【hdu2289】Cup——二分

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分类专栏：题解心得二分文章标签： hdu 二分

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本文介绍了一道关于计算圆台中水高度的问题。通过给定的圆台底部和顶部半径、圆台高度及水体积，利用圆台体积公式和三角关系，采用二分查找法求解水的高度。

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题目：

Cup

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6303 Accepted Submission(s): 1994

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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描述：已知圆台的地面半径和高，里面水的体积，求水的高

题解：知道圆台体积公式和三角关系（得到高和底面半径的关系）就好办了，二分枚举就行。也可以推出来公式直接算

代码：

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define PI acos(-1.0)
#define exp 1e-9
double solve(double r,double R,double h,double H)
{
    double u = h/H*(R-r) + r;
    return PI/3*(r*r+r*u+u*u)*h;
}
int main()
{
    int t;
    double r,R,H,V,mid,vv,f,l;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf%lf%lf",&r,&R,&H,&V);
        f=0;
        l=100;
        while(l-f>exp)
        {
            mid=(l+f)/2;
            vv=solve(r,R,mid,H);
            if(fabs(vv-V)<=exp)
                break;
            else if(vv>V)
                l=mid-exp;
            else
                f=mid+exp;
        }
        printf("%.6lf\n",mid);
    }
    return 0;
}