I have a word (can't provide it here) and its encrypted code:
1C6FC813B5D83F96BB5A4B62F421225E060AC2FCD3A2C4C31B029D92C96444AB6F943DE9B6D83764C8AC5DF743DF9A1FFAC14686B531BF2CDA551F561
I should find the encryption algorithm used for it. I tried SHA-512 and generated me the same code as below but by including some zeroes.
I tried many online SHA generators and they all provided the same below encrypted code.
1c6fc813b5d83f960bb5a4b62f421225e060ac2fcd3a2c4c31b029d92c96444ab6f943de09b6d83764c8ac5df743df9a1ffac146860b05031bf2cd0a551f5601
Any idea about how some zeros are excluded?
答案 0 :(得分：1)
这是以十六进制显示的相同SHA-512哈希，但有以下异常：在顶部哈希中，十六进制中所有字节都有一个前导data = np.empty((2999986, 3))
for i in range(len(k_neighboors)):
w, v = PCA(k_neighboors[i])
data[i] = v[:,2]
break #: I break the loop in order to don't have to wait for ever.
data
Out[64]:
array([[ 0.10530792, 0.01028906, 0.99438643],
[ 0. , 0. , 0. ],
[ 0. , 0. , 0. ],
...,
[ 0. , 0. , 0. ],
[ 0. , 0. , 0. ],
[ 0. , 0. , 0. ]])
它被删除，可能二进制到十六进制的转换不正确。格式可能是0而不是%x。
示例：
%02x输出：
5524066＆lt; - 注意失踪的
printf("%x%x%x%x\n", 0x55, 0x02, 0x40, 0x66); printf("%02x%02x%02x%02x\n", 0x55, 0x02, 0x40, 0x66);
55024066