HDOJ 2289 Cup

最新推荐文章于 2018-11-16 09:30:00 发布

原创最新推荐文章于 2018-11-16 09:30:00 发布 · 340 阅读

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本文介绍了一种解决特定形状容器内液体高度计算的问题。利用数学原理，通过已知的容器顶部半径、底部半径、容器总高度及液体体积，采用二分查找法精确计算出容器中液体的实际高度。

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Cup

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

题目大意：有一个倒置圆台状的容器（类似于碗），里面有一些水，目前知道谁的体积、顶部半径、底部半径和整个容器的高，求里面水的高度。

解题思路：由数学知识可知，圆台的体积V = πh(R ^ 2 + R * r + r ^ 2) / 3，再根据相似三角形可以推出水面半径为r + h * (R - r) / H，在0到H之间二分h，验证二分得到的体积与输入的体积是否相等即可。这道题二分半径的话应该会WA，还有读题要仔细，这个圆台是倒着放的，我一直以为是正着放的，WA了好多次，以后一定要注意理解题意！

代码如下：

#include <cstdio>
#include <cmath>

#define PI acos(-1.0)

#define EPS 1e-10

double r,R,H,V;

double getV(double ur,double h)
{
	return PI * h * (ur * ur + ur * r + r * r) / 3.0;
}



int main()
{
	int t;
	scanf("%d",&t);
	double lb,ub,mid;
	while(t--){
		scanf("%lf %lf %lf %lf",&r,&R,&H,&V);
		lb = 0,ub = 100;
		while(ub - lb > EPS){
			mid = (lb + ub) / 2.0;
			if(getV(r + mid * (R - r) / H,mid) > V)
                ub = mid;
            else
                lb = mid;
		}
		printf("%.6f\n",mid);
	}
	return 0;
}