Toxophily 物理问题

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.<br>We all like toxophily.<br><br>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?<br><br>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. <br>

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.<br>Technical Specification<br><br>1. T ≤ 100.<br>2. 0 ≤ x, y, v ≤ 10000. <br>

Output

For each test case, output the smallest answer rounded to six fractional digits on a separated line.<br>Output "-1", if there's no possible answer.<br><br>Please use radian as unit. <br>

Sample Input

   
   
   
   

    
    
    
    3<br>0.222018 23.901887 121.909183<br>39.096669 110.210922 20.270030<br>138.355025 2028.716904 25.079551<br>
   
   
   
   

Sample Output

   
   
   
   

    
    
    
    1.561582<br>-1<br>-1<br>
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    这道题非常简单，本质上就是解一个二次方程求最小值罢了。
   
   
   
   
   
   
   
   

    
    
    
    由题目我们可以得到这：
   
   
   
   
   
   
   
   

    
    
    
    x = v*cosa*t
   
   
   
   
   
   
   
   

    
    
    
    y = vtsina - 1/2(g*t^2)
   
   
   
   
   
   
   
   

    
    
    
    两个式子消去t得到 gt^2 tana^2 - 2v^2xtana + 2v^2 + gx^2 = 0
   
   
   
   
   
   
   
   

    
    
    
    这是一元二次方程组，变量为tana，求最小值，也是角的最小值。
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    #include<iostream>
#include<cmath>
#include<stdio.h>


using namespace std;
#define p acos(-1.0)
#define g 9.8
double x,y,v,t,Min=0,Max=p/2,a,b,c,t1,t2,q,e1,e2;
int main()
{
     int n;
     scanf("%d",&n);
     while(n--)
     {
    scanf("%lf%lf%lf",&x,&y,&v);
    a = g*x*x;
    b = -2*(v*v*x);
    c = 2*v*v*y+g*x*x;
    q = b*b - 4*a*c;


    t1 = (-b + (sqrt(q)))/(2*a);
    t2 = (-b - (sqrt(q)))/(2*a);


    e1 = atan(t1);
    e2 = atan(t2);


    if(e1>=Min&&e1<=Max&&e2>=Min&&e2<=Max)
    {
        if(e1 < e2)
            printf("%.6lf\n",e1);
        else
            printf("%.6lf\n",e2);
    }
    else if(e1>=Min&&e1<=Max)
    {
       printf("%.6lf\n",e1);
    }
    else if(e2>=Min&&e2<=Max)
    {
        printf("%.6lf\n",e2);
    }
    else
        printf("%d\n",-1);
     }




    return 0;
}