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. 

 

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. 

 

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. 

 

3

0.222018 23.901887 121.909183

39.096669 110.210922 20.270030

138.355025 2028.716904 25.079551 

 

1.561582

-1

-1 

 

第一种是通过数学公式求解，另外一种是三分一次高度所相应的倾斜角，再二分符合条件的倾斜角

公式法：

有题目能够知道：x,y,v都是已知条件

设vx=v*cos(α),vy=v*sin(α),同一时候从P(0,0)点到达目标点花了t时间,重力加速度为G=9.8.

∴x=vx*t,y=vy*t-1/2*G*t².

消掉vx,vy,t能够转换为y=v*sin(α)*x/(v*cos(α))-1/2*g*x²/(v²*cos(α)²).

∴将sin(α)/cos(α)=tan(α);

∴y=v*x*tan(α)-(1/2*g*x²/v²)*((sin(α)²+cos(α)²)/cos(α)²);

∴y=v*x*tan(α)-(1/2*g*x²/v²)*(1+tan(α)²);

∴将其进行整理能够得到:g*x²*tan(α)²-2*v²*x*tan(α)+2*v²y+g*x²=0;

∴能够得到△=b²-4*a*c;

∴令a=g*x²,b=-2*v²*x,c=2*v²y+g*x².

又∵x1=(-b+(b²-4*a*c)½)/(2*a),x2=(-b-(b²-4*a*c)½)/(2*a).

∴能够通过上述公式将tan(α)求出,然后就是通过atan((tan(α)))将α求出

接着检查α是否符合条件就能够了。

/*

Author: 2486

Memory: 1616 KB		Time: 0 MS

Language: C++		Result: Accepted

*/

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

using namespace std;

const double PI=acos(-1);

const double G=9.8;

int T;

double x,y,v;

int main() {

    scanf("%d",&T);

    while(T--) {

        scanf("%lf%lf%lf",&x,&y,&v);

        double a=G*x*x,b=-2.0*v*v*x,c=2.0*v*v*y+G*x*x;

        double posi=(-b+sqrt(b*b-4.0*a*c))/2.0/a;

        double ne=(-b-sqrt(b*b-4.0*a*c))/2.0/a;

        posi=atan(posi),ne=atan(ne);

        if(posi>=0&&posi<=PI/2.0&&ne>=0&&ne<=PI/2.0) {

            printf("%.6lf\n",posi>ne?

ne:posi);

        } else if(ne>=0&&ne<=PI/2.0) {

            printf("%.6lf\n",ne);

        } else if(posi>=0&&posi<=PI/2.0) {

            printf("%.6lf\n",posi);

        } else printf("-1\n");

    }

    return 0;

}

三分二分方法

/*

Author: 2486

Memory: 1628 KB		Time: 0 MS

Language: C++		Result: Accepted

*/

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

using namespace std;

const double PI=acos(-1);

const double eps=1e-10;

int T;

double x,y,v;

double C(double m) {

    double vx=v*cos(m),vy=v*sin(m);

    return (vy*x)/vx-9.8*(x/vx*x/vx)/2.0;

}

bool B(double m) {

    double vx=v*cos(m),vy=v*sin(m);

    return (vy*x)/vx-9.8*(x/vx*x/vx)/2.0>=y;

}

int main() {

    scanf("%d",&T);

    while(T--) {

        scanf("%lf%lf%lf",&x,&y,&v);

        double lb=0,ub=PI/2.0;

        ///////////////求出最大高度所相应的倾斜度////////////////

        while(ub-lb>eps) {

            double mid=(ub+lb)/2.0;

            double mmid=(ub+mid)/2.0;

            if(C(mid)>C(mmid)) {

                ub=mmid;

            } else lb=mid;

        }

        if(C(ub)<y) {

            printf("-1\n");

            continue;

        }

        ///////////////////////////////

        lb=0;

        ////////////////求出无限接近目标的倾斜度///////////////

        while(ub-lb>eps) {

            double mid=(ub+lb)/2.0;

            if(B(mid)) {

                ub=mid;

            } else lb=mid;

        }

        ///////////////////////////////

        printf("%.6lf\n",ub);

    }

    return 0;

}