PAT (Advanced Level) Practice 1028

最新推荐文章于 2025-07-24 20:33:53 发布

原创最新推荐文章于 2025-07-24 20:33:53 发布 · 200 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#c++ #c语言

PTA(A) 专栏收录该内容

4 篇文章

订阅专栏

本文描述了一个关于记录计算机房进出人员的问题，通过分析输入文件中的一系列签到和签退记录，找出当天第一个解锁门和最后一个锁门的人。文章提供了一个C语言实现的解决方案，通过遍历所有记录并比较时间戳来确定目标人员。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1006 Sign In and Sign Out (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

该题和PTA Basic level 1028类似，不再详述思路

#include <cstdio>
struct{
    char id[20];
    int early;
    int late;
}unlocked,locked,tmp;
int main()
{
    int m;
    char time1[10],time2[10];
    
    unlocked.early = 240000;
    locked.late = 0;
    scanf("%d", &m);
    for(int i = 0;i < m;i++)
    {
        scanf("%s%s%s", tmp.id, time1, time2);
        tmp.early = (time1[0]-'0')*100000+(time1[1]-'0')*10000+(time1[3]-'0')*1000+(time1[4]-'0')*100+(time1[6]-'0')*10+time1[7]-'0';
        tmp.late = (time2[0]-'0')*100000+(time2[1]-'0')*10000+(time2[3]-'0')*1000+(time2[4]-'0')*100+(time2[6]-'0')*10+time2[7]-'0';
        if(tmp.early < unlocked.early) unlocked = tmp;
        if(tmp.late > locked.late) locked = tmp;
    }
    printf("%s %s", unlocked.id, locked.id);
    
    return 0;
}

时间复杂度O(n)