[SHCTF 2023 校外赛道] reverse

week1

ez_asm

想不到第1题是个汇编，咱也不知道拿啥能弄成c，不过这题也不难，直接能看懂，关键部分。

取出异或0x1e然后保存，再取出-0xa再保存。

.text:0000000000401566                               loc_401566:                             ; CODE XREF: main+65↓j
.text:0000000000401566 8B 45 FC                      mov     eax, [rbp+var_4]         计数器 指针
.text:0000000000401569 48 98                         cdqe
.text:000000000040156B 48 8D 15 AE 1A 00 00          lea     rdx, flag                       
.text:0000000000401572 0F B6 04 10                   movzx   eax, byte ptr [rax+rdx]  取出1个
.text:0000000000401576 83 F0 1E                      xor     eax, 1Eh                 xor 1e
.text:0000000000401579 89 C1                         mov     ecx, eax
.text:000000000040157B 8B 45 FC                      mov     eax, [rbp+var_4]
.text:000000000040157E 48 98                         cdqe
.text:0000000000401580 48 8D 15 99 1A 00 00          lea     rdx, flag                      
.text:0000000000401587 88 0C 10                      mov     [rax+rdx], cl            保存
.text:000000000040158A 8B 45 FC                      mov     eax, [rbp+var_4]
.text:000000000040158D 48 98                         cdqe
.text:000000000040158F 48 8D 15 8A 1A 00 00          lea     rdx, flag                      
.text:0000000000401596 0F B6 04 10                   movzx   eax, byte ptr [rax+rdx]  再取出1个
.text:000000000040159A 83 E8 0A                      sub     eax, 0Ah                 -a 
.text:000000000040159D 89 C1                         mov     ecx, eax
.text:000000000040159F 8B 45 FC                      mov     eax, [rbp+var_4]
.text:00000000004015A2 48 98                         cdqe
.text:00000000004015A4 48 8D 15 75 1A 00 00          lea     rdx, flag                      
.text:00000000004015AB 88 0C 10                      mov     [rax+rdx], cl            保存
.text:00000000004015AE FF 45 FC                      inc     [rbp+var_4]

>>> a = 'nhuo[M`7mc7uhc$7midgbTf`7`$7%#ubf7 ci5Y'
>>> bytes([ (i+0xa)^0x1e for i in a.encode()])
b'flag{It_is_als0_impor@nt_t0_13arn_4sm!}'

easy_re

IDA打开，逻辑很简单，高低位互换

int __cdecl main(int argc, const char **argv, const char **envp)
{
  char Str[52]; // [rsp+20h] [rbp-40h] BYREF
  int v5; // [rsp+54h] [rbp-Ch]
  int j; // [rsp+58h] [rbp-8h]
  int i; // [rsp+5Ch] [rbp-4h]

  _main();
  printf("Input your flag:");
  scanf("%s", Str);
  v5 = strlen(Str);
  for ( i = 0; i < v5; ++i )
    Str[i] = (Str[i] >> 4) | (16 * Str[i]);
  for ( j = 0; j < v5; ++j )
  {
    if ( Str[j] != des[j] )
    {
      printf("Wrong!");
      exit(0);
    }
  }
  printf("Right!");
  return 0;
}

>>> a = '66C61676B7452797F54703F53703C66733F5478656F52696E6162797F507270326C633D6D7'
>>> bytes([int(a[i:i+2][::-1],16) for i in range(0, len(a),2)])
b'flag{Try_t0_s0lv3_the_binary_pr0bl3m}'

seed

10字节随机数的key异或

  printf("Input your flag:");
  scanf("%s", v6);
  srand(seed);
  for ( i = 0; i <= 9; ++i )
  {
    v3 = rand() % 255;
    v5[i] = v3;
  }
  for ( j = 0; j <= 44; ++j )
    *((_BYTE *)v6 + j) ^= v5[j % 10];
  for ( k = 0; k <= 44; ++k )
  {
    if ( *((unsigned __int8 *)v6 + k) != des[k] )
    {
      printf("Wrong!");
      exit(0);
    }
  }
  printf("Right!");

