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Encryption/decryption expert needed for DRM

I need help to modify a DRM removal python script to work as my need. The task requires python coding skills as well as knowledge related to encryption/decryption.

Навыки: Python, Архитектура ПО, Linux, Java, Программирование на С

О работодателе:
( 10 отзыв(-а, -ов) ) Ho Chi Minh, Vietnam

ID проекта: #32729219

6 фрилансеров(-а) готовы выполнить эту работу в среднем за $84

mzdesmag

Hi, I see that you need some assistant in Encryption/decryption expert needed for DRM I have 5+ experience on Python Developing and I have experience of Building Management, Distributed, Database Applications. with Ma Больше

$75 USD за 3 дней(-я)
(19 отзывов(-а))
5.2
walide726

Hello sir, I have been doing this kind of python projects many times. I am a Python programmer, and I can help you to achieve your goal, basically I will use RSA, MD5 or AES algorithms for encryption. and decreptyon is Больше

$60 USD за 3 дней(-я)
(5 отзывов(-а))
2.5
Work12345x

Hi, I am a very talented software programmer with 13+ years of development experience (5+ years professional work experience). I am a results-oriented professional and possess experience using cutting-edge development Больше

$50 USD за 3 дней(-я)
(7 отзывов(-а))
2.6
vladilavsuhovoy1

Hey! I am professional Python coder. I am familiar with Python and I have a lot of work experiences in C Programming, Software Architecture, Java, Python and Linux. I can start right away. I want to discuss for this Больше

$150 USD за 5 дней(-я)
(3 отзывов(-а))
2.6
Samidullo

Hi there I read your post and I really want to work with you, if it is possible. I am studying in Cyber security field. so I can complete your project in short time I work as a AI engineer, you can check my profile t Больше

$80 USD за 7 дней(-я)
(2 отзывов(-а))
1.6
sohailkhan2k22

Yes i can do your job by using python , i can use base64 base32 to encrypt or decrypt , please provide me your project detail , so we can discuss it properly Thank you

$90 USD за 4 дней(-я)
(0 отзывов(-а))
0.0