Biker's Trip Odometer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3061    Accepted Submission(s): 2103

Problem Description
Most bicycle speedometers work by using a Hall Effect sensor fastened to the front fork of the bicycle. A magnet is attached to one of the spokes on the front wheel so that it will line up with the Hall Effect switch once per revolution of the wheel. The speedometer monitors the sensor to count wheel revolutions. If the diameter of the wheel is known, the distance traveled can be easily be calculated if you know how many revolutions the wheel has made. In addition, if the time it takes to complete the revolutions is known, the average speed can also be calculated. For this problem, you will write a program to determine the total distance traveled (in miles) and the average speed (in Miles Per Hour) given the wheel diameter, the number of revolutions and the total time of the trip. You can assume that the front wheel never leaves the ground, and there is no slipping or skidding.
 
Input
Input consists of multiple datasets, one per line, of the form:
diameter revolutions time
The diameter is expressed in inches as a floating point value. The revolutions is an integer value. The time is expressed in seconds as a floating point value. Input ends when the value of revolutions is 0 (zero).
 
Output
For each data set, print:
Trip #N: distance MPH
Of course N should be replaced by the data set number, distance by the total distance in miles (accurate to 2 decimal places) and MPH by the speed in miles per hour (accurate to 2 decimal places). Your program should not generate any output for the ending case when revolutions is 0.
Constants
For p use the value: 3.1415927. There are 5280 feet in a mile. There are 12 inches in a foot. There are 60 minutes in an hour. There are 60 seconds in a minute. There are 201.168 meters in a furlong.
 
Sample Input
26 1000 5
27.25 873234 3000
26 0 1000
 
Sample Output
Trip #1: 1.29 928.20
Trip #2: 1179.86 1415.84
 
Source
 
Recommend
 
 #include <stdio.h>
#define pi 3.1415927 int main()
{
double a,c;
int b,k=;
while(scanf("%lf %d %lf",&a,&b,&c),b)
{
printf("Trip #%d: ",k++);
printf("%.2lf %.2lf\n",*pi*a*b///,*pi*a*b*////c);
}
return ;
}

简单题,轻松水过

hdu_1038_Biker's Trip Odometer_201311021643的更多相关文章

  1. Lesson 4 An existing trip

    Text I have just received a letter from my brother,Tim. He is in Australia. He has been there for si ...

  2. dp or 贪心 --- hdu : Road Trip

    Road Trip Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KB Total submit users: 29 ...

  3. 【poj1041】 John's trip

    http://poj.org/problem?id=1041 (题目链接) 题意 给出一张无向图,求字典序最小欧拉回路. Solution 这鬼畜的输入是什么心态啊mdzz,这里用vector储存边, ...

  4. 1301. The Trip

    A number of students are members of a club that travels annually to exotic locations. Their destinat ...

  5. 三分 --- POJ 3301 Texas Trip

    Texas Trip Problem's Link:   http://poj.org/problem?id=3301 Mean: 给定n(n <= 30)个点,求出包含这些点的面积最小的正方形 ...

  6. 烟大 Contest1024 - 《挑战编程》第一章:入门 Problem C: The Trip(水题)

    Problem C: The Trip Time Limit: 1 Sec  Memory Limit: 64 MBSubmit: 19  Solved: 3[Submit][Status][Web ...

  7. hdu 3018 Ant Trip 欧拉回路+并查集

    Ant Trip Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem ...

  8. URAL 1004 Sightseeing Trip(最小环)

    Sightseeing Trip Time limit: 0.5 secondMemory limit: 64 MB There is a travel agency in Adelton town ...

  9. Codeforces Round #365 (Div. 2) Mishka and trip

    Mishka and trip 题意: 有n个城市,第i个城市与第i+1个城市相连,他们边的权值等于i的美丽度*i+1的美丽度,有k个首都城市,一个首都城市与每个城市都相连,求所有边的权值. 题解: ...

随机推荐

  1. 过河 2005年NOIP全国联赛提高组(离散化+dp)

    1105 过河 2005年NOIP全国联赛提高组  时间限制: 1 s  空间限制: 128000 KB  题目等级 : 钻石 Diamond       题目描述 Description 在河上有一 ...

  2. Unity项目 - 吃豆人Pacman

    项目展示 Github项目地址:Pacman 涉及知识 切片制作 Animations 状态机设置,any state切换,重写状态机 按键读取进行整数距离的刚体移动 用射线检测碰撞性 渲染顺序问题 ...

  3. VUE修改每个页面title

    //index.js routes: [ { name:'home', path: '/home/:openname', component: Home, meta: { title: '首页' } ...

  4. mysql select 操作优先级

    单表查询操作 select filed1,filed2... form table where ... group by ... having .... order by ... limit ... ...

  5. 【转】 Java 集合系列07之 Stack详细介绍(源码解析)和使用示例

    概要 学完Vector了之后,接下来我们开始学习Stack.Stack很简单,它继承于Vector.学习方式还是和之前一样,先对Stack有个整体认识,然后再学习它的源码:最后再通过实例来学会使用它. ...

  6. 忘记Oracle密码

    1./as sysdba 2.然后你忘记密码的用户名例如Scott alter user scott identified by root 3.exit 4.sqlplus 重新登录

  7. [Windows Server 2012] 手工破解MySQL密码

    ★ 欢迎来到[护卫神·V课堂],网站地址:http://v.huweishen.com★ 护卫神·V课堂 是护卫神旗下专业提供服务器教学视频的网站,每周更新视频.★ 本节我们将带领大家:破解MySQL ...

  8. RadioButtonList的兩種實現方式

    一種是修改ItemTemplate,即ListBoxItem裏面的内容 <ListBox ItemsSource="{Binding}"> <ListBox.It ...

  9. Web 服务器与应用服务器的区别是什么?

    不太严谨的说法:web服务器就是负责接收用户的Request,然后响应html等给客户浏览器.应用服务器处理一些业务逻辑等. 作者:luo链接:https://www.zhihu.com/questi ...

  10. std::vector遍历

    std::vector是我在标准库中实用最频繁的容器.总结一下在遍历和创建vector时需要注意的一些地方. 在不考虑线程安全问题的前提下,在C++11中有五种遍历方式. 方式一 for (size_ ...