Overview

This problem involves moving a robot on circularly arranged seats and summing up the hunger values of the visited seats after each movement.

Analysis

Essence of the Problem

This problem is a simple simulation problem that involves updating positions on a circular seat arrangement and aggregating the values of visited seats.

Key Insights

1. Handling Circular Arrays: Since the seats are arranged in a circle, moving R from seat \(N\) returns to seat \(1\), and moving L from seat \(1\) goes to seat \(N\). This can be concisely expressed using modular arithmetic (mod).

2. Index Conversion: In the problem statement, seat numbers range from \(1\) to \(N\), but in programming, array indices range from \(0\) to \(N-1\), so it’s convenient to manage using 0-indexed notation.

Understanding Through a Concrete Example

For example, when \(N = 4\), \(A = [10, 20, 30, 40]\), \(S = \) RRL:

Instruction	Position Before	Position After	Recorded Hunger
R	0 (seat 1)	1 (seat 2)	\(A[1] = 20\)
R	1 (seat 2)	2 (seat 3)	\(A[2] = 30\)
L	2 (seat 3)	1 (seat 2)	\(A[1] = 20\)

Total: \(20 + 30 + 20 = 70\)

A Naive Approach Is Sufficient

In this problem, \(Q \leq 10^5\), so processing each instruction one by one is fast enough at \(O(Q)\). No special optimization techniques are needed.

Algorithm

1. Initialization: Set the current position pos to \(0\) (corresponding to seat 1) and initialize the total total to \(0\)
2. Processing Each Instruction:
   • For R: Move clockwise with pos = (pos + 1) % N
   • For L: Move counterclockwise with pos = (pos - 1) % N
3. Recording Hunger: Add the hunger value A[pos] corresponding to the position pos after movement to total
4. Output: Output total after processing all instructions

Points About Modular Arithmetic

• (pos + 1) % N: When position is \(N-1\), it wraps back to \(0\) instead of becoming \(N\)
• (pos - 1) % N: In Python, modulo of negative numbers is handled correctly, so when position is \(0\), it becomes \(N-1\)

Complexity

• Time Complexity: \(O(N + Q)\)
  • \(O(N)\) for reading array \(A\)
  • \(O(Q)\) for processing \(Q\) instructions, each taking \(O(1)\)
• Space Complexity: \(O(N)\)
  • \(O(N)\) for storing array \(A\)

Implementation Notes

1. 0-indexed vs 1-indexed: The problem statement uses 1-indexed notation, but managing with 0-indexed in the program makes array access more natural.

2. Python’s Modulo Operation: In Python, the % operator always returns a non-negative result for negative numbers, so (pos - 1) % N works correctly. Caution is needed in languages like C++.

3. Reading Input: Since the string \(S\) may contain a trailing newline, we remove it using strip().

   Source Code

def main():
    N, Q = map(int, input().split())
    A = list(map(int, input().split()))
    S = input().strip()
    
    pos = 0  # 0-indexed, initially at seat 1 (index 0)
    total = 0
    
    for cmd in S:
        if cmd == 'R':
            pos = (pos + 1) % N
        else:  # cmd == 'L'
            pos = (pos - 1) % N
        total += A[pos]
    
    print(total)

if __name__ == "__main__":
    main()

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