HDOJ-----1084结构体排序

“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
一共五道题，每行第一个数表示做题数目，做五题得一百分，做0题得50分，4题，3题，2题，1题先做完的那一半人分别得95,85,75,65分，其余得90,80,70,60分
按顺序输出每个人得分
ps：四个人都做四道题，前两个人得95分，三个人做四道题，只有一个人得95分。
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct node{int a, b;}s[1010];node x[1010], y[1010], p[1010], q[1010];bool cmp(node m, node n){return m.b < n.b;}int main(){int a, b, m1, m2, m3, m4, c1, c2, c3, n1, n2, n3, n4;while(~scanf("%d", &a) && a > 0){n1 = n2 = n3 = n4 = 0;for(int i = 0; i < a; i++){scanf("%d%d:%d:%d", &b, &c1, &c2, &c3);s[i].a = b;if(b == 5){s[i].b = 100;}if(b == 4){s[i].b = 90;x[n1].a = i;x[n1++].b = c1 * 3600 + c2 * 60 + c3;}if(b == 3){s[i].b = 80;y[n2].a = i;y[n2++].b = c1 * 3600 + c2 * 60 + c3;}if(b == 2){s[i].b = 70;p[n3].a = i;p[n3++].b = c1 * 3600 + c2 * 60 + c3;}if(b == 1){s[i].b = 60;q[n4].a = i;q[n4++].b = c1 * 3600 + c2 * 60 + c3;}if(b == 0){s[i].b = 50;}}sort(x, x + n1, cmp);sort(y, y + n2, cmp);sort(p, p + n3, cmp);sort(q, q + n4, cmp);for(int i = 0; i < n1 / 2; i++){s[x[i].a].b += 5;}for(int i = 0; i < n2 / 2; i++){s[y[i].a].b += 5;}for(int i = 0; i < n3 / 2; i++){s[p[i].a].b += 5;}for(int i = 0; i < n4 / 2; i++){s[q[i].a].b += 5;}for(int i = 0; i < a; i++){printf("%d\n", s[i].b);}printf("\n");}return 0;}