What Is Your Grade?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4032    Accepted Submission(s): 1198


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
 

 

Author
lcy
 
  
#include<stdio.h>
#include<algorithm>
#include<stdlib.h>//qsort
#include<iostream>
#include<string.h>
using namespace std;
struct Node
{
    int num;//输入的原始编号
    int n;//正确解题数
    int fen;//分数
    int time;//用掉的时间
}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);
        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;

}

文章来源: http://www.cnblogs.com/kuangbin/archive/2011/11/01/2232037.html