Factors of an integer: Difference between revisions
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Test: |
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<lang funl> |
<lang funl>for x <- [103, 316, 519, 639, 760] |
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for x <- [103, 316, 519, 639, 760] |
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println( 'The set of factors of ' + x + ' is ' + factors(x) )</lang> |
println( 'The set of factors of ' + x + ' is ' + factors(x) )</lang> |
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Revision as of 18:41, 17 July 2014
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 positive integer. These factors are the positive integers by which the number being factored can be divided to yield a positive integer result (though the concepts function correctly for zero and negative integers, the set of factors of zero has countably infinite members, and the factors of negative integers can be obtained from the factors of related positive numbers without difficulty; this task does not require handling of either of these cases). Note that even prime numbers will have at least two factors; ‘1’ and themselves.
See also:
0815
<lang 0815> <:1:~>|~#:end:>~x}:str:/={^:wei:~%x<:a:x=$~ =}:wei:x<:1:+{>~>x=-#:fin:^:str:}:fin:{{~% </lang>
ACL2
<lang Lisp>(defun factors-r (n i)
(declare (xargs :measure (nfix (- n i))))
(cond ((zp (- n i))
(list n))
((= (mod n i) 0)
(cons i (factors-r n (1+ i))))
(t (factors-r n (1+ i)))))
(defun factors (n)
(factors-r n 1))</lang>
ActionScript
<lang ActionScript>function factor(n:uint):Vector.<uint> { var factors:Vector.<uint> = new Vector.<uint>(); for(var i:uint = 1; i <= n; i++) if(n % i == 0)factors.push(i); return factors; }</lang>
Ada
<lang Ada>with Ada.Text_IO; with Ada.Command_Line; procedure Factors is
Number : Positive; Test_Nr : Positive := 1;
begin
if Ada.Command_Line.Argument_Count /= 1 then
Ada.Text_IO.Put (Ada.Text_IO.Standard_Error, "Missing argument!");
Ada.Command_Line.Set_Exit_Status (Ada.Command_Line.Failure);
return;
end if;
Number := Positive'Value (Ada.Command_Line.Argument (1));
Ada.Text_IO.Put ("Factors of" & Positive'Image (Number) & ": ");
loop
if Number mod Test_Nr = 0 then
Ada.Text_IO.Put (Positive'Image (Test_Nr) & ",");
end if;
exit when Test_Nr ** 2 >= Number;
Test_Nr := Test_Nr + 1;
end loop;
Ada.Text_IO.Put_Line (Positive'Image (Number) & ".");
end Factors;</lang>
Aikido
<lang aikido>import math
function factor (n:int) {
var result = []
function append (v) {
if (!(v in result)) {
result.append (v)
}
}
var sqrt = cast<int>(Math.sqrt (n))
append (1)
for (var i = n-1 ; i >= sqrt ; i--) {
if ((n % i) == 0) {
append (i)
append (n/i)
}
}
append (n)
return result.sort()
}
function printvec (vec) {
var comma = ""
print ("[")
foreach v vec {
print (comma + v)
comma = ", "
}
println ("]")
}
printvec (factor (45)) printvec (factor (25)) printvec (factor (100))</lang>
ALGOL 68
Note: The following implements generators, eliminating the need of declaring arbitrarily long int arrays for caching. <lang algol68>MODE YIELDINT = PROC(INT)VOID;
PROC gen factors = (INT n, YIELDINT yield)VOID: (
FOR i FROM 1 TO ENTIER sqrt(n) DO
IF n MOD i = 0 THEN
yield(i);
INT other = n OVER i;
IF i NE other THEN yield(n OVER i) FI
FI
OD
);
[]INT nums2factor = (45, 53, 64);
FOR i TO UPB nums2factor DO
INT num = nums2factor[i]; STRING sep := ": "; print(num);
- FOR INT j IN # gen factors(num, # ) DO ( #
- (INT j)VOID:(
print((sep,whole(j,0)));
sep:=", "
- OD # ));
print(new line)
OD</lang> Output:
+45: 1, 45, 3, 15, 5, 9
+53: 1, 53
+64: 1, 64, 2, 32, 4, 16, 8
APL
<lang APL> factors←{(0=(⍳⍵)|⍵)/⍳⍵}
factors 12345
1 3 5 15 823 2469 4115 12345
factors 720
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 30 36 40 45 48 60 72 80 90 120 144 180 240 360 720</lang>
AutoHotkey
<lang AutoHotkey>msgbox, % factors(45) "`n" factors(53) "`n" factors(64)
Factors(n) { Loop, % floor(sqrt(n))
{ v := A_Index = 1 ? 1 "," n : mod(n,A_Index) ? v : v "," A_Index "," n//A_Index
}
Sort, v, N U D,
Return, v
}</lang>
Output: 1,3,5,9,15,45 1,53 1,2,4,8,16,32,64
AutoIt
<lang AutoIt>;AutoIt Version: 3.2.10.0 $num = 45 MsgBox (0,"Factors", "Factors of " & $num & " are: " & factors($num)) consolewrite ("Factors of " & $num & " are: " & factors($num)) Func factors($intg)
$ls_factors=""
For $i = 1 to $intg/2
if ($intg/$i - int($intg/$i))=0 Then
$ls_factors=$ls_factors&$i &", "
EndIf Next Return $ls_factors&$intg
EndFunc</lang>
Output: Factors of 45 are: 1, 3, 5, 9, 15, 45
AWK
<lang AWK>
- syntax: GAWK -f FACTORS_OF_AN_INTEGER.AWK
BEGIN {
print("enter a number or C/R to exit")
} { if ($0 == "") { exit(0) }
if ($0 !~ /^[0-9]+$/) {
printf("invalid: %s\n",$0)
next
}
n = $0
printf("factors of %s:",n)
for (i=1; i<=n; i++) {
if (n % i == 0) {
printf(" %d",i)
}
}
printf("\n")
} </lang>
output:
enter a number or C/R to exit invalid: -1 factors of 0: factors of 1: 1 factors of 2: 1 2 factors of 11: 1 11 factors of 64: 1 2 4 8 16 32 64 factors of 100: 1 2 4 5 10 20 25 50 100 factors of 32766: 1 2 3 6 43 86 127 129 254 258 381 762 5461 10922 16383 32766 factors of 32767: 1 7 31 151 217 1057 4681 32767
BASIC
This example stores the factors in a shared array (with the original number as the last element) for later retrieval.
