PAT甲级真题 1006 Sign In and Sign Out (25分) C++实现（同时求最大、最小值，比较string）

最新推荐文章于 2022-08-12 23:25:22 发布

原创最新推荐文章于 2022-08-12 23:25:22 发布 · 231 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #字符串

PAT 专栏收录该内容

209 篇文章

订阅专栏

本文介绍了一种简单而有效的方法来确定谁是最早到达和最后离开办公室的人。通过比较每个员工的签到和签退时间，算法能够快速找出每天开门和关门的员工，确保了办公室的安全管理。

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 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

题目大意：给了id对应的到来时间和离开时间，求最早来的和最晚走的。

思路

可直接用大于>和小于<比较string，用两个临时变量分别记录最小时间和最大时间，再用两个临时变量记录相应的id即可。

代码

#include <iostream>
using namespace std;

int main(){
    int n;
    cin >> n;
    string minId = "";
    string maxId = "";
    string minTime = "23:59:59";
    string maxTime = "00:00:00";
    for (int i=0; i<n; i++){
        string id, t1, t2;
        cin >> id >> t1 >> t2;
        if (t1 < minTime){
            minTime = t1;
            minId = id;
        }
        if (t2 > maxTime){
            maxTime = t2;
            maxId = id;
        }
    }
    cout << minId << " " << maxId << endl;
    return 0;
}