What Is Your Grade?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9580 Accepted Submission(s): 2940

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

5 06:30:17

4 07:31:27

4 08:12:12

4 05:23:13

5 06:30:17

-1

Sample Output

100

不想多解释，水题一道，题没看错就行，这么一道水题我竟wa了很多次，很是忧桑。。。

直接附上代码

 #include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define N 110

using namespace std;

struct st
{
    int id, x, h, y;
} node[N];

int cmp(const void *a, const void *b)
{
    st *s1 = (st *)a, *s2 = (st *)b;
    if(s1->x == s2->x)
        return s1->h - s2->h;
    else
        return s2->x - s1->x;
}

int cmp1(const void *a, const void *b)
{
    return *(int *)a - *(int *)b;
}

int main()
{
    int i, a1, a2, a3, a4, n, h, m, s;
    while(scanf("%d", &n), n != -1)
    {
        memset(node, 0, sizeof(node));
        a1 = a2 = a3 = a4 = 0;
        for(i = 0 ; i < n ; i++)
        {
            scanf("%d", &node[i].x);
            scanf("%d:%d:%d", &h, &m, &s);
            node[i].h  = h * 3600 + m * 60 + s;
            node[i].id = i;
            if(node[i].x == 1)
                a1++;
            else if(node[i].x == 2)
                a2++;
            else if(node[i].x == 3)
                a3++;
            else if(node[i].x == 4)
                a4++;
        }
        qsort(node, n, sizeof(node[0]), cmp);
        a1 /= 2, a2 /= 2, a3 /= 2, a4 /= 2;
        for(i = 0 ; i < n ; i++)
        {
            if(node[i].x == 5)
                node[i].y = 100;
            else if(node[i].x == 4)
            {
                if(a4 != 0)
                {
                    node[i].y = 95;
                    a4--;
                }
                else
                    node[i].y = 90;
            }
            else if(node[i].x == 3)
            {
                if(a3 != 0)
                {
                    node[i].y = 85;
                    a3--;
                }
                else
                    node[i].y = 80;
            }
            else if(node[i].x == 2)
            {
                if(a2 != 0)
                {
                    node[i].y = 75;
                    a2--;
                }
                else
                    node[i].y = 70;
            }
            else if(node[i].x == 1)
            {
                if(a1 != 0)
                {
                    node[i].y = 65;
                    a1--;
                }
                else
                    node[i].y = 60;
            }
            else
                node[i].y = 50;
        }
        qsort(node, n, sizeof(node[0]), cmp1);
        for(i = 0 ; i < n ; i++)
            printf("%d
", node[i].y);
        printf("
");

    }
    return 0;
}