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     }

 

posted @ 2013-11-08 07:14  龚细军  阅读(372)  评论(0)  编辑  收藏  举报