Overview

This is a problem where you need to examine the scores of all students in all classes and count how many students scored at or above the passing threshold of \(K\) points.

Analysis

This problem is a simple aggregation problem where you check all given scores and count those that satisfy the condition.

Key Observations

1. Understanding the input format: Each line starts with the number of students \(A_i\), followed by \(A_i\) scores
2. Check all students without omission: Process all classes in order and check the scores of all students in each class
3. Judgment condition: Since the score must be “\(K\) points or higher”, we check whether \(B_{i,j} \geq K\) is satisfied

Is a Naive Approach Sufficient?

• The total number of students is at most \(10^6\)
• Checking whether each student scored \(K\) or higher can be done in \(O(1)\)
• Therefore, simply checking all students is efficient enough

However, care must be taken with the input reading method. In Python, calling input() line by line can be slow for large amounts of input.

Algorithm

1. First, read \(N\) (number of classes) and \(K\) (passing threshold)
2. Initialize a variable count to 0 to count the number of passing students
3. For each class \(i\) (\(i = 1, 2, \ldots, N\)):
   • Read the number of students \(A_i\) in that class
   • Read the scores for \(A_i\) students in sequence
   • For each score \(B_{i,j}\), if \(B_{i,j} \geq K\), increment count by 1
4. Output the final count

Concrete Example

Sample input:

2 60
3 55 70 80
2 59 60

• Class 1: 3 students with scores 55, 70, 80 → 2 students scored 60 or higher: 70, 80
• Class 2: 2 students with scores 59, 60 → 1 student scored 60 or higher: 60
• Total: \(2 + 1 = 3\) students

Complexity

• Time complexity: \(O\left(\sum_{i=1}^{N} A_i\right)\)
  • Because we check each student’s score once
  • At most \(10^6\) operations
• Space complexity: \(O\left(\sum_{i=1}^{N} A_i\right)\)
  • Because we read the input data all at once
  • Can be reduced to \(O(\max A_i)\) if processed line by line

Implementation Notes

1. Fast input processing: By reading all input at once with sys.stdin.read() and splitting with split(), you can process large amounts of input efficiently

2. Index management: Keep the input data as an array and manage the current reading position with an idx variable. This allows flexible processing of each line’s structure (number of students first, then scores)

3. Counting method: By immediately judging and counting each score as it’s read, there’s no need to store all scores (memory efficient)

   Source Code

import sys

def main():
    input_data = sys.stdin.read().split()
    idx = 0
    N = int(input_data[idx])
    idx += 1
    K = int(input_data[idx])
    idx += 1
    
    count = 0
    for _ in range(N):
        A = int(input_data[idx])
        idx += 1
        for _ in range(A):
            B = int(input_data[idx])
            idx += 1
            if B >= K:
                count += 1
    
    print(count)

if __name__ == "__main__":
    main()

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