Diary of dishes | Blue Bridge Cup - hexadecimal to octal (hand torn version) with hexadecimal conversion notes

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Never thought of it , A few days after returning home for the Chinese New Year break , I will do some questions on New Year's Eve because of boredom , Then write an explanation .

List of articles

• • 【 subject 】
  • 【 My code 】
  • 【 Binary conversion notes 】
  • 【 reflection 】

【 subject 】

Topic link ：http://lx.lanqiao.cn/problem.page?gpid=T51

【 My code 】

My idea is ： First Hexadecimal turn Binary system , Then turn octal , Here, two dictionaries are directly created to store data . The code needs to pay attention to that the binary matches upside down , And when there are less than three numbers, you need to make up 0, Everyone else should be able to see .

book1 = {
    
    '0': '0000',
    '1': '0001',
    '2': '0010',
    '3': '0011',
    '4': '0100',
    '5': '0101',
    '6': '0110',
    '7': '0111',
    '8': '1000',
    '9': '1001',
    'A': '1010',
    'B': '1011',
    'C': '1100',
    'D': '1101',
    'E': '1110',
    'F': '1111'
}
book2 = {
    
    '000': '0',
    '001': '1',
    '010': '2',
    '011': '3',
    '100': '4',
    '101': '5',
    '110': '6',
    '111': '7'
}

n = int(input())
sn = ['' for _ in range(n)]
for i in range(n):
    sn[i] = input()
    
for s in sn: # Get binary form number 
    num = ''
    for j in s:
        num += book1[j]
        
    ss = ''
    t = ''
    count = 0
    for k in num[::-1]:
        t = k+t
        count += 1
        if count == 3: # Count in groups of three , When three are converted later 
            count = 0
            ss = book2[t] + ss
            t = ''
            
    if count == 1:# There are two situations when there are less than three , Fill in the front respectively 0, Be careful not to write +=
        t = '00'+t
        ss = book2[t] + ss
    if count == 2:
        t = '0' + t
        ss = book2[t]+ss
    print(int(ss))

【 Binary conversion notes 】

Use Python Built in function for binary conversion , It's not hard to find that they are all First convert to decimal Then convert it to the corresponding hexadecimal , Let's give a simple example , Others will not be repeated ：

# 8 Turn into the system 16 Base number ：8->10->16
n=input()
print(hex(int(n,8)))

 Input ：1010
 Output ：0x208

If you don't want to take the first two marks , The following methods can be used ：

• Method 1 ： section , Take the result from the third character

# 10 Base to zero 2 Base number 
n=int(input())
print(bin(n)[2:]) #  Slicing operation 

 Input ：10
 Output ：12

• Method 2 ： Use format Function to format ,format Function has its own conversion function ,yyds

n=input()
print("{:b}".format(int(n,8)))
#  First the 8 The base number is converted to 10 Base number ,
#  And then in format Add a... To the slot of b, Equivalent to realizing bin Function functions 
#  But the result is without 0b Prefixed 

 Input ：1010
 Output ：1000001000

 Other bases ：（n On the right is the original base ）
print("{:o}".format(int(n,16))) #  Convert other types to 8 Base number 
print("{:x}".format(int(n,8))) #  Convert other types to 16 Base number 

【 reflection 】

Happy New Year! ！

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