luoganttcc

摘要： Python 实现冒泡 nums = [3,1,2]n=len(nums)for i in range(n-1): for j in range(n-i-1): if nums[j]>nums[j+1]: ... 阅读全文

posted @ 2022-08-19 22:57 luoganttcc 阅读(5) 评论(0) 推荐(0)

摘要： Python 实现冒泡 nums = [3,1,2]n=len(nums)for i in range(n-1): for j in range(n-i-1): if nums[j]>nums[j+1]: ... 阅读全文

posted @ 2022-08-19 22:57 luoganttcc 阅读(6) 评论(0) 推荐(0)

摘要： nums = [3,1,2]n=len(nums)for i in range(n-1): for j in range(n-... 阅读全文

posted @ 2022-08-19 22:57 luoganttcc 阅读(9) 评论(0) 推荐(0)

摘要： nums = [3,1,2]n=len(nums)for i in range(n-1): for j in range(n-... 阅读全文

posted @ 2022-08-19 22:57 luoganttcc 阅读(16) 评论(0) 推荐(0)

摘要： python 递归方式实现斐波那契数列 import time t1=time.time()def factorial(n): if n==1 or n==2: return 1 else: re... 阅读全文

posted @ 2022-08-19 22:57 luoganttcc 阅读(14) 评论(0) 推荐(0)

摘要： python 递归方式实现斐波那契数列 import time t1=time.time()def factorial(n): if n==1 or n==2: return 1 else: re... 阅读全文

posted @ 2022-08-19 22:57 luoganttcc 阅读(5) 评论(0) 推荐(0)

摘要： import time t1=time.time()def factorial(n): if n==1 or n==2: ... 阅读全文

posted @ 2022-08-19 22:57 luoganttcc 阅读(20) 评论(0) 推荐(0)

摘要： import time t1=time.time()def factorial(n): if n==1 or n==2: ... 阅读全文

posted @ 2022-08-19 22:57 luoganttcc 阅读(22) 评论(0) 推荐(0)

摘要：递归方法求n! def jiec(n): if n==0: return 1 else: return n*jiec(n-1)ss=jiec(10)print(ss) 3628800 ... 阅读全文

posted @ 2022-08-19 22:57 luoganttcc 阅读(6) 评论(0) 推荐(0)

摘要：递归方法求n! def jiec(n): if n==0: return 1 else: return n*jiec(n-1)ss=jiec(10)print(ss) 3628800 ... 阅读全文

posted @ 2022-08-19 22:57 luoganttcc 阅读(3) 评论(0) 推荐(0)