根据波峰波谷转化成010的二进制格式“110011011011001100001110011111110111010111011000010101110101010110011011101011101110110111011110011111101”
然后你会发现他的位数不够在每组前面+0然后转化为ASCII码
l='110011011011001100001110011111110111010111011000010101110101010110011011101011101110110111011110011111101'
h=[]
j=''
for i in range(0,len(l),7):
h.append('0'+l[i:i+7])
for g in h :
j+=chr(int(g,2))
print(j)
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