HDU2298：Toxophily（三分 + 二分）

Toxophily

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2068    Accepted Submission(s): 1147

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

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

30.222018 23.901887 121.90918339.096669 110.210922 20.270030138.355025 2028.716904 25.079551

 

Sample Output

1.561582-1-1

 

Source

The 4th Baidu Cup final

题意：二维面上给定目标坐标和初速度，求在坐标原点(0,0)，发射出去能经过目标点的最小的角度。

思路：先用三分求出子弹经过横坐标x时所能达到的最大高度，再在此范围内二分出答案，目标在y轴上需要特判。设角度为θ，y = v*sinθ*t - 1/2*g*t*t，t = x / v*cosθ。

# include <iostream># include <cstdio># include <cmath># define pi acos(-1)using namespace std;double x, y, v;double fun(double a){    double t = x/(v*cos(a));    return v*sin(a)*t - 4.9*t*t;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%lf%lf%lf",&x,&y,&v);        double l=0, r=pi/2, mid, mmid;        if(x == 0)        {            if(v*v/19.6 >= y)                printf("%.6f\n",r);            else                puts("-1");            continue;        }        for(int i=0; i<100; ++i)        {            mid = (l+r)/2;            mmid = (mid+r)/2;            if(fun(mid)>fun(mmid))                r = mmid;            else                l = mid;        }        if(fun(l) < y)        {            puts("-1");            continue;        }        l = 0;        for(int i=0; i<100; ++i)        {            mid = (l+r)/2;            if(fun(mid)>y)                r = mid;            else                l = mid;        }        printf("%.6f\n",r);    }    return 0;}