Grid Path

Research checked: official Codeforces editorial, CF Step hints, and the supplied statement. The official tutorial text is basically blank, but the official solution confirms the $2m+1$ matrix compression. The proof below is from the statement, not copied.

Shape of a path

The chip can move left, right, and down, but never up. Therefore the visited rows are exactly $1,2,\dots,t$ for some $1\le t\le n$. In one fixed row, horizontal movement visits a contiguous block of columns, so that row contributes an interval $[l_i,r_i]$.

The first row interval must contain column $1$, hence $l_1=1$. For two consecutive visited rows, the chip moves down through some column, so the two intervals must intersect:

$[l_i,r_i]\cap[l_{i+1},r_{i+1}]\ne \emptyset.$

Conversely, any such interval sequence is realizable: in each row, enter through a column in the previous intersection, sweep enough left/right to visit the whole interval, then go down through a column in the next intersection. Re-visits do not matter because the path is a set. So we only need to count interval sequences of length at most $n$.

A naive interval DP has $O(m^2)$ states. That is too big for matrix exponentiation. Worse, if we transition by choosing a down-column, we count the same path set multiple times. That is the trap. Do not count witnesses; count sets.

The ticket compression

Use 0-indexed columns in the implementation. Define two kinds of ticket states for each column $x$:

• $S_{x,1}$: the next interval must contain $x$.
• $S_{x,0}$: the next interval must start at $x$.

There is also a stopped state $F$.

When we choose a current row interval $[l,r]$, we emit the following tickets for the next row:

• $S_{l,1}$,
• $S_{x,0}$ for every $l<x\le r$.

Why does this work? Take the previous interval $[l,r]$ and some candidate next interval $[L,R]$.

If $L\le l\le R$, then $[L,R]$ contains $l$, so it is accepted by $S_{l,1}$. No start-ticket $S_{x,0}$ with $x>l$ can also accept it, because its left endpoint is $L\le l$.

Otherwise, if the intervals still intersect, then $l<L\le r$. In that case the unique accepting ticket is $S_{L,0}$, because the next interval starts at $L$.

If neither case holds, the intervals do not intersect. So every intersecting next interval is accepted exactly once. That one-word exactly is doing all the heavy lifting here.

Building the transition matrix

Let $K=2m+1$. States $0\dots 2m-1$ are ticket states, with index $2x+f$ for $S_{x,f}$, and state $K-1$ is $F$.

From ticket $S_{i,f}$, enumerate every interval $[l,r]$ accepted by it:

• if $f=1$, then $l\le i\le r$;
• if $f=0$, then $l=i$ and $r\ge i$.

For every such interval $[l,r]$:

• add one transition to $F$, meaning the path stops after this row;
• for every $j\in[l,r]$, add one transition to $S_{j,[j=l]}$, meaning this row interval emits its tickets for the next row.

Also add a self-loop $F\to F$, so after the path stops, the remaining unused rows just pad the matrix walk.

The start state is $S_{0,1}$, because the first row interval must contain column $0$; since no column is left of $0$, that means it starts at $0$, exactly as required.

Each non-stopped matrix step chooses the interval for one row. If the path uses $t$ rows, the last chosen interval transitions to $F$, then $F$ loops for the remaining $n-t$ steps. Therefore the answer is

$(T^n)_{S_{0,1},F}.$

Equivalently, start with a row vector $v$ containing $1$ at $S_{0,1}$ and output $(vT^n)_F$.

Correctness proof

Lemma 1. Every valid chip path corresponds to a sequence of intervals $[l_1,r_1],\dots,[l_t,r_t]$ with $l_1=0$ and consecutive intervals intersecting.

Proof. Rows are visited in order and cannot be skipped before the last visited row, because the only vertical move is down by one row. In a row, moving left/right visits a contiguous interval. Moving from row $i$ to row $i+1$ happens in one column belonging to both row intervals, so they intersect. The first interval contains the starting column $0$, hence starts at $0$. $\square$

Lemma 2. Every such interval sequence is realizable by the chip.

Proof. For each adjacent pair, choose any column in their intersection as the down-column. In a row, starting from the entry column, the chip can walk left and right enough to visit the whole interval and finish at the chosen down-column. The last row can be swept similarly and then stop. Revisiting cells changes no set. $\square$

Lemma 3. For a fixed interval $[l,r]$, the emitted tickets accept exactly the intervals that intersect $[l,r]$, and each accepted interval is accepted once.

Proof. Let the next interval be $[L,R]$. If $L\le l\le R$, then $S_{l,1}$ accepts it, and no $S_{x,0}$ for $x>l$ can accept because $L\le l$. If not, intersection is possible exactly when $l<L\le r$, and then the unique accepting ticket is $S_{L,0}$. All other intervals are disjoint. $\square$

Lemma 4. Matrix walks from $S_{0,1}$ to $F$ in exactly $n$ steps are in bijection with valid path sets.

Proof. By construction, each non-stopped transition chooses exactly one row interval accepted by the previous ticket. By Lemma 3, this is exactly the same as choosing the next interval intersecting the previous interval, without duplicate down-column counts. A transition to $F$ marks the final visited row, and the self-loop on $F$ pads unused lower rows. Lemmas 1 and 2 connect these interval sequences exactly to chip path sets. $\square$

Therefore the algorithm outputs the number of possible paths modulo $mod$.

Edge cases

For $n=1$, only the first row exists, and the answer is $m$: intervals $[1,1],[1,2],\dots,[1,m]$. The matrix gives this in one step.

For $m=1$, the only choice is how many rows of the single column are visited, so the answer is $n\bmod mod$. The same transition matrix degenerates correctly.

The modulus is arbitrary, not necessarily prime. We only use addition and multiplication, so no modular inverse nonsense shows up.

Complexity

There are $K=2m+1\le 301$ states. Building the transition matrix is $O(m^4)$ loosely, tiny here. Matrix exponentiation costs

$O(K^3\log n)$

time and $O(K^2)$ memory. The implementation multiplies a row vector by powers of the matrix, saving the extra full result-matrix multiplications. Products fit in unsigned 64-bit if we reduce the accumulator every 16 terms, since $mod\le 10^9$.