内接有矩形的矩形的面积比

给定两个矩形，X的长宽比为a：b ，Y的长宽比为c：d 。只要边的比例保持不变，就可以调整两个矩形的大小。任务是将第二个矩形放置在第一个矩形内，以使至少一个边相等，并且两个矩形的边重叠，并求出(第二个矩形所占据的空间):(第一个矩形所占的空间)的比率。
例子：

Input: a = 1, b = 1, c = 3, d = 2
Output: 2:3
The dimensions can be 3X3 and 3X2.

Input: a = 4, b = 3, c = 2, d = 2
Output: 3:4
The dimensions can be 4X3 and 3X3

方法：如果我们使矩形的一侧相等，则所需的比率将是另一侧的比率。
考虑2种情况：

• a * d
• b * c

由于将比率的两边相乘不会改变其值。首先尝试使a和c相等，可以通过将(a：b)乘以lcm / a和(c：d)乘以lcm / c使其等于lcm。乘法之后，(b：d)的比率将是必需的答案。可以通过将b和d除以gcd(b，d)来降低该比率。
下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
 
// Function to find the ratio
void printRatio(int a, int b, int c, int d)
{
    if (b * c > a * d) {
        swap(c, d);
        swap(a, b);
    }
 
    // LCM of numerators
    int lcm = (a * c) / __gcd(a, c);
 
    int x = lcm / a;
    b *= x;
 
    int y = lcm / c;
    d *= y;
 
    // Answer in reduced form
    int k = __gcd(b, d);
    b /= k;
    d /= k;
 
    cout << b << ":" << d;
}
 
// Driver code
int main()
{
    int a = 4, b = 3, c = 2, d = 2;
 
    printRatio(a, b, c, d);
 
    return 0;
}

Java

// Java implementation of above approach
 
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)
          return b;
        if (b == 0)
          return a;
        
        // 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 the ratio
 static void printRatio(int a, int b, int c, int d)
{
    if (b * c > a * d) {
        int temp = c;
        c =d;
        d =c;
        temp =a;
        a =b;
        b=temp;
     
    }
 
    // LCM of numerators
    int lcm = (a * c) / __gcd(a, c);
 
    int x = lcm / a;
    b *= x;
 
    int y = lcm / c;
    d *= y;
 
    // Answer in reduced form
    int k = __gcd(b, d);
    b /= k;
    d /= k;
 
    System.out.print( b + ":" + d);
}
 
    // Driver code
    public static void main (String[] args) {
        int a = 4, b = 3, c = 2, d = 2;
 
    printRatio(a, b, c, d);
    }
}
   
// This code is contributed by inder_verma..

Python3

import math
# Python3 implementation of above approach
 
# Function to find the ratio
def printRatio(a, b, c, d):
    if (b * c > a * d):
        swap(c, d)
        swap(a, b)
     
    # LCM of numerators
    lcm = (a * c) / math.gcd(a, c)
 
    x = lcm / a
    b = int(b * x)
 
    y = lcm / c
    d = int(d * y)
 
    # Answer in reduced form
    k = math.gcd(b,d)
    b =int(b / k)
    d = int(d / k)
 
    print(b,":",d)
 
# Driver code
if __name__ == '__main__':
    a = 4
    b = 3
    c = 2
    d = 2
 
    printRatio(a, b, c, d)
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of above approach
 
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)
        return b;
        if (b == 0)
        return a;
         
        // 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 the ratio
static void printRatio(int a, int b, int c, int d)
{
    if (b * c > a * d) {
        int temp = c;
        c =d;
        d =c;
        temp =a;
        a =b;
        b=temp;
     
    }
 
    // LCM of numerators
    int lcm = (a * c) / __gcd(a, c);
 
    int x = lcm / a;
    b *= x;
 
    int y = lcm / c;
    d *= y;
 
    // Answer in reduced form
    int k = __gcd(b, d);
    b /= k;
    d /= k;
 
    Console.WriteLine( b + ":" + d);
}
 
// Driver code
 
    public static void Main () {
        int a = 4, b = 3, c = 2, d = 2;
 
    printRatio(a, b, c, d);
    }
}
 
// This code is contributed by inder_verma..

PHP

 $b)
        return __gcd($a - $b, $b);
    return __gcd($a, $b - $a);
}
 
// Function to find the ratio
function printRatio($a, $b, $c, $d)
{
    if ($b * $c > $a * $d)
    {
        $temp = $c;
        $c = $d;
        $d = $c;
         
        $temp = $a;
        $a = $b;
        $b = $temp;
    }
     
    // LCM of numerators
    $lcm = ($a * $c) / __gcd($a, $c);
     
    $x = $lcm / $a;
    $b *= $x;
     
    $y = $lcm / $c;
    $d *= $y;
     
    // Answer in reduced form
    $k = __gcd($b, $d);
    $b /= $k;
    $d /= $k;
     
    echo $b . ":" . $d;
}
 
// Driver code
$a = 4; $b = 3; $c = 2; $d = 2;
 
printRatio($a, $b, $c, $d);
 
// This code is contributed
// by Akanksha Rai
?>

Javascript

// Javascript implementation of above approach
 
// Recursive function to return 
// gcd of a and b 
function __gcd(a, b) 
{ 
 
    // Everything divides 0 
    if (a == 0) 
        return b; 
    if (b == 0) 
        return a; 
       
    // 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 the ratio
function printRatio(a, b, c, d)
{
     if (b * c > a * d) 
    {
        temp = c;
        c = d;
        d = c;
           
        temp = a;
        a = b;
        b = temp;
    }
 
    // LCM of numerators
    let lcm = (a * c) / __gcd(a, c);
 
    let x = lcm / a;
    b *= x;
 
    let y = lcm / c;
    d *= y;
 
    // Answer in reduced form
    let k = __gcd(b, d);
    b /= k;
    d /= k;
 
    document.write(b + ":" + d);
}
 
// Driver code
 
    let a = 4, b = 3, c = 2, d = 2;
    printRatio(a, b, c, d);
 
 
// This code is contributed by Manoj
 

输出：

3:4