Note that this will error out if you pass 32767 (or higher). <lang qbasic>DECLARE SUB factor (what AS INTEGER)
REDIM SHARED factors(0) AS INTEGER
DIM i AS INTEGER, L AS INTEGER
INPUT "Gimme a number"; i
factor i
PRINT factors(0); FOR L = 1 TO UBOUND(factors)
PRINT ","; factors(L);
NEXT PRINT
SUB factor (what AS INTEGER)
DIM tmpint1 AS INTEGER DIM L0 AS INTEGER, L1 AS INTEGER
REDIM tmp(0) AS INTEGER REDIM factors(0) AS INTEGER factors(0) = 1
FOR L0 = 2 TO what
IF (0 = (what MOD L0)) THEN
'all this REDIMing and copying can be replaced with:
'REDIM PRESERVE factors(UBOUND(factors)+1)
'in languages that support the PRESERVE keyword
REDIM tmp(UBOUND(factors)) AS INTEGER
FOR L1 = 0 TO UBOUND(factors)
tmp(L1) = factors(L1)
NEXT
REDIM factors(UBOUND(factors) + 1)
FOR L1 = 0 TO UBOUND(factors) - 1
factors(L1) = tmp(L1)
NEXT
factors(UBOUND(factors)) = L0
END IF
NEXT
END SUB</lang>
Sample outputs:
Gimme a number? 17 1 , 17 Gimme a number? 12345 1 , 3 , 5 , 15 , 823 , 2469 , 4115 , 12345 Gimme a number? 32765 1 , 5 , 6553 , 32765 Gimme a number? 32766 1 , 2 , 3 , 6 , 43 , 86 , 127 , 129 , 254 , 258 , 381 , 762 , 5461 , 10922 , 16383 , 32766
Batch File
Command line version: <lang dos>@echo off set res=Factors of %1: for /L %%i in (1,1,%1) do call :fac %1 %%i echo %res% goto :eof
- fac
set /a test = %1 %% %2 if %test% equ 0 set res=%res% %2</lang>
Outputs:
>factors 32767 Factors of 32767: 1 7 31 151 217 1057 4681 32767 >factors 45 Factors of 45: 1 3 5 9 15 45 >factors 53 Factors of 53: 1 53 >factors 64 Factors of 64: 1 2 4 8 16 32 64 >factors 100 Factors of 100: 1 2 4 5 10 20 25 50 100
Interactive version: <lang dos>@echo off set /p limit=Gimme a number: set res=Factors of %limit%: for /L %%i in (1,1,%limit%) do call :fac %limit% %%i echo %res% goto :eof
- fac
set /a test = %1 %% %2 if %test% equ 0 set res=%res% %2</lang>
Outputs:
>factors Gimme a number:27 Factors of 27: 1 3 9 27 >factors Gimme a number:102 Factors of 102: 1 2 3 6 17 34 51 102
BBC BASIC
<lang bbcbasic> INSTALL @lib$+"SORTLIB"
sort% = FN_sortinit(0, 0)
PRINT "The factors of 45 are " FNfactorlist(45)
PRINT "The factors of 12345 are " FNfactorlist(12345)
END
DEF FNfactorlist(N%)
LOCAL C%, I%, L%(), L$
DIM L%(32)
FOR I% = 1 TO SQR(N%)
IF (N% MOD I% = 0) THEN
L%(C%) = I%
C% += 1
IF (N% <> I%^2) THEN
L%(C%) = (N% DIV I%)
C% += 1
ENDIF
ENDIF
NEXT I%
CALL sort%, L%(0)
FOR I% = 0 TO C%-1
L$ += STR$(L%(I%)) + ", "
NEXT
= LEFT$(LEFT$(L$))</lang>
Output:
The factors of 45 are 1, 3, 5, 9, 15, 45 The factors of 12345 are 1, 3, 5, 15, 823, 2469, 4115, 12345
bc
<lang bc>/* Calculate the factors of n and return their count.
* This function mutates the global array f[] which will * contain all factors of n in ascending order after the call! */
define f(n) {
auto i, d, h, h[], l, o
/* Local variables:
* i: Loop variable.
* d: Complementary (higher) factor to i.
* h: Will always point to the last element of h[].
* h[]: Array to hold the greater factor of the pair (x, y), where
* x * y == n. The factors are stored in descending order.
* l: Will always point to the next free spot in f[].
* o: For saving the value of scale.
*/
/* Use integer arithmetic */ o = scale scale = 0
/* Two factors are 1 and n (if n != 1) */ f[l++] = 1 if (n == 1) return(1) h[0] = n
/* Main loop */
for (i = 2; i < h[h]; i++) {
if (n % i == 0) {
d = n / i
if (d != i) {
h[++h] = d
}
f[l++] = i
}
}
/* Append the values in h[] to f[] */
while (h >= 0) {
f[l++] = h[h--]
}
scale = o return(l)
}</lang>
C
<lang c>#include <stdio.h>
- include <stdlib.h>
typedef struct {
int *list; short count;
} Factors;
void xferFactors( Factors *fctrs, int *flist, int flix ) {
int ix, ij;
int newSize = fctrs->count + flix;
if (newSize > flix) {
fctrs->list = realloc( fctrs->list, newSize * sizeof(int));
}
else {
fctrs->list = malloc( newSize * sizeof(int));
}
for (ij=0,ix=fctrs->count; ix<newSize; ij++,ix++) {
fctrs->list[ix] = flist[ij];
}
fctrs->count = newSize;
}
Factors *factor( int num, Factors *fctrs) {
int flist[301], flix;
int dvsr;
flix = 0;
fctrs->count = 0;
free(fctrs->list);
fctrs->list = NULL;
for (dvsr=1; dvsr*dvsr < num; dvsr++) {
if (num % dvsr != 0) continue;
if ( flix == 300) {
xferFactors( fctrs, flist, flix );
flix = 0;
}
flist[flix++] = dvsr;
flist[flix++] = num/dvsr;
}
if (dvsr*dvsr == num)
flist[flix++] = dvsr;
if (flix > 0)
xferFactors( fctrs, flist, flix );
return fctrs;
}
int main(int argc, char*argv[]) {
int nums2factor[] = { 2059, 223092870, 3135, 45 };
Factors ftors = { NULL, 0};
char sep;
int i,j;
for (i=0; i<4; i++) {
factor( nums2factor[i], &ftors );
printf("\nfactors of %d are:\n ", nums2factor[i]);
sep = ' ';
for (j=0; j<ftors.count; j++) {
printf("%c %d", sep, ftors.list[j]);
sep = ',';
}
printf("\n");
}
return 0;
}</lang>
Prime factoring
<lang C>#include <stdio.h>
- include <stdlib.h>
- include <string.h>
/* 65536 = 2^16, so we can factor all 32 bit ints */ char bits[65536];
typedef unsigned long ulong; ulong primes[7000], n_primes;
typedef struct { ulong p, e; } prime_factor; /* prime, exponent */
void sieve() { int i, j; memset(bits, 1, 65536); bits[0] = bits[1] = 0; for (i = 0; i < 256; i++) if (bits[i]) for (j = i * i; j < 65536; j += i) bits[j] = 0;
/* collect primes into a list. slightly faster this way if dealing with large numbers */ for (i = j = 0; i < 65536; i++) if (bits[i]) primes[j++] = i;
n_primes = j; }
int get_prime_factors(ulong n, prime_factor *lst) { ulong i, e, p; int len = 0;
for (i = 0; i < n_primes; i++) { p = primes[i]; if (p * p > n) break; for (e = 0; !(n % p); n /= p, e++); if (e) { lst[len].p = p; lst[len++].e = e; } }
return n == 1 ? len : (lst[len].p = n, lst[len].e = 1, ++len); }
int ulong_cmp(const void *a, const void *b) { return *(const ulong*)a < *(const ulong*)b ? -1 : *(const ulong*)a > *(const ulong*)b; }
int get_factors(ulong n, ulong *lst) { int n_f, len, len2, i, j, k, p; prime_factor f[100];
n_f = get_prime_factors(n, f);
len2 = len = lst[0] = 1; /* L = (1); L = (L, L * p**(1 .. e)) forall((p, e)) */ for (i = 0; i < n_f; i++, len2 = len) for (j = 0, p = f[i].p; j < f[i].e; j++, p *= f[i].p) for (k = 0; k < len2; k++) lst[len++] = lst[k] * p;
qsort(lst, len, sizeof(ulong), ulong_cmp); return len; }
int main() { ulong fac[10000]; int len, i, j; ulong nums[] = {3, 120, 1024, 2UL*2*2*2*3*3*3*5*5*7*11*13*17*19 };
sieve();
for (i = 0; i < 4; i++) { len = get_factors(nums[i], fac); printf("%lu:", nums[i]); for (j = 0; j < len; j++) printf(" %lu", fac[j]); printf("\n"); }
return 0; }</lang>output
3: 1 3 120: 1 2 3 4 5 6 8 10 12 15 20 24 30 40 60 120 1024: 1 2 4 8 16 32 64 128 256 512 1024 3491888400: 1 2 3 4 5 6 7 8 9 10 11 ...(>1900 numbers)... 1163962800 1745944200 3491888400
C++
<lang Cpp>#include <iostream>
- include <vector>
- include <algorithm>
- include <iterator>
std::vector<int> GenerateFactors(int n) {
std::vector<int> factors;
factors.push_back(1);
factors.push_back(n);
for(int i = 2; i * i <= n; ++i)
{
if(n % i == 0)
{
factors.push_back(i);
if(i * i != n)
factors.push_back(n / i);
}
}
std::sort(factors.begin(), factors.end()); return factors;
}
int main() {
const int SampleNumbers[] = {3135, 45, 60, 81};
for(size_t i = 0; i < sizeof(SampleNumbers) / sizeof(int); ++i)
{
std::vector<int> factors = GenerateFactors(SampleNumbers[i]);
std::cout << "Factors of " << SampleNumbers[i] << " are:\n";
std::copy(factors.begin(), factors.end(), std::ostream_iterator<int>(std::cout, "\n"));
std::cout << std::endl;
}
}</lang>
C#
C# 3.0 <lang csharp>using System; using System.Linq; using System.Collections.Generic;
public static class Extension {
public static List<int> Factors(this int me)
{
return Enumerable.Range(1, me).Where(x => me % x == 0).ToList();
}
}
class Program {
static void Main(string[] args)
{
Console.WriteLine(String.Join(", ", 45.Factors()));
}
}</lang>
C# 1.0 <lang csharp>static void Main(string[] args) { do { Console.WriteLine("Number:"); Int64 p = 0; do { try { p = Convert.ToInt64(Console.ReadLine()); break; } catch (Exception) { }
} while (true);
Console.WriteLine("For 1 through " + ((int)Math.Sqrt(p)).ToString() + ""); for (int x = 1; x <= (int)Math.Sqrt(p); x++) { if (p % x == 0) Console.WriteLine("Found: " + x.ToString() + ". " + p.ToString() + " / " + x.ToString() + " = " + (p / x).ToString()); }
Console.WriteLine("Done."); } while (true); }</lang>
Example output:
Number: 32434243 For 1 through 5695 Found: 1. 32434243 / 1 = 32434243 Found: 307. 32434243 / 307 = 105649 Done.
