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

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

1

5 06:30:17

-1

 

Sample Output

100

90

90

95

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 ;
     }