Overview

This is a problem of determining whether the total time required for \(N\) preparatory tasks fits within the time (in minutes) from the preparation start time \(S\) to the departure deadline time \(T\).

Analysis

The key to solving this problem is comparing the “available time” and the “required time” in the same unit (minutes).

1. Calculating the available time: Takahashi can prepare from hour \(S\) to hour \(T\). The duration between these times is \((T - S)\) hours. Since \(1\) hour is \(60\) minutes, converting to minutes gives \((T - S) \times 60\) minutes. Since the constraints guarantee \(S \leq T\), this value will never be negative.

2. Calculating the total required time: There are \(N\) tasks, and each task \(i\) takes \(A_i\) minutes. Since the tasks are performed consecutively, the total required time is simply the sum of all \(A_i\). As stated in the problem, “the completion time does not change depending on the order,” so there is no need to consider the order of tasks.

3. Judgment condition: If “total required time \(\leq\) available time,” then the preparation can be finished by the deadline of \(T\) hours \(0\) minutes, so the answer is Yes; otherwise, it is No.

Algorithm

1. Read \(N, S, T\) and each task duration \(A_1, A_2, \dots, A_N\) from the input.
2. Calculate the available time: available_minutes = (T - S) * 60
3. Calculate the total task time: total_required_minutes = sum(A)
4. Determine whether total_required_minutes <= available_minutes and output the result.

Complexity

• Time complexity: \(O(N)\)
  • This is because we need to traverse the list once to compute the sum of \(N\) task durations.
• Space complexity: \(O(N)\)
  • \(O(N)\) memory is used to store the \(N\) input task durations in a list.

Implementation Notes

• Unit consistency: When converting time (hours) to minutes, make sure not to forget to multiply by \(60\).

• Computing the sum: In Python, you can concisely calculate the sum of a list using the sum() function.

  Source Code

import sys

def solve():
    # 入力を取得
    input_data = sys.stdin.read().split()
    if not input_data:
        return
    
    # N, S, T を取得
    n = int(input_data[0])
    s = int(input_data[1])
    t = int(input_data[2])
    
    # 各作業にかかる時間 A_i を取得
    a = list(map(int, input_data[3:]))
    
    # 利用可能な時間を分単位で計算
    available_minutes = (t - s) * 60
    
    # 必要な合計時間を計算
    total_required_minutes = sum(a)
    
    # 判定と出力
    if total_required_minutes <= available_minutes:
        print("Yes")
    else:
        print("No")

if __name__ == "__main__":
    solve()

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