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Output: 24
(a)
INPUT bookTitle, newStatus
FOR i FROM 0 TO 99
IF TITLES[i] == bookTitle THEN
STATUS[i] = newStatus
ENDIF
ENDFOR
(b)
FOR i FROM 0 TO 99
IF STATUS[i] == "Available" THEN
OUTPUT TITLES[i] + " by " + AUTHORS[i]
ENDIF
ENDFOR
(c)
INPUT selectedAuthor
FOR i FROM 0 TO 99
IF AUTHORS[i] == selectedAuthor THEN
OUTPUT TITLES[i] + ": " + STATUS[i]
ENDIF
ENDFOR
(a)
FOR i FROM 0 TO 49
IF Salaries[i] >= 50000 THEN
BONUS[i] = Salaries[i] * 0.10
ELSE
BONUS[i] = Salaries[i] * 0.05
ENDIF
ENDFOR
(b)
In parallel arrays, each index corresponds to the same employee across all arrays. For example, the data at index 0 in the Names[], Ages[], Salaries[], and BONUS[] arrays all belong to the first employee, ensuring that the information is correctly matched.
(c)
FOR i FROM 0 TO 49
IF Ages[i] < 30 AND Salaries[i] > 40000 THEN
OUTPUT Names[i] + ", Age: " + Ages[i]
ENDIF
ENDFOR
(d)
INPUT maxAge, minSalary
FOR i FROM 0 TO 49
IF Ages[i] <= maxAge AND Salaries[i] >= minSalary THEN
OUTPUT Names[i] + ", Salary: " + Salaries[i]
ENDIF
ENDFOR
(a)
maxPopulation = 0
cityWithMaxPopulation = ""
FOR i FROM 0 TO 99
IF Population[i] > maxPopulation THEN
maxPopulation = Population[i]
cityWithMaxPopulation = Cities[i]
ENDIF
ENDFOR
OUTPUT cityWithMaxPopulation
(b)
highEmissionCities = []
FOR i FROM 0 TO 99
perCapitaEmission = CO2Emissions[i] / Population[i]
IF perCapitaEmission > 1000 THEN
APPEND Cities[i] TO highEmissionCities
ENDIF
ENDFOR
OUTPUT highEmissionCities
(c)
totalFines = 0
FOR i FROM 0 TO 99
perCapitaEmission = CO2Emissions[i] / Population[i]
IF perCapitaEmission > 999 AND perCapitaEmission < 2000 THEN
fine = 100 * Population[i]
ELSE IF perCapitaEmission > 1999 AND perCapitaEmission < 3000 THEN
fine = 125 * Population[i]
ELSE IF perCapitaEmission > 2999 THEN
fine = 150 * Population[i]
ENDIF
totalFines = totalFines + fine
ENDFOR
OUTPUT totalFines
So:
Output: 10, 20
| A | B | C | B > C | B mod 4 = 0 | Output |
|---|---|---|---|---|---|
| 1 | 10 | 5 | True | ||
| 4 | 8 | True | |||
| 6 | 6 | ||||
| True | |||||
| 7 | 6 | False | 7 | ||
| 10 | 4 | False |
(a)
INPUT playerName, pointsScored
FOR i FROM 0 TO 14
IF NAMES[i] == playerName THEN
SCORES[i] = SCORES[i] + pointsScored
ENDIF
ENDFOR
(b)
FOR i FROM 0 TO 14
IF POSITIONS[i] == "Guard" AND SCORES[i] > 100 THEN
OUTPUT NAMES[i] + " - " + POSITIONS[i]
ENDIF
ENDFOR
a) That's a function that calls itself.
