HDU 2298（物理题，纯公式计算）

Toxophily

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 186    Accepted Submission(s): 106

Problem 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
   
   
   
   
   
   
   
   

    
    
    
    x=v*t*cosα，y=v*t*sinα-g*t^2/2;
   
   
   
   
   
   
   
   

    
    
    
    消去x,得：y=v*sinα*x/(v*cosα)-g*(x^2/(v^2*conα^2))/2;
   
   
   
   
   
   
   
   

    
    
    
    最后得到：g*x^2tanα^2-2*v^2*x*tanα+2*v^2*y+g*x^2=0;
   
   
   
   
   
   
   
   

    
    
    
    a=g*x^2; b=-2*v^2*x; c=2*v^2*y+g*x^2;
   
   
   
   
   
   
   
   

    
    
    
    dt=b*b-4*a*c;
   
   
   
   
   
   
   
   

    
    
    
    tanα=(-b+(-)sqrt(dt))/2*a;
   
   
   
   
   
   
   
   

    
    
    
    最后判断α是否满足~
   
   
   
   
   
   
   
   

    
    
    #include <iostream>
#include <math.h>
using namespace std;
#define PI acos(-1.0)
#define g 9.8

double x,y,v,dt,tan_jd1,tan_jd2,jd1,jd2,a,b,c,MAX=PI/2,MIN=0;

int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		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;
			dt=b*b-4*a*c;
			tan_jd1=(-b+sqrt(dt))/(2*a);
			tan_jd2=(-b-sqrt(dt))/(2*a);
			jd1=atan(tan_jd1);
			jd2=atan(tan_jd2);
			if((jd1>=MIN && jd1<=MAX) && (jd2>=MIN && jd2<=MAX))
				printf("%.6lf/n",jd1<jd2?jd1:jd2);
			else if(jd1>=MIN && jd1<=MAX)
				printf("%.6lf/n",jd1);
			else if(jd2>=MIN && jd2<=MAX)
				printf("%.6lf/n",jd2);
			else
				printf("-1/n");
		}
	}
	return 0;
}