Time Limit: 1 Second Memory Limit: 32768 KB
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
题解：找出函数，两种方法，一种求导直接求，另一种是三分；
#include<stdio.h>
#include<math.h>
int main(){
int T;
double H,h,D;
//公式：D-x+H-(H-h)*D/x;
//求导：-1+（H-h）*D/(x*x)
//导数等于0，求得极值x=sqrt((H-h)*D)
//影子在地面最长时x=(H-h)*D/H;
scanf("%d",&T);
while(T--){
scanf("%lf%lf%lf",&H,&h,&D);
double jz=sqrt((H-h)*D);
double ans;
if(jz<=(H-h)*D/H)ans=D-(H-h)*D/H;
else if(jz>=D)ans=h;
else ans=D-jz+H-(H-h)*D/jz;
printf("%.3lf\n",ans);
}
return ;
}
三分：
#include<stdio.h>
#include<math.h>
double H,h,D;
double getl(double x){
return D-x+H-(H-h)*D/x;
}
void sanfen(){
double l=(H-h)*D/H,m,mm,r=D;//l从地面最长开始
while(r-l>1e-){
m=(l+r)/;
mm=(m+r)/;
if(getl(m)>=getl(mm))r=mm;
else l=m;
}
printf("%.3lf\n",getl(l));
}
int main(){
int T;
scanf("%d",&T);
while(T--){
scanf("%lf%lf%lf",&H,&h,&D);
sanfen();
}
return ;
}