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[CTF] 2019CISCN 部分wp

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yechen123 发表于 2019-4-23 20:52 回帖奖励


毕竟是国赛,难度挺大的,种类也杂。

比赛第一天就出了一道逆向题,被大佬各种秒。
第二天倒是挺多题目,但是下午才能安心解题,emmmm。

附上我们队的wp,所有题目也包含在里边了。
[Asm] 纯文本查看 复制代码
https://shimo.im/docs/RcFy1DWUUK0F9nOh/read


0x0 easyGo

代码审计,发现代码。


得到?#22336;?#34920;
[Asm] 纯文本查看 复制代码
6789_-abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ012345


看到?#25945;?#19978;各位大佬的解题速度,目测题目仅仅简单的魔改base64。
再翻代码,发?#31181;?#35201;函数sub_495150。主要判?#19979;?#36753;都在里边了。

[Asm] 纯文本查看 复制代码
lea     rax, aTgrbtxmzgd6zha ;
mov     [rsp+100h+var_F8], rax
mov     [rsp+100h+var_F0], 38h
call    sub_43F800


得到?#29992;?#21518;的?#22336;?br />
[Asm] 纯文本查看 复制代码
tGRBtXMZgD6ZhalBtCUTgWgZfnkTgqoNsnAVsmUYsGtCt9pEtDEYsql3


解密得到
flag{92094daf-33c9-431e-a85a-8bfbd5df98ad}

0x1 bbvvmm
打开发现。

SM4算法无疑。
username经过SM4后再base64?#29992;?#21518;输出,先解base64。


然后网上套脚本解密。(出处:https://blog.csdn.net/songdawww/article/details/79112548
[Asm] 纯文本查看 复制代码
#-*-coding:utf-8-*-
"""
SM4 GM
@author: Dawei
"""
from print_log import *
import copy

#Expanded SM4 S-boxes    Sbox table: 8bits input convert to 8 bits output
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,
]

# System parameter
FK = [0xa3b1bac6,0x56aa3350,0x677d9197,0xb27022dc]

# fixed parameter
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
]

ENCRYPT = 0
DECRYPT = 1


def GET_UINT32_BE(key_data):
    tmp_data = int((key_data[0] << 24) | (key_data[1] << 16) | (key_data[2] << 8) | (key_data[3]))
    return tmp_data


def PUT_UINT32_BE(n):
    return [int((n>>24)&0xff), int((n>>16)&0xff), int((n>>8)&0xff), int((n)&0xff)]


# rotate shift left marco definition
def SHL(x, n):
    xx = int(int(x << n) & 0xffffffff)
    return xx


def ROTL(x, n):
    xx = SHL(x, n)
    yy = xx | int((x >> (32 - n)) & 0xffffffff)
    return yy


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


# look up in SboxTable and get the related value.
# args:    [in] inch: 0x00~0xFF (8 bits unsigned value).
def sm4Sbox(idx):
    return SboxTable[idx]


# Calculating round encryption key.
# args:    [in] a: a is a 32 bits unsigned value;
# return: sk[i]: i{0,1,2,3,...31}.
def sm4CalciRK(ka):
    b = [0, 0, 0, 0]
    a = PUT_UINT32_BE(ka)
    b[0] = sm4Sbox(a[0])
    b[1] = sm4Sbox(a[1])
    b[2] = sm4Sbox(a[2])
    b[3] = sm4Sbox(a[3])
    bb = GET_UINT32_BE(b[0:4])
    rk = bb ^ (ROTL(bb, 13)) ^ (ROTL(bb, 23))
    return rk


# private F(Lt) function:
# "T algorithm" == "L algorithm" + "t algorithm".
# args:    [in] a: a is a 32 bits unsigned value;
# return: c: c is calculated with line algorithm "L" and nonline algorithm "t"
def sm4Lt(ka):
    b = [0, 0, 0, 0]
    a = PUT_UINT32_BE(ka)
    b[0] = sm4Sbox(a[0])
    b[1] = sm4Sbox(a[1])
    b[2] = sm4Sbox(a[2])
    b[3] = sm4Sbox(a[3])
    bb = GET_UINT32_BE(b[0:4])
    c = bb ^ (ROTL(bb, 2)) ^ (ROTL(bb, 10)) ^ (ROTL(bb, 18)) ^ (ROTL(bb, 24))
    return c

