What Does This Program Do? - Looping

Contest 2 adds loops to the WDTPD problems. You’ll see FOR loops and WHILE loops in pseudocode. The skill is tracking variables across iterations without losing count.

ACSL loop syntax

FOR loop

FOR i = 1 TO 5 STEP 2
    (statements)
NEXT

Rules:

STEP is optional. If omitted, the step is $+ 1$
The end value is inclusive - the loop runs when i equals the end value
STEP can be negative (counting down)

Example: FOR i = 1 TO 5 STEP 2 runs with $i = 1, 3, 5$ (three iterations).

Example: FOR i = 10 TO 1 STEP -3 runs with $i = 10, 7, 4, 1$ (four iterations).

WHILE loop

WHILE condition
    (statements)
END WHILE

The condition is checked before each iteration. If false initially, the body never executes.

Try it: trace a FOR loop

Watch how sum accumulates across each iteration.

Code

1sum = 02FOR i = 1 TO 4 STEP 13    sum = sum + i4NEXT5OUTPUT: sum

Variables

Name	Value
sum	0

Step 1 of 10

The iteration table

For loop problems, make a table with one row per iteration. This is non-negotiable for accuracy.

Example:

INPUT: n = 5
s = 0
FOR i = 1 TO n
    s = s + i * i
NEXT
OUTPUT: s

Iteration	i	i*i	s
-	-	-	$0$
$1$	$1$	$1$	$1$
$2$	$2$	$4$	$5$
$3$	$3$	$9$	$14$
$4$	$4$	$16$	$30$
$5$	$5$	$25$	$55$

Answer: 55 (sum of squares from $1$ to $5$ )

Nested loops

When loops are inside other loops, the inner loop runs fully for each iteration of the outer loop.

Example:

count = 0
FOR i = 1 TO 3
    FOR j = 1 TO i
        count = count + 1
    NEXT
NEXT
OUTPUT: count

Outer i	Inner j values	count after inner loop
$1$	$1$	$1$
$2$	$1, 2$	$3$
$3$	$1, 2, 3$	$6$

Answer: 6

Tip: For nested loops, track the outer loop iteration and summarize the inner loop’s effect. Don’t try to track every single inner iteration unless necessary.

WHILE loops with changing conditions

WHILE loops are trickier because the number of iterations isn’t obvious upfront.

Example:

INPUT: x = 100
count = 0
WHILE x > 1
    x = INT(x / 2)
    count = count + 1
END WHILE
OUTPUT: count

Iteration	x (before)	INT(x/2)	x (after)	count
1	100	50	50	1
2	50	25	25	2
3	25	12	12	3
4	12	6	6	4
5	6	3	3	5
6	3	1	1	6

Now x = 1, so x > 1 is false. Loop ends.

Answer: 6

Recognizing the pattern: This counts how many times you can halve a number before reaching 1 - it’s essentially $⌊ lo g_{2} (100)⌋ = 6$ .

Try it: trace a WHILE loop

This program extracts and sums the digits of a number.

Code

1n = 1232sum = 03WHILE n > 04    digit = n MOD 105    sum = sum + digit6    n = INT(n / 10)7END WHILE8OUTPUT: sum

Variables

Name	Value
n	123
sum	0

Step 1 of 15

Accumulator patterns

Most loop problems follow one of these patterns:

Sum accumulator:

s = 0
FOR i = 1 TO n
    s = s + (something involving i)
NEXT

Product accumulator:

p = 1
FOR i = 1 TO n
    p = p * (something involving i)
NEXT

(This is factorial when the “something” is just i.)

Counter:

count = 0
FOR i = 1 TO n
    IF (condition involving i) THEN
        count = count + 1
    END IF
NEXT

Max/min finder:

max = A[0]
FOR i = 1 TO n-1
    IF A[i] > max THEN
        max = A[i]
    END IF
NEXT

Recognizing these patterns helps you predict the answer and verify your trace.

The MOD trick in loops

ACSL loves using MOD inside loops. Common patterns:

Digit extraction (right to left):

n = 1234
WHILE n > 0
    digit = n MOD 10
    n = INT(n / 10)
END WHILE

This processes digits $4, 3, 2, 1$ in that order.

Checking divisibility:

count = 0
FOR i = 1 TO 100
    IF i MOD 3 == 0 AND i MOD 5 == 0 THEN
        count = count + 1
    END IF
NEXT

This counts numbers from $1$ to $100$ divisible by both $3$ and $5$ .

Common mistakes

Off-by-one errors. ACSL FOR loops are inclusive on both ends. FOR i = 1 TO 5 runs $5$ times, not $4$ .
Forgetting the STEP. FOR i = 10 TO 1 with no STEP has step $+ 1$ , so it never executes ( $10$ is already past $1$ going up).
WHILE condition checked before each iteration. If the condition is false at the start, the loop body never runs.
Variable modified inside the loop affecting the loop condition. In a WHILE loop, if the loop body changes the variable in the condition, trace carefully.
INT() is the floor function. $INT (x)$ returns the greatest integer $\leq x$ . So $INT (7/2) = 3$ , but $INT (- 7/2) = - 4$ (not $- 3$ ).

Contest strategy

Build an iteration table every time. No exceptions.
For FOR loops, count the iterations first $(end - start) / step + 1$
For WHILE loops, trace until the condition fails
If a loop has many iterations, look for a pattern after 3-4 rows
Typical time: 90-120 seconds per problem