1006 Sign In and Sign Out (25)(25 分)

At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in's and out's, you are supposed to find the ones who have unlocked and locked the door on that day.

Input Specification:

Each input file contains one test case. Each case contains the records for one day. The case starts with a positive integer M, which is the total number of records, followed by M lines, each in the format:

ID_number Sign_in_time Sign_out_time

where times are given in the format HH:MM:SS, and ID number is a string with no more than 15 characters.

Output Specification:

For each test case, output in one line the ID numbers of the persons who have unlocked and locked the door on that day. The two ID numbers must be separated by one space.

Note: It is guaranteed that the records are consistent. That is, the sign in time must be earlier than the sign out time for each person, and there are no two persons sign in or out at the same moment.

Sample Input:

3
CS301111 15:30:28 17:00:10
SC3021234 08:00:00 11:25:25
CS301133 21:45:00 21:58:40

Sample Output:

SC3021234 CS301133
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<vector>
#include<stack>
#define inf 0x3f3f3f3f
using namespace std;
int compare(int x,int y,int z,int a,int b,int c)
{
if(x==a)
{
if(y==b)
{
if(z>=c)
return ;
else return ;
}
else if(y>b)
return ;
else return ;
}
else if(x>a)
return ;
else
return ;
} int main()
{
int n;
while(cin>>n)
{
string mi,ma;
int ai=,bi=,ci=;
int aa=,ba=,ca=;
for(int i=;i<=n;i++)
{
string s;
cin>>s;
int a,b,c;
scanf("%d:%d:%d",&a,&b,&c);
if(compare(ai,bi,ci,a,b,c)==)
{
mi=s;
ai=a;
bi=b;
ci=c;
}
scanf("%d:%d:%d",&a,&b,&c);
if(compare(aa,ba,ca,a,b,c)==)
{
ma=s;
aa=a;
ba=b;
ca=c;
}
}
cout<<mi<<" "<<ma;
}
return ;
}
 

1006 Sign In and Sign Out (25)(25 分)思路:普通的时间比较题。。。的更多相关文章

  1. PAT 甲级 1006 Sign In and Sign Out (25)(25 分)

    1006 Sign In and Sign Out (25)(25 分) At the beginning of every day, the first person who signs in th ...

  2. 1006 Sign In and Sign Out (25 分)

    1006 Sign In and Sign Out (25 分) At the beginning of every day, the first person who signs in the co ...

  3. PAT甲 1006. Sign In and Sign Out (25) 2016-09-09 22:55 43人阅读 评论(0) 收藏

    1006. Sign In and Sign Out (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue ...

  4. pat 1006 Sign In and Sign Out(25 分)

    1006 Sign In and Sign Out(25 分) At the beginning of every day, the first person who signs in the com ...

  5. PAT (Advanced Level) Practice 1006 Sign In and Sign Out (25 分) 凌宸1642

    PAT (Advanced Level) Practice 1006 Sign In and Sign Out (25 分) 凌宸1642 题目描述: At the beginning of ever ...

  6. 1006. Sign In and Sign Out (25)

    At the beginning of every day, the first person who signs in the computer room will unlock the door, ...

  7. pat 1006 Sign In and Sign Out (25)

    At the beginning of every day, the first person who signs in the computer room will unlock the door, ...

  8. 1006 Sign In and Sign Out (25)(25 point(s))

    problem At the beginning of every day, the first person who signs in the computer room will unlock t ...

  9. 1006.Sign in and Sign out(25)—PAT 甲级

    At the beginning of every day, the first person who signs in the computer room will unlock the door, ...

随机推荐

  1. 使用游标、存储过程、pivot 三种方法导入数据

    --使用游标循环 if (exists (select * from sys.objects where name = 'Base_RecordTend_Test')) drop proc Base_ ...

  2. 网页宽高clientWidth clientHeight获得数值不对的问题

    当网页内容撑不满一屏时,通过以下代码获得整个网页高度会有问题 document.body.clientHeight;document.body.clientWidth; 得到的宽高不对,可能是因为ht ...

  3. [置顶] kubernetes1.7新特性:PodDisruptionBudget控制器变化

    背景概念 在Kubernetes中,为了保证业务不中断或业务SLA不降级,需要将应用进行集群化部署.通过PodDisruptionBudget控制器可以设置应用POD集群处于运行状态最低个数,也可以设 ...

  4. Linux 释放物理内存和虚拟内存

    1.查看内存占用情况 $ free -m -h total used free shared buff/cache available Mem: .7G .0G .9G 385M 780M .0G S ...

  5. iOS中求出label中文字的行数和每一行的内容

    今天遇到一个需求,需要计算label中文字的行数.想了好久也没想到好的解决方法,就在网上找了下.结果发现一篇文章是讲这个的.这部分代码不但能够求出一个label中文字行数,更厉害的是能够求出每一行的内 ...

  6. yum安装php5.5,php5.6和php7.0

    本文主要介绍在CentOS系统下的php多个版本的安装使用 1.清理系统上的旧版本php 1)查询已安装的php软件 rpm -qa|grep php* yum list installed | gr ...

  7. CentOS 7.4搭建Kubernetes 1.8.5集群

    环境介绍 角色 操作系统 IP 主机名 Docker版本 master,node CentOS 7.4 192.168.0.210 node210 17.11.0-ce node CentOS 7.4 ...

  8. Android Dialog 创建上下文菜单

    Android Dialog中的listview创建上下文菜单 listView.setOnCreateContextMenuListener(new OnCreateContextMenuListe ...

  9. discuz数据库迁移,改密码后,相关配置文件修改

    网上看到这篇文章,觉得有用就收藏下 网站系统需要修改的位置有两处 Discuz 和 UC-center ①路径:/wwwroot/config/config_global.php 这个根据你网站安装的 ...

  10. WebLogic11g-创建域(Domain)及基本配置

    转:http://www.codeweblog.com/weblogic11g-%e5%88%9b%e5%bb%ba%e5%9f%9f-domain-%e5%8f%8a%e5%9f%ba%e6%9c% ...