（写公式，解方程）LightOJ - 1062

LightOJ - 1062                    

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. Ay 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 exactlyc 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 ofx,y, andc.

Output

For each case, output the case number and the width of the street in feet. Errors less than10^-6 will be ignored.

Sample Input

4

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

题意：看图，求出？部分即可；

思路：

第一步：写出正确的求解公式，和解方程类似（公式右边只有常数 V，其余的都在左边，f（mid）的比较对象就是 V）；

第二步：灵活套用二分公式，解出方程；

#include<cstdio>#include<iostream>#include<cmath>#include<cstring>#include<algorithm>using namespace std;double c,y,x,a;double f(double a){return c/sqrt(y*y-a*a)+c/sqrt(x*x-a*a);//找出正确的公式； }int main(){int t,p=0;scanf("%d",&t);while(t--){scanf("%lf %lf %lf",&x,&y,&c);double left=0.0,right=min(x,y),mid,k;while(right-left>=0.0000000001)//精度 {mid=(left+right)/2;if(f(mid)==1){break;}else if(f(mid)>1){right=mid;//right=mid-0.0000001; }else    left=mid;//left=mid-0.0000001; }printf("Case %d: %lf\n",++p,mid);}return 0;}