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This article is part of the Scaler Topics Fortnightly Contest - 27

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Solution Approach

First, it’s important to observe that in order to maintain a certain number of options indefinitely, this number must be at least 2 and divide A. Let’s identify the smallest such number, denoted as d. If d is less than or equal to B, we can consistently vote for the first d options evenly. However, if d is greater than B, each round would inevitably reduce the number of remaining options until only one is left. Therefore, the answer is 1 if and only if d is greater than B.

To efficiently find the value of d, we recognize that d is the smallest divisor of A, excluding 1. We can determine d through various methods, one of which involves checking all numbers from 2 up to the square root of A. If no divisors are found, then A is a prime number, and d is equal to A. This approach results in a solution with a time complexity of O(sqrt(A)).

While the previous solution is effective, it may not be fast enough in certain languages like Python. To optimize it, we can employ the sieve of Eratosthenes to find the smallest divisor, leading to a pre-computation time complexity of O(A log A) or even faster, with O(1) time complexity to answer each test case

Time complexity(A) Space complexity: O(A)