lightoj1062【二分】

来源:互联网 发布:海美迪网络机顶盒价格 编辑:程序博客网 时间:2024/04/28 15:02

B - 
Crossed Ladders
Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu
Submit Status Practice LightOJ 1062

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 xy, 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

题意:如图所示给出x,y,c求出?的长度

解题思路:二分
#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<algorithm>using namespace std;double MIN(double a,double b){return a<b?a:b;}int main(){int t,k=1;double x,y,c;scanf("%d",&t);while(t--){scanf("%lf%lf%lf",&x,&y,&c);double left=0,right=MIN(x,y);int size=100;while(size--){double mid=(left+right)/2.0;double angle1=acos(mid/x);double angle2=acos(mid/y);double f=c/tan(angle1)+c/tan(angle2);if(f<mid)left=mid;else right=mid;}printf("Case %d: %.7lf\n",k++,left);}return 0;}
0 0
原创粉丝点击