Editorial

Let $S=\sum_{i=1}^{N}A_i$ . One strength- $K$ particle adds $K^2$ in total, so only strengths satisfying $K^2\mid S$ are candidates.

Let $X_i$ be the particle count at orbital $i$ , with indices cycling through $1,\ldots,N$ . Compute $C_{i} = A_{i - 1} - 2 A_{i} +_{}$ . Then . Split the indices into cycles of length . Integrate once, determine the initial slope from the cyclic condition, and integrate again. Every cycle permits adding an arbitrary integer constant to its base solution.

Let $B$ be the sum of the base solutions and let $L$ be the sum of their negated minima. The required particle count is $P=S/K^2$ . The candidate is valid exactly when $m\mid(P-B)$ and .

With $D$ candidate strengths, the time complexity is $O(ND+N)$ and the space complexity is $O(N)$ .

Proof

The second difference of the triangular kernel leaves $-2$ at its center and $1$ at both positions $K$ steps away. On a cycle, writing $p_t=y_t-y_{t-1}$ gives . The slopes close exactly when , and uniquely fixes the initial slope. A nonintegral initial slope is impossible.

In subtask 1, a nonzero particle has peak $K\leq\max A_i\leq3$ , leaving only three strengths.

For subtask 2, let $d$ be the common difference. If $d=0$ , check . If , we must have . Let , . The base-solution sum is and the minimum nonnegative total is . Thus check and , taking per candidate.

OrBiTal (hard)

Editorial