题目描述

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.

样例

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

算法1

浮点数二分

参考文献

http://poj.org/problem?id=2507

分析过程

C++ 代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>

using namespace std;

double x, y, c;

bool check(double mid)
{
    double ry = sqrt(y * y - mid * mid);
    double rx = sqrt(x * x - mid * mid);

    double h = rx * ry / (rx + ry);

    if (h < c) return true; //h<c，间距取大了，会让梯子下滑，这样交点高度就小了
    else return false;
}

int main()
{
    scanf("%lf%lf%lf", &x, &y, &c);

    double l, r;
    if (x < y) l = 0, r = x; //取小的边长,作为右边界,因为梯子不能趴地上吧
    else l = 0, r = y;

    while (r - l > 1e-5) //答案保留小数点后3位，这里多取2位即可  POJ原题数据1e-4也能过
    {
        double mid = (l + r) / 2;

        if(check(mid)) r = mid;
        else l = mid;
    }

    printf("%.3lf\n", l);

    return 0;
}