字节凯撒爆破

python 复制代码
# a=[43,33,44,42,54,116,124,46,46,121,46,120,126,121,43,43,127,41,123,127,120,121,120,124,116,40,47,127,43,47,121,46,124,126,121,125,125,48]
a=b'1RWLQJbWKDWbDHVWKHWLFbHGXFDWLRQbSOD\VbDQbLPSRUWDQWbUROHbLQbVKDSLQJbDbPRUHbEHDXWLIXObPLQGBIODJ^EHOLHYHBLQB\RXUVHOI`BSDUWLDOBNH\sBLVB(&\'(&)'
import re

for x in range(256):
    flag=''
    flag_xor= ''
    for each in a:
        # print(chr(int(each,16)-x%256),end='')
        tmp=chr((each+x)%256)
        xor = chr((each ^ x)%256)
        flag+=tmp
        flag_xor+=xor
    rule=re.compile('flag.*?\{.*?\}',flags=re.IGNORECASE)
    #flag格式应为   flag{123qwe}
    if  rule.findall(flag):
        flag=rule.findall(flag)
        print(flag,'\n','ascii increase',x)
    if  rule.findall( flag_xor):
        regres=rule.findall( flag_xor)
        print(flag_xor,'\n','xor-mode',x)
    # print(flag)
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