PAT Advanced 1006. Sign In and Sign Out (25) (C语言实现)

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题目

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 , which is the total number
of records, followed by 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

思路

基本和PAT Basic 1028. 人口普查(20)（C语言实现）思路一致。

找到最早签到和最晚签出的即可。

代码

最新代码@github，欢迎交流

#include <stdio.h>
#include <string.h>

int main()
{
    int N, HH, MM, SS;
    int firsttime = 86400, lasttime = -1, time;
    char firstname[16], lastname[16], name[16];

    scanf("%d", &N);
    for(int i = 0; i < N; i++)
    {
        scanf("%s %d:%d:%d", name, &HH, &MM, &SS);
        time = (HH * 60 + MM) * 60 + SS;
        if(time < firsttime)
        {
            firsttime = time;
            strcpy(firstname, name);
        }

        scanf("%d:%d:%d", &HH, &MM, &SS);
        time = (HH * 60 + MM) * 60 + SS;
        if(time > lasttime)
        {
            lasttime = time;
            strcpy(lastname, name);
        }
    }

    printf("%s %s", firstname, lastname);

    return 0;
}