B. Crossed ladders

B. Crossed ladders

Time Limit: 1000ms

Memory Limit: 32768KB

64-bit integer IO format: %I64d      Java class name:Main

SubmitStatus PID: 24431

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. Ay 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 exactlyc feet from the ground. How wide is the street?

Input

Each line of input contains three positive floating point numbers giving the values ofx, 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 1012.619429 8.163332 310 10 310 10 1

Sample Output

26.0337.0008.000

9.798

题意：两栋楼之间有两个梯子，如图中的虚线所示，一个梯子的长度为x，另一个梯子的长度为y，两个梯子的交点离地面的高度为c，问两栋楼之间的距离。
这是一个几何题。设宽度为w，交点距左楼距离为a，则根据三角形相似可以推出：
把第二个式子代入第一个式子可以得出一个等式，把方程左边写成关于w的函数，求导可得是减函数。因为w一定小于x，y中的最小值（三角形的直角边小于斜边），二分求出答案即可。

#include<stdio.h>#include<math.h>#include<cstdio>#include<algorithm>using namespace std;#define min(a,b) a <= b ? a : b;double x,y,c;double we(double w){    return 1-c/sqrt(y*y-w*w)-c/sqrt(x*x-w*w);}int main(){    while(scanf("%lf%lf%lf",&x,&y,&c)!=-1)    {        double r,l=0;        r=min(x,y);        double mid;        while(r-l>1e-8)        {            mid=(l+r)/2;            if(we(mid)>0)            {                l=mid;            }            else            {                r=mid;            }        }        printf("%.3lf\n",mid);    }}