# private F function:
# Calculating and getting encryption/decryption contents.
# args:    [in] x0: original contents;
# args:    [in] x1: original contents;
# args:    [in] x2: original contents;
# args:    [in] x3: original contents;
# args:    [in] rk: encryption/decryption key;
# return the contents of encryption/decryption contents.
def sm4F(x0, x1, x2, x3, rk):
    return (x0 ^ sm4Lt(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):
        self.sm4_setkey(key_data, mode)

    def sm4_setkey(self, key, mode):
        MK = [0, 0, 0, 0]
        k = [0]*36
        MK[0] = GET_UINT32_BE(key[0:4])
        MK[1] = GET_UINT32_BE(key[4:8])
        MK[2] = GET_UINT32_BE(key[8:12])
        MK[3] = GET_UINT32_BE(key[12:16])
        k[0:4] = XOR(MK[0:4], FK[0:4])
        for i in range(32):
            k[i + 4] = k[i] ^ (sm4CalciRK(k[i + 1] ^ k[i + 2] ^ k[i + 3] ^ CK[i]))
            self.sk[i] = k[i + 4]
        self.mode = mode
        if mode == DECRYPT:
            for idx in range(16):
                t = self.sk[idx]
                self.sk[idx] = self.sk[31 - idx]
                self.sk[31 - idx] = t

    def sm4_one_round(self, sk, in_put):
        out_put = []
        ulbuf = [0]*36
        ulbuf[0] = GET_UINT32_BE(in_put[0:4])
        ulbuf[1] = GET_UINT32_BE(in_put[4:8])
        ulbuf[2] = GET_UINT32_BE(in_put[8:12])
        ulbuf[3] = GET_UINT32_BE(in_put[12:16])
        for idx in range(32):
            ulbuf[idx + 4] = sm4F(ulbuf[idx], ulbuf[idx + 1], ulbuf[idx + 2], ulbuf[idx + 3], sk[idx])

        out_put += PUT_UINT32_BE(ulbuf[35])
        out_put += PUT_UINT32_BE(ulbuf[34])
        out_put += PUT_UINT32_BE(ulbuf[33])
        out_put += PUT_UINT32_BE(ulbuf[32])
        return out_put

    def sm4_crypt_ecb(self, input_data):
        # SM4-ECB block encryption/decryption
        length = len(input_data)
        i = 0
        output_data = []
        while length > 0:
            output_data += self.sm4_one_round(self.sk, input_data[i:i+16])
            i += 16
            length -= 16
        return output_data

    def sm4_crypt_cbc(self, iv, input_data):
        #SM4-CBC buffer encryption/decryption
        length = len(input_data)
        i = 0
        output_data = []
        tmp_input = [0]*16
        if self.mode == ENCRYPT:
            while length > 0:
                tmp_input[0:16] = XOR(input_data[i:i+16], iv[0:16])
                output_data += self.sm4_one_round(self.sk, tmp_input[0:16])
                iv = copy.deepcopy(output_data[i:i+16])
                i += 16
                length -= 16
        else:
            while length > 0:
                output_data += self.sm4_one_round(self.sk, input_data[i:i+16])
                output_data[i:i+16] = XOR(output_data[i:i+16], iv[0:16])
                iv = copy.deepcopy(input_data[i:i + 16])
                i += 16
                length -= 16
        return output_data


def sm4_crypt_ecb(mode, key, data):
    sm4_d = Sm4()
    sm4_d.sm4_set_key(key, mode)
    en_data = sm4_d.sm4_crypt_ecb(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__":
    log_init()

    key_data = [0xDA,0x98,0xF1,0xDA,0x31,0x2A,0xB7,0x53,0xA5,0x70,0x3A,0xB,0xFD,0x29,0xd,0xD6]
    input_data = [0xEF,0x46,0x8D,0xBA,0xF9,0x85,0xB2,0x50,0x9C,0x9E,0x20,0x0C,0xF3,0x52,0x5A,0xB6]
    #input_data = [0x55]*16
    #iv_data = [0]*16

    sm4_d = Sm4()
    # sm4_d.sm4_set_key(key_data, ENCRYPT)#ENCRYPT ?#29992;?#26159;0 解密是1

    # en_data = sm4_d.sm4_crypt_ecb(input_data)
    # print_data(en_data, "en_data:")

    sm4_d.sm4_set_key(key_data, DECRYPT)##ENCRYPT ?#29992;?#26159;0 解密是1
    de_data = sm4_d.sm4_crypt_ecb(input_data)#解密
    print_data(de_data, "de_data:")

得到
[Asm] 纯文本查看 复制代码
0x36 0x32 0x36 0x31 0x36 0x34 0x37 0x32 0x36 0x35 0x37 0x32 0x33 0x31 0x33 0x32

经过变化后得到username
[Asm] 纯文本查看 复制代码
badrer12


password是一个6位的?#22336;?#20018;,然后进入vm。


发现会跟"xyz{|}"异或,但是好像没有跟别的?#22336;?#20018;对比,尝试提交,成功。

[Asm] 纯文本查看 复制代码
username:badrer12
password:xyz{|}


因为nc.send的时候会把\n发过去,所?#26434;胮wntools发包。
[Asm] 纯文本查看 复制代码
from pwn import *
context.log_level="debug"
#p=process("./bbvvmm")
p=remote("39.97.228.196",10001)
p.recv()
p.sendline("badrer12")
p.recv()
p.send("xyz{|}")
p.interactive()


