uva 10566 Crossed ladders（二分 + 几何）

最新推荐文章于 2025-04-05 21:48:32 发布

wdmzsly

最新推荐文章于 2025-04-05 21:48:32 发布

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分类专栏：二分查找

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本文介绍了一道关于两根梯子交叉的数学问题，并通过二分查找法结合几何原理求解街道宽度。输入梯子长度及交叉高度，输出街道宽度，示例包含具体输入输出示例。

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Crossed ladders

Time Limit: 1000ms

Memory Limit: 32768KB

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 10
12.619429 8.163332 3
10 10 3
10 10 1

Sample Output

26.033
7.000
8.000
9.798
第一次做二分加几何的题目，心里很激动啊，可惜比赛的时候一直在找能直接得到答案的几何关系
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#define eps 1e-9
using namespace std;

double x,y,c;

double Binary_search(double mid){
    return 1-c/sqrt(x*x-mid*mid)-c/sqrt(y*y-mid*mid);
}

int main(){
    double l,r,mid;
    while(cin>>x>>y>>c){
        l=0.00;
        r=min(x,y);
        while(r-l>eps){
            mid=(l+r)/2;
            if(Binary_search(mid)>0)
                l=mid;
            else
                r=mid;
        }
        printf("%.3lf\n",mid);
    }
    return 0;
}