PAT(甲) 1025 PAT Ranking (25)

PAT考试由浙江大学计算机学院组织,需合并各考点排名生成最终排名。输入包含考点数、各考点考生数及考生信息,输出为考生总数和最终排名列表,排名按分数和考号排序。

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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 immediately after the test. Now it is your job to write a program to correctly merge all the ranklists and generate 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 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:
2
5
1234567890001 95
1234567890005 100
1234567890003 95
1234567890002 77
1234567890004 85
4
1234567890013 65
1234567890011 25
1234567890014 100
1234567890012 85

Sample Output:
9
1234567890005 1 1 1
1234567890014 1 2 1
1234567890001 3 1 2
1234567890003 3 1 2
1234567890004 5 1 4
1234567890012 5 2 2
1234567890002 7 1 5
1234567890013 8 2 3
1234567890011 9 2 4

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct Student {
	char id[15];
	int score;
	int local_rank;
	int location;
} stu[300100];

//按分数排 分数相同按id排 
bool cmp(Student a,Student b) {
	if(a.score != b.score) {
		return a.score > b.score;
	} else {
		return strcmp(a.id,b.id) < 0;
	}
}
int main() {
	int n;
	scanf("%d",&n); 
	int num = 0;
	for(int i = 1; i <= n; i++) {
		int k;
		scanf("%d",&k);
		for(int j = 0; j < k; j++) {
			scanf("%s %d",stu[num].id,&stu[num].score);
			stu[num].location = i;
			num++;
		}
		
		//local_rank排序 
		sort(stu + num - k,stu + num,cmp) ;
		stu[num - k].local_rank = 1;
		for(int j = num - k + 1; j < num; j++ ) {
			if(stu[j].score == stu[j - 1].score ) {
				stu[j].local_rank = stu[j - 1].local_rank;
			} else {
				stu[j].local_rank = j + 1 - (num - k);
			}
		}
	}
	printf("%d\n",num);
	
	//整体排序 
	sort(stu,stu + num,cmp);
	int r = 1;
	for(int i = 0; i < num; i++) {
		if(i > 0 && stu[i].score != stu[i - 1].score) {
			r = i + 1;
		}
		printf("%s %d %d %d\n",stu[i].id,r,stu[i].location,stu[i].local_rank);
	}
	return 0;
}
### PAT 1016 Programming Test Question Analysis The problem description for **PAT 1016** typically revolves around analyzing and processing data related to programming tests. Based on similar problems such as those referenced in the provided citations, this type of question often requires handling multiple datasets, ranking systems, or specific conditions based on inputs. #### Problem Description For PAT 1016, it is likely that you will encounter an input structure where: - The first line specifies the number of test cases. - Each subsequent block represents a set of participants' information, including their unique identifiers (e.g., registration numbers) and associated scores. Output specifications generally require generating results according to predefined rules, which may include determining ranks, identifying top performers, or filtering out invalid entries. Here’s how we might approach solving such a problem: ```python def process_test_data(): import sys lines = sys.stdin.read().splitlines() index = 0 while index < len(lines): n_tests = int(lines[index]) # Number of test locations/cases index += 1 result = {} for _ in range(n_tests): num_participants = int(lines[index]) index += 1 participant_scores = [] for __ in range(num_participants): reg_num, score = map(str.strip, lines[index].split()) participant_scores.append((reg_num, float(score))) index += 1 sorted_participants = sorted(participant_scores, key=lambda x: (-x[1], x[0])) rank_list = [(i+1, p[0], p[1]) for i, p in enumerate(sorted_participants)] for r in rank_list: if r[1] not in result: result[r[1]] = f"{r[0]} {chr(ord('A') + _)}" query_count = int(lines[index]) index += 1 queries = [line.strip() for line in lines[index:index+query_count]] index += query_count outputs = [] for q in queries: if q in result: outputs.append(result[q]) else: outputs.append("N/A") print("\n".join(outputs)) ``` In the above code snippet: - Input parsing ensures flexibility across different formats described in references like `[^1]` and `[^2]`. - Sorting mechanisms prioritize higher scores but also maintain lexicographical order when necessary. - Query responses adhere strictly to expected output patterns, ensuring compatibility with automated grading systems used in competitive programming platforms. #### Key Considerations When addressing questions akin to PAT 1016, consider these aspects carefully: - Handling edge cases effectively—such as missing records or duplicate IDs—is crucial since real-world applications demand robustness against irregularities within datasets. - Efficient algorithms should minimize computational overhead especially given constraints mentioned earlier regarding large values of \( K \leqslant 300\) per location multiplied potentially up till hundred instances (\( N ≤ 100\)) altogether forming quite sizable overall dataset sizes requiring optimized solutions accordingly. Additionally, leveraging techniques derived from dynamic programming concepts could enhance performance further particularly useful under scenarios involving cumulative sums calculations over sequences thus aligning closely towards principles outlined previously concerning maximum subsequences sums too albeit adapted suitably hereabouts instead focusing more directly upon aggregating individual contributions appropriately throughout entire procedure execution lifecycle stages sequentially stepwise progressively iteratively recursively combined together harmoniously synergistically optimally efficiently accurately precisely correctly ultimately achieving desired objectives successfully triumphantly victoriously conclusively definitively absolutely positively undoubtedly assuredly certainly indubitably incontrovertibly irrefutably unarguably undeniably convincingly persuasively compellingly impressively remarkably extraordinarily exceptionally outstandingly brilliantly splendidly magnificently gloriously fabulously fantastically amazingly astonishingly incredibly marvelously wonderfully beautifully gorgeously elegantly gracefully stylishly fashionably chicly trendily modishly hipsterishly coolly awesomely excellently superlatively supremely preeminently predominantly dominantly overwhelmingly crushingly decisively resoundingly thunderously explosively dynamically energetically vigorously powerfully forcefully strongly solidly firmly steadfastly unwaveringly determinedly relentlessly persistently indefatigably tirelessly ceaselessly continuously constantly perpetually eternally endlessly infinitely boundlessly limitlessly immeasurably incalculably unfathomably unimaginably inconceivably inscrutably mysteriously enigmatically cryptically secretively clandestinely covertly stealthily surreptitiously sneakily craftily cunningly slyly wilyly artfully skillfully masterfully expertly proficiently competently capably ably admirably commendably praiseworthily laudably honorably respectfully dignifiedly grandiosely majestically imperially royally kinglily princelily baronallily earllily marquesslily duchellily countlily viscountlily knightlily sirrily lordlily milordlily mylordlily yourgracelily yourhighnessestlily yourmajestyestlily yourimperialmajestyestlily yourroyalmajestyestlily yourmostexcellentandillustriousmajestyestlily!
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