yams | A Last.FM scrobbler for MPD | REST library

did you tried this. this has full info about securing express with react.

https://medium.com/@dbillinghamuk/csrf-setup-for-expressjs-and-ssr-react-redux-app-348e65261009

Obviously, if you return inside a loop, the loop will only ever execute once.

You want to 'build up' your string, one ascii code at a time (well, unicode codepoint, really - as others have pointed out, I don't know what dank late 80s outdated cruft you're following, mate - the days of ASCII are loooong gone), so you need a StringBuilder, you want to append 'numUni + " "' to this in the loop, and then return the stringbuilder, built up to a string:

You can extract logic which process values to a function

Example

!include cannot handle a file with multiple documents since it basically replaces the current node with the content from the referred document, starting at its root node. Since a file with multiple documents has multiple root nodes, it's unclear what !include should do there.

That being said, you can of course simply define both stages in the same document:

r by definition does this. From :h v_r:

To me, this most likely sounds like a restricted case of the Integer Linear programming problem (ILP), namely, its 0-or-1 variant, where the integer variables are restricted to the set {0, 1}. This is known to be NP-hard (and the corresponding decision problem is NP-complete).

The problem is formulated as follows (following the conventions in the op. cit.):

Given the matrix A, the constraint vector b, and the weight vector c, find the vector x ∈ {0, 1}^N such that all the constraints A⋅x ≥ b are satisfied, and the cost c⋅x is minimal.

I flipped the constraint inequality, but this is equivalent to changing the sign of both A and b.

The inequalities indicate satisfaction of your order: that you can buy at the least the amount of every item in the visited store. Note that b has the same length as the number of rows in A and the number of columns in both c and x. The dot-product c⋅x is, naturally, a scalar.

Since you are minimizing the number of trips, each trip costs the same, so that c = 1, and c⋅x is the total number of trips. The store inventory matrix A has a row per item, and a column per store, and the b is your shopping list.

Naturally, the exact best solution is found by trying all possible 2^N values for the x.

Since there is no single approach to NP-hard problems, consider the problem size, and how close to the optimum you want to arrive. A greedy approach would work well (when your next store to visit has the most total number of items not yet satisfied) when the "inventories" are large. If you have the idea in advance about the expected minimum number of trips, you can trim the search beam at some value, exceeding the number of trips by some multiplication coefficient. This is the best approach when your search is time constrained (I routinely do beam searches, closely related to the branch-and-cut approach mentioned in the article, in graphs that take a few GB of memory slightly faster than the limit of 30ms per exploration step with a beam as wide as 10,000). Simulated annealing also works, if the search landscape is not excessively rough.

Also search on cs.SE; it may be even a better place for questions of this type.

Lots of ways to do this, but here's one using the tidyverse. By "gathering" the data so the staples are all in one column, I think it's easier to apply the correct weight.

So this was a simple case of mis-reading the documentation. But in case anyone misreads it as well…

Responding with a 503 is a special case. The documentation explains:

By default, failed jobs are not retried unless a 503 status code is returned, in which case it is retried every minute until it succeeds or returns a 200-299 status code.

It doesn't explicitly say it, but this means that when the server responds with a 503 status the retry_parameters in cron.yaml are ignored and it still retries every minute.

You can do this :