Cup

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2356 Accepted Submission(s): 741

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

100 100 100 3141562

Sample Output

99.999024

Source

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二分

题目大意：一个圆台型的杯子，告诉杯子的上底和下底的半径、杯子的高度，以及水的体积，要求水的高度。

算法分析：

水平面的半径u可以根据直角梯形的面积求出：

(r+u)*h + (u+R)*(H-h) = (r+R)*H

得出(R-r)*h = (u-r)*H

得 u = r + (R-r)*h/H

圆台体积计算公式：V= π*h*( R^2 + R*u + u^2 ) / 3

#include
      
       #include
       
        #include
        
         #include
         
          using namespace std;#define eps 1e-8double PI=acos(-1);double r,R,H,V;double cal(double x){    double u=(R-r)*x/H+r;    return PI*x*(pow(r,2)+pow(u,2)+r*u)/3;}int main(){    //freopen("input.txt","r",stdin);    int t;    scanf("%d",&t);    while(t--){        scanf("%lf%lf%lf%lf",&r,&R,&H,&V);        double left=0,right=H;        double mid;        while(right-left>eps){            mid=(right+left)/2;            if(cal(mid)-V>eps)                right=mid;            else                left=mid;        }        printf("%.6lf\n",mid);    }    return 0;}