medusa | An open source interactive disassembler | Reverse Engineering library
kandi X-RAY | medusa Summary
kandi X-RAY | medusa Summary
An open source interactive disassembler
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QUESTION
I'm trying to create models from my Oracle database 10g using Entity Framework Core 5 dbcontext scaffold
:
ANSWER
Answered 2021-May-05 at 06:07The fine manual says minimum oracle 11.2 for that provider
Upgrade your db or change provider
QUESTION
I am just starting to master modular applications and I have a problem with the logback.
Before modularity was introduced into the project, Logback successfully worked with the same settings.
As soon as I start working with the LoggerContext and add the line requires ch.qos.logback.classic
to the module-info.java
, the application crashes.
Crash occurs in the line receiving the logger:
...ANSWER
Answered 2021-Apr-14 at 14:28It is a bug in logback 1.3.0-alpha5
Add
QUESTION
I am reading File that contains
...ANSWER
Answered 2021-Apr-12 at 17:05This code reads your file and stores the data in the std::vector
Info
using stringstreams and other things to parse.
QUESTION
I have generated three classes from XSD schema:
...ANSWER
Answered 2021-Feb-17 at 12:50We are missing the medusa
initialization code, but you should do something like this:
QUESTION
I have a text file with many rows that generally follow the patterns shown below and I'd like to extract the segments numbered 1-4 in the image below. I cannot do it with Excel because the punctuation is not sufficiently consistent so I'd like to use RegEx.
I am looking for 4 distinct RegEx expressions, corresponding to the 4 items.
What I have so far:
(.+?(?=/))
gets me everything up to the/
but I can't figure out how to split it in the Yellow and Cyan sections(?<=\/\s)(.*)
gets me everything after the/
but includes the Mintmark portion
Here is a good sample of the file contents:
...ANSWER
Answered 2021-Jan-26 at 20:49You could use a single pattern with 4 capturing groups.
QUESTION
I have ansible role which checks the latest backups from my databases (cassandra and elasticsearch) so i have this output
...ANSWER
Answered 2021-Jan-05 at 15:19If the output of the ansible-playbook other/check_backup_status.yml -i hosts/mvd/prod/hosts.yml
command goes to the standard output, I can redirect that output to awk in the next way:
QUESTION
I have an online store and am trying to display five similar products on the product page. There is a many-to-many relationship between products and tags which I'm trying to query (join table) using INNER JOIN
. Here is the SQL query I have thus far:
ANSWER
Answered 2020-Nov-21 at 18:30You can group by product and sort the results by the number of matched tags:
QUESTION
ANSWER
Answered 2020-Oct-22 at 09:30I think a lot of what you did, specifically around data wrangling, was not necessary, especially since you called d3.hierarchy()
and d3.cluster()
afterwards. I've replaced this with d3.stratify
(which deals with hierarchical data that is not yet in the right format).
I've also replaced d3.cluster
with d3.tree()
because it was unclear to me why you'd want to use d3.cluster
here. Your data has multiple parents, multiple roots and even floating nodes, and d3 is not meant to deal with that. My workaround has been to attach pseudonodes to every level, so as to make sure that there is only one node and that all nodes are at the right level at all times. To make sure the links were drawn correctly, I've written a custom getLinks
function, that can deal with multiple parents.
I've also written a custom link generator that draws the links somewhat in the way that you want them. d3 doesn't offer much of flexibility here, but you can use the source code for inspiration.
Edit
I've changed the logic to be more focused on which "partners" got a child, so both links to the same child are on the same level - like in your picture. I've also drawn the nodes based on how many partners they have, and have given every link an offset so the lines are more distinct.
I've sorted the nodes so that the real pro-creators are at the top (Zeus), which gives a more balanced and less crowded view.
QUESTION
this.statues.shift();
is isolating the correct element and removing it from the array. When this happens though, person.stoned needs to be false. Any pointers? Thank you.
Code:
...ANSWER
Answered 2020-Jul-18 at 22:14You need to save shifted object and set its property, like that:
QUESTION
This is a question from the 2019 Australian Informatics Olympiad Competition:
After the success of your latest research project in mythical DNA, you have gained the attention of a most diabolical creature: Medusa. Medusa has snakes instead of hair. Each of her snakes’ DNA is represented by an uppercase string of letters. Each letter is one of S, N, A, K or E. Your extensive research shows that a snake’s venom level depends on its DNA. A snake has venom level x if its DNA:
• has exactly 5x letters • begins with x copies of the letter S
• then has x copies of the letter N
• then has x copies of the letter A
• then has x copies of the letter K
• ends with x copies of the letter E.
For example, a snake with venom level 1 has DNA SNAKE, while a snake that has venom level 3 has DNA SSSNNNAAAKKKEEE. If a snake’s DNA does not fit the format described above, it has a venom level of 0. Medusa would like your help making her snakes venomous, by deleting zero or more letters from their DNA. Given a snake’s DNA, can you work out the maximum venom level this snake could have?
Is it possible using binary search to obtain an algorithm with complexity O(nlogn)?
...ANSWER
Answered 2020-Jul-14 at 14:05Yes you can use binary search to search for maximum venom level.
initially: l=0
r=n
when for some m=(l+r+1)/2
we can check in O(n)
time if we can obtain such venom level m, just by taking first m
letters S
then next m letters N
and so on. If there's not enough letters we update our search interval to r=m-1
otherwise l=m
Example: suppose input is SNAKESSSNNAAKKE
- Binary search initial values:
l=0
r=15
l=0
r=15
m = (l+r+1)/2 = 8
- we cant get venom value of 8 so
r=m-1 = 7
l=0
r=7
m = (l+r+1)/2 = 4
- we cant get venom value of 4 so
r=m-1 = 3
l=0
r=3
m = (l+r+1)/2 = 2
- we can get venom value of 2 so
l=m = 2
l=2
r=3
m = (l+r+1)/2 = 3
- we cant get venom value of 3 so
r=m-1 = 2
- now since
l==r
we terminate binary search and conclude that answer is2
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