PAT-A-1039. Course List for Student (25)
来源:互联网 发布:mac os x10.10iso镜像 编辑:程序博客网 时间:2024/05/23 01:18
1039. Course List for Student (25)
Zhejiang University has 40000 students and provides 2500 courses. Now given the student name lists of all the courses, you are supposed to output the registered course list for each student who comes for a query.
Input Specification:
Each input file contains one test case. For each case, the first line contains 2 positive integers: N (<=40000), the number of students who look for their course lists, and K (<=2500), the total number of courses. Then the student name lists are given for the courses (numbered from 1 to K) in the following format: for each course i, first the course index i and the number of registered students Ni (<= 200) are given in a line. Then in the next line, Ni student names are given. A student name consists of 3 capital English letters plus a one-digit number. Finally the last line contains the N names of students who come for a query. All the names and numbers in a line are separated by a space.
Output Specification:
For each test case, print your results in N lines. Each line corresponds to one student, in the following format: first print the student's name, then the total number of registered courses of that student, and finally the indices of the courses in increasing order. The query results must be printed in the same order as input. All the data in a line must be separated by a space, with no extra space at the end of the line.
Sample Input:11 54 7BOB5 DON2 FRA8 JAY9 KAT3 LOR6 ZOE11 4ANN0 BOB5 JAY9 LOR62 7ANN0 BOB5 FRA8 JAY9 JOE4 KAT3 LOR63 1BOB55 9AMY7 ANN0 BOB5 DON2 FRA8 JAY9 KAT3 LOR6 ZOE1ZOE1 ANN0 BOB5 JOE4 JAY9 FRA8 DON2 AMY7 KAT3 LOR6 NON9Sample Output:
ZOE1 2 4 5ANN0 3 1 2 5BOB5 5 1 2 3 4 5JOE4 1 2JAY9 4 1 2 4 5FRA8 3 2 4 5DON2 2 4 5AMY7 1 5KAT3 3 2 4 5LOR6 4 1 2 4 5NON9 0
#include<iostream>#include<cstdio>#include<vector>#include<algorithm>#include<stdio.h>using namespace std;const int M = 26 * 26 * 26 * 10 + 1;vector<int> vi[M];int Getid(char name[]){ int ans = 0; for (int i = 0; i < 3; i++) { ans = ans * 26 + (name[i] - 'A'); } ans = ans * 10 + (name[3] - '0'); return ans;}int main(){ int n, k; char name[5]; scanf("%d %d", &n, &k); //cin >> n >> k; int courseid, num; for (int i = 0; i < k; i++) { scanf("%d%d", &courseid, &num); //cin >> courseid >> num; for (int j = 0; j < num; j++) { scanf("%s", name); //cin >> name; int id = Getid(name); vi[id].push_back(courseid); } } for (int i = 0; i < n; i++) { scanf("%s", name); //cin >> name; int id = Getid(name); sort(vi[id].begin(), vi[id].end()); printf("%s", name); printf(" %d", vi[id].size()); for (int j = 0; j < vi[id].size(); j++) { printf(" %d", vi[id][j]); } printf("\n"); } return 0;}
- PAT-A-1039. Course List for Student
- PAT A 1039. Course List for Student (25)
- PAT(A) - 1039. Course List for Student (25)
- PAT-A-1039. Course List for Student (25)
- 1039. Course List for Student (25)-PAT
- PAT 1039. Course List for Student (25)
- PAT 1039. Course List for Student (25)
- PAT 1039. Course List for Student (25)
- PAT 1039. Course List for Student (25)
- 【PAT】1039. Course List for Student (25)
- PAT A 1047.Student List for Course (25)
- PAT A 1048.Student List for Course (25)
- PAT(A) - 1047. Student List for Course (25)
- PAT-A-1047. Student List for Course (25)
- PAT 1039. Course List for Student
- PAT 1039. Course List for Student
- 【PAT】1039. Course List for Student
- PAT--1039. Course List for Student
- 线程池的submit和execute的区别
- C语言链表使用详解
- FEP: 测试 docker matrix, 获得 CPU, MEMORY, 网络性能的 统计信息
- Android 分组ListView+索引条
- 利用jsoup爬取网页信息
- PAT-A-1039. Course List for Student (25)
- java设计模式之建造者模式
- 如何指定html中select控件的宽度?
- activity之间及fragment之间切换的动画效果
- 构造KD树,并用KD树求最近邻点
- YII框架的依赖注入容器
- 网页添加icon小图标
- SSM搭建(整合)+用户模块(登录和注销)实现
- iOS进阶(二)Objective-C底层原理