PKU 2507 [二分]

通过三角形相似的性质，设要求的边为w，推出以下方程  c * sqrt(x * x - w * w) + c * sqrt(y*y -w*w) - sqrt(x * x - w * w) * sqrt(y * y - w * w)

w的上届是x,y中较小值(斜边大于直角边)，然后二分该方程。 

  

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

inline double cal(double x,double y,double c,double w)
{
    return c * sqrt(x * x - w * w) + c * sqrt(y*y -w*w) - sqrt(x * x - w * w) * sqrt(y * y - w * w);
}
int main()
{
    double x,y,c;
    double up;
    double down;
    double ans,mid;
    while(scanf("%lf%lf%lf",&x,&y,&c)!=EOF)
    {
        down = 0;
        up  = x < y ? x : y;
        while(true)
        {
             mid = (up+down)/2.0;
             ans = cal(x,y,c,mid);
            if(fabs(ans)<=0.000001)
            {
                break;
            }
            else if(ans>0)
            {
                up = mid;
            }
            else
            {
                down = mid;
            }
        }
        printf("%.3f\n",mid);
    }
    return 0;
}