1004Toxophily

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.<br>We all like toxophily.<br><br>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?<br><br>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. <br>

 

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.<br>Technical Specification<br><br>1. T ≤ 100.<br>2. 0 ≤ x, y, v ≤ 10000. <br>

 

Output

For each test case, output the smallest answer rounded to six fractional digits on a separated line.<br>Output "-1", if there's no possible answer.<br><br>Please use radian as unit. <br>

 

Sample Input

    
    
    
    

     
     
     
     3<br>0.222018 23.901887 121.909183<br>39.096669 110.210922 20.270030<br>138.355025 2028.716904 25.079551<br>
    
    
    
    

 

Sample Output

    
    
    
    

     
     
     
     1.561582<br>-1<br>-1<br>
    
    
    
    

 

Source

The 4th Baidu Cup final

简单题意：

  给出目标的坐标和初速度，编写一个程序求出最小的角度使射出的箭能够达到给定的点。如果不能达到，返回-1，如果能到达，输出最小的角度。

解题思路形成过程：

  需要用到物理知识，所以，首先建立物理模型，如果箭能够达到最小的角度，那么向上的速度为0，当（vy-0）/9.8<t时，此时的解必定不是最小的。然后运用二分法，即可得到结果。

感想：

  ACM和物理知识结合在一起的经典。

AC代码：

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#define PI acos(-1.0)
#define G 9.8
#define eps 1e-12
using namespace std;
double X,Y,v,sita;
double cal(double x)
{
    double t;
    t=X/(v*cos(x));
    return v*sin(x)*t-0.5*G*t*t;
}
double triplediv()
{
    int i;
    double mid1,mid2,left,right,h1,h2;
    left=0,right=0.5*PI;
    for(i=1;i<=100;i++)
    {
        mid1=(2*left+right)/3;
        mid2=(left+2*right)/3;
        h1=cal(mid1);
        h2=cal(mid2);
        if(h1>h2)
        {
            right=mid2;
        }
        else
        {
            left=mid1;
        }
    }
    sita=left;
    return cal(left);
}
void doublediv(double maxy)
{
    int i;
    double left,right,mid,h;
    left=0,right=sita;
    for(i=1;i<=100;i++)
    {
        mid=(left+right)/2;
        h=cal(mid);
        if(h>Y)
            right=mid;
        else
            left=mid;
    }
    printf("%.6lf\n",left);
}
int main()
{
    int t;
    double maxy;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf%lf",&X,&Y,&v);
        maxy=triplediv();
        if(maxy<Y)
            printf("-1\n");
        else
            doublediv(maxy);
    }
    return 0;
}