Cup（二分查找）

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1022 Accepted Submission(s): 365

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

/*解题思路
主要考了二分查找 不过需要再次之前需要利用圆台的体积公式V＝　PI * h*(R*R +r*r + R*r)(h为圆台的高，R为圆台上面的半径，r为圆台下面的半径) 通过相似三角形可以得出m = (R -r) *h/H + r
然后带入体积公式 在求出来h就可以了
*/

#include<algorithm>#include<iostream>#include<math.h>#include<string.h>#define PI  3.1415926535898using namespace std;double r,R,H,V;double f(double h){    double m = (R-r)*h/H + r;    return PI * h * (m * m + r * r + m * r)/3 - V;}int main(){    int ncase;    while(cin>>ncase){        while(ncase--){            cin>>r>>R>>H>>V;            double x1 = 0;            double x2 = H;            double mid ;            double y1 = f(x1);            double y2 = f(x2);            while(fabs(y1-y2)>=0.000001){                mid = (x1 + x2)/2;                if(f(mid)>=0){                    x2 = mid;                }else                x1 = mid;                y1 = f(x1);                y2 = f(x2);            }            printf("%.6f\n",mid);        }    }}