CUMTCTF-2021-3月赛

RE

一、rookie
用IDA打开后有大量代码混淆，但都是没用的代码有用的只有一点

写脚本反推

a = 'Q0TMR@HRZg~rn?7}]E_<A\x7FTGV+nQJ^X&'
b = ''
import base64
for i in range(32):
    x = i ^ ord(a[i])
    b += chr(x)
c = base64.b64decode(b)
print(c)

得到flag，CUMTCTF{rook13?R0O7kIT!}

二、fake
使用ILSpy反编译得到项目文件，用vs2019打开得到源代码

写脚本反推flag

a = [95,6,174,99,71,28,254,55,90,181,151,89,130,193,211,104,201,31,146,81,214,
     240,91,211,137,82,140,70,19,37]
array = [0]*256
key = 'IS_THi5_r3al_FLAg?'
f = ''
for i in range(256):
     array[i] = i
num2 = len(key)
num = 0
for i in range(256):
     num = (num+array[i]+ord(key[i%num2]))%256
     b = array[i]
     array[i] = array[num]
     array[num] = b
num3 = 0
num = 0
for i in range(30):
     num3 = (num3+1)%256
     num = (num+array[num3])%256
     b2 = array[num3]
     array[num3] = array[num]
     array[num] = b2
     item = a[i] ^ array[(array[num3]+array[num]) % 256]
     f += chr(item)
print(f)

得出flag，CUMTCTF{ThI$_IS_7H3_Re@l_fl@g}

Crypto

一、Easy_xor
用记事本打开只有一串代码

AQ0eFhsHBCM2IysqHSA8MHl+Oz0gHS82HRghJyYu

结合题目BXS*10可能是key，以及xor
写出脚本

a = 'AQ0eFhsHBCM2IysqHSA8MHl+Oz0gHS82HRghJyYu'
b = 'BXS'*10
f = ''
import base64
c = base64.b64decode(a)
for i in range(30):
     x = ord(b[i]) ^ c[i]
     f += chr(x)
print(f)

得到flag，CUMTCTF{easy_xor!-yes_we_@re~}