hdu 1038 Biker's Trip Odometer(水题)
题目链接:http://acm.hdu.edu.cn/showproblem.php?
pid=1038
Biker's Trip Odometer
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.
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).
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.
26 1000 5
27.25 873234 3000
26 0 1000
Trip #1: 1.29 928.20
Trip #2: 1179.86 1415.84
pid=1064" target="_blank">1064
1084 1031 1027求路程和速度。
题目中直径给的是inche英尺,给的时间是second秒。可是终于要求的是路程是多少mile英里。速度求的是每hour 小时多少mile英里。仅仅要把单位换算清楚就能够了,还有就是直径和时间是浮点型,圈数是整型。
#include <iostream>
#include <cstdio> using namespace std; #define PI 3.1415927 int main()
{
double d,t;
int q;
double s;
double v;
int flag=1;
while (~scanf("%lf%d%lf",&d,&q,&t))
{
if (q==0)
break;
s=PI*d*q/(5280*12);
t/=3600;
v=s/t;
printf ("Trip #%d: %.2lf %.2lf\n",flag++,s,v);
}
return 0;
}
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