kzt38 ta55h 9n32b z543f 79s36 5t22f za5d3 4z4tr 9t62e 7i92a ft92h 6s4se 8be4e 6s5ya hytbh 39t3z t7ett nkae5 eb6i6 aek53 kai7d $PAR Almost ripe for a run Get a free trial at: https://t.co/NbEHDcudFV https://t.co/INkTUk1wk6 |

$PAR Almost ripe for a run Get a free trial at: https://t.co/NbEHDcudFV https://t.co/INkTUk1wk6

2021.12.07 22:04 ShortAlgo $PAR Almost ripe for a run Get a free trial at: https://t.co/NbEHDcudFV https://t.co/INkTUk1wk6

$PAR Almost ripe for a run Get a free trial at: https://t.co/NbEHDcudFV https://t.co/INkTUk1wk6 submitted by ShortAlgo to UltraAlgo [link] [comments]


2021.12.07 22:04 danran84 danran84's IGS Rep Page

PLEASE DO NOT DELETE THIS Users may feel free to add any other rep here, steamtrades.com, /SGS rep, /GCXRep, etc but understand that the only fully accepted rep on /indiegameswap is on /IGSRep. Traders may reject trades for any suspicion and should feel free to contact the mods for background checks
You may only edit things beneath this line or total number of complete trades. Insert what you like here.Examples: Other Rep pages, your steam/orgin profile, your favorite type of game, favorite quote, picture, anything! This is your area!!
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2021.12.07 22:04 Suo9876 Fix your game EA (Player Career)

I’ve won the Championship League 3 years in a row and it still says it’s my debut after each match 🤦‍♂️
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2021.12.07 22:04 danish_princess Among Us cookies for my son's birthday this week

Among Us cookies for my son's birthday this week submitted by danish_princess to Baking [link] [comments]


2021.12.07 22:04 Tenacioussourd AMC LETS GOOOOOO I WANNA HEAR THE APES 💪🏽

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2021.12.07 22:04 TheoT37 What starter did you pick/plan to pick for BDSP?

Poll for personal research/knowledge. What starter did you pick for either Brilliant Diamond or Shining Pearl? I've always picked Turtwig in the original diamond (Chimchar once) but just curious what the trend is for the remakes, as I'm getting a copy soon.
View Poll
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2021.12.07 22:04 New_York_Joe It's not Criterion but Touch of Evil coming from Kino on 4K with 3 different cuts of the film

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2021.12.07 22:04 LeilaJoyy test

test
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2021.12.07 22:04 Nmalacane25 What was the worst Christmas gift you ever received?

submitted by Nmalacane25 to AskReddit [link] [comments]


2021.12.07 22:04 ShortAlgo $PARR Almost ripe for a run Get a free trial at: https://t.co/NbEHDcudFV https://t.co/Ti3VP9k78O

$PARR Almost ripe for a run Get a free trial at: https://t.co/NbEHDcudFV https://t.co/Ti3VP9k78O submitted by ShortAlgo to UltraAlgo [link] [comments]


2021.12.07 22:04 Hewasnumber1st I feel like I’m being annoying

Recently went out with this girl and have been trying to text just when I feel like it. Whenever I think of something to send her, my mind immediately gets rid of the idea and then I just don’t end up saying anything.
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2021.12.07 22:04 MrMetalhead3029483 Black cat paws are the best

Black cat paws are the best submitted by MrMetalhead3029483 to blackcats [link] [comments]


2021.12.07 22:04 yepok2222 Is it possible to compliment someone who drinks cocktails from a can? Could you try?

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2021.12.07 22:04 Noble_Mushtak [2021 Day 7] Applying Slope Trick from Competitive Programming

I've seen a few interesting solutions to 2021 Day 7 which try to be more efficient than just brute force, like this dynamic programming solution or using the median/mean, but I haven't seen anyone use the "slope trick" technique from competitive programming. I don't think slope trick is the best way to solve this problem, but I wanted to introduce slope trick to some people in this community because slope trick is an interesting method that can be used to solve a few different algo problems, not just today's problem. If you want to read a separate tutorial on slope trick, I would recommend reading this CodeForces blog post, but I will be explaining slope trick as much as necessary in order to explain my solution, and my explanation will be slightly different from the one in this blog post.
So, what's slope trick? For our purposes, slope trick is a novel way of representing functions that are "slope trickable," i.e. (1) continuous, (2) concave up, and (3) piecewise linear.

