【笨方法学PAT】1047 Student List for Course (25 分)

本文介绍了一个学生选课系统的实现过程,系统能够处理大量学生和课程数据,通过输入学生的选课信息,输出每门课程的选课人数及学生名单。采用STL中的vector容器存储数据,确保了数据处理的效率。

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一、题目

Zhejiang University has 40,000 students and provides 2,500 courses. Now given the registered course list of each student, you are supposed to output the student name lists of all the courses.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 numbers: N (≤40,000), the total number of students, and K (≤2,500), the total number of courses. Then N lines follow, each contains a student's name (3 capital English letters plus a one-digit number), a positive number C (≤20) which is the number of courses that this student has registered, and then followed by C course numbers. For the sake of simplicity, the courses are numbered from 1 to K.

Output Specification:

For each test case, print the student name lists of all the courses in increasing order of the course numbers. For each course, first print in one line the course number and the number of registered students, separated by a space. Then output the students' names in alphabetical order. Each name occupies a line.

Sample Input:

10 5
ZOE1 2 4 5
ANN0 3 5 2 1
BOB5 5 3 4 2 1 5
JOE4 1 2
JAY9 4 1 2 5 4
FRA8 3 4 2 5
DON2 2 4 5
AMY7 1 5
KAT3 3 5 4 2
LOR6 4 2 4 1 5

Sample Output:

1 4
ANN0
BOB5
JAY9
LOR6
2 7
ANN0
BOB5
FRA8
JAY9
JOE4
KAT3
LOR6
3 1
BOB5
4 7
BOB5
DON2
FRA8
JAY9
KAT3
LOR6
ZOE1
5 9
AMY7
ANN0
BOB5
DON2
FRA8
JAY9
KAT3
LOR6
ZOE1

二、题目大意

给出选课人数和课程数目,然后再给出每个人的选课情况,请针对每门课程输出选课人数以及所有选该课的学生姓名。

三、考点

STL

四、注意

1、使用vector<vector<string>> vec;

2、如果使用vector<set<string>> vec会超时。

五、代码

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<string>
#include<set>
#include<vector>
#include<algorithm>
using namespace std;
vector<vector<string>> vec;
bool cmp(string s1, string s2) {
	return s1 < s2;
}
int main() {
	//read
	int n, k;
	scanf("%d %d", &n, &k);
	vec.resize(k + 1);
	while (n--) {
		string s;
		int m;
		s.resize(4);
		scanf("%s %d", &s[0], &m);
		int id;
		while (m--) {
			scanf("%d", &id);
			vec[id].push_back(s);
		}
	}

	//output
	for (int i = 1; i <= k; ++i) {
		printf("%d %d\n", i, vec[i].size());
		sort(vec[i].begin(), vec[i].end(), cmp);
		for (int j = 0; j < vec[i].size(); ++j)
			printf("%s\n", vec[i][j].c_str());
	}

	system("pause");
	return 0;
}
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