hdu cup

Cup

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 163 Accepted Submission(s): 57

Problem Description

The WHU ACM Team has a big cup, with which every member drinks water. Now, we know the volume of the water in the cup, can you tell us it height?

The radius of the cup's top and bottom circle is known, the cup's height is also known.

 

Input

The input consists of several test cases. The first line of input contains an integer T, indicating the num of test cases.
Each test case is on a single line, and it consists of four floating point numbers: r, R, H, V, representing the bottom radius, the top radius, the height and the volume of the hot water.

Technical Specification

1. T ≤ 20.
2. 1 ≤ r, R, H ≤ 100; 0 ≤ V ≤ 1000,000,000.
3. r ≤ R.
4. r, R, H, V are separated by ONE whitespace.
5. There is NO empty line between two neighboring cases.

 

Output

For each test case, output the height of hot water on a single line. Please round it to six fractional digits.

 

Sample Input

1
100 100 100 3141562

 

Sample Output

99.999024

本题可以用数学解法，也可以用2分搜索，但用数学方法解效率更高。

数学方法：cup的形状可能有两种，圆柱型，或是上大下小的圆台型。如果是圆柱形则直接用体积除以底面积即得出高；

如果是圆台型，可以将圆台下边补全成为一个圆锥体，求出补全那部分的体积，按照比例关系体积之比是高之比的3次方。

公式：

小圆锥的高：h0 = H*r/(R-r);//补全的那个小圆锥的高

小圆锥的体积：v0 = (pi*h0*r*r)/3;//补全的那个小圆锥的体积

根据公式：((v0+v)/v0)^(1/3) = (h+h0)/h0
 求出 h .

注意，如果水的体积大于杯子的体积那么求出的h将大于H，此时最大的h应该为杯子子的高度。

 

#include <iostream>
#include <stdio.h>
#include <math.h>
using namespace std;
double pi = acos(-1.0);

//知识点：圆台的体积公式V=pi*（r*r+r*R+R*R)*h/3
int main()
{
    int t;
    double r,R,H,V,h,h0,v0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf%lf%lf",&r,&R,&H,&V);
        if(r==R)
        {
            h = V/(pi*r*r);
        }
        else if(r<R)
        {
            h0 = H*r/(R-r);//补全的那个小圆锥的高
            v0 = (pi*h0*r*r)/3;//补全的那个小圆锥的体积
            h = h0*(pow((v0+V)/v0,1.0/3)-1);//公式原形：((v0+v)/v0)^(1/3) = (h+h0)/h0
        }
         if(h>H)h = H;
         printf("%.6f\n",h);
    }
    return 0;
}