FAC1003 — Programming Revision Questions

Source style: Adapted from the Revision (Part B) slides. Each question gives a real-world scenario, specifies required sub-functions (void/non-void, pass-by-value, recursion, pointers), and expects a complete C++ program with example output.

Section A — Modular Functions (Non-Void + Void)

Q1 — Electricity Bill Calculator

A utility company wants a C++ program to calculate residential electricity bills. The company charges:

Usage (kWh)	Rate (RM/kWh)
First 200 kWh	0.218
Next 100 kWh (201–300)	0.334
Above 300 kWh	0.516

The company also applies a fixed monthly charge of RM5.00 for all customers. Customers who use more than 600 kWh receive an additional 5% surcharge on the total bill (excluding the fixed charge).

Task:

Write a complete C++ program that does the following:

Ask the user to enter:
- Customer name
- Total electricity usage in kWh (as a double)
Use a non-void function named calculateCharge(double usage) that:
- Returns the total consumption charge based on the tiered rates above
- (The fixed charge and surcharge are applied separately — not inside this function)
Use a non-void function named calculateSurcharge(double charge) that:
- Returns 5% of the charge if usage exceeds 600 kWh
- Returns 0 otherwise
Use a void function named displayBill(string name, double usage, double charge, double surcharge) that:
- Displays the customer name
- Shows a breakdown of usage, tiered charge, fixed charge (RM5.00), surcharge, and total amount
- Formats all monetary values to 2 decimal places

All values are passed by value.

Example output:

Enter customer name: Aisha
Enter total electricity usage (kWh): 450

--- Electricity Bill ---
Customer Name: Aisha
Usage: 450.00 kWh
Consumption Charge: RM140.20
Fixed Charge: RM5.00
Surcharge: RM0.00 (usage ≤ 600 kWh)
Total Amount: RM145.20

Example output (with surcharge):

Enter customer name: Beng
Enter total electricity usage (kWh): 720

--- Electricity Bill ---
Customer Name: Beng
Usage: 720.00 kWh
Consumption Charge: RM272.72
Fixed Charge: RM5.00
Surcharge: RM13.64 (5% of consumption charge)
Total Amount: RM291.36

Q2 — Loan Repayment Calculator

A bank offers personal loans with the following terms:

Loan amounts between RM1,000 and RM50,000
Annual interest rate: 4.5% for loans ≤ RM10,000, 3.9% for loans > RM10,000
Loan tenure: 6, 12, 24, or 36 months

The monthly payment is calculated using the formula:

monthlyPayment = (loanAmount + totalInterest) / tenureMonths

where totalInterest = loanAmount × annualRate × (tenureMonths / 12).

Write a complete C++ program that does the following:

Ask the user to enter:
- Customer name
- Loan amount (RM)
- Loan tenure in months (6, 12, 24, or 36 — the program MUST validate this)
Use a non-void function named getInterestRate(double amount) that:
- Returns the applicable annual interest rate based on loan amount
- Returns -1.0 if the loan amount is outside the valid range
Use a non-void function named calculateMonthlyPayment(double amount, double rate, int tenure) that:
- Computes and returns the monthly payment using the formula above
Use a void function named displayLoanSummary(string name, double amount, int tenure, double rate, double monthly) that:
- Displays a complete loan summary table
- Shows: loan amount, annual rate, tenure, total interest paid, total repayment, and monthly payment
- Formats all monetary values to 2 decimal places

Validation rules:

If the user enters an invalid loan amount (< 1000 or > 50000), the program should display "Invalid loan amount. Must be between RM1,000 and RM50,000." and exit.
If the user enters an invalid tenure (not 6, 12, 24, or 36), the program should display "Invalid tenure. Must be 6, 12, 24, or 36 months." and keep asking until a valid value is entered.

Example output:

Enter customer name: Chan
Enter loan amount (RM): 15000
Enter loan tenure (months): 24

--- Loan Summary ---
Customer: Chan
Loan Amount: RM15000.00
Annual Interest Rate: 3.90%
Tenure: 24 months
Total Interest: RM1170.00
Total Repayment: RM16170.00
Monthly Payment: RM673.75

Section B — Recursion

Q3 — Digit Sum & Digital Root

Write a C++ program that processes a positive integer entered by the user using recursion. The program has two tasks:

Task A — Sum of Digits: Compute the sum of all digits of the number recursively. Task B — Digital Root: Keep summing the digits until a single digit remains (the digital root).

