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发表于 2016-5-31 21:20:29
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本帖最后由 Looker 于 2016-5-31 21:29 编辑
##暴力破解方法
s = '0000000096c73e97'
s1 = '\x00\x00\x00\x00\x96\xc7\x3e\x97'
strtemp = ''
for i in range(0, len(s)-1, 2):
if s[i:i+2] == '00':
strtemp += '\x00'
if s[i:i+2] == '96':
strtemp += '\x96'
if s[i:i+2] == 'c7':
strtemp += '\xc7'
if s[i:i+2] == '3e':
strtemp += '\x3e'
if s[i:i+2] == '97':
strtemp += '\x97'
###......
###......
###......
###暴力破解16 * 16 = 256条if语句,外加一条else语句用于判断溢出
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print(s1)
print(strtemp)
print(s1 == strtemp) |
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