计算矩形中的正方形数

给定amxn矩形，其中有多少个正方形?

例子 ：

Input:  m = 2, n = 2
Output: 5
There are 4 squares of size 1x1 + 1 square of size 2x2.

Input: m = 4, n = 3
Output: 20
There are 12 squares of size 1x1 + 
          6 squares of size 2x2 + 
          2 squares of size 3x3.

让我们首先针对m = n(即正方形)解决这个问题：
对于m = n = 1，输出：1
对于m = n = 2，输出：4 +1 [4的大小为1×1 + 1的大小为2×2]
对于m = n = 3，输出：9 + 4 + 1 [9的大小为1×1 + 4的大小为2×2 + 1的大小为3×3]
对于m = n = 4，输出16 + 9 + 4 +1 [尺寸1×1的16 +尺寸2×2的9 +尺寸3×3的4 +尺寸4×4的1]
通常，它似乎是n ^ 2 +(n-1)^ 2 +…1 = n(n + 1)(2n + 1)/ 6

当m不等于n时，让我们解决这个问题：
让我们假设m <= n

通过上面的解释，我们知道amxm矩阵的平方数为m(m + 1)(2m + 1)/ 6

当我们添加一列时会发生什么，即mx(m + 1)矩阵中的平方数是多少?

当我们添加一列时，增加的平方数为m +(m-1)+…+ 3 + 2 +1
[大小为1×1的m平方+大小为2×2 +的(m-1)平方+…+大小为mxm的1平方]
等于m(m + 1)/ 2

因此，当我们添加(nm)列时，增加的平方总数为(nm)* m(m + 1)/ 2。
因此，平方总数为m(m + 1)(2m + 1)/ 6 +(nm)* m(m + 1)/ 2。
使用相同的逻辑，我们可以证明n <= m。

所以总的来说

Total number of squares = m x (m+1) x (2m+1)/6 + (n-m) x m x (m+1)/2 

when n is larger dimension

对矩形使用上述逻辑，我们还可以证明一个正方形中的正方形数为n(n + 1)(2n + 1)/ 6

下面是上述公式的实现。

C++

// C++ program to count squares
// in a rectangle of size m x n
#include
using namespace std;
 
// Returns count of all squares
// in a rectangle of size m x n
int countSquares(int m, int n)
{
// If n is smaller, swap m and n
if (n < m)
    swap(m, n);
 
// Now n is greater dimension,
// apply formula
return m * (m + 1) * (2 * m + 1) /
     6 + (n - m) * m *(m + 1) / 2;
}
 
// Driver Code
int main()
{
int m = 4, n = 3;
cout << "Count of squares is "
     << countSquares(m, n);
}

C

// C program to count squares
// in a rectangle of size m x n
 
#include 
 
// Returns count of all squares
// in a rectangle of size m x n
int countSquares(int m, int n)
{
  int temp;
// If n is smaller, swap m and n
  if (n < m)
  {
      temp=n;
      n=m;
      m=temp;
  }
  // Now n is greater dimension,
  // apply formula
  return m * (m + 1) * (2 * m + 1) /
      6 + (n - m) * m *(m + 1)/ 2;
}
 
// Driver Code
int main()
{
    int m = 4, n = 3;
    printf("Count of squares is %d",countSquares(m, n));
}
 
// This code is contributed by Hemant Jain.

Java

// Java program to count squares
// in a rectangle of size m x n
 
class GFG
{
    // Returns count of all squares
    // in a rectangle of size m x n
    static int countSquares(int m, int n)
    {
    // If n is smaller, swap m and n
    if (n < m)
    {
        // swap(m, n)
        int temp = m;
        m = n;
        n = temp;
    }
         
     
    // Now n is greater dimension,
    // apply formula
    return m * (m + 1) * (2 * m + 1) /
        6 + (n - m) * m * (m + 1) / 2;
    }
     
    // Driver Code
    public static void main(String[] args)
    {
        int m = 4, n = 3;
        System.out.println("Count of squares is " +
                            countSquares(m, n));
    }
}

Python3

# Python3 program to count squares
# in a rectangle of size m x n
 
# Returns count of all squares
# in a rectangle of size m x n
def countSquares(m, n):
     
    # If n is smaller, swap m and n
    if(n < m):
        temp = m
        m = n
        n = temp
         
    # Now n is greater dimension,
    # apply formula
    return ((m * (m + 1) * (2 * m + 1) /
           6 + (n - m) * m * (m + 1) / 2))
 
# Driver Code
if __name__=='__main__':
    m = 4
    n = 3
    print("Count of squares is "
         ,countSquares(m, n))
 
# This code is contributed by mits.

C#

// C# program to count squares in a rectangle
// of size m x n
using System;
 
class GFG {
     
    // Returns count of all squares in a
    // rectangle of size m x n
    static int countSquares(int m, int n)
    {
    // If n is smaller, swap m and n
    if (n < m)
    {
        // swap(m,n)
        int temp = m;
        m = n;
        n = temp;
    }
             
    // Now n is greater dimension, apply
    // formula
    return m * (m + 1) * (2 * m + 1) / 6 +
               (n - m) * m * (m + 1) / 2;
    }
     
    // Driver method
    public static void Main()
    {
        int m = 4, n = 3;
         
        Console.WriteLine("Count of squares is "
                          + countSquares(m, n));
    }
}
 
//This code is contributed by vt_m.

