Hints : A Simple GCD Problem (Hard Version)

As in C1, preserving all subarray GCDs reduces to preserving each consecutive-pair GCD $g_i = \gcd(a_i, a_{i+1})$, and $a’i$ must be a multiple of $L_i = \operatorname{lcm}(g{i-1}, g_i)$. Moreover, setting every $a’_i = L_i$ (multiplier 1) always preserves GCDs. In C2, $b_i$ is independent of $a_i$ — so even positions where $a_i = L_i$ might be changeable. What new possibilities does this open?

Answer to Hint 1: Two cases arise. If $a_i \neq L_i$ (meaning $a_i / L_i \geq 2$): as long as $L_i \leq b_i$, we can set $a’_i = L_i$ — a “free” safe operation, exactly like C1. If $a_i = L_i$: we need a different multiple of $L_i$, namely $L_i \cdot p$ for some $p \geq 2$, with $L_i \cdot p \leq b_i$. This is possible when $b_i \geq 2L_i$. But using a non-trivial multiplier ($p > 1$) creates a new difficulty. What could go wrong?

Answer to Hint 2: We still need $\gcd(a’i, a’{i+1}) = g_i$ exactly. Writing $a’i = L_i \cdot k_i$, the “excess factor” visible to the pair $(i, i+1)$ from position $i$ is $k_i \cdot (L_i / g_i)$, and from position $i+1$ is $k{i+1} \cdot (L_{i+1} / g_i)$. For the GCD to stay exactly $g_i$, these two excess factors must be coprime. When $k_i = 1$, this is automatically satisfied. But when $k_i > 1$, its prime factors could clash with the neighbor’s fixed or chosen factors. What type of $k_i$ is most “efficient” to consider?

Answer to Hint 3: It suffices to try prime multipliers. A composite multiplier $k$ introduces all of $k$’s prime factors into the excess, gaining nothing over just using one prime. A single prime $p$ is the cleanest choice: it introduces exactly one new factor to check for coprimality. Define $u_i = L_i / g_i$ (the fixed excess toward the right) and $v_i = L_i / g_{i-1}$ (toward the left). When choosing prime $p$ as the multiplier at position $i$, which primes are “forbidden”?

Answer to Hint 4: A prime $p$ at position $i$ adds $p$ to the excess factor on both sides. For the left pair $(i-1, i)$: the left neighbor’s fixed excess includes $u_{i-1}$. If $p \mid u_{i-1}$, then even with $k_{i-1} = 1$, the excess factors share prime $p$, violating coprimality — so $p$ is forbidden. Symmetrically, for the right pair: $p$ is forbidden if $p \mid v_{i+1}$. How many forbidden primes can there be, and is this manageable?

Answer to Hint 5: $u_{i-1}$ and $v_{i+1}$ each divide values bounded by $\sim 10^9$, so each has at most about 9 distinct prime factors. Together, at most $\sim 18$ primes are forbidden. Among the first few dozen primes ($2, 3, 5, 7, \ldots$), there will always be plenty of valid choices. Collecting up to $\sim 25$ non-forbidden primes $p$ with $L_i \cdot p \leq b_i$ gives a small set of candidate multipliers per position. Now: with multiple positions each having a small candidate set, how do you pick compatible choices to maximize total operations?

Answer to Hint 6: Model it as a DP over positions. For each position $i$, enumerate a small set of states $(k, c)$: the multiplier $k$ and whether an operation was performed ($c \in {0,1}$). The transition from position $i-1$ to $i$ checks compatibility: multipliers $k_{i-1}$ and $k_i$ are compatible if $k_{i-1} = 1$, $k_i = 1$, or $\gcd(k_{i-1}, k_i) = 1$. (The forbidden-prime filtering already handles the fixed excess parts $u$ and $v$.) What is the complexity?

Answer to Hint 7: Each position has at most $\sim 27$ states (keep original, change to $L_i$, and up to $\sim 25$ prime multipliers). The DP transition checks all pairs of consecutive states: $O(27^2)$ per position. Total: $O(n \cdot 27^2 + \text{sieve})$, where the sieve precomputes small primes up to $\sim \lfloor \max(b_i / L_i) \rfloor$. The answer is $\max_s \mathrm{dp}[n][s]$.