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!
A test case starting with a negative integer terminates the input and this test case should not to be processed.
这破题WA了9次才过 ⊙﹏⊙b汗
#include <cstdio> #include <algorithm> 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; }