CTF 逆向入门(二)：RC4流密码、索引置换、位运算与累加异或

def rc4_init(key):
    S = list(range(256))          # 初始化 S 盒 [0, 1, 2, ..., 255]
    j = 0
    for i in range(256):
        j = (j + S[i] + ord(key[i % len(key)])) % 256
        S[i], S[j] = S[j], S[i]   # 交换
    return S

def rc4_crypt(S, data):
    i = j = 0
    result = []
    for byte in data:
        i = (i + 1) % 256
        j = (j + S[i]) % 256
        S[i], S[j] = S[j], S[i]
        t = (S[i] + S[j]) % 256
        result.append(byte ^ S[t])  # 明文/密文与密钥流异或
    return bytes(result)

def rc4_init(s_box, key, key_len):
    """KSA: 用密钥初始化 256 字节的 S 盒"""
    k = [0] * 256
    for i in range(256):
        s_box[i] = i
        k[i] = key[i % key_len]
    j = 0
    for i in range(256):
        j = (j + s_box[i] + ord(k[i])) % 256
        s_box[i], s_box[j] = s_box[j], s_box[i]

def rc4_crypt(s_box, data, data_len, key, key_len):
    """PRGA: 生成密钥流并与 data 异或（加密解密通用）"""
    rc4_init(s_box, key, key_len)
    i = j = 0
    for k in range(data_len):
        i = (i + 1) % 256
        j = (j + s_box[i]) % 256
        s_box[i], s_box[j] = s_box[j], s_box[i]
        t = (s_box[i] + s_box[j]) % 256
        data[k] ^= s_box[t]

# 解密
s_box = [0] * 257
data = [0xe8, 0x2b, 0x33, 0x25, 0xb2, 0x55,
        0xe9, 0x0d, 0x5d, 0xaa, 0x69, 0xfd, 0x1b,
        0x47, 0xd1, 0x7c, 0xa6, 0xff, 0x52, 0xe1,
        0x6c, 0xe8, 0x4c]
key = "SakuraiCora"

rc4_crypt(s_box, data, len(data), key, len(key))
print(''.join(chr(i) for i in data))

# 题目加密逻辑（从反编译的 .pyc 还原）
def enc(input_str):
    tmp = ""
    for i in range(len(input_str)):
        tmp += input_str[table[i]]
    return tmp

# 校验：enc(flag) == "REla{PSF!!fg}!Y_SN_1_0U"

encode = "REla{PSF!!fg}!Y_SN_1_0U"
table = [7, 8, 1, 2, 4, 5, 13, 16, 20, 21, 0, 3, 22, 19, 6, 12, 11, 18, 9, 10, 15, 14, 17]

# 解密：把密文字符放回原始位置
flag = [''] * len(encode)
for i in range(len(encode)):
    flag[table[i]] = encode[i]

print("".join(flag))
# 输出: flag{PRY!F!SS_1_Y0U!!}

text = [198, 232, 816, 200, 1536, 300, 6144, 984,
        51200, 570, 92160, 1200, 565248, 756,
        1474560, 800, 6291456, 1782, 65536000]

flag = ''
for i in range(1, len(text) + 1):
    if i % 2 != 0:          # 奇数位：右移还原（等价于除以 2^i）
        flag += chr(text[i-1] >> i)
    else:                    # 偶数位：整除还原
        flag += chr(text[i-1] // i)

print(flag)

for (i = 1; i < len; ++i)
    __b[i] ^= __b[i - 1];

# 密文（从 IDA 中提取的字节序列）
enc = "f\nk\fw&O.@\x11x\rZ;U\x11p\x19F\x1Fv\"M#D\x0Eg\x06h\x0FG2O"
enc_ord = [ord(ch) for ch in enc]

flag = [0] * 33
flag[0] = enc_ord[0]  # 第一个字节直接还原

for i in range(1, 33):
    flag[i] = enc_ord[i] ^ enc_ord[i-1]  # 前密解后密

print(''.join(chr(x) for x in flag))