Chapel
Inspired by the Clojure solution: <lang chapel>iter factors(n) { for i in 1..floor(sqrt(n)):int { if n % i == 0 then { yield i; yield n / i; } } }</lang>
Clojure
<lang lisp>(defn factors [n] (filter #(zero? (rem n %)) (range 1 (inc n))))
(print (factors 45))</lang>
(1 3 5 9 15 45)
Improved version. Considers small factors from 1 up to (sqrt n) -- we increment it because range does not include the end point. Pair each small factor with its co-factor, flattening the results, and put them into a sorted set to get the factors in order. <lang lisp>(defn factors [n]
(into (sorted-set)
(mapcat (fn [x] [x (/ n x)])
(filter #(zero? (rem n %)) (range 1 (inc (Math/sqrt n)))) )))</lang>
Same idea, using for comprehensions. <lang lisp>(defn factors [n]
(into (sorted-set)
(reduce concat
(for [x (range 1 (inc (Math/sqrt n))) :when (zero? (rem n x))]
[x (/ n x)]))))</lang>
CoffeeScript
<lang coffeescript># Reference implementation for finding factors is slow, but hopefully
- robust--we'll use it to verify the more complicated (but hopefully faster)
- algorithm.
slow_factors = (n) ->
(i for i in [1..n] when n % i == 0)
- The rest of this code does two optimizations:
- 1) When you find a prime factor, divide it out of n (smallest_prime_factor).
- 2) Find the prime factorization first, then compute composite factors from those.
smallest_prime_factor = (n) ->
for i in [2..n] return n if i*i > n return i if n % i == 0
prime_factors = (n) ->
return {} if n == 1
spf = smallest_prime_factor n
result = prime_factors(n / spf)
result[spf] or= 0
result[spf] += 1
result
fast_factors = (n) ->
prime_hash = prime_factors n
exponents = []
for p of prime_hash
exponents.push
p: p
exp: 0
result = []
while true
factor = 1
for obj in exponents
factor *= Math.pow obj.p, obj.exp
result.push factor
break if factor == n
# roll the odometer
for obj, i in exponents
if obj.exp < prime_hash[obj.p]
obj.exp += 1
break
else
obj.exp = 0
return result.sort (a, b) -> a - b
verify_factors = (factors, n) ->
expected_result = slow_factors n
throw Error("wrong length") if factors.length != expected_result.length
for factor, i in expected_result
console.log Error("wrong value") if factors[i] != factor
for n in [1, 3, 4, 8, 24, 37, 1001, 11111111111, 99999999999]
factors = fast_factors n console.log n, factors if n < 1000000 verify_factors factors, n</lang>
output <lang coffeescript>> coffee factors.coffee 1 [ 1 ] 3 [ 1, 3 ] 4 [ 1, 2, 4 ] 8 [ 1, 2, 4, 8 ] 24 [ 1, 2, 3, 4, 6, 8, 12, 24 ] 37 [ 1, 37 ] 1001 [ 1, 7, 11, 13, 77, 91, 143, 1001 ] 11111111111 [ 1, 21649, 513239, 11111111111 ] 99999999999 [ 1,
3, 9, 21649, 64947, 194841, 513239, 1539717, 4619151, 11111111111, 33333333333, 99999999999 ]</lang>
Common Lisp
We iterate in the range 1..sqrt(n) collecting ‘low’ factors and corresponding ‘high’ factors, and combine at the end to produce an ordered list of factors.
<lang lisp>(defun factors (n &aux (lows '()) (highs '()))
(do ((limit (isqrt n)) (factor 1 (1+ factor)))
((= factor limit)
(when (= n (* limit limit))
(push limit highs))
(nreconc lows highs))
(multiple-value-bind (quotient remainder) (floor n factor)
(when (zerop remainder)
(push factor lows)
(push quotient highs)))))</lang>
D
Procedural Style
<lang d>import std.stdio, std.math, std.algorithm;
T[] factors(T)(in T n) pure nothrow {
if (n == 1)
return [n];
T[] res = [1, n];
T limit = cast(T)real(n).sqrt + 1;
for (T i = 2; i < limit; i++) {
if (n % i == 0) {
res ~= i;
immutable q = n / i;
if (q > i)
res ~= q;
}
}
return res.sort().release;
}
void main() {
writefln("%(%s\n%)", [45, 53, 64, 1111111].map!factors);
}</lang>
- Output:
[1, 3, 5, 9, 15, 45] [1, 53] [1, 2, 4, 8, 16, 32, 64] [1, 239, 4649, 1111111]
Functional Style
<lang d>import std.stdio, std.algorithm, std.range;
auto factors(I)(I n) {
return iota(1, n + 1).filter!(i => n % i == 0);
}
void main() {
36.factors.writeln;
}</lang>
- Output:
[1, 2, 3, 4, 6, 9, 12, 18, 36]
E
<lang e>def factors(x :(int > 0)) {
var xfactors := []
for f ? (x % f <=> 0) in 1..x {
xfactors with= f
}
return xfactors
}</lang>
Ela
Using higher-order function
<lang ela>open list
factors m = filter (\x -> m % x == 0) [1..m]</lang>
Using comprehension
<lang ela>factors m = [x \\ x <- [1..m] | m % x == 0]</lang>
Erlang
<lang erlang>factors(N) ->
[I || I <- lists:seq(1,trunc(N/2)), N rem I == 0]++[N].</lang>
F#
If number % divisor = 0 then both divisor AND number / divisor are factors.
So, we only have to search till sqrt(number).
Also, this is lazily evaluated. <lang fsharp>let factors number = seq {
for divisor in 1 .. (float >> sqrt >> int) number do
if number % divisor = 0 then
yield divisor
yield number / divisor
}</lang>
Factor
USE: math.primes.factors
( scratchpad ) 24 divisors .