用flag{头可以出来5个，后边的就只能一个个试，看着合适再试下一个

a = bytes.fromhex('402928E9C204A4ED9F535F753CD1CD2BA8C48969152116EFD72792DFCA535F2A3CD1CE03A3EFA578161A2DE1C4')
from pwn import xor

flag = b'flag{Give_'
key = xor(flag, a[:len(flag)])

for i in range(256):
    tkey = key + bytes([i])
    m = xor(tkey.ljust(10, b'\x00'),a)
    if all([0x20<=m[v]<=0x7e for v in range(len(flag),len(a),10)]):
        print(i, m)

#flag{Give_y0u_the_se3d_and_D0_you_w@nt_t0_do}

signin

这才是签到，跳过

ez_math

代码是公式，这种很常见，用z3

print("Please input flag:")
flag = input()
if len(flag)!=42:
	print("Check your length!")
	exit()

l=[]
for i in range(6):  #7字符1段，分6段
	s=""
	for j in flag[i*7:i*7+7]:
		s+=hex(ord(j))[2:]
	l.append(int(s,16))
if (
(593*l[0] + 997*l[1] + 811*l[2] + 258*l[3] + 829*l[4] + 532*l[5])== 0x5b8e0aef71d34ff43 and \
(605*l[0] + 686*l[1] + 328*l[2] + 602*l[3] + 695*l[4] + 576*l[5])== 0x551a262360964ef7f and \
(373*l[0] + 512*l[1] + 449*l[2] + 756*l[3] + 448*l[4] + 580*l[5])== 0x49d158a5657d6931c and \
(560*l[0] + 635*l[1] + 422*l[2] + 971*l[3] + 855*l[4] + 597*l[5])== 0x625568d5abbabf4f3 and \
(717*l[0] + 507*l[1] + 388*l[2] + 925*l[3] + 324*l[4] + 524*l[5])== 0x50ee0c025e70e3c23 and \
(312*l[0] + 368*l[1] + 884*l[2] + 518*l[3] + 495*l[4] + 414*l[5])== 0x40e735f8aa2815f65):
	print("Good job!")
else:
	print("Wrong\nTry again!!!")
	exit()

from z3 import *

l = [Int(f'l_{i}') for i in range(6)]
s = Solver()
s.add((593*l[0] + 997*l[1] + 811*l[2] + 258*l[3] + 829*l[4] + 532*l[5])== 0x5b8e0aef71d34ff43)
s.add((605*l[0] + 686*l[1] + 328*l[2] + 602*l[3] + 695*l[4] + 576*l[5])== 0x551a262360964ef7f)
s.add((373*l[0] + 512*l[1] + 449*l[2] + 756*l[3] + 448*l[4] + 580*l[5])== 0x49d158a5657d6931c)
s.add((560*l[0] + 635*l[1] + 422*l[2] + 971*l[3] + 855*l[4] + 597*l[5])== 0x625568d5abbabf4f3)
s.add((717*l[0] + 507*l[1] + 388*l[2] + 925*l[3] + 324*l[4] + 524*l[5])== 0x50ee0c025e70e3c23)
s.add((312*l[0] + 368*l[1] + 884*l[2] + 518*l[3] + 495*l[4] + 414*l[5])== 0x40e735f8aa2815f65)

s.check()
d = s.model()

flag = b''
for i in l:
    flag+=long_to_bytes(d[i].as_long())

