Overview

Given the movements of Takahashi and Aoki on a number line over \(T\) minutes, the problem asks us to count the number of integer time points \(i\) (\(1 \leq i \leq T\)) at which their coordinates coincide.

Analysis

The key insight in this problem is to focus not on the absolute coordinates of the two people, but on the “difference between their coordinates”.

1. Tracking the Coordinate Difference

Let \(x_i\) be Takahashi’s coordinate at time \(i\) and \(y_i\) be Aoki’s coordinate. The condition for them to meet is \(x_i = y_i\), i.e., \(x_i - y_i = 0\). By setting the initial difference (at time 0) as \(diff = X - Y\) and updating this \(diff\) according to each minute’s movement, we can determine whether the two are at the same location.

2. How the Difference Changes with Movement

Consider how \(diff\) changes over a one-minute interval. Let \(\Delta a\) be Takahashi’s movement and \(\Delta b\) be Aoki’s movement (L corresponds to \(-1\), R to \(+1\), S to \(0\)). The new difference can be computed as: $\(diff_{new} = diff_{old} + (\Delta a - \Delta b)\)$

For example: - If Takahashi moves R (+1) and Aoki moves L (-1): the difference changes by \(+1 - (-1) = +2\). - If both move in the same direction (e.g., both L): the change in difference is \(-1 - (-1) = 0\), so the relative distance does not change.

3. Important Notes

• Time 0 is not counted: As stated in the problem, “time 0 is not included as a target,” so even if \(X = Y\) initially, it should not be included in the count.
• Intermediate positions are ignored: Even if they pass each other during a one-minute movement, it is not considered a “meeting.” We only check coordinates at integer time points \(i\).

Algorithm

1. Compute the initial coordinate difference diff = X - Y and prepare a counter meet_count = 0.
2. For each time step \(i = 1\) to \(T\), repeat the following:
   • Check the \(i\)-th character of strings \(A\) and \(B\).
   • Add to or subtract from diff according to Takahashi’s movement.
   • Add to or subtract from diff according to Aoki’s movement (since Aoki moving in the positive direction reduces the difference, it becomes a subtraction).
   • If diff == 0, increment meet_count by 1.
3. Output the final meet_count.

Complexity

• Time Complexity: \(O(T)\) The processing completes in a single loop over the string of length \(T\), making it very fast. Although \(T \leq 10^6\), a simple loop in Python is sufficient to run within the time limit.
• Space Complexity: \(O(T)\) \(O(T)\) memory is used to store the input movement path strings.

Implementation Tips

• Fast I/O: Since \(T\) can be large, in Python it is efficient to read all input at once using sys.stdin.read().split().

• Using the zip function: When iterating over the two strings \(A\) and \(B\) simultaneously, using zip(a_moves, b_moves) allows for concise code.

• Optimizing conditional branches: When both make the same movement (ma == mb), diff does not change, so skipping that processing can provide a slight speedup.

  Source Code

import sys

def solve():
    # 入力を一括で読み込み、空白で分割して取得します。
    # T: 時間, X: 高橋君の初期座標, Y: 青木君の初期座標, A: 高橋君の移動, B: 青木君の移動
    input_data = sys.stdin.read().split()
    if len(input_data) < 5:
        return
    
    t = int(input_data[0])
    x = int(input_data[1])
    y = int(input_data[2])
    a_moves = input_data[3]
    b_moves = input_data[4]
    
    # 高橋君と青木君の現在の座標の差 (高橋 - 青木) を管理します。
    # 初期時刻 (時刻 0) の差を計算します。
    diff = x - y
    meet_count = 0
    
    # 各時刻 i (1 から T) における移動を順番に処理します。
    # zip を使用して文字列 A と B の各文字を同時に取り出します。
    for ma, mb in zip(a_moves, b_moves):
        # 高橋君と青木君の移動が異なる場合のみ、座標の差を更新します。
        # 同じ移動（例：共に 'L'）をした場合、相対的な差は変わりません。
        if ma != mb:
            # 高橋君の移動による差の変化
            if ma == 'L':
                diff -= 1
            elif ma == 'R':
                diff += 1
            
            # 青木君の移動による差の変化
            # (高橋 - 青木) なので、青木君が正の方向に動くと差は減少し、負の方向に動くと増加します。
            if mb == 'L':
                diff += 1
            elif mb == 'R':
                diff -= 1
        
        # 時刻 i における座標が等しい（差が 0）かどうかを判定します。
        if diff == 0:
            meet_count += 1
            
    # 出会った回数を出力します。
    sys.stdout.write(str(meet_count) + '\n')

if __name__ == '__main__':
    solve()

This editorial was generated by gemini-3-flash-thinking.