Program to find the number of men initially

A certain number of men can do a certain piece of work in D days. If there were m more men engaged in the work then the work can be done in d days less. The task is to find how many men were there initially.
Examples:

Input: D = 5, m = 4, d = 4 
Output: 1
Input: D = 180, m = 30, d = 20 
Output: 240 

Approach: Let the initial number of men be M and days be D
Amount of work completed M men in D days will be M * D
i.e. Work Done = M * D …(1) 
If there are M + m men then the same amount of work is completed in D – d days. 
i.e. Work Done = (M + m) * (D – d) …(2) 
Equating equations 1 and 2, 

M * D = (M + m) * (D – d) 
M * D = M * (D – d) + m * (D – d) 
M * D – M * (D – d) = m * (D – d) 
M * (D – (D – d)) = m * (D – d) 
M = m * (D – d) / d

Below is the implementation of the above approach: 

C++

// C++ implementation of above approach.
#include <bits/stdc++.h>
using namespace std;

// Function to return the
// number of men initially
int numberOfMen(int D, int m, int d)
{

    int Men = (m * (D - d)) / d;

    return Men;
}

// Driver code
int main()
{
    int D = 5, m = 4, d = 4;

    cout << numberOfMen(D, m, d);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

// Function to return the
// number of men initially
static int numberOfMen(int D, int m, int d)
{

    int Men = (m * (D - d)) / d;

    return Men;
}

// Driver code
public static void main(String args[])
{
    int D = 5, m = 4, d = 4;

    System.out.println(numberOfMen(D, m, d));

}
}
// This code is contributed by Arnab Kundu

Python3

# Python implementation of above approach.

# Function to return the
# number of men initially
def numberOfMen(D, m, d):
    Men = (m * (D - d)) / d;

    return int(Men);


# Driver code
D = 5; m = 4; d = 4;

print(numberOfMen(D, m, d));


# This code contributed by Rajput-Ji

C#

//  C# implementation of the approach
using System;

class GFG
{

// Function to return the
// number of men initially
static int numberOfMen(int D, int m, int d)
{

    int Men = (m * (D - d)) / d;

    return Men;
}

// Driver code
public static void Main()
{
    int D = 5, m = 4, d = 4;

    Console.WriteLine(numberOfMen(D, m, d));

}
}

// This code is contributed by anuj_67..

Javascript

<script>

// Javascript implementation of above approach.

// Function to return the
// number of men initially
function numberOfMen(D, m, d)
{

    var Men = (m * (D - d)) / d;

    return Men;
}

// Driver code
var D = 5, m = 4, d = 4;
document.write(numberOfMen(D, m, d));

</script>

Output:

1

Time Complexity: O(1)

Auxiliary Space: O(1)