E - What Is Your Grade?

E - What Is Your Grade?

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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

//题目意思搞了好久才懂，不懂英语真难玩啊。
有点类似ACM排名
第一行代表有多少个人，然后是每个人的成绩，做了几个题，用时多少。做了5个，不管用时，都是100，少做1个扣10分，没做不管用时都是50分，做了相同题目的，用时排在一半之前加5分，例如 4个人都做4个，前两个 95分，后两个 90 分， 3个人做 4 个，只有第一个 95 分，后两个90分。

//可以dp，可以模拟，然后我果断选了模拟。。。
有几重循环，竟然还是 0ms 有点惊讶

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 #include <algorithm>
 using namespace std;

 struct People
 {
     int num;
     int t;
     int h,m,s;
     int fen;
 }people[];

 int cmp(People a,People b)
 {
     if (a.t!=b.t)
         return a.t>b.t;

     if (a.h!=b.h)   //用时少在前面
         return a.h<b.h;
     if (a.m!=b.m)
         return a.m<b.m;
     return a.s<b.s;
 }

 int main()
 {
     int n;
     int i,j,k;
     while (scanf("%d",&n)!=EOF)
     {
         if (n<) break;

         for (i=;i<=n;i++)
         {
             scanf("%d %d:%d:%d",&people[i].t,&people[i].h,&people[i].m,&people[i].s);
             people[i].num=i;
         }
         sort(people+,people+n+,cmp);

         for (i=;i<=n;i++)
         {
             if (people[i].t!=)
                 break;
             else
                 people[i].fen=;
         }

         int star=i;

         for (i=;i>=;i--)//做题数
         {
             for (j=star;j<=n;j++)   //找到做题数相同的一块  str - (j-1)
                 if (people[j].t!=i)
                     break;

             int mid=(star+j-)/;
             if ((star+j-)%==)
                 mid--;

             for (k=star;k<j;k++)
             {
                 if (k<=mid)
                 people[k].fen=+i*+;
                 else
                 people[k].fen=+i*;
             }
             star=j;
             if (star>n) break;
         }

         for (j=star;j<=n;j++) people[j].fen=;

         if (n==&&people[].t!=&&people[].t!=) people[].fen+=;

         for (i=;i<=n;i++)
         {
             for (j=;j<=n;j++)
             {
                 if (people[j].num==i)
                 {
                     printf("%d\n",people[j].fen);
                     break;
                 }
             }
         }
         printf("\n");
     }
     return ;
 }