Blue Bridge Cup real question: one question with clear code, master three codes

Distance Blue Bridge Cup 56 God  

The purpose of learning algorithm is to improve yourself  

Thank a station for explaining the three codes Portal Original code Inverse code Complement code _ Bili, Bili _bilibili

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  Problem analysis ： Many friends must be like Xiao Zheng Yes Explanation of the title stem ： give Bytes as the value of a signed integer Don't understand at the beginning Why there are negative numbers ？ For Xiaobai like Xiao Zheng , One byte, eight bits , The range of data represented [00000000,11111111] namely [0,255] Where did the negative number come from ? 

  Actually The above understanding is based on Unsigned data , actually , There is another kind of numerical data called Signed data .

There are three representations of signed data ： Original code , Inverse code , Complement code

As long as you know the original code   The latter two codes can be easily solved

8 Bit source code highest 0 A positive sign ,1 A minus sign The range of the last seven digits is [0,127]

therefore 8 Range of bit source code [-127,127] So there are two steps to determine the original code of a number ：1： The sign determines the highest position 2： The absolute value of the number determines the last seven digits

Understand the original code The inverse code is easy to understand ： The inverse code is based on the original code , Except for the highest position （ Sign bit ）, The rest is reversed

Understand the inverse code Complement is easy to understand ： Complement is based on inverse On the basis of irony +1

And these three codes , Complement is the most important Because in the computer system , All values are represented and stored by complements .

So the following focuses on the method of calculating complement （ This question is also based on complement ）

First study negative numbers ：

for instance ： seek -1 Complement So we know -1 The original code of is 10000001

that -1 The inverse of is 11111110 that -1 The complement of is 11111111

It's that simple . Then for negative numbers [-127,-1] Can follow the above similar requirements -1 The method of calculation

And then to -128, We Regulations Its complement is 10000000

Then study positive numbers ：

A complement to a positive number = The inverse of a positive number = The original code of a positive number     Sum up 8 The representation range of bit complement [-128,127]

therefore , After knowing the three codes （ The most important thing is complement ）, Return to the topic , Right now Bytes as the value of a signed integer Is the concept of clear ？ In fact, let's find the complement of each integer  

Each line of Chinese characters has 32 Bytes make up ,16*16 The pixel ： a line 2 Bytes , A byte 8 position , form 16 That's ok

So print it out ： So the question is what is the power of nine ？

s="""4 0 4 0 4 0 4 32 -1 -16 4 32 4 32 4 32 4 32 4 32 8 32 8 32 16 34 16 34 32 30 -64 0 
16 64 16 64 34 68 127 126 66 -124 67 4 66 4 66 -124 126 100 66 36 66 4 66 4 66 4 126 4 66 40 0 16 
4 0 4 0 4 0 4 32 -1 -16 4 32 4 32 4 32 4 32 4 32 8 32 8 32 16 34 16 34 32 30 -64 0 
0 -128 64 -128 48 -128 17 8 1 -4 2 8 8 80 16 64 32 64 -32 64 32 -96 32 -96 33 16 34 8 36 14 40 4 
4 0 3 0 1 0 0 4 -1 -2 4 0 4 16 7 -8 4 16 4 16 4 16 8 16 8 16 16 16 32 -96 64 64 
16 64 20 72 62 -4 73 32 5 16 1 0 63 -8 1 0 -1 -2 0 64 0 80 63 -8 8 64 4 64 1 64 0 -128 
0 16 63 -8 1 0 1 0 1 0 1 4 -1 -2 1 0 1 0 1 0 1 0 1 0 1 0 1 0 5 0 2 0 
2 0 2 0 7 -16 8 32 24 64 37 -128 2 -128 12 -128 113 -4 2 8 12 16 18 32 33 -64 1 0 14 0 112 0 
1 0 1 0 1 0 9 32 9 16 17 12 17 4 33 16 65 16 1 32 1 64 0 -128 1 0 2 0 12 0 112 0 
0 0 0 0 7 -16 24 24 48 12 56 12 0 56 0 -32 0 -64 0 -128 0 0 0 0 1 -128 3 -64 1 -128 0 0"""
a=s.split('\n')

dict={}

def reverse(str):
    ans=''
    for i in str:
        ans+='0' if int(i)==1 else '1'
    return ans

for j in range(-128,128):
    if j>=0:
        dict[j]=(8-len(bin(j)[2:]))*'0'+bin(j)[2:]
    elif j==-128:
        dict[j]='10000000'
    else:
        tmp=reverse((7-len(bin(abs(j))[2:]))*'0'+bin(abs(j))[2:])# Except for sign bit negation 
        tmp_add=bin(int(tmp,2)+1)[2:]#+1
        dict[j]='1'+(7-len(tmp_add))*'0'+tmp_add
    
for i in a:
    tmp=list(map(int,i.split()))
    start=0#end=31
    while start<=30:
        s=dict[tmp[start]]+dict[tmp[start+1]]
        print(s)
        start+=2
    print('\n')
    

Because the length is too long , Just put one ‘ Nine ’ Okay , Specifically, you can run the code by yourself

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