{ 1 2 3 4 6 8 12 24 }
FALSE
<lang false>[1[\$@$@-][\$@$@$@$@\/*=[$." "]?1+]#.%]f: 45f;! 53f;! 64f;!</lang>
Forth
This is a slightly optimized algorithm, since it realizes there are no factors between n/2 and n. The values are saved on the stack and - in true Forth fashion - printed in descending order. <lang Forth>: factors dup 2/ 1+ 1 do dup i mod 0= if i swap then loop ;
- .factors factors begin dup dup . 1 <> while drop repeat drop cr ;
45 .factors 53 .factors 64 .factors 100 .factors</lang>
Fortran
<lang fortran>program Factors
implicit none integer :: i, number write(*,*) "Enter a number between 1 and 2147483647" read*, number
do i = 1, int(sqrt(real(number))) - 1
if (mod(number, i) == 0) write (*,*) i, number/i
end do
! Check to see if number is a square
i = int(sqrt(real(number)))
if (i*i == number) then
write (*,*) i
else if (mod(number, i) == 0) then
write (*,*) i, number/i
end if
end program</lang>
FunL
Function to compute set of factors: <lang funl>def factors( n ) = {d | d <- 1..n if d|n}</lang>
Test: <lang funl>for x <- [103, 316, 519, 639, 760]
println( 'The set of factors of ' + x + ' is ' + factors(x) )</lang>
- Output:
The set of factors of 103 is {1, 103}
The set of factors of 316 is {158, 4, 79, 1, 2, 316}
The set of factors of 519 is {1, 3, 173, 519}
The set of factors of 639 is {9, 639, 71, 213, 1, 3}
The set of factors of 760 is {8, 19, 4, 40, 152, 5, 10, 76, 1, 95, 190, 760, 20, 2, 38, 380}
GAP
<lang gap># Built-in function DivisorsInt(Factorial(5));
- [ 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 ]
- A possible implementation, not suitable to large n
div := n -> Filtered([1 .. n], k -> n mod k = 0);
div(Factorial(5));
- [ 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 ]
- Another implementation, usable for large n (if n can be factored quickly)
div2 := function(n)
local f, p; f := Collected(FactorsInt(n)); p := List(f, v -> List([0 .. v[2]], k -> v[1]^k)); return SortedList(List(Cartesian(p), Product));
end;
div2(Factorial(5));
- [ 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 ]</lang>
Go
Trial division, no prime number generator, but with some optimizations. It's good enough to factor any 64 bit integer, with large primes taking several seconds. <lang go>package main
import "fmt"
func main() {
printFactors(-1) printFactors(0) printFactors(1) printFactors(2) printFactors(3) printFactors(53) printFactors(45) printFactors(64) printFactors(600851475143) printFactors(999999999999999989)
}
func printFactors(nr int64) {
if nr < 1 {
fmt.Println("\nFactors of", nr, "not computed")
return
}
fmt.Printf("\nFactors of %d: ", nr)
fs := make([]int64, 1)
fs[0] = 1
apf := func(p int64, e int) {
n := len(fs)
for i, pp := 0, p; i < e; i, pp = i+1, pp*p {
for j := 0; j < n; j++ {
fs = append(fs, fs[j]*pp)
}
}
}
e := 0
for ; nr & 1 == 0; e++ {
nr >>= 1
}
apf(2, e)
for d := int64(3); nr > 1; d += 2 {
if d*d > nr {
d = nr
}
for e = 0; nr%d == 0; e++ {
nr /= d
}
if e > 0 {
apf(d, e)
}
}
fmt.Println(fs)
fmt.Println("Number of factors =", len(fs))
}</lang> Output:
Factors of -1 not computed Factors of 0 not computed Factors of 1: [1] Number of factors = 1 Factors of 2: [1 2] Number of factors = 2 Factors of 3: [1 3] Number of factors = 2 Factors of 53: [1 53] Number of factors = 2 Factors of 45: [1 3 9 5 15 45] Number of factors = 6 Factors of 64: [1 2 4 8 16 32 64] Number of factors = 7 Factors of 600851475143: [1 71 839 59569 1471 104441 1234169 87625999 6857 486847 5753023 408464633 10086647 716151937 8462696833 600851475143] Number of factors = 16 Factors of 999999999999999989: [1 999999999999999989] Number of factors = 2
Groovy
A straight brute force approach up to the square root of N: <lang groovy>def factorize = { long target ->
if (target == 1) return [1L]
if (target < 4) return [1L, target]
def targetSqrt = Math.sqrt(target)
def lowfactors = (2L..targetSqrt).grep { (target % it) == 0 }
if (lowfactors == []) return [1L, target]
def nhalf = lowfactors.size() - ((lowfactors[-1] == targetSqrt) ? 1 : 0)
[1] + lowfactors + (0..<nhalf).collect { target.intdiv(lowfactors[it]) }.reverse() + [target]
}</lang>
Test: <lang groovy>((1..30) + [333333]).each { println ([number:it, factors:factorize(it)]) }</lang> Output:
[number:1, factors:[1]] [number:2, factors:[1, 2]] [number:3, factors:[1, 3]] [number:4, factors:[1, 2, 4]] [number:5, factors:[1, 5]] [number:6, factors:[1, 2, 3, 6]] [number:7, factors:[1, 7]] [number:8, factors:[1, 2, 4, 8]] [number:9, factors:[1, 3, 9]] [number:10, factors:[1, 2, 5, 10]] [number:11, factors:[1, 11]] [number:12, factors:[1, 2, 3, 4, 6, 12]] [number:13, factors:[1, 13]] [number:14, factors:[1, 2, 7, 14]] [number:15, factors:[1, 3, 5, 15]] [number:16, factors:[1, 2, 4, 8, 16]] [number:17, factors:[1, 17]] [number:18, factors:[1, 2, 3, 6, 9, 18]] [number:19, factors:[1, 19]] [number:20, factors:[1, 2, 4, 5, 10, 20]] [number:21, factors:[1, 3, 7, 21]] [number:22, factors:[1, 2, 11, 22]] [number:23, factors:[1, 23]] [number:24, factors:[1, 2, 3, 4, 6, 8, 12, 24]] [number:25, factors:[1, 5, 25]] [number:26, factors:[1, 2, 13, 26]] [number:27, factors:[1, 3, 9, 27]] [number:28, factors:[1, 2, 4, 7, 14, 28]] [number:29, factors:[1, 29]] [number:30, factors:[1, 2, 3, 5, 6, 10, 15, 30]] [number:333333, factors:[1, 3, 7, 9, 11, 13, 21, 33, 37, 39, 63, 77, 91, 99, 111, 117, 143, 231, 259, 273, 333, 407, 429, 481, 693, 777, 819, 1001, 1221, 1287, 1443, 2331, 2849, 3003, 3367, 3663, 4329, 5291, 8547, 9009, 10101, 15873, 25641, 30303, 37037, 47619, 111111, 333333]]
Haskell
Using D. Amos module Primes [1] for finding prime factors <lang Haskell>import HFM.Primes(primePowerFactors) import Data.List
factors = map product.
mapM (uncurry((. enumFromTo 0) . map .(^) )) . primePowerFactors</lang>
List comprehension
Naive, functional, no import <lang Haskell>factors_naive n = [i | i <-[1..n], (mod n i) == 0]</lang> <lang Haskell>factors_naive 6 [1,2,3,6] </lang>
Factor, cofactor. Rearrange a list of tuples to a sorted list <lang Haskell>import Data.List tuple_to_list lt = (fst lt) ++ (snd lt) factors_co n = sort (tuple_to_list(unzip
[ (j, (div n j)) | j <-
[i | i <-
[1..truncate (sqrt (fromIntegral n))]
, (mod n i) == 0]] ))
</lang> <lang Haskell>factors_co 6 [1,2,3,6] </lang>
HicEst
<lang hicest> DLG(NameEdit=N, TItle='Enter an integer')
DO i = 1, N^0.5 IF( MOD(N,i) == 0) WRITE() i, N/i ENDDO
END</lang>
Icon and Unicon
<lang Icon>procedure main(arglist) numbers := arglist ||| [ 32767, 45, 53, 64, 100] # combine command line provided and default set of values every writes(lf,"factors of ",i := !numbers,"=") & writes(divisors(i)," ") do lf := "\n" end
link factors</lang> Sample Output:
factors of 32767=1 7 31 151 217 1057 4681 32767 factors of 45=1 3 5 9 15 45 factors of 53=1 53 factors of 64=1 2 4 8 16 32 64 factors of 100=1 2 4 5 10 20 25 50 100
J
J has a primitive, q: which returns its prime factors. <lang J>q: 40
2 2 2 5</lang>
Alternatively, q: can produce provide a table of the exponents of the unique relevant prime factors <lang J> __ q: 420 2 3 5 7 2 1 1 1</lang>
With this, we can form lists of each of the potential relevant powers of each of these prime factors <lang J> ((^ i.@>:)&.>/) __ q: 420 ┌─────┬───┬───┬───┐ │1 2 4│1 3│1 5│1 7│ └─────┴───┴───┴───┘</lang>
From here, it's a simple matter (*/&>@{) to compute all possible factors of the original number
<lang J>factrs=: */&>@{@((^ i.@>:)&.>/)@q:~&__
factrs 40 1 5 2 10 4 20 8 40</lang>
However, a data structure which is organized around the prime decomposition of the argument can be hard to read. So, for reader convenience, we should probably arrange them in a monotonically increasing list:
<lang J> factors=: [: /:~@, */&>@{@((^ i.@>:)&.>/)@q:~&__
factors 420
1 2 3 4 5 6 7 10 12 14 15 20 21 28 30 35 42 60 70 84 105 140 210 420</lang>
A less efficient, but concise variation on this theme:
<lang J> ~.,*/&> { 1 ,&.> q: 40 1 5 2 10 4 20 8 40</lang>
This computes 2^n intermediate values where n is the number of prime factors of the original number.
