hdu 1084 排序

#include <stdio.h>
#include <iostream>
#include <string>
#include <stdlib.h>
using namespace std;
#define mark student
#define que solve
struct	student
{
	int solve;
	int time;
	int grade;
	int set;
};
int cmp(const void *a, const void *b )
{
	if((*(student *)a).solve != (*(student *)b).solve)
		return (*(mark *)b).que-(*(mark *)a).que;
	return (*(mark *)a).time-(*(mark *)b).time;
}
int main()
{
	int n;
	while( ~scanf("%d", &n)	&& n!= -1)
	{
		student list[110];
		student *p[110];
		int num[6];
		memset(num, 0, sizeof(num) );
		for(int i=0; i<n; i++)
		{
			int hour, minute, second;
			scanf("%d %d:%d:%d", &list[i].solve, &hour, &minute, &second);
			list[i].time = hour*60*60 +minute * 60 +second;
			num[list[i].solve] ++;
			list[i].set =i;
		}
		qsort(list, n, sizeof(list[0]), cmp);
		int count =0;
		for(int i=5; i>=0; i--)
		{
			if(i == 0)
				for(int i=0; i<num[0]; i++)		list[count++].grade = 50;
			else if(i == 5)
				for(int i=0; i<num[5]; i++)		list[count++].grade = 100;
			else
			{
				for(int j=1; j<=num[i]; j++)
				{
					if(j<=num[i]/2)
						list[count++].grade = (i+5)*10+5;
					else
						list[count++].grade = (i+5)*10;
				}
			}
		}
		for(int i=0; i<n; i++)
		{
			for(int j=0; j<n; j++)
			{
				if(list[j].set == i)	printf("%d\n", list[j].grade);
			}
		}
		printf("\n");
	}
	return 0;
}
/*
Problem Description
“Point, point, life of student!”
This is a ballad（歌谣）well known in colleges, and you must care about your score in this exam too. How many points can you get? Now, I told you the rules which are used in this course.
There are 5 problems in this final exam. And I will give you 100 points if you can solve all 5 problems; of course, it is fairly difficulty for many of you. If you can solve 4 problems, you can also get a high score 95 or 90 (you can get the former(前者) only when your rank is in the first half of all students who solve 4 problems). Analogically（以此类推）, you can get 85、80、75、70、65、60. But you will not pass this exam if you solve nothing problem, and I will mark your score with 50.
Note, only 1 student will get the score 95 when 3 students have solved 4 problems.
I wish you all can pass the exam! 
Come on!
Input
Input contains multiple test cases. Each test case contains an integer N (1<=N<=100, the number of students) in a line first, and then N lines follow. Each line contains P (0<=P<=5 number of problems that have been solved) and T（consumed time）. You can assume that all data are different when 0<p.
A test case starting with a negative integer terminates the input and this test case should not to be processed.
Output
Output the scores of N students in N lines for each case, and there is a blank line after each case.
Sample Input
4
5 06:30:17
4 07:31:27
4 08:12:12
4 05:23:13
1
5 06:30:17
-1
Sample Output
100
90
90
95
100
*/

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