题目链接:异性相吸
题目:
密文:

key:

查看两个文件的二进制


相同为0,不同的为1
m='0000011100011111000000000000001100001000000001000001001001010101000000110001000001010100010110000100101101011100010110000100101001010110010100110100010001010010000000110100010000000010010110000100011000000110010101000100011100000101010101100100011101010111010001000001001001011101010010100001010000011011'
key='0110000101110011011000010110010001110011011000010111001101100100011000010111001101100100011000010111001101100100011000010111001101100100011000010111001101100100011000010111001101100100011000010111001101100100011000010111001101100100011000010111001101100100011100010111011101100101011100110111000101100110'
flag = ''
for i in range(len(m)):
if m[i] == key[i]:
flag += '0'
else:
flag += '1'
print(flag)
把得到的字符串用二进制查看,得到flag

flag:flag{ea1bc0988992276b7f95b54a7435e89e}
该博客讲述了如何通过异或操作解密给定的密文,并将结果转换为二进制字符串。最终,通过观察二进制模式,得到了隐藏的flag。这是一个涉及到密码学和二进制操作的问题。
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