  1. Continuous means there are no gaps in the function, i.e. you can't have functions that jump instantaneously from 2 to 3.
  2. Concave up means the function is always non-decreasing in slope, i.e. |x| and x2 are both concave up functions because they never decrease in slope. Often times, you can tell if a function is concave up if it has a right-side-up U or V shape.
  3. Piecewise linear means that there are only a finite number of places where the function changes slope, and the function is a completely straight line in between the places where it changes slope.
Here is a Desmos graph with some slope trickable functions. You'll notice these functions are either just sums of linear functions and absolute value functions, which is because f(x)=x and f(x)=|x| are the simplest slope trickable functions.
The naive way to represent these functions is as a bunch of linear functions stitched together. For example, you might represent f(x)=|x| as y=-x over x < 0 and y=x over x >= 0. And you could have more complicated piecewise-linear functions like:
 { -3x-1 over x < -1 f(x) = { -x+1 over -1 <= x < 1 { 2x-2 over x > 1 
And this means that you have to keep track of the slope and y-intercept both before and after every slope change in the function. And this makes it really cumbersome to work with slope trickable functions, because you have to do bookkeeping to make sure the slope trickable function always remains continuous and that you are accounting for all the slope changes correctly, etc.
Instead, slope trick represents slope trickable functions using two pieces: (1) A function `y=mx+b` representing what `f(x)` is for all `x < M`, where `M` is the location of the first slope change in `f(x)` and (2) A set of pairs representing all the places where `f(x)` changes slope, and how much the slope changes. For example, f(x)=|x| becomes
 y=-x, slope_changes=[(x=0, slope_change=+2)] 
Because |x|=-x for all x < 0, and then |x| goes from slope -1 to +1 at x=0. And the above piecewise-linear function becomes:
y=-3x+1, slope_changes=[(x=-1, slope_change=+2), (x=1, slope_change=+3)] 
And now it becomes much easier to do operations, because we only have one slope and y-intercept to keep track of, and the rest is just keep track of when and by how much the slope changes. For example, if we wanted to add the above two slope trickable functions, we just add the two functions on the left and merge the slope changes, like this:
y=-4x+1, slope_changes=[(x=-1, slope_change=+2), (x=0, slope_change=+2), (x=1, slope_change=+3)] 
Now, in this Advent of Code problem, we want to find the minimum of the function c(x), which represents the cost of moving all the crabs to position x. Let [p1, ..., pn] represent the initial positions of all the crabs. For any 1 <= i <= n, what's the cost of moving crab i to position x? It is just |x-pi|. This is clearly a slope trickable function, which can be represented as
y=-x+pi, slope_changes=[(x=i, slope_change=+2)] 
And c(x) is just the sum of all of these functions, so c(x) can also be represented using slope trick. For example, let's say we have four crabs, initially at positions [3,0,1,2]. Then, c(x) is the sum of the following functions:
y=-x+3, slope_changes=[(x=3, slope_change=+2)] y=-x+0, slope_changes=[(x=0, slope_change=+2)] y=-x+1, slope_changes=[(x=1, slope_change=+2)] y=-x+2, slope_changes=[(x=2, slope_change=+2)] 
So c(x) is:
y=-4x+6, slope_changes=[(x=0, slope_change=+2), (x=1, slope_change=+2), (x=2, slope_change=+2), (x=3, slope_change=+3)] 
Now, we want to find the minimum of c(x). Since c(x) is slope trickable, we know c(x) is concave up, which means c(x) has exactly one minimum, at the location where c(x) changes from negative to non-negative slope (i.e. the location where c(x) stops decreasing and starts increasing). So we can just loop through all the slope changes, find the location x where the slope goes from negative to positive, and output c(x). Here is the Scala code for this algorithm (the positions argument represents a sorted list of all the initial locations of the crabs):
