Hints: Grid Path

The path is the set of cells visited, not the sequence. So two sequences that visit the same set of cells count as one path. What structure does this set have? Can the chip ever “jump” to a cell that is not adjacent to the current set?

Answer to Hint 1: The chip only moves left, right, or down. So once we are in a row, we can only extend within that row (left/right) and then go down. In each row, the visited cells form a contiguous segment (a set of consecutive columns). So the path is uniquely described by: for each row $i$ we visit, an interval $[l_i, r_i]$ of columns.

Answer to Hint 2: So you can represent the “state” after processing some rows by the current row and the column interval $[l, r]$ that has been visited in that row. From state $(row, l, r)$, you can go down (same interval), or extend the interval one step left ($l \to l-1$) or one step right ($r \to r+1$) and then possibly go down. So the number of paths is the number of ways to evolve the interval as we go from row 1 to row $n$.

Answer to Hint 3: Because we only care about the set of cells and the order of moves doesn’t matter for the set, we can think: in each row we have an interval $[l, r]$. From $(l, r)$ we can (1) move down only → next row same interval; (2) extend left → $(l-1, r)$ then eventually down; (3) extend right → $(l, r+1)$ then eventually down. So the transitions from state $(l, r)$ to the “next row” state can be encoded: we get to $(l’, r’)$ if we can reach a state in the next row that has interval $[l’, r’]$ (which could be $[l,r]$, $[l-1,r]$, $[l,r+1]$, etc. depending on moves). So we have a state space of column intervals $[l, r]$ with $1 \le l \le r \le m$.

Answer to Hint 4: The number of possible intervals is $O(m^2)$. So we can define a state vector of size $O(m^2)$ (or a clever encoding of size $O(m)$). One row transition corresponds to a linear transformation on this state vector: from each $(l, r)$ we can transition to a few $(l’, r’)$ in the “next row.” So the number of paths over $n$ rows is the sum of entries of $T^n \cdot v_0$, where $T$ is the transition matrix and $v_0$ is the initial state (e.g. only $(1,1)$ at the start).

Answer to Hint 5: $n$ can be up to $10^8$, so we cannot iterate row by row. We need matrix exponentiation: compute $T^n$ in $O(\mathrm{states}^3 \log n)$ time. With $O(m^2)$ states this would be too slow. So we need a smaller state space.

Answer to Hint 6: Observe that from the starting cell $(1,1)$, the interval in row 1 is always $[1,1]$. When we move, we only extend to the left (column 0 not allowed, so we only extend right in column index) or to the right. So we can encode the state by: “left extent” and “right extent” relative to start, or by the pair $(l, r)$ with a single index: e.g. map $(l, r)$ to one dimension. The model solution uses a state space of size $w = 2m - 1$ or similar (about $2m$ states), so the matrix is $w \times w$ and exponentiation is $O(w^3 \log n)$, which is feasible for $m \le 150$.

Answer to Hint 7: The encoding can be: one “type” of state for “current interval is $[1, r]$” (prefix from column 1) and another for “current interval is $[l, m]$” (suffix to column $m$), or a combined indexing so that each reachable interval $[l, r]$ corresponds to one state index. Build the transition matrix $T$: for each state $i$, which states $j$ can we reach in one row step? Then $dp^{(n)} = T^n \cdot dp^{(0)}$, and the answer is the entry corresponding to the “full” or accepting state (e.g. any state that can represent having visited some set; often the state that has the same interval we end with after all moves).

Answer to Hint 8: Implementation: (1) Encode states (e.g. $0$ = interval $[1,1]$; states for $[1, r]$ for $r \ge 2$; states for $[l, m]$ for $l < m$; possibly one “full row” state). (2) For each possible interval $[l, r]$ in a row, determine which intervals we can have in the next row (same, extend left, extend right) and fill $T$. (3) Use binary exponentiation: while (n) { if (n & 1) dp = T * dp; T = T * T; n /= 2; }. (4) Output the entry of $dp$ for the “final” state, modulo $mod$.

Answer to Hint 9: Use unsigned 32-bit (or 64-bit) for the matrix entries and do modular arithmetic with $mod$ (which can be up to $10^9$). When multiplying two matrices, use 128-bit or careful casting to avoid overflow. The final answer is the number of distinct sets of cells reachable from $(1,1)$, which equals the sum of the number of paths that end in each possible state (or the single state that corresponds to “any path” depending on how you define the accepting state).