C243: examination ranking

Problem description

There is 5 Problem sum 1 Additional questions , The highest score per question 20 branch , The total score is calculated as the sum of the scores of all topics . Give the data of a group of candidates , Rank them from high to low according to the total score , If the total score is the same, the one with higher score according to the additional questions will be preferred .

Enter description

The first line is an integer N, Indicates the number of candidates （N Less than 100）, Back N Behavioral candidate data , Each line contains the candidate's name （ Length not exceeding 20 Characters ） as well as 6 An integer separated by spaces , It represents the scores of questions 1 to 5 and additional questions respectively （ Last item ）.

The output shows that

Output sorting results , The name of each candidate 、 Total score 、 Additional questions score , Separate... By spaces .

sample input

3
Jony 18 20 20 20 20 20
Kavin 20 20 20 20 20 18
Kaku 15 15 15 15 15 15

sample output

Jony 118 20
Kavin 118 18
Kaku 90 15

#include<stdio.h>
int main()
{
    
	int n,i,j;
	struct stu
	{
    
		char m[20];
		int f[6];
		int z=0;
	}a[100],t;
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
    
		scanf("%s",a[i].m);
		for(j=0;j<6;j++)
		{
    
			scanf("%d",&a[i].f[j]);
			a[i].z+=a[i].f[j];
		}
	}
	for(i=0;i<n-1;i++)
	{
    
		for(j=0;j<n-1-i;j++)
		{
    
			if(a[j].z<a[j+1].z)
			{
    
				t=a[j];a[j]=a[j+1];a[j+1]=t;
			}
			else if(a[j].z==a[j+1].z)
			{
    
				if(a[j].f[5]<a[j+1].f[5])
				{
    
					t=a[j];a[j]=a[j+1];a[j+1]=t;
				}
			}
		}
	}
	for(i=0;i<n;i++)
	{
    
		printf("%s %d %d",a[i].m,a[i].z,a[i].f[5]);
		if(i!=n-1) printf("\n");
	}
	return 0;
}