hdu 2603 Wiskey's Power(数学题)，待改正

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本文介绍了一个物理模拟问题，即计算呕吐物从特定高度抛出并碰撞天花板后的落地坐标。问题包含速度、角度等参数，并提供了AC代码实现。

1、http://acm.hdu.edu.cn/showproblem.php?pid=2603

2、题目大意

一个物品重m，从3米处上抛，知道上抛的角度，3.5米处有隔板，撞到隔板后以相同角度往下走，求最终落地的坐标

3、这是一道解物理方程的题目，但是不知道哪里错了

题目：

Wiskey's Power

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 393 Accepted Submission(s): 180

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/s ², 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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4、AC代码：

#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<string.h>
int main()
{
    double m,v,a;
    double vx1,vy1,h1=0.5,h2=3.5,vx,vy,vz,t1,t2,g=9.87,x1,x2;
    while(scanf("%lf%lf%lf",&m,&v,&a)!=EOF)
    {
        vx1=v*cos(a*3.1415926/180);
        vy1=v*sin(a*3.1415926/180);
        if(vy1*vy1/(2*g)<=h1)
        {
            t1=vy1/g;
            t2=sqrt(2*h2/g);
            printf("%.3f\n",vx1*(t1+t2));
        }
        else
        {
            t1=(vy1-sqrt(vy1*vy1-2*g*h1))/g;
            vx=vx1;
            vy=vy1-g*t1;
            x1=vx*t1;
            t2=(-vy+sqrt(vy*vy+2*g*h2))/g;
            x2=vx*t2;
            printf("%.3f\n",x1+x2);
        }
    }
    return 0;
}
/*
100 200 50
200 10 35
*/