POJ 1905 Expanding Rods二分

                                  Expanding Rods

Time Limit:500MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status

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. 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.

Sample Input

3

1000 100 0.0001

150 10 0.00006

10 0 0.001

Sample Output

Case 1: 61.3289915

Case 2: 2.2502024857

Case 3: 0

一根某种材料做的直杆被夹在两面墙之间，当他受热时长度会变长，就会因两面墙的挤压而向上隆起。长度变化函数为 L'=(1+n*C)*L，给定L，C，n，求向上拱起的高度H。

要利用数学公式：

设弦长为L0(即原长)，弧长为L1=(1+n*C)*l0，目标值为h，半径为R，弧所对圆心角为2θ(弧度制)。
可以得到以下方程组：
圆的弧长公式：L1=2θR
三角函数公式：L0=2*R*sinθ，变换得θ=arcsin(L0/(2*R))
勾股定理：R^2=(R-h)^2+(0.5*L0)^2，变换得L0^2+4*h^2=8*h*R
合并①②式得到
L1=2*R*arcsin(L0/(2*R))
半径R可以由③式变换得到
R=(L0^2+4*h^2)/(8*h)
可以用二分枚举h的值，计算出R和L1，与题目中L1进行比较。

#include 
#include 
int main()
{int t,k=1; 
    double L0,L1,N,C;
    double high,low,h,r,L2;
    scanf("%d",&t);
    while(t--)
    {scanf("%lf%lf%lf",&L0,&N,&C);
        L1=(1+N*C)*L0;
        high=0.5*L0;
        low=0.0;
        while (high-low>0.0000001)
        {
            h=(high+low)/2;
            r=(L0*L0+4*h*h)/(8*h);
            L2=2*r*asin(L0/(2*r));
            if (L1