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[10];//time
            int p;//problem
            int s;//score
    }a[120];
    while((scanf("%d",&n))&&n!=-1)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d%s",&a[i].p,&a[i].t);
            a[i].s=50+10*a[i].p;
        }
        int x=1;
        while(x<5)
        {
            char time[120][10]={"99:99:99"};
            int num=0;
            
            for(int i=0;i<n;i++)
            {
                if(a[i].p==x)
                {
                    strcpy(time[num++],a[i].t);
                }
            }
            for(int i=0;i<num/2;i++)
            {
                for(int j=i+1;j<num;j++)
                {
                    if(strcmp(time[i],time[j])>0)
                    {
                        char p[10];
                        strcpy(p,time[i]);
                        strcpy(time[i],time[j]);
                        strcpy(time[j],p);
                    }
                }
            }
            for(int i=0;i<n;i++)
            {
                if(a[i].p==x&&strcmp(a[i].t,time[num/2-1])<=0)
                {
                    a[i].s+=5;
                }
            }
            x++;
        }
        for(int i=0;i<n;i++)
        {
            printf("%d\n",a[i].s);
        }
        printf("\n");
    }
}