Crossed Ladders - UVa 10566 二分几何

最新推荐文章于 2020-04-16 18:43:24 发布

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本文介绍了一道经典的几何问题——交叉梯子问题。题目要求已知两个梯子的长度及交点距地面的高度，求解街道宽度。文章通过二分查找算法解决了这一问题，并提供了完整的代码实现。

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Crossed Ladders
Input: Standard Input

Output: Standard Output

Time Limit: 1 Second

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?

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

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 Output for Sample Input

30 40 10

12.619429 8.163332 3

10 10 3

10 10 1

26.033

7.000

8.000

9.798

题意：给定两个梯子的长度x和y，求交点的高度为c时，梯子底部的距离是多少。

思路：用二分的方法，当已知这个距离的时候，求高度就很简单了。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
double x,y,c,eps=1e-8;
double solve(double BD)
{
    double D,B,ED;
    D=acos(BD/x);
    B=acos(BD/y);
    ED=BD*sin(B)/sin(M_PI-D-B);
    return ED*sin(D);
}
int main()
{
    double l,r,mi;
    while(~scanf("%lf%lf%lf",&x,&y,&c))
    {
        l=0;r=min(x,y);
        while(r-l>eps)
        {
            mi=(l+r)/2;
            if(solve(mi)>c)
              l=mi;
            else
              r=mi;
        }
        printf("%.3f\n",l);
    }
}