POJ 1905 Expanding Rods(二分)

http://poj.org/problem?id=1905

Expanding Rods Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 10655 Accepted: 2724 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 Source Waterloo local 2004.06.12

如图，原来长L的直棒变成长L'的弧，求偏移量（就是弧中点到弦的距离，实在不知道是哪个术语>_<）

高中数学问题，画个弦、弧所对应的扇形（数形结合），列出关于弧所对圆心角的方程（化归转移），发现是个超越方程，不会解，变成函数（函数方程），用二分求零点，然后再求出偏移量

本题要注意精度（PI和角度），当n*c=0时直接输出0.000，用MinGW GCC输出用%f

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<algorithm>
#include<ctime>
#include<cctype>
#include<cmath>
#include<string>
#include<cstring>
#include<queue>
#include<vector>
#define sqr(x) (x)*(x)
#define INF 0x1f1f1f1f
#define PI 3.14159265358979
#define LL long long
#define mm 

using namespace std;

double l,n,c,cital,citar,l_,a;	//cital,citar 角度的下界与上界，l_弧长，a偏移量 

double ans(double x)
{
	return l_/2/x*(1-cos(x));	//算偏移量，公式自己推 
	
}

double fun(double x)	
{
	return x/sin(x)-l_/l;		//二分函数 
}

int main()
{
	while(scanf("%lf %lf %lf",&l,&n,&c)&&(l!=-1)&&(n!=-1)&&(c!=-1))
	{
		l_=(1+n*c)*l;
		
		cital=0;
		citar=PI;
		
		if (n*c<1e-8)
		{
			printf("0.000\n");
			continue;
		}
		
		while (cital<citar)
		{
			double mid=(cital+citar)/2;
			double y=fun(mid);
			
			if (fabs(y)<1e-12)
			{
				a=ans(mid);
				break;
			}
			else
			{
				if (y>0)
				{
					citar=mid;
				
				}
				else
				{
					cital=mid;
				}
				if (fabs(ans(citar)-ans(cital))<1e-5)
				{
						a=ans(citar);
						break;
				}
				
			}
		} 
		printf("%.3f\n",a);
		
	}
	
	return 0;
}