1006 Sign In and Sign Out (25 分) PAT 甲级简单题

ゞ浴缸里的玫瑰 2024-02-19 20:53 18阅读 0赞

1006 Sign In and Sign Out (25 分)

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

细节太多，需要注意

我的代码：

#include<iostream>
    #include<cstring>
    using namespace std;
    int main()
    {
    	int m;
    	string in,out;
    	string temp;
    	int ih,im,is,oh,om,os;
    	int h1,m1,s1,h2,m2,s2;
    	scanf("%d",&m);
    	for(int i=0;i<m;i++)
    	{
    		int i_f=0,o_f=0;
    		cin>>temp;
    		scanf("%2d:%2d:%2d %2d:%2d:%2d",&h1,&m1,&s1,&h2,&m2,&s2);
    		if(i==0)
    		{
    			ih=h1,im=m1,is=s1,oh=h2,om=m2,os=s2;
    			in=temp;
    			out=temp;
    		}
    		else
    		{
    			if(h1<ih)
    			i_f=1;
    			else
    			if(h1==ih&&m1<im)
    			i_f=1;
    			else
    			if(h1==ih&&m1==im&&s1<is)
    			i_f=1;
    			if(h2>oh)
    			o_f=1;
    			else
    			if(h2==oh&&m2>om)
    			o_f=1;
    			else
    			if(h2==oh&&m2==om&&s2>os)
    			o_f=1;
    		}
    		if(i_f==1)
    		{
    			ih=h1,im=m1,is=s1;
    			in=temp;
    		}
    		if(o_f==1)
    		{
    			oh=h2,om=m2,os=s2;
    			out=temp;
    		}
    		
    	}
    	cout<<in<<" "<<out<<endl;
    	return 0;
    }

别人的代码：

解题思路

读入M。  
进行M次循环。设置开门时间char t\_open\[\] = 235959，关门时间char t\_close\[\] = 000000。开门人char id\_open\[16\]，关门人id\_close\[16\]。  
1.读入身份号码，上班时间和下班时间。  
2.使用strcmp将上班时间与t\_open进行比较，如果上班时间小于t\_open，那么strcpy(t\_open,上班时间)。strcpy(id\_open, id)。  
3.使用strcmp将下班时间与t\_close进行比较，如果下班时间大于t\_close，那么strcpy(t\_close, 下班时间)。strcpy(id\_close, id)。

输出结果。

#include <iostream>
    #include <cstdio>
    #include <cstring>
    
    int main(void)
    {
        setvbuf(stdin, new char[1 << 20], _IOFBF, 1 << 20);
        setvbuf(stdout, new char[1 << 20], _IOFBF, 1 << 20);
        int m;
        scanf("%d", &m);
        char id_open[16], id_close[16], t_open[] = "23:59:59", t_close[] = "00:00:00";
        char id[16], arrive[9], depart[9];
        for (int i = 0; i < m; i++) {
            scanf("%s %s %s", id, arrive, depart);
            if (strcmp(t_open, arrive) > 0) {
                strcpy(id_open, id);
                strcpy(t_open, arrive);
            }
            if (strcmp(t_close, depart) < 0) {
                strcpy(id_close, id);
                strcpy(t_close, depart);
            }
        }
        printf("%s %s\n", id_open, id_close);
        return 0;
    }