Hdu1084 What Is Your Grade?

What Is Your Grade?

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

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

 

题意：答题，根据学生的solve数目和solve时间给学生打分，solve5道题的得100分，solve nothing的得50分。4道题的得95或90，3道题的得85或80，2道题的得75或70，。。。

在答1，2,3,4道题的学生中时间排在前面一半的学生得高分，注意：只有一个学生解答一个数目的题目，他得高分；3个同学解答了同样数目的题，只有排在最前面的第一个得高分，剩下的两个都得低分。输出也要按输入的顺序。

 

代码：

#include<stdio.h>#include<string.h>struct node{int  solve;char time[12];int score;}s[110];int num[6];int n;int main(){while(~scanf("%d",&n) && n>0){int i,j;memset(num,0,sizeof(num));for (i = 0; i < n; i++){int x;scanf("%d%s",&s[i].solve,s[i].time);x = s[i].solve*10 + 50;s[i].score = x;if(s[i].solve < 5&&s[i].solve > 0) {++num[s[i].solve];}}int low;for (i = 0; i < n; i++){low = 0;if(num[s[i].solve]){for (j = 0; j < n; j++){if( s[i].solve == s[j].solve && strcmp(s[i].time,s[j].time) > 0) low++;}if(num[s[i].solve]==1 || low < num[s[i].solve]/2)s[i].score +=5;}printf("%d\n",s[i].score);}printf("\n");}return 0;}