Overview

This is a problem to determine whether the total time required for preparation work fits within the available time before departure.

Analysis

The key insights for this problem are as follows:

1. The order of tasks does not affect the result: The problem statement clearly states “the completion time does not change depending on the order.” In other words, regardless of the order in which you perform the tasks, the total time to complete all tasks is simply “the sum of each task’s duration.”

2. Align the units of time: The departure deadline \(T\) and start time \(S\) are given in “hours,” but each task duration \(A_i\) is given in “minutes.” To make a comparison, we need to align the units.

3. Reduces to a simple comparison problem: Ultimately, this problem reduces to the following comparison:

   • If total preparation time ≤ available time, output Yes
   • Otherwise, output No

Concrete Example

For example, if \(S = 9\) (start at 9 o’clock), \(T = 12\) (depart by 12 o’clock), and task durations are \([30, 60, 45]\) minutes: - Available time: \((12 - 9) \times 60 = 180\) minutes - Required time: \(30 + 60 + 45 = 135\) minutes - Since \(135 \leq 180\), we can make it in time → Yes

Algorithm

1. Calculate the total time for all tasks: \(\text{total\_time} = \sum_{i=1}^{N} A_i\)
2. Calculate the available time before departure in minutes: \(\text{available\_time} = (T - S) \times 60\)
3. If \(\text{total\_time} \leq \text{available\_time}\), output Yes; otherwise, output No

total_time = A[1] + A[2] + ... + A[N]
available_time = (T - S) × 60

if total_time ≤ available_time:
    output "Yes"
else:
    output "No"

Complexity

• Time complexity: \(O(N)\)
  • \(O(N)\) to calculate the sum of all elements in array \(A\)
• Space complexity: \(O(N)\)
  • Size of the array to store the input

Implementation Notes

• Don’t forget unit conversion: Remember to multiply \((T - S)\) by \(60\) to convert hours to minutes.
• Python’s sum() function: The sum of all elements in a list can be concisely calculated with sum(A).
• Boundary condition: Since the problem states that arriving at “exactly \(T\) hours \(0\) minutes” is acceptable, use <= (including equality) for the inequality.

Source Code

N, S, T = map(int, input().split())
A = list(map(int, input().split()))

total_time = sum(A)
available_time = (T - S) * 60

if total_time <= available_time:
    print("Yes")
else:
    print("No")

This editorial was generated by claude4.5opus.