Overview

This is a problem where you count the number of cars, out of \(N\) used cars, that satisfy both a given price range (at least \(L\) and at most \(R\)) and a mileage upper limit (at most \(M\)).

Analysis

The key to this problem is accurately organizing the conditions for determining whether each car is a “candidate car.”

1. Price condition: \(L \leq P_i \leq R\)
2. Mileage condition: \(K_i \leq M\)

We count the cars that satisfy both conditions simultaneously.

Looking at the constraints, the number of cars \(N\) is at most \(10^5\). If we take a straightforward approach (linear search) of checking each car one by one to see if it satisfies the conditions, the number of computations is around \(10^5\). In typical competitive programming time limits (1–2 seconds), approximately \(10^7 \sim 10^8\) computations are possible, so this approach is well within the time limit.

There is no need to introduce any special algorithms or data structures — this can be solved with a simple implementation using loops (for statements) and conditional branching (if statements).

Algorithm

We proceed with the following steps:

1. Read \(N, L, R, M\) from input.
2. Initialize a variable candidate_count to \(0\) to track the number of “candidate cars.”
3. For each of the \(N\) cars, repeat the following:
   • Read the price \(P_i\) and mileage \(K_i\) of the \(i\)-th car.
   • If \(L \leq P_i\) and \(P_i \leq R\), and \(K_i \leq M\), increment candidate_count by \(1\).
4. Output the final value of candidate_count.

Complexity

• Time complexity: \(O(N)\)
  • Since we perform one check for each of the \(N\) cars, the time is proportional to the number of cars.
• Space complexity: \(O(N)\)
  • In the provided code, sys.stdin.read().split() is used to read all input at once and store it as a list, so memory consumption is proportional to the input size.
  • (If you read and process one line at a time, this can be reduced to \(O(1)\).)

Implementation Notes

• Efficient input: Since \(N\) can be as large as \(10^5\), repeatedly calling input() in Python may result in long execution times. By using sys.stdin.read().split() to read everything at once, input can be processed efficiently.

• Chained comparison operators: In Python, you can chain comparison operators like L <= P <= R. This allows you to write conditions concisely.

  Source Code

import sys

def solve():
    # 全ての入力を一度に読み込み、空白で分割します
    input_data = sys.stdin.read().split()
    if not input_data:
        return
    
    # 最初の4つの値（N, L, R, M）を取得します
    N = int(input_data[0])
    L = int(input_data[1])
    R = int(input_data[2])
    M = int(input_data[3])
    
    candidate_count = 0
    
    # 各車の情報をループで確認します
    # 車の情報はインデックス 4 以降に 2 つずつ格納されています
    for i in range(N):
        P = int(input_data[4 + 2 * i])
        K = int(input_data[5 + 2 * i])
        
        # 条件：価格が L 以上 R 以下、かつ走行距離が M 以下
        if L <= P <= R and K <= M:
            candidate_count += 1
            
    # 結果を出力します
    print(candidate_count)

if __name__ == '__main__':
    solve()

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