def solvePart1(positions: Vector[Long]): Long = { //slope*x+yIntercept is the sum of -x+p1, -x+p2, ..., -x+pn over all positions p1, ..., pn //Therefore, slope is -(number of positions) and yIntercept is the sum of all the positions var slope = -positions.length var yIntercept = positions.sum for (pos <- positions) { //Use current slope and y-intercept to calculate c(pos) val fuelNeeded = slope*pos+yIntercept //Increment slope by 2 slope += 2 //If we went from a negative to a non-negative slope, that means this point is our minimum if (slope >= 0) { return fuelNeeded } //Calculate new y-intercept for next position yIntercept = fuelNeeded-slope*pos } throw new AssertionError("This point will never be reached!") } 
I really like this solution because (1) it's pretty short, even though it took a lot of thinking to come up with this and (2) once we've done this slope trick analysis, it's pretty easy to see why the median is the answer to Part 1. Because the slope of c(x) begins at -N (where N is the number of positions) and increases by 2 at every position. And the answer is where the slope goes from negative to non-negative, so clearly, the answer will be at the N/2th position, which is the median.
Now, how do we apply slope trick to part 2? Let's ask the same question we asked before: For any 1 <= i <= n, what's the cost of moving crab i to position x? Now, this cost is 1+2+...+|x-pi|=|x-pi|*(|x-pi|+1)/2. Now, this does not look like a linear function, it looks sort of quadratic, so how does slope trick apply? Remember that we only care about the value of this function at integer values of x, so instead of representing this function as a smooth quadratic function, we can represent it as a piecewise linear function with a bunch of abrupt slope changes at every integer. Moreover, if the maximum initial position of a crab is MAX_P, then we know that the answer is going to be in the interval [0, MAX_P], so we don't care about slope changes at x < 0 or x > MAX_P. Therefore, we only care about a finite number of slope changes: x=0, x=1, ..., x=MAX_P. And since we only have a finite number of slope changes, we can still use slope trick to represent this function.
Next, how do we represent |x-pi|*(|x-pi|+1)/2 using slope trick? First, we need to come up with y=mx+b, where m and b are the slope and y-intercept of this function for x < 0. The y-intercept is just the function evaluated at 0, which is pi*(pi+1)/2. And the slope is the slope of the function right before 0, i.e. the slope of the function from x=-1 to 0. And this slope represents the difference in cost between moving the crab to position x=0 and moving the crab to position x=-1. This would be the (pi+1)th step of moving the crab, so the difference in cost is -(pi+1). Ergo, our y=mx+b is y=(-pi-1)*x+[pi*(pi+1)/2].
Next, we need to figure out our slope changes. First, consider x^* < pi. What is the slope of the function from x=x^*-1 to x=x^* vs the slope from x=x^* to x=x^*+1? Well, the slope of the function over these intervals represents the difference in costs, which represents one step of motion of the crab. And the step of motion from x=x^*-1 to x=x^* is more costly by 1 unit than the step of motion from x=x^* to x=x^*+1, because we're moving farther away from the crab and every step of motion costs 1 more unit than the previous step of motion. Ergo, the slope from x=x^* to x=x^*+1 is less negative by 1 unit than the slope from x=x^* to x=x^*+1, so the slope change at x^* is -(-1)=+1 (the double negative comes from "less negative"). You can make an analogous argument to show the slope change at x^* is +1 for all x^* > pi.
And finally, what is the slope change at x=pi. Well, from x=pi-1 to x=pi, the function goes from 1 to 0, i.e. the slope is -1. And from x=pi to x=pi+1, the function goes from 0 to 1, i.e. the slope is +1. So the slope change at x=pi is +1-(-1)=+2. Thus, there is a slope change of +2 at x=pi and a slope change of +1 everywhere else. To make things more uniform, I will say there is a slope change of +1 at every integer x=0, ..., x=MAX_P, and there is an "extra" slope change of +1 at x=pi.
