暑期集训训练3练习题B - Toxophily（HDU2298）

最新推荐文章于 2018-07-22 12:42:00 发布

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本文介绍了一种计算弓箭射击水果所需角度的方法。通过输入水果坐标及箭矢初速等参数，利用数学公式求解最小射角。文章提供了一段C++代码实现，并通过样例验证了算法的有效性。

B - Toxophily

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 2298

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

这道题主要是需要推两个公式：

水平方向上：v*cos(a)*t=x;

竖直方向上：v*sin(a)*t-g*t*t/2.0=y;

得到：y=x*tan(a)-g*x*x/(v*v*cos(a)cos(a)*2.0);

可以证明这个y是随着角度a呈凹曲线变化的，虽然我也不知道怎么证明。

所以我的代码如下：

#include <iostream>
#include <cmath>
#include <iomanip>
#include <cstdio>
using namespace std;
const double PI=acos(-1.0);
const double EPS=1e-8;
const double g=9.8;
double x,y,v;

double f(double a)
{
    double t=x/(v*cos(a));
    return v*sin(a)*t-g*t*t/2.0;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int T;
    cin>>T;
    while(T--)
    {
        cin>>x>>y>>v;
        double lo=0,hi=PI/2,mid,mmid;
        while(lo+EPS<hi)
        {
            mid=lo+(hi-lo)/2.0;
            mmid=mid+(hi-mid)/2.0;
            //cout<<mid<<" "<<mmid<<endl;
            if(f(mid)>f(mmid)) hi=mmid;
            else lo=mid;
            //cout<<hi<<" "<<lo<<endl;
        }
        if(f(mid)+EPS<y){cout<<"-1"<<endl;continue;}
        lo=0,hi=mid;
        while(lo+EPS<hi)
        {
            mid=lo+(hi-lo)/2.0;
            if(f(mid)+EPS>=y) hi=mid;
            else lo=mid;
        }
        cout<<fixed<<setprecision(6)<<hi<<endl;
    }
    return 0;
}