#flag{N0_One_kn0ws_m@th_B3tter_Th@n_me!!!!}

ez_apk

jadx打开是base58变表，密文和码表不在layout目录里，发现外部有多个dex文件，打开class3.dex找到码表和密文

从classes3.dex找到密文和码表，base58变表解密
enc ='5TAYhycAPT1aAd535TGdWYQ8CvfoRjErGEreqhDpqv1LydTqd3mxuK2hhUp9Pws3u9mq6eX'
code1 = '9LfnoVpi1HrzBSKxhNFeyY745R2g3QmqsTCZJuDvcMdkE8wPGbUXajtAW6'
#flag{Jue_1_ju3_Y0ung_and_G0at_1s_go0d_for_yOuR_body}

week2

签到题？

在堆数字128运算后XXX后边不详，先动调得到这些数据

可以看到右下角栈里的Arglist的数据，显然是可显示的字符，导出后得到，显示base64

ZmxhZ3thMTBlN2NjYy1iODAyLWUzZWItYzg1OWE3LTMwOTQwZTIyNmR9

not gcc 未完成

pycode

给的python字节码，一点点手搓。看明白了再反一下就行

import base64 

flag = '....'
value = ''
output = ''

#24
for i in range(1000):
    w = 1024
    x = w%3
    y = w//9
    x = x*y
    w -= z     

#94
for i in range(10000):
    w = 20 
    x = w%6 
    y = w//3 
    z = x*y 
    w += z 

#166
for i in range(1000):
    w = 1024
    x = w%3 
    y = w//9 
    x = x*y 
    w -= z 

#238
for i in range(10000):
    w = 20 
    x = w%6 
    y = w//3 
    z = x*y
    w += z 

#310
for i in range(len(flag)):
    temp = flag[i]
    temp = chr(ord(temp)^8)    
    value += temp

for i in range(len(flag)):
    temp = value[i]
    temp = chr(ord(temp)+3)
    output += temp 

obfuscated_output = base64.b64encode(output.decode())
obfuscated_output = obfuscated_output[::-1]
obfuscated_output = obfuscated_output.replace('0','t')
obfuscated_output = obfuscated_output.replace('c','4')
obfuscated_output = obfuscated_output.replace('+','-')
print(obfuscated_output)

#------------------------------------------------------------
from itertools import product 
from base64 import * 

enc = '==AeAF3M-tzO-giQ-AUQosDQ9tGK7MDPuhC47tDNB5Tb8Yn4sdW4'
enc = enc[::-1].replace('-','+')
a = [('0','t') if v=='t' else ('c','4') if v=='4' else (v) for v in enc]


for k in product(*a):
    try:
        #print(''.join(k))
        b = b64decode(''.join(k).encode())
        #print(b)
        c =bytes([(i-3)^8 for i in b])
        print(c)
    except:
        pass 
        
b'flag{1b36920e-c180-b250-6537-30238f5}'

run!润

从关键代码可以看出有左右上下和上下层，是个三维迷宫，先得到数据

数据在这

取出来打印出来看，然后手搓

from pwn import u32 

#每层map 2*32位 8层从上到下
#在memset下断点,动调得到正确的map数据
dat = open('run.dat', 'rb').read()
print(dat.hex())
dat = ''.join([bin(u32(dat[i:i+4]))[2:].zfill(32) for i in range(0, len(dat), 4)])

for i in range(0, len(dat),8):
    if i%64==0: print()
    print(dat[i:i+8])

从后向前走每步编上号，然后从前向后得到串

输入串运行得到flag

C:\2023_ctf\2023_SHCTF\week2_r>run.exe
This is a easy puzzle.
Input your route:
ssdddssuuuwwwwqqqdddduussaauuuaaaaassssqddddddduuwwwaasusssdd
Right!Here are you flag:flag{7a30a122cbf428239147d86fcb3a2a37}

