【PAT题解集合】

1006 Sign In and Sign Out

题目详情 - 1006 Sign In and Sign Out (pintia.cn)

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
1

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
1
2
3
4

Sample Output:

SC3021234 CS301133
1

思路

1. 我们可以用四个变量分别记录最早来的人的 id 和 time 以及最晚走的人的 id 和 time ，在输入的过程中进行更新。
2. 每次输入一个人的信息，都与 open_time 和 close_time 进行比较，如果有更早的或者更晚的就进行更新。注意，第一个人要直接更新，因为它之前没有人所以无法进行比较。

代码

#include
using namespace std;

int main()
{
    //存储开门即最早和关门即最晚的信息，
    string open_id, open_time;
    string close_id, close_time;

    //输入有多少人
    int m;
    cin >> m;

    for (int i = 0; i < m; i++)
    {
        string id, in_time, out_time;
        cin >> id >> in_time >> out_time;

        //判断是否是来的最早的人
        if (!i || in_time < open_time)
        {
            open_id = id;
            open_time = in_time;
        }

        //判断是否是走的最晚的人
        if (!i || out_time > close_time)
        {
            close_id = id;
            close_time = out_time;
        }
    }

    cout << open_id << " " << close_id << endl;

    return 0;
}
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