Another less efficient approach, in which remainders are examined up to the square root, larger factors obtained as fractions, and the combined list nubbed and sorted might be: <lang J>factorsOfNumber=: monad define
Y=. y"_ /:~ ~. ( , Y%]) ( #~ 0=]|Y) 1+i.>.%:y
)
factorsOfNumber 40
1 2 4 5 8 10 20 40</lang>
Another approach:
<lang J>odometer =: #: i.@(*/) factors=: (*/@:^"1 odometer@:>:)/@q:~&__</lang>
See http://www.jsoftware.com/jwiki/Essays/Odometer
Java
<lang java5>public static TreeSet<Long> factors(long n) {
TreeSet<Long> factors = new TreeSet<Long>();
factors.add(n);
factors.add(1L);
for(long test = n - 1; test >= Math.sqrt(n); test--)
if(n % test == 0)
{
factors.add(test);
factors.add(n / test);
}
return factors;
}</lang>
JavaScript
<lang javascript>function factors(num) {
var n_factors = [], i;
for (i = 1; i <= Math.floor(Math.sqrt(num)); i += 1)
if (num % i === 0)
{
n_factors.push(i);
if (num / i !== i)
n_factors.push(num / i);
}
n_factors.sort(function(a, b){return a - b;}); // numeric sort
return n_factors;
}
factors(45); // [1,3,5,9,15,45] factors(53); // [1,53] factors(64); // [1,2,4,8,16,32,64]</lang>
Julia
<lang julia>function factors(n)
f = [one(n)]
for (p,e) in factor(n)
f = reduce(vcat, f, [f*p^j for j in 1:e])
end
return length(f) == 1 ? [one(n), n] : sort!(f)
end</lang> Example output:
julia> factors(45)
6-element Array{Int64,1}:
1
3
5
9
15
45
K
<lang K> f:{d:&~x!'!1+_sqrt x;?d,_ x%|d}
f 1
1
f 3
1 3
f 120
1 2 3 4 5 6 8 10 12 15 20 24 30 40 60 120
f 1024
1 2 4 8 16 32 64 128 256 512 1024
f 600851475143
1 71 839 1471 6857 59569 104441 486847 1234169 5753023 10086647 87625999 408464633 716151937 8462696833 600851475143
#f 3491888400 / has 1920 factors
1920
/ Number of factors for 3491888400 .. 3491888409 #:'f' 3491888400+!10
1920 16 4 4 12 16 32 16 8 24</lang>
LFE
Using List Comprehensions
This following function is elegant looking and concise. However, it will not handle large numbers well: it will consume a great deal of memory (on one large number, the function consumed 4.3GB of memory on my desktop machine): <lang lisp> (defun factors (n)
(list-comp ((<- i (when (== 0 (rem n i))) (lists:seq 1 (trunc (/ n 2))))) i))
</lang>
Non-Stack-Consuming
This version will not consume the stack (this function only used 18MB of memory on my machine with a ridiculously large number): <lang lisp> (defun factors (n)
"Tail-recursive prime factors function." (factors n 2 '()))
(defun factors
((1 _ acc) (++ acc '(1))) ((n _ acc) (when (=< n 0)) #(error undefined)) ((n k acc) (when (== 0 (rem n k))) (factors (div n k) k (cons k acc))) ((n k acc) (factors n (+ k 1) acc)))
</lang>
Output in the REPL: <lang lisp> > (factors 10677106534462215678539721403561279) (104729 104729 104729 98731 98731 32579 29269 1) </lang>
Liberty BASIC
<lang lb>num = 10677106534462215678539721403561279 maxnFactors = 1000 dim primeFactors(maxnFactors), nPrimeFactors(maxnFactors) global nDifferentPrimeNumbersFound, nFactors, iFactor
print "Start finding all factors of ";num; ":"
nDifferentPrimeNumbersFound=0 dummy = factorize(num,2) nFactors = showPrimeFactors(num) dim factors(nFactors) dummy = generateFactors(1,1) sort factors(), 0, nFactors-1 for i=1 to nFactors
print i;" ";factors(i-1)
next i
print "done"
wait
function factorize(iNum,offset)
factorFound=0
i = offset
do
if (iNum MOD i)=0 _
then
if primeFactors(nDifferentPrimeNumbersFound) = i _
then
nPrimeFactors(nDifferentPrimeNumbersFound) = nPrimeFactors(nDifferentPrimeNumbersFound) + 1
else
nDifferentPrimeNumbersFound = nDifferentPrimeNumbersFound + 1
primeFactors(nDifferentPrimeNumbersFound) = i
nPrimeFactors(nDifferentPrimeNumbersFound) = 1
end if
if iNum/i<>1 then dummy = factorize(iNum/i,i)
factorFound=1
end if
i=i+1
loop while factorFound=0 and i<=sqr(iNum)
if factorFound=0 _
then
nDifferentPrimeNumbersFound = nDifferentPrimeNumbersFound + 1
primeFactors(nDifferentPrimeNumbersFound) = iNum
nPrimeFactors(nDifferentPrimeNumbersFound) = 1
end if
end function
function showPrimeFactors(iNum)
showPrimeFactors=1
print iNum;" = ";
for i=1 to nDifferentPrimeNumbersFound
print primeFactors(i);"^";nPrimeFactors(i);
if i<nDifferentPrimeNumbersFound then print " * "; else print ""
showPrimeFactors = showPrimeFactors*(nPrimeFactors(i)+1)
next i
end function
function generateFactors(product,pIndex)
if pIndex>nDifferentPrimeNumbersFound _
then
factors(iFactor) = product
iFactor=iFactor+1
else
for i=0 to nPrimeFactors(pIndex)
dummy = generateFactors(product*primeFactors(pIndex)^i,pIndex+1)
next i
end if
end function</lang>
Outcome: <lang lb>Start finding all factors of 10677106534462215678539721403561279: 10677106534462215678539721403561279 = 29269^1 * 32579^1 * 98731^2 * 104729^3 1 1 2 29269 3 32579 4 98731 5 104729 6 953554751 7 2889757639 8 3065313101 9 3216557249 10 3411966091 11 9747810361 12 10339998899 13 10968163441 14 94145414120981 15 99864835517479 16 285308661456109 17 302641427774831 18 317573913751019 19 321027175754629 20 336866824130521 21 357331796744339 22 1020878431297169 23 1082897744693371 24 1148684789012489 25 9295070881578575111 26 9859755075476219149 27 10458744358910058191 28 29880090805636839461 29 31695334089430275799 30 33259198413230468851 31 33620855089606540541 32 35279725624365333809 33 37423001741237879131 34 106915577231321212201 35 113410797903992051459 36 973463478356842592799919 37 1032602289299548955255621 38 1095333837964291484285239 39 3129312029983540559911069 40 3319420643851943354153471 41 3483202590619213772296379 42 3694810384914157044482761 43 11197161487859039232598529 44 101949856624833767901342716951 45 108143405156052462534965931709 46 327729719588146219298926345301 47 364792324112959639158827476291 48 10677106534462215678539721403561279 done</lang>
Logo
<lang logo>to factors :n
output filter [equal? 0 modulo :n ?] iseq 1 :n
end
show factors 28 ; [1 2 4 7 14 28]</lang>
Lua
<lang lua>function Factors( n )
local f = {}
for i = 1, n/2 do
if n % i == 0 then
f[#f+1] = i
end
end
f[#f+1] = n
return f
end</lang>
Maple
<lang Maple> numtheory:-divisors(n); </lang>
Mathematica / Wolfram Language
<lang Mathematica>Factorize[n_Integer] := Divisors[n]</lang>
MATLAB / Octave
<lang Matlab> function fact(n);
f = factor(n); % prime decomposition
K = dec2bin(0:2^length(f)-1)-'0'; % generate all possible permutations
F = ones(1,2^length(f));
for k = 1:size(K)
F(k) = prod(f(~K(k,:))); % and compute products
end;
F = unique(F); % eliminate duplicates
printf('There are %i factors for %i.\n',length(F),n);
disp(F);
end;
</lang>
Output:
>> fact(12)
There are 6 factors for 12.
