1047 Student List for Course (25 分) 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: E…

Zhejiang University has 40000 students and provides 2500 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.…

1047. Student List for Course (25) Zhejiang University has 40000 students and provides 2500 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…

1047 Student List for Course (25 分)   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:…

1047 Student List for Course 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 inp…

1047. Student List for Course (25) 时间限制 400 ms 内存限制 64000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Zhejiang University has 40000 students and provides 2500 courses. Now given the registered course list of each student, you are supposed to output…

https://pintia.cn/problem-sets/994805342720868352/problems/994805433955368960 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…

简单题. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<map> #include<queue> #include<string> #include<vector> using namespace std; +; struct X { ]; vect…

一开始是建立了course[2501][40001]数组,存储每节课的学生编号然后for循环两层输出,但这样复杂度为O(2500*40000),也很明显导致最后时间超时后来发现最多40000学生,每个学生最多选20门课,那么总共也就40000*20所以直接就存储学生-课程的信息,然后排个序,按照课程从小到大,课程一样的话则按字典序然后从头扫一遍即可,复杂度O(80000)不过要注意一点,有些课可能并没有出现,所以要做个判断,输出x 0. #include <iostream> #include…

题意: 输入两个正整数N和K(N<=40000,K<=2500),接下来输入N行,每行包括一个学生的名字和所选课程的门数,接着输入每门所选课程的序号.输出每门课程有多少学生选择并按字典序输出学生的名字. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ]; vector<]; int main(){ ios::sync_with_stdio…