문제 설명
Reversing Basic Challenge #9
이 문제는 사용자에게 문자열 입력을 받아 정해진 방법으로 입력값을 검증하여 correct 또는 wrong을 출력하는 프로그램이 주어집니다.
해당 바이너리를 분석하여 correct를 출력하는 입력값을 찾으세요!
획득한 입력값은 DH{} 포맷에 넣어서 인증해주세요.
예시) 입력 값이 Apple_Banana일 경우 flag는 DH{Apple_Banana}
암호화시키는 enc.py
byte_7FF663F04000 = [0x7E, 0x7D, 0x9A, 0x8B, 0x25, 0x2D, 0xD5, 0x3D,\
0x03, 0x2B, 0x38, 0x98, 0x27, 0x9F, 0x4F, 0xBC,\
0x2A, 0x79, 0x00, 0x7D, 0xC4, 0x2A, 0x4F, 0x58]
byte_7FF663F04020 = [0x63, 0x7C, 0x77, 0x7B, 0xF2, 0x6B, 0x6F, 0xC5,\
0x30, 0x01, 0x67, 0x2B, 0xFE, 0xD7, 0xAB, 0x76,\
0xCA, 0x82, 0xC9, 0x7D, 0xFA, 0x59, 0x47, 0xF0,\
0xAD, 0xD4, 0xA2, 0xAF, 0x9C, 0xA4, 0x72, 0xC0,\
0xB7, 0xFD, 0x93, 0x26, 0x36, 0x3F, 0xF7, 0xCC,\
0x34, 0xA5, 0xE5, 0xF1, 0x71, 0xD8, 0x31, 0x15,\
0x04, 0xC7, 0x23, 0xC3, 0x18, 0x96, 0x05, 0x9A,\
0x07, 0x12, 0x80, 0xE2, 0xEB, 0x27, 0xB2, 0x75,\
0x09, 0x83, 0x2C, 0x1A, 0x1B, 0x6E, 0x5A, 0xA0,\
0x52, 0x3B, 0xD6, 0xB3, 0x29, 0xE3, 0x2F, 0x84,\
0x53, 0xD1, 0x00, 0xED, 0x20, 0xFC, 0xB1, 0x5B,\
0x6A, 0xCB, 0xBE, 0x39, 0x4A, 0x4C, 0x58, 0xCF,\
0xD0, 0xEF, 0xAA, 0xFB, 0x43, 0x4D, 0x33, 0x85,\
0x45, 0xF9, 0x02, 0x7F, 0x50, 0x3C, 0x9F, 0xA8,\
0x51, 0xA3, 0x40, 0x8F, 0x92, 0x9D, 0x38, 0xF5,\
0xBC, 0xB6, 0xDA, 0x21, 0x10, 0xFF, 0xF3, 0xD2,\
0xCD, 0x0C, 0x13, 0xEC, 0x5F, 0x97, 0x44, 0x17,\
0xC4, 0xA7, 0x7E, 0x3D, 0x64, 0x5D, 0x19, 0x73,\
0x60, 0x81, 0x4F, 0xDC, 0x22, 0x2A, 0x90, 0x88,\
0x46, 0xEE, 0xB8, 0x14, 0xDE, 0x5E, 0x0B, 0xDB,\
0xE0, 0x32, 0x3A, 0x0A, 0x49, 0x06, 0x24, 0x5C,\
0xC2, 0xD3, 0xAC, 0x62, 0x91, 0x95, 0xE4, 0x79,\
0xE7, 0xC8, 0x37, 0x6D, 0x8D, 0xD5, 0x4E, 0xA9,\
0x6C, 0x56, 0xF4, 0xEA, 0x65, 0x7A, 0xAE, 0x08,\
0xBA, 0x78, 0x25, 0x2E, 0x1C, 0xA6, 0xB4, 0xC6,\
0xE8, 0xDD, 0x74, 0x1F, 0x4B, 0xBD, 0x8B, 0x8A,\
0x70, 0x3E, 0xB5, 0x66, 0x48, 0x03, 0xF6, 0x0E,\
0x61, 0x35, 0x57, 0xB9, 0x86, 0xC1, 0x1D, 0x9E,\
0xE1, 0xF8, 0x98, 0x11, 0x69, 0xD9, 0x8E, 0x94,\
0x9B, 0x1E, 0x87, 0xE9, 0xCE, 0x55, 0x28, 0xDF,\
0x8C, 0xA1, 0x89, 0x0D, 0xBF, 0xE6, 0x42, 0x68,\
0x41, 0x99, 0x2D, 0x0F, 0xB0, 0x54, 0xBB, 0x16]
#input = "ABCDEFGHIJKLMNOPQRSTUVW"
input = "ABCDEFG"
output = []
def ror(n, m):
shift = n >> m
src = n << (8 - m)
src &= 255
return shift | src
def xor(a1, a2):
if type(a1) is str:
a1 = ord(a1)