1 2 3 4 6 12
>> fact(28)
There are 6 factors for 28.
1 2 4 7 14 28
>> fact(64)
There are 7 factors for 64.
1 2 4 8 16 32 64
>>fact(53)
There are 2 factors for 53.
1 53
Maxima
The builtin divisors function does this.
<lang maxima>(%i96) divisors(100);
(%o96) {1,2,4,5,10,20,25,50,100}</lang>
Such a function could be implemented like so: <lang maxima>divisors2(n) := map( lambda([l], lreduce("*", l)),
apply( cartesian_product,
map( lambda([fac],
setify(makelist(fac[1]^i, i, 0, fac[2]))),
ifactors(n))));</lang>
Mercury
Mercury is both a logic language and a functional language. As such there are two possible interfaces for calculating the factors of an integer. This code shows both styles of implementation. Note that much of the code here is ceremony put in place to have this be something which can actually compile. The actual factoring is contained in the predicate factor/2 and in the function factor/1. The function form is implemented in terms of the predicate form rather than duplicating all of the predicate code.
The predicates main/2 and factor/2 are shown with the combined type and mode statement (e.g. int::in) as is the usual case for simple predicates with only one mode. This makes the code more immediately understandable. The predicate factor/5, however, has its mode broken out onto a separate line both to show Mercury's mode statement (useful for predicates which can have varying instantiation of parameters) and to stop the code from extending too far to the right. Finally the function factor/1 has its mode statements removed (shown underneath in a comment for illustration purposes) because good coding style (and the default of the compiler!) has all parameters "in"-moded and the return value "out"-moded.
This implementation of factoring works as follows:
- The input number itself and 1 are both considered factors.
- The numbers between 2 and the square root of the input number are checked for even division.
- If the incremental number divides evenly into the input number, both the incremental number and the quotient are added to the list of factors.
This implementation makes use of Mercury's "state variable notation" to keep a pair of variables for accumulation, thus allowing the implementation to be tail recursive. !Accumulator is syntax sugar for a *pair* of variables. One of them is an "in"-moded variable and the other is an "out"-moded variable. !:Accumulator is the "out" portion and !.Accumulator is the "in" portion in the ensuing code.
Using the state variable notation avoids having to keep track of strings of variables unified in the code named things like Acc0, Acc1, Acc2, Acc3, etc.
fac.m
<lang Mercury>:- module fac.
- - interface.
- - import_module io.
- - pred main(io::di, io::uo) is det.
- - implementation.
- - import_module float, int, list, math, string.
main(!IO) :-
io.command_line_arguments(Args, !IO),
list.filter_map(string.to_int, Args, CleanArgs),
list.foldl((pred(Arg::in, !.IO::di, !:IO::uo) is det :-
factor(Arg, X),
io.format("factor(%d, [", [i(Arg)], !IO),
io.write_list(X, ",", io.write_int, !IO),
io.write_string("])\n", !IO)
), CleanArgs, !IO).
- - pred factor(int::in, list(int)::out) is det.
factor(N, Factors) :-
Limit = float.truncate_to_int(math.sqrt(float(N))),
factor(N, 2, Limit, [], Unsorted),
list.sort_and_remove_dups([1, N | Unsorted], Factors).
- - pred factor(int, int, int, list(int), list(int)).
- - mode factor(in, in, in, in, out) is det.
factor(N, X, Limit, !Accumulator) :-
( if X > Limit
then true
else ( if 0 = N mod X
then !:Accumulator = [X, N / X | !.Accumulator]
else true ),
factor(N, X + 1, Limit, !Accumulator) ).
- - func factor(int) = list(int).
%:- mode factor(in) = out is det. factor(N) = Factors :- factor(N, Factors).
- - end_module fac.</lang>
Use and output
Use of the code looks like this:
$ mmc fac.m && ./fac 100 999 12345678 booger factor(100, [1,2,4,5,10,20,25,50,100]) factor(999, [1,3,9,27,37,111,333,999]) factor(12345678, [1,2,3,6,9,18,47,94,141,282,423,846,14593,29186,43779,87558,131337,262674,685871,1371742,2057613,4115226,6172839,12345678])
МК-61/52
П9 1 П6 КИП6 ИП9 ИП6 / П8 ^ [x] x#0 21 - x=0 03 ИП6 С/П ИП8 П9 БП 04 1 С/П БП 21
MUMPS
<lang MUMPS>factors(num) New fctr,list,sep,sqrt If num<1 Quit "Too small a number" If num["." Quit "Not an integer" Set sqrt=num**0.5\1 For fctr=1:1:sqrt Set:num/fctr'["." list(fctr)=1,list(num/fctr)=1 Set (list,fctr)="",sep="[" For Set fctr=$Order(list(fctr)) Quit:fctr="" Set list=list_sep_fctr,sep="," Quit list_"]"
w $$factors(45) ; [1,3,5,9,15,45] w $$factors(53) ; [1,53] w $$factors(64) ; [1,2,4,8,16,32,64]</lang>
NetRexx
<lang NetRexx>/* NetRexx ***********************************************************
- 21.04.2013 Walter Pachl
- 21.04.2013 add method main to accept argument(s)
- /
options replace format comments java crossref symbols nobinary class divl
method main(argwords=String[]) static
arg=Rexx(argwords)
Parse arg a b
Say a b
If a= Then Do
help='java divl low [high] shows'
help=help||' divisors of all numbers between low and high'
Say help
Return
End
If b= Then b=a
loop x=a To b
say x '->' divs(x)
End
method divs(x) public static returns Rexx
if x==1 then return 1 /*handle special case of 1 */
lo=1
hi=x
odd=x//2 /* 1 if x is odd */
loop j=2+odd By 1+odd While j*j<x /*divide by numbers<sqrt(x) */
if x//j==0 then Do /*Divisible? Add two divisors:*/
lo=lo j /* list low divisors */
hi=x%j hi /* list high divisors */
End
End
If j*j=x Then /*for a square number as input */
lo=lo j /* add its square root */
return lo hi /* return both lists */</lang>
Output:
java divl 1 10 1 -> 1 2 -> 1 2 3 -> 1 3 4 -> 1 2 4 5 -> 1 5 6 -> 1 2 3 6 7 -> 1 7 8 -> 1 2 4 8 9 -> 1 3 9 10 -> 1 2 5 10
Nimrod
<lang nimrod>import intsets, math, algorithm
proc factors(n): seq[int] =
var fs = initIntSet()
for x in 1 .. int(sqrt(float(n))):
if n mod x == 0:
fs.incl(x)
fs.incl(n div x)
result = @[] for x in fs: result.add(x) sort(result, system.cmp[int])
echo factors(45)</lang>
Niue
<lang Niue>[ 'n ; [ negative-or-zero [ , ] if
[ n not-factor [ , ] when ] else ] n times n ] 'factors ;
[ dup 0 <= ] 'negative-or-zero ; [ swap dup rot swap mod 0 = not ] 'not-factor ;
( tests ) 100 factors .s .clr ( => 1 2 4 5 10 20 25 50 100 ) newline 53 factors .s .clr ( => 1 53 ) newline 64 factors .s .clr ( => 1 2 4 8 16 32 64 ) newline 12 factors .s .clr ( => 1 2 3 4 6 12 ) </lang>
Oberon-2
Oxford Oberon-2 <lang oberon2> MODULE Factors; IMPORT Out,SYSTEM; TYPE LIPool = POINTER TO ARRAY OF LONGINT; LIVector= POINTER TO LIVectorDesc; LIVectorDesc = RECORD cap: INTEGER; len: INTEGER; LIPool: LIPool; END;
