注意：这道题需要脱壳，而且需要改特征值，详细请看
[LitCTF 2024]hello_upx——入土为安的第一天-CSDN博客
脱完壳发现有256这个特殊的数，是rc4类型的题，最后有一个异或
a = "9F041CEFA92386B6F56F27B96155FD42" aa = "5E4C614B08306D1502BC6854D68716D4" # Splitting strings into pairs of hexadecimal characters b = [a[i:i+2] for i in range(0,len(a),2)] c = b[::-1] d1 = ''.join(c) b = [aa[i:i+2] for i in range(0,len(aa),2)] c = b[::-1] d2 = ''.join(c) print(d1) print(d2) # Concatenating d1 and d2 encode = d1 + d2 print(encode) # Converting concatenated string to list of integers ddd = [int(encode[i:i+2],16) for i in range(0,len(encode),2)] # Initialize flag list flag = [] # Initializing variables v22 = [0]*256 v20 = [0]*256 key = [ord(i) for i in "justfortest"] print(key) # Initialization of S-box v22 and v20 for i in range(256): v22[i] = i v20[i] = key[i % len(key)] v5 = 0 for i in range(256): v8 = v22[i] v5 = (v8 + v20[i] + v5) % 256 v22[i], v22[v5] = v22[v5], v22[i] v9 = 0 v10 = 0 # Encryption process for i in range(len(ddd)): v10 = (v10 + 1) % 256 v11 = v22[v10] v9 = (v11 + v9) % 256 v22[v10], v22[v9] = v22[v9], v22[v10] # Swap flag.append(ddd[i] ^ v22[(v22[v10] + v22[v9]) % 256]) # Printing encrypted flag in hexadecimal hex_flag = [hex(f) for f in flag] print(hex_flag) # Attempting to decode encrypted flag to string (handle UnicodeDecodeError) try: decrypted_text = ''.join(chr(f) for f in flag) print("Decrypted text:", decrypted_text) except UnicodeDecodeError: print("Decryption result contains non-printable characters.") # Printing encrypted flag as ASCII characters (for demonstration, may cause errors) for i in range(len(flag)): try: print(chr(flag[i]), end='') except ValueError: print("[Non-printable]", end='')
