HDU 5761 Rower Bo

求解船到河岸最短时间问题

最新推荐文章于 2025-09-10 22:21:01 发布

转载最新推荐文章于 2025-09-10 22:21:01 发布 · 55 阅读

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原文链接：http://www.cnblogs.com/pk28/p/5708833.html

文章标签：

#java

本文介绍了一种计算船只从特定位置出发到达河岸原点所需最短时间的方法。考虑了水流速度和船只相对于水的速度，通过数学公式计算得出结果，并提供了具体的编程实现。

Rower Bo

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 502 Accepted Submission(s): 146
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

Input

There are several test cases. (no more than 1000)

For each test case,there is only one line containing three 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

Sample Input

2 3 3 2 4 3

Sample Output

Infinity

1.1428571429

/* ***********************************************
Author        :guanjun
Created Time  :2016/7/26 15:54:55
File Name     :p310.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10010
#define cle(a) memset(a,0,sizeof(a))
#define pi 3.1415926
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
priority_queue<int,vector<int>,greater<int> >pq;
struct Node{
    int x,y;
};
struct cmp{
    bool operator()(Node a,Node b){
        if(a.x==b.x) return a.y> b.y;
        return a.x>b.x;
    }
};

bool cmp(int a,int b){
    return a>b;
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    double a,v1,v2;
    while(cin>>a>>v1>>v2){
        if(a==0){
            printf("%.7f\n",0);continue;
        }
        if(v1<=v2){
            puts("Infinity");
        }
        else{
            printf("%.7f\n",a*v1/(v1*v1-v2*v2));
        }
    }
    return 0;
}