代写LP002、代做Data Structures、代写Java程序、代做Java编程调试Matlab程序|代做SPSS

Homework 1 - Sorting Student RecordsLP002 Data StructuresOctober 31, 20181 IntroductionYou are asked to implement a tiny Java program to manage Student Records for our university. Inparticular, your program should be able to read the score reports of students from a given file. Morespecifically, your MSSR shall provide end users with efficient and effective sorting the given largenumber of student records.2 SpecificationYour MSSR system shall provide users with searching and sorting on a large number of student scorerecords.2.1 Read score recordsThe score report has the following format:(a) The first line contains the number of records, says N (N ≥ 1).(b) Each of the next N lines contains a comma separated list of student’s StudentID, StudentName, Course Code and his/her obtained Score. StudentID starting with ‘s’ occupies totally 11 characters. Student Name is a varied string from 5 to 14 characters. Course Code is a fixed 5-character string. Score is an integer ranging from 0 to 100.(c) You may assume there is no space before and after the comma delimiter and there is no trailingspace after each line of data.(d) Note that you SHOULD NOT change the input data file format.The following is an example of data file:15s1109853011,JackieChan,CS001,85s0907451244,FranklinPapadias,MA002,65s0308893477,MirandaLambert,HM467,35s4045128481,SophiaPam,CM604,75s0812340023,PhilipsChiu,CS001,71As shown in the above example, there are 5 students in this score report.2.2 Store recordsAfter reading the score reports successfully, your program should be able to store student scores. Inparticular, your program should store the course information as follows:Student:Student IDStudent NameScoreCode CodeNote: You need to design a class to sore the student record. You should specify enough space for each field defined in your class (data structure).2.3 SortingYou are required to display the report according to StudentID, student name, course codeand the score in either an ascending order or a descending order. For example as shown as follows,the students are listed in ascending order lexicographically according to their StudentID.output:s0308893477,MirandaLambert,HM467,35s081代写LP002留学生作业、代做Data Structures作业、代写Java程序作业、代做Java编程作业 调试Mat2340023,PhilipsChiu,CS001,71s0907451244,FranklinPapadias,MA002,65s1109853011,JackieChan,CS001,85s4045128481,SophiaPam,CM604,75Note: To specify which field to be sorted and the sorting ways (ascending or descending), you can letuser input it as a parameter.An example input for specifying the sorting parameter:2Please choose the sorting field: 1(1) StudentID; (2) Student Name; (3) Course Code; (4) ScorePlease choose the sorting ways: A(A)scending order; (D)escending order2.4 PerformanceYour system shall be efficient in both searching and sorting. In particular, the running time for a singlequery of your system shall be less than 20 seconds with the number of records N is less than 100,000.Every extra second leads to subtracting one point in your score (note that the time for readingrecords will not be counted in the performance). Therefore, please try to call faster algorithms (e.g.quick sort, merge sort and binary search trees, etc.) to implement your system.3 SubmissionYour Submission1. A soft copy of your source program including all the components.2. readme file (readme.txt) which contains:(a) file list(b) file description(c) methods of compilation(d) method of execution(e) known bugs of your systemA good example:/ Assignment 1 Name : Alex Fung Student ID: 1009853X I011 YYYY Section:LP002 email: [email protected] Description: This file contains the ThreeSum class, which counts the number of tuples equal to 0./public class ThreeSum{/Description: Brute force method to compute the the number of3 tuples equal to 0 by checking condition: a[i] + a[j] + a[k] == 0 Parameters: a: the array Return: count: the number satisfying the condition/public static int count(int[] a){int N = a.length; // The size of arrayint count = 0; // the resultfor (int i = 0; i for (int j = i+1; j for (int k = j+1; k if (a[i] + a[j] + a[k] == 0)count++;return count;}/ A testing client reads an array from standard input and prints the number of tuples equal to zero./public static void main(String[] args){int[] a = StdArrayIO.readInt1D();StdOut.println(count(a));}}转自：http://ass.3daixie.com/2018112339684262.html