HDU 5761 Rower Bo（积分）

题目链接：HDU 5761

题面：

Rower Bo

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 917    Accepted Submission(s): 320
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 v1 ,and the speed of the water flow is v2 .He will adjust the direction of v1 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,v1,v2 .

0≤a≤100 , 0≤v1,v2,≤100 , a,v1,v2 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

 

题意：

    一艘船从(0，a)点出发，希望到达(0,0)点，其每时每刻都受到水流水平向右的速度v2，以及其本身始终朝向（0，0）点的速度v1，求到达(0,0)点所需时间。

解题：

     因为v1有向下的分速度，故肯定会到达x轴，但达到x轴能否到达(0,0)点，是看v1,v2的大小关系，如果v1<=v2，那么是肯定到不了的，输出Infinity，但a=0是特殊情况，直接输出0。队友比赛时写了个模拟程序，奈何精度不够或T，但却为我们猜公式提供了遍历，我们当时是猜出来的.....

【官方题解】

1010 Rower Bo

首先这个题微分方程强解显然是可以的，但是可以发现如果设参比较巧妙就能得到很方便的做法。

先分解v1v​1​​，

设船到原点的距离是$r$，容易列出方程

drdt=v2cosθ−v1​dt​​dr​​=v​2​​cosθ−v​1​​

dxdt=v2−v1cosθ​dt​​dx​​=v​2​​−v​1​​cosθ

上下界都是清晰的，定积分一下：

0−a=v2∫0Tcosθdt−v1T0−a=v​2​​∫​0​T​​cosθdt−v​1​​T

0−0=v2T−v1∫0Tcosθdt0−0=v​2​​T−v​1​​∫​0​T​​cosθdt

直接把第一个式子代到第二个里面

v2T=v1v2(−a+v1T)v​2​​T=​v​2​​​​v​1​​​​(−a+v​1​​T)

T=v1av12−v22T=​v​1​​​2​​−v​2​​​2​​​​v​1​​a​​

这样就很Simple地解完了，到达不了的情况就是v1<v2v​1​​<v​2​​（或者$a>0$且v1=v2v​1​​=v​2​​）。