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

 

Sample Output

 

Source

 #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/)*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/*PI*H*(R*R+r*r+R*r);

         double left= , right=H , mid = H/ ,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)/;

                 vv = g(f(mid),mid);

             }

             printf("%.6lf\n",mid);

         }

     }

     return ;

 }