HDOJ1084 What Is Your Grade?

What Is Your Grade?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 8302    Accepted Submission(s): 2547

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

这破题WA了9次才过 ⊙﹏⊙b汗

#include <cstdio>
#include <algorithm>
using std::sort;

struct Node{
	int pos, num, s, val;
} stu[102];
int arr[6];

bool cmp1(Node a, Node b)
{
	if(a.num == b.num) return a.s < b.s;
	return a.num > b.num;
}

bool cmp2(Node a, Node b)
{
	return a.pos < b.pos;
}

int main()
{
	int n, h, m, s, num;
	while(scanf("%d", &n) == 1 && n > 0){
		for(int i = 1; i < 6; ++i) arr[i] = 0;
		for(int i = 0; i < n; ++i){
			scanf("%d %d:%d:%d", &num, &h, &m, &s);
			s += m * 60 + h * 3600;
			
			stu[i].pos = i;
			stu[i].num = num;
			stu[i].s = s;
			stu[i].val = 100 - (5 - num) * 10;
			++arr[num];
		}
		
		sort(stu, stu + n, cmp1);
		
		for(int i = 4, pos = 0; i; --i){
			if(arr[i]){
				while(stu[pos].num != i) ++pos;
				if(arr[i] == 1) stu[pos++].val += 5;			
				for(int j = 0; j < arr[i] / 2; ++j)
					stu[pos++].val += 5;
			}
		}
		
		sort(stu, stu + n, cmp2);
		
		for(int i = 0; i < n; ++i)
			printf("%d
", stu[i].val);
		printf("
");
	}
	return 0;
}