HDU

Problem C

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 29   Accepted Submission(s) : 12

Problem 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 1012.619429 8.163332 310 10 310 10 1

 

Sample Output

26.0337.0008.0009.798

 

其实这是一道几何题来着（要用到二分）：

ac代码：

#include<iostream>  
#include<cstdio>
#include<cmath>  //sqrt什么的要用
#define min(a,b) a <= b ? a : b;
using namespace std;
double x, y, c;
double get1(double w)
{
    return 1 - c / sqrt(x * x - w * w) - c / sqrt(y * y - w * w);//无奈高数学得

   //不怎么样，所有这里推了半天，其实利用相似三角形马上可以求解；
}
int main()
{
    while(~scanf("%lf%lf%lf",&x,&y,&c))
    {
        double l = 0, mid, r = min(x,y);
        while(r - l > 1e-8)
        {
            mid = (l + r) / 2;
            if(get1(mid) > 0)
                l = mid;
            else
                r = mid;
        }
        printf("%.3f\n",mid);  //小心，别用%.3lf,就是过不了！！！
    }
    return 0;
}

若能留下您宝贵的建议，

万分感谢~~~