def is_legitmate_code(code: str) -> bool: code = [str(i) for i in code if i != ' '] code_letters = ["A", "B", "Z", "T", "X"] if code[0] not in code_letters or len(code) <= 1: return False try: temp = code[1:] n = [int(i) for i in temp] except: return...
回答于 2021-09-23 22:23
转换成像素矩阵,在B矩阵中找与A矩阵相同的位置 "在B矩阵中找与A矩阵相同的位置" 这要用什么算法?
回答于 2021-02-28 12:06