HDU 1036 Average is not Fast Enough! （水题）

Average is not Fast Enough!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6475    Accepted Submission(s): 2730

Problem Description

A relay is a race for two or more teams of runners. Each member of a team runs one section of the race. Your task is to help to evaluate the results of a relay race. 

You have to process several teams. For each team you are given a list with the running times for every section of the race. You are to compute the average time per kilometer over the whole distance. That's easy, isn't it? 
So if you like the fun and challenge competing at this contest, perhaps you like a relay race, too. Students from Ulm participated e.g. at the "SOLA" relay in Zurich, Switzerland. For more information visit http://www.sola.asvz.ethz.ch/ after the contest is over.

 

Input

The first line of the input specifies the number of sections n followed by the total distance of the relay d in kilometers. You may safely assume that 1 <= n <= 20 and 0.0 < d < 200.0. Every following line gives information about one team: the team number t (an integer, right-justified in a field of width 3) is followed by the n results for each section, separated by a single space. These running times are given in the format "h:mm:ss" with integer numbers for the hours, minutes and seconds, respectively. In the special case of a runner being disqualified, the running time will be denoted by "-:--:--". Finally, the data on every line is terminated by a newline character. Input is terminated by EOF.

 

Output

For each team output exactly one line giving the team's number t right aligned in a field of width 3, and the average time for this team rounded to whole seconds in the format "m:ss". If at least one of the team's runners has been disqualified, output "-" instead. Adhere to the sample output for the exact format of presentation.

 

Sample Input

  

   2 12.5
  5 0:23:21 0:25:01
 42 0:23:32 -:--:--
  7 0:33:20 0:41:35
  

 

Sample Output

  

     5: 3:52 min/km
 42: -
  7: 6:00 min/km
  

 

Source

University of Ulm Local Contest 2001

题目大意：给出一队赛跑的队员数以及赛跑的路程，求出所有队员所用时间总和的平均速度；如样例：2表示每队有两名队员，12.5表示路程，5、42、7表示队伍编号（无用），后面的时间分别表示每个队员所用的时间，“-”则表示队员不合格。

这一题看上去很复杂，我刚拿到题的时候在稿纸上算了很多遍，总以为题意给的是开始时间和结束时间，后来又仔细读了一下题目，才了解了题目的意思；

总体来说是一道水题，只需要将所有时间相加然后除以路程即可，因为存在不合格情况，所以不能直接定义int型，而是用char型进行间接转换；

最大的难点在于格式输出，留的空格数以及空格补不补0是最需要注意的，否则很容易就会PE。

#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
int main(){
    int n;
    double mile;
    scanf("%d%lf",&n,&mile);
    int number;
    char s[7];
    while(~scanf("%d",&number)){
        printf("%3d:",number);//注意输出格式
        double sum=0;
        bool flag=true;
        for(int i=0;i<n;i++){
            scanf("%s",s);
            if(s[0]=='-'&&flag==true)//判断是否合格
                flag=false;
            else
                sum+=(s[0]-'0')*3600+((s[2]-'0')*10+s[3]-'0')*60+(s[5]-'0')*10+(s[6]-'0');//将char型转化为int型并进行求和
        }
        if(flag){
            int x=(sum/mile+0.5)/60;//将结果进行四舍五入求出m
            int y=(int)(sum/mile+0.5)%60;//将结果四舍五入求出s
            printf("%2d:%02d min/km\n",x,y);//注意输出格式
        }
        else
            printf(" -\n");
    }
    return 0;
}