PROCEDURE New(cap: INTEGER): LIVector; VAR v: LIVector; BEGIN NEW(v); v.cap := cap; v.len := 0; NEW(v.LIPool,cap); RETURN v END New;
PROCEDURE (v: LIVector) Add(x: LONGINT); VAR newLIPool: LIPool; BEGIN IF v.len = LEN(v.LIPool^) THEN (* run out of space *) v.cap := v.cap + (v.cap DIV 2); NEW(newLIPool,v.cap); SYSTEM.MOVE(SYSTEM.ADR(v.LIPool^),SYSTEM.ADR(newLIPool^),v.cap * SIZE(LONGINT)); v.LIPool := newLIPool END; v.LIPool[v.len] := x; INC(v.len) END Add;
PROCEDURE (v: LIVector) At(idx: INTEGER): LONGINT; BEGIN RETURN v.LIPool[idx]; END At;
PROCEDURE Factors(n:LONGINT): LIVector;
VAR
j: LONGINT;
v: LIVector;
BEGIN
v := New(16);
FOR j := 1 TO n DO
IF (n MOD j) = 0 THEN v.Add(j) END;
END;
RETURN v
END Factors;
VAR v: LIVector; j: INTEGER; BEGIN v := Factors(123); FOR j := 0 TO v.len - 1 DO Out.LongInt(v.At(j),4);Out.Ln END; Out.Int(v.len,6);Out.String(" factors");Out.Ln END Factors. </lang> Output:
1
3
41
123
4 factors
Objeck
<lang objeck>use IO; use Structure;
bundle Default {
class Basic {
function : native : GenerateFactors(n : Int) ~ IntVector {
factors := IntVector->New();
factors-> AddBack(1);
factors->AddBack(n);
for(i := 2; i * i <= n; i += 1;) {
if(n % i = 0) {
factors->AddBack(i);
if(i * i <> n) {
factors->AddBack(n / i);
};
};
};
factors->Sort();
return factors;
}
function : Main(args : String[]) ~ Nil {
numbers := [3135, 45, 60, 81];
for(i := 0; i < numbers->Size(); i += 1;) {
factors := GenerateFactors(numbers[i]);
Console->GetInstance()->Print("Factors of ")->Print(numbers[i])->PrintLine(" are:");
each(i : factors) {
Console->GetInstance()->Print(factors->Get(i))->Print(", ");
};
"\n\n"->Print();
};
}
}
}</lang>
OCaml
<lang ocaml>let rec range = function 0 -> [] | n -> range(n-1) @ [n]
let factors n =
List.filter (fun v -> (n mod v) = 0) (range n)</lang>
Oz
<lang oz>declare
fun {Factors N}
Sqr = {Float.toInt {Sqrt {Int.toFloat N}}}
Fs = for X in 1..Sqr append:App do
if N mod X == 0 then
CoFactor = N div X
in
if CoFactor == X then %% avoid duplicate factor
{App [X]} %% when N is a square number
else
{App [X CoFactor]}
end
end
end
in
{Sort Fs Value.'<'}
end
in
{Show {Factors 53}}</lang>
PARI/GP
<lang parigp>divisors(n)</lang>
Pascal
<lang pascal>program Factors; var
i, number: integer;
begin
write('Enter a number between 1 and 2147483647: ');
readln(number);
for i := 1 to round(sqrt(number)) - 1 do
if number mod i = 0 then
write (i, ' ', number div i, ' ');
// Check to see if number is a square
i := round(sqrt(number));
if i*i = number then
write(i)
else if number mod i = 0 then
write(i, number/i);
writeln;
end.</lang> Output:
Enter a number between 1 and 2147483647: 49 1 49 7 Enter a number between 1 and 2147483647: 353435 1 25755 3 8585 5 5151 15 1717 17 1515 51 505 85 303 101 255
Perl
<lang perl>sub factors {
my($n) = @_;
return grep { $n % $_ == 0 }(1 .. $n);
} print join ' ',factors(64);</lang>
Perl 6
<lang perl6>sub factors (Int $n) {
sort uniq
map { $^x, $n div $^x },
grep { $n %% $^x },
1 .. sqrt $n;
}</lang>
If we don't bother about performance at all we can make the code quite shorter by reviewing all integers from 1 to $n:
<lang perl6>sub factors (Int $n) { grep $n %% *, 1 .. $n }</lang>
PHP
<lang PHP>function GetFactors($n){
$factors = array(1, $n);
for($i = 2; $i * $i <= $n; $i++){
if($n % $i == 0){
$factors[] = $i;
if($i * $i != $n)
$factors[] = $n/$i;
}
}
sort($factors);
return $factors;
}</lang>
PicoLisp
<lang PicoLisp>(de factors (N)
(filter
'((D) (=0 (% N D)))
(range 1 N) ) )</lang>
PL/I
<lang PL/I>do i = 1 to n;
if mod(n, i) = 0 then put skip list (i);
end;</lang>
PowerShell
Straightforward but slow
<lang powershell>function Get-Factor ($a) {
1..$a | Where-Object { $a % $_ -eq 0 }
}</lang>
This one uses a range of integers up to the target number and just filters it using the Where-Object cmdlet. It's very slow though, so it is not very usable for larger numbers.
A little more clever
<lang powershell>function Get-Factor ($a) {
1..[Math]::Sqrt($a) `
| Where-Object { $a % $_ -eq 0 } `
| ForEach-Object { $_; $a / $_ } `
| Sort-Object -Unique
}</lang> Here the range of integers is only taken up to the square root of the number, the same filtering applies. Afterwards the corresponding larger factors are calculated and sent down the pipeline along with the small ones found earlier.
ProDOS
Uses the math module: <lang ProDOS>editvar /newvar /value=a /userinput=1 /title=Enter an integer: do /delimspaces %% -a- >b printline Factors of -a-: -b- </lang>
Prolog
Simple Brute Force Implementation <lang Prolog> brute_force_factors( N , Fs ) :-
integer(N) , N > 0 , setof( F , ( between(1,N,F) , N mod F =:= 0 ) , Fs ) .
</lang>
A Slightly Smarter Implementation <lang Prolog> smart_factors(N,Fs) :-
integer(N) , N > 0 , setof( F , factor(N,F) , Fs ) .
factor(N,F) :-
L is floor(sqrt(N)) , between(1,L,X) , 0 =:= N mod X , ( F = X ; F is N // X ) .
</lang>
Not every Prolog has between/3: you might need this:
<lang Prolog>
between(X,Y,Z) :-
integer(X) , integer(Y) , X =< Z , between1(X,Y,Z) .
between1(X,Y,X) :-
X =< Y .
between1(X,Y,Z) :-
X < Y , X1 is X+1 , between1(X1,Y,Z) .
</lang>
Output:
?- N=36 ,( brute_force_factors(N,Factors) ; smart_factors(N,Factors) ). N = 36, Factors = [1, 2, 3, 4, 6, 9, 12, 18, 36] ; N = 36, Factors = [1, 2, 3, 4, 6, 9, 12, 18, 36] . ?- N=53,( brute_force_factors(N,Factors) ; smart_factors(N,Factors) ). N = 53, Factors = [1, 53] ; N = 53, Factors = [1, 53] . ?- N=100,( brute_force_factors(N,Factors);smart_factors(N,Factors) ). N = 100, Factors = [1, 2, 4, 5, 10, 20, 25, 50, 100] ; N = 100, Factors = [1, 2, 4, 5, 10, 20, 25, 50, 100] . ?- N=144,( brute_force_factors(N,Factors);smart_factors(N,Factors) ). N = 144, Factors = [1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144] ; N = 144, Factors = [1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144] . ?- N=32765,( brute_force_factors(N,Factors);smart_factors(N,Factors) ). N = 32765, Factors = [1, 5, 6553, 32765] ; N = 32765, Factors = [1, 5, 6553, 32765] . ?- N=32766,( brute_force_factors(N,Factors);smart_factors(N,Factors) ). N = 32766, Factors = [1, 2, 3, 6, 43, 86, 127, 129, 254, 258, 381, 762, 5461, 10922, 16383, 32766] ; N = 32766, Factors = [1, 2, 3, 6, 43, 86, 127, 129, 254, 258, 381, 762, 5461, 10922, 16383, 32766] . 38 ?- N=32767,( brute_force_factors(N,Factors);smart_factors(N,Factors) ). N = 32767, Factors = [1, 7, 31, 151, 217, 1057, 4681, 32767] ; N = 32767, Factors = [1, 7, 31, 151, 217, 1057, 4681, 32767] .
