Factors of an integer: Difference between revisions
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p 45.factors</lang> |
p 45.factors</lang> |
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=={{header|Tcl}}== |
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<lang tcl>proc factors {n} { |
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set factors {1} |
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for {set i 2} {$i <= sqrt($n)} {incr i} { |
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if {$n % $i == 0} { |
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lappend factors $i [expr {$n / $i}] |
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} |
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} |
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return [lsort -unique -integer $factors] |
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} |
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puts [factors 64] |
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puts [factors 45] |
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puts [factors 53]</lang> |
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output |
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<pre>1 2 4 8 16 32 |
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1 3 5 9 15 |
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1</pre> |
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Revision as of 17:08, 15 August 2009
You are encouraged to solve this task according to the task description, using any language you may know.
Basic Data Operation
This is a basic data operation. It represents a fundamental action on a basic data type.
You may see other such operations in the Basic Data Operations category, or:
Integer Operations
Arithmetic |
Comparison
Boolean Operations
Bitwise |
Logical
String Operations
Concatenation |
Interpolation |
Comparison |
Matching
Memory Operations
Pointers & references |
Addresses
Compute the factors of a number.
See also Prime decomposition
Clojure
<lang lisp>(defn factors [n] (filter #(zero? (rem n %)) (range 1 n)))
(print (factors 45))</lang>
(1 3 5 9 15)
Python
<lang python>>>> def factors(n): return [i for i in range(1,n//2) if not n%i]
>>> factors(45) [1, 3, 5, 9, 15]</lang>
Ruby
<lang ruby>class Integer
def factors() (1..self - 1).select { |n| (self % n).zero? } end
end p 45.factors</lang>
[1, 3, 5, 9, 15]
As we only have to loop up to , we can write <lang ruby>class Integer
def factors()
2.upto(Math.sqrt(self)).select {|i| (self % i).zero?} \
.inject([1]) {|f, i| f << i << self/i} \
.sort
end
end p 45.factors</lang>
Tcl
<lang tcl>proc factors {n} {
set factors {1}
for {set i 2} {$i <= sqrt($n)} {incr i} {
if {$n % $i == 0} {
lappend factors $i [expr {$n / $i}]
}
}
return [lsort -unique -integer $factors]
} puts [factors 64] puts [factors 45] puts [factors 53]</lang> output
1 2 4 8 16 32 1 3 5 9 15 1