计算在大小为 N 的正方形中内接的大小为 K 的正方形

给定两个整数N和K ，任务是找到大小为K的正方形内接在大小为N的正方形中的数量。

例子：

Input: N = 4, K = 2
Output: 9
Explanation:
There are 9 squares of size 2 inscribed in a square of size 4.

Input: N = 5, K = 3
Output: 9
Explanation:
There are 9 squares of size 3 inscribed in a square of size 5.

方法：解决问题的关键观察是大小为N的正方形中的正方形总数为(N * (N + 1)* (2*N + 1)) / 6 。因此，从大小为N的正方形中可能出现的大小为K的正方形总数为：

$(N - K + 1) * (N - K + 1)$

下面是上述方法的实现：

C++

// C++ implementation of the
// above approach
 
#include 
using namespace std;
 
// Function to calculate the number
// of squares of size K in a square
// of size N
int No_of_squares(int N, int K)
{
    // Stores the number of squares
    int no_of_squares = 0;
 
    no_of_squares
        = (N - K + 1) * (N - K + 1);
 
    return no_of_squares;
}
 
// Driver Code
int main()
{
    // Size of the
    // bigger square
    int N = 5;
 
    // Size of
    // smaller square
    int K = 3;
    cout << No_of_squares(N, K);
    return 0;
}

Java

// Java implementation of the
// above approach
import java.util.*;
class GFG{
 
// Function to calculate the
// number of squares of size
// K in a square of size N
static int No_of_squares(int N,
                         int K)
{
  // Stores the number
  // of squares
  int no_of_squares = 0;
 
  no_of_squares = (N - K + 1) *
                  (N - K + 1);
 
  return no_of_squares;
}
 
// Driver Code
public static void main(String[] args)
{
  // Size of the
  // bigger square
  int N = 5;
 
  // Size of
  // smaller square
  int K = 3;
  System.out.print(No_of_squares(N, K));
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 implementation of the
# above approach
 
# Function to calculate the
# number of squares of size
# K in a square of size N
def No_of_squares(N, K):
   
    # Stores the number
    # of squares
    no_of_squares = 0;
    no_of_squares = (N - K + 1) * (N - K + 1);
    return no_of_squares;
 
# Driver Code
if __name__ == '__main__':
   
    # Size of the
    # bigger square
    N = 5;
 
    # Size of
    # smaller square
    K = 3;
    print(No_of_squares(N, K));
 
# This code is contributed by 29AjayKumar

C#

// C# implementation of the
// above approach
using System;
 
class GFG{
 
// Function to calculate the
// number of squares of size
// K in a square of size N
static int No_of_squares(int N, int K)
{
     
    // Stores the number
    // of squares
    int no_of_squares = 0;
     
    no_of_squares = (N - K + 1) *
                    (N - K + 1);
     
    return no_of_squares;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Size of the
    // bigger square
    int N = 5;
     
    // Size of
    // smaller square
    int K = 3;
     
    Console.Write(No_of_squares(N, K));
}
}
 
// This code is contributed by Amit Katiyar

Javascript

 
// JavaScript program for
// the above approach
 
// Function to calculate the
// number of squares of size
// K in a square of size N
function No_of_squares(N, K)
{
  // Stores the number
  // of squares
  let no_of_squares = 0;
  
  no_of_squares = (N - K + 1) *
                  (N - K + 1);
  
  return no_of_squares;
}
 
// Driver code
 
    // Size of the
  // bigger square
  let N = 5;
  
  // Size of
  // smaller square
  let K = 3;
  document.write(No_of_squares(N, K));
   
  // This code is contributed by splevel62.

输出：

9

时间复杂度： O(1)
辅助空间： O(1)

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