Authur's box

初始化S和K

S加K交换，很符合RC4加密

从汇编里找到对应的k

解密，后部加了异或的RC4


s = [i for i in range(256)]
k = (b'h3rE1Sy0UkEy'*30)[:256]
v3 = 0
for i in range(256):
    v3 = (s[i]+v3+k[i])%256
    s[i],s[v3] = s[v3],s[i]

v6 = 0
v7 = 0
#byte_408A80
ekey = [0]*42
for i in range(42):
    v6 = (v6+1)%256 
    v7 = (v7 + s[v6])%256 
    s[v6],s[v7] = s[v7],s[v6]
    ekey[i] = s[(s[v6]+s[v7])%256]^0x22

str2 = bytes.fromhex('3894E7FE4C620212054AFD714D13EC165437909BD9E9D2EDED5AD4D815596C3DB1DA43CB5A7396F996B7')
from pwn import xor 
print(xor(str2, bytes(ekey)))

week3

java是最棒的语言吗

chacha20加密

public class ChaCha20 {
    public static void main(String[] strArr) {
        Scanner scanner = new Scanner(System.in);
        System.out.println("input your flag:");
        String nextLine = scanner.nextLine();
        if (Arrays.equals(encrypt(nextLine.getBytes(StandardCharsets.UTF_8), "Shctf_Welcomes_Have_4_good_t1me_".getBytes(), "HsehrcOedfgs".getBytes()), hexStringToBytes("ce43283af73d106815fe5293b474f5309d44063c7fde19533300c60603dfe528d19aee2f6db615191e45"))) {
            System.out.println("right!");
        } else {
            System.out.println("error!");
        }
    }

    private static byte[] encrypt(byte[] bArr, byte[] bArr2, byte[] bArr3) {
        int[] chachaInit = chachaInit(bArr2, bArr3);
        byte[] bArr4 = new byte[bArr.length];
        byte[] bArr5 = new byte[64];
        for (int i = 0; i < bArr.length; i += 64) {
            chachaBlock(chachaInit, bArr5);
            for (int i2 = 0; i2 < 64 && i + i2 < bArr.length; i2++) {
                bArr4[i + i2] = (byte) (bArr[i + i2] ^ bArr5[i2]);
            }
            chachaInit[12] = chachaInit[12] + 1;
        }
        return bArr4;
    }

    private static int[] chachaInit(byte[] bArr, byte[] bArr2) {
        int[] iArr = new int[16];
        iArr[0] = 1634760805;
        iArr[1] = 857760878;
        iArr[2] = 2036477234;
        iArr[3] = 1797285236;
        for (int i = 0; i < 8; i++) {
            iArr[4 + i] = bytesToIntLittleEndian(bArr, i * 4);
        }
        iArr[12] = 0;
        iArr[13] = 0;
        iArr[14] = bytesToIntLittleEndian(bArr2, 0);
        iArr[15] = bytesToIntLittleEndian(bArr2, 4);
        return iArr;
    }

    private static void chachaBlock(int[] iArr, byte[] bArr) {
        int[] copyOf = Arrays.copyOf(iArr, 16);
        for (int i = 0; i < 10; i++) {
            chachaDoubleRound(copyOf);
        }
        for (int i2 = 0; i2 < 16; i2++) {
            intToBytesLittleEndian(iArr[i2] + copyOf[i2], bArr, i2 * 4);
        }
    }

    private static void chachaDoubleRound(int[] iArr) {
        quarterRound(iArr, 0, 4, 8, 12);
        quarterRound(iArr, 1, 5, 9, 13);
        quarterRound(iArr, 2, 6, 10, 14);
        quarterRound(iArr, 3, 7, 11, 15);
        quarterRound(iArr, 0, 5, 10, 15);
        quarterRound(iArr, 1, 6, 11, 12);
        quarterRound(iArr, 2, 7, 8, 13);
        quarterRound(iArr, 3, 4, 9, 14);
    }

    private static void quarterRound(int[] iArr, int i, int i2, int i3, int i4) {
        iArr[i] = iArr[i] + iArr[i2];
        iArr[i4] = rotateLeft(iArr[i4] ^ iArr[i], 16);
        iArr[i3] = iArr[i3] + iArr[i4];
        iArr[i2] = rotateLeft(iArr[i2] ^ iArr[i3], 12);
        iArr[i] = iArr[i] + iArr[i2];
        iArr[i4] = rotateLeft(iArr[i4] ^ iArr[i], 8);
        iArr[i3] = iArr[i3] + iArr[i4];
        iArr[i2] = rotateLeft(iArr[i2] ^ iArr[i3], 7);
    }

