矩形中最大面积的平方数

给定边m和n的矩形。将矩形切成较小的相同块，以使每个块都是一个正方形，其边长应尽可能大，而矩形的剩余部分应没有。这样的正方形的印刷数量。

例子：

Input: 9 6
Output: 6
Rectangle can be cut into squares of size 3.

Input: 4 2
Output: 2
Rectangle can be cut into squares of size 2.

方法：任务是将矩形切成边长为s的正方形，而矩形的剩余部分不留，因此s必须将m和n都除。同样，正方形的边应尽可能大，因此，s应该是m和n的最大公约数。
因此， s = gcd(m，n) 。
要找到切成矩形的正方形的数量，要做的任务是将矩形的面积除以大小为s的正方形的面积。

C++

// C++ code for calculating the
// number of squares
#include 
using namespace std;
  
// Function to find number of squares
int NumberOfSquares(int x, int y)
{
    // Here in built c++ gcd function is used
    int s = __gcd(x, y);
  
    int ans = (x * y) / (s * s);
  
    return ans;
}
  
// Driver code
int main()
{
    int m = 385, n = 60;
  
    // Call the function NumberOfSquares
    cout << NumberOfSquares(m, n);
  
    return 0;
}

Java

// Java code for calculating 
// the number of squares
import java.io.*;
  
class GFG
{
    // Recursive function to
    // return gcd of a and b
    static int __gcd(int a, int b)
    {
        // Everything divides 0 
        if (a == 0 || b == 0)
        return 0;
      
        // base case
        if (a == b)
            return a;
      
        // a is greater
        if (a > b)
            return __gcd(a - b, b);
        return __gcd(a, b - a);
    } 
  
  
// Function to find 
// number of squares
static int NumberOfSquares(int x, 
                           int y)
{
    // Here in built c++ 
    // gcd function is used
    int s = __gcd(x, y);
  
    int ans = (x * y) / (s * s);
  
    return ans;
}
  
// Driver Code
public static void main (String[] args) 
{
    int m = 385, n = 60;
  
    // Call the function
    // NumberOfSquares
    System.out.println(NumberOfSquares(m, n));
}
}
  
// This code is contributed by anuj_67.

Python3

# Python3 code for calculating 
# the number of squares
  
# Recursive function to
# return gcd of a and b
def __gcd(a, b):
      
    # Everything divides 0 
    if (a == 0 or b == 0):
        return 0;
  
    # base case
    if (a == b):
        return a;
  
    # a is greater
    if (a > b):
        return __gcd(a - b, b);
    return __gcd(a, b - a);
  
# Function to find 
# number of squares
def NumberOfSquares(x, y):
      
    # Here in built PHP
    # gcd function is used
    s = __gcd(x, y);
  
    ans = (x * y) / (s * s);
  
    return int(ans);
  
# Driver Code
m = 385;
n = 60;
  
# Call the function
# NumberOfSquares
print(NumberOfSquares(m, n));
  
# This code is contributed 
# by mit

C#

// C# code for calculating 
// the number of squares
using System;
  
class GFG
{
      
    // Recursive function to
    // return gcd of a and b
    static int __gcd(int a, int b)
    {
        // Everything divides 0 
        if (a == 0 || b == 0)
        return 0;
      
        // base case
        if (a == b)
            return a;
      
        // a is greater
        if (a > b)
            return __gcd(a - b, b);
        return __gcd(a, b - a);
    } 
  
  
// Function to find 
// number of squares
static int NumberOfSquares(int x, 
                           int y)
{
    // Here in built c++ 
    // gcd function is used
    int s = __gcd(x, y);
  
    int ans = (x * y) / 
              (s * s);
  
    return ans;
}
  
// Driver Code
static public void Main ()
{
int m = 385, n = 60;
  
// Call the function
// NumberOfSquares
Console.WriteLine(NumberOfSquares(m, n));
}
}
  
// This code is contributed by ajit

PHP

 $b)
        return __gcd($a - $b, $b);
    return __gcd($a, $b - $a);
} 
  
// Function to find 
// number of squares
function NumberOfSquares($x, $y) 
{
    // Here in built PHP
    // gcd function is used
    $s = __gcd($x, $y);
  
    $ans = ($x * $y) / 
           ($s * $s);
  
    return $ans;
}
  
// Driver Code
$m = 385;
$n = 60;
  
// Call the function
// NumberOfSquares
echo (NumberOfSquares($m, $n));
  
// This code is contributed 
// by akt_mit
?>

输出：

924