poj 2501 Average Speed

Average Speed

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4842		Accepted: 2168

Description

You have bought a car in order to drive from Waterloo to a big city. The odometer on their car is broken, so you cannot measure distance. But the speedometer and cruise control both work, so the car can maintain a constant speed which can be adjusted from time to time in response to speed limits, traffic jams, and border queues. You have a stopwatch and note the elapsed time every time the speed changes. From time to time you wonder, "how far have I come?". To solve this problem you must write a program to run on your laptop computer in the passenger seat.

Input

Standard input contains several lines of input: Each speed change is indicated by a line specifying the elapsed time since the beginning of the trip (hh:mm:ss), followed by the new speed in km/h. Each query is indicated by a line containing the elapsed time. At the outset of the trip the car is stationary. Elapsed times are given in non-decreasing order and there is at most one speed change at any given time.

Output

For each query in standard input, you should print a line giving the time and the distance travelled, in the format below.

Sample Input

00:00:01 100
00:15:01
00:30:01
01:00:01 50
03:00:01
03:00:05 140

Sample Output

00:15:01 25.00 km
00:30:01 50.00 km
03:00:01 200.00 km

Source

Waterloo local 2001.09.22

 

分析：

模拟，注意一开始可能就问，或者一开始赋予0的速度。

getchar()判断是询问还是更新。

具体代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<stack>
using namespace std;
double trans(string s){
    return ((s[0]-'0')*10+s[1]-'0')*3600+((s[3]-'0')*10+s[4]-'0')*60+(s[6]-'0')*10+s[7]-'0';
}
int main(){
    string s;
    double nowspeed=0,nowtime=0,speed,time,havdrive=0;
    while(cin>>s){
        time=trans(s);
        if(getchar()==' '){//速度更新,已行驶的路程更新,基准时间更新
            cin>>speed;
            havdrive+=(time-nowtime)/3600*nowspeed;
            nowtime=time;
            nowspeed=speed;
        }
        else{
            cout<<s;
            printf(" %.2f km\n",havdrive+(time-nowtime)/3600*nowspeed);//写成printf(" %.2f km\n",havdrive+(time-nowtime)/3600*nowspeed)就错了！！
        }
    }
    return 0;
}