Overview

This problem asks us to find the number of open lockers after performing divisor-related operations on \(N\) lockers. The conclusion is that the answer is the count of perfect squares up to \(N\) (i.e., \(\lfloor\sqrt{N}\rfloor\)).

Analysis

Key Insight: How Many Times Each Locker is Toggled

Locker \(k\) is toggled during operation \(i\) when \(k\) is a multiple of \(i\), which means when \(i\) is a divisor of \(k\).

Therefore, locker \(k\) is toggled exactly as many times as the number of divisors of \(k\).

Determining the Final State

• Initially, all lockers are closed
• If toggled an odd number of times → open
• If toggled an even number of times → closed

This means lockers with an odd number of divisors are open.

Condition for an Odd Number of Divisors

Normally, divisors come in pairs. For example, the divisors of \(12\) pair up as \((1, 12), (2, 6), (3, 4)\), so the number of divisors is even.

However, only for perfect squares, we have \(\sqrt{k} \times \sqrt{k} = k\), and \(\sqrt{k}\) exists alone without a pair.

Example: The divisors of \(9\) are \(1, 3, 9\), and since \(3 \times 3 = 9\), \(3\) stands alone. There are \(3\) divisors (odd).

Conclusion

The locker numbers that are open among those up to \(N\) are \(1, 4, 9, 16, 25, \ldots\) — the perfect squares.

The count of perfect squares up to \(N\) is \(\lfloor\sqrt{N}\rfloor\).

Why a Naive Approach Won’t Work

Since \(N \leq 10^{18}\) is extremely large, approaches that simulate each locker with \(O(N)\) or \(O(N \log N)\) complexity are far too slow. Using the mathematical insight above, we need to solve this in \(O(1)\).

Algorithm

1. Read input \(N\)
2. Calculate \(\lfloor\sqrt{N}\rfloor\) (in Python, use math.isqrt(N))
3. Output the result

Complexity

• Time complexity: \(O(1)\) (only computing the square root)
• Space complexity: \(O(1)\)

Implementation Notes

• Why use math.isqrt(): Since \(N\) can be as large as \(10^{18}\), using int(math.sqrt(N)) may cause errors due to floating-point precision issues. math.isqrt() computes the integer square root exactly using integer arithmetic, making it safe for large numbers.

• Verification with a concrete example: For \(N = 10\)

  • Locker \(1\): divisors \(\{1\}\) → \(1\) time → open
  • Locker \(2\): divisors \(\{1, 2\}\) → \(2\) times → closed
  • Locker \(3\): divisors \(\{1, 3\}\) → \(2\) times → closed
  • Locker \(4\): divisors \(\{1, 2, 4\}\) → \(3\) times → open
  • Locker \(9\): divisors \(\{1, 3, 9\}\) → \(3\) times → open

The open lockers are \(1, 4, 9\) — that’s \(3\) lockers = \(\lfloor\sqrt{10}\rfloor = 3\) ✓

Source Code

import math

N = int(input())
print(int(math.isqrt(N)))

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