Overview

This is a problem of finding the minimum number of folds needed to fold a rope of length \(L\) in half repeatedly until it fits within a storage case of length \(W\).

Analysis

The operation in this problem computes \(\lceil x/2 \rceil\) for the current length \(x\). That is, if \(x\) is odd, it becomes \((x+1)/2\), and if \(x\) is even, it becomes \(x/2\).

A straightforward approach would be to start from \(L\) and repeat the folding operation until the length becomes \(W\) or less. Although the maximum value of \(L\) is \(10^{18}\), which is very large, a simple loop still runs fast enough. This is because the value is roughly halved with each operation, so the number of required operations is at most about 60 (since \(2^{60} \approx 10^{18}\)).

Algorithm

1. First check whether \(L \leq W\); if so, return 0
2. If \(L > W\), initialize the current length to \(L\) and set a counter to 0
3. While the current length is greater than \(W\), repeat the following:
   • Update the current length to \(\lceil x/2 \rceil\) (computed as \((x + 1) // 2\))
   • Increment the counter by 1
4. Output the value of the counter

Complexity

• Time complexity: \(O(\log L)\)
  • Since the length is roughly halved with each operation, the number of iterations is approximately \(\log_2 L\)
• Space complexity: \(O(1)\)
  • Only a constant number of variables are used

Implementation Notes

• The entire computation can be done using integer arithmetic only (no floating-point numbers needed)

• The folding operation can be expressed exactly as \((x + 1) // 2\)

• Even for large numbers (\(10^{18}\)), the process is efficient since only about 60 operations are needed

• Be careful not to forget the initial condition check (\(L \leq W\))

  Source Code

def main():
    import sys
    data = sys.stdin.readline().split()
    L = int(data[0])
    W = int(data[1])
    
    if L <= W:
        print(0)
        return
        
    count = 0
    current = L
    while current > W:
        current = (current + 1) // 2
        count += 1
        
    print(count)

if __name__ == "__main__":
    main()

This editorial was generated by deepseekv3.