    private static int rotateLeft(int i, int i2) {
        return (i << i2) | (i >>> (32 - i2));
    }

    private static int bytesToIntLittleEndian(byte[] bArr, int i) {
        return ((bArr[i + 3] & 255) << 24) | ((bArr[i + 2] & 255) << 16) | ((bArr[i + 1] & 255) << 8) | (bArr[i] & 255);
    }

    private static void intToBytesLittleEndian(int i, byte[] bArr, int i2) {
        bArr[i2] = (byte) (i & 255);
        bArr[i2 + 1] = (byte) ((i >>> 8) & 255);
        bArr[i2 + 2] = (byte) ((i >>> 16) & 255);
        bArr[i2 + 3] = (byte) ((i >>> 24) & 255);
    }

    private static byte[] hexStringToBytes(String str) {
        int length = str.length();
        byte[] bArr = new byte[length / 2];
        for (int i = 0; i < length; i += 2) {
            bArr[i / 2] = (byte) ((Character.digit(str.charAt(i), 16) << 4) + Character.digit(str.charAt(i + 1), 16));
        }
        return bArr;
    }

    private static String bytesToHexString(byte[] bArr) {
        StringBuilder sb = new StringBuilder();
        int length = bArr.length;
        for (int i = 0; i < length; i++) {
            sb.append(String.format("%02x", Byte.valueOf(bArr[i])));
        }
        return sb.toString();
    }
}

key和密文会给了。这种解释性的语言似乎没有多少隐私。

k1 = "Shctf_Welcomes_Have_4_good_t1me_"
k2 = "HsehrcOedfgs"
enc = bytes.fromhex("ce43283af73d106815fe5293b474f5309d44063c7fde19533300c60603dfe528d19aee2f6db615191e45")

def encrypt(m,k1,k2):
    chacha = chachainit(k1,k2)
    a4 = [0]*len(m)
    a5 = [0]*64
    for i in range(0, len(m),64):
        block(chacha, b5)
        for i2 in range(64):
            if i2>=len(m): break 
            a4[i+i2] = m[i+i2]^a5[i2]
        chacha[12] += 1 
    return a4 

def chachainit(k1,k2):
    a = [0]*16
    a[0] = 1634760805
    a[1] = 857760878
    a[2] = 2036477234
    a[3] = 1797285236
    for i in range(8):
        a[i+4] = u32(p32(a[i])[::-1])
    a[14] = u32(k2[:4]);
    a[15] = u32(k2[4:]);
    return a

ststst

带SMC的程序，先按原程序patch再反编译

msg = open('ststst','rb').read()
a = msg[:0x696]+bytes([i^0xc3 for i in msg[0x696:0x763]])+ msg[0x763:]
open('st2','wb').write(a)

打开新生成的文件，发现是个tea加密

RC4和tea是逆向最爱用的，tea可以随便写左右加key，可以保证每次都不一样

__int64 __fastcall main(int a1, char **a2, char **a3)
{
  int i; // [rsp+4h] [rbp-3Ch]
  _QWORD s1[6]; // [rsp+10h] [rbp-30h] BYREF

  s1[5] = __readfsqword(0x28u);
  puts("plz input u fl4g:");
  __isoc99_scanf("%32s", s1);
  sub_400763();
  for ( i = 0; i <= 3; ++i )
    sub_400696(&s1[i], &unk_601080);
  if ( !memcmp(s1, &unk_6010A0, 0x20uLL) )
    puts("yeh~");
  else
    puts("oh,no");
  return 0LL;
}

