C170: retest screening

Problem description

After the announcement of the results of the first postgraduate entrance examination, you need to be right m Sort the scores of students , Screen out the former who can enter the retest n Famous student .
The sorting rule is to sort according to the total score first , If the total score is the same, it will be sorted according to the score of English single subject , When the total score and English score are the same, the person with small test number is in the front .
Now we give this m The results of the first postgraduate entrance examination of students , Please screen out those who can enter the re examination n Students and output... In order of ranking from high to low .

Enter description

Input is m+1 That's ok , The first line is two integers m and n, It indicates the total number of people and the number of people who can enter the retest ,m and n Space between ,0<n<m<200.
Next is m Row data , Each line contains three pieces of information , Each represents a student's test number （ Length not exceeding 20 String ）、 Total score （ Less than 500 The integer of ） And English single subject scores （ Less than 100 The integer of ）, These three items are separated by spaces .

The output shows that

Output the items entering the retest in the order of ranking from high to low n Information about students .

sample input

5 3
XD20160001 330 65
XD20160002 330 70
XD20160003 340 60
XD20160004 310 80
XD20160005 360 75

sample output

XD20160005 360 75
XD20160003 340 60
XD20160002 330 70

#include<stdio.h>
int main()
{
    
	int m,n,i,j;
	struct stu
	{
    
		char h[20];
		int a;
		int e;
	}l[200],t;
	scanf("%d %d",&m,&n);
	for(i=0;i<m;i++)
	{
    
		scanf("%s %d %d",l[i].h,&l[i].a,&l[i].e);
	}
	for(i=0;i<m-1;i++)
	{
    
		for(j=0;j<m-1-i;j++)
		{
    
			if(l[j].a<l[j+1].a)
			{
    
				t=l[j];l[j]=l[j+1];l[j+1]=t;
			}
			else if(l[j].a==l[j+1].a)
			{
    
				if(l[j].e<l[j+1].e)
				{
    
					t=l[j];l[j]=l[j+1];l[j+1]=t;
				}
				else if(l[j].e==l[j+1].e)
				{
    
					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++)
	{
    
		printf("%s %d %d",l[i].h,l[i].a,l[i].e);
		if(i!=n-1) printf("\n");
	}
	return 0;
}