結果

問題 No.683 Two Operations No.3
ユーザー qwewe
提出日時 2025-05-14 13:05:18
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 45 ms / 2,000 ms
コード長 3,886 bytes
コンパイル時間 173 ms
コンパイル使用メモリ 82,100 KB
実行使用メモリ 52,224 KB
最終ジャッジ日時 2025-05-14 13:07:08
合計ジャッジ時間 1,501 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge1
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ファイルパターン 結果
other AC * 16
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ソースコード

diff # 

import sys

# Set a higher recursion depth limit for potentially deep recursive calls.
# The problem constraints (A, B up to 10^18) might lead to deep recursion paths
# if one coordinate decreases slowly (by 1).
# The optimization for states on axes (X=0 or Y=0) should mitigate the worst cases,
# but a higher limit is safer. If this causes issues on the judge (e.g., memory),
# an iterative approach or further mathematical insight might be needed.
try:
   # Set a reasonably large limit, e.g., 100,000.
   # Default is often 1000, which might be too low.
   # If the actual path length is bounded polylogarithmically, even ~2500 could be enough.
   # Let's use 100,000 as a safer value.
   sys.setrecursionlimit(100000)
except Exception:
   # If setting recursion limit fails (e.g., due to OS limits or security restrictions),
   # we just proceed with the default limit. This might fail for some test cases.
   pass 

# Use a dictionary for memoization to store results of computed states (X, Y).
# This avoids recomputing results for the same state, turning exponential complexity
# into pseudo-polynomial (related to the number of reachable states).
memo = {}

def can_reach_zero(X, Y):
    """
    Recursive function with memoization to check if state (X, Y) can reach (0, 0)
    using reverse operations.
    """
    # Base case: If either coordinate is 0, we can reach (0, 0).
    # This is because states (N, 0) for N > 0 can only transition to (N-1, 0) via RevOp2, eventually reaching (0, 0).
    # States (0, M) for M > 0 can only transition to (0, M-1) via RevOp1, eventually reaching (0, 0).
    # This also correctly handles the target state (0, 0) itself.
    if X == 0 or Y == 0: 
        return True 
    
    # Base case: If coordinates become negative (should not happen from non-negative A, B with reverse ops), it's an invalid path.
    if X < 0 or Y < 0:
        return False
        
    # Check memoization table. If state result is already computed, return it.
    state = (X, Y)
    if state in memo:
        return memo[state]

    # Optimization: If both X and Y are odd positive integers, we are stuck.
    # Neither reverse operation can be applied, because RevOp1 requires X even,
    # and RevOp2 requires Y even. Since X > 0 and Y > 0, this state cannot reach (0, 0).
    if X % 2 != 0 and Y % 2 != 0:
         memo[state] = False
         return False

    # Variable to track if we found any path to (0, 0) from the current state.
    found_path = False

    # Try applying Reverse Operation 1: (X, Y) -> (X/2, Y-1)
    # This is possible if X is even. Since we are past the X=0 check, X > 0.
    # Also Y >= 1 is required, which is true since we are past the Y=0 check.
    if X % 2 == 0:
        # Recursively call for the resulting state. If it can reach (0, 0), set found_path to True.
        if can_reach_zero(X // 2, Y - 1):
            found_path = True
    
    # Try applying Reverse Operation 2: (X, Y) -> (X-1, Y/2)
    # This is possible if Y is even. Since we are past the Y=0 check, Y > 0.
    # Also X >= 1 is required, which is true since we are past the X=0 check.
    # We only explore this path if the RevOp1 path didn't already find a solution.
    # This is a short-circuiting optimization.
    if not found_path and Y % 2 == 0:
         # Recursively call for the resulting state. If it can reach (0, 0), set found_path to True.
         if can_reach_zero(X - 1, Y // 2):
             found_path = True
    
    # Store the result for the current state in the memoization table and return it.
    memo[state] = found_path
    return found_path

# Read non-negative integers A and B from input.
A, B = map(int, sys.stdin.readline().split())

# Call the function starting from the target state (A, B).
# Print "Yes" if (A, B) can reach (0, 0), otherwise print "No".
if can_reach_zero(A, B):
    print("Yes")
else:
    print("No")
0