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. For each case, the first line contains 2 numbers: N (<=40000), the total number of students, and K (<=2500), 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
思路:建立2500个课程的vector,存放学生的姓名(最开始的时候用了list,超时;不要用c++的cin和cout)
#include <iostream>
#include"stdio.h"
#include"stdlib.h"
#include"vector"
#include"string"
#include"algorithm" using namespace std; int main()
{
long N;
int K,C;
cin>>N>>K;
scanf("%ld%d",&N,&N); int order;
vector<string> l[2501]; string name;
for(int i=1;i<=N;i++){
cin>>name;
scanf("%d",&C);
for(int j=1;j<=C;j++){
scanf("%d",&order);
l[order].push_back(name);
}
} for(int i=1;i<=K;i++){
int size = l[i].size();
printf("%d %d\n",i,size);
if(size)
{
sort(l[i].begin(),l[i].end()); vector<string>::iterator iter;
for(iter=l[i].begin();
iter<l[i].end();
iter++){
printf("%s\n",(*iter).c_str());
}
}
}
return 0;
}

1047. Student List for Course (25)的更多相关文章

  1. PAT 解题报告 1047. Student List for Course (25)

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

  2. PAT 甲级 1047 Student List for Course (25 分)(cout超时,string scanf printf注意点,字符串哈希反哈希)

    1047 Student List for Course (25 分)   Zhejiang University has 40,000 students and provides 2,500 cou ...

  3. 1047 Student List for Course (25分)

    Zhejiang University has 40,000 students and provides 2,500 courses. Now given the registered course ...

  4. PAT (Advanced Level) 1047. Student List for Course (25)

    简单题. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> ...

  5. PAT甲级题解-1047. Student List for Course (25)-排序

    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 ...

  6. 【PAT甲级】1047 Student List for Course (25 分)

    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 ...

  7. PAT 1047 Student List for Course[一般]

    1047 Student List for Course (25 分) Zhejiang University has 40,000 students and provides 2,500 cours ...

  8. 1047 Student List for Course ——PAT甲级真题

    1047 Student List for Course Zhejiang University has 40,000 students and provides 2,500 courses. Now ...

  9. PAT 甲级 1047 Student List for Course

    https://pintia.cn/problem-sets/994805342720868352/problems/994805433955368960 Zhejiang University ha ...

随机推荐

  1. 把浏览器的私有模式添加到VS中

    题记:在用VS进行Web开发的时候,常常希望VS的调试不会对浏览器造成固定的影响,那么使用浏览器的私有模式来启动就很有必要. 前几天SCOTT HANSELMAN分享了一个开发Web应用程序的小技巧, ...

  2. Java Security: Illegal key size or default parameters?

    来自:http://stackoverflow.com/questions/6481627/java-security-illegal-key-size-or-default-parameters I ...

  3. hihoCoder 1391 Countries【预处理+排序+优先队列】2016北京网络赛

    题目:http://hihocoder.com/problemset/problem/1391 题目大意: A和B两个国家互射导弹,每个国家都有一个防御系统,在防御系统开启的时间内可以将到达本国的导弹 ...

  4. php 常见的问题

    1. this指针错误的引用变量($(php)<->*(c)) $this->inputData right wrong 2. json_encode(array) 不一定按数组关键 ...

  5. Maven项目在Eclipse中调试 Debug

    废话不说一路跟图走. 断点会进入到如下页面点击Edit Source Lookup Path 如下图操作 成功进入Debug模式

  6. Uva10635 LCS

    题目链接:http://vjudge.net/contest/137498#problem/G 题意:有两个长度为p+1和q+1的序列,每个序列的中的各个元素互不相同,且都是1~n^2之间的整.两个序 ...

  7. 安装PyMysql的基本步骤

    X:\Users\**>c: c:\>cd python c:\Python>python ez_setup.py Downloading https://pypi.io/packa ...

  8. 伪Acmer的推理(dfs/bfs)

    时间限制:1000MS  内存限制:65535K 提交次数:12 通过次数:9 收入:32 题型: 编程题   语言: C++;C Description 现在正是期末,在复习离散数学的Acmer遇到 ...

  9. oracle通过sequence定义触发器支持id的自增

    sequence:s_author_id,从1开始步幅是1,最大值是999999999 -- Create sequence create sequence S_AUTHOR_ID minvalue ...

  10. [转]HTML5本地存储——Web SQL Database

    在HTML5 WebStorage介绍了html5本地存储的Local Storage和Session Storage,这两个是以键值对存储的解决方案,存储少量数据结构很有用,但是对于大量结构化数据就 ...