Light Bulb（三分）

ZOJ Problem Set - 3203

Light Bulb

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；

法一：

 1 #include<stdio.h>
 2 #include<math.h>
 3 int main(){
 4     int T;
 5     double H,h,D;
 6     //公式：D-x+H-(H-h)*D/x;
 7      //求导：-1+（H-h）*D/(x*x)
 8      //导数等于0，求得极值x=sqrt((H-h)*D)
 9      //影子在地面最长时x=(H-h)*D/H;
10      scanf("%d",&T);
11      while(T--){
12          scanf("%lf%lf%lf",&H,&h,&D);
13          double jz=sqrt((H-h)*D);
14          double ans;
15          if(jz<=(H-h)*D/H)ans=D-(H-h)*D/H;
16          else if(jz>=D)ans=h;
17          else ans=D-jz+H-(H-h)*D/jz;
18          printf("%.3lf\n",ans);
19      }
20     return 0;
21 }

三分：

 1 #include<stdio.h>
 2 #include<math.h>
 3 double H,h,D;
 4 double getl(double x){
 5     return D-x+H-(H-h)*D/x;
 6 }
 7 void sanfen(){
 8     double l=(H-h)*D/H,m,mm,r=D;//l从地面最长开始
 9     while(r-l>1e-10){
10         m=(l+r)/2;
11         mm=(m+r)/2;
12         if(getl(m)>=getl(mm))r=mm;
13         else l=m;
14     }
15     printf("%.3lf\n",getl(l));
16 }
17 int main(){
18     int T;
19     scanf("%d",&T);
20     while(T--){
21         scanf("%lf%lf%lf",&H,&h,&D);
22         sanfen();
23     }
24     return 0;
25 }