Overview

Given the initial state of each light and the number of times each switch is pressed, determine the final ON/OFF state of each light.

Analysis

In this problem, each light can be processed independently, so we can consider them individually.

The key observation is that the state of a light is determined by the parity (even or odd) of the number of switch presses \(K_i\): - If the switch is pressed an even number of times, the light’s state does not change (it returns to its original state) - If the switch is pressed an odd number of times, the light’s state is toggled

For example, if the initial state is Yes (ON) and \(K_i = 3\) (odd), the final state will be No (OFF).

By utilizing this property, we don’t need to actually press the switch \(K_i\) times — we only need to check whether \(K_i\) is odd. This allows us to efficiently handle even very large values of \(K_i\) (up to \(10^9\)).

A naive approach (e.g., toggling the state \(K_i\) times in a loop) will not finish in time when \(K_i\) is large (TLE).

Algorithm

1. For each light, repeat the following:
   • If the initial state \(S_i\) is Yes, set it to ON; if No, set it to OFF
   • If the number of presses \(K_i\) is odd, toggle the current state
   • Record the final state
2. Output the final state of all lights

Complexity

• Time complexity: \(O(N)\)
• Space complexity: \(O(N)\)

Implementation Notes

• After reading the input as a string, convert it to an integer (K = int(K))

• Toggling the state can be easily done using the not operator

• Rather than outputting everything at once, it is more efficient to store each result in a list and join them with newlines at the end

  Source Code

N = int(input())
results = []

for _ in range(N):
    S, K = input().split()
    K = int(K)
    
    # 初期状態が "Yes" (ON) かどうか
    is_on = (S == "Yes")
    
    # K回押した後の状態: Kが偶数なら変化なし、奇数なら反転
    if K % 2 == 1:
        is_on = not is_on
    
    # 最終状態を結果に追加
    results.append("Yes" if is_on else "No")

print("\n".join(results))

This editorial was generated by qwen3-coder-480b.