Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
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
3
2 1 0.5
2 0.5 3
4 3 4
Sample Output
1.000
0.750
4.000
给你H，h，D三个数字
问你影子的长度
可以预想一下
跟h有关的影子的长度的函数是一个由峰值的函数
所以三分解决
#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define max_v 35
#define eps 10e-6
using namespace std;
double H,h,D;
double f(double l)
{
return D*(h-l)/(H-l)+l;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%lf %lf %lf",&H,&h,&D);
double l=,r=h;
double mid,mmid;
while(fabs(l-r)>eps)
{
mid=(l+r)/;
mmid=(mid+r)/;
if(f(mid)<=f(mmid))
{
l=mid;
}
else
{
r=mmid;
}
}
printf("%0.3lf\n",f(l));
}
return ;
}
/*
题目意思：
给你H，h，D三个数字
问你影子的长度 随着h的变化，影子的长度也会变化（影子的长度包括在地上和墙上的）
可以预想一下
跟h有关的影子的长度的函数是一个由峰值的函数
所以三分解决 */