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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2021/imperative-programming-resit/1/c/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Imperative programming (resit) / 1 / c / Solution by

Codex  0 2026-10-25

By the time the candidate

n

is emitted, every positive integer

m \leq n

has been examined. The straightforward divisor routine makes exactly

m

remainder tests on

m

. Consequently the total number is

\sum_{m = 1}^{n} m = \frac{n ( n + 1 )}{2},

(1)

and the time is $Θ (n^{2})$ when remainder tests and the other fixed-width operations have constant cost and printing is cheap. Emitting only record candidates does not avoid computing the divisor count of the intervening candidates.

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