安洵杯 2019——crackMe

题目思路：

输入flag到byte_2BA1E4
修改base64的表，大小写互换
以where_are_u_now?作为16位密钥产生轮密钥
通过SM4加密算法，生成加密后的16位字节数
然后通过变种base64（非魔改），生成字符串
相邻两位位置进行交换得到最终的结果

解题过程：
首先这道题是通过hook的方式来调用的各组函数，其中有反debug的内容，没看出是怎么反的所以直接用到了一个技巧，修改push内容然后修改eip到函数跳转地址即可。

按正常执行流程走，结束完base64的变表操作，如下图，然后在红线处进行eip跳转

到密钥产生函数

然后跳到SM加密函数，其中dword_2BA218是轮密钥，byte_2BA1E4是我们输入的值，byte_2BA180是输出的16个字节

然后是最终变换，函数区域，他先将结果Str2进行了邻位互换，然后进行的base64的变换

在这里的base64他是做了改动的，最终index值他做了一个+24%64的运算，如下图

而且因为16位不能整除3，最后两位补成叹号，如下图

大致流程出来了，写代码就可以了，首先之前base64脚本进行一下变化，并且打印出原始的16个字节
代码：

def mo64_decode(a):
    tb = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789+/"                       #魔改的表
    c=''
    for i in range(len(a)):
        if(tb.index(a[i])<24):
            b = str(bin(tb.index(a[i])+64-24)).replace("0b", "").zfill(6)
        else:
            b=str(bin(tb.index(a[i])-24)).replace("0b","").zfill(6)        #6 base64   5 base32   4 base16
        c+=str(b)
    m=""
    i=0;arr=[]
    for i in range(0,len(c),8):
        m+=chr(int(c[i:i+8],2))
        arr.append(int(c[i:i+8],2))
    print(arr)
    return m
def mo64_encode(a):
    tb='abcdefghijklmnopqrstuvwxyzABCDEFQRSTUVWXYPGHIJKLMNOZ0123456789+/'
    c=''
    for i in range(len(a)):
        b=str(bin(ord(a[i])).replace("0b","")).zfill(8)
        c+=str(b)
    m=""
    i=0
    for i in range(0,len(c),6):
        m+=tb[int(c[i:i+6],2)]
    return m
ch="U1ATIOpkOyWSvGm/YOYFR4"                        #此处已交换

mo64_decode(ch)

产生数组[89, 208, 149, 41, 13, 242, 64, 6, 20, 244, 141, 39, 105, 6, 135, 78]
嫁接到百度到了sm4算法中去，得到结果
sm4算法：

# S盒
SboxTable = \
    [
        0xd6, 0x90, 0xe9, 0xfe, 0xcc, 0xe1, 0x3d, 0xb7, 0x16, 0xb6, 0x14, 0xc2, 0x28, 0xfb, 0x2c, 0x05,
        0x2b, 0x67, 0x9a, 0x76, 0x2a, 0xbe, 0x04, 0xc3, 0xaa, 0x44, 0x13, 0x26, 0x49, 0x86, 0x06, 0x99,
        0x9c, 0x42, 0x50, 0xf4, 0x91, 0xef, 0x98, 0x7a, 0x33, 0x54, 0x0b, 0x43, 0xed, 0xcf, 0xac, 0x62,
        0xe4, 0xb3, 0x1c, 0xa9, 0xc9, 0x08, 0xe8, 0x95, 0x80, 0xdf, 0x94, 0xfa, 0x75, 0x8f, 0x3f, 0xa6,
        0x47, 0x07, 0xa7, 0xfc, 0xf3, 0x73, 0x17, 0xba, 0x83, 0x59, 0x3c, 0x19, 0xe6, 0x85, 0x4f, 0xa8,
        0x68, 0x6b, 0x81, 0xb2, 0x71, 0x64, 0xda, 0x8b, 0xf8, 0xeb, 0x0f, 0x4b, 0x70, 0x56, 0x9d, 0x35,
        0x1e, 0x24, 0x0e, 0x5e, 0x63, 0x58, 0xd1, 0xa2, 0x25, 0x22, 0x7c, 0x3b, 0x01, 0x21, 0x78, 0x87,
        0xd4, 0x00, 0x46, 0x57, 0x9f, 0xd3, 0x27, 0x52, 0x4c, 0x36, 0x02, 0xe7, 0xa0, 0xc4, 0xc8, 0x9e,
        0xea, 0xbf, 0x8a, 0xd2, 0x40, 0xc7, 0x38, 0xb5, 0xa3, 0xf7, 0xf2, 0xce, 0xf9, 0x61, 0x15, 0xa1,
        0xe0, 0xae, 0x5d, 0xa4, 0x9b, 0x34, 0x1a, 0x55, 0xad, 0x93, 0x32, 0x30, 0xf5, 0x8c, 0xb1, 0xe3,
        0x1d, 0xf6, 0xe2, 0x2e, 0x82, 0x66, 0xca, 0x60, 0xc0, 0x29, 0x23, 0xab, 0x0d, 0x53, 0x4e, 0x6f,
        0xd5, 0xdb, 0x37, 0x45, 0xde, 0xfd, 0x8e, 0x2f, 0x03, 0xff, 0x6a, 0x72, 0x6d, 0x6c, 0x5b, 0x51,
        0x8d, 0x1b, 0xaf, 0x92, 0xbb, 0xdd, 0xbc, 0x7f, 0x11, 0xd9, 0x5c, 0x41, 0x1f, 0x10, 0x5a, 0xd8,
        0x0a, 0xc1, 0x31, 0x88, 0xa5, 0xcd, 0x7b, 0xbd, 0x2d, 0x74, 0xd0, 0x12, 0xb8, 0xe5, 0xb4, 0xb0,
        0x89, 0x69, 0x97, 0x4a, 0x0c, 0x96, 0x77, 0x7e, 0x65, 0xb9, 0xf1, 0x09, 0xc5, 0x6e, 0xc6, 0x84,
        0x18, 0xf0, 0x7d, 0xec, 0x3a, 0xdc, 0x4d, 0x20, 0x79, 0xee, 0x5f, 0x3e, 0xd7, 0xcb, 0x39, 0x48,
    ]

