Output Table: ID: 123 ; Name: Enan ; CGPA: 2.75 ; Exams: 80, 70, 85, 75 ; Average Mark: 80 ID: 131 ; Name: Akash ; CGPA: 2.72 ; Exams: 80, 70, 65, 75; Average Mark: 75 ID: 132 ; Name: Shovon ; CGPA: 2.7; Exams: 75, 75, 75; Average Mark: 75 ID: 141 ; Name: Tanvir ; CGPA: 3.3 ; Exams: 80, 90, 85, 80, 90; Average Mark: 88.33 ID: 142 ; Name: Junayet ; CGPA: 2.5 ; Exams: 75, 80, 70 ; Average Mark: 75 Note: The Student list will be printed in ascending order based on student ID. Average Mark: You have to take best 3 exam marks. (From the number of exams were taken) For example, you have to take 5 best exam marks if the number of exams is 7. And then you have to take the average. In our example, the number of exams was 5. That's why the average number was based on 3 best exam marks.
Output Table: ID: 123 ; Name: Enan ; CGPA: 2.75 ; Exams: 80, 70, 85, 75 ; Average Mark: 80 ID: 131 ; Name: Akash ; CGPA: 2.72 ; Exams: 80, 70, 65, 75; Average Mark: 75 ID: 132 ; Name: Shovon ; CGPA: 2.7; Exams: 75, 75, 75; Average Mark: 75 ID: 141 ; Name: Tanvir ; CGPA: 3.3 ; Exams: 80, 90, 85, 80, 90; Average Mark: 88.33 ID: 142 ; Name: Junayet ; CGPA: 2.5 ; Exams: 75, 80, 70 ; Average Mark: 75 Note: The Student list will be printed in ascending order based on student ID. Average Mark: You have to take best 3 exam marks. (From the number of exams were taken) For example, you have to take 5 best exam marks if the number of exams is 7. And then you have to take the average. In our example, the number of exams was 5. That's why the average number was based on 3 best exam marks.
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Question
Solve in C++
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Transcribed Image Text:Show the output according to the output table.
Output Table:
ID: 123 ; Name: Enan ; CGPA: 2.75 ; Exams: 80, 70, 85, 75 ; Average Mark: 80
ID: 131 ; Name: Akash ; CGPA: 2.72 ; Exams: 80, 70, 65, 75; Average Mark: 75
ID: 132 ; Name: Shovon ; CGPA: 2.7 ; Exams: 75, 75, 75; Average Mark: 75
ID: 141 ; Name: Tanvir ; CGPA: 3.3 ; Exams: 80, 90, 85, 80, 90; Average Mark: 88.33
ID: 142 ; Name: Junayet ; CGPA: 2.5 ; Exams: 75, 80, 70; Average Mark: 75
Note: The Student list will be printed in ascending order based on student ID.
Average Mark: You have to take best 3 exam marks. (From the number of exams were
taken)
For example, you have to take 5 best exam marks if the number of exams is 7. And then
you have to take the average. In our example, the number of exams was 5. That's why
the average number was based on 3 best exam marks.
Hint: Similar example for student class header file
class Student
{
private:
int id;
string name;
float cgpa;
unsortedType<int> marks ;
public:
Student();
Student(int , string, float, unsortedType<int>);
Transcribed Image Text:Suppose, you are asked to make a system where you can put some information of the students
of CSE225. You have to take inputs according to the input table
Input Table:
Number of students in CSE225?
Number of Exams?
Student ID:
141
Student Name:
Tanvir
Student CGPA:
3.3
Number of Exams attended :
Exam 1:
80
Exam 2:
90
Exam 3:
85
Exam 4:
80
Exam 5:
90
Student ID:
123
Student Name:
Enan
Student CGPA:
2.75
Number of Exams attended :
4
Exam 1:
80
Exam 2:
70
Exam 3:
85
Exam 4:
75
Student ID:
142
Student Name:
Junayet
Student CGPA:
2.5
Number of Exams attended :
3
Exam 1:
75
Еxam 2:
80
Exam 3:
70
Student ID:
131
Student Name:
Akash
Student CGPA:
2.72
Number of Exams attended :
Exam 1:
80
Exam 2:
70
Exam 3:
65
Exam 4:
75
Student ID:
132
Student Name:
Shovon
Student CGPA:
2.7
Number of Exams attended :
3
Exam 1:
75
Exam 2:
75
Exam 3:
75
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