一道CTF题,首先要爆破sha256的前四位,于是初学python的我写了这样的代码:
sha256enc="a8c0939f9abe212c5fa3714d317212a9ef52a6c5b8fff4a6e4ae1cec851fe370" key="NSkpCmlLih6f0f0s" strlist=string.letters+string.digits strlist=''.join(strlist) for a in strlist: for b in strlist: for c in strlist: for d in strlist: code = a+b+c+d encinfo = hashlib.sha256(code+key).hexdigest() if encinfo == sha256enc: print code print encinfo break
哇,这么长的四重循环,看的就很难受,但是向大佬学会了一招生成笛卡尔积的招式:
# sha256碰撞,得出xxxx的值 code = '' strlist = itertools.product(string.letters + string.digits, repeat=4) for i in strlist: code = i[0] + i[1] + i[2] + i[3] encinfo = hashlib.sha256(code + key).hexdigest() if encinfo == sha256enc: print code break
这样就好看多了,原理如下:
一、笛卡尔积:itertools.product(*iterables[, repeat])
直接对自身进行笛卡尔积:
import itertools for i in itertools.product('ABCD', repeat = 2): print (''.join(i),end=' ')
输出结果:
AA AB AC AD BA BB BC BD CA CB CC CD DA DB DC DD
print (”.join(i))这个语句可以让结果直接排列到一起
end=’ ‘可以让默认的输出后换行变为一个空格
两个元组进行笛卡尔积:
import itertools a = (1, 2, 3) b = ('A', 'B', 'C') c = itertools.product(a,b) for i in c: print(i,end=' ')
输出结果:
(1, ‘A’) (1, ‘B’) (1, ‘C’) (2, ‘A’) (2, ‘B’) (2, ‘C’) (3, ‘A’) (3, ‘B’) (3, ‘C’)
二、排列:itertools.permutations(iterable[, r])
import itertools for i in itertools.permutations('ABCD', 2): print (''.join(i),end=' ')
输出结果:
AB AC AD BA BC BD CA CB CD DA DB DC
三、组合:itertools.combinations(iterable, r)
import itertools for i in itertools.combinations('ABCD', 3): print (''.join(i))
输出结果:
ABC
ABD
ACD
BCD
四、组合(包含自身重复):itertools.combinations_with_replacement(iterable, r)
import itertools for i in itertools.combinations_with_replacement('ABCD', 3): print (''.join(i),end=' ')
输出结果:
AAA AAB AAC AAD ABB ABC ABD ACC ACD ADD BBB BBC BBD BCC BCD BDD CCC CCD CDD DDD
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