easy_re-ucsc

str和v7异或

def xor_decrypt(input_str, key):  
    """  
    用于异或解密字符串。  
  
    :param input_str: 原始字符串  
    :param key: 异或密钥（整数）  
    :return: 解密后的字符串  
    """    return ''.join(chr(ord(char) ^ key) for char in input_str)  
  
  
# 示例数据  
encrypted_str = "n=<;:h2<'?8:?'9hl9'h:l>'2>>2>hk=>;:?"  
  
key = 10  # 密钥为9  
# for key in range(10, 150):  
  
# 解密  
decrypted_str = xor_decrypt(encrypted_str, key)  
print(f"解密结果: {decrypted_str}")  
#  
# encrypted_str = "]U[du~|t@{z@wj.}.~q@gjz{z@wzqW~/b;"  
# print( ''.join(chr(ord(c) ^ 0x1F) for c in encrypted_str))

simplere-ucsc

用010把CTF改成UPX

就能脱下来了

经过一个base58再经过一个异或

Base58换了表

import math  
  
# ── 1) 先把你给出的密文贴进来（末尾 0x00 是字符串结束）：  
buf2 = bytes([  
    0x72,0x7A,0x32,0x48,0x34,0x4E,0x3F,0x3A,0x42,0x33,  
    0x47,0x69,0x75,0x63,0x7C,0x7D,0x77,0x62,0x65,0x64,  
    0x7B,0x6F,0x62,0x50,0x73,0x2B,0x68,0x6C,0x67,0x47,  
    0x69,0x15,0x42,0x75,0x65,0x40,0x76,0x61,0x56,0x41,  
    0x11,0x44,0x7F,0x19,0x65,0x4C,0x40,0x48,0x65,0x60,  
    0x01,0x40,0x50,0x01,0x61,0x6F,0x69,0x57,  
    0x00  # 末尾结束符  
])  
  
b = buf2[:-1]  
L = len(b)  # 58  
Str_bytes = bytearray(L)  
for k in range(L):  
    Str_bytes[k] = b[L-1-k] ^ (L-k)  
  
Str = Str_bytes.decode('ascii')  
print("Recovered Str =", Str)  
tb58 = "wmGbyFp7WeLh2XixZUYsS5cVv1ABRrujdzQ4Kfa6gP8HJN3nTCktqEDo9M"  
N = 0  
for c in Str:  
    N = N * 58 + tb58.index(c)  
  
blen = (N.bit_length() + 7)//8  
flag = N.to_bytes(blen, 'big').decode('ascii')  
print("FLAG =", flag)

ez_debug

下断点v5就是flag