DSACTF7月re

DSACTF—re部分解

隐秘的角落

go语言是这种风格，不好定位主函数

用字符串定位

找到key为thisiskkk

密文也找到了

发现在初始化中对密文有操作

所以EXP

import hashlib
m = hashlib.md5()
enc=[   0xD8, 0xE5, 0x85, 0xBE, 0xE7, 0xF8, 0x58, 0x75, 0x95, 0x65, 
  0x85, 0xE3, 0xA6, 0x47, 0x59, 0xB9, 0x14, 0x6F, 0x33, 0xB5, 
  0xCA, 0x84, 0x0B, 0xE7, 0x92, 0x0E, 0xD2, 0xFD, 0x64, 0x18, 
  0x96, 0xD0, 0x0F, 0x5E,0x44, 0x3E ]#
for i in range(len(enc)):
    enc[i]^=0x23
v7 = 0
v9 = 0
v8 = 0
key="thisiskkk"
key1=[]
for i in range(len(key)):
    key1.append(ord(key[i]))
v12=[0]*256
for i in range(256):
    v12[i]=i
for j in range(256):
    v3 = v7 + v12[j]
    v7 = (key1[j % len(key1)] + v3) % 256
    v12[j],v12[v7]=v12[v7],v12[j]
   
for k in range(len(enc)):
    v9 = (v9 + 1) % 256
    v8 = (v8 + v12[v9]) % 256
    v12[v9],v12[v8]=v12[v8] ,v12[v9] 
    enc[k] ^= v12[(v12[v8] + v12[v9]) % 256]
print(bytes(enc)) 
m.update(bytes(enc))
print(m.hexdigest())
#DASCTF{9e1963bbbb1285b993c862a5a6f12604}

EZGO

这题划分为密码题更合适

go语言不好定位主函数，从字符串下手

跟着交叉应用来到main

// main.main
void __cdecl main_main()
{
    
  __int128 v0; // xmm0
  __int64 v1; // rcx
  __int64 v2; // rax
  __int64 i; // rdx
  int v4; // ebx
  __int64 v5; // rdx
  __int64 v6; // [rsp+10h] [rbp-120h]
  __int64 v7; // [rsp+18h] [rbp-118h]
  __int64 v8; // [rsp+20h] [rbp-110h]
  __int64 v9; // [rsp+28h] [rbp-108h]
  __int64 v10; // [rsp+30h] [rbp-100h]
  __int64 v11; // [rsp+38h] [rbp-F8h]
  __int64 v12; // [rsp+40h] [rbp-F0h]
  __int64 v13; // [rsp+48h] [rbp-E8h]
  __int64 v14; // [rsp+58h] [rbp-D8h]
  __int64 v15; // [rsp+60h] [rbp-D0h]
  __int64 v16[3]; // [rsp+68h] [rbp-C8h] BYREF
  char v17; // [rsp+80h] [rbp-B0h] BYREF
  __int64 v18; // [rsp+88h] [rbp-A8h]
  __int64 *v19; // [rsp+90h] [rbp-A0h]
  __int64 v20[2]; // [rsp+98h] [rbp-98h] BYREF
  __int64 v21[2]; // [rsp+A8h] [rbp-88h] BYREF
  __int64 v22[2]; // [rsp+B8h] [rbp-78h] BYREF
  __int64 v23[2]; // [rsp+C8h] [rbp-68h] BYREF
  __int128 v24; // [rsp+D8h] [rbp-58h]
  __int64 v25[2]; // [rsp+E8h] [rbp-48h] BYREF
  __int128 v26; // [rsp+F8h] [rbp-38h]
  __int64 v27; // [rsp+108h] [rbp-28h] BYREF
  __int64 v28; // [rsp+110h] [rbp-20h]
  __int64 v29[2]; // [rsp+118h] [rbp-18h]

  while ( (unsigned __int64)&v17 <= *(_QWORD *)(*(_QWORD *)NtCurrentTeb()->NtTib.ArbitraryUserPointer + 16LL) )
    runtime_morestack_noctxt();
  v19 = (__int64 *)runtime_newobject((__int64)&RTYPE_string);
  v22[0] = (__int64)&RTYPE__ptr_string;
  v22[1] = (__int64)v19;
  fmt_Fscanf(
    (__int64)&go_itab__ptr_os_File_comma__ptr_io_Reader,
    qword_59D908,
    (__int64)"%s",
    2LL,
    (__int64)v22,
    1LL,
    1LL);
  if ( v19[1] != 40 )
    goto LABEL_10;
  LOBYTE(v27) = 0;
  v28 = 0LL;
  v0 = 0LL;
  *(_OWORD *)v29 = 0LL;
  v1 = *v19;
  v18 = *v19;
  v2 = v19[1];
  v15 = v2;
  for ( i = 0LL; i < v2; i = v14 )
  {
    
