思路：比赛的时候看到样例答案就是8/7，然后猜了几发公式...

正解：

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。

，

，容易列出方程

上下界都是清晰的，定积分一下：

直接把第一个式子代到第二个里面

（或者且）。

#include<cmath>
#include<cstring>
int main()
{
    double a,v1,v2;
    while(scanf("%lf%lf%lf",&a,&v1,&v2)==3)
    {
        if(a==0)
        {
            printf("0.0000000000\n");
            continue;
        }
        if(v1<=v2)
        {
            printf("Infinity\n");
            continue;
        }
        double ans=a*v1,p=v1*v1-v2*v2;
        ans/=p;
        printf("%.10lf\n",ans);
    }

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

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