HDU-2289 Cup

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

学习：pi的精度巧妙用法；3.0而不是3 ；二分法解题思想。写函数会更简单一些。

                       Cup

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3020    Accepted Submission(s): 943

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

#include<stdio.h>
#include<math.h>
#define pi acos(-1.0)//pi的精度巧妙用法。
int main()
{
	double R,r,H,h,v,r1,h1,h2,r2,r3,v2,v3,v1;
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lf%lf%lf%lf",&r,&R,&H,&v);
		h1=0;
		h2=H;
		r1=R-(R-r)*(H-h1)/H;
		v1=1/3.0*pi*h1*(r*r+r1*r1+r*r1);
		if(v1==v)
		{	
			printf("%lf\n",h1);
			break;
		}
		r2=R-(R-r)*(H-h2)/H;
		v2=1/3.0*pi*h1*(r*r+r2*r2+r*r2);//3.0而不是3，
		if(v2==v)
		{
			printf("%lf\n",h2);
			break;
		}
	    while(h2-h1>1e-10)//二分法解题思想。
		{
			h=(h1+h2)/2;
			r3=R-(R-r)*(H-h)/H;
			v3=1/3.0*pi*h*(r*r+r3*r3+r*r3);
		    if(v3>v)
				h2=h;
			else
				h1=h;
		} 
		printf("%lf\n",h);
	}
	return 0;
}