SnailN2

摘要： break_out = False while 1: s = [] for i in range(3): x = int(input('请输入一个数:\n')) if x == -1: # 设计一个退出循环的条件 break_out = True break s.append(x) if break 阅读全文

摘要： def SlowSnail(n): while n != 1: # 循环保证递归 for index in range(2, n + 1): if n % index == 0: n //= index # n 等于 n//index if n == 1: print(index) else: pr 阅读全文