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

 

 

这个题可以说是基数排序的一个简单版，但是我写了将近俩个小时才写出来，实在是水啊。。

 

 

#include<stdio.h>
#include<string.h>
void Exchange(int f[100],int n,int m)
{
    int temp;
    temp=f[n];
    f[n]=f[m];
    f[m]=temp;
}
void Sort(int f[100],char time[100][10],int k)
{
    for(int i=0;i<k;i++)
    {
        int p=k-1;
        for(int j=k-1;j>i;j--)
        {
            if(strcmp(time[f[j]],time[f[j-1]])<0)
            {
                Exchange(f,j,j-1);
                p=j;
            }
        }
        if(p==k-1) break;
    }
}
void Score(int score[100],char time[100][10],int solved[100],int n)
{
    int f[100];
    for(int i=1;i<5;i++)
    {
        int k=0;
        for(int j=0;j<n;j++)
        {
            if(solved[j]==i)
            {
                f[k]=j;
                k++;
            }
        }
        Sort(f,time,k);         //对时间进行排序
        for(j=0;j<k/2;j++)
            score[f[j]]=5;
    }
}

void main()
{
    int n;
    char time[100][10];
    int solved[100];
    int score[100];
    while(1)
    {
        scanf("%d",&n);
        if(n<0) return;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&solved[i]);
            scanf("%s",time[i]);
            score[i]=0;
        }
        Score(score,time,solved,n);
        for(i=0;i<n;i++)
        {
            switch (solved[i])
            {
                 case 5 : printf("100\n"); break;
                 case 4 : printf("%d\n",90+score[i]); break;
                 case 3 : printf("%d\n",80+score[i]); break;
                 case 2 : printf("%d\n",70+score[i]); break;
                 case 1 : printf("%d\n",60+score[i]); break;
                 default : printf("50\n");
            }
        }
        printf("\n");
    }
}