VladMetz's KeyGenMeByVladMetz

Author:
VladMetz

Language:
C/C++

Upload:
2022-01-17 20:43

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Platform:
Windows

Difficulty:
6.0

Quality:
4.0

Arch:
x86

Downloads:
19

Size:
181.00 KB

Writeups:
0

Comments:
9

Description

Hi, I decided to test how difficult it is to find the key to my creation, I think it will not be easy, although I could be wrong, it is forbidden to change anything in the program, the file is clean, good luck

Comments (9)
Writeups (0)

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demberto on 2022-01-27 12:52: Key is 6 chars long, that's all I can figure out, apart from some crazy instructions below I never saw before

Temagen on 2022-01-29 03:49: can help me with solution

VladMetz on 2022-01-29 11:31: I was wrong it is not keygen , it is ctf, I will give you a hint the flag is more than 6 characters, where did you stop?

Slava on 2022-02-05 05:22: can you say what compiler you've used? And am I supposed to get a string for a right flag? So far I've only found "Invalid"

GreeZW on 2022-02-05 18:47: int main(int argc, const char **argv, const char **envp) { int Input_Key; // ecx unsigned int D; // edi void **E; // esi int F; // ecx int *G; // edx void **H; // eax int I; // esi float J; // xmm0_4 int K; // esi int L; // ecx double M; // xmm0_8 double N; // xmm1_8 float O; // xmm5_4 const char *Strings; // edx float P; // xmm5_4 void **Q; // eax int R; // [esp+8h] [ebp-80h] BYREF int S; // [esp+2Ch] [ebp-5Ch] int T; // [esp+30h] [ebp-58h] int U; // [esp+34h] [ebp-54h] int V; // [esp+38h] [ebp-50h] int W; // [esp+3Ch] [ebp-4Ch] int X; // [esp+40h] [ebp-48h] int Y; // [esp+44h] [ebp-44h] int Z; // [esp+48h] [ebp-40h] float AB; // [esp+4Ch] [ebp-3Ch] float CD; // [esp+50h] [ebp-38h] int EF; // [esp+54h] [ebp-34h] int GH; // [esp+58h] [ebp-30h] bool IJ; // [esp+5Fh] [ebp-29h] void *Unit[4]; // [esp+60h] [ebp-28h] BYREF unsigned int A; // [esp+70h] [ebp-18h] unsigned int B; // [esp+74h] [ebp-14h] int C; // [esp+84h] [ebp-4h] Unit[0] = 0; A = 0; B = 15; C = -1; Calc_Bad_String(Unit, &Unknow_Value0, 0); C = 1; Put_String(Input_Key, "Input key: "); Calc_Pass(Dword_Unknow, Unit); D = A; E = (void **)Unit[0]; if ( A = 6 ) { F = 0; IJ = B = 16; G = &R; do { H = Unit; if ( F = 8 ) { ++G; if ( IJ ) H = E; *G = *((char *)H + F); } else { if ( IJ ) H = E; *(&S + F) = *((char *)H + F); } ++F; } while ( F = 0.00000011920929 ) Strings = "Invalid\n"; Put_String(L, Strings); Close_Console((int)"pause"); E = (void **)Unit[0]; } if ( B = 16 ) { Q = E; if ( B + 1 = 4096 ) { E = (void **)*(E - 1); if ( (unsigned int)((char *)Q - (char *)E - 4) 31 ) Invalid_Parameter_No_Info_No_Return(); } Free_Unit(E); } return 0; } This is the main app. Use float never seen that any var is by my sh**t Logical is not good but give a look

VladMetz on 2022-02-06 12:22: Slava, message valid flag - Yes, you higher! GreeZW , you didn't show all the logic of the application, yours is unlikely to help in any way

VladMetz on 2022-02-06 12:26: Slava, I opened it in x32dbg

survivalizeed on 2024-04-17 21:55: I have the entire routine reversed now. The key is 16 chars long and is split in half and then processed. Only thing which seems rather impossible is inverting it. I will try my best now ...

survivalizeed on 2024-04-17 22:08: For anyone looking for the complete cleaned up routine (heavy spoiler) int keypart1[8] = {}; int keypart2[8] = {}; for (int i = 0; i 7) { keypart2[i - 8] = input[i]; } else { keypart1[i] = (int)input[i]; } } int sum = keypart1[0] + keypart1[1] + keypart1[2] + keypart1[3] + keypart1[4] + keypart1[5] + keypart1[6] + keypart1[7]; int sum_pow = pow(keypart1[0], 2) + pow(keypart1[1], 2) + pow(keypart1[2], 2) + pow(keypart1[3], 2) + pow(keypart1[4], 2) + pow(keypart1[5], 2) + pow(keypart1[6], 2) + pow(keypart1[7], 2); double added_log = 0; for (int i = 0; i = 0.00000011920929) { std::cout

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crackmes.one

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VladMetz's KeyGenMeByVladMetz