## Count: Combination, sum of 4 integers = 100 [on hold]

Suppose we need 4 integers x, y, z, w that satisfy a property: sum (x, y, z, w) = 100
How many possible unique answers are there? Please explain the answers. I really have trouble understanding the nature of this problem. Thank you so much!

I became curious.
Are you counting well on PPD statistics?
Thank you very much

## How do I filter and count lodash.groupby in a typoscript array?

I want the statistics data from echartsjs with & # 39; center & # 39; Show.
Demo is: Enter the link description here

For example
I have an array,
T has some features, such as start time, end time and more …

``````class T{
public start: string;
public end: string;
}
``````

I use lodash.groupBy get a dictionary,

``````const groups = groupBy(data, function (item) {
return Math.abs(moment(item.end, 'YYYY-MM-DD').diff(moment(item.start.From, 'YYYY-MM-DD'), 'days'));
});
``````

I will build an xaxis array

`````` private  buildDateRangeXdata(): Array{
let xdata = Array();
xdata.push('0-7');
xdata.push('8-15');
xdata.push('16-30');
xdata.push('31-45');
xdata.push('46-60');
xdata.push('61+');
return xdata;
}
``````

I do not know how to find the data in xaxis compliant groups.
how can i do it?

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## Spells – Does a familiar action count on your own initiative?

In our group, we have always had a familiar action "at the same time" with their Master on a single Initiation Tally (the Master's).

But if you consider that

• A confidant can act independently
• has – in most cases because of his own skill – another initiatory modifier than his master
• can use the "delay action" or the "standby action" which automatically leads to a new position in the boot order

A confidant should rather have their own initiative count.

The problem now is – assuming the familiar is not worn, sitting on a shoulder, etc. – that it can not move together with his master, if he acts on his own initiative. Even if the familiar and the master act in immediate succession, someone has to move first.

If master and confidant want to stay together within 5 feet, the only way to accomplish this would be to move no more than 5 feet per turn, otherwise they would lose contact.

This would have consequences, especially with regard to the trusted share spells ability of the familiar.

Share Spells (PHB): … If the spell or effect has a duration other than "Instant", it no longer affects the known "if" it moves farther than 5 feet away and will not affect the familiar even if it returns to the Master before the end of the duration. (Highlighting me)

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## Count records from multiple columns by eliminating null values ​​in the Hive table

I'm using the following command to get the sum of records from 8 columns, but get zero in O / P [see below].

Command part 1

command part 2

output

How can this be fixed?

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## Performance – Count terribly slow from the sub-selection

I have the following problem.
This query in my database will run in 2 seconds

``````select
events.*,
events.start AS eventFrom,
events.title AS eventTitle,
event_types.name AS eventTypeName,
COUNT(DISTINCT event_speakers.person_id) AS speakersCnt,
COUNT(DISTINCT event_guests_registered.id) AS registeredCnt,
COUNT(DISTINCT event_guests_attended.id) AS attendedCnt,
COUNT(DISTINCT event_partnerships.id) AS partnershipsCnt,
(
SELECT SUM(amount)
WHERE event_id = events.id
) AS sponsorshipsAmount, IF(events.published_at,1,0) AS eventPublished from `events`

left join `event_speakers` on `event_speakers`.`event_id` = `events`.`id`
left join `event_guests_registered` on `event_guests_registered`.`event_id` = `events`.`id`
left join `event_guests_attended` on `event_guests_attended`.`event_id` = `events`.`id`
left join `event_partnerships` on `event_partnerships`.`event_id` =2 `events`.`id`
left join `event_types` on `event_types`.`id` = `events`.`event_type_id`

where `end` < NOW()
group by `events`.`id`
``````

But if I want to count it as a number of rows, adding select count () may affect performance
Query looks like this

``````select count(1) from (
select
events.id AS eventID,
events.start AS eventFrom,
events.title AS eventTitle,
event_types.name AS eventTypeName,
COUNT(DISTINCT event_speakers.person_id) AS speakersCnt,
COUNT(DISTINCT event_guests_registered.id) AS registeredCnt,
COUNT(DISTINCT event_guests_attended.id) AS attendedCnt,
COUNT(DISTINCT event_partnerships.id) AS partnershipsCnt,
(
SELECT SUM(amount)
WHERE event_id = events.id
) AS sponsorshipsAmount, IF(events.published_at,1,0) AS eventPublished from `events`

left join `event_speakers` on `event_speakers`.`event_id` = `events`.`id`
left join `event_guests_registered` on `event_guests_registered`.`event_id` = `events`.`id`
left join `event_guests_attended` on `event_guests_attended`.`event_id` = `events`.`id`
left join `event_partnerships` on `event_partnerships`.`event_id` = `events`.`id`
left join `event_types` on `event_types`.`id` = `events`.`event_type_id`

where `end` < NOW()
group by `events`.`id`) count_row_table
``````

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## So you count "+1" when a certain date occurs

Hello everybody!

I have this code, which shows me a future working day, so no weekends. I also want to exclude personalized holidays. In that case, the same thing happens (I think count ++).

For example, I want to exclude this data: ("2019-11-6", "2019-11-13") How can I integrate this into the code? I need this to indicate a future delivery date on a business day, excluding weekends and some custom data.

