python-递归，二分查找

# print(list("胡辣汤"))

# lst = ["河南话", "四川话", "东北", "山东", "上海"]
#
# r = reversed(lst)
# print(list(r))

# huiwen = "不是上海自来水来自海上"

# s = huiwen[::-1]
# it = reversed(huiwen) # 返回的是迭代器
# s = ""
# for el in it:
#     s += el
# print(s)


# lst = ["河南话", "四川话", "东北", "山东", "上海"]
# s = slice(3,5) # 切片. 麻烦
# print(lst[s])

# s = "我叫{name}, 我来自{home}, 我喜欢干{hobby}".format(name="周杰伦", home="台湾", hobby="唱歌")
# print(s)
#
# name = "周杰伦"
# home="台湾"
# hobby="唱歌"
#
# print(f"我叫{name}, 我来自{home}, 我喜欢干{hobby}")


# s = "门神"
# s1 = s.center(20)
# print(s1)
#
# print(format("门神", "^20"))
# print(format("门神", "<20"))
# print(format("门神", ">20"))

# print(format(3, 'b'))  # ⼆二进制  binary  0b11
# print(format(65, 'c'))  # 转换成unicode字符  ascii
# print(format(11, 'd'))  # ⼗十进制
# print(format(11, 'o'))  # ⼋八进制  0o    13
# print(format(11, 'x'))  # ⼗六进制(⼩写字⺟)
# print(format(11, 'X'))  # ⼗六进制(⼤写字⺟)
# print(format(11, 'n'))  # 和d⼀样
# print(format(11))  # 和d⼀样



# print(format(123456789, 'e')) # 科学计数法. 默认保留6位小数  表示大的数据
# print(format(123456789, '.2e')) # 科学计数法. 保留2位小数(⼩写)
# print(format(123456789, '0.2E')) # 科学计数法. 保留2位⼩数(⼤写)

# 浮点计数法
# print(format(1.23456789, 'f')) # 浮点数计数法. 保留6位⼩数
# print(format(1.23456789, '0.2f')) # ⼩浮点数计数法. 保留2位⼩数
# print(format(1, '0.10f')) # ⼩浮点数计数法. 保留10位⼩数
# print(format(1.23456789e+10000, 'F')) # ⼩浮点数计数法.

# a = 0b11 # 数字3
# print(type(a))
# print(a)

# print(ord("中"))  # 20013
# print(chr(20014))

# for i in range(65536):
#     print(chr(i), end= " ")


# print(ascii("a")) # 判断出你给的信息是否是ascii


# 转义字符\n 换行\t tab 制表符\r 回车\"   双引号\'  单引号\\    \
print('你好, 我叫周杰伦.我来自\\n台湾. 我老婆叫昆凌')

repr() # 还原字符串最官方的效果str:python的字符串repr: 所有的字符串print(repr("你好. \\n我叫\周杰伦"))  # python -> repr() -> cpython c里面. 报错

# print(r"\n\t范德萨发生\r\b\ffdsaf")  # 原封不动的显示字符串
# 函数自己调用自己

# def func():
#     print("我是递归")
#     func()
#
# func()   # 官方最大1000,你永远跑不到1000, 我实测998

# while 1:
#     print("我不是递归")


# 树形结构的遍历
# import os
#
# def func(lujing, n): # "d:/a/"
#     lst = os.listdir(lujing) # 打开文件夹. 列出该文件夹内的所有文件名
#     for el in lst: # el是文件的名字.  b, c
#         # 还原文件路径
#         path = os.path.join(lujing, el) # "d:/a/b"
#         if os.path.isdir(path): # 判断路径是否是文件夹
#             print("..." * n,el) # 显示文件夹的名字
#             func(path, n + 1)  # 在来一次  ################
#         else:
#             print("\t" * n,el) # 显示文件
#
# func("d:/a", 0)

# def func(lujing, n): # d:/a/b
#     lst = os.listdir(lujing)
#     for el in lst: # b, c
#         # 路径
#         path = os.path.join(lujing, el) # 拼接路径 d:/a/b
#         # 判断是文件还是文件夹
#         if os.path.isdir(path):
#             print("\t" * n, el)
#             func(path, n+1)
#         else:
#             f = open(path, mode="wb")
#             f.write(b'1')
#             print("\t" * n, el)
#
#
# func("d:/a/", 0)

# 二分法查找主要的作用就是查找元素
# lst = [1,3,5,7,12,36,68,79] # 数据集 百万级数据
# num = int(input("请输入你要查找的元素信息:"))
#
# for el in lst:
#     if num == el:
#         print("存在")
#         break
# else:
#     print("不存在")
# len(lst)
#  0 1 2 3 4  5  6  7
# [1,3,5,7,12,36,68,79]
# n = 28 # 3
# 规则. 掐头结尾取中间, 必须是有序序列,

# 100000000


# 二分法查找 (需要你明白和掌握)
# lst = [1,3,5,7,12,36,68,79]
# n = int(input("请输入一个数"))
#
# left = 0
# right = len(lst) - 1
#
# while left <= right:
#     mid = (left + right)//2
#     if n > lst[mid]:
#         left = mid + 1
#     elif n < lst[mid]:
#         right = mid - 1
#     else:
#         print("存在")
#         break
#
# else:
#     print("不存在")

 

# 递归
# def func(n, lst):
#
#     left = 0
#     right = len(lst) - 1
#     if lst != []:
#         mid = (left + right)//2
#         if n > lst[mid]:
#             func(n, lst[mid+1:]) # 改变列表
#         elif n < lst[mid]:
#             func(n, lst[:mid])
#         else:
#             print("找到了")
#             return
#     else:
#         print("没找到")
#         return
#
# n = int(input("请输入你要查找的数:"))
# func(n, [1,3,5,7,12,36,68,79]) # 78

# 递归二 (需要你明白和掌握)
# def func(n, lst, left, right): # 递归找到什么是可以变的. 什么是不可以变的
#     if left <= right:
#         mid = (left + right) // 2
#         if n > lst[mid]:
#             left = mid + 1
#             return func(n, lst, left, right)
#         elif n < lst[mid]:
#             right = mid - 1
#             return func(n, lst, left, right) # 递归如果有返回值. 所有调用递归的地方必须写return
#         else:
#             print("找到了")
#             return mid  # 难点
#     else:
#         print("找不到")
#         return -1
#
# n = int(input("请输入你要查找的数:"))
# lst = [1,3,55,98,37,41,2,5,1,4]
# ret = func(n, lst, 0, len(lst)-1) # 78
# print(ret)

 

# 最快的查询
 1 lst = [1,3,55,98,37,41,2,5,1,4]

new_lst = []
for i in range(99):
    new_lst.append(0)


for i in lst:    # 1,3,55,98
    new_lst[i] = 1

print(new_lst)


i = int(input('请输入你要找的数据'))
if new_lst[i] == 0: # 1次
    print("不存在")
else:
    print("存在")