1025 PAT Ranking （25 分）

1025 PAT Ranking （25 分）

Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhejiang University. Each test is supposed to run simultaneously in several places, and the ranklists will be merged（合并） immediately after the test. Now it is your job to write a program to correctly merge all the ranklists and generate（产生） the final rank.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive number N (≤100), the number of test locations. Then N ranklists follow, each starts with a line containing a positive integer K (≤300), the number of testees, and then K lines containing the registration number (a 13-digit number) and the total score of each testee. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first print in one line the total number of testees. Then print the final ranklist in the following format:

registration_number final_rank location_number local_rank

The locations are numbered from 1 to N. The output must be sorted in nondecreasing（不减少的） order of the final ranks. The testees with the same score must have the same rank, and the output must be sorted in nondecreasing order of their registration numbers.

Sample Input:

251234567890001 951234567890005 1001234567890003 951234567890002 771234567890004 8541234567890013 651234567890011 251234567890014 1001234567890012 85

Sample Output:

91234567890005 1 1 11234567890014 1 2 11234567890001 3 1 21234567890003 3 1 21234567890004 5 1 41234567890012 5 2 21234567890002 7 1 51234567890013 8 2 31234567890011 9 2 4

参考代码：

#include#include#includeusing namespace std;struct student{ char id[15]; //准考证号 int score; //分数 int local_rank; //考场内排名 int location_number; //考场号}stu[30010];bool cmp(student a, student b){ if(a.score != b.score) return a.score > b.score;//先按分数高低排 else return strcmp(a.id, b.id) < 0; //分数相同，按照准考证号排}int main(int argc, char const *argv[]){ int N,n; scanf("%d", &N); //考场数 int num = 0; for (int i = 1; i <= N; ++i) { scanf("%d", &n); //考场内人数 for (int j = 0; j < n; ++j) { scanf("%s %d", stu[num].id, &stu[num].score); stu[num].location_number = i; //该考生的考场号为i num++; //总考生数加一 } sort(stu + num - n, stu + num, cmp); //将该考场的考生排序 stu[num- n].local_rank = 1; //将该考场的第一local_rank赋值为1 for (int j = num - n + 1 ; j < num; ++j) { if(stu[j].score == stu[j - 1].score){ //如果考场与前一位考生分数相同 stu[j].local_rank = stu[j - 1].local_rank; }else{ stu[j].local_rank = j + 1 - (num - n); // 该考场的考生人数 } } } printf("%d\n", num); //输出总考生数 sort(stu,stu + num ,cmp); //将所有考生排序 int r = 1; //当前考生的排名 for (int i = 0; i < num; ++i) { if(r > 0 && stu[i].score != stu[i - 1].score){ //当前考生与上一名考生分数不同时，r更新为r+1 r = i + 1; } printf("%s ", stu[i].id); printf("%d %d %d\n", r, stu[i].location_number, stu[i].local_rank); } return 0;}

版权声明：本文内容由网络用户投稿，版权归原作者所有，本站不拥有其著作权，亦不承担相应法律责任。如果您发现本站中有涉嫌抄袭或描述失实的内容，请联系我们jiasou666@gmail.com 处理，核实后本网站将在24小时内删除侵权内容。