1047. Student List for Course
来源:互联网 发布:冰与火之歌结局 知乎 编辑:程序博客网 时间:2024/05/29 11:47
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.
#include <stdio.h>#include <stdlib.h>#include <string.h>#include <vector>#include <algorithm>using namespace std;#define N 400001char name[N][5];bool cmp(int a,int b);int main (){ int n,c,i; scanf("%d %d",&n,&c); vector<int> course[c+1]; int cid,j,k; for( i=0;i<n;i++) { scanf("%s %d",name[i],&k); for( j=0;j<k;j++) { scanf("%d",&cid); course[cid].push_back(i); } } for( i=1;i<=c;i++) { printf("%d %d\n",i,course[i].size()); sort(course[i].begin(),course[i].end(),cmp); for( j=0;j<course[i].size();j++) printf("%s\n",name[course[i][j]]); } system("pause"); return 0; }bool cmp(int a,int b){ return strcmp(name[a],name[b])<0; }
- 1047. Student List for Course
- 1047. Student List for Course
- 1047. Student List for Course
- 1047.Student List for Course
- 1047. Student List for Course
- 1047. Student List for Course
- Course List for Student
- Student List for Course
- 1047. Student List for Course (25)
- 1047. Student List for Course (25)-PAT
- 1047. Student List for Course (25)
- 【PAT】1047. Student List for Course (25)
- 1047. Student List for Course (25)
- 1047. Student List for Course (25)
- 1047. Student List for Course (25)
- 1047. Student List for Course (25)
- PAT 1047. Student List for Course
- PAT 1047. Student List for Course
- socket通信模型
- 第1章1节《MonkeyRunner源码剖析》概述:前言(原创)
- apt-get: Could not resolve 'archive.ubuntu.com'
- FlappyBird开发总结(二)——场景
- Codeforces Round #295 (Div. 2) C. DNA Alignment(数学水题)
- 1047. Student List for Course
- Android基础笔记(一)-快速入门
- Linux共享库.so文件的命名和动态链接
- JAVAEE------css层叠样式表知识点总结
- event
- MongoDB使用学习(一)-基础知识
- 关于编码遇到的一些问题
- Android模拟器无线网络怎么配置网络连通
- Java状态模式(State模式)