HDU 2603 Wiskey's Power（物理题）

Wiskey's Power

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 662    Accepted Submission(s): 309

Problem Description

Come back school from the 33^rdACM / ICPC Asia ChenDu, everyone is exhausted, in particular the captain Wiskey. As saying that night, Wiskey drink a lot of wine, just as he blurred, fall to sleep. All of a sudden, Wiskey felt a slight discomfort in the chest, and then vomiting, vomitted all over. The next day, Wiskey is extremely sluggish, vomit still on the train.

Your task is to calculate the coordinates of vomit.
We assume that the quality of vomit is m, and its size can be ignored.
As the figure below:

The vomit start from the blue point A, whose speed is V, and its angle with X-axis is a. If the vomit hit the ceiling, then its value of the speed don't changed and if before the collision the angle of the speed with X-axis is b, then after the collision the angle of the speed with X-axis is b , too.
Ignore air resistance, acceleration due to gravity g = 9.87m/s², calculate and output Q.
(you can assume that the vomit will not fall to the higher berth)

 

Input

Each line will contain three numbers V , m and a (0 <= a < 90)described above.
Process to end of file.

Output

For each case, output Q in one line, accurate up to 3 decimal places.

Sample Input

100 100 45

Sample Output

3.992

Author

WhereIsHeroFrom

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

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题解：一个喝醉的人，在3m高的地方呕吐。。。（我都想吐。。。）他是斜向上吐的，可能会吐到天花板。。。

          问你这个人吐了多远的水平距离。

          物理题啊。套公式即可。

          这是Wiskey's Power ？？？？（我又想笑。。。）

AC代码：

#include
#include
#include
#define PI acos(-1.0)
#define g  9.87

int main()
{
	double v,m,a;
	double H2=0.5; int H1=3;
	while(~scanf("%lf%lf%lf",&v,&m,&a))
	{
		double Vx=v*cos(a*PI/180); //水平速度 
		double Vy=v*sin(a*PI/180);  //垂直速度 
		double x,t1,t2,t;
		if(Vy*Vy/(2*g)>H2)		//吐得超过天花板。。。。 
		{   
	        double t1=2*(Vy-sqrt(Vy*Vy-2*H2*g))/g; //吐上天花板时：上升和下落 
            double t2=(sqrt(Vy*Vy+2*g*H1)-Vy)/g; // 下落 
            t=t1+t2;  
		}
		else   //吐不到天花板 
        {
            t=2*Vy/g+(sqrt(Vy*Vy+2*g*H1)-Vy)/g;
        }
        x=t*Vx; 
		printf("%.3lf\n",x);
	}
	return 0;
}