Toxophily HDU

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

发射炮弹，计算抛物线能否击中某个坐标，如果能则输出最小角度。
直接三分得出的角度无法保证是最小的，而且如果判断条件是坐标是否相符也无法解决不可能击中的情况，所以要先三分找射击高度最高的情况下的角度，如果大于y坐标，则再从0到r二分找能击中的角度。

#include <stdio.h>#include <math.h>double x,y,v,sy,vx,vy,ty,tx,t;using namespace std;const double PI=acos(-1.0);bool check(double r){    vx=v*cos(r);    vy=v*sin(r);    tx=x/vx;    sy=vy*tx-0.5*9.8*tx*tx;    return (sy>y);}double ym(double r){    vx=v*cos(r);    vy=v*sin(r);    tx=x/vx;    sy=vy*tx-0.5*9.8*tx*tx;    return sy;}int main(){    int T;    double low,high,mid,mmid,cm,cmm;    scanf("%d",&T);    while(T--){        scanf("%lf%lf%lf",&x,&y,&v);        low=0;high=PI/2;        while(high-low>1e-10){            mid=(low+high)/2;            mmid=(mid+high)/2;            cm=ym(mid);            cmm=ym(mmid);           // printf("%.6f %.6f %.6f %.6f\n",mid,cm,mmid,cmm);            if(cm<cmm){                low=mid;            }else{                high=mmid;            }        }        if(cm<y){            printf("-1\n");        }else{            low=0,high=mid;            while(high-low>1e-10){                mid=(low+high)/2;                if(check(mid)){                    high=mid;                }else{                    low=mid;                }            }            printf("%.6lf\n",high);        }    }    return 0;}