# 常数FK
FK = [0xa3b1bac6, 0x56aa3350, 0x677d9197, 0xb27022dc];
ENCRYPT = 0;
DECRYPT = 1

# 固定参数CK
CK = \
    [
        0x00070e15, 0x1c232a31, 0x383f464d, 0x545b6269,
        0x70777e85, 0x8c939aa1, 0xa8afb6bd, 0xc4cbd2d9,
        0xe0e7eef5, 0xfc030a11, 0x181f262d, 0x343b4249,
        0x50575e65, 0x6c737a81, 0x888f969d, 0xa4abb2b9,
        0xc0c7ced5, 0xdce3eaf1, 0xf8ff060d, 0x141b2229,
        0x30373e45, 0x4c535a61, 0x686f767d, 0x848b9299,
        0xa0a7aeb5, 0xbcc3cad1, 0xd8dfe6ed, 0xf4fb0209,
        0x10171e25, 0x2c333a41, 0x484f565d, 0x646b7279
    ]


def padding(data):  # 填充
    print("plaintext:\t", bytes(data))
    file_data_list = list(data)

    lenth = len(file_data_list)
    # print ("data lenth:", lenth)
    remainder = lenth % 16
    if remainder != 0:
        i = 16 - remainder  # i为需要填充的位数
        # print ("padding numbers = ", i)
        for j in range(i):
            file_data_list.append(i)  # 填充 char 0-(i-1)
    #if remainder == 0:
        #for k in range(16):
            #file_data_list.append(0x08)  # 刚好的话 填充0x08
    print("after PKCS5 padding:", file_data_list)
    return file_data_list


def list_4_8_to_int32(key_data):  # 列表4个8位，组成32位
    return int((key_data[0] << 24) | (key_data[1] << 16) | (key_data[2] << 8) | (key_data[3]))


def n32_to_list4_8(n):  # 把n分别取32位的每8位放入列表
    return [int((n >> 24) & 0xff), int((n >> 16) & 0xff), int((n >> 8) & 0xff), int((n) & 0xff)]


# 循环左移
def shift_left_n(x, n):
    return int(int(x << n) & 0xffffffff)


def shift_logical_left(x, n):
    return shift_left_n(x, n) | int((x >> (32 - n)) & 0xffffffff)  # 两步合在一起实现了循环左移n位


def XOR(a, b):
    return list(map(lambda x, y: x ^ y, a, b))


