1006. Sign In and Sign Out (25)

最新推荐文章于 2022-08-12 23:25:22 发布

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本文介绍了一种通过记录人员签到和签退时间来确定每日最先开门和最后关门者的系统实现方法。该方法使用C++语言编程，并通过字符串比较找出最早签到和最晚签退的时间及对应的ID号。

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 <cstdio>
#include <string.h>
using namespace std;
struct Person{
    char id[16];
    char signIn[10];
    char signOut[10];
};
int main()
{
    freopen("in.txt","r",stdin);
    int m;
    scanf("%d",&m);
    Person* p=new Person[m];
    for(int i=0;i<m;i++){
        scanf("%s %s %s",p[i].id,p[i].signIn,p[i].signOut);
    }
    //找出登录时间最小的那个
    int minIndex=0;
    char* min=p[0].signIn;
    for(int i=1;i<m;i++){
        if(strcmp(p[i].signIn,min)<0){
            minIndex=i;
            min=p[i].signIn;
        }
    }
    //找出登出时间最大的那个
    int maxIndex=0;
    char* max=p[0].signOut;
    for(int i=1;i<m;i++){
        if(strcmp(p[i].signOut,max)>0){
            maxIndex=i;
            max=p[i].signOut;
        }
    }
    printf("%s %s",p[minIndex].id,p[maxIndex].id);
    return 0;
}