For example, let's consider the [3,0,1,2] example from before. Then, we would represent the |x-pi|*(|x-pi|+1)/2 functions using slope trick as follows:
y=-4x+6, slope_changes=[(x=0, slope_change=+1), (x=1, slope_change=+1), (x=2, slope_change=+1), (x=3, slope_change=+1), (x=3, slope_change=+1)] y=-1x+0, slope_changes=[(x=0, slope_change=+1), (x=0, slope_change=+1), (x=1, slope_change=+1), (x=2, slope_change=+1), (x=3, slope_change=+1)] y=-2x+1, slope_changes=[(x=0, slope_change=+1), (x=1, slope_change=+1), (x=1, slope_change=+1), (x=2, slope_change=+1), (x=3, slope_change=+1)] y=-3x+3, slope_changes=[(x=0, slope_change=+1), (x=1, slope_change=+1), (x=2, slope_change=+1), (x=2, slope_change=+1), (x=3, slope_change=+1)] 
Now, c(x) is the sum of |x-pi|*(|x-pi|+1)/2 over all positions pi. So to represent c(x) in slope trick, we first need to find y=mx+b by taking the sum of (-pi-1)*x+[pi*(pi+1)/2] over all positions pi. The slope m becomes -(sum of all positions)-N (where N is the number of positions) and the y-intercept b can be calculated with a simple loop taking the sum of all pi*(pi+1)/2. Next, we need to merge all the slope changes. For every 0 <= x <= MAX_P, the slope change at x is N, to account for the +1 slope change from every function |x-pi|*(|x-pi|+1)/2 that occurs at every integer, plus the number of positions equal to x, to account for the "extra" slope change of +1 at x=pi in |x-pi|*(|x-pi|+1)/2.
Now that we have c(x) in slope trick, we can follow the same algorithm of finding the position where c(x) goes from negative to non-negative slope that we did in Part 1. Here is the Scala implementation:
def solvePart2(positions: Vector[Long]): Long = { val maxPos = positions.max //At the beginning of each iteration of the below for loop, idx is the index pointing to the least element in positions which is greater than or equal to pos //This allows us to easily count the number of positions equal to pos var idx = 0 //Calculate initial slope and y-intercept var slope = -positions.sum-positions.length var yIntercept = positions.map(pos => (pos*(pos+1))/2).sum for (pos <- Range(0, maxPos.toInt+1, 1)) { //Use current slope and y-intercept to calculate c(pos) val fuelNeeded = slope*pos+yIntercept //Increment slope by N slope += positions.length //Increment slope by number of positions equal to pos while ((idx < positions.length) && (positions(idx) == pos)) { slope += 1 idx += 1 } //If we went from a negative to a non-negative slope, that means this point is our minimum if (slope >= 0) { return fuelNeeded } //Calculate new y-intercept for next position yIntercept = fuelNeeded-slope*pos } throw new AssertionError("This point will never be reached!") } 
And this gives us an O(N log N+MAX_P) solution to Part 2 (the N log N comes from sorting, which happens before this function is called), which isn't the best complexity possible (I believe the solution which uses the mean is O(N)), but is definitely better than O(N * MAX_P) which happens when you do naive brute force! And also, just like in Part 1, this slope trick analysis gives us an understanding why the answer is close to the mean: Because if we do some calculations, we'll find that the slope from x=x^* to x=x^*+1 is N*x^*+z(x^*)-S where S is the sum of all positions in the array and z(x^*) is the number of positions in the array less than or equal to x^*. We want to find the minimum integer x^* such that the slope N*x^*+z(x^*)-S >= 0, and we know that 0 <= z(x^*) <= N, so by solving the inequality, we get x^*=ceil((S-z)/N) for some 0 <= z <= N, i.e. ceil((S-N)/N) <= x^* <= ceil(S/N). And since ceil((S-N)/N)=ceil(S/N)-1, this means either x^*=ceil(S/N)-1 or x^*=ceil(S/N), which are the two closest integers to the exact mean S/N.
If you'd like to see my full Scala solution, you can find it on my GitHub here. Have fun, and thanks for reading!
submitted by Noble_Mushtak to adventofcode [link] [comments]


