Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Special Judge
Problem Description
There is a river on the Cartesian coordinate system,the river is flowing along the x-axis direction.
Rower Bo is placed at $(0,a)$ at first.He wants to get to origin $(0,0)$ by boat.Boat speed relative to water is $v_1$,and the speed of the water flow is $v_2$.He will adjust the direction of $v_1$ to origin all the time.
Your task is to calculate how much time he will use to get to origin.Your answer should be rounded to four decimal places.
If he can't arrive origin anyway,print "Infinity" (without quotation marks).
Input
There are several test cases. (no more than 1000)
For each test case,there is only one line containing three integers $a,v_1,v_2$.
$0\le a\le 100, 0\le v1,v2,\le 100, a,v_1,v_2$ are integers
Output
For each test case,print a string or a real number.
If the absolute error between your answer and the standard answer is no more than $10^{-4}$, your solution will be accepted.
Sample Input
2 3 3
2 4 3
Sample Output
Infinity
1.1428571429
Source
2016 Multi-University Training Contest 3
这题在现场做的时候试图直接解微分方程，但微分方程推出来却发现没法解，但题解上说，直接解微分方程可以搞，可能是我推微分方程的姿势不对，有待研究，微积分快忘干净了，好惨。
我现场的思路是：
考虑直角坐标下的运动学方程：
\begin{align} \frac{dx}{dt} &= v_2-\frac{x}{\sqrt{x^2+y^2}}v_1 \\ \frac{dy}{dt} &= -\frac{y}{\sqrt{x^2+y^2}}v_1 \end{align}
注意到如果将运动学方程写成直角坐标$(r,\theta)$的形式会更方便：
由$x=r\cos{\theta}, y=r\sin{\theta}, r=\sqrt{x^2+y^2}$, 上面两式可化成：
\begin{align} \cos{\theta} \, \dot{r} -\dot{\theta}\sin{\theta} \, r &= v_2-v_1\cos{\theta} \\ \sin{\theta} \, \dot{r} + \cos{\theta} \, \dot{\theta} r &= -v_1\sin{\theta} \end{align}
由上述两方程可解出$\dot{r}, r$：
\begin{align} \dot{r} &= v_2\cos{\theta}-v_1 \\ r &= -\frac{v_2\sin{\theta}}{\dot{\theta}} \end{align}
将上两式代入
\begin{align} \dot{r}=\frac{dr}{dt} \end{align}
得
\begin{align} \frac{v_2\sin{\theta} \, \ddot{\theta}}{\dot{\theta}^2}-v_2\cos{\theta} = v_2\cos{\theta}-v_1 \end{align}
要从这个微分方程解出$\theta=\theta(t)$比较困难，我目前还解不出。
这个微分方程是可解的：
令
\[ \frac{d\theta}{dt} = f \]
则
\[ \frac{d^2\theta}{dt^2} = \frac{df}{dt} =\frac{df}{d\theta}\frac{d\theta}{dt} = \frac{df}{d\theta}f \]
将上两式代入原方程，得
\[ \frac{df}{f} = \frac{2v_2\cos{\theta} - v_1}{v_2 \sin{\theta}} d\theta = (2\cot{\theta} -\frac{v_1}{v_2}\csc{\theta}) d\theta \]
积分得
\[ \ln{f} = 2\ln{| \sin{\theta} |} - \frac{v_1}{v_2}\ln{| \csc{\theta} -\cot{\theta} |} +C_1 \]
即
\[ f(\theta) = C_2\frac{\sin^2{\theta}} {|\csc{\theta}-\cot{\theta}|^{ \frac {v_1} {v2} } } \]
但到这一步貌似也没什么用，就算把右边的关于$\theta$的积分积出来，还要再求反函数才能得到$\theta (t)$。再次好惨。。
可能如果选择解$r=r(t)$会简单一点，再试试吧。。逃。。。
题解上给出的做法是：
将
\[ \frac{dr}{dt} = v_2\cos{\theta}-v_1 \]
\[ \frac{dx}{dt} = v_2-\frac{x}{\sqrt{x^2+y^2}}v_1 = v_2 - v_1 \cos{\theta}\]
实际上第一式（$\frac{dr}{dt}$的表达式）可直接写出来，不需要像上面那样去推导。由于题目已说明船相对于水流的速度时刻指向原点，那么自然有$ \frac{dr}{dt} = v_2\cos{\theta}-v_1 $
两式对$t$积分，从时刻$0$积到（到达原点的）时刻$T$,这两个定积分就写成：
\[ 0-a = - v_1T + v_2\int_{0}^{T}{\cos{\theta} \, \text{d}\theta} \]
\[0-0 = v_2T - v_1 \int_{0}^{T}{\cos{\theta} \, \text{d}\theta} \]
这样便可解出
\[ T = \frac{v_1a}{v_1^2 - v_2^2} \]
从而不能到达原点的情况是 $a>0$且$v_1\le v_2$。