foj 1164 Average is not Fast Enough!

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 this 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

Sample Output

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

/*
一开始连题目都没看懂，最后知道大致题意为：每个时间是跑总距离的1/n 然后求平均值 min/km 
思路：先把时间转化成秒，去计算，注意这边要四舍五入，网上查的。不然会出错。还有注意到double和int类型的计算

*/
#include<iostream>
using namespace std;
int main()
{
	int n,t;
	double d,sum_s=0;
	int min,s;
	char h,m1,m2,s1,s2;
	bool flag=true;
	int i;
	scanf("%d%lf",&n,&d);
	while(scanf("%d",&t)!=EOF)
	{
		flag=true;
		sum_s=0;
		for(i=0;i<n;i++)
		{
			getchar();//注意回车以及空格
			scanf("%c:%c%c:%c%c",&h,&m1,&m2,&s1,&s2);
			if(h=='-') flag =false;
			if(!flag) continue;
			else 
				s=(h-'0')*3600+((m1-'0')*10+(m2-'0'))*60+(s1-'0')*10+(s2-'0');
			sum_s+=s; 
		}
		if(flag)
		{
			s=1.0*sum_s/d+0.5;//四舍五入
			printf("%3d: %d:%02d min/km\n",t,s/60,s%60);
		}
		else printf("%3d: -\n",t);
	}
	return 0;
}

收获：这题主要是输入问题，其他问题不大，如果使用%c就要注意回车和空格的存储。还有对于 EOF 的使用。四舍五入的写法。

凭着一腔热血，去努力奋斗吧！