UVa 10566 - Crossed Ladders 【二分】

最新推荐文章于 2020-04-16 18:43:24 发布

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最新推荐文章于 2020-04-16 18:43:24 发布

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原文链接：http://www.cnblogs.com/cniwoq/p/6770902.html

本文介绍了一道关于两根梯子交叉的数学问题，并提供了一种通过二分法求解街道宽度的有效算法实现。该问题涉及到梯子长度、交叉点高度及街道宽度之间的几何关系。

Crossed Ladders

Time Limit:2000MS Memory Limit: 32768KB 64bit IO Format: %lld & %lluDescription

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

Input starts with an integer T (≤ 10), denoting the number of test cases.

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

Output

For each case, output the case number and the width of the street in feet. Errors less than 10^-6 will be ignored.

Sample Input

30 40 10

12.619429 8.163332 3

10 10 3

10 10 1

Sample Output

Case 1: 26.0328775442

Case 2: 6.99999923

Case 3: 8

Case 4: 9.797958971

列出方程组，利用二分求解精确值。

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;

double x, y, c;
double ans(double z) {
    return 1 - c/sqrt(x*x - z*z) - c/sqrt(y*y - z*z);
}
int main() {
    int t;
    scanf("%d", &t);
    int cnt = 0;
    while(t--){
        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(ans(mid) > 0) l = mid;
            else  r = mid;
        }
        printf("Case %d: %.8lf\n", ++cnt, mid);
    }
    return 0;
}