Function to return the number of
// uncoloured cubes
int uncolouredCubes(int H)
{
 
    // If H is less than 2, then
    // returning 0
    if (H < 2) {
        return 0;
    }
 
    int n = H - 2;
 
    // Sum of the squares of first
    // n natural numbers
    return (n * (n + 1) * (2 * n + 1)) / 6;
}
 
// Driver Code
int main()
{
    int H = 3;
    cout << uncolouredCubes(H);
}

Java

// Java program for the above problem
import java.util.*;
public class GFG
{
   
// Function to return the number of
// uncoloured cubes
static int uncolouredCubes(int H)
{
   
    // If H is less than 2, then
    // returning 0
    if (H < 2) {
        return 0;
    }
    int n = H - 2;
 
    // Sum of the squares of first
    // n natural numbers
    return (n * (n + 1) * (2 * n + 1)) / 6;
}
 
// Driver Code
public static void main(String args[])
{
    int H = 3;
    System.out.println(uncolouredCubes(H));
}
}
 
// This code is contributed by Samim Hossain Mondal

Python3

# Python program for the above problem
 
# Function to return the number of
# uncoloured cubes
def uncolouredCubes(H):
 
  # If H is less than 2, then
  # returning 0
    if (H < 2):
        return 0;
 
    n = H - 2;
 
    # Sum of the squares of first
    # n natural numbers
    return (n * (n + 1) * (2 * n + 1)) / 6;
 
# Driver Code
H = 3;
print((int)(uncolouredCubes(H)));
 
# This code is contributed by Saurabh Jaiswal.

C＃

// C# program for the above problem
using System;
class GFG
{
// Function to return the number of
// uncoloured cubes
static int uncolouredCubes(int H)
{
    // If H is less than 2, then
    // returning 0
    if (H < 2) {
        return 0;
    }
    int n = H - 2;
 
    // Sum of the squares of first
    // n natural numbers
    return (n * (n + 1) * (2 * n + 1)) / 6;
}
 
// Driver Code
public static void Main()
{
    int H = 3;
    Console.Write(uncolouredCubes(H));
}
}
// This code is contributed by Samim Hossain Mondal

Javascript

// Javascript program for the above problem
 
// Function to return the number of
// uncoloured cubes
function uncolouredCubes(H)
{
 
  // If H is less than 2, then
  // returning 0
  if (H < 2) {
    return 0;
  }
 
  let n = H - 2;
 
  // Sum of the squares of first
  // n natural numbers
  return (n * (n + 1) * (2 * n + 1)) / 6;
}
 
// Driver Code
 
let H = 3;
document.write(uncolouredCubes(H));
 
// This code is contributed by gfgking.

输出

1

时间复杂度： O(1)
辅助空间： O(1)