Programa para encontrar la cabeza de salida en una carrera

Dada la ventaja que A da a B y C en una carrera de 100 metros. La tarea es encontrar la ventaja inicial que B puede darle a C en la misma carrera.

Ejemplos: 

Input: B = 10 meters, C = 28 meters
Output: 20 meters
B can give C a start of 20 meters.

Input: B = 20 meters, C = 50 meters
Output: 62 meters
B can give C a start of 62 meters.

Acercarse:  

Total de metros en una carrera = 100 metros. 
A está delante de B por 10 metros. Cuando A completó sus 100 metros B completó sus 90 metros. 
De manera similar, A está por delante de C por 28 metros. Cuando A completó sus 100 metros C completó sus 72 metros. 
Ahora, cuando B completó sus 90 metros, C completó sus 72 metros. 
Entonces, cuando B completó sus 100 metros, C completó sus 80 metros. 
–> (( C * 100) / B) 
–> (( 72 * 100) / 90) es decir, 80 metros 
Entonces B puede darle a C un comienzo de 20 metros 
 

A continuación se muestra la implementación del enfoque anterior: 

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the B start to C
int Race(int B, int C)
{
    int result = 0;
 
    // When B completed it's 100 meter
    // then Completed meters by C is
    result = ((C * 100) / B);
 
    return 100 - result;
}
 
// Driver Code.
int main()
{
    int B = 10, C = 28;
 
    // When A completed it's 100 meter
    // Then completed meters of B and C is
    B = 100 - B;
    C = 100 - C;
 
    cout << Race(B, C) << " meters";
 
return 0;
}

// Java implementation of above approach
public class GFG
{
     
// Function to find the B start to C
static int Race(int B, int C)
{
    int result = 0;
 
    // When B completed it's 100 meter
    // then Completed meters by C is
    result = ((C * 100) / B);
 
    return 100 - result;
}
 
// Driver Code
public static void main(String[] args)
{
    int B = 10;
    int C = 28;
     
    // When A completed it's 100 meter
    // Then completed meters of B and C is
    B = 100 - B;
    C = 100 - C;
     
    System.out.println(Race(B, C) + " meters");
}
}
 
// This code is contributed
// by ChitraNayal

# Python 3 implementation
# of above approach
 
# Function to find the
# B start to C
def Race(B, C):
 
    result = 0;
 
    # When B completed it's 100 meter
    # then Completed meters by C is
    result = ((C * 100) // B)
 
    return 100 - result
 
# Driver Code
if __name__ == "__main__":
    B = 10
    C = 28
     
    # When A completed it's 100 meter
    # Then completed meters of B and C is
    B = 100 - B;
    C = 100 - C;
     
    print(str(Race(B, C)) + " meters")
 
# This code is contributed
# by ChitraNayal

// C# implementation of above approach
using System;
class GFG
{
 
// Function to find the B start to C
static int Race(int B, int C)
{
    int result = 0;
 
    // When B completed it's 100 meter
    // then Completed meters by C is
    result = ((C * 100) / B);
 
    return 100 - result;
}
 
// Driver Code
public static void Main()
{
    int B = 10;
    int C = 28;
     
    // When A completed it's 100 meter
    // Then completed meters of B and C is
    B = 100 - B;
    C = 100 - C;
     
    Console.Write(Race(B, C) + " meters");
}
}
 
// This code is contributed
// by ChitraNayal

<?php
// PHP implementation of above approach
 
// Function to find the B start to C
function Race($B, $C)
{
    $result = 0;
 
    // When B completed it's 100 meter
    // then Completed meters by C is
    $result = (($C * 100) / $B);
 
    return 100 - $result;
}
 
// Driver Code
$B = 10;
$C = 28;
 
// When A completed it's 100 meter
// Then completed meters of B and C is
$B = 100 - $B;
$C = 100 - $C;
 
echo Race($B, $C) . " meters";
 
// This code is contributed
// by ChitraNayal
?>

<script>
 
// Javascript implementation of above approach
 
// Function to find the B start to C
function Race(B, C)
{
    var result = 0;
     
    // When B completed it's 100 meter
    // then Completed meters by C is
    result = ((C * 100) / B);
 
    return 100 - result;
}
 
// Driver Code
var B = 10, C = 28;
 
// When A completed it's 100 meter
// Then completed meters of B and C is
B = 100 - B;
C = 100 - C;
 
document.write(Race(B, C) + " meters");
 
// This code is contributed by itsok
 
</script>

20 meters