- Questions for the qualification round
1. Tourist
Overall, Alex will want to first see attractions 1 to N in order, then see them all a second time in the same order, then all a third time in the same order, and so on. If we write out an infinite sequence of attractions 1, 2, ..., N-1, N, 1, 2, ..., N-1, N, 1, ..., then Alex will see the first K attractions in this list on his first visit, the next K on his second visit, and so on. This means that, on his Vth visit, Alex will see the (K*(V-1)+i)th attraction in the list for i=1..K. We can convert each of these values into its corresponding attraction index between 1 and N using modulus, sort these indices in increasing order, and then output their corresponding attraction names in that order.
2. Interception
Writing out how the polynomial will be evaluated once the order of operations is reversed, we end up with a series of N+2 terms all exponentiated together (right-associatively): (P_N * x) ^ ((N * P_{N-1}) * x) ^ ... ^ ((2 + P_1) * x) ^ ((1 + P_0) * x) ^ (0). The only way for such an expression to potentially evaluate to 0 is if the first term is equal to 0, as a^b ≠ 0 when a is a non-zero real number (unless b = -infinity, which isn't possible here). Since P_N is guaranteed to be non-zero, this means that the polynomial can only possibly evaluate to 0 when x = 0. Now, when x = 0, it’s clear that each of the N+2 terms above is also equal to 0. So, we’re interested in evaluating the expression 0^0^...^0^0, with N+2 0’s. We can observe an alternating pattern based on the number of 0’s: 0^0 = 1, 0^0^0 = 0^1 = 0, 0^0^0^0 = 1, 0^0^0^0^0 = 0, and so on. So, this expression evaluates to 0 when the number of 0’s is odd, and to 1 when the number of 0’s is even. It follows that the polynomial has a single x-intercept at x = 0 when N is odd, and no x-intercepts when N is even, independent of its coefficients.