Overview

Given the scores \(A_i\) of \(N\) students, the problem asks us to count how many students have an absolute difference from the reference score \(S\) that is at most \(T\) (i.e., students satisfying \(|A_i - S| \leq T\)).

Analysis

What this problem requires is determining whether each student’s score \(A_i\) satisfies the following condition: $\(|A_i - S| \leq T\)$

Removing the absolute value sign, this is equivalent to the score falling within the following range: $\(S - T \leq A_i \leq S + T\)$

Checking the constraints, the number of students \(N\) is at most \(2 \times 10^5\). Since the check for each student can be done with a simple comparison, a straightforward approach of checking all students one by one (linear scan) is more than sufficient to run within the time limit.

Algorithm

1. Read \(N, S, T\) and the list of scores \(A\) from the input.
2. Initialize a variable count to \(0\) to keep track of the number of passing students.
3. Loop through the scores of all \(N\) students one by one.
   • For each score \(A_i\), compute \(|A_i - S|\) using the absolute value function (abs() in Python).
   • If this value is at most \(T\), increment count by \(1\).
4. Output the final value of count.

Complexity

• Time Complexity: \(O(N)\)
  • For \(N\) students, we perform one check per student, so the processing time is proportional to \(N\). Even when \(N = 2 \times 10^5\), modern computers can handle this in a few milliseconds to a few tens of milliseconds.
• Space Complexity: \(O(N)\)
  • If all input scores are stored in a list, the memory usage is proportional to the number of scores.

Implementation Notes

• Fast Input: Since \(N\) can be as large as \(2 \times 10^5\), calling input() repeatedly in Python may lead to long execution times. By using sys.stdin.read().split() to read all input at once, processing can be done more efficiently.

• Absolute Value Calculation: While you could use conditional branching (if statements) to check whether \(A_i - S\) is positive or negative, as per the mathematical definition, using the built-in abs() function provided by the programming language allows for more concise code.

  Source Code

import sys

def solve():
    # 入力を一括で読み込み、スペース区切りで分割する
    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])
    
    # 生徒の点数リストを取得
    a_list = input_data[3:]
    
    count = 0
    # 各生徒の点数について判定を行う
    for i in range(n):
        a_i = int(a_list[i])
        # |A_i - S| <= T かどうかをチェック
        if abs(a_i - s) <= t:
            count += 1
            
    # 結果を出力
    print(count)

if __name__ == "__main__":
    solve()

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