hdu5761Rower Bo+数学积分

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

​dt/dr​​=v​2​​cosθ−v​1​​

dx/dt=v2−v1cosθ

上下界都是清晰的，定积分一下：

相对于Y轴方向：0−a=v2∫T0cosθdt−v1T
相对于x轴方向：0−0=v2T−v1∫T0cosθdt

直接把第一个式子代到第二个里面

v2T=v1v2(−a+v1T)v​2​​T

T=v1av12−v22​​

#include<cstdio>#include<cstring>#include<cmath>#include<map>using namespace std;#define LL long longint main(){    int a,v1,v2;    while(scanf("%d %d %d",&a,&v1,&v2)!=EOF){        if(a==0) printf("0.0000000000\n");        else if(v1<=v2) printf("Infinity\n");        else{            printf("%.10f\n",1.0*a*v1/(v1*v1-v2*v2));        }    }    return 0;}