For example:

Number: 4723
Digit sum: 4 + 7 + 2 + 3 = 16
Digital root: 1 + 6 = 7

Requirements:

Ask the user to enter a positive integer (no validation needed — assume valid input).
Use a recursive non-void function int sumOfDigits(int n) that:
- Returns the sum of all digits of n
- Base case: if n == 0, return 0
- Recursive case: return (n % 10) + sumOfDigits(n / 10)
Use a recursive non-void function int digitalRoot(int n) that:
- If n < 10, return n
- Otherwise, return digitalRoot(sumOfDigits(n))
Use a void function displayDigitAnalysis(int original) that:
- Displays the number entered
- Shows the step-by-step digit sum calculation: each digit separated by +
- Shows the sum of digits
- Shows the digital root with its step-by-step reduction

Example output:

Enter a positive integer: 4723

--- Digit Analysis ---
Number: 4723
Digit Sum: 4 + 7 + 2 + 3 = 16
Digital Root: 16 → 1 + 6 = 7
Final Digital Root: 7

Example output (single digit):

Enter a positive integer: 9

--- Digit Analysis ---
Number: 9
Digit Sum: 9 = 9
Digital Root: 9 (already a single digit)
Final Digital Root: 9

Q4 — Recursive Prime Factorisation

Write a C++ program that recursively decomposes a positive integer into its prime factors.

Prime factorisation is the process of breaking a number into the product of its prime factors:

12 = 2 × 2 × 3
30 = 2 × 3 × 5
100 = 2 × 2 × 5 × 5
97 = 97 (prime — no factorisation)

Requirements:

Ask the user to enter a positive integer greater than 1. If the user enters a number ≤ 1, display "Please enter a number greater than 1." and ask again until valid input is given.
Use a recursive non-void function bool isPrime(int n, int divisor = 2) that:
- Returns true if n is prime, false otherwise
- Base cases: if n <= 2, return n == 2; if n % divisor == 0, return false
- Recursive case: return isPrime(n, divisor + 1)
- Only check divisors up to sqrt(n) — stop recursion when divisor * divisor > n
Use a recursive void function void factorise(int n, bool first = true) that:
- If n == 1, return (done)
- If n is prime, print n and return
- Otherwise, find the smallest divisor d of n (starting from 2)
- Print d, then recursively call factorise(n / d, false)
- Print × between factors — the first parameter controls whether a × is printed before the factor
Use a void function void displayFactorisation(int n) that:
- Calls factorise(n) to print the prime factors
- Prints n = followed by the factorisation
- If the number is prime, prints n = n (prime) instead

Example output:

Enter a positive integer (> 1): 84

84 = 2 × 2 × 3 × 7

Example output (prime):

Enter a positive integer (> 1): 97

97 = 97 (prime)

Example output (invalid → retry):

Enter a positive integer (> 1): 1
Please enter a number greater than 1.
Enter a positive integer (> 1): -5
Please enter a number greater than 1.
Enter a positive integer (> 1): 60

60 = 2 × 2 × 3 × 5

Section C — Pointers & Dynamic 2D Arrays

Q5 — Temperature Anomaly Tracker

A climate research centre records daily temperature anomalies across multiple weather stations. Each station takes measurements at different times of day. Write a C++ program that processes this data using dynamic 2D arrays and pointer notation only.

Requirements:

The program shall first prompt the user to enter:
- The number of weather stations
- The number of daily readings (each station has the same number of readings)
Based on these values, dynamically allocate a 2D array of doubles. The memory MUST be allocated and accessed strictly using pointer notation (i.e., *(*(arr + i) + j), NOT arr[i][j]).
The program shall then read the readings (temperature anomalies in °C) for each station across all daily readings.
For each station, compute:
- Mean temperature anomaly
- Maximum anomaly (highest positive value)
- Minimum anomaly (lowest negative value — or 0 if all readings are positive)
- Range (max - min)
Classify each station based on its mean anomaly:

Mean Anomaly (°C) Classification

> +2.0 Extreme Warming

+0.5 to +2.0 Moderate Warming

-0.5 to +0.5 Stable

-2.0 to -0.5 Moderate Cooling

< -2.0 Extreme Cooling
Use a void function void displayStationReport(double* anomalies, int numReadings, int stationIndex) that:
- Takes a pointer to a single row of the 2D array (one station's data)
- Computes and displays the mean, max, min, range, and classification for that station
Memory cleanup: Deallocate the 2D array after processing.

Mean Anomaly (°C)	Classification
> +2.0	Extreme Warming
+0.5 to +2.0	Moderate Warming
-0.5 to +0.5	Stable
-2.0 to -0.5	Moderate Cooling
< -2.0	Extreme Cooling

Hint: To allocate a 2D array using pointers:

double** data = new double*[numStations];
for (int i = 0; i < numStations; i++)
    data[i] = new double[numReadings];

But all access must use pointer arithmetic: *(*(data + i) + j).

