What Is Your Grade?

What Is Your Grade?    HDU - 1084

“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

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct student{int nu,q,u,gr;//nu代表最对的题，q代表输入时的顺序，u代表做对相同数量的题的同学中按时间的顺序int ti;       //gr代表成绩，ti代表时间}s[105];bool cmp(student a,student b){if(a.nu!=b.nu)   //最对题的数量相等时比较时间排序return a.nu>b.nu;elsereturn a.ti<b.ti;}bool cmp2(student a,student b){return a.q<b.q;}int main(){int n,a[6],f,i;char sti[10];while( ~scanf("%d",&n) && n > 0){memset(a,0,sizeof(a));for( i = 0; i < n; i++){scanf("%d%s",&s[i].nu,sti);a[s[i].nu]++;//a数组记录做对相同数量的题的人数s[i].ti = ((sti[0]-'0')*10 + sti[1]-'0')*3600 + ((sti[3]-'0')*10 + sti[4]-'0')*60 + ((sti[6]-'0')*10 + sti[7]-'0');//时间都统一用秒表示，s[i].q = i;}sort(s,s+n,cmp);//按得分排序f = 1;for( i = 0; i < n; i++){s[i].u=f;//f代表在同类数量题的同学中的顺序if(s[i].nu != s[i + 1].nu) f = 1;else f++;if(s[i].nu == 5) s[i].gr=100;if(s[i].nu == 4) {if(s[i].u <= a[4]/2){s[i].gr=95;}else s[i].gr=90;}if(s[i].nu == 3) {if(s[i].u <= a[3]/2){s[i].gr=85;}else s[i].gr=80;}if(s[i].nu == 2) {if(s[i].u <= a[2]/2){s[i].gr=75;}else s[i].gr=70;}if(s[i].nu == 1) {if(s[i].u <= a[1]/2){s[i].gr=65;}else s[i].gr=60;}if(s[i].nu == 0) {s[i].gr=50;}}sort(s,s+n,cmp2);//最后再按记录顺序排序，按原顺序输出for( i = 0; i < n; i++){printf("%d\n",s[i].gr);}printf("\n");}return 0;}