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Toxophily

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 76   Accepted Submission(s) : 37

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

 题目大意：一个人的射出箭初始速度一定，让你选择一个角度使恰好击中制定目标。若无法击中则输出-1.

 解题思路：1随着角度的变化，在目标x的对应的y为先增大后减小，可以运用三分法进行迭代，最后求出近似解。

               2直接用物理方法用三角函数表示出关系，直接用cmath的函数进行一次计算即可。

注意事项：三分法要判断是否能够打中目标

ps：用数学方法太容易了，只是推出公式可能acm中可欲不可求的。

#include<iostream>#include<cmath>#include<cstdio>using namespace std;int main(){    const double g=9.8;    double x,y,v,s,s1,a,b;    int n;    scanf("%d",&n);    while(n--)    {        scanf("%lf%lf%lf",&x,&y,&v);        s=sqrt(x*x+y*y);        a=atan(y/x);        s1=v*v*(1-sin(a))/g/cos(a)/cos(a);        if(s1<s)printf("-1\n");        else        {            b=(asin(s*g*cos(a)*cos(a)/v/v+sin(a))-a)/2;            b=b+a;            printf("%lf\n",b);        }    }}