[DDCTF2018](╯°□°）╯︵ ┻━┻

(╯°□°）╯︵ ┻━┻
50pt

(╯°□°）╯︵ ┻━┻

d4e8e1f4a0f7e1f3a0e6e1f3f4a1a0d4e8e5a0e6ece1e7a0e9f3baa0c4c4c3d4c6fbb9b2b2e1e2b9b9b7b4e1b4b7e3e4b3b2b2e3e6b4b3e2b5b0b6b1b0e6e1e5e1b5fd

首先看它的长度，有134，那就排除md5和sha1加密。可能是十六进制的字符串。直接转字符串是不可读的，转成十进制[212, 232, 225, 244, 160, 247, 225, 243, 160, 230, 225, 243, 244, 161, 160, 212, 232, 229, 160, 230, 236, 225, 231, 160, 233, 243, 186, 160, 196, 196, 195, 212, 198, 251, 185, 178, 178, 225, 226, 185, 185, 183, 180, 225, 180, 183, 227, 228, 179, 178, 178, 227, 230, 180, 179, 226, 181, 176, 182, 177, 176, 230, 225, 229, 225, 181, 253]发现都大于128,所以都减128,最后转ascii码得到flag.

key='d4e8e1f4a0f7e1f3a0e6e1f3f4a1a0d4e8e5a0e6ece1e7a0e9f3baa0c4c4c3d4c6fbb9b2b2e1e2b9b9b7b4e1b4b7e3e4b3b2b2e3e6b4b3e2b5b0b6b1b0e6e1e5e1b5fd'
hex=[]
flag=''
for i in range(0,len(key)-1,2):
    zdh=key[i:i+2]
    hex.append(zdh)
for i in range(len(hex)):
    zdh1=hex[i]
    flag+=chr(int(int(zdh1,16))-128)
print(flag)

flag{922ab9974a47cd322cf43b50610faea5}

发现都大于128,所以都减128,最后转ascii码得到flag.