Factors of an integer: Difference between revisions
mNo edit summary |
mNo edit summary |
||
| Line 1: | Line 1: | ||
{{Task|Basic language learning}}{{basic data operation}} |
{{Task|Basic language learning}}{{basic data operation}} |
||
| ⚫ | |||
| ⚫ | |||
Compute the factors of a number. |
|||
=={{header|Clojure}}== |
=={{header|Clojure}}== |
||
| Line 34: | Line 37: | ||
end |
end |
||
p 45.factors</lang> |
p 45.factors</lang> |
||
| ⚫ | |||
| ⚫ | |||
Revision as of 16:56, 15 August 2009
You are encouraged to solve this task according to the task description, using any language you may know.
Basic Data Operation
This is a basic data operation. It represents a fundamental action on a basic data type.
You may see other such operations in the Basic Data Operations category, or:
Integer Operations
Arithmetic |
Comparison
Boolean Operations
Bitwise |
Logical
String Operations
Concatenation |
Interpolation |
Comparison |
Matching
Memory Operations
Pointers & references |
Addresses
Compute the factors of a number.
Clojure
<lang lisp>(defn factors [n] (filter #(zero? (rem n %)) (range 1 n)))
(print (factors 45))</lang>
(1 3 5 9 15)
Python
<lang python>>>> def factors(n): return [i for i in range(1,n//2) if not n%i]
>>> factors(45) [1, 3, 5, 9, 15]</lang>
Ruby
<lang ruby>class Integer
def factors() (1..self - 1).select { |n| (self % n).zero? } end
end p 45.factors</lang>
[1, 3, 5, 9, 15]
As we only have to loop up to , we can write <lang ruby>class Integer
def factors()
2.upto(Math.sqrt(self)).select {|i| (self % i).zero?} \
.inject([1]) {|f, i| f << i << self/i} \
.sort
end
end p 45.factors</lang>