b)
So:
(a)
sum = 0
for row from 0 TO 7
for col from 0 TO 5
sum = sum + X[row][col]
end for
end for
return "Sum of all integers: " + sum
(b)
sum = 0
count = 0
for row from 0 TO 7
for col from 0 TO 5
sum = sum + X[row][col]
count = count + 1
end for
end for
average = sum / count
return "Average of all integers: " + average
(a)
MIN = WEIGHTS.getNext()
WEIGHTS.resetNext()
loop while WEIGHTS.hasNext()
TEMP = WEIGHTS.getNext()
if TEMP < MIN then
MIN = TEMP
end if
end loop
return MIN
(b)
WEIGHTS.resetNext()
TOTAL = 0
COUNT = 0
loop while WEIGHTS.hasNext()
TEMP = WEIGHTS.getNext()
TOTAL = TOTAL + TEMP
COUNT = COUNT + 1
end loop
AVERAGE = TOTAL / COUNT
return AVERAGE
(a)
test(InvoiceID)
N = length(PAID)
for i from 0 to N-1
IF PAID[i] == InvoiceID then
return "Paid"
end if
end for
return "Not paid"
end test
(b)
N = length(PAID)
FLAG = true
loop while FLAG = true
FLAG = false
for i from 0 to N-2
if PAID[i] > PAID[i + 1] then
TEMP = PAID[i]
PAID[i] = PAID[i + 1]
PAID[i + 1] = TEMP
FLAG = true
end if
end loop
end loop
(a) A queue follows the First In, First Out (FIFO) principle, ensuring that the first patient to check in is the first one to be processed. This guarantees fairness in the order of check-ins and prevents newer patients from skipping ahead. Using a queue maintains an organized and predictable system.
(b) A stack follows the Last In, First Out (LIFO) principle, meaning the last patient to check in would be processed first. This is unfair to earlier patients, as newer arrivals would be treated before those who have been waiting longer. In real-world scenarios like hospital queues, a FIFO structure (queue) ensures fairness.
(c)
| NUM | SUM | Output |
|---|---|---|
| 5 | ||
| 4 | ||
| 3 | ||
| 2 | ||
| 1 | ||
| 1 | ||
| 2 | 3 | |
| 3 | 6 | |
| 4 | 10 | |
| 5 | 15 | 15 |
(d) A recursive function relies on the concept of function calls being stored in a call stack. Each recursive call pushes a new function instance onto the stack. When a base case is reached, the function calls are popped from the stack in reverse order.
loop for Count from 0 to 7
input NUMBER
if NUMBER mod 2 = 0 then
if NUMBER mod 4 = 0 then
output NUMBER
end if
end if
end loop
TITLES[], AUTHORS[], and STATUS[]. The status of the books changes based on library activities.| Title | Author | Status |
|---|---|---|
| The Great Gatsby | F. Scott Fitzgerald | Available |
| To Kill a Mockingbird | Harper Lee | Checked Out |
| 1984 | George Orwell | Available |
STATUS[] array based on the book titles that have been checked out or returned. Assume you are given the book title and the new status (either "Available" or "Checked Out").| Names |
|---|
| John Smith |
| Jane Doe |
| Emily Johnson |
| Ages |
|---|
| 28 |
| 34 |
| 25 |
| Salaries |
|---|
| 55,000 |
| 48,000 |
| 60,000 |
BONUS[].
X = 2
Y = 10
loop while Y > X
X = X + 2
output(Y - X)
Y = Y - 2
end loop
| City | Population | Total CO2 Emissions (metric kg) |
|---|---|---|
| New York | 8,336,817 | 60,000,000 |
| Los Angeles | 3,979,576 | 40,000,000 |
| Chicago | 2,693,976 | 30,000,000 |
func(X)
if X > 1
then
return func(X - 1) + func(X - 2)
else
return X
end if
end func
loop for Index from 0 to 7
input NUMBER
if NUMBER mod 2 = 0 then
if NUMBER mod 5 = 0 then
output NUMBER
end if
end if
end loop
A = 1
B = 10
C = 5
loop while B > C
A = A + 3
B = B - 2
if B mod 4 = 0 then
C = C + 1
end if
output(A + B - C)
end loop
NAMES[], POSITIONS[], and SCORES[]. Each player's score changes as the season progresses.SCORES[] array when a player scores some points in a new game. Assume you are given the player's name and the points they scored in a game. Add these points to their current season score.
fibonacci(n)
if n <= 1
then
return n
else
return fibonacci(n - 1) + fibonacci(n - 2)
end if
end fibonacci
| 2 | 3 | 7 | 0 | 1 | 9 |
| 1 | 5 | 9 | 4 | 3 | 2 |
| 9 | 0 | 4 | 6 | 2 | 4 |
| . | . | . | . | . | . |
| . | . | . | . | . | . |
| 7 | 7 | 6 | 9 | 0 | 3 |
| 230 | 310 | 17 | 315 | 29 | 111 | 821 | 109 |
NUM = 5
loop while NUM > 0
stack.push(NUM)
NUM = NUM - 1
end loop
SUM = 0
loop while not stack.isEmpty()
NUM = stack.pop()
SUM = SUM + NUM
end loop
output SUM
| NUM | SUM | Output |
|---|---|---|
| 5 |