__int64 __fastcall sub_400696(unsigned int *a1, _DWORD *a2)
{
  __int64 result; // rax
  unsigned int v3; // [rsp+10h] [rbp-10h]
  unsigned int v4; // [rsp+14h] [rbp-Ch]
  int v5; // [rsp+18h] [rbp-8h]
  unsigned int i; // [rsp+1Ch] [rbp-4h]

  v3 = *a1;
  v4 = a1[1];
  v5 = 0;
  for ( i = 0; i <= 0x1F; ++i )
  {
    v5 -= 1640531527;
    v3 += (v4 + v5) ^ (16 * v4 + *a2) ^ ((v4 >> 5) + a2[1]);
    v4 += (v3 + v5) ^ (16 * v3 + a2[2]) ^ ((v3 >> 5) + a2[3]);
  }
  *a1 = v3;
  result = v4;
  a1[1] = v4;
  return result;
}

from pwn import u32,p32
from ctypes import *
'''
  for ( i = 0; i <= 0x1F; ++i )
  {
    v5 -= 1640531527;
    v3 += (v4 + v5) ^ (16 * v4 + *a2) ^ ((v4 >> 5) + a2[1]);
    v4 += (v3 + v5) ^ (16 * v3 + a2[2]) ^ ((v3 >> 5) + a2[3]);
  }
'''
def decrypt(v,k):
    v0 = c_uint32(v[0])
    v1 = c_uint32(v[1])
    delta = 0x9e3779b9
    sum1 = c_uint32((delta) * 32)
    for i in range(32):       
        v1.value -= (sum1.value + v0.value) ^ (k[2] + (v0.value << 4)) ^ (k[3] + (v0.value >> 5))
        v0.value -= (sum1.value + v1.value) ^ (k[0] + (v1.value << 4)) ^ (k[1] + (v1.value >> 5))
        sum1.value -= delta       
    return p32(v0.value) + p32(v1.value)

key = [0x1234567, 0x89ABCDEF, 0xFEDCBA98, 0x76543210]
enc = bytes.fromhex('69258FDBE383CD4080E633A044A6F7FF173A0C6966B821B6A7E2E73492A610AD')
enc = [u32(enc[i:i+4]) for i in range(0,32,4)]

msg = [decrypt(enc[2*i: 2*i+2], key) for i in range(4)]
print(msg)
print(b''.join(msg))
#5ef846656801c9b9714388d2ccd98cdd

