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Put numbers into different piles


Put numbers into different piles

By : bravo
Date : November 17 2020, 11:52 AM
I wish this help you You have several problems.
First of all, your n variable is not initialized. That should trigger a compiler error.
code :
if(all_dishes.at(i)>*piles.at(j).end())
if(all_dishes.at(i) > piles.at(j).back())


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Minimal stone piles

Minimal stone piles


By : Ming Fan
Date : March 29 2020, 07:55 AM
I wish did fix the issue. bestval is just set to -1 initially and is updated the first time around the loop to d. After that, bestval is updated again each time that d is a better value (aka smaller difference in weights) than the current bestval.
The key code is here...
code :
if bestval == -1 or d < bestval:
    bestval = d 
JQuery.on() piles up

JQuery.on() piles up


By : user3181918
Date : March 29 2020, 07:55 AM
I wish this help you I guess you want to achieve behaviour where even with multiple clicks, only 1 event listener is present. Although the following it's not ideal, but it's the most simple workaround. Use .off function to remove previous event listener.
code :
$(document).off('aCustomEventName');
$(document).on('aCustomEventName', function(){...});
$('#set').on('click', function () {
    $(document).off('aCustomEventName');
    $(document).on('aCustomEventName', function (event) {
        alert('event caught');
    });
});

$('#trigger').on('click', function () {
    $(document).trigger("aCustomEventName");
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div>
    Click on Set Listener button a few times then click on the Trigger Event button..
</div>

<button id="set">Set Listener</button>
<button id="trigger">Trigger Event</button>
Applescript to resort piles of numbers

Applescript to resort piles of numbers


By : bVelope
Date : March 29 2020, 07:55 AM
Any of those help To find and replace strings in AppleScript the native way is using text item delimiters. There are a fixed number of values separated by spaces (or tabs) on each line, using text item delimiters, text itemsand string concatenation we can solve your problem.
I've added an addition linefeed in front and at the back of the string to show that lines that doesn't contain 4 words are ignored.
code :
set theString to "
v 0.186472 0.578063 1.566364
v -0.186472 0.578063 1.566364
v 0.335649 0.578063 1.771483
"
set theLines to paragraphs of theString

set oldTIDs to AppleScript's text item delimiters


repeat with i from 1 to count theLines
    set AppleScript's text item delimiters to {space, tab}
    if (count of text items of item i of theLines) = 4 then
        set theNumbers to text items 2 thru -1 of item i of theLines
        set AppleScript's text item delimiters to ", "
        set item i of theLines to "(" & (theNumbers as string) & "),"
    else
        set item i of theLines to missing value
    end if
end repeat

set theLines to text of theLines
set AppleScript's text item delimiters to linefeed
set newString to theLines as string
set AppleScript's text item delimiters to oldTIDs
return newString
Sorting 3 piles of numbers row by row left to right

Sorting 3 piles of numbers row by row left to right


By : Duane
Date : March 29 2020, 07:55 AM
it should still fix some issue Hey I need to sort out 3 lists in a special way after they have been mixed up and placed like in a deck row by row (Link to the full question): , Just zip them:
code :
>>> a=[1,2,3]
>>> b=[4,5,6]
>>> c=[7,8,9]

>>> a, b, c = map(list, zip(b, a, c))  # in the order specified: b, a, c

>>> a
[4, 1, 7]
>>> b
[5, 2, 8]
>>> c
[6, 3, 9]
>>> a[:], b[:], c[:] = zip(b, a, c)
>>> a=[1,2,3,4]
>>> b=[5,6,7,8]
>>> c=[9,10,11,12]

>>> tmp = b + a + c   # concatenate them in the order

>>> # distribute them: every third element, starting with 0 (a), 1 (b) and 2 (c)
>>> a, b, c = tmp[::3], tmp[1::3], tmp[2::3]  

>>> a
[5, 8, 3, 10]
>>> b 
[6, 1, 4, 11]
>>> c
[7, 2, 9, 12]
Algorithms for distributing natural numbers in to equal piles

Algorithms for distributing natural numbers in to equal piles


By : Prodip Biswas
Date : March 29 2020, 07:55 AM
Does that help I would argue that there is no efficient way to solve this problem because it is a NP-hard problem.
Proof:
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