ICB Securities Trading Company Assistant Programmer Exam 11 July 2026 — Question Solution | আইসিবি সিকিউরিটিজ ট্রেডিং কোম্পানি সহকারী প্রোগ্রামার নিয়োগ পরীক্ষা ২০২৬ — প্রশ্ন সমাধান

Approach (C). Read the file line by line into an array of structures, compute each student's total and average while reading, grade the student from the average (each subject is out of 100), pick the three highest totals, and finally average each of the four mark columns.

Part	Technique
(a) total, average	sum of 4 marks; average = total ⁄ 4 in float
(b) grade	if–else if ladder on the average: ≥80 A+, ≥70 A, ≥60 B, ≥50 C, below 50 F
(c) top three	3 passes of selection sort (largest total moves to the front)
(d) best subject	column-wise average of Marks1…Marks4, keep the index of the maximum

flowchart TD
    A["Open student.txt (stop if NULL)"] --> B["fscanf ID, Name, 4 marks"]
    B --> C["total = sum, avg = total / 4"]
    C --> D["grade from avg"]
    D --> B
    B -->|"EOF"| E["Print table (a)(b)"]
    E --> F["3 selection passes: top three (c)"]
    F --> G["Column averages, pick maximum (d)"]

Program

#include <stdio.h>
#include <string.h>

#define MAX 1000
#define SUB 4

typedef struct {
    char id[20];
    char name[50];
    float marks[SUB];
    float total;
    float avg;
    char grade[3];
} Student;

/* (b) grade from the average mark */
void assignGrade(Student *s)
{
    if (s->avg >= 80)      strcpy(s->grade, "A+");
    else if (s->avg >= 70) strcpy(s->grade, "A");
    else if (s->avg >= 60) strcpy(s->grade, "B");
    else if (s->avg >= 50) strcpy(s->grade, "C");
    else                   strcpy(s->grade, "F");
}

int main(void)
{
    Student st[MAX];
    int n = 0, i, j, k;
    FILE *fp = fopen("student.txt", "r");

    if (fp == NULL) {
        printf("Cannot open student.txt\n");
        return 1;
    }

    /* read: StudentID Name Marks1 Marks2 Marks3 Marks4 */
    while (n < MAX && fscanf(fp, "%19s %49s %f %f %f %f", st[n].id, st[n].name,
                  &st[n].marks[0], &st[n].marks[1],
                  &st[n].marks[2], &st[n].marks[3]) == 6) {
        /* (a) total and average */
        st[n].total = 0;
        for (j = 0; j < SUB; j++)
            st[n].total += st[n].marks[j];
        st[n].avg = st[n].total / SUB;
        assignGrade(&st[n]);                 /* (b) */
        n++;
    }
    fclose(fp);

    if (n == 0) {
        printf("No records found\n");
        return 0;
    }

    /* (a) + (b) report */
    printf("%-6s %-8s %6s %7s %5s\n", "ID", "Name", "Total", "Average", "Grade");
    for (i = 0; i < n; i++)
        printf("%-6s %-8s %6g %7.2f %5s\n", st[i].id, st[i].name,
               st[i].total, st[i].avg, st[i].grade);

    /* (c) top three by total: partial selection sort, descending */
    for (i = 0; i < 3 && i < n; i++) {
        int max = i;
        for (k = i + 1; k < n; k++)
            if (st[k].total > st[max].total)
                max = k;
        if (max != i) {
            Student tmp = st[i];
            st[i] = st[max];
            st[max] = tmp;
        }
    }
    printf("\nTop three students:\n");
    for (i = 0; i < 3 && i < n; i++)
        printf("%d. %s %s Total = %g (Grade %s)\n", i + 1,
               st[i].id, st[i].name, st[i].total, st[i].grade);

    /* (d) subject with the highest average mark */
    float subAvg[SUB] = {0};
    int best = 0;
    for (j = 0; j < SUB; j++) {
        for (i = 0; i < n; i++)
            subAvg[j] += st[i].marks[j];
        subAvg[j] /= n;
        if (subAvg[j] > subAvg[best])
            best = j;
    }
    printf("\nSubject averages:");
    for (j = 0; j < SUB; j++)
        printf(" Marks%d = %.2f", j + 1, subAvg[j]);
    printf("\nHighest average: Marks%d (%.2f)\n", best + 1, subAvg[best]);

    return 0;
}

Sample student.txt

101 Rahim 85 78 92 88
102 Karim 65 72 58 70
103 Nusrat 90 88 95 91
104 Tanvir 45 52 40 48
105 Sadia 75 80 68 77
106 Arif 55 60 49 62

Output (real run)

ID     Name      Total Average Grade
101    Rahim       343   85.75    A+
102    Karim       265   66.25     B
103    Nusrat      364   91.00    A+
104    Tanvir      185   46.25     F
105    Sadia       300   75.00     A
106    Arif        226   56.50     C

Top three students:
1. 103 Nusrat Total = 364 (Grade A+)
2. 101 Rahim Total = 343 (Grade A+)
3. 105 Sadia Total = 300 (Grade A)

Subject averages: Marks1 = 69.17 Marks2 = 71.67 Marks3 = 67.00 Marks4 = 72.67
Highest average: Marks4 (72.67)

Trace of a sample run

• Rahim: 85 + 78 + 92 + 88 = 343, average 343 ⁄ 4 = 85.75 → ≥ 80 → A+. Karim: 265 ⁄ 4 = 66.25 → ≥ 60 → B. Tanvir: 185 ⁄ 4 = 46.25 → below 50 → F.
• Top three: pass 1 finds the largest total 364 (Nusrat) and swaps it to index 0; pass 2 finds 343 (Rahim); pass 3 finds 300 (Sadia). Only 3 passes are needed, not a full sort.
• Subject averages: Marks1 = 415 ⁄ 6 = 69.17, Marks2 = 430 ⁄ 6 = 71.67, Marks3 = 402 ⁄ 6 = 67.00, Marks4 = 436 ⁄ 6 = 72.67 → Marks4 is the subject with the highest average.

