Long Journey

The obvious simulation is dead on arrival: $m$ can be $10^{12}$. The actual state space is tiny, because traps only care about position modulo small numbers and time modulo $n$.

Rephrase The Rule

Use phases $0,1,\dots,n-1$. Phase $p$ means the current turn activates traps in cells $x$ with

$x\bmod a_p=b_p.$

On a turn, you first choose stay or move right by one. Then phase $p$ activates. So the cell you land on must be safe for phase $p$:

$x\bmod a_p\ne b_p.$

This includes the final move into cell $m$. That special bold sentence in the statement is just saying: no, you cannot reach $m$ while simultaneously getting blasted by its trap. Nice try.

Periodicity

Let

$L=\operatorname{lcm}(a_1,a_2,\dots,a_n).$

Because every $a_i\le 10$, $L\le \operatorname{lcm}(2,3,4,5,6,7,8,9,10)=2520$.

For every phase, whether a cell is safe depends only on its residue modulo $L$. So the trip from cell $kL$ to cell $(k+1)L$ has exactly the same trap layout for every $k$. The only missing information is the current time phase.

That suggests compressing one whole block of length $L$ into a matrix.

Segment Transfer

For any $D\le L$, define

$go_D[s][e]$

as the minimum number of turns needed to advance exactly $D$ cells, starting at residue $0$ with next phase $s$, and ending with next phase $e$.

To compute this, run BFS on states $(j,p)$:

• $j$ is how many cells have been advanced inside this segment, with $0\le j\le D$.
• $p$ is the next phase.

From $(j,p)$, one turn later the phase becomes $(p+1)\bmod n$.

You have two possible actions:

• stay at $j$, allowed only if residue $j\bmod L$ is safe for phase $p$;
• move to $j+1$, allowed only if $j<D$ and residue $(j+1)\bmod L$ is safe for phase $p$.

Every edge costs one turn, so BFS gives shortest times.

Why finite BFS is enough: if you ever reach the same pair $(j,p)$ again later, the future is identical but you are strictly slower. Keeping the later copy would be pure cope.

Block Exponentiation

Let

$m=qL+r.$

Compute:

• $B=go_L$, the transfer for one full block;
• $T=go_r$, the transfer for the final leftover cells.

Matrix multiplication is min-plus multiplication:

$(A\otimes B)[i][j]=\min_k(A[i][k]+B[k][j]).$

Then $B^q$ tells us the minimum turns needed to cross $q$ complete blocks for every possible start/end phase.

We start before turn $1$, so the starting phase is $0$. The answer is

$\min_{p,e} (B^q[0][p]+T[p][e]).$

If all such values are infinite, the journey is impossible.

Correctness Sketch

First, the BFS transition exactly matches one legal turn: choose stay/move, then check whether the landed cell is safe for the active phase. Therefore $go_D$ contains exactly the shortest valid ways to advance $D$ cells.

Second, every valid path from $0$ to $m$ crosses the cells $L,2L,3L,\dots,qL$ in order. Between two consecutive multiples of $L$, the residue pattern is identical to every other full block, and the only carried state is the next phase. So the journey decomposes into $q$ independent full-block transfers plus one leftover transfer.

Third, min-plus multiplication is exactly chaining transfers: pick the intermediate phase that minimizes total time. Binary exponentiation just chains the same full-block transfer $q$ times quickly.

So the algorithm finds the true minimum number of turns, or correctly reports impossibility.

Complexity

For one transfer distance $D$, BFS uses $O(nD)$ states for each starting phase, so $O(n^2D)$ time. We compute it for $D=L$ and $D=r$.

Matrix exponentiation costs $O(n^3\log q)$.

With $n\le 10$ and $L\le 2520$, this is very comfortably within limits.