if type(a2) is str:
a2 = ord(a2)
ret = a1 ^ a2
return ret
def convert_to_int(a1):
if type(a1) is str:
return ord(a1)
return a1
def sub_7FF663F010A0(a1):
a1 = list(a1)
a1 = [ord(char) for char in a1]
# print(a1)
v5 = "I_am_KEY"
ret = 0
v2 = a1[ret]
# print(v2)
for i in range(0, 16, 1):
for j in range(0, 8, 1):
v2 = ror( convert_to_int(a1[(j+1)& 7]) + byte_7FF663F04020[xor(v5[j], v2)] & 0xff, 5)
a1[(j+1) & 7] = v2
print("v2: " + hex(v2))
# print([hex(char) for char in a1])
print("====")
#0xd, 0x79, 0x3d, 0x7f
# print(a1)
# print([hex(char) for char in a1])
for i in range(0, len(a1), 1):
output.append(a1[i])
print(output)
return ret
def sub_7FF663F01000(a1):
v3 = len(a1)
if ((v3+1) % 8) != 0:
return False
a1 = a1 + "\x00"
for i in range(0, v3+1, 8):
sub_7FF663F010A0(a1[i:i+8])
print(a1[i:i+8])
return True
print(sub_7FF663F01000(input))
for i in range(0, len(output), 1):
print(hex(output[i]), end=' ')
복호화시키는 (CTF Flag를 얻어내는) solve.py
byte_7FF663F04000 = [0x7E, 0x7D, 0x9A, 0x8B, 0x25, 0x2D, 0xD5, 0x3D,\
0x03, 0x2B, 0x38, 0x98, 0x27, 0x9F, 0x4F, 0xBC,\
0x2A, 0x79, 0x00, 0x7D, 0xC4, 0x2A, 0x4F, 0x58]
byte_7FF663F04020 = [0x63, 0x7C, 0x77, 0x7B, 0xF2, 0x6B, 0x6F, 0xC5,\
0x30, 0x01, 0x67, 0x2B, 0xFE, 0xD7, 0xAB, 0x76,\
0xCA, 0x82, 0xC9, 0x7D, 0xFA, 0x59, 0x47, 0xF0,\
0xAD, 0xD4, 0xA2, 0xAF, 0x9C, 0xA4, 0x72, 0xC0,\
0xB7, 0xFD, 0x93, 0x26, 0x36, 0x3F, 0xF7, 0xCC,\
0x34, 0xA5, 0xE5, 0xF1, 0x71, 0xD8, 0x31, 0x15,\
0x04, 0xC7, 0x23, 0xC3, 0x18, 0x96, 0x05, 0x9A,\
0x07, 0x12, 0x80, 0xE2, 0xEB, 0x27, 0xB2, 0x75,\
0x09, 0x83, 0x2C, 0x1A, 0x1B, 0x6E, 0x5A, 0xA0,\
0x52, 0x3B, 0xD6, 0xB3, 0x29, 0xE3, 0x2F, 0x84,\
0x53, 0xD1, 0x00, 0xED, 0x20, 0xFC, 0xB1, 0x5B,\
0x6A, 0xCB, 0xBE, 0x39, 0x4A, 0x4C, 0x58, 0xCF,\
0xD0, 0xEF, 0xAA, 0xFB, 0x43, 0x4D, 0x33, 0x85,\
0x45, 0xF9, 0x02, 0x7F, 0x50, 0x3C, 0x9F, 0xA8,\
0x51, 0xA3, 0x40, 0x8F, 0x92, 0x9D, 0x38, 0xF5,\
0xBC, 0xB6, 0xDA, 0x21, 0x10, 0xFF, 0xF3, 0xD2,\
0xCD, 0x0C, 0x13, 0xEC, 0x5F, 0x97, 0x44, 0x17,\
0xC4, 0xA7, 0x7E, 0x3D, 0x64, 0x5D, 0x19, 0x73,\
0x60, 0x81, 0x4F, 0xDC, 0x22, 0x2A, 0x90, 0x88,\
0x46, 0xEE, 0xB8, 0x14, 0xDE, 0x5E, 0x0B, 0xDB,\
0xE0, 0x32, 0x3A, 0x0A, 0x49, 0x06, 0x24, 0x5C,\
0xC2, 0xD3, 0xAC, 0x62, 0x91, 0x95, 0xE4, 0x79,\
