hdu 2289(数学+几何+二分)

Cup

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4857    Accepted Submission(s): 1547

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

Source

The 4th Baidu Cup final

 题目分析:圆台的公式 V = PI/3*(R*R+r*R+r*r)*h
因为杯子形状不变,所以高度和r满足一个函数关系,所以枚举其中其中一个,然后二分另一个即可

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
//#define PI 3.1415926535898
//#define eps 1e-8

using namespace std;

int t;

double r,R,v,H;
double PI = acos(-1);
double eps = 1e-8;

double f ( double h )
{
    double u = (R-r)*h/H+r;
    return PI*h*( pow(u,2)+pow(r,2)+r*u)/3;
}
/*
double search ( double v )
{
    double left = 0 , right = H , mid;
    while ( right - left > eps )
    {
        mid = ( left + right ) / 2.0;
        if ( f(mid) - v > eps ) right = mid;
        else left = mid;
    }
    return mid;
}
*/
int main ( )
{
    scanf ( "%d" , &t );
    while ( t-- )
    {
        scanf ( "%lf%lf%lf%lf" , &r , &R , &H , &v );
        double left = 0 , right = H , mid;
        while ( right - left > eps )
        {
            mid = ( right + left )/2;
            if ( f(mid)-v > eps ) right = mid;
            else left = mid;
        }
        printf ( "%.6lf\n" , mid  );
    }
    return 0;
}