Key points

• fscanf(...) == 6 stops cleanly at end of file or on a malformed line; fopen is checked for NULL.
• %49s reads a one-word name, as in the record format; if names may contain spaces, keep the file comma-separated and read each line with fgets + sscanf.
• Ties keep file order: the strict > in part (c) and part (d) keeps the earlier student or subject.
• Marks are read as float, so the average is never truncated by integer division.
• i < 3 && i < n keeps part (c) safe when the file has fewer than three students.
• If grades are wanted per subject, call the same ladder on each mark instead of the average.

Complexity (n students, 4 subjects)

• (a)(b) one pass: O(n). (c) 3 selection passes: O(3n) = O(n), better than a full O(n²) or O(n log n) sort. (d) O(4n) = O(n).
• Overall time O(n); space O(n) for the array.

Exam tip: state the assumption that each mark is out of 100 so the grade table is applied to the average, and always check both fopen and the fscanf return value.

Language: C++ (C has no classes; the same design works in Java with extends and super(...)).

OOP principles used

• Encapsulation: ID, name and basic salary are protected, so main() cannot change them directly; they are set only through the constructor.
• Inheritance: PermanentEmployee is an Employee (public inheritance) and reuses its data and display().
• Constructor chaining: the derived constructor passes the values to Employee(id, name, basic) in its initializer list.
• Polymorphism: display() is virtual and is overridden to add the allowance and total salary.

classDiagram
    class Employee {
        #int employeeId
        #string employeeName
        #double basicSalary
        +Employee(int, string, double)
        +display() void
    }
    class PermanentEmployee {
        +PermanentEmployee(int, string, double)
        +calculateAllowance() double
        +calculateTotalSalary() double
        +display() void
    }
    Employee <|-- PermanentEmployee

Program

#include <iostream>
#include <iomanip>
#include <string>
using namespace std;

// (a) Base class
class Employee {
protected:                       // visible to derived classes, hidden from main()
    int employeeId;
    string employeeName;
    double basicSalary;

public:
    Employee(int id, const string &name, double basic)
        : employeeId(id), employeeName(name), basicSalary(basic) {}

    virtual ~Employee() {}

    virtual void display() const {
        cout << "ID: " << employeeId
             << ", Name: " << employeeName
             << ", Basic: " << basicSalary << endl;
    }
};

// (b) Derived class
class PermanentEmployee : public Employee {
public:
    PermanentEmployee(int id, const string &name, double basic)
        : Employee(id, name, basic) {}          // base-class constructor

    double calculateAllowance() const {
        return basicSalary * 0.20;               // 20% of basic
    }

    double calculateTotalSalary() const {
        return basicSalary + calculateAllowance();
    }

    void display() const override {              // employee info + total salary
        Employee::display();
        cout << "  Allowance (20%): " << calculateAllowance()
             << ", Total Salary: " << calculateTotalSalary() << endl;
    }
};

// (c) main()
int main() {
    cout << fixed << setprecision(2);

    // (i) three PermanentEmployee objects
    PermanentEmployee emp[3] = {
        PermanentEmployee(101, "Rahim", 45000),
        PermanentEmployee(102, "Karim", 52000),
        PermanentEmployee(103, "Nusrat", 48500)
    };

    // (ii) details and total salary of each employee
    for (int i = 0; i < 3; i++)
        emp[i].display();

    // (iii) employee with the highest total salary
    int best = 0;
    for (int i = 1; i < 3; i++)
        if (emp[i].calculateTotalSalary() > emp[best].calculateTotalSalary())
            best = i;

    cout << "\nHighest total salary:" << endl;
    emp[best].display();
    return 0;
}

Output (real run)

ID: 101, Name: Rahim, Basic: 45000.00
  Allowance (20%): 9000.00, Total Salary: 54000.00
ID: 102, Name: Karim, Basic: 52000.00
  Allowance (20%): 10400.00, Total Salary: 62400.00
ID: 103, Name: Nusrat, Basic: 48500.00
  Allowance (20%): 9700.00, Total Salary: 58200.00

Highest total salary:
ID: 102, Name: Karim, Basic: 52000.00
  Allowance (20%): 10400.00, Total Salary: 62400.00

Trace

Employee	Basic	Allowance = 0.20 × Basic	Total = Basic + Allowance
101 Rahim	45000	9000	54000
102 Karim	52000	10400	62400
103 Nusrat	48500	9700	58200

The search loop starts with best = 0 (54000), compares 62400 > 54000 → best = 1, then 58200 > 62400 is false, so Karim (62400) is printed as the highest total salary.

Mapping to the marks

• (a) Employee: protected members, constructor with initializer list, display().
• (b) PermanentEmployee: calculateAllowance(), calculateTotalSalary(), overridden display() that calls Employee::display() and then prints the total.
• (c) main(): an array of three objects, a loop that displays each, and a max-search on calculateTotalSalary().

Why protected, not private: a derived class can read protected members directly. With private members, calculateAllowance() could not use basicSalary, and the base class would need a public getter such as getBasicSalary().

Complexity: each method is O(1); displaying and finding the highest of k employees is O(k) time (here k = 3) and O(k) space.

Exam tip: call the base constructor in the derived initializer list, and make display() virtual so a base pointer or reference still prints the total salary.