easyre

把串异或后再当成python运行，这个值爆破得到23

import base64
import marshal
import sympy as sp
encoded_data = b'#`VVVVVVVVVVVVVVVVVVVVVSVVVVFVVVV_YZVVVVMVU|VNFV@pU|V{xUMVYvVzBSMVDSVFRVMFDSV\\VQMV@\x7fVAxPMFU{V@BPp`]vU%B_MF]eVy]VMFY|UxZUVFUbTPBSMVrSVFRVMV\x7fCVT|]N`^VVVVVVVVVVVVVVVpVVVVPVVVVF`VVV_GFVVVVsVU\'V@FUp`PSVO\'TMV].V$FUMVPSVBFVOC".U_`SqV]/UU|VQ`U/V_`RsV]/V^ZUQpVMVUtVMVR@V_\'SqV]/Vo|VqV]/UU|VVpU/Vy`RGVU/Vy`SGVUoPPFTUVU.U_\'SsVXSV_\'QqVQRVQ&pqFM/UPFSQ`U|VENVqFE/V$`TqVFMVUtVMVR@V_\'SqV]/Vo|VqV]/UU|VVpU/Vy`RGVU/Vy`SGVU/Vy`TqVFMV_`TqVZMVUtVMVR@VU|VqFs/UvVRqVM/U\'RVxFRUV_QfqVACVT|RCb|VVFVV!FVVVVSgVFVVVT|Q%pEdvOY\'%pAnN@"yMsxSuPAb%p{~rOE{NO]nNOyvUzQ`tPAbMT|^%pYeMO{vTOUdN@{bsPA#sYxUB.xUvcxUvAx\\N%{`vPAnsPA#sYxRN%\x7f\x7ftcxUv!|Vtp/VVVS!UzM&u~"`rsx[tzZ\'O%AbN$]"t_FUVVVVto`VVVVVVF`UUV^ZVDVU_V^^VFNTTVRZVEVUPpRNVEVTt\x7fRVVVUmT`VVVPA#N@&`uPAqv%A"tnxVVVSN{U!ez%M\'!&&VP ez!UZmA.\'X"g^\'/NUcvXd.TPRTTD!&UB\\`dT.R}Q{!QQUdr~UguyU&sTU"u$An^PMdN@t!rpA&sPNcXQxSr@Am@p]bu\'#gT_^EVVVVtp|VVVUvU@YxM@Ye%pA`tz{bsYxQv@"`sOCvUzAbN%.|MsxRMzo\x7fM&x]M@"}ty{`sPA|tp/VVVUnS`VVV_^GVVVVt\x7fVVVVSvTSocu%E&uPB<VFVVV_ZFVVVVTUFRVFFTTVRZVpxTTVR\\Vp**'
xor_key = int(input('Plz input key (0<key<100):'))
x = sp.symbols('x')
f = x ** 2 + x + 1
integral_value = sp.integrate(f, (x, 1, xor_key))
check_value = 13024
if integral_value * 3 == check_value:
    xor_decoded_data = bytes((lambda .0: [ byte ^ xor_key for byte in .0 ])(encoded_data))
    decoded_data = base64.b64decode(xor_decoded_data)
    code_obj = marshal.loads(decoded_data)
    exec(code_obj)
else:
    print('Wrong!!')
    return None

转出再反编译

encoded_data = b'#`VVVVVVVVVVVVVVVVVVVVVSVVVVFVVVV_YZVVVVMVU|VNFV@pU|V{xUMVYvVzBSMVDSVFRVMFDSV\\VQMV@\x7fVAxPMFU{V@BPp`]vU%B_MF]eVy]VMFY|UxZUVFUbTPBSMVrSVFRVMV\x7fCVT|]N`^VVVVVVVVVVVVVVVpVVVVPVVVVF`VVV_GFVVVVsVU\'V@FUp`PSVO\'TMV].V$FUMVPSVBFVOC".U_`SqV]/UU|VQ`U/V_`RsV]/V^ZUQpVMVUtVMVR@V_\'SqV]/Vo|VqV]/UU|VVpU/Vy`RGVU/Vy`SGVUoPPFTUVU.U_\'SsVXSV_\'QqVQRVQ&pqFM/UPFSQ`U|VENVqFE/V$`TqVFMVUtVMVR@V_\'SqV]/Vo|VqV]/UU|VVpU/Vy`RGVU/Vy`SGVU/Vy`TqVFMV_`TqVZMVUtVMVR@VU|VqFs/UvVRqVM/U\'RVxFRUV_QfqVACVT|RCb|VVFVV!FVVVVSgVFVVVT|Q%pEdvOY\'%pAnN@"yMsxSuPAb%p{~rOE{NO]nNOyvUzQ`tPAbMT|^%pYeMO{vTOUdN@{bsPA#sYxUB.xUvcxUvAx\\N%{`vPAnsPA#sYxRN%\x7f\x7ftcxUv!|Vtp/VVVS!UzM&u~"`rsx[tzZ\'O%AbN$]"t_FUVVVVto`VVVVVVF`UUV^ZVDVU_V^^VFNTTVRZVEVUPpRNVEVTt\x7fRVVVUmT`VVVPA#N@&`uPAqv%A"tnxVVVSN{U!ez%M\'!&&VP ez!UZmA.\'X"g^\'/NUcvXd.TPRTTD!&UB\\`dT.R}Q{!QQUdr~UguyU&sTU"u$An^PMdN@t!rpA&sPNcXQxSr@Am@p]bu\'#gT_^EVVVVtp|VVVUvU@YxM@Ye%pA`tz{bsYxQv@"`sOCvUzAbN%.|MsxRMzo\x7fM&x]M@"}ty{`sPA|tp/VVVUnS`VVV_^GVVVVt\x7fVVVVSvTSocu%E&uPB<VFVVV_ZFVVVVTUFRVFFTTVRZVpxTTVR\\Vp**'

from pwn import xor 
import dis
from base64 import b64decode 

msg = xor(bytes([23]), encoded_data)
msg = b64decode(msg)
open('easyre_1.pyc','wb').write(msg)
dis.dis(msg)

