Overview

This problem asks us to find the coordinates of robots at time \(T\) as they travel back and forth along a corridor, reflecting off walls. By mathematically handling the reflection behavior using a “folding” technique, we can compute the answer for each robot in \(O(1)\).

Analysis

Consider the case without walls

If there were no walls, the coordinate of robot \(i\) at time \(T\) would simply be \(X_i + V_i \times T\). However, in reality, the robot reflects at both ends of the interval \([0, L]\), so this value may go out of range.

Simulating reflections is inefficient

Since \(T\) can be up to \(10^9\) and \(|V_i|\) can also be up to \(10^9\), the number of reflections can be enormous. Simulating reflections one by one is far too slow.

The folding technique

The key observation here is that reflective motion within the interval \([0, L]\) can be understood as a folding pattern with period \(2L\).

Consider the coordinate \(p = X_i + V_i \times T\) assuming no walls. The robot travels back and forth as \(0 \to L \to 0 \to L \to \cdots\), which corresponds to “zigzag folding” of coordinates on the number line with period \(2L\).

Specifically:

1. Compute the remainder \(r\) of \(p\) divided by \(2L\) (\(0 \leq r < 2L\))
2. If \(r \leq L\), the answer is \(r\)
3. If \(r > L\), the answer is \(2L - r\) (folding back)

Concrete example (when \(L = 5\)): - \(p = 7\) → \(r = 7 \bmod 10 = 7\) → Since \(7 > 5\), the answer is \(10 - 7 = 3\) - This corresponds to starting from coordinate \(0\), going to \(5\) (+5), then reflecting and going back \(2\) (-2), ending at coordinate \(3\).

Handling reflection at walls at time \(0\)

As stated in the problem, if a robot starts on a wall and has velocity directed toward that wall, it immediately reverses at time \(0\). This is handled in the code by negating \(V\).

Algorithm

For each robot \(i\), do the following:

1. Handle initial reversal: If \(X_i = 0\) and \(V_i < 0\), set \(V_i \leftarrow -V_i\). If \(X_i = L\) and \(V_i > 0\), set \(V_i \leftarrow -V_i\).
2. Compute the wall-free coordinate: \(p = X_i + V_i \times T\)
3. Determine the coordinate by folding:
   • \(r = p \bmod 2L\) (Python’s % operator returns a non-negative remainder even for negative numbers)
   • If \(r \leq L\), the answer is \(r\); otherwise, the answer is \(2L - r\)

Complexity

• Time complexity: \(O(N)\) (\(O(1)\) computation per robot)
• Space complexity: \(O(N)\) (for storing the output)

Implementation Notes

• Behavior of Python’s % operator: In Python, a % m (with \(m > 0\)) always returns a value in the range \([0, m)\), even when \(a\) is negative. This allows us to correctly handle cases where \(p\) is negative without any additional case distinctions.

• Special case when \(L = 0\): When \(L = 0\), all robots can only be at coordinate \(0\), so we handle this separately to avoid division by \(2L\).

• Integer arithmetic only: As guaranteed by the problem, the answer is always an integer, so we can compute everything exactly using only integer arithmetic without floating-point operations.

  Source Code

import sys

def solve():
    input_data = sys.stdin.buffer.read().split()
    idx = 0
    N = int(input_data[idx]); idx += 1
    L = int(input_data[idx]); idx += 1
    T = int(input_data[idx]); idx += 1
    
    out = []
    for i in range(N):
        X = int(input_data[idx]); idx += 1
        V = int(input_data[idx]); idx += 1
        
        # Handle initial reflection at walls at time 0
        if X == 0 and V < 0:
            V = -V
        elif X == L and V > 0:
            V = -V
        
        # Position without walls
        pos = X + V * T
        
        # Reflect into [0, L] using the folding method
        # The robot bounces in a "zigzag" pattern with period 2L
        # Map pos into [0, 2L) cycle, then fold
        if L == 0:
            out.append(0)
            continue
        
        # We need to fold pos into [0, L]
        # First, normalize pos modulo 2L into [0, 2L)
        cycle = 2 * L
        pos_mod = pos % cycle  # Python's modulo ensures non-negative result for positive cycle
        
        if pos_mod <= L:
            result = pos_mod
        else:
            result = cycle - pos_mod
        
        out.append(result)
    
    sys.stdout.write('\n'.join(map(str, out)) + '\n')

solve()

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