二分查找解单调方程..

二分查找解单调方程..

Crossed ladders
Time Limit:1000MS  Memory Limit:65536K
Total Submit:1837 Accepted:605

Description
A narrow street is lined with tall buildings. An x foot long ladder is rested at the base of the building on the right side of the street and leans on the building on the left side. A y foot long ladder is rested at the base of the building on the left side of the street and leans on the building on the right side. The point where the two ladders cross is exactly c feet from the ground. How wide is the street?

Input
Each line of input contains three positive floating point numbers giving the values of x, y, and c.

Output
For each line of input, output one line with a floating point number giving the width of the street in feet, with three decimal digits in the fraction.

Sample Input

30 40 10
12.619429 8.163332 3
10 10 3
10 10 1

Sample Output

26.033
7.000
8.000
9.798

Source
The UofA Local 2000.10.14

My Code:

#include  < iostream >
#include  < algorithm >
#include  < cmath >
#include  < iomanip >
using   namespace  std;

void  round( double  x,  double  y,  double  h) 
{
     double  s1, s2, s;
     double  t, beg, end;
     if  (x  ==  y)  {
        s  =  sqrt(x * x - 4 * h * h);
    }   else   {
         if  (y  <  x)  {
            swap(x, y);
        }
        beg  =   0 ;
        end  =  x;
        s  =  (beg  +  end)  /   2 ;
        t  =   1   /  sqrt(x * x - s * s)  +   1   /  sqrt(y * y - s * s);
         while  (fabs(t - 1 / h) > 0.000000001 )  {
             if  (t  <   1 / h)  {
                beg  =  s;
                s  =  (beg  +  end)  /   2 ;
            }   else   if  (t  >   1 / h)  {
                end  =  s;
                s  =  (beg  +  end)  /   2 ;
            }   else   {
                 break ;
            }
            t  =   1   /  sqrt(x * x - s * s)  +   1   /  sqrt(y * y - s * s);
        }
    }
    cout  <<  setiosflags(ios:: fixed )  <<  setprecision( 3 )  <<  s  <<  endl;
}

int  main() 
{ 
     double  x, y, h;
     while  (cin  >>  x  >>  y  >>  h)  {
        round(x, y, h);
    }
     return   0 ;
}