又是个rc4

def rc4_encrypt(key, plaintext):
    S = list(range(256))
    j = 0
    for i in range(256):
        j = (j + S[i] + key[i % len(key)]) % 256
        S[i],S[j] = S[j],S[i]
    i = j = 0
    ciphertext = []
    for char in plaintext:
        i = (i + 1) % 256
        j = (j + S[i]) % 256
        S[i],S[j] = S[j],S[i]
        k = S[(S[i] + S[j]) % 256]
        ciphertext.append(char ^ k)
    print(bytes(ciphertext))

key = b'example_key'
check = b'\xd8\x94\x1e\xab\x9bft\xeb]@\x1b\xba\xe6\xe8\x133W\xdd\x0e\xe6\x924\xf1\x80mh\xeb=\x08a\x02\t.\xb5\x05B\xb0\xb0/D\x8cY'
rc4_encrypt(key, check)

喵喵喵

分别从汇编得到密文和反编译里得到大概的key

这是AES+base64只是key和iv未知，大概是上边这个串。用go_parser翻过来的程序参数基本看不出来，看汇编这块可以知道key是023_S0_e4sy_m1ao 然后用解密后的明文异或一下flag{得到iv(这个串的前边16字节：SHCTF_2023_S0_e4)

网站上解密：flag{GO1an6_REAl1Y_eaSy_10R_D3V3lop_aND_quIckIY_RuN_65601185ae5b}

crackme

lua的字节码，与python，java类似，用unluac.jar解包，这个效果非常不错。

print("please input your flag:")
flag = io.read()
code = {}
secret = {
  54,57,566,532,1014,1,7,508,10,12,498,494,6,24,14,20,489,492,0,10,490,498,517,539,21,528,517,530,543,9,13,0,4,51,562,518,526,7,9,12,5,3,513,575,514,6,519,513,556,31,1,594,117,15
}
l = string.len(flag)
-- 序号从1开始
for i = 1, l do
  num = ((string.byte(flag, i) + i) % 333 + 444) % 555 - 1
  table.insert(code, num)
end
for i = 1, l do
  x = i - 1
  if i + 2 >= l then
    code[i] = code[i % l + 1] ~ code[(i + 1) % l + 1]
  else
    code[i] = code[(i + 1) % l] ~ code[(i + 2) % l]
  end
end
for i = 1, l do
  if secret[i] ~= code[i] then
    print("Incorrect")
    return
  end
end
print("You win,flag is", flag)

lua的数组0是基本不用的，这给反编译带来很大麻烦，已经习惯了从0开始，以后如果出中文的会更更更不方便。

enc = [54,57,566,532,1014,1,7,508,10,12,498,494,6,24,14,20,489,492,0,10,490,498,517,539,21,528,517,530,543,9,13,0,4,51,562,518,526,7,9,12,5,3,513,575,514,6,519,513,556,31,1,594,117,15]
flag = [546,553,543,550,16]+[0]*len(enc)
for i in range(2, len(enc)):
    flag[i] = flag[i-1]^enc[i-2]

for i in range(1,len(enc)+1):
    for j in range(0x20,0x7f):
        if ((j+i)%333+444)%555-1 == flag[i-1]:
            print(chr(j), end= '')
            break 
#flag{C000ngr4tulat1ons!Y0u_Cr4cked_m3_991468a8de6a!!!}