Toxophily-数论以及二分三分

G - Toxophily

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  HDU 2298

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

        
        
        
        

         
         
         
         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 
        
        
        
        

 

第一种是通过数学公式求解，第二种是三分一次高度所对应的倾斜角，再二分符合条件的倾斜角

公式法：

有题目可以知道：x,y,v都是已知条件
设vx=v*cos(α),vy=v*sin(α),同时从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可以转换为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);
∴将其进行整理可以得到:g*x²*tan(α)²-2*v²*x*tan(α)+2*v²y+g*x²=0;
∴可以得到△=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).
∴可以通过上述公式将tan(α)求出,然后就是通过atan((tan(α)))将α求出
接着检查α是否符合条件就可以了。

/*
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;
        ///////////////求出最大高度所对应的倾斜度////////////////
        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;
}