LightOJ 1062 Crossed Ladders（大楼距离，二分）

最新推荐文章于 2017-02-17 19:05:43 发布

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本文针对两个梯子交叉于狭窄街道的问题，通过已知梯子长度及交叉点高度，运用数学几何原理，推算出街道宽度的具体算法实现。示例输入输出展示了不同情况下的精确计算结果。

1062 - Crossed Ladders（链接到题目）

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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

Output for Sample Input

30 40 10

12.619429 8.163332 3

10 10 3

10 10 1

Case 1: 26.0328775442

Case 2: 6.99999923

Case 3: 8

Case 4: 9.797958971

题意：

知道两根木棍的长度 x , y 以及两根木棍交点距离地面的高度，求两栋大楼的距离

思路：

以后用几何画板演示。

参考代码：

#include<stdio.h>
#include<math.h>

double MIN(double x,double y) {
	return x<y?x:y;
}

int main() {
	double  x,y,c;
	int t,v=1;
	scanf("%d",&t);
	while(t--) {
		scanf("%lf%lf%lf",&x,&y,&c);
		int Q=64;
		double l=0,r=MIN(x,y);//取较小的为保证运算数据的有效性 
		while(Q--) {
			double mid=(l+r)/2.0;
			double a1=acos(mid/x);//木棍 x 和地面的夹角 
			double a2=acos(mid/y);//木棍 y 和地面的夹角 
			double d=c/tan(a1)+c/tan(a2);//枚举两栋楼的距离 
			if(d<mid) {
				l=mid;
			} else {
				r=mid;
			}
		}
		printf("Case %d: %lf\n",v,l);
		v++;
	}
	return 0;
}