poj 1905 Expanding Rods（二分）

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 0.000115000 10 0.0000610 0 0.001-1 -1 -1

Sample Output

61.329225.0200.000

Source

看似是个几何问题应该列方程组求问题，但是却无法求出，因此根据题上所说的H的范围，下线是0，上线就是L/2;

根据公式，最后我们可以得出

  double r=(4*mid*mid+l*l)/(8*mid);
      double s=2*r*asin(l/(2*r));

然后二分枚举找出H的值

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>#include<math.h>#include<string>#include<iomanip>using namespace std;typedef long long ll;const double esp=1e-5;double l,n,c;int main(){    while(cin>>l>>n>>c)    {        if(l<0&&n<0&&c<0)            break; //如果l,n,c都小于0输入结束        double L=(1+n*c)*l;  //求出弧的长度        double last=0.5*l;        double mid;        double first=0.0;        while(last-first>esp)        {            mid=(first+last)/2;            double r=(4*mid*mid+l*l)/(8*mid);            double s=2*r*asin(l/(2*r));            if(s<L)  first=mid;            else last=mid;        }        double h=mid;          cout<<fixed<<setprecision(3)<<h<<endl; 最后我在这里有了疑问，后来测试了一下，setprecision()函数在第四位小数是5的时候不进位，而用printf就会四舍五入，因此以后再做类似的题还是先把浮点数转化为整数，这样会容易的多，避免的四舍五入时精度不准    }    return 0;}