HDU 1804——What Is Your Grade?

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

题目大意：意思大概是给学生打分数，做出的时间越短，题目越多，分数越高。

解决办法：题目不是很难，只要读懂题目，能对时间等数据进行妥善处理，就可以解决了

接下来是代码：

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

struct Node
{
    int num;
    int n;//题目
    int fen;//分数
    int time;//时间
}node[110];

int cmp1(const void *a,const void *b)
{
    struct Node *c=(Node *)a;
    struct Node *d=(Node *)b;
    if(c->n == d->n)return c->time - d->time;
    return d->n - c->n;
}

int cmp2(const void *a,const void *b)
{
    struct Node *c=(Node *)a;
    struct Node *d=(Node *)b;
    return c->num - d->num;
}

int main()
{
    int n,i,j,k;
    int sum[10];
    int tol[10];
    int h,m,s;
    while(scanf("%d",&n))
    {
        if(n<0)break;
        memset(sum,0,sizeof(sum));
        memset(tol,0,sizeof(tol));
        for(i=0;i<n;i++)
        {
            scanf("%d %d:%d:%d",&node[i].n,&h,&m,&s);
            node[i].num=i;
            node[i].time=h*3600+m*60+s;
            sum[node[i].n]++;
        }
        for(i=4;i>0;i--)
        {
            tol[i]=tol[i+1];
            tol[i]+=sum[i+1];
        }
        qsort(node,n,sizeof(node[0]),cmp1);//qsort对时间排序更为有效
        for(i=0;i<n;i++)
        {
            node[i].fen=node[i].n*10+50;
            if(node[i].n>=1&&node[i].n<5&&(((i+1)-tol[node[i].n])<=sum[node[i].n]/2))
                node[i].fen+=5;
        }
        qsort(node,n,sizeof(node[0]),cmp2);
        for(i=0;i<n;i++)
        {
            printf("%d\n",node[i].fen);
        }
        printf("\n");
    }
    return 0;

}