HDU 2289 Cup （二分搜索）

最新推荐文章于 2021-05-31 16:14:06 发布

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本博客介绍如何利用给定的圆台型杯子的上底和下底半径、高度以及水的体积，计算水的高度。通过数学推导和编程实现，详细解释了求解过程。

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Cup

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7016 Accepted Submission(s): 2158

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

题意：

题目大意：一个圆台型的杯子，告诉杯子的上底和下底的半径、杯子的高度，以及水的体积，要求水的高度。

/*
水平面的半径u可以根据直角梯形的面积求出：
(r+u)*h + (u+R)*(H-h) = (r+R)*H
得出(R-r)*h = (u-r)*H
得 u = r + (R-r)*h/H
圆台体积计算公式：V= π*h*( R^2 + R*u + u^2 ) / 3
*/

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>

using namespace std;

#define exp 1e-8  ///精确值（10-^8）
double PI= acos(-1);///圆周率的精确值
double r,R,H,V;
double gent(double x)
{
    double u=(R-r)*x/H+r;///水面的半径
    return PI*x*(pow(r,2)+pow(u,2)+r*u)/3;///水的体积
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf%lf%lf",&r,&R,&H,&V);
        double left=0;
        double right=H;
        double mid;
        ///while(right-left>=exp)///终止条件
       for(int i=0;i<100;i++) ///终止条件也可以用100次循环（可达到10-^30负精确）
        {
            mid=(left+right)/2;
            if(gent(mid)-V>exp)///以中间高度求得的体积如果小于V则高度在左侧
                right=mid;
            else
                left=mid; ///否则在右侧
        }

        printf("%.6lf\n",mid);

    }
    return 0;
}