找出给定饮酒量时杯子是否为空

给定一个直径为D厘米的圆柱玻璃。玻璃杯中的初始水位等于距底部H厘米。你以每秒M毫升的速度喝水。但是如果你不喝玻璃杯里的水，水位会以每秒N厘米的速度增加。任务是找出清空玻璃杯所需的时间，或者找出是否可以清空玻璃杯。
例子：

Input: D = 1, H = 1, M = 1, N = 1
Output: 3.65979
Input: D = 1, H = 2, M = 3, N = 100
Output: -1

方法：这是一道几何题。众所周知，玻璃的面积是pie * r ² ，其中r表示半径即(D / 2) 。因此，要找到每秒消耗水的速率，请将给定的体积（已知 1 毫升等于 1 立方厘米）除以面积。
如果该值小于将水倒入玻璃杯中的速度，如果没有喝醉，则答案为否，否则玻璃杯可能是空的。
要找到时间，除以h / (v / (pie * r ² ) – e) ，这就是杯子变空的时间。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
double pie = 3.1415926535897;
 
// Function to return the time when
// the glass will be empty
double findsolution(double d, double h,
                    double m, double n)
{
    double k = (4 * m) / (pie * d * d);
 
    // Check the condition when the
    // glass will never be empty
    if (n > k)
        return -1;
 
    // Find the time
    double ans = (h / (k - n));
    return ans;
}
 
// Driver code
int main()
{
    double d = 1, h = 1, m = 1, n = 1;
 
    cout << findsolution(d, h, m, n);
 
    return 0;
}

Java

// Java implementation of the approach
 
class GFG
{
 
static double pie = 3.1415926535897;
 
// Function to return the time when
// the glass will be empty
static double findsolution(double d, double h,
                    double m, double n)
{
    double k = (4 * m) / (pie * d * d);
 
    // Check the condition when the
    // glass will never be empty
    if (n > k)
        return -1;
 
    // Find the time
    double ans = (h / (k - n));
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    double d = 1, h = 1, m = 1, n = 1;
 
    System.out.printf("%.5f",findsolution(d, h, m, n));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
pie = 3.1415926535897
 
# Function to return the time when
# the glass will be empty
def findsolution(d, h, m, n):
 
    k = (4 * m) / (pie * d * d)
 
    # Check the condition when the
    # glass will never be empty
    if (n > k):
        return -1
 
    # Find the time
    ans = (h / (k - n))
    return round(ans, 5)
 
# Driver code
d = 1
h = 1
m = 1
n = 1
 
print(findsolution(d, h, m, n))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
static double pie = 3.1415926535897;
 
// Function to return the time when
// the glass will be empty
static double findsolution(double d, double h,
                           double m, double n)
{
    double k = (4 * m) / (pie * d * d);
 
    // Check the condition when the
    // glass will never be empty
    if (n > k)
        return -1;
 
    // Find the time
    double ans = (h / (k - n));
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    double d = 1, h = 1, m = 1, n = 1;
 
    Console.Write("{0:F5}",
            findsolution(d, h, m, n));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

输出：

3.65979

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