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[]-'')*+s[]-'')*+((s[]-'')*+s[]-'')*+(s[]-'')*+s[]-'';
 }
 int main(){
     string s;
     double nowspeed=,nowtime=,speed,time,havdrive=;
     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)/*nowspeed;
             nowtime=time;
             nowspeed=speed;
         }
         else{
             cout<<s;
             printf(" %.2f km\n",havdrive+(time-nowtime)/*nowspeed);//写成printf(" %.2f km\n",havdrive+(time-nowtime)/3600*nowspeed)就错了！！
         }
     }
     return ;
 }