0xE7, 0xC8, 0x37, 0x6D, 0x8D, 0xD5, 0x4E, 0xA9,\
0x6C, 0x56, 0xF4, 0xEA, 0x65, 0x7A, 0xAE, 0x08,\
0xBA, 0x78, 0x25, 0x2E, 0x1C, 0xA6, 0xB4, 0xC6,\
0xE8, 0xDD, 0x74, 0x1F, 0x4B, 0xBD, 0x8B, 0x8A,\
0x70, 0x3E, 0xB5, 0x66, 0x48, 0x03, 0xF6, 0x0E,\
0x61, 0x35, 0x57, 0xB9, 0x86, 0xC1, 0x1D, 0x9E,\
0xE1, 0xF8, 0x98, 0x11, 0x69, 0xD9, 0x8E, 0x94,\
0x9B, 0x1E, 0x87, 0xE9, 0xCE, 0x55, 0x28, 0xDF,\
0x8C, 0xA1, 0x89, 0x0D, 0xBF, 0xE6, 0x42, 0x68,\
0x41, 0x99, 0x2D, 0x0F, 0xB0, 0x54, 0xBB, 0x16]
input = byte_7FF663F04000
# input = [0xc6, 0x3e, 0x2d, 0xfc, 0x4e, 0xfa, 0x74, 0xbc, 0xb9, 0x92, 0x9c, 0xe6, 0x60, 0x2e, 0xc7, 0x17, 0x44, 0x1c, 0x29, 0x56, 0x2a, 0x5c, 0x6d, 0xaa]
# input = [0xec, 0xc1, 0xb8, 0x35, 0x1a, 0xc5, 0x17, 0x92]
print(len(input))
output = []
def ror(n, m):
shift = n >> m
src = n << (8 - m)
src &= 255
return shift | src
def rol(n, m):
shift = n << m
shift &= 255
src = n >> (8 - m)
return shift | src
def xor(a1, a2):
if type(a1) is str:
a1 = ord(a1)
if type(a2) is str:
a2 = ord(a2)
ret = a1 ^ a2
return ret
def convert_to_int(a1):
if type(a1) is str:
return ord(a1)
return a1
def sub_7FF663F010A0(a1):
# a1 = list(a1)
# a1 = [ord(char) for char in a1]
# print(a1)
v5 = "I_am_KEY"
ret = 0
v2 = a1[0]
for i in range(15, -1, -1):
for j in range(7, -1, -1):
# v2 = rol( convert_to_int(a1[(j+1)& 7]) + byte_7FF663F04020[xor(v5[j], v2)] & 0xff, 5)
v2 = (rol(convert_to_int(a1[(j+1)& 7]) & 0xff, 5) - byte_7FF663F04020[xor(v5[j], a1[j& 7])]) & 0xff
a1[(j+1) & 7] = v2
print("v2: " + hex(v2))
print("====")
for i in range(0, len(a1), 1):
output.append(a1[i])
print(output)
# for i in range(len(output)):
# print(chr(output[i]))
return ret
def sub_7FF663F01000(a1):
v3 = len(a1)
# if ((v3+1) % 8) != 0:
# return False
a1.append("\x00")
for i in range(0, v3, 8):
sub_7FF663F010A0(a1[i:i+8])
print("AAA")
print(a1[i:i+8])
return True
print(sub_7FF663F01000(input))
for i in range(0, len(output), 1):
print(hex(output[i]), end=' ')
print("\n")
for i in range(0, len(output), 1):
print(chr(output[i]), end='')
Result
.... 0x52 0x65 0x76 0x65 0x72 0x73 0x65 0x5f 0x5f 0x79 0x6f 0x75 0x72 0x5f 0x5f 0x62 0x72 0x61 0x69 0x6e 0x5f 0x3b 0x29 0x0 Reverse__your__brain_;)
FLAG
Reverse__your__brain_;)