jinzi0113@163.com's FirstCrackMe

Author:
jinzi0113@163.com

Language:
C/C++

Upload:
2022-10-18 10:12

Download

Platform:
Windows

Difficulty:
1.6

Quality:
3.7

Arch:
x86

Downloads:
8

Writeups:
2

Comments:
13

Description

My First CrackMe,This's Very Easy!

Comments (13)
Writeups (2)

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MountainMan on 2022-10-21 03:44: [Click to reveal]

jinzi0113@163.com on 2022-10-21 09:39: MountainMan You need to install the vc runtime

HarpoonHarry on 2022-10-28 15:27: [Click to reveal]

sultan on 2022-11-10 09:25: [Click to reveal]

xyz on 2022-11-12 12:27: was the objective to just it to say you got it or to get the actual login details? xd

aovzerk on 2022-11-13 12:00: [Click to reveal]

Tvorojok on 2022-11-18 19:13: Gen pass - https://prnt.sc/UAwoxS-PzHP6 Valid - https://prnt.sc/TEjs3XmlphRU

testesdrives on 2022-11-20 15:45: [Click to reveal]

Orkada on 2022-11-22 23:44: 0xc000007b error

wimm on 2022-12-07 19:15: ``` 00F8FD64 "D``R`!bh" ```

SidneyJob on 2022-12-10 08:00: flag{Danof4edx_flag} - this is not flag! If you want solve this task you need to get string "Congratulations, you did it." with out patching programm

featherfx on 2023-03-21 17:58: put a breakpoint on the string comparison (call ), type in a random username like bob and a password,press enter and the program should break, ECX should hold the password u typed and EAX should hold the password it's expecting for the username bob, restart program enter bob again and paste password. Congrats u cracked it

lilly on 2023-05-04 19:12: I have the feeling many have a really difficult solution. But it basicly only need 1 letter. (thats not a strong password XD)

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Solution by SidneyJob on 2022-12-13 05:27:

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Solution by kontrol on 2023-04-30 23:04:

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jinzi0113@163.com's FirstCrackMe