PAT A 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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本篇介绍了一个简单的程序设计问题，即如何通过记录找出每天最早进入和最晚离开计算机房的人。输入包括每个人的ID及进出时间，输出则是这两人的ID。

原题：

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

即求最早入和最晚出的学号，直接模拟就可以了。

代码把数据都存了再处理，比较浪费空间，其实依次扫过去就可以了……

代码：

#include <iostream>
using namespace std;

int get_time();	//输入时间，转化为秒

int main()
{
	int m,i;
	int *time_in,*time_out;	//入、出时间
	char **num;	//学号
	int time_in_min=0,time_out_max=0;	//最早进入，最晚离开的序号

	cin>>m;
	time_in=new int[m];
	time_out=new int[m];
	num=new char*[m];
	for(i=0;i<m;i++)
		num[i]=new char[16];

	for(i=0;i<m;i++)	//输入数据
	{
		cin>>num[i];
		time_in[i]=get_time();
		time_out[i]=get_time();
		if(time_in[i]<time_in[time_in_min])
			time_in_min=i;
		if(time_out[i]>time_out[time_out_max])
			time_out_max=i;
	}

	cout<<num[time_in_min]<<" "<<num[time_out_max];

	for(i=0;i<m;i++)
		delete[] num[i];
	delete [] num;
	delete [] time_in;
	delete [] time_out;

	return 0;
}

int get_time()	//获取时间，以秒为单位
{
	int t1,t2;
	cin>>t1;
	t1*=60;
	cin.get();	//为跳过时间中的":"分割符，需要一个cin.get()
	cin>>t2;
	t1=(t1+t2)*60;
	cin.get();
	cin>>t2;
	t1+=t2;
	return t1;
}