POJ 1905：Expanding Rods 求函数的二分

Expanding Rods

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 13780		Accepted: 3563

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

题意是一根长度为L的木棒，加热n度之后长度变为L'=(1+n*C)*L，其中C是木棒的热膨胀系数。给出L,n,C。求木棒在加热前后中心点的距离是多少。

四个方程：(假设要求的距离为x)

L'=(1+n*C)*L

θr = 1/2*L'

sinθ= 1/2*L/r

r*r = (1/2*L)*(1/2*L) + (r-x)*(r-x)

这个题目有意思的地方在于导出x非常困难，所以只能枚举x的值，然后能够得到r的值，然后去带入方程2 3，去和正确值比较。导出最终的函数是一个单调性函数，所以直接二分。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

#define esp 1e-5

double L, N, C, S;

bool check(double mid)
{
	double ans = 2 * asin((L / 2) / ((L*L + 4 * mid*mid) / (8 * mid)))*((L*L + 4 * mid*mid) / (8 * mid));
	return ans >= S;
}

int main()
{
	//freopen("i.txt","r",stdin);
	//freopen("o.txt","w",stdout);
	
	double left, right, mid;

	while (cin>>L>>N>>C)
	{
		if (L + N + C < 0 )
			break;
		S = (1 + N*C)*L;
		left = 0.0;
		right = L/2;

		while (right - left > esp)
		{
			mid = (right + left) / 2;
			if (check(mid))
			{
				right = mid;
			}
			else
			{
				left = mid;
			}
		}
		printf("%.3lf
",right);
	}
	return 0;
}