c++ - Smallest number that is evenly divisible by all of the numbers from 1 to 20? -

-

I had this problem [], but very poor programming method, looking at code in C ++, 

  # include & lt; Iostream & gt; using namespace std; // lowest divisble number up to 20 int () {int number = 20, flag = 0; While (flag == 0) {if ((2%) == 0 & amp; amp; & amp; amp; & amp; amp; & amp; amp; amp; amp; amp; amp; amp ; And amp; (number% 4) == 0 & amp; amp; amp; amp; amp;) == 0 & amp; (Number% 6) == 0 & amp; Amp; Amp; (Amp; 7%) == 0 & amp; Amp; Amp; (Number% 8) == 0 & amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; = 0 & amp; Amp; (Num% 10) == 0 & amp; Amp; Amp; Amp; Amp; (Amp; number% 11) == 0 & amp; (Number% 12) == 0 & amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; & Amp;; & Amp; Amp; Amp; Amp; (Number% 14) == 0 & amp; Amp; Amp; Amp; Amp; Amp; (Amp; %% 15) == 0 & amp; (Number% 16) == 0 & amp; (Number% 17) == 0 & amp; & Amp; (Number% 18) == 0 & amp; Amp; Amp; (Number% 19) == 0 & amp; Amp; Amp; Amp; Amp; Amp; & Amp; (Number% 20) == 0) {flag = 1; The court's & lt; & Lt; "The lowest divisible number is up to 20" & lt; & Lt; The number & lt; & Lt; Endl; } Num ++; }}

I was solving it in C ++ and was trapped in a loop, how would this step solve? ...

  • Think
    • P> I do not know how to use the control structures, so this step 

      if ((num% 2) == 0 & amp; (number% 3) == 0 ; Amp; amp; (num% 4) == 0 & amp; amp; & amp; amp; amp; & amp; amp; amp; amp; amp; amp; amp; amp ; & Amp; (Number% 6) == 0 & amp; 7) == 0 & amp; (Num% 8) == 0 & amp; Amp; Amp; Amp; (Amp; 9% 9) == 0 & amp; Amp; Amp; Amp; Amp; Amp; Amp; & Amp; (Number% 10) == 0 & amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; & Amp; == 0 & amp; amp; (Number% 12) == 0 & amp; Amp; Amp; Amp; (Amp; number% 13) == 0 & amp; (Num% 14) == 0 & amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp; Amp & amp ;; 0 & amp; Amp; Amp; (Number% 16) == 0 & amp; Amp; Amp; Amp; (Number% 17) == 0 & amp; (Number% 18) == 0 & amp; (Number% 19) == 0 & amp; ; & Amp; (Number% 20) == 0) `

    How to code it properly?

    The answer to this problem is: 

      Abhilash @ Abhilash: ~ $ ./a.out The lowest division Ible number 20 is 232792560

A quick way to answer this problem is, number theory Other answers are signs of how to do it. This answer is a better way to write the status if in your code.

If you want to change only the long term situation, you can better express it to the loop: 

  if ((num% 2) == 0 & amp; (num% 3) == 0 & amp; (num% 4) == 0 & amp; amp;; (number% 5) == 0 & amp; amp; amp; ; 6%) == 0 & amp; amp; amp; (num% 7) == 0 & amp; amp; & amp; amp; amp; amp; amp; amp; amp; amp; amp ; Amp; amp; amp; amp; amp; amp; amp; amp; amp; amp; amp;% number 9) == 0 & amp; amp; ; (Number% 10) == 0 & amp; Amp; Amp; (Amp; number% 11) == 0 & amp; Amp; Amp; (Number% 12) == 0 & amp; Amp; 13) == 0 & amp; (Num% 14) == 0 & amp; Amp; Amp; Amp; (Amp; number% 15) == 0 & amp; Amp; Amp; Amp; Amp; (Num% 16) == 0 & amp; Amp; Amp; Amp; Amp; & Amp; (Number% 17) == 0 & amp; Amp; (Number% 18) == 0 & amp; Amp; Amp; Amp; (Number% 19) == 0 & amp; (Number% 20) == 0) {...}

becomes: 

  {int divisor; Break (num% divisor! = 0) for (divisor = 2; divisor & lt; = 20; divisor ++); If (Separator! = 21) {...}}

The style is not very good, but I think this is what you wanted.


Comments

Post a Comment

Popular posts from this blog

oracle - The fastest way to check if some records in a database table? -

-
I have a big table to work with. I have to see that there are some records whose parent_id is equal to my passing value. Currently I do this by implementing "my_table using select count (*) where parent_id =: id"; If the result> 0, it means that they exist. Because this is a very large table, and I do not care what the actual number of records present, I just want to know if it exists, so I think count (*) Bit is disabled. How do I apply this requirement most quickly? I'm using Oracle 10. # Tips for Hibernation & amp; Tricks It suggests writing in such a way: Integer number = (integer) session .createQuery ("Select Count (*) to ...."). UniqueResult (); I do not know what is the magic of the unique rule? Why does it fast? To compare, select "1 from mytable where parent_id = passId and rowrum", which is more efficient? If you are not interested in the number of records, then there is an EXISTS query one: Select 'y'
Read more

php - multilevel menu with multilevel array -

-
Just ask someone to solve this problem? I want to create a multilevel menu I can not find the right solution generated with multilevel array, it always gets an array_push error My study came from the original idea But still my Can not match the need. This is my MySQL [code] ----- --------------------- --------------- | ID | PARENT_ID | Name | Link | Seq | 1 | 0 | Dashboad | Dashboard / | 1 | 2 | 0 | Menu 1 | Menu 1 / | 2 | 3 | 0 | Menu 2 | Menu 2 / | 3 | 4 | 0 | Menu 3 | Menu 3 / | 4 | 5 | 2 | Joint Add / | 1 | 6 | 3 | Joint Add / | 1 | 7 | 2 | Edit | Edit / | 2 | 8 | 4 | Joint Add / | 1 ------------------------------------------ [/ Code] I want to do something like this in my array array ('dashboard' = & gt; array ('id' = & gt; '' 'Dame' = & gt; 'Dashboard', 'Title' = & gt; 'Dashboard', 'MatchMin' = & gt; Base_URL (). 'Dashboard /', 'Active' => = & Gt; NULL), &
Read more

jQuery UI: Datepicker month format -

-
I am trying to change the format of the month that shows in the popup (not in the date that the picker actually I do not have any choice anywhere to do this, do I like to use a sub-string () or something else? What I normally display is Trying to change: Class = "ui-datepicker-title"> January & lt; / span & gt; ... & lt; / div & gt; Div>
Read more