zoukankan      html  css  js  c++  java
  • PAT 甲级 1047 Student List for Course

    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 of all the courses.

    Input Specification:

    Each input file contains one test case. For each case, the first line contains 2 numbers: N (≤), the total number of students, and K (≤), 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 (≤) 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

    代码:

    #include <bits/stdc++.h>
    using namespace std;
    
    int N, K;
    
    struct Node{
      string name;
      int num;
    }node[40010];
    
    struct Class{
      vector<string> id;
      int cnt;
    }cclass[2510];
    
    int main() {
      
      scanf("%d%d", &N, &K);
      for(int i = 1; i <= N; i ++) {
        
        cin >> node[i].name;
        scanf("%d", &node[i].num);
        
        for(int j = 1; j <= node[i].num; j ++) {
          int x;
          scanf("%d", &x);
          cclass[x].cnt ++;
          cclass[x].id.push_back(node[i].name);
        }
      }
      
      for(int i = 1; i <= K; i ++) {
        printf("%d %d
    ", i, cclass[i].cnt);
        sort(cclass[i].id.begin(), cclass[i].id.end());
        for(int j = 0; j < cclass[i].cnt; j ++)
          printf("%s
    ", cclass[i].id[j].c_str());
      }
      return 0;
    }
    

      

  • 相关阅读:
    制作一个命令式的 React 弹出层组件 (适用 React Native)
    React 中的 onInput/onChange
    防抖和节流及对应的React Hooks封装
    React Native选择器组件-react-native-slidepicker
    React Portal
    Quartz学习 之 Jobs 和 Triggers
    Quartz学习 之 关键接口
    Quartz学习 之 入门
    JAVA NIO 原理探秘 --- Socket
    JAVA面试题
  • 原文地址:https://www.cnblogs.com/zlrrrr/p/9928128.html
Copyright © 2011-2022 走看看