Overview

This problem asks for the minimum number of operations needed to reduce a rope of length \(L\) centimeters to \(W\) centimeters or less, by repeatedly performing the operation “halve the length (rounding up fractions).”

Analysis

The key point of this problem is that “the rope’s length decreases exponentially with each operation.”

1. Simulating the Operation

If the current rope length is \(x\), the length after one fold is \(\lceil x / 2 \rceil\). For example, consider the case \(L=10, W=2\): - Step 0: 10cm - Step 1: \(\lceil 10 / 2 \rceil = 5\)cm (still greater than \(W\)) - Step 2: \(\lceil 5 / 2 \rceil = 3\)cm (still greater than \(W\)) - Step 3: \(\lceil 3 / 2 \rceil = 2\)cm (\(W\) or less achieved!) Therefore, the answer is 3 operations.

2. Constraints and Computational Complexity

The rope length \(L\) can be as large as \(10^{18}\), which is a very large value, but each operation approximately halves the length. Since \(2^{60}\) exceeds \(10^{18}\), for any input, \(L\) will reach 1 in at most about 60 operations. Therefore, naively simulating the operations one at a time is more than fast enough.

3. How to Compute the Ceiling Division

Integer division in programming (/ or // in most languages) generally performs floor division (rounding down). To compute the ceiling \(\lceil x / 2 \rceil\) of a positive integer \(x\) divided by 2 using only integer arithmetic, the following formula is convenient: - (x + 1) // 2

Algorithm

The solution is obtained by the following steps:

1. Read the inputs \(L, W\).
2. Initialize a variable count to 0 to record the number of folds.
3. While \(L > W\), repeat the following (using a while loop):
   • Update \(L\) to \((L + 1) // 2\).
   • Increment count by 1.
4. Output the final value of count.

If \(L \leq W\) from the beginning, the while loop condition is not satisfied, so count remains 0, which correctly handles this case as specified in the problem statement.

Computational Complexity

• Time Complexity: \(O(\log L)\) Since \(L\) is halved with each operation, the number of loop iterations is proportional to the logarithm of \(L\). When \(L=10^{18}\), the maximum number of iterations is about 60.
• Space Complexity: \(O(1)\) Only the input values and a counting variable are stored, so the algorithm runs with a constant amount of memory.

Implementation Notes

• Handling Large Numbers: Python natively supports large integers (arbitrary-precision integers), so values as large as \(10^{18}\) can be computed directly.

• Loop Condition: Since we repeat “until the length becomes \(W\) or less,” the continuation condition is L > W.

  Source Code

import sys

def solve():
    # 入力を読み込む
    line = sys.stdin.readline()
    if not line:
        return
    L, W = map(int, line.split())

    count = 0
    # ロープの長さ L が W を超えている間、折りたたみを繰り返す
    while L > W:
        # L / 2 の切り上げを計算する
        L = (L + 1) // 2
        count += 1
    
    # 結果を出力
    print(count)

if __name__ == "__main__":
    solve()

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This editorial was generated by gemini-3-flash-thinking.