hdu2289 Cup 二分搜索

Cup

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7972    Accepted Submission(s): 2421

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

1100 100 100 3141562

 

Sample Output

99.999024

 

Source

居然是二分。。。。

一下子接受不了吧.jpg

好吧我也是看讨论版才知道的。。。。

感觉解方程的题也能这么玩，那就是完全考数学了啊

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <vector>#include <cmath>#include <map>#include <set>using namespace std;const double pi = 3.14159265358979323846264338327950288419716939937511;const double eps = 1e-8;int main(){    int t;    scanf("%d",&t);    while(t--){        double r,R,H,V;        scanf("%lf%lf%lf%lf",&r,&R,&H,&V);        double left = 0.0f,right = H;        while(right-left>eps){            double middle = (left+right)/2;            double now = middle/H*(R-r) + r;            double killme = pi*middle*(r*r+now*now+r*now)/3;            if(killme > V || abs(killme - V)<eps){                right = middle;            }else{                left = middle + eps;            }        }        //cout<<pi*left*(R*R+pow(left/H*r,2)+R*(left/H*r))/3<<endl;        printf("%lf\n",left);    }    return 0;}