hdu 1084 What Is Your Grade(水题)

What Is Your Grade?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 9540    Accepted Submission(s): 2934


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

 

对于qsort函数中的cmp函数不怎么熟导致多次WA


回顾一下cmp函数：inline int MyCmp(const void* a, const void* b) 都是有两个元素 作为参数，返回一个int 值， 如果 比较函数返回大于0，qsort就认为 a>b , 如果比较函数返回等于0 qsort就认为a 和b 这两个元素相等，返回小于零
qsort就认为 ab),你比较函数却返回一个 -1 （小于零的）那么qsort认为a<本文中排序都是采用的从小到大排序>

代码：

#include <stdio.h>

#include <stdlib.h>

#include <string.h>
struct Node{
   int grade;//最后分数
   int No;       
//同组中的排名
   int t;   
//作出题目
   int time;
   int bsNo;  
//原始顺序号
}a[120];
int cmp1(constvoid *a,constvoid
*b){      //如果作出题目一样，以time为准小－》大
   struct
Node *c=(structNode *)a;
   struct
Node *d=(structNode *)b;
   if(c->t==d->t)r
4000
eturn c->time-d->time;
   return d->t - c->t;
}

int cmp2(const
void *a,const
void *b)   //返回初始顺序
{
   struct
Node *c=(structNode *)a;
   struct
Node *d=(structNode *)b;
   return c->bsNo-d->bsNo;
}
int main()
{
   int rs[6];
   int N;
   while (scanf("%d",&N)!=EOF&&N>=0)
{
       memset(rs,
0,sizeof(rs));
       for (int i=0; i<N; i++) {
           int m,s,h;
           scanf("%d %d:%d:%d",&a[i].t,&h,&m,&s);
           a[i].time = h*3600+m*60+s;
            rs[a[i].t]++;
           a[i].bsNo=i;
        }
       qsort(a, N,sizeof(a[0]),cmp1);
       a[0].No=1;
       for (int i=1; i<N; i++) {
           a[i].No=1;
           if(a[i].t==a[i-1].t)
               a[i].No+=a[i-1].No;
        }
       for (int i=0; i<N; i++) {
           a[i].grade=a[i].t*10+50;
           if((a[i].t>=1&&a[i].t<5)&&a[i].No<=rs[a[i].t]/2)
               a[i].grade+=5;
        }
       qsort(a, N,sizeof(a[0]),cmp2);
       for (int i=0; i<N; i++) {
           printf("%d\n",a[i].grade);
        }
       printf("\n");
    }
   return
0;
}