hdoj 2289 Cup

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 <cstdio>
    #include <cmath>
    #define pi acos(-1.0)//圆周率
    #define exp 1e-9 //精度

    double  gets(double h, double H, double r, double R)
    {
        double k = h / H * (R - r) + r; //根据三角形的相识原理求出当前高度的上底的圆的半径;
        return pi / 3.0 *(r * r + r * k + k * k) * h;//圆台的体积等于三分之一高乘于上底和下底的面积在加上上底和下底的乘积的开方;
    }

    int main()
    {
       int T;
       double r,R,h,H,V;
       scanf("%d",&T);
       while(T--)
       {
           scanf("%lf %lf %lf %lf",&r,&R,&H,&V);
           double ua,ub;
           ua = 0;
           ub = H;
           while(ua < ub)
           {
               h = (ua + ub) / 2.0;
               double sum = gets(h,H,r,R);
        if(sum < V) ua = h + exp;//当前的水体积小于原本的体积,则高度不够,缩小区间
               else ub = h - exp;
           }
           printf("%.6lf\n",ua);
       }
       return 0;
    }