Cup（二分）

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=2289

hdu_2289:Cup

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5747    Accepted Submission(s): 1807

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

 题解：这是一道经典的二分的题，要注意在二分的时候对精度的判断和对特殊情况的考虑比如水溢出的情况。在推算公式的时候也要注意分母上的数不要有可能等于0 ，这就要求用更加合理的方法推算公式，这个题就是要用倾斜角来算比较好，这样分母上就是H不可能等于0；

代码：

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
#define PI acos(-1)
#define eps 1e-8
double r , R , H , rr ,v,vv;
double f(double h)
{
    return h*(R-r)/H+r;
}
double g(double rr,double h)
{
    return (1.0/3)*h*PI*(rr*rr+r*r+rr*r);
}
int main()
{
    int t; 
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf %lf %lf %lf",&r,&R,&H,&v);
        double tm = 1.0/3*PI*H*(R*R+r*r+R*r);
        double left=0 , right=H , mid = H/2 ,vv = g(f(mid),mid);
        if(tm<=v){
            printf("%.6lf\n",H);
            continue;//注意是continue还是break，因为这个哇了
        }
        else{
            while(fabs(vv-v) > eps){
                if(vv>v) 
                    right = mid;
                else 
                    left = mid ;
                mid = (left+right)/2;
                vv = g(f(mid),mid);
            }
            printf("%.6lf\n",mid);
        }
    }
    return 0;
}