HDU 2298（物理題，純公式計算）

Toxophily

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

Problem DescriptionThe 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.  

 

InputThe 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.  

 

OutputFor 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

1.561582-1-1x=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;最後得到：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;最後判斷α是否滿足~#include <iostream>#include <math.h>using namespace std;#define PI acos(-1.0)#define g 9.8double 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);elseprintf("-1/n");}}return 0;}