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：

 1     #include<stdio.h>

 2     #include<string.h>

 3     #include<stdlib.h>

 4     struct nod

 5     {

 6      int num;

 7      int tol;

 8     }start[102];

 9 

10     int grand[6][2]={{50,50},{65,60},{75,70},{85,80},{95,90},{100,100}};

11      int pos[6][52],tag[6];

12     

13     int main()

14     {

15         int p,i,j;

16         int hh,mm,ss,cnt=1,step;

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

18         while(scanf("%d",&p),p!=-1)

19         {

20          memset(tag,0,sizeof(tag));

21          memset(pos,0,sizeof(pos));

22          for(i=0;i<p;i++)

23           {

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

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

26            start[i].tol=hh*3600+mm*60+ss;

27            step=0;

28            /*插入法排序*/

29            while(pos[start[i].num][step]!=0&&pos[start[i].num][step]<start[i].tol)

30                step++;

31            int gg=0;

32            while(pos[start[i].num][gg]!=0)

33            {

34                gg++;

35            }

36          /*往后移，使用插入法*/

37            while(gg>step)

38            {

39              pos[start[i].num][gg]=pos[start[i].num][gg-1];

40              gg--;

41            }

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

43           }

44            bool flag ;

45             for(i=0;i<p;i++)

46              {

47                 flag=false;

48                 for(j=0;j<tag[start[i].num]/2;j++)

49                 {

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

51                    {

52                       printf("%d\n",grand[start[i].num][0]);

53                       flag=true;

54                       break;

55                    }

56                 }

57                 if(!flag)  printf("%d\n",grand[start[i].num][1]);

58              }

59              putchar(10);

60         }

61       return 0;

62     }