N个人共同完成工作所需的最短时间

鉴于没有。 N个人的工作时间的总和，以完成某项工作。任务是找到所有一起工作时将花费的小时数。
例子：

Input: n = 2, a = 6.0, b = 3.0
Output: 2 Hours

Input: n = 3, a = 6.0, b = 3.0, c = 4.0
Output: 1.33333 Hours

解决方案：

• 如果某人可以在“ n”天内完成一件工作，那么在一天之内，该人将进行“ 1 / n”工作。
• 同样，如果一个人可以在’m’天内完成一件工作，那么在一天之内，该人将执行’1 / m’工作。
• 很快…。对于其他人。

因此，由N个人在1天内完成的总工作量为

1/n + 1/m + 1/p…… + 1/z 
Where n, m, p ….., z are the number of days taken by each person respectively.
The result of the above expression will be the part of work done by all person together in 1 day, let’s say a / b. 
To calculate the time taken to complete the whole work will be b / a.

考虑两个人的例子：

Time taken by 1st person to complete a work = 6 hours
Time taken by 2nd person to complete the same work = 2 hours

Work done by 1st person in 1 hour = 1/6
Work done by 2nd person in 1 hour = 1/2
So, total work done by them in 1 hour is
=> 1 / 6 + 1/ 2 
=> (2 + 6) / (2 * 6)
=> 8 / 12

So, to complete the whole work, the time taken will be 12/8.

C++

// C++ implementation of above approach
#include 
using namespace std;
 
// Function to calculate the time
float calTime(float arr[], int n)
{
 
    float work = 0;
    for (int i = 0; i < n; i++)
        work += 1 / arr[i];
 
    return 1 / work;
}
 
// Driver Code
int main()
{
    float arr[] = { 6.0, 3.0, 4.0 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << calTime(arr, n) << " Hours";
 
    return 0;
}

Java

// Java implementation
// of above approach
import java.io.*;
 
class GFG
{
     
// Function to calculate the time
static double calTime(double arr[], int n)
{
    double work = 0;
    for (int i = 0; i < n; i++)
        work += 1 / arr[i];
 
    return 1 / work;
}
 
// Driver Code
public static void main (String[] args)
{
    double arr[] = { 6.0, 3.0, 4.0 };
    int n = arr.length;
     
    System.out.println(calTime(arr, n) +
                              " Hours");
}
}
 
// This code is contributed
// by inder_verma.

Python3

# Python3 implementation of
# above approach
 
# Function to calculate the time
def calTime(arr, n):
 
    work = 0
    for i in range(n):
        work += 1 / arr[i]
 
    return 1 / work
     
# Driver Code
arr = [ 6.0, 3.0, 4.0 ]
n = len(arr)
 
print(calTime(arr, n), "Hours")
 
# This code is contributed
# by Sanjit_Prasad

C#

// C# implementation
// of above approach
using System;
class GFG
{
     
// Function to calculate the time
static double calTime(double []arr,
                      int n)
{
    double work = 0;
    for (int i = 0; i < n; i++)
        work += 1 / arr[i];
 
    return Math.Round(1 / work, 5);
}
 
// Driver Code
public static void Main ()
{
    double []arr = { 6.0, 3.0, 4.0 };
    int n = arr.Length;
     
    Console.Write(calTime(arr, n) +
                         " Hours");
}
}
 
// This code is contributed by Smitha

Javascript

// javascript implementation
// of above approach
 
    // Function to calculate the time
    function calTime(arr , n)
    {
        var work = 0;
        for (i = 0; i < n; i++)
            work += 1 / arr[i];
 
        return 1 / work;
    }
 
    // Driver Code
     
    var arr = [ 6.0, 3.0, 4.0 ];
    var n = arr.length;
 
    document.write(calTime(arr, n).toFixed(5) + " Hours");
 
// This code is contributed by Rajput-Ji.

输出：

1.33333 Hours

注意：这里的输入数组包含小时，可以是几天，几分钟……等等。