What Is Your Grade? Time Limit: 2000/1000 MS (
Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8302 Accepted Submission(s): 2547
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
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
这破题WA了9次才过 ⊙?⊙b汗
#include
#include
using std::sort; struct Node{ int pos, num, s, val; } stu[102]; int arr[6]; bool cmp1(Node a, Node b) { if(a.num == b.num) return a.s < b.s; return a.num > b.num; } bool cmp2(Node a, Node b) { return a.pos < b.pos; } int main() { int n, h, m, s, num; while(scanf("%d", &n) == 1 && n > 0){ for(int i = 1; i < 6; ++i) arr[i] = 0; for(int i = 0; i < n; ++i){ scanf("%d %d:%d:%d", &num, &h, &m, &s); s += m * 60 + h * 3600; stu[i].pos = i; stu[i].num = num; stu[i].s = s; stu[i].val = 100 - (5 - num) * 10; ++arr[num]; } sort(stu, stu + n, cmp1); for(int i = 4, pos = 0; i; --i){ if(arr[i]){ while(stu[pos].num != i) ++pos; if(arr[i] == 1) stu[pos++].val += 5; for(int j = 0; j < arr[i] / 2; ++j) stu[pos++].val += 5; } } sort(stu, stu + n, cmp2); for(int i = 0; i < n; ++i) printf("%d\n", stu[i].val); printf("\n"); } return 0; }
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