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 }