joj 2228 Crossed ladders 二分法求解方程

joj 2228 Crossed ladders 二分法求解方程

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 Specification

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

Output Specification

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

joj测试数据很弱，可能在其他oj上过不了，根据三角形相似得出以下关于w的方程，解w，第一次用二分法，不错。
f(w) = c - sqrt((y*y-w*w)*(x*x-w*w))/(sqrt(y*y-w*w) + sqrt(x*x -w*w))

#include < iostream >
#include < cstdlib >
#include < iomanip >
#include < math.h >
using   namespace  std;
// f(w) = c - sqrt((y*y-w*w)*(x*x-w*w))/(sqrt(y*y-w*w) + sqrt(x*x -w*w))
   double  func( double  x, double  y, double  c)
  {
     double  wdown = 0.0 ,wup,w,fw;
     if (x > y)
    wup = y;
     else
    wup = x;
    w = wup / 2.0 ;
     while ( 1 )
    {
        fw = c - sqrt((y * y - w * w) * (x * x - w * w)) / (sqrt(y * y - w * w)  +  sqrt(x * x  - w * w));
         if (fw > 0.0001 )
         {
                wup = w;
                w = (wup + wdown) / 2 ;
         }
         else
        {
             if (fw <- 0.0001 )
            {
                wdown = w;
                w = (wup + wdown) / 2 ;
            }
             else
            {
                 return  w;
            }
        }
    }
        
   }
  int  main()          
{   // freopen("s.txt","r",stdin);
   // freopen("key.txt","w",stdout);
   double  x,y,c;
while ( cin >> x >> y >> c)
{
    cout << fixed << setprecision( 3 ) << func(x,y,c) << endl;
}

   // system("PAUSE");
   return     0 ;
  }