矩形中最大面积的正方形数

给定一个边为 m 和 n 的矩形。将矩形切成相同的小块，使每块都是一个正方形，具有最大可能的边长，没有矩形的剩余部分。打印形成的此类正方形的数量。
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：任务是将矩形切成正方形，边长为 s，没有剩余的矩形部分，因此s必须同时划分m和n 。此外，正方形的边应该是最大可能的，因此，s 应该是 m 和 n 的最大公约数。
所以， s = gcd(m, n) 。
要找到矩形被切割成的正方形数，要完成的任务是将矩形的面积除以大小为 s 的正方形面积。

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
?>

Javascript

输出：