0x2 where_u_are

这题目的wp是kaller师傅@zhukai055 写的。


编译成o文件:
main函数中

关键的两步。
f22:找到?#22336;?#22312;table_32的位置,f23?#22909;看?#30340;pos转成5bit的二进制,放在偏移0x11e0的内存上。

f24:把二进制按单双数的位置拆开,分别用 累加除法判断,结果分别放在0x12b0 0x12b8。
最后在main?#20449;?#26029; 结果。


写脚本爆破这段的时候,结果一直不为整数,可能是浮点数在IDA反编译的时候出了问题。
[Asm] 纯文本查看 复制代码
0.0001>=v7 -25 >=-0.0001 #v7=0x12b0
0.0001>=(v7+8) -175 >=-0.0001
#这里的精确度也是试出来的。

附C脚本
[Asm] 纯文本查看 复制代码
int _tmain(int argc, _TCHAR* argv[])
{
	char b32_table[] = "0123456789bcdefghjkmnpqrstuvwxyz";
	unsigned int v6; // [rsp+44h] [rbp-17Ch]
	char *v7; // [rsp+B8h] [rbp-108h]
	char *v8; // [rsp+C0h] [rbp-100h]
	double v9; // [rsp+D0h] [rbp-F0h]
	double v10; // [rsp+D8h] [rbp-E8h]
	double v11; // [rsp+E0h] [rbp-E0h]
	double v12; // [rsp+E8h] [rbp-D8h]
	double v13; // [rsp+F0h] [rbp-D0h]
	double v14; // [rsp+F8h] [rbp-C8h]
	int Sz_1[20] = {0};
	int Sz_2[20] = { 0 };
	int Sz_3[40] = { 0 };
	v13 = -90.0;
	v14 = 90.0;

	int Sz[21] = { 0 };
	for (int i = 0; i < 0xfffff; i++)//爆破20位二进制
	{
		for (int j = 0; j < 20; j++)//int转20bit数组
		{

			Sz[j] = i >> j & 1;
			//cout << Sz[j];
		}
		v9 = 0.0;
		v10 = 0.0;
		v11 = -180.0;
		v12 =180.0;
		for (int j = 0; j < 20; ++j)
		{
			
			if (Sz[j] == 1)
			{
				v11 = v10;
				v10 = (v10 + v12) / 2.0;
			}
			else if (Sz[j] == 0)
			{
				v12 = v10;
				v10 = (v11 + v10) / 2.0;
			}

			
		}
		if (v10 - 175 >= -0.0001&&v10 - 175 <= 0.0001)
		{
			for (int j = 0; j < 20; j++)
			{

				Sz_1 [j]= Sz[j];
				cout << Sz_1[j];
			}
			cout << endl;
		}
	}
	for (int i = 0; i < 0xfffff; i++)
	{
		for (int j = 0; j < 20; j++)
		{

			Sz[j] = i >> j & 1;
			//cout << Sz[j];
		}
		v9 = 0.0;
		v10 = 0.0;
		v11 = -90.0;
		v12 = 90.0;
		for (int j = 0; j < 20; ++j)
		{

			if (Sz[j] == 1)
			{
				v11 = v10;
				v10 = (v10 + v12) / 2.0;
			}
			else if (Sz[j] == 0)
			{
				v12 = v10;
				v10 = (v11 + v10) / 2.0;
			}


		}
		if (v10 - 25 >= -0.0001&&v10 - 25 <= 0.0001)
		{
			for (int j = 0; j < 20; j++)
			{

				Sz_2[j] = Sz[j];
				cout << Sz_2[j];
			}
			cout << endl ;
		}
		
		
	}
	for (int i = 0; i < 20; i++)
	{
		Sz_3[i * 2+1] = Sz_2[i];
		Sz_3[i * 2 ] = Sz_1[i];

	}
	for (int i = 0; i < 40; i++)
	{
		cout << Sz_3[i];

	}
	cout << endl;
	int Sz_4[8] = { 0 };
	for (int i = 0; i < 8; i++)
	{
		for (int j = 0; j < 5; j++)
		{
			cout << Sz_3[5 * i + j] << " " <<( Sz_3[5 * i + j] << (5 - j-1) )<< endl;
			Sz_4[i] = Sz_4[i] + (Sz_3[5 * i + j] << (5 - j-1));
			
		}
		cout << Sz_4[i] << endl;
		
	}
	system("pause");
	cout << "flag{";
	for (int i = 0; i < 8; i++)
	{
		
		cout  <<b32_table[Sz_4[i]];

	}
	cout << "}";
	
	system("pause");
	return 0;
}

flag{xukqnpp5}

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沙发
13180707766 发表于 2019-4-23 22:16
厉害!!!!
3#
dongfang155 发表于 2019-4-23 22:35
4#
学士天下 发表于 2019-4-24 08:18
5#
pcjy 发表于 2019-4-24 09:38
原来这么骚。。。发不全,根本没在意这个我r
6#
JY. 发表于 2019-4-24 15:17
原来这么骚。。。发不全,根本没在意
7#
衣者?#31455;?#36208;刀口 发表于 2019-4-25 07:59

91xinpian会员的能解开吗,哈哈哈
8#
weeew 发表于 2019-4-25 08:35
看不懂  路过支持一个
9#
lchhhhhh 发表于 2019-4-27 15:42
来支持师傅了
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