| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 4994 | Accepted: 2245 |
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
有趣的一题~
【题意】：给出时间和速度，求距离
总结：
1、时间的输入方法：scanf("%d:%d:%d",&h,&m,&s)，注意积累！
2、关于空格的的输入控制使用char ch = getchar()，同时它还作为了本题的一个是否输出的标识控制的条件。
3、时间的输出方法：printf("%02d:%02d:%02d...
#include<iostream> #include<stdio.h> using namespace std; int cal(int h,int k,int m) { return h*3600+k*60+m; } int main() { int h,k,m; double v=0; double ans=0; int pre=0; while(scanf("%d:%d:%d",&h,&k,&m)!=EOF) { char ch=getchar(); if(ch==' ') { int a; scanf("%d",&a); int now=cal(h,k,m); ans+=(now-pre)*v; pre=now; v=a/3.6; } else { printf("%02d:%02d:%02d %.2f km ",h,k,m,(ans+(cal(h,k,m)-pre)*v)/1000); } } return 0; }