Light Bulb--zoj3203(三分法）

Light Bulb

---

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

 1 #include<stdio.h>

 2 #include<math.h>

 3 double hs(double H,double h,double D,double X)

 4 {

 5     return D*(h-X)/(H-X)+X;//定义求长度的函数，这里的X是人在墙上的投影长度

 6 }

 7 int main()

 8 {

 9     int n;

10     scanf("%d",&n);

11     while(n--)

12     {

13         double a,b,c;

14         scanf("%lf%lf%lf",&a,&b,&c);

15         double mid,midmid,l=0,r=b;

16         do

17         {

18             mid=(l+r)/2;

19             midmid=(mid+r)/2;

20             if(hs(a,b,c,mid)>hs(a,b,c,midmid))

21             r=midmid;

22             else

23             l=mid;

24         }while(r-l>1e-9);

25         printf("%.3lf\n",hs(a,b,c,mid));

26     }

27     return 0;

28 }

 

主要的思想是三分法求函数的最大值，不过要注意精度我精确到1e-6都wa了，也是醉了！

建议以后都要精确到1e-9！