C171: attendance system

Problem description

The laboratory uses the attendance system to check the attendance of students . The attendance system will record the time of each student entering and leaving the laboratory every time in a day .
Each student has a unique number , Each attendance record includes the student's number , Entry time 、 Time to leave .
Give the attendance records of all students for one day , Please count the time each student works in the laboratory , And give a statistical table of one day according to the working hours from long to short , When the working hours are the same, they are sorted according to the number from small to large .

Enter description

The first line of input contains an integer n, Indicates the number of attendance records .1≤n≤100, The number of students is no more than 100 The positive integer .
Next is n That's ok , Each line is an attendance record , Each record includes the student number k, Entry time t1 And departure time t2 Three items .
t1 and t2 The format is “hh:mm”, Two digit hours and two digit minutes . for example 14:20 It means 2:20 p.m , All times are 24 hourly , And they are all within the same day .

The output shows that

Output statistics sorted by working hours and student numbers . The statistics table contains several rows , Attendance records of each student , It consists of student number and total working hours , The total working time is in minutes .

sample input

5
3 08:00 11:50
1 09:00 12:00
3 13:50 17:30
1 14:00 18:00
2 17:00 24:00

sample output

3 450
1 420
2 420

#include<stdio.h>
int main()
{
    
	int n,i,j,j1,j2,c1,c2,k;
	struct stu
	{
    
		int h;
		int z;
	}l[100],t;
	scanf("%d",&n);
	k=n;
	for(i=0;i<n;i++)
	{
    
		scanf("%d %d:%d %d:%d",&l[i].h,&j1,&j2,&c1,&c2);
		l[i].z=c1*60+c2-j1*60-j2;
	}
	for(i=1;i<n;i++)
	{
    
		for(j=0;j<i;j++)
		{
    
			if(l[i].h==l[j].h)
			{
    
				l[j].z+=l[i].z;
				l[i].z=0;
				l[i].h=0;
			}
		}
	}
	for(i=0;i<n-1;i++)
	{
    
		for(j=0;j<n-1-i;j++)
		{
    
			if(l[j].z<l[j+1].z)
			{
    
				t=l[j];l[j]=l[j+1];l[j+1]=t;
			}
			else if(l[j].z==l[j+1].z)
			{
    
				if(l[j].h>l[j+1].h)
				{
    
					t=l[j];l[j]=l[j+1];l[j+1]=t;
				}
			}
		}
	}
	for(i=0;i<n;i++)
	{
    
		if(l[i].h!=0) printf("%d %d\n",l[i].h,l[i].z);
	}
	return 0;
}