HDU 1084 What Is Your Grade

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
#include <cstdio>
#include <cstring>
#define N 110
struct student
{
	int solve, score;
	char time[10];
}stu[N];
int sum[6];
int n;
void input()
{
	memset(sum, 0, sizeof(sum));
	for (int i = 1; i <= n; i++)
	{
		scanf("%d%*c%s", &stu[i].solve, stu[i].time);
		sum[stu[i].solve]++;

		if (stu[i].solve == 5) stu[i].score = 100;
		else if (stu[i].solve == 0) stu[i].score = 50;
		else stu[i].score = 0;
	}
	for (int i = 1; i <= 4; i++) sum[i] /= 2;
}
void rank()
{
	for (int i = 1; i <= 4; i++)
	{
		for (int j = 1; j <= sum[i]; j++)
		{
			int mini;
			char mint[20] = "99:99:99";
			for (int k = 1; k <= n; k++)
				if (stu[k].solve == i && !stu[k].score && strcmp(stu[k].time, mint) < 0)
				{
					strcpy(mint, stu[k].time);
					mini = k;
				}
			stu[mini].score = 10 * (i + 5) + 5;
		}
		for (int k = 1; k <= n; k++)
			if (stu[k].solve == i && !stu[k].score) stu[k].score = 10 * (i + 5);
	}
}
void output()
{
	for (int i = 1; i <= n; i++) printf("%d\n", stu[i].score);
	printf("\n");
}
int main()
{
	while (scanf("%d", &n), ~n)
	{
		input();
		rank();
		output();
	}
	return 0;
}