HDU 2289 cup

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

题意：一个杯子，r，R，H，V分别代表下底半径，上底半径，杯子的高，杯中热水的体积。计算杯中热水的高度。

分析：该题可以用二分做。

代码：

#include<cstdio>#include <math.h>#include <algorithm>#define ep 1e-9#define pi acos(-1.0)using namespace std;double tj(double r,double R,double h,double H){    double u= h*(R-r)/H + r;    return (pi*h*(r*r+u*u+r*u))/3;}int main(){    int t;    scanf("%d",&t);    double up,d,mid,tv;  //up为上端点，d为下端点，tv为当前的假设高度的热水体积    double r,R,H,V;    while(t--)    {        scanf("%lf%lf%lf%lf",&r,&R,&H,&V);        up=H,d=0;        while(up-d>ep)        {            mid=(up+d)/2;            tv=tj(r,R,mid,H);            if(fabs(V-tv)<=ep) //当假设体积和题给的热水体积相等时，此时的高度即为所求

            break;            else if(tv<V)                d=mid+ep;            else                up=mid-ep;        }        printf("%.6lf\n",mid);    }    return 0;}