Cup

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 486    Accepted Submission(s): 150

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知識點：圓臺的體積公式V=pi*（r*r+r*R+R*R)*h/3

#include <iostream> #include <cmath> using namespace std; #define PI acos(-1.0) double GetVolume(double r,double R,double h,double H) { double u = h/H*(R-r) + r;//中部小圓臺上底半徑,根據比例可以求 return PI/3*(r*r+r*u+u*u)*h; } int main() { int t; double r,R,H,V,s,S,mids,midh,midv,f,l; scanf("%d",&t); while(t--) { scanf("%lf%lf%lf%lf",&r,&R,&H,&V); s=PI*r*r; S=PI*R*R; f=0;l=100; while(l-f>1e-9) { midh=(l+f)/2; midv=GetVolume(r,R,midh,H); if(fabs(midv-V)<=1e-9) break; else if(midv>V) l=midh-1e-9; else f=midh+1e-9; } printf("%.6lf/n",midh); } return 0; }