Yes
import sys
xor_keys = [ 5, 0x12, 0xFA, 0x34, 0xE1, 0x7C, 0x38, 0x5A, 0x61,
0x30, 0xA, 0x5A, 0xD1, 0x10, 0xCA, 0x72, 0x15, 0x51,
0xA5, 0x64, 1, 0x5F, 0x31, 0x41, 0x7A, 0x25, 0x2B, 0x8A, 0x3F,
0x1F, 0x9A ];
def decrypt(string_length, bytes):
str_length = string_length & ~(string_length >> 31)
counter = 0
while True:
key = xor_keys[counter]
if counter > 31:
counter = 0
bytes[counter] ^= key
counter += 1
str_length -= 1
if str_length == 0:
break
return str(bytes)
n = 0x37401D8C1B977D66
length = 8
print(decrypt(length, bytearray(n.to_bytes(16, sys.byteorder))))