POJ 2501 Average Speed --from lanshui_Yang

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

      题目大意 很简单 ，但有几点需要注意 ：                                                                          第一 ，第一行 没有速度 要也要 输出 “0.00 km”；                                                    第二 ，输入数据时 ，用 scanf或是 cin 会 RE ！！ 要用 gets（）~                                      第三 ，这一点比较奇怪 ，我写的这个程序，用G++交能AC ，但用C++交 会 RE~                     具体讲解请看代码：                                                                        
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<cmath>
using namespace std;
int hh[100500],mm[100500],ss[100500];
char s[1000];
double speed[100500],lu[100500];
int main()
{
    int i = 1,j;
    memset(s,'\0',sizeof(s));
    while (gets(s))
    {
        hh[i]=(s[0]-'0')*10 + (s[1]-'0');
        mm[i]=(s[3]-'0')*10 + (s[4]-'0');
        ss[i]=(s[6]-'0')*10 + (s[7]-'0');
        lu[1] = 0;
        int sumt;
        if(s[8]==' ')
        {
            int l = strlen(s);
            int k;
            speed[i]=0;
            for(k=9;k<l;k++)
            {
                speed[i]*=10;
                speed[i]+=(s[k]-'0');
            }
            if(i > 1)
            {
                sumt = (hh[i]-hh[i-1])*3600 + (mm[i]-mm[i-1])*60 + (ss[i]-ss[i-1]);
                lu[i] = sumt * speed[j]/3.6 + lu[i-1];
            }
            j=i;
            i++;
            memset(s,'\0',sizeof(s));
            continue;
        }
        else
        {
            if(i > 1)
            {
                sumt = (hh[i]-hh[i-1])*3600 + (mm[i]-mm[i-1])*60 + (ss[i]-ss[i-1]);
                lu[i] = sumt * speed[j]/3.6 + lu[i-1];
                printf("%02d:%02d:%02d %.2f km\n",hh[i],mm[i],ss[i],lu[i]/1000);
            }
            else
            printf("%02d:%02d:%02d %.2f km\n",hh[i],mm[i],ss[i],lu[i]/1000);
            memset(s,'\0',sizeof(s));
            i++;
        }
    }
    return 0;
}