>>>bin(10)

'0b1010'

>>> bin(20)

'0b10100'

# -*- coding: UTF-8 -*-

 

# 獲取用戶(hù)輸入十進(jìn)制數(shù)

dec = int(input("輸入數(shù)字："))

 

print("轉(zhuǎn)換為八進(jìn)制為：", oct(dec))

print("轉(zhuǎn)換為十六進(jìn)制為：", hex(dec))

#-*- coding: utf-8
from decimal import Decimal
def bTod(n, pre=4):
  '''
  把一個(gè)帶小數(shù)的二進(jìn)制數(shù)n轉(zhuǎn)換成十進(jìn)制
  小數(shù)點(diǎn)后面保留pre位小數(shù)
  '''
  string_number1 = str(n) #number1 表示二進(jìn)制數(shù)，number2表示十進(jìn)制數(shù)
  decimal = 0 #小數(shù)部分化成二進(jìn)制后的值
  flag = False 
  for i in string_number1: #判斷是否含小數(shù)部分
    if i == '.':
      flag = True
      break
  if flag: #若二進(jìn)制數(shù)含有小數(shù)部分
    string_integer, string_decimal = string_number1.split('.') #分離整數(shù)部分和小數(shù)部分
    for i in range(len(string_decimal)):
      decimal += 2**(-i-1)*int(string_decimal[i]) #小數(shù)部分化成二進(jìn)制
    number2 = int(str(int(string_integer, 2))) + decimal
    return round(number2, pre)
  else: #若二進(jìn)制數(shù)只有整數(shù)部分
    return int(string_number1, 2)#若只有整數(shù)部分 直接一行代碼二進(jìn)制轉(zhuǎn)十進(jìn)制 python還是騷 
 
def dTob(n, pre=4):
  '''
  把一個(gè)帶小數(shù)的十進(jìn)制數(shù)n轉(zhuǎn)換成二進(jìn)制
  小數(shù)點(diǎn)后面保留pre位小數(shù)
  '''
  string_number1 = str(n) #number1 表示十進(jìn)制數(shù)，number2表示二進(jìn)制數(shù)
  flag = False 
  for i in string_number1: #判斷是否含小數(shù)部分
    if i == '.':
      flag = True
      break
  if flag:
    string_integer, string_decimal = string_number1.split('.') #分離整數(shù)部分和小數(shù)部分
    integer = int(string_integer)
    decimal = Decimal(str(n)) - integer
    l1 = [0,1]
    l2 = []
    decimal_convert = ""
    while True: 
      if integer == 0: break
      x,y = divmod(integer, 2) #x為商，y為余數(shù) 
      l2.append(y)
      integer = x
    string_integer = ''.join([str(j) for j in l2[::-1]]) #整數(shù)部分轉(zhuǎn)換成二進(jìn)制 
    i = 0
    while decimal != 0 and i < pre: 
      result = int(decimal * 2) 
      decimal = decimal * 2 - result 
      decimal_convert = decimal_convert + str(result) 
      i = i + 1
    string_number2 = string_integer + '.' + decimal_convert
    return float(string_number2)
 
   
  else: #若十進(jìn)制只有整數(shù)部分
    l1 = [0,1]
    l2 = []
    while True: 
      if n == 0: break
      x,y = divmod(n, 2) #x為商，y為余數(shù) 
      l2.append(y)
      n = x
    string_number = ''.join([str(j) for j in l2[::-1]]) 
    return int(string_number)