关键是自己没有读懂题目而已，不过还好，终于给做出来了......
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 }