poj 2501

Average Speed

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4714 Accepted: 2104

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 10000:15:0100:30:0101:00:01 5003:00:0103:00:05 140

Sample Output

00:15:01 25.00 km00:30:01 50.00 km03:00:01 200.00 km

第一行可以没有速度!!!!卑鄙之极……
#include <cstdio>#include <iostream>#include <string.h>#include <stdlib.h>#include <fstream>#include <math.h>#include <queue>#include <algorithm>#define INF 0x3f3f3f3fusing namespace std;int main(){//freopen ("input.txt","r",stdin);int h,m,s,th=0,tm=0,ts=0;double d;double pd=0;char c;double v=0;while(cin>>h>>c>>m>>c>>s){if(c=(getchar())==' '){pd+=((h-th)*60*60+(m-tm)*60+(s-ts))*(v/3600.0);cin>>v;th=h,tm=m,ts=s;}else{d=pd+((h-th)*60*60+(m-tm)*60+(s-ts))*(v/3600.0);printf("%02d:%02d:%02d %.2f km\n",h,m,s,d);}}return 0;}