Given a container that has X liters of milk. Y liters of milk is drawn out and replaced with Y liters of water. This operation is done Z times. The task is to find out the quantity of milk left in the container.
Examples:
Input: X = 10 liters, Y = 2 liters, Z = 2 times Output: 6.4 liters Input: X = 25 liters, Y = 6 liters, Z = 3 times Output: 10.97 liters
Formula:-
Below is the required implementation:
C++
// C++ implementation using above formula #include <bits/stdc++.h> using namespace std;
// Function to calculate the Remaining amount. float Mixture( int X, int Y, int Z)
{ float result = 0.0, result1 = 0.0;
// calculate Right hand Side(RHS).
result1 = ((X - Y) / ( float )X);
result = pow (result1, Z);
// calculate Amount left by
// multiply it with original value.
result = result * X;
return result;
} // Driver Code int main()
{ int X = 10, Y = 2, Z = 2;
cout << Mixture(X, Y, Z) << " litres" ;
return 0;
} |
Java
// Java code using above Formula. import java.io.*;
class GFG
{ // Function to calculate the // Remaining amount. static double Mixture( int X, int Y, int Z)
{ double result1 = 0.0 , result = 0.0 ;
// calculate Right hand Side(RHS).
result1 = ((X - Y) / ( float )X);
result = Math.pow(result1, Z);
// calculate Amount left by
// multiply it with original value.
result = result * X;
return result;
} // Driver Code public static void main(String[] args)
{ int X = 10 , Y = 2 , Z = 2 ;
System.out.println(( float )Mixture(X, Y, Z) +
" litres" );
} } // This code is contributed // by Naman_Garg |
Python 3
# Python 3 implementation using # above formula # Function to calculate the # Remaining amount. def Mixture(X, Y, Z):
result = 0.0
result1 = 0.0
# calculate Right hand Side(RHS).
result1 = ((X - Y) / X)
result = pow (result1, Z)
# calculate Amount left by
# multiply it with original value.
result = result * X
return result
# Driver Code if __name__ = = "__main__" :
X = 10
Y = 2
Z = 2
print ( "{:.1f}" . format (Mixture(X, Y, Z)) +
" litres" )
# This code is contributed by ChitraNayal |
C#
// C# code using above Formula. using System;
class GFG
{ // Function to calculate the // Remaining amount. static double Mixture( int X,
int Y, int Z)
{ double result1 = 0.0, result = 0.0;
// calculate Right hand Side(RHS).
result1 = ((X - Y) / ( float )X);
result = Math.Pow(result1, Z);
// calculate Amount left by
// multiply it with original value.
result = result * X;
return result;
} // Driver Code public static void Main()
{ int X = 10, Y = 2, Z = 2;
Console.WriteLine(( float )Mixture(X, Y, Z) +
" litres" );
} } // This code is contributed // by Akanksha Rai(Abby_akku) |
PHP
<?php // PHP implementation of above formula // Function to calculate the // Remaining amount. function Mixture( $X , $Y , $Z )
{ $result = 0.0;
$result1 = 0.0;
// calculate Right hand Side(RHS).
$result1 = (( $X - $Y ) / $X );
$result = pow( $result1 , $Z );
// calculate Amount left by
// multiply it with original value.
$result = $result * $X ;
return $result ;
} // Driver Code $X = 10;
$Y = 2;
$Z = 2;
echo Mixture( $X , $Y , $Z ), " litres" ;
// This code is contributed // by Sanjit_Prasad ?> |
Javascript
<script> // Javascript implementation using above formula // Function to calculate the Remaining amount. function Mixture(X, Y, Z)
{ var result = 0.0, result1 = 0.0;
// calculate Right hand Side(RHS).
result1 = ((X - Y) / X);
result = Math.pow(result1, Z);
// calculate Amount left by
// multiply it with original value.
result = result * X;
return result;
} // Driver Code var X = 10, Y = 2, Z = 2;
document.write( Mixture(X, Y, Z).toFixed(1) + " litres" );
</script> |
Output:
6.4 litres
Time Complexity: O(log(n))
Auxiliary Space: O(1)