zoukankan      html  css  js  c++  java
  • 1047 Student List for Course (25 分)

    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:

    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

    #include<iostream>
    #include<vector>
    #include<algorithm>
    #include<queue>
    #include<string>
    #include<map>
    #include<set>
    #include<stack>
    #include<string.h>
    #include<cstdio>
    #include <unordered_map>
    #include<cmath>
    
    using namespace std;
    
    
    int main()
    {
        int n,k;
        scanf("%d%d",&n,&k);
        vector<vector<string>> vt(k+1);
        char name[10];
        int c;
        int temp;
        for(int i=0;i<n;i++)
        {
            scanf("%s%d",name,&c);
            for(int j=0;j<c;j++)
            {
                scanf("%d",&temp);
                vt[temp].push_back(name);
            }
        }
        for(int i=1;i<k+1;i++)
        {
            printf("%d %d
    ",i,vt[i].size());
            sort(vt[i].begin(),vt[i].end());
            for(auto &name:vt[i])
                printf("%s
    ",name.c_str());
        }
        return 0;
    }
  • 相关阅读:
    SQL server 分页方法小结
    在电脑上测试手机网站全攻略
    android批量插入数据效率对比
    表格细边框的两种CSS实现方法
    作为一个非纯粹的优质码农,应该有怎么样的心态?
    C#注册表读写完整操作类
    SQL Server默认1433端口修改方法
    学习编程一年多的体会
    mac上virtualbox创建vm需要注意启动顺序
    git diff patch方法
  • 原文地址:https://www.cnblogs.com/zhanghaijie/p/10327657.html
Copyright © 2011-2022 走看看