Description
The recreation center of WHU ACM Team has indoor billiards, Ping Pang, chess and bridge, toxophily, deluxe ballrooms KTV rooms, fishing, climbing, and so on.
We all like toxophily.
Bob is hooked on toxophily recently. Assume that Bob is at point (0,0) and he wants to shoot the fruits on a nearby tree. He can adjust the angle to fix the trajectory. Unfortunately, he always fails at that. Can you help him?
Now given the object's coordinates, please calculate the angle between the arrow and x-axis at Bob's point. Assume that g=9.8N/m.
We all like toxophily.
Bob is hooked on toxophily recently. Assume that Bob is at point (0,0) and he wants to shoot the fruits on a nearby tree. He can adjust the angle to fix the trajectory. Unfortunately, he always fails at that. Can you help him?
Now given the object's coordinates, please calculate the angle between the arrow and x-axis at Bob's point. Assume that g=9.8N/m.
Input
The input consists of several test cases. The first line of input consists of an integer T, indicating the number of test cases. Each test case is on a separated line, and it consists three floating point numbers: x, y, v. x and y indicate the coordinate of the fruit. v is the arrow's exit speed.
Technical Specification
1. T ≤ 100.
2. 0 ≤ x, y, v ≤ 10000.
Technical Specification
1. T ≤ 100.
2. 0 ≤ x, y, v ≤ 10000.
Output
For each test case, output the smallest answer rounded to six fractional digits on a separated line.
Output "-1", if there's no possible answer.
Please use radian as unit.
Output "-1", if there's no possible answer.
Please use radian as unit.
Sample Input
3 0.222018 23.901887 121.909183 39.096669 110.210922 20.270030 138.355025 2028.716904 25.079551
Sample Output
1.561582 -1 -1
这道题主要是需要推两个公式:
水平方向上:v*cos(a)*t=x;
竖直方向上:v*sin(a)*t-g*t*t/2.0=y;
得到:y=x*tan(a)-g*x*x/(v*v*cos(a)cos(a)*2.0);
可以证明这个y是随着角度a呈凹曲线变化的,虽然我也不知道怎么证明。
所以我的代码如下:
#include <iostream>
#include <cmath>
#include <iomanip>
#include <cstdio>
using namespace std;
const double PI=acos(-1.0);
const double EPS=1e-8;
const double g=9.8;
double x,y,v;
double f(double a)
{
double t=x/(v*cos(a));
return v*sin(a)*t-g*t*t/2.0;
}
int main()
{
//freopen("in.txt","r",stdin);
int T;
cin>>T;
while(T--)
{
cin>>x>>y>>v;
double lo=0,hi=PI/2,mid,mmid;
while(lo+EPS<hi)
{
mid=lo+(hi-lo)/2.0;
mmid=mid+(hi-mid)/2.0;
//cout<<mid<<" "<<mmid<<endl;
if(f(mid)>f(mmid)) hi=mmid;
else lo=mid;
//cout<<hi<<" "<<lo<<endl;
}
if(f(mid)+EPS<y){cout<<"-1"<<endl;continue;}
lo=0,hi=mid;
while(lo+EPS<hi)
{
mid=lo+(hi-lo)/2.0;
if(f(mid)+EPS>=y) hi=mid;
else lo=mid;
}
cout<<fixed<<setprecision(6)<<hi<<endl;
}
return 0;
}