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. That means you have to calculate h as in the picture.

Input

Input starts with an integer T (≤ 20), denoting the number of test cases.

Each case contains three non-negative real numbers: the initial length of the rod in millimeters L, the temperature change in degrees n and the coefficient of heat expansion of the material C. Input data guarantee that no rod expands by more than one half of its original length. All the numbers will be between 0 and 1000 and there can be at most 5 digits after the decimal point.

Output

For each case, print the case number and the displacement of the center of the rod in single line. Errors less than 10^-6 will be ignored.

思路：L‘ = p*r ——p为弧度

r = （L/2）/sin（p/2）

然后公式可以化为 （2×L^‘） /L=(p)/sin(p/2);

然后右边求导可知道右边随p单调增，然后二分p即可。

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<string.h>
 5 #include<queue>
 6 #include<stack>
 7 #include<set>
 8 #include<math.h>
 9 using namespace std;
10 typedef long long LL;
11 double pi=acos(-1);
12 int main(void)
13 {
14     int i,j,k;
15     scanf("%d",&k);
16     int s;
17     double L ,n,c;
18     for(s=1; s<=k; s++)
19     {
20         scanf("%lf %lf %lf",&L,&n,&c);
21         double l=0;
22         double r=2*pi;
23         int ans=0;
24         double ll=(1+n*c)*L;
25         while(ans<=100)
26         {
27             double mid=(l+r)/2;
28             double ac=2*ll*sin(1.0*mid/2);
29             double ak=L*mid;
30             if(ac>=ak)
31             {
32                 l=mid;
33             }
34             else r=mid;
35             ans++;
36         }
37         double rr=(1.0*L/2)/(sin(1.0*r/2));
38         double hh=rr*cos(1.0*r/2);
39         double h=rr-hh;
40         printf("Case %d:",s);
41         printf(" %.6f\n",h);
42     }
43     return 0;
44 }