HDOJ 1084 What Is Your Grade?

What Is Your Grade？

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

Total Submission(s): 10834    Accepted Submission(s): 3371

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

45 06:30:174 07:31:274 08:12:124 05:23:1315 06:30:17-1

 

Sample Output

100909095100

 

根据时间的不同分数也不一样，这种题都知道怎么写，但是怎么简化代码很关键。下面的代码有借鉴别人的。很巧妙而且简洁。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct node {    int x;    char s[10];} a[120];int main() {    int n;    while(scanf("%d",&n),n!=-1) {        for(int i=0; i<n; i++) {            scanf("%d %s",&a[i].x,&a[i].s);        }        int cnt[5]= {0};        for(int i=4; i>0; i--) {            for(int j=0; j<n; j++) {                if(a[j].x==i)                    cnt[i]++;            }        }        for(int i=0; i<n; i++) {            if(a[i].x==5)                printf("100\n");            else if(a[i].x==0)                printf("50\n");            else {                int t=0;                for(int j=0; j<n; j++) {                    if(a[j].x==a[i].x&&strcmp(a[j].s,a[i].s)<0)                        t++;                }                if(t>=cnt[a[i].x]/2)                    printf("%d\n",100-(5-a[i].x)*10);                else                    printf("%d\n",100-(5-a[i].x)*10+5);            }        }        printf("\n");    }    return 0;}