Example output:

Enter number of weather stations: 3
Enter number of daily readings: 4

Station 1, Reading 1: 1.2
Station 1, Reading 2: 2.5
Station 1, Reading 3: 3.1
Station 1, Reading 4: 2.8
Station 2, Reading 1: -0.5
Station 2, Reading 2: -1.2
Station 2, Reading 3: 0.3
Station 2, Reading 4: -0.8
Station 3, Reading 1: 3.5
Station 3, Reading 2: 4.2
Station 3, Reading 3: 3.9
Station 3, Reading 4: 4.8

--- Station Report ---
Station 1: Mean = 2.40°C | Max = 3.10°C | Min = 1.20°C | Range = 1.90°C | Moderate Warming
Station 2: Mean = -0.55°C | Max = 0.30°C | Min = -1.20°C | Range = 1.50°C | Moderate Cooling
Station 3: Mean = 4.10°C | Max = 4.80°C | Min = 3.50°C | Range = 1.30°C | Extreme Warming

Q6 — Sparse Matrix Compression (Dynamic Pointer Structure)

[!warning] Hard This question combines dynamic memory allocation, pointers, and structured programming. It is more difficult than a typical exam question — treat it as a stretch exercise.

A sparse matrix is a matrix where most elements are zero. Instead of storing all the zeros, we can use a compressed representation: store only the non-zero values along with their row and column positions.

Write a C++ program that converts a regular 2D matrix into a compressed sparse representation using dynamic arrays and pointer notation.

Requirements:

Ask the user to enter the dimensions of a matrix:
- Number of rows
- Number of columns
Dynamically allocate a 2D integer array (int**) and read its elements from the user.
After reading the full matrix, count the number of non-zero elements.
Dynamically allocate three 1D arrays using pointers:
- int* values — stores the non-zero values
- int* rowIndices — stores the row index of each non-zero value
- int* colIndices — stores the column index of each non-zero value
The size of each array should be exactly the number of non-zero elements.
Use a void function void compressMatrix(int** matrix, int rows, int cols, int*& values, int*& rowIndices, int*& colIndices, int& nonZeroCount) that:
- Scans the matrix row by row
- For each non-zero element, appends its value, row index, and column index to the respective dynamic arrays
- The function parameters values, rowIndices, colIndices are passed by reference (they will be allocated inside the function)
Use a void function void displayMatrix(int** matrix, int rows, int cols) that:
- Displays the full matrix in grid format
Use a void function void displayCompressed(int* values, int* rowIndices, int* colIndices, int nonZeroCount) that:
- Displays the compressed representation as a table
Use a void function void reconstructAndDisplay(int* values, int* rowIndices, int* colIndices, int nonZeroCount, int rows, int cols) that:
- Reconstructs the matrix from the compressed representation
- Displays it — it should match the original matrix exactly
Memory cleanup: Deallocate ALL dynamically allocated memory (the original 2D array AND the three compressed arrays).

Example output:

Enter number of rows: 3
Enter number of columns: 4

Row 0:
  Col 0: 0
  Col 1: 0
  Col 2: 5
  Col 3: 0
Row 1:
  Col 0: 0
  Col 1: 0
  Col 2: 0
  Col 3: 0
Row 2:
  Col 0: 3
  Col 1: 0
  Col 2: 0
  Col 3: 8

Original Matrix:
  0  0  5  0
  0  0  0  0
  3  0  0  8

Compressed Representation (Non-zero elements: 3):
Value   Row   Col
  5      0     2
  3      2     0
  8      2     3

Reconstructed Matrix:
  0  0  5  0
  0  0  0  0
  3  0  0  8

Matrices match!

Example output (all zeros):

Enter number of rows: 2
Enter number of columns: 2

Row 0:
  Col 0: 0
  Col 1: 0
Row 1:
  Col 0: 0
  Col 1: 0

Original Matrix:
  0  0
  0  0

No non-zero elements found. Matrix is entirely zero.

FAC1003 — Programming Revision Questions — Answer Scheme — Complete C++ solutions
FAC1003 - Programming II
FAC1003 — Final Exam Scope 2025-2026
FAC1003 Tutorial 14 — OOP & C++ Solutions

FAC1003 — Programming Revision Questions (Functions, Recursion, Pointers)

FAC1003 — Programming Revision Questions

Section A — Modular Functions (Non-Void + Void)

Q1 — Electricity Bill Calculator

Q2 — Loan Repayment Calculator

Section B — Recursion

Q3 — Digit Sum & Digital Root

Q4 — Recursive Prime Factorisation

Section C — Pointers & Dynamic 2D Arrays

Q5 — Temperature Anomaly Tracker

Q6 — Sparse Matrix Compression (Dynamic Pointer Structure)

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