HDU--1084 What Is Your Grade?

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

 AC代码：

感觉难度太大啊....

#include<stdio.h>
#include<algorithm>
using namespace std;
struct stu
{
    int num;
    int sj;
    int id;   
}p[100];
bool cmp(stu x,stu y)
{
    if(x.num!=y.num)
      return x.num>y.num;
    else return x.sj<y.sj;            
}
int main()
{   int n,h,f,s,i,a[101];
    while(scanf("%d",&n)!=EOF&&n>=0)
        {int t,j,num1;
         for(i=0;i<n;i++)
           { scanf("%d %d:%d:%d",&p[i].num,&h,&f,&s);
             p[i].sj=h*3600+f*60+s;
             p[i].id=i;
                                          
             a[i]=100-(5-p[i].num)*10;    //p的id号和a中的分数对应
           }
         sort(p,p+n,cmp);
         
         for(i=0;i<n;i++)//这个for要好好理解
            {t=p[i].num;
             
             j=i;
             num1=0;
             for(i;p[i].num==t&&i<n;i++)
             num1++;
             i--; 
             if(p[i].num==5||p[i].num==0) continue;
             if(num1>1)
             num1/=2;    
             while(num1--)
                {a[p[j++].id]+=5;//没有这个P的id号直接输出有错误，a数组里面的分数顺序和p排序不对应
                                 
                }
            }
         for(i=0;i<n;i++)
            printf("%d
",a[i]); 
        printf("
");    
        }
    
    
    
return 0;    
 }