Toxophily

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 99 Accepted Submission(s): 56

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

30.222018 23.901887 121.90918339.096669 110.210922 20.270030138.355025 2028.716904 25.079551

Sample Output

1.561582-1-1

分析：

网上这道题的解法几乎都是2分法效率还可以，但是用数学方法解题可以实现0ms通过。

用公式，根据正交分解坐标系，得出方程的通式。

想 x^2*g/(2*v^2)*tan^2(ß) - x*tan(ß) +y + x^2*g/(2*v^2) = 0;

即：a = g*pow(x,2)/(2*pow(v,2));

b = -x;

c = y + g*pow(x,2)/(2*pow(v,2));

根据求根公式求出根。

注意讨论：

（1） x==0&&y==0时，ß = 0；

（2） x==0&&y>0时，ß=90；

（3）方程无解时 ß=-1；

（4）方程的解为负数时，ß=-1；（0<=ß<=90）。

#include 
         
           #include 
          
            #include 
           
             using namespace std; int main() {
        int t; double a,b,c,angle,z; double x,y,v,g = 9.8,T,ans1,ans2; scanf("%d",&t); while(t--) {
        scanf("%lf%lf%lf",&x,&y,&v); if(x==0&&y==0) printf("0\n"); else if(x==0&&y>0) printf("90\n"); else {
        a = g*pow(x,2)/(2*pow(v,2)); b = -x; c = y+a; T = pow(b,2) - 4*a*c; angle = 0; if(T<0) printf("-1\n"); else {
        ans1 = ((-b)+pow(T,1.0/2))/(2*a); ans2 = ((-b)-pow(T,1.0/2))/(2*a); if(ans1>=0) angle = atan(ans1); if(ans2>=0) {
        z =  atan(ans2); if(z

http://www.cnblogs.com/newpanderking/archive/2011/08/25/2153590.html