2021.12.07 22:04 throwawayzy126483 Quantum Frequency of this group.

Long story short is scientific studies have shown that intention collapses a particular quantum frequency. Blessed chocolate experiments, intention in water etc.
Group meditation is proven also etc.
Anyway just go on Google Scholar and look for people like Dean Radin.
What I’m suggesting is we put a shit ton of intention into this subreddit and make it one where remote viewing is really easy, happens amazingly easily.
Anyway I think we’ve got too much quantum info from sceptics and newbies in this place so those of us who know need to get the subreddit back and build it on a quantum level.
I for one know that this subreddit is where everyone comes to have a perfect remote viewing session. Anyone agree?
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2021.12.07 22:04 YeetGod11011 Rate my Present & Future Arsenal Squad.

Rate my Present & Future Arsenal Squad. submitted by YeetGod11011 to fut [link] [comments]


2021.12.07 22:04 Otherwise_Ranger_686 Join the ForTheBoys 18+ 🍑 Discord Server!

Join the ForTheBoys 18+ 🍑 Discord Server! submitted by Otherwise_Ranger_686 to AstridWet [link] [comments]


2021.12.07 22:04 Careless_Sail4116 22 [M4F] Germany - A Latino in Baden-Württemberg

Moving to Europe was a dream that I had ever since I was a little kid and I still can't believe it!
But moving to a new country also means starting from scratch, so someone who was already lonely only feels lonelier.
I'm a person who always tries to show how much you mean to me. I love writing long texts (and sometimes even letters!) describing how grateful I am for having you in my life. It doesn't matter if I have to wake up early, I'll always stay up late if it means getting to talk to you a little more :)
To me cuddles are the best thing in the world. Nothing beats that feeling of safety, of being cared for, of having someone that is there for you and above all, that feeling of knowing, that in that moment everything is alright.
I'd love to hear all you have to say, all about your passions (it's so amazing to hear someone talking exited about a subject, you can kind of hear the happiness in their voice!), all about your day or whatever topic you want to talk about.
I like writing good morning messages and sending a puppy/kitty video. Mornings are hard, sometimes it's hard to get out of bed, so why not have at least one good thing, no matter how small it is, to start your day?
I'll do everything and anything to get a smile from you :)
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2021.12.07 22:04 EntranceUpper Star Wars in VR is too cool!

Star Wars in VR is too cool! submitted by EntranceUpper to VRGaming [link] [comments]


2021.12.07 22:04 pixieO Have you used UIMA AS for text processing? With a number of python-based architectures currently available (spacy, sparkNLP...), is there any reason to invest in learning to use UIMA?

I would love to hear your experiences with UIMA and if you think there is any future for UIMA. I am wondering if all UIMA based systems will be soon replaced by something in Python.
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2021.12.07 22:04 NORDLAN We'll make sure Biden's Capital gains tax hike doesn't pass: U.S. Chamber of Commerce CEO

We'll make sure Biden's Capital gains tax hike doesn't pass: U.S. Chamber of Commerce CEO submitted by NORDLAN to RepublicanValues [link] [comments]


2021.12.07 22:04 trey2811 Does anyone know the Kohls Admin command to give cash?

I've looked in the commands and also looked online. I can't find the command. Does anyone know it?

Thanks.
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2021.12.07 22:04 engsuki trend #artvsartist2021 on twitter

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2021.12.07 22:04 SlightlyHazy Knoxville, Let's Go!!!

Just bought tickets for Knoxville in February! So stoked! This will be my first Big Something show and I cannot wait!!!
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2021.12.07 22:04 AggressiveCompany523 NBA Highlights - Nuggets vs. Bulls! Bulls comeback on the Nuggets for the W!

NBA Highlights - Nuggets vs. Bulls! Bulls comeback on the Nuggets for the W! submitted by AggressiveCompany523 to Selfpromote [link] [comments]


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