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Hi guys
Need a little help again. Say I have an unordered list, and I want it to always have an even number of li's in it. How can I use jQuery to count the number of li's, and add one blank one at the end, if the number is odd?
If you're wondering why...
Started by Sam on
, 4 posts
by 4 people.
Answer Snippets (Read the full thread at stackoverflow):
;"); } });
This should add one list item to every unordered list with an odd number of list items..
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I want set some constraint to the serial type,it only produce even or odd numbers.
Started by yjfuk on
, 3 posts
by 3 people.
Answer Snippets (Read the full thread at stackoverflow):
Or 2 for producing either odd or even number:
Odd
CREATE SEQUENCE odd_seq INCREMENT BY 2 START.
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What is the fastest way to find if a number is even or odd?
Started by aks on
, 12 posts
by 12 people.
Answer Snippets (Read the full thread at stackoverflow):
Int is_odd(int num) { return num & 1; }
Check the least significant bit:
if (number & 0x01) { // It's odd } else { // It's even....
Bool is_odd = number & 1;
Check to see if the last bit is 1.
Be counted as even though.
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Ask your Facebook Friends
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For (unsigned int i = 1; i <= 100; i++) { if (i & 0x 1) { std::cout << i<<","; } }
why does (and how): if( i & 0x 1 ) figure out the odd number?
Started by Tom on
, 10 posts
by 10 people.
Answer Snippets (Read the full thread at stackoverflow):
So to find an odd number you have to see if the integer you're working with has (1,2,4,8,16 ..)....
Is that enough explanation?
When we look at numbers in base 10, it's easy to tell (-2,-1,0,1....).
The number is odd.
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Hi,
i want to generate a pseudo-random bool stream based on a modulo operation on another stream of integers (say X), so the operation would be
return ( X % 2);
The only problem is that X is a stream of integers that always ends in 1, so for instance ...
Started by oneAday on
, 4 posts
by 4 people.
Answer Snippets (Read the full thread at stackoverflow):
Use the penultimate bits instead:
return (x & 0x2) >> 1;
So say the next number from your.
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I have to write a program which reads in 3 numbers (using input boxes), and depending on their values it should write one of these messages:
All 3 numbers are odd OR All 3 numbers are even OR 2 numbers are odd and 1 is even OR 1 number is odd and 2 are...
Started by appreciation on
, 9 posts
by 9 people.
Answer Snippets (Read the full thread at stackoverflow):
So keeping count) ....
Every number is either ever or odd.
I don't think two variables are REQUIRED.
2 - You need two variables: one to hold the number of odd with alphazero.
Have to do:
1 - Get 3 numbers from the user.
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I'd like to find elisp's analog of:
sum(n for n in numbers if n % 2) # Python numbers.select { |n| n % 2 != 0 }.inject { |a, b| a + b } # Ruby
Imperative way:
(defun oddp (number) (not (= (mod number 2) 0))) (defun sum-odd-with-dolist (list) (let ((acc...
Started by J.F. Sebastian on
, 3 posts
by 3 people.
Answer Snippets (Read the full thread at stackoverflow):
I doubt ....
They come as standard with Emacs Lisp and there's nothing wrong with using them .
Apply '+ (delq nil (mapcar (lambda (x) (and (= 1 (% x 2)) x)) '(1 2 3 4 5))))
The idiomatic way to do it is to use the functions and macros in the cl packages .
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The problem is to print natural nos. 1,2,...n such that the parent process prints all odd numbers and child all even numbers using POSIX signals. Output should be: Parent : 1 Child : 2 Parent : 3 and so on...any suggestions?
Started by Sushant on
, 9 posts
by 7 people.
Answer Snippets (Read the full thread at stackoverflow):
*/ printf("Child: %lu\n", c */ for (;;) { /* Print ... .
; /* Child's first number will always be 2 */ if (c > n) exit(0); /* If c > n we have nothing to do */ do number, flush for proper output sequencing when output is not a terminal.
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M = [[1,2,3], [4,5,6], [7,8,9]] col2 = [row[1] + 1 for row in M if row[1] % 2 == 0] print (col2)
Output: [3, 9]
I'm expecting it to filter out the odd numbers, but it does the opposite.
Started by Nimbuz on
, 5 posts
by 5 people.
Answer Snippets (Read the full thread at stackoverflow):
M = [[1,2,3, but you are appending 3 to the result
when row[1]==5 , it is odd, so you filter it out
and when row[1]==8.
The modulus of any number divided by 2 is 0 or 1, 1 when it is odd.
To == 1 from == 0.
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I have a list of textual entries that a user can enter into the database and I need to validate these inputs with Regular Expressions because some of them are complex. One of fields must have gaps in the numbers (i.e., 10, 12, 14, 16...). My question ...
Started by James on
, 4 posts
by 4 people.
Answer Snippets (Read the full thread at stackoverflow):
So the Regex for odd number runs could be:
"^(\s*\d*[13579]\s*,)*(\s*\d*[13579]\s*)$"
Replace [13579Odd Numbers....
Or odd number can be tested by only looking at the last digit, which need to be even or odd too.
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