zoukankan      html  css  js  c++  java
  • HDU-1036 Average is not Fast Enough!

    Average is not Fast Enough!

    http://acm.hdu.edu.cn/showproblem.php?pid=1036

    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
     
    #include<stdio.h>
    int main()
    {
        int n;
        double d;
        int num;
        char h,m1,m2,s1,s2;
        scanf("%d",&n);
        scanf("%lf",&d);
        while(scanf("%d",&num)!=EOF)
        {
            
            printf("%3d: ",num);
            bool flag=true;
            int sumtime=0;
            for(int i=0;i<n;i++)
            {getchar();
                scanf("%c:%c%c:%c%c",&h,&m1,&m2,&s1,&s2);
                if(h=='-') flag=false;
                if(flag==false)continue;
                sumtime+=(h-'0')*3600+((m1-'0')*10+(m2-'0'))*60+(s1-'0')*10+(s2-'0');
            }    
            if(flag==false)printf("-
    ");
            else
            {
                double t1=sumtime/d;
                int t2=(int)(t1+0.5);
                printf("%d:%02d min/km
    ",t2/60,t2%60);
                
            }    
        }    
        return 0;
    }    
  • 相关阅读:
    属性,类方法,静态方法,Python2和3方法
    类的继承
    面向对象空间和组合
    面向对象
    内置函数和匿名函数
    一个有点意思的习题
    APUE学习笔记——10.18 system函数 与waitpid
    Linux服务器静态路由配置
    APUE学习笔记——11 线程同步、互斥锁、自旋锁、条件变量
    APUE学习笔记——11 线程基础
  • 原文地址:https://www.cnblogs.com/cancangood/p/4726421.html
Copyright © 2011-2022 走看看