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最新推荐文章于 2017-08-14 12:41:57 发布

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探讨了在物理模型中，一艘船如何调整其相对于水流的速度以最短时间抵达目标地点的问题。通过数学公式推导出船抵达目的地所需的时间。

Rower Bo

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1327 Accepted Submission(s): 506
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

题意：物理..一条船过河（位于y轴到原点），速度时刻朝向原点...英语大渣竟然没看懂这句话，给船速水速求时间，

思路：大物...V分1，V分2列方程： (V船cosα-V水)dt = 0,(V船-V水cosα)dt = 0；解方程..

因为要约掉cosα所以 V分2不能列垂直方向上的方程（垂直出来sinα）

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>

using namespace std;

int main()
{
    double a,x,y;
    while(cin>>a>>x>>y)
    {
        if(a)
        {
            if(x > y)
                printf("%.6f\n",a/(x-(y*y/x)));
            else
                cout<<"Infinity"<<endl;
        }
        else
            cout<<0.000000<<endl;
    }
}