1137 - Expanding Rods

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Time Limit:0.5 second(s)

Memory Limit:32 MB

When a thin rod of lengthLis heatedndegrees, it expands to a new lengthL' = (1+n*C)*L, whereCis 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 calculatehas in the picture.

Input

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

Each case contains three non-negative real numbers: the initial length of the rod in millimetersL, the temperature change in degreesnand the coefficient of heat expansion of the materialC. 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 than10^-6will be ignored.

Sample Input	Output for Sample Input
3 1000 100 0.0001 150 10 0.00006 10 0 0.001	Case 1: 61.3289915 Case 2: 2.2502024857 Case 3: 0

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SPECIAL THANKS: JANE ALAM JAN (DESCRIPTION, SOLUTION, DATASET, PICTURES)

思路：L' = p*r ——p為弧度

r = （L/2）/sin（p/2）

然后公式可以化為 （2×L^'） /L=(p)/sin(p/2);

然后右邊求導可知道右邊隨p單調增，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
",h);
42     }
43     return 0;
44 }

油！油！you@

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