E - What Is Your Grade?

E - What Is Your Grade?

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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

//题目意思搞了好久才懂，不懂英语真难玩啊。
有点类似ACM排名
第一行代表有多少个人，然后是每个人的成绩，做了几个题，用时多少。做了5个，不管用时，都是100，少做1个扣10分，没做不管用时都是50分，做了相同题目的，用时排在一半之前加5分，例如 4个人都做4个，前两个 95分，后两个 90 分， 3个人做 4 个，只有第一个 95 分，后两个90分。

//可以dp，可以模拟，然后我果断选了模拟。。。
有几重循环，竟然还是 0ms 有点惊讶

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

struct People
{
    int num;
    int t;
    int h,m,s;
    int fen;
}people[101];

int cmp(People a,People b)
{
    if (a.t!=b.t)
        return a.t>b.t;

    if (a.h!=b.h)   //用时少在前面
        return a.h<b.h;
    if (a.m!=b.m)
        return a.m<b.m;
    return a.s<b.s;
}

int main()
{
    int n;
    int i,j,k;
    while (scanf("%d",&n)!=EOF)
    {
        if (n<0) break;

        for (i=1;i<=n;i++)
        {
            scanf("%d %d:%d:%d",&people[i].t,&people[i].h,&people[i].m,&people[i].s);
            people[i].num=i;
        }
        sort(people+1,people+n+1,cmp);

        for (i=1;i<=n;i++)
        {
            if (people[i].t!=5)
                break;
            else
                people[i].fen=100;
        }

        int star=i;

        for (i=4;i>=1;i--)//做题数
        {
            for (j=star;j<=n;j++)   //找到做题数相同的一块  str - (j-1)
                if (people[j].t!=i)
                    break;

            int mid=(star+j-1)/2;
            if ((star+j-1)%2==0)
                mid--;
            
            for (k=star;k<j;k++)
            {
                if (k<=mid)
                people[k].fen=50+i*10+5;
                else
                people[k].fen=50+i*10;
            }
            star=j;
            if (star>n) break;
        }
        
        for (j=star;j<=n;j++) people[j].fen=50;

        if (n==1&&people[1].t!=5&&people[1].t!=0) people[1].fen+=5;

        for (i=1;i<=n;i++)
        {
            for (j=1;j<=n;j++)
            {
                if (people[j].num==i)
                {
                    printf("%d\n",people[j].fen);
                    break;
                }
            }
        }
        printf("\n");
    }
    return 0;
}