（HDU - 2298）Toxophily

最新推荐文章于 2022-06-11 08:23:47 发布

原创最新推荐文章于 2022-06-11 08:23:47 发布 · 330 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#hdu

简单物理专栏收录该内容

1 篇文章

订阅专栏

（HDU - 2298）Toxophily

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

Total Submission(s): 2310 Accepted Submission(s): 1275

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

T ≤ 100.
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

题目大意：一个小球从(0,0)处与x轴正向呈一定角度θ以初速度v发射，目标物的坐标为(x,y)，问这个角度存在吗？存在的话最小角度是多少？

思路：将小球速度分解为x轴方向和y轴方向，由物理学速度公式可得： $①v*cosθ*t=x;②v*sinθ*t-1/2*g*t^2=y$ 联立二式，消去t可能方程（利用 $cosθ^2+sinθ^2=1$ ） $③g*x^2/(2*v^2)tanθ^2-x*tanθ+g*x^2/(2*v^2)+y=0$ .于是原题等价于方程③在区间[0,π/2]有无解，有解取最小解。

#include<cstdio>
#include<cmath>
using namespace std;

const double g=9.8;
const double pi=acos(-1.0);

int main()
{
    int T;
    double x,y,v;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lf%lf%lf",&x,&y,&v);
        double a=(g*x*x)/(2*v*v),b=-x,c=(g*x*x)/(2*v*v)+y;
        double t=b*b-4*a*c;
        if(t<0)
        {
            printf("-1\n");
            continue;
        }
        double x1=(-b-sqrt(t))/(2*a),x2=(-b+sqrt(t))/(2*a);
        x1=atan(x1);
        x2=atan(x2);
        if(x1>=0&&x2<=pi/2) printf("%.6f\n",x1);
        else if(x1>=0&&x1<=pi/2&&x2>pi) printf("%.6f\n",x1);
        else if(x1>0&&x2>=0&&x2<=pi/2) printf("%.6f\n",x2);
        else printf("-1\n");
    }
    return 0;
}