给定总共N + 1条管道所需的时间，其中N条管道用于填充水箱，而一条管道用于清空水箱。任务是计算如果同时打开所有N + 1个管道，水箱将被充满的时间。
例子：
Input: n = 2,
pipe1 = 12 hours, pipe2 = 14 hours,
emptypipe = 30 hours
Output: 8 hours
Input: n = 1,
pipe1 = 12 hours
emptypipe = 18 hours
Output: 36 hours 方法：
- 如果pipe1可以在’n’小时内填充一个水箱，则在1小时内，pipe1将能够填充’1 / n’水箱。
- 同样，如果pipe2可以在’m’小时内填充一个水箱，那么在一小时内，pipe2将能够填充’1 / m’水箱。
- 很快…。用于其他管道。
因此，在1小时内用N条管道填充储水池的总工作量为
1/n + 1/m + 1/p…… + 1/z
Where n, m, p ….., z are the number of hours taken by each pipes respectively.
The result of the above expression will be the part of work done by all pipes together in 1 hours, let’s say a / b.
To calculate the time taken to fill the cistern will be b / a.
考虑两个管道的示例：
Time taken by 1st pipe to fill the cistern = 12 hours
Time taken by 2nd pipe to fill the cistern = 14 hours
Time taken by 3rd pipe to empty the cistern = 30 hours
Work done by 1st pipe in 1 hour = 1/12
Work done by 2nd pipe in 1 hour = 1/14
Work done by 3nd pipe in 1 hour = – (1/30) as it empty the pipe.
So, total work done by all the pipes in 1 hour is
=> ( 1 / 12 + 1/ 14 ) – (1 / 30)
=> ((7 + 6 ) / (84)) – (1 / 30)
=> ((13) / (84)) – (1 / 30)
=> 51 / 420
So, to Fill the cistern time required will be 420 / 51 i.e 8 hours Approx.
下面是上述方法的实现：
C++
// C++ implementation of above approach
#include
using namespace std;
// Function to calculate the time
float Time(float arr[], int n, int Emptypipe)
{
float fill = 0;
for (int i = 0; i < n; i++)
fill += 1 / arr[i];
fill = fill - (1 / (float)Emptypipe);
return 1 / fill;
}
// Driver Code
int main()
{
float arr[] = { 12, 14 };
float Emptypipe = 30;
int n = sizeof(arr) / sizeof(arr[0]);
cout << floor(Time(arr, n, Emptypipe)) << " Hours";
return 0;
} Java
// Java implementation of
// above approach
import java.io.*;
class GFG
{
// Function to calculate the time
static float Time(float arr[], int n,
float Emptypipe)
{
float fill = 0;
for (int i = 0; i < n; i++)
fill += 1 / arr[i];
fill = fill - (1 / (float)Emptypipe);
return 1 / fill;
}
// Driver Code
public static void main (String[] args)
{
float arr[] = { 12, 14 };
float Emptypipe = 30;
int n = arr.length;
System.out.println((int)(Time(arr, n,
Emptypipe)) + " Hours");
}
}
// This code is contributed
// by inder_verma.Python3
# Python3 implementation of
# above approach
# Function to calculate the time
def Time(arr, n, Emptypipe) :
fill = 0
for i in range(0,n) :
fill += (1 / arr[i])
fill = fill - (1 / float(Emptypipe))
return int(1 / fill)
# Driver Code
if __name__=='__main__':
arr = [ 12, 14 ]
Emptypipe = 30
n = len(arr)
print((Time(arr, n, Emptypipe))
, "Hours")
# This code is contributed by
# Smitha Dinesh SemwalC#
// C# implementation of
// above approach
using System;
class GFG
{
// Function to calculate the time
static float Time(float []arr, int n,
float Emptypipe)
{
float fill = 0;
for (int i = 0; i < n; i++)
fill += 1 / arr[i];
fill = fill - (1 / (float)Emptypipe);
return 1 / fill;
}
// Driver Code
public static void Main ()
{
float []arr = { 12, 14 };
float Emptypipe = 30;
int n = arr.Length;
Console.WriteLine((int)(Time(arr, n,
Emptypipe)) +
" Hours");
}
}
// This code is contributed
// by inder_verma.PHP
Javascript
8 Hours