关键是自己没有读懂题目而已，不过还好，终于给做出来了......

What Is Your Grade?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7101 Accepted Submission(s): 2186

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

5 06:30:17

-1

Sample Output

100

Author

lcy

水体，没有涉及太大的算法，插入法就过了。。

代码：

     #include<stdio.h>

     #include<string.h>

     #include<stdlib.h>

     struct nod

     {

      int num;

      int tol;

     }start[];

     int grand[][]={{,},{,},{,},{,},{,},{,}};

      int pos[][],tag[];

     int main()

     {

         int p,i,j;

         int hh,mm,ss,cnt=,step;

         /*freopen("test.in","r",stdin);*/

         while(scanf("%d",&p),p!=-)

         {

          memset(tag,,sizeof(tag));

          memset(pos,,sizeof(pos));

          for(i=;i<p;i++)

           {

            scanf("%d %d:%d:%d",&start[i].num,&hh,&mm,&ss);

            tag[start[i].num]++;

            start[i].tol=hh*+mm*+ss;

            step=;

            /*插入法排序*/

            while(pos[start[i].num][step]!=&&pos[start[i].num][step]<start[i].tol)

                step++;

            int gg=;

            while(pos[start[i].num][gg]!=)

            {

                gg++;

            }

          /*往后移，使用插入法*/

            while(gg>step)

            {

              pos[start[i].num][gg]=pos[start[i].num][gg-];

              gg--;

            }

            pos[start[i].num][gg]=start[i].tol;

           }

            bool flag ;

             for(i=;i<p;i++)

              {

                 flag=false;

                 for(j=;j<tag[start[i].num]/;j++)

                 {

                    if(pos[start[i].num][j]==start[i].tol)

                    {

                       printf("%d\n",grand[start[i].num][]);

                       flag=true;

                       break;

                    }

                 }

                 if(!flag)  printf("%d\n",grand[start[i].num][]);

              }

              putchar();

         }

       return ;

     }

of

HDUOJ----(1084)What Is Your Grade?

What Is Your Grade?