Expanding Rods(二分)

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 10287		Accepted: 2615

Description

When a thin rod of length L is heated n degrees, it expands to a new length L'=(1+n*C)*L, where C is the coefficient of heat expansion. 
When a thin rod is mounted on two solid walls and then heated, it expands and takes the shape of a circular segment, the original rod being the chord of the segment. 

Your task is to compute the distance by which the center of the rod is displaced. 

Input

The input contains multiple lines. Each line of input contains three non-negative numbers: the initial lenth of the rod in millimeters, the temperature change in degrees and the coefficient of heat expansion of the material. Input data guarantee that no rod expands by more than one half of its original length. The last line of input contains three negative numbers and it should not be processed.

Output

For each line of input, output one line with the displacement of the center of the rod in millimeters with 3 digits of precision. 

Sample Input

1000 100 0.0001

15000 10 0.00006

10 0 0.001

-1 -1 -1

Sample Output

61.329

225.020

0.000

题意：一个固定在两面墙之间的杆，受热会变长从而弯曲，问弯曲前后杆中点间的距离；
思路：直接二分弯曲弧度，二分区间为(0，pi),推导出弧度与弯曲后长度LL之间的关系，注意精度还有特判，当L,n,c有一个为0时输出0；

 1 #include<stdio.h>

 2 #include<string.h>

 3 #include<math.h>

 4 const double eps = 1e-12;

 5 

 6 int main()

 7 {

 8     double L,n,c;

 9     while(~scanf("%lf %lf %lf",&L,&n,&c))

10     {

11         if(L<0 && n<0 && c<0)

12             break;

13         if(L == 0 || n == 0 || c == 0)

14         {

15             printf("0.000\n");

16             continue;

17         }

18 

19         double LL = (1+n*c)*L;

20         double min = 0,max = acos(-1.0),mid;

21 

22         while(max-min > eps)//二分弧度

23         {

24             mid = (max+min)/2;

25             if(2*LL/L > mid/sin(mid/2))

26                 min = mid;

27             else max = mid;

28         }

29         printf("%.3lf\n",LL/mid * (1-cos(mid/2)));

30     }

31 }

View Code