(三分法)Light Bulb

Light Bulb

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 62   Accepted Submission(s) : 37

Problem Description

Compared to wildleopard's wealthiness, his brother mildleopard is rather poor. His house is narrow and he has only one light bulb in his house. Every night, he is wandering in his incommodious house, thinking of how to earn more money. One day, he found that the length of his shadow was changing from time to time while walking between the light bulb and the wall of his house. A sudden thought ran through his mind and he wanted to know the maximum length of his shadow.

Input

The first line of the input contains an integer T (T <= 100), indicating the number of cases.

Each test case contains three real numbers H, h and D in one line. H is the height of the light bulb while h is the height of mildleopard. D is distance between the light bulb and the wall. All numbers are in range from 10^-2 to 10³, both inclusive, and H - h >= 10^-2.

Output

For each test case, output the maximum length of mildleopard's shadow in one line, accurate up to three decimal places..

Sample Input

32 1 0.52 0.5 34 3 4

Sample Output

1.0000.7504.000

 

Source

The 6th Zhejiang Provincial Collegiate Programming Contest

 

题目大意:求人落在地上和墙上的影子的最长长度是多少?

分析:分情况:

1.影子全部落在地上.

2.影子部分落在地上,部分落在墙上. 那么三分.

三分法步骤:

1.求出目标函数f的表达式, 求它的二阶导数,判断它是否为凸函数或凹函数.

2.用L,R,mid1,mid2四点将区间分成三段, 每次舍掉三段中离极值点较远的部分再进行三分, 直到mid1和mid2近似相等.

#include<cstdio>#include<cmath>#include<iostream>#include<algorithm>using namespace std;const double EPS = 1e-8;double H, h, D;double f(double x){return -x + H + D + (h - H)*D / x;}double ternary_search(double L, double R){while (R - L > EPS){double mid1 = L + (R-L) / 3;double mid2 = R - (R-L) / 3;if (f(mid1) < f(mid2))L = mid1;elseR = mid2;}return f((L + R) / 2);}int main(){int T;scanf("%d", &T);while (T--){scanf("%lf%lf%lf", &H, &h, &D);double ans = h*D / H;ans = max(ans, ternary_search((H - h)*D / H, D));printf("%.3lf\n", ans);}return 0;}