Hdoj 2289.Cup 题解

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

Source

The 4th Baidu Cup final

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思路

高精度二分的题目，注意圆台的体积公式为\(V=\frac{1}{3}\pi h(R^2+r^2+Rr)\)

不断二分查找最接近的h就好了，详见代码

代码

#include<bits/stdc++.h>
#define M_PI 3.14159265358979323846
using namespace std;
double r,R,H,V;
const double eps = 1e-8;
double cal(double h)
{
	double t = r + (R-r)*h/H;//求出h高度对应的截面的圆的半径
	return M_PI*h*(t*t + r*r + t*r)/3;
}
int main()
{
	int t;
	cin >> t;
	while(t--)
	{
		cin >> r >> R >> H >> V;
		double ans = 0;
		double l = 0, r = 100.0;
		double mid;
		while(r-l>=eps)
		{
			mid = (l+r)/2;
			if(cal(mid)<V)
				l = mid;
			else
				r = mid;
		}
		printf("%.6lf\n",l);
	}
	return 0;
}