hdoj 1084 What Is Your Grade?



http://acm.hdu.edu.cn/showproblem.php?pid=1084

What Is Your Grade?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8691    Accepted Submission(s): 2681

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
  
  

   要注意红色部分，不是frist，开始看成frist错了好几次
  
  


  
  

   
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
struct Subject
{
    int N;
    int P;
    char TIME[20];
    int score;
}man[200];
int cmp_P_T(const void *a,const void *b)
{
    struct Subject *c=(Subject *)a;
    struct Subject *d=(Subject *)b;
    if(c->P!=d->P)
        return d->P-c->P;
    else
        return strcmp(c->TIME,d->TIME);
}
int cmp_N(const void *a,const void *b)
{
    return ((Subject *)a)->N-((Subject *)b)->N;
}
int main()
{
    int i,j,k,N,a[10],b[10],c[10];
    while(scanf("%d",&N)&&N!=-1)
    {
        for(i=0;i<N;i++)
        {
            scanf("%d %s",&man[i].P,man[i].TIME);
            man[i].N=i;
        }
       // man[i].P=-1;//man[i].P必须和前面的不一样，所以man[i].P=-1
        qsort(man,N,sizeof(man[0]),cmp_P_T);

        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(c,0,sizeof(c));
        for(i=0,j=0,a[0]=man[0].P;i<N;i++)//将出现得分的情况放到数组a中，如得分为5,5,5,4,3,3,2.则a数组为5,4,3,2.
        if(man[i].P!=a[j])
        a[++j]=man[i].P;

        for(i=0,j=0;i<N;i++)//将得分出现的次数放到b数组中
        if(man[i].P==a[j])
        b[j]+=1;
        else
        b[++j]+=1;

        for(i=0;i<=j;i++)//得分出现的次数的一般
        if(b[i]>1)
        c[i]=b[i]/2;
        else
        c[i]=b[i];

        //for(i=0;i<=j;i++)
        //printf("%d %d %d#\n",a[i],b[i],c[i]);

        for(i=0,j=0,k=1;i<N;i++)
        {
            if(man[i].P==5)
            man[i].score=100;

            if(man[i].P==4&&k<=c[j])
            man[i].score=95;
            if(man[i].P==4&&k>c[j])
            man[i].score=90;

            if(man[i].P==3&&k<=c[j])
            man[i].score=85;
            if(man[i].P==3&&k>c[j])
            man[i].score=80;

            if(man[i].P==2&&k<=c[j])
            man[i].score=75;
            if(man[i].P==2&&k>c[j])
            man[i].score=70;

            if(man[i].P==1&&k<=c[j])
            man[i].score=65;
            if(man[i].P==1&&k>c[j])
            man[i].score=60;

            if(man[i].P==0)
            man[i].score=50;

            k++;
            if(k>b[j])
            {
                j+=1;
                k=1;
            }
        }

        //for(i=0;i<N;i++)
        //printf("%d %s %d#\n",man[i].P,man[i].TIME,man[i].score);
        qsort(man,N,sizeof(man[0]),cmp_N);
        for(i=0;i<N;i++)
        printf("%d\n",man[i].score);
        printf("\n");


    }
    return 0;
}