物理数学题

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.

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.

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.

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

#include 
#include 
#define pi 4*atan(1)
double x,y,v;
double cal()
{
    double a=9.8*x*x;
    double b=x*2*v*v;
    double c=y*2*v*v+9.8*x*x;
    double dt=b*b-4*a*c;
    double x1=(b+sqrt(dt))/2/a;
    double x2=(b-sqrt(dt))/2/a;
    x1=atan(x1);
    x2=atan(x2);
    if(x12&&x1>0&&x22&x2>0)
        return x1else if(x12&&x1>0)
    return x1;
    else if(x22&&x2>0)
        return x2;
    return -1;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf%lf",&x,&y,&v);
        if(cal()!=-1)
            printf("%.6f\n",cal());
        else
            printf("-1\n");
    }
    return 0;
}

Reflect:
1，斜抛+一元二次方程的根；
2，记住斜抛的坐标系是竖直的二维坐标系，不是三维~；
3，对于一元二次方程的根，可以先直接算，然后在判断数据的合法性，而不用先判断是否存在在去算；