HDOJ 2289 Cup(数学二分)

Cup

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5487    Accepted Submission(s): 1746

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



    
    
    
    纯是考数学的啊(逃 ac代码： 
    
    
    
    
#include<stdio.h>
#include<string.h>
#include<math.h>
#define PI acos(-1.0)
#include<iostream>
#include<algorithm>
using namespace std;
double reallyv(double a,double b,double c,double d)
{
	double x=(c/d)*(b-a)+a;
	double v;
	v=(a*a+x*x+a*x)*PI*c/3;
	return v;
}
int main()
{
	int n;
	double low,high,mid,r,R,H,V,rv;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%lf%lf%lf%lf",&r,&R,&H,&V);
		low=0;
		high=H;
		mid=(low+high)/2;
		rv=reallyv(r,R,mid,H);
		while(fabs(rv-V)>1e-6)
		{
			if(fabs(high-low)<=1e-6)//不加超时
			break;
			if(rv>V)
			high=mid;
			else
			low=mid;
			mid=(low+high)/2;
			rv=reallyv(r,R,mid,H);
		}
		printf("%.6lf\n",mid);
	}
	return 0;
}