1006. Sign In and Sign Out (25)[C语言]

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本文介绍了一个签到与签退系统的实现方案，该系统能够记录每天第一个进入和最后一个离开的人，通过比较不同人员的签到和签退时间来确定开门和关门的人选。

1006. Sign In and Sign Out (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 <stdio.h>
#include <stdlib.h>

typedef struct User{
	char name[25];
	char login_in[12];
	char login_out[12];
}user;

static user users[500000] ;
static int n  = 0;

int compare(char u1[] , char u2[]){
	int i  = 0 ;
	for( i = 0 ; i < 12 ; i ++ ){
		if(u1[i] > u2[i]){
			return 1;
		}else if(u1[i] < u2[i]){
			return -1;
		}
	}
	return 0 ;
}

int main(){
	
	int i ; 
	scanf("%d" , &n);
	if(n <=0) return 0;
	int fin = 0 ;
	int fout = 0 ;
	for(i = 0 ; i < n ; i++ ){
		scanf("%s",users[i].name);
		scanf("%s",users[i].login_in);
 		scanf("%s",users[i].login_out);
 		if(i > 0){
 			if(compare(users[fin].login_in , users[i].login_in ) == 1 ){
 				fin = i ;
			}
 			if(compare(users[fout].login_out , users[i].login_out ) == -1 ){
 				fout = i;
			 }
		 }
	}
	printf("%s %s",users[fin].name , users[fout].name) ;

	return 0;
	
}