    v4 = *(unsigned __int8 *)(v1 + i);
    if ( (unsigned int)v4 >= 0x80 )
    {
    
      v4 = runtime_decoderune(v1, v2, i);
      v5 = v8;
    }
    else
    {
    
      v5 = i + 1;
    }
    v14 = v5;
    v11 = math_big_nat_shl(v28, v29[0], v29[1], v28, v29[0], v29[1], 8LL);
    v29[0] = v12;
    v29[1] = v13;
    v28 = v11;
    LOBYTE(v25[0]) = 0;
    v25[1] = 0LL;
    v26 = 0LL;
    v6 = math_big__ptr_Int_SetInt64((__int64)v25, v4);
    math_big__ptr_Int_Add((__int64)&v27, (__int64)&v27, v6);
    v2 = v15;
    v1 = v18;
    v0 = 0LL;
  }
  LOBYTE(v23[0]) = 0;
  v23[1] = 0LL;
  v24 = v0;
  math_big__ptr_Int_SetString(
    (__int64)v23,
    (__int64)"13145309456454850877228433642468099885703532627357198144609408341691751453534987676043709654743561019039155"
             "6347148927592380050533193934285571983556924577144473815598516557161",
    174LL,
    10LL);
  math_big__ptr_Int_Mul((__int64)&v27, (__int64)&v27, (__int64)&v27);
  math_big__ptr_Int_Mod((__int64)&v27, (__int64)&v27, (__int64)v23);
  v9 = math_big_nat_itoa(v28, v29[0], v29[1], v27, 10LL);
  if ( (unsigned __int64)runtime_slicebytetostring((__int64)v16, v9, v10) == 173
    && runtime_memequal(
         v7,
         (__int64)"335292815327342949386143410478703216167666281141823200936009909834563601227041859559210120519180804495"
                  "87733939007294096845300395098833835443815283246602601870001850089370636",
         173LL) )
  {
    
    v21[0] = (__int64)&RTYPE_string;
    v21[1] = (__int64)&off_514208;
    fmt_Fprintln((__int64)&go_itab__ptr_os_File_comma__ptr_io_Writer, qword_59D910, (__int64)v21, 1LL, 1LL);
  }
  else
  {
    
LABEL_10:
    v20[0] = (__int64)&RTYPE_string;
    v20[1] = (__int64)&off_514218;
    fmt_Fprintln((__int64)&go_itab__ptr_os_File_comma__ptr_io_Writer, qword_59D910, (__int64)v20, 1LL, 1LL);
  }
}

应该读取字符串，判断其长度，

放在结构体里面比较合适

就是这个有点牛马，根据数组寻址公式，正在结构体中正好是len的位置，不知道为啥不翻译成v19->len!=40

果然还是伪代码

接着对变量进行改名，ida起的名字不好看

对循环的把握

v4是我们输入的的可见字符，不可能超过128所以if里面的直接不看

对v11进行查看，发现后面没有用到的地方，直接不看

这一段没用，不看

上网查询得知，SetInt64,把v4变为数字，返回v25=v4,

而add就更明显，参数1=参数2+参数3，也就是v27=v27+v6（也是v4）

Int-Setstring, 参数1以参数4为基数存储参数3位的参数2的字符串，由于参数4是10

我们理解v23=int(这些字符串)

Mul就是v27=v27*v27

mod就是v27=v27%v23

itoa把int转为字符串类型与下面的字符串进行比较

我一开始用z3解，跑不出，用angr结果电脑环境出问题了。

其实是Rabin算法，这才弄出来

EXP

from libnum import *
import gmpy2
c=33529281532734294938614341047870321616766628114182320093600990983456360122704185955921012051918080449587733939007294096845300395098833835443815283246602601870001850089370636
n=131453094564548508772284336424680998857035326273571981446094083416917514535349876760437096547435610190391556347148927592380050533193934285571983556924577144473815598516557161
e=2
q=17489158711316178659
p=7516261744453902635364442762653073356746063224482072262455102025715350278471780391042196223686233375846890331396948280463168691132631674699134296333350979
inv_p = gmpy2.invert(p, q)
inv_q = gmpy2.invert(q, p)
mp = pow(c, (p + 1) // 4, p)
mq = pow(c, (q + 1) // 4, q)
a = (inv_p * p * mq + inv_q * q * mp) % n
b = n - int(a)
c = (inv_p * p * mq - inv_q * q * mp) % n
d = n - int(c)
aa=[a,b,c,d]
for i in aa:
    if b"DASCTF" in n2s(int(i)): 
       print(n2s(int(i)))