Programa para resolver el Problema de Alligation

Escribe un programa para encontrar la proporción en la que un comerciante mezclará dos tipos de arroz por valor de Rs. kg y Rs. kg, por lo que el costo promedio de la mezcla es Rs. kg.
Ejemplos : 

Input : X = 50, Y = 70, Z = 65
Output : Ratio = 1:3

Input : X = 1000, Y = 2000, Z = 1400
Output : Ratio = 3:2

De acuerdo con la regla de Alligation , la relación de los pesos de dos elementos mezclados será inversamente proporcional a la desviación de los atributos de estos dos elementos del atributo promedio de la mezcla resultante.
 

w1 / w2 = (d - m) / (m - c)

El siguiente programa ilustra el enfoque anterior:
 

#include <bits/stdc++.h>
using namespace std;
 
// Function to find the ratio of two mixtures
void alligation(float x, float y, float m)
{
    // Find the cheaper among x and y
    float c = (x <= y) ? x : y;
    // Find the dearer among x and y
    float d = (x >= y) ? x : y;
 
    // Find ratio r1:r2
    int r1 = d - m;
    int r2 = m - c;
 
    // Convert the ration into simpler form
    int gcd = __gcd(r1, r2);
 
    cout << r1 / gcd << ":" << r2 / gcd;
}
 
// Driver code
int main()
{
    float x, y, z;
    x = 50;
    y = 70;
    z = 65;
 
    alligation(x, y, z);
 
    return 0;
}

// Java implementation of the
// above approach.
import java.util.*;
 
class solution
{
 
static float __gcd(float a, float b)
{
    float dividend,divisor;
     
    // a is greater or equal to b
    if(a>=b)
    dividend = a;
    else
    dividend = b;
     
    // b is greater or equal to a
    if(a<=b)
    divisor = a;
    else
    divisor = b;
     
while(divisor>0)
{
float remainder = dividend % divisor;
dividend = divisor;
divisor = remainder;
 
}
return dividend;
}
// Function to find the ratio of two mixtures
static void alligation(float x, float y, float m)
{
    // Find the cheaper among x and y
    float c;
    if (x <= y)
    c = x;
    else
    c = y;
    // Find the dearer among x and y
    float d ;
    if (x >= y)
    d = x;
    else
    d = y;
 
    // Find ratio r1:r2
    float r1 = d - m;
    float r2 = m - c;
 
    // Convert the ration into simpler form
    float gcd = __gcd(r1, r2);
 
    System.out.println((int)(r1 / gcd)+":"+(int)(r2 / gcd));
}
 
// Driver code
public static void main(String args[])
{
    float x, y, z;
    x = 50;
    y = 70;
    z = 65;
 
    alligation(x, y, z);
}
}
 
// This code is contributed by
// Shashank_sharma

# Python 3 implementation of the
# above approach.
from math import gcd
 
# Function to find the ratio
# of two mixtures
def alligation(x, y, m):
     
    # Find the cheaper among x and y
    if (x <= y):
        c = x
    else:
        c = y
         
    # Find the dearer among x and y
    if (x >= y):
        d = x
    else:
        d = y
 
    # Find ratio r1:r2
    r1 = d - m
    r2 = m - c
 
    # Convert the ration into simpler form
    __gcd = gcd(r1, r2)
 
    print(r1 // __gcd, ":", r2 // __gcd)
 
# Driver code
if __name__ == '__main__':
    x = 50
    y = 70
    z = 65
 
    alligation(x, y, z)
 
# This code is contributed by
# Surendra_Gangwar

// C# implementation of the
// 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 of
    // two mixtures
    static void alligation(float x,
                           float y, float m)
    {
        // Find the cheaper among x and y
        float c = (x <= y) ? x : y;
         
        // Find the dearer among x and y
        float d = (x >= y) ? x : y;
     
        // Find ratio r1:r2
        int r1 = (int)(d - m);
        int r2 = (int)(m - c);
     
        // Convert the ration into
        // simpler form
        int gcd = __gcd(r1, r2);
     
        Console.Write(r1 / gcd + ":" +
                      r2 / gcd);
    }
     
    // Driver code
    public static void Main()
    {
        float x, y, z;
        x = 50;
        y = 70;
        z = 65;
     
        alligation(x, y, z);
    }
}
 
// This code is contributed
// by Akanksha Rai

<?php
// PHP implementation of the
// above approach.
function __gcd($a, $b)
{
    $dividend; $divisor;
     
    // a is greater or equal to b
    if($a >= $b)
        $dividend = $a;
    else
        $dividend = $b;
     
    // b is greater or equal to a
    if($a <= $b)
        $divisor = $a;
    else
        $divisor = $b;
     
    while($divisor > 0)
    {
        $remainder = $dividend % $divisor;
        $dividend = $divisor;
        $divisor = $remainder;
    }
    return $dividend;
}
 
// Function to find the ratio of
// two mixtures
function alligation($x, $y, $m)
{
    // Find the cheaper among x and y
    if ($x <= $y)
        $c = $x;
    else
        $c = $y;
     
    // Find the dearer among x and y
    if ($x >= $y)
        $d = $x;
    else
        $d = $y;
 
    // Find ratio r1:r2
    $r1 = $d - $m;
    $r2 = $m - $c;
 
    // Convert the ration into
    // simpler form
    $gcd = __gcd($r1, $r2);
 
    echo (int)($r1 / $gcd) . ":" .
         (int)($r2 / $gcd);
}
 
// Driver code
$x = 50;
$y = 70;
$z = 65;
 
alligation($x, $y, $z);
 
// This code is contributed by
// Mukul Singh
?>

<script>
    // Javascript implementation of the 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 of
    // two mixtures
    function alligation(x, y, m)
    {
        // Find the cheaper among x and y
        let c = (x <= y) ? x : y;
           
        // Find the dearer among x and y
        let d = (x >= y) ? x : y;
       
        // Find ratio r1:r2
        let r1 = (d - m);
        let r2 = (m - c);
       
        // Convert the ration into
        // simpler form
        let gcd = __gcd(r1, r2);
       
        document.write(parseInt(r1 / gcd, 10) + ":" + parseInt(r2 / gcd, 10));
    }
     
    let x, y, z;
    x = 50;
    y = 70;
    z = 65;
 
    alligation(x, y, z);
 
// This code is contributed by mukesh07.
</script>

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