Javascript

 
// javascript program to count squares
// in a rectangle of size m x n
 
// Returns count of all squares
// in a rectangle of size m x n
function countSquares( m,  n)
{
 
// If n is smaller, swap m and n
if (n < m)
    [m, n] = [n, m];
 
// Now n is greater dimension,
// apply formula
return m * (m + 1) * (2 * m + 1) /
    6 + (n - m) * m *(m + 1) / 2;
}
 
// Driver Code
    let m = 4;
    let n = 3;
document.write("Count of squares is "+countSquares(n, m));
 
// This code is contributed by jana_sayantan.

输出 ：

Count of Squares is 20

替代解决方案：

1. 让我们取m = 2，n = 3；
2. 边1的平方数将是6，因为将有两种情况，一种是沿水平(2)的1单元边的正方形，第二种是沿垂直(3)的1单元边的正方形。给我们2 * 3 = 6平方
3. 当侧面为2个单位时，一种情况是两个单位的边沿的正方形，仅沿水平方向放置一个，第二个案例为垂直的两个位置。因此，平方数= 2
4. 因此我们可以推断出，大小为1 * 1的平方数将为m * n。大小为2 * 2的平方数将为(n-1)(m-1)。因此，大小为n的平方数将为1 *(m-n + 1)。

平方总数的最终公式将是n *(n + 1)(3m-n + 1)/ 6 。

C++

// C++ program to count squares
// in a rectangle of size m x n
#include 
using namespace std;
 
// Returns count of all squares
// in a rectangle of size m x n
int countSquares(int m, int n)
{
 
    // If n is smaller, swap m and n
    if (n < m) {
        int temp = m;
        m = n;
        n = temp;
    }
 
    // Now n is greater dimension,
    // apply formula
    return n * (n + 1) * (3 * m - n + 1) / 6;
}
 
// Driver Code
int main()
{
    int m = 4, n = 3;
    cout << "Count of squares is " << countSquares(m, n);
}
 
// This code is contributed by 29AjayKumar

C

// C program to count squares
// in a rectangle of size m x n
 
#include 
 
// Returns count of all squares
// in a rectangle of size m x n
int countSquares(int m, int n)
{
 
    // If n is smaller, swap m and n
    if (n < m)
    {
        int temp = m;
        m = n;
        n = temp;
    }
 
    // Now n is greater dimension,
    // apply formula
    return n * (n + 1) * (3 * m - n + 1) / 6;
}
 
// Driver Code
int main()
{
    int m = 4, n = 3;
    printf("Count of squares is %d",countSquares(m, n));
}
 
// This code is contributed by Hemant Jain

Java

// Java program to count squares
// in a rectangle of size m x n
import java.util.*;
 
class GFG
{
 
    // Returns count of all squares
    // in a rectangle of size m x n
    static int countSquares(int m, int n)
    {
 
        // If n is smaller, swap m and n
        if (n < m)
        {
            int temp = m;
            m = n;
            n = temp;
        }
 
        // Now n is greater dimension,
        // apply formula
        return n * (n + 1) * (3 * m - n + 1) / 6;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int m = 4;
        int n = 3;
        System.out.print("Count of squares is " +
                             countSquares(m, n));
    }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to count squares
# in a rectangle of size m x n
 
# Returns count of all squares
# in a rectangle of size m x n
def countSquares(m, n):
     
    # If n is smaller, swap m and n
    if(n < m):
        temp = m
        m = n
        n = temp
         
    # Now n is greater dimension,
    # apply formula
    return n * (n + 1) * (3 * m - n + 1) // 6
 
# Driver Code
if __name__=='__main__':
    m = 4
    n = 3
    print("Count of squares is",
           countSquares(m, n))
 
# This code is contributed by AnkitRai01

C＃

// C# program to count squares
// in a rectangle of size m x n
using System;
 
class GFG
{
 
    // Returns count of all squares
    // in a rectangle of size m x n
    static int countSquares(int m, int n)
    {
 
        // If n is smaller, swap m and n
        if (n < m)
        {
            int temp = m;
            m = n;
            n = temp;
        }
 
        // Now n is greater dimension,
        // apply formula
        return n * (n + 1) * (3 * m - n + 1) / 6;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int m = 4;
        int n = 3;
        Console.Write("Count of squares is " +
                          countSquares(m, n));
    }
}
 
// This code is contributed by Rajput-Ji

Java脚本

 
// Javascript program to count squares
// in a rectangle of size m x n
 
// Returns count of all squares
// in a rectangle of size m x n
function countSquares(m , n)
{
     
    // If n is smaller, swap m and n
    if (n < m)
    {
        var temp = m;
        m = n;
        n = temp;
    }
 
    // Now n is greater dimension,
    // apply formula
    return n * (n + 1) * (3 * m - n + 1) / 6;
}
 
// Driver Code
var m = 4;
var n = 3;
 
document.write("Count of squares is " +
               countSquares(m, n));
 
// This code is contributed by shikhasingrajput
 

输出 ：

Count of Squares is 20