CDZSC_2015寒假新人（1）——基础 i

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

 

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 using namespace std;

 int main()
 {
     int n;
     struct
     {
             char t[];//time
             int p;//problem
             int s;//score
     }a[];
     while((scanf("%d",&n))&&n!=-)
     {
         for(int i=;i<n;i++)
         {
             scanf("%d%s",&a[i].p,&a[i].t);
             a[i].s=+*a[i].p;
         }
         int x=;
         while(x<)
         {
             char time[][]={"99:99:99"};
             int num=;

             for(int i=;i<n;i++)
             {
                 if(a[i].p==x)
                 {
                     strcpy(time[num++],a[i].t);
                 }
             }
             for(int i=;i<num/;i++)
             {
                 for(int j=i+;j<num;j++)
                 {
                     if(strcmp(time[i],time[j])>)
                     {
                         char p[];
                         strcpy(p,time[i]);
                         strcpy(time[i],time[j]);
                         strcpy(time[j],p);
                     }
                 }
             }
             for(int i=;i<n;i++)
             {
                 if(a[i].p==x&&strcmp(a[i].t,time[num/-])<=)
                 {
                     a[i].s+=;
                 }
             }
             x++;
         }
         for(int i=;i<n;i++)
         {
             printf("%d\n",a[i].s);
         }
         printf("\n");
     }
 }