HDU 5761 Rower Bo（多校3）

Rower Bo

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1087    Accepted Submission(s): 401
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

 題解：

代碼：

#include <algorithm>
#include <iostream>
#include <numeric>
#include <cstring>
#include <iomanip>
#include <string>
#include <vector>
#include <cstdio>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#define LL long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const LL Max = (1LL<<32) - 1;
const double esp = 1e-6;
const double PI = 3.1415926535898;
const int INF = 0x3f3f3f3f;
using namespace std;

int main(){
    int a,v1,v2;
    while(~scanf("%d %d %d",&a,&v1,&v2)){
        if(a == 0)
            printf("0\n");
        else if(v1 <= v2)
            printf("Infinity\n");
        else
            printf("%.10f\n",(double)v1*a / (v1*v1-v2*v2));
    }
    return 0;
}