Overview

The positions of the stalls return to the exact same arrangement every time \(N\) time units pass.
Using this periodicity, we compress the state into “current time \(\bmod N\), Takahashi’s position, and the set of stalls already purchased from,” then find the shortest time using breadth-first search (BFS).

Analysis

Naively thinking, at each time step Takahashi can:

• Move left
• Stay in place
• Move right

for a maximum of 3 possible actions.
Exhaustively searching all possibilities up to time \(T\) would result in \(3^T\) cases, which is far too slow.

So, what should we use as our state?

1. Stall movement has period \(N\)

The position of the \(i\)-th stall at time \(t\) is

\(((S_i + tD_i - 1) \bmod N) + 1\)

If we replace time with \(t+N\):

\(((S_i + (t+N)D_i - 1) \bmod N) + 1\)

Since \(ND_i \equiv 0 \pmod N\), this is the same as the original position.
Therefore, each stall’s position repeats every \(N\) time units.
In other words, the entire arrangement of stalls can be determined by looking at only time \(\bmod N\).

2. Information needed to determine the future

To determine what actions are possible from a given point onward, only the following 3 pieces of information are needed:

• The remainder of the current time \(t \bmod N\)
• Which section Takahashi is currently in
• Which stalls have already been purchased from

Conversely, if these 3 are the same, the future is completely identical.
This is because the positions of stalls not yet purchased are determined by the time remainder, and stalls already purchased can be ignored.

3. The “purchased set” can be stored as a bitmask

Since \(M \leq 10\), the set of purchased stalls can be managed with a bitmask.

• bit \(i\) is 1: stall \(i\) has been purchased from
• bit \(i\) is 0: not yet purchased

The state where all purchases are complete is full = (1<<M)-1.

4. Buy from all stalls at the same location simultaneously

In this problem, if there are multiple unpurchased stalls in the same section, all of them are bought at the same time.
There is no option to buy from only some of them.

Therefore, for each

• time remainder \(r\)
• section \(p\)

if we precompute the bitmask representing “the set of stalls at that place at that time,” then the purchased set for the next state is simply

\(\text{new\_mask} = \text{mask} \;|\; \text{buy\_at}[r][p]\)

5. BFS gives the shortest time

Each transition corresponds to “advancing time by 1.”
This means we have a graph where all edge costs are 1, so using BFS, the first time we reach the goal is the minimum.

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Algorithm

Define the state as follows:

• \(r = t \bmod N\)
• \(p =\) Takahashi’s current position
• \(mask =\) set of purchased stalls

Here, the state represents “immediately after the purchase processing at time \(t\) is complete.”
The initial state is:

• Time \(0\)
• Position is section \(1\)
• Nothing has been purchased yet

Note that, as stated in the problem, no purchases occur at time 0, so the initial mask remains 0.

Precomputation 1: Stalls available at each time and section

Converting to 0-indexed, for each \(r=0,1,\dots,N-1\) and each position \(p=0,1,\dots,N-1\):

• When the time remainder is \(r\)
• And the section is \(p\)

we build the bitmask buy_at[r][p] of stalls present there.

Precomputation 2: Takahashi’s possible destinations

From position \(p\), there are at most 3 positions reachable at the next time step:

• Stay
• Left
• Right

At boundaries, we ensure we don’t go out of range.

BFS

We put the initial state into the queue and perform BFS.
From state \((r,p,mask)\), at the next time step:

1. Takahashi moves to \(q\)
2. The time remainder becomes \(nr=(r+1)\bmod N\)
3. Buy from all stalls at section \(q\)

So the transition destination is:

\((nr, q, mask \;|\; buy\_at[nr][q])\)

We explore this with BFS, and when the purchased set equals the full set of stalls, that time is the answer.
If we cannot reach that state, output -1.

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Complexity

• Time complexity: \(O(N^2 2^M)\)
• Space complexity: \(O(N^2 2^M)\)

Details

The number of states is

\(N \times N \times 2^M\)

Since there are at most 3 transitions from each state, the overall computation is sufficiently fast.
Under the given constraints, the maximum is approximately

\(100 \times 100 \times 2^{10} \approx 10^7\)

states.

Implementation Notes

• In implementation, using 0-indexed makes the stall position calculation simpler:
  \(p = (S_i + rD_i) \bmod N\)
• Be careful that no purchases occur at time 0. You must not apply buy_at[0][0] to the initial state.
• Storing buy_at[r][p] as a bitmask allows simultaneous purchases to be processed with a single OR operation.
• In the code:
  • (time remainder, position) are combined into base
  • state = (base << M) | mask

encoding everything into a single integer, and visited is managed with a bytearray for speed.

Source Code

import sys
from collections import deque

def main():
    input = sys.stdin.readline
    N, M = map(int, input().split())

    S = [0] * M
    D = [0] * M
    for i in range(M):
        s, d = map(int, input().split())
        S[i] = s - 1
        D[i] = d % N

    full = (1 << M) - 1
    NN = N * N

    # buy_at_base[res * N + pos] = bitmask of stalls at that position at that time residue
    buy_at_base = [0] * NN
    for r in range(N):
        offset = r * N
        for i in range(M):
            p = (S[i] + r * D[i]) % N
            buy_at_base[offset + p] |= 1 << i

    # transitions from base state (time residue, position) to next time residue positions
    trans = [None] * NN
    for r in range(N):
        nr_offset = ((r + 1) % N) * N
        row = r * N
        for p in range(N):
            lst = [nr_offset + p]
            if p > 0:
                lst.append(nr_offset + p - 1)
            if p + 1 < N:
                lst.append(nr_offset + p + 1)
            trans[row + p] = tuple(lst)

    total_states = NN * (1 << M)
    visited = bytearray(total_states)

    # state encoding: (base << M) | mask
    start = 0  # time 0, position 0, mask 0
    visited[start] = 1
    q = deque([start])

    time = 0
    shift = M
    buy = buy_at_base
    tr = trans
    vis = visited
    target = full

    while q:
        for _ in range(len(q)):
            state = q.popleft()
            mask = state & target
            base = state >> shift

            for nb in tr[base]:
                nm = mask | buy[nb]
                ns = (nb << shift) | nm
                if not vis[ns]:
                    if nm == target:
                        print(time + 1)
                        return
                    vis[ns] = 1
                    q.append(ns)
        time += 1

    print(-1)

if __name__ == "__main__":
    main()

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This editorial was generated by gpt-5.4-high.