CSU-ACM2017暑期训练5-三分 D

题目：

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

题意：一个果子在平面坐标的(x，y)处，你人在(0,0)处丢出一个速度为v的石子，问：使得石子能扔到果子的扔石子最小角度为多少(初速度与x轴的夹角)

思路：高中物理推出公式：y = v*sinθ*t -1/2*g*t*t，t = x / v*cosθ. 根据此公式三分求得可以达到果子的高度及以上高度的最大角度，然后在二分求得满足要求的最小角度

代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<list>#include<deque>#include<map>#include<queue>using namespace std;typedef long long ll;const double PI = acos(-1.0);const double eps = 1e-6;const int INF = 1000000000;const int maxn = 100;int T;double x,y,v;double l,r;/*    y = v*sinθ*t - 1/2*g*t*t，t = x / v*cosθ。*/double cal(double th)//th为的弧度{    double t=x/(v*cos(th));    return v*sin(th)*t-4.9*t*t;}int main(){    scanf("%d",&T);while(T--)    {        scanf("%lf%lf%lf",&x,&y,&v);        l=0,r=PI/2;        if(x==0)        {            if(v*v/19.6>=y)//v^2=2gy                printf("%.6f",PI/2);            else                printf("-1\n");            continue;        }        for(int i=0;i<=100;i++)          {            double mid,midmid;            mid=(l+r)/2.0;            midmid=(mid+r)/2.0;            if(cal(mid)>cal(midmid))                r=midmid;            else                l=mid;        }        if(cal(r)<y)        {            printf("-1\n");            continue;        }        l=0;        for(int i=0;i<=100;i++)        {            double mid=(l+r)/2.0;            if(cal(mid)>y)                r=mid;            else                l=mid;        }        printf("%.6f\n",r);    }    return 0;}