I have read and tested many manuals, but found nothing that works.

``````jQuery(function(\$) {
var monthNames = ("gennaio", "febbraio", "marzo", "aprile", "maggio", "giugno", "luglio", "agosto", "settembre", "ottobre", "novembre", "dicembre");
var dayNames = ("domenica", "lunedì", "martedì", "mercoledì", "giovedì", "venerdì", "sabato")

var endDate = "",
count = 0;
var someDate = new Date();
someDate.setDate(someDate.getDate());

endDate = new Date(someDate.setDate(someDate.getDate() + 1));
if (endDate.getDay() != 0 && endDate.getDay() != 6) {
count++;
}
}

\$('#Date').html(dayNames(endDate.getDay()) + ' ' + endDate.getDate() + ' ' + monthNames(endDate.getMonth()) + ' ' + endDate.getFullYear());
});``````

Code (JavaScript):

The code works well for excluding weekends, but I also want to help add custom holiday dates.

Thank you so much!

## Algorithms – Matrix element count in O (1) space

Check that the number of nonzero elements in each line is the maximum $$2$$ can be done in $$O (1)$$ Place in a row after the other by simply counting these elements.

So the only interesting part is checking the state about the columns.
This can be done in $$O (M)$$ Spaces by holding a counter for each column. In detail the counter $$c_i in {0,1,2 }$$ combined with $$i$$The -th column indicates the number of rows displayed so far that contain a nonzero item in the column $$i$$, Once a row with a nonzero item in the column $$i$$ seen and $$c_i$$ was already $$2$$you can return "false" immediately. If the entire matrix is ​​processed without this event, you can return true.

Unfortunately, your problem can not be solved $$x = o (M)$$ Bits.
Suppose this were the case and consider some matrices $$M$$ of columns and $$N = 2 million + 1$$ Lines, in the form described below.

The $$i$$-th of the first $$M$$ Lines have zeros in all columns $$j neq i$$while $$i$$-th column contains either a $$0$$ or a $$1$$,

The number of different choices for the first $$M$$ Lines is $$2 ^ M$$and yet there is at most $$2 ^ x <2 ^ M$$ (for big enough $$M$$) various possible states of the $$x$$ Bits of memory. This means that there are (at least) two different possibilities $$R$$ and $$R #$$ of the first $$M$$ Lines that leave the algorithm in the same internal state.

In other words, for every choice of $$(M + 1)$$-th to $$(2 million + 1)$$-ten lines, the algorithm must behave the same, regardless of whether the set of the first $$M$$ Rows was $$R$$ or $$R #$$,

Since $$R$$ and $$R #$$ are unique, there must be (at least) one index $$i$$ so that
the element $$r_i$$ by doing $$i$$-th column of $$i$$-th row of $$R$$ is different from the element $$r & # 39; i$$ by doing $$i$$-th column of $$i$$-th row of $$R #$$, Without limiting the general public $$r_i = 0$$ and $$r & # 39; i = 1$$,

To the $$i = 1, points, M$$, let them $$(M + i)$$-th line is the line that contains a single line $$1$$ by doing $$i$$-th column. Finally, choose the $$(2 million + 1)$$-th row to have exactly one $$1$$ by doing $$i$$-th column.

This leads to a contradiction: on the one hand, if the amount of the first $$M$$ Lines is $$R$$The algorithm must return true because there is at most one non-zero element per line and no more than two non-zero elements per column. on the other hand, if the first sentence $$M$$ Lines is $$R #$$the algorithm must return "false" since column $$i$$ Has $$3$$ Non-zero elements (in $$i$$-th, $$(M + i)$$-th and $$(2 million + 1)$$-th row).

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## postgresql – Count the number of records per month in the specified date range  Hello, I need to generate a query in which I count how many records per date service area each month exist. The dates in which there are no records with this service code this month show 0.
as I show in the table. They are how I expect the data (the table is created manually).

The second picture is a diagram of the DB
PD. In the service table I was wrong in the field Service_Name is service_name

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## Why does this Google Sheets formula count empty cells?

I'm new to Google Sheets, but use it for simple tasks only. Nevertheless, I came across a problem that I can not seem to solve. I start with a picture of my test google page below: If you look at the picture above, this is a screenshot of a test sheet that shows the problem I can not solve. Column A uses the formula = IF (E3 <> ", COUNTA (\$ E \$ 3: E3)," ") to count and number the values ​​in column E."

I then copied and pasted the values ​​from column A in column C and tried to perform the same calculation in column B in column C, with some changes made to the values. The same formula (now in column B) counts the values, but does not skip the empty cells anymore. Does anyone know why that is? I suspect it has something to do with the fact that Edit> Paste> Paste Only Values ​​does not work the way I expected, because there is now something in the empty columns of Column C that I can not see. Or maybe my understanding of the formula is off?

Summary: Why does the formula: = IF (E3 <> ", COUNTA (\$ E \$ 3: E3)," ") work between column A and column E, but not in the same way between column B and column C? I want it to count the values ​​AND skip the empty cells. This image is a small version of a much larger Google leaf, but the problem is the same. I need to be able to edit a column and then re-count / renumber the values.

Thanks!