# s盒查找
def sbox(idx):
    return SboxTable[idx]


def extended_key_LB(ka):  # 拓展密钥算法LB
    a = n32_to_list4_8(ka)  # a是ka的每8位组成的列表
    b = [sbox(i) for i in a]  # 在s盒中每8位查找后，放入列表b，再组合成int bb
    bb = list_4_8_to_int32(b)
    rk = bb ^ (shift_logical_left(bb, 13)) ^ (shift_logical_left(bb, 23))
    return rk


def linear_transform_L(ka):  # 线性变换L
    a = n32_to_list4_8(ka)
    b = [sbox(i) for i in a]
    bb = list_4_8_to_int32(b)  # bb是经过s盒变换的32位数
    return bb ^ (shift_logical_left(bb, 2)) ^ (shift_logical_left(bb, 10)) ^ (shift_logical_left(bb, 18)) ^ (
    shift_logical_left(bb, 24))  # 书上公式


def sm4_round_function(x0, x1, x2, x3, rk):  # 轮函数
    return (x0 ^ linear_transform_L(x1 ^ x2 ^ x3 ^ rk))


class Sm4(object):
    def __init__(self):
        self.sk = [0] * 32
        self.mode = ENCRYPT

    def sm4_set_key(self, key_data, mode):  # 先算出拓展密钥
        self.extended_key_last(key_data, mode)

    def extended_key_last(self, key, mode):  # 密钥扩展算法
        MK = [0, 0, 0, 0]
        k = [0] * 36
        MK[0] = list_4_8_to_int32(key[0:4])
        MK[1] = list_4_8_to_int32(key[4:8])
        MK[2] = list_4_8_to_int32(key[8:12])
        MK[3] = list_4_8_to_int32(key[12:16])
        k[0:4] = XOR(MK, FK)
        for i in range(32):
            k[i + 4] = k[i] ^ (extended_key_LB(k[i + 1] ^ k[i + 2] ^ k[i + 3] ^ CK[i]))
        self.sk = k[4:]  # 生成的32轮子密钥放到sk中

        self.mode = mode
        if mode == DECRYPT:  # 解密时rki逆序
            self.sk.reverse()

    def sm4_one_round(self, sk, in_put):  # 一轮算法 ，4个32位的字=128bit=16个字节（8*16）
        item = [list_4_8_to_int32(in_put[0:4]), list_4_8_to_int32(in_put[4:8]), list_4_8_to_int32(in_put[8:12]),
                list_4_8_to_int32(in_put[12:16])]  # 4字节一个字，把每4个字节变成32位的int
        x = item

        for i in range(32):
            temp = x[3]
            x[3] = sm4_round_function(x[0], x[1], x[2], x[3], sk[i])  # x[3]成为x[4]
            x[0] = x[1]
            x[1] = x[2]
            x[2] = temp

            print("%dround----->" % (i + 1), "key:%-12d\n" % sk[i], "result：", x)
        res = x
        # res = reduce (lambda x, y: [x[1], x[2], x[3], sm4_round_function (x[0], x[1], x[2], x[3], y)],sk, item) #32轮循环加密
        res.reverse()
        rev = map(n32_to_list4_8, res)
        out_put = []
        [out_put.extend(_) for _ in rev]
        return out_put

    def encrypt(self, input_data):
        # 块加密
        output_data = []
        tmp = [input_data[i:i + 16] for i in range(0, len(input_data), 16)]  # 输入数据分块
        [output_data.extend(each) for each in map(lambda x: self.sm4_one_round(self.sk, x), tmp)]
        return output_data


def encrypt(mode, key, data):
    sm4_d = Sm4()
    sm4_d.sm4_set_key(key, mode)
    en_data = sm4_d.encrypt(data)
    return en_data


def sm4_crypt_cbc(mode, key, iv, data):
    sm4_d = Sm4()
    sm4_d.sm4_set_key(key, mode)
    en_data = sm4_d.sm4_crypt_cbc(iv, data)
    return en_data


if __name__ == "__main__":
    data = "1234567812345678"
    print("plaintext:   ",data)
    data = bytes(data, 'ascii')
    key_data = []
    print("16 bytes key:")
    tmp = "where_are_u_now?"
    [key_data.append(ord(i)) for i in tmp]
    for i in range(16):
        key_data[i] = key_data[i]
    print(key_data)

    sm4_d = Sm4()  # 创建一个Sm4对象
    sm4_d.sm4_set_key(key_data, ENCRYPT)  # 加密模式
    padding_data = padding(data)  # 明文填充
    en_data = sm4_d.encrypt(padding_data)  # 加密
    en_data = [89, 208, 149, 41, 13, 242, 64, 6, 20, 244, 141, 39, 105, 6, 135, 78]
    print("ciphertext:\t", en_data)
    sm4_d.sm4_set_key(key_data, DECRYPT)  # 解密模式
    print("\ndecode:")
    de_data = sm4_d.encrypt(en_data)  # 解密
    # print(file_data)
    print("plaintext：", de_data)
    print("after decode：", end='')
    [print(chr(i), end='') for i in de_data]        #打印flag

得到结果
SM4foRExcepioN?!

sm4算法就保存了，至于学习的话，懂得大致原理就行，不是专门密码的不用太深