POJ - 2507 Crossed Ladders

Crossed Ladders

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

Description

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

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

数学知识：设左边的墙与x的交点距地面为a，右边的墙与y的交点为b，街道宽为z，x与y的交点距左面墙为d。

则由相似三角形得：c/a=(z-d)/z；c/b=d/z。消去中间变量d得：c/a+c/b=1，即(a+b)/a*b=c，构造函数F=(a+b)/(a*b)-c=0,而a，b可通过x,y,z求出。再用二分法求解即可

#include<cstdio>  #include<cmath>#define eps 1e-7//精度 double x,y,z,c;double F(double z){return sqrt((x*x-z*z)*(y*y-z*z))/(sqrt(x*x-z*z)+sqrt(y*y-z*z))-c;}//定义F函数 int main(){    int t,ans=1;    scanf("%d",&t);      while(t--){        scanf("%lf%lf%lf",&x,&y,&c);        double l=0,r=x<y?x:y,mid=(l+r)/2;//区间右端点取x，y中最小值         printf("Case %d: ",ans++);        while(r-l>eps){        if(F(mid)>0)//F(mid)>0说明mid比实际的z要小，所以舍弃左半部分区间         l=mid;elser=mid;mid=(l+r)/2;}printf("%lf\n",mid);    }     return 0;  }