PureBasic
<lang PureBasic>Procedure PrintFactors(n)
Protected i, lim=Round(sqr(n),#PB_Round_Up)
NewList F.i()
For i=1 To lim
If n%i=0
AddElement(F()): F()=i
AddElement(F()): F()=n/i
EndIf
Next
;- Present the result
SortList(F(),#PB_Sort_Ascending)
ForEach F()
Print(str(F())+" ")
Next
EndProcedure
If OpenConsole()
Print("Enter integer to factorize: ")
PrintFactors(Val(Input()))
Print(#CRLF$+#CRLF$+"Press ENTER to quit."): Input()
EndIf</lang> Output can look like
Enter integer to factorize: 96 1 2 3 4 6 8 12 16 24 32 48 96
Python
Naive and slow but simplest (check all numbers from 1 to n): <lang python>>>> def factors(n):
return [i for i in range(1, n + 1) if not n%i]</lang>
Slightly better (realize that there are no factors between n/2 and n): <lang python>>>> def factors(n):
return [i for i in range(1, n//2 + 1) if not n%i] + [n]
>>> factors(45) [1, 3, 5, 9, 15, 45]</lang>
Much better (realize that factors come in pairs, the smaller of which is no bigger than sqrt(n)): <lang python>>>> from math import sqrt >>> def factor(n):
factors = set()
for x in range(1, int(sqrt(n)) + 1):
if n % x == 0:
factors.add(x)
factors.add(n//x)
return sorted(factors)
>>> for i in (45, 53, 64): print( "%i: factors: %s" % (i, factor(i)) )
45: factors: [1, 3, 5, 9, 15, 45] 53: factors: [1, 53] 64: factors: [1, 2, 4, 8, 16, 32, 64]</lang>
More efficient when factoring many numbers: <lang python>def factor(n): a, r = 1, [1] while a * a < n: a += 1 if n % a: continue b, f = 1, [] while n % a == 0: n //= a b *= a f += [i * b for i in r] r += f if n > 1: r += [i * n for i in r] return r</lang>
R
<lang R>factors <- function(n) {
if(length(n) > 1)
{
lapply(as.list(n), factors)
} else
{
one.to.n <- seq_len(n)
one.to.n[(n %% one.to.n) == 0]
}
} factors(60)</lang>
1 2 3 4 5 6 10 12 15 20 30 60
<lang R>factors(c(45, 53, 64))</lang>
[[1]] [1] 1 3 5 9 15 45 [[2]] [1] 1 53 [[3]] [1] 1 2 4 8 16 32 64
Racket
<lang Racket>
- lang racket
- a naive version
(define (naive-factors n)
(for/list ([i (in-range 1 (add1 n))] #:when (zero? (modulo n i))) i))
(naive-factors 120) ; -> '(1 2 3 4 5 6 8 10 12 15 20 24 30 40 60 120)
- much better
- use `factorize' to get prime factors and construct the
- list of results from that
(require math) (define (factors n)
(sort (for/fold ([l '(1)]) ([p (factorize n)])
(append (for*/list ([e (in-range 1 (add1 (cadr p)))] [x l])
(* x (expt (car p) e)))
l))
<))
(naive-factors 120) ; -> same
- to see how fast it is
(define huge 1200034005600070000008900000000000000000) (time (length (factors huge)))
- I get 42ms for getting a list of 7776 numbers
- but actually the math library comes with a `divisors' function that
- does the same, except even faster
(divisors 120) ; -> same
(time (length (divisors huge)))
- And this one clocks at 17ms
</lang>
REALbasic
<lang vb>Function factors(num As UInt64) As UInt64()
'This function accepts an unsigned 64 bit integer as input and returns an array of unsigned 64 bit integers
Dim result() As UInt64
Dim iFactor As UInt64 = 1
While iFactor <= num/2 'Since a factor will never be larger than half of the number
If num Mod iFactor = 0 Then
result.Append(iFactor)
End If
iFactor = iFactor + 1
Wend
result.Append(num) 'Since a given number is always a factor of itself
Return result
End Function</lang>
REXX
optimized version
This REXX version has no effective limits on the number of digits in the number to be factored [by adjusting the number of digits (precision)].
This REXX version also supports negative integers and zero.
It also indicates primes in the output as well as the number of factors.
<lang rexx>/*REXX pgm displays divisors of any (negative/zero/positive) integers.*/
@.=left(,7); @.1='{unity}'; @.2='[prime]' /*unity & prime tags.*/
parse arg low high inc . /*get optional args. */
high=word(high low 20,1); low=word(low 1,1); inc=word(inc 1,1) /*opts*/
w=length(high)+1; numeric digits max(9,w) /*'nuff digs for // */
say center('n',1+w) '#divisors' center('divisors',60) /*header. */
say copies('─',1+w) '─────────' copies('─' ,60) /*separator.*/
do n=low to high by inc; divs=divisors(n); #=words(divs); p=@.#
if divs=='infinite' then #='∞'; if n<1 then p=@.. /*handle N<1*/
say right(n,w)" " center('['#"]",9) "──► " p ' ' divs
end /*n*/ /* [↑] process a range of ints.*/
exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────DIVISORS subroutine─────────────────*/ divisors: procedure; parse arg x; x=abs(x); if x==1 then return 1; b=x if x==0 then return 'infinite'; odd=x//2 a=1 /* [↓] use only EVEN|ODD integers*/
do j=2+odd by 1+odd while j*j<x /*divide by all integers up to √x*/
if x//j==0 then do; a=a j; b=x%j b; end /*add divs to α&ß lists if ÷*/
end /*j*/ /* [↑] % is REXX integer divide*/
/* [↓] adjust for square. _ */
if j*j==x then return a j b /*Was X a square? If so, add √x.*/
return a b /*return divisors (both lists). */</lang>
output when the input used is: -6 200
n #divisors divisors
───── ───────── ────────────────────────────────────────────────────────────
-6 [4] ──► 1 2 3 6
-5 [2] ──► 1 5
-4 [3] ──► 1 2 4
-3 [2] ──► 1 3
-2 [2] ──► 1 2
-1 [1] ──► 1
0 [∞] ──► infinite
1 [1] ──► {unity} 1
2 [2] ──► [prime] 1 2
3 [2] ──► [prime] 1 3
4 [3] ──► 1 2 4
5 [2] ──► [prime] 1 5
6 [4] ──► 1 2 3 6
7 [2] ──► [prime] 1 7
8 [4] ──► 1 2 4 8
9 [3] ──► 1 3 9
10 [4] ──► 1 2 5 10
11 [2] ──► [prime] 1 11
12 [6] ──► 1 2 3 4 6 12
13 [2] ──► [prime] 1 13
14 [4] ──► 1 2 7 14
15 [4] ──► 1 3 5 15
16 [5] ──► 1 2 4 8 16
17 [2] ──► [prime] 1 17
18 [6] ──► 1 2 3 6 9 18
19 [2] ──► [prime] 1 19
20 [6] ──► 1 2 4 5 10 20
21 [4] ──► 1 3 7 21
22 [4] ──► 1 2 11 22
23 [2] ──► [prime] 1 23
24 [8] ──► 1 2 3 4 6 8 12 24
25 [3] ──► 1 5 25
26 [4] ──► 1 2 13 26
27 [4] ──► 1 3 9 27
28 [6] ──► 1 2 4 7 14 28
29 [2] ──► [prime] 1 29
30 [8] ──► 1 2 3 5 6 10 15 30
31 [2] ──► [prime] 1 31
32 [6] ──► 1 2 4 8 16 32
33 [4] ──► 1 3 11 33
34 [4] ──► 1 2 17 34
35 [4] ──► 1 5 7 35
36 [9] ──► 1 2 3 4 6 9 12 18 36
37 [2] ──► [prime] 1 37
38 [4] ──► 1 2 19 38
39 [4] ──► 1 3 13 39
40 [8] ──► 1 2 4 5 8 10 20 40
41 [2] ──► [prime] 1 41
42 [8] ──► 1 2 3 6 7 14 21 42
43 [2] ──► [prime] 1 43
44 [6] ──► 1 2 4 11 22 44
45 [6] ──► 1 3 5 9 15 45
46 [4] ──► 1 2 23 46
47 [2] ──► [prime] 1 47
48 [10] ──► 1 2 3 4 6 8 12 16 24 48
49 [3] ──► 1 7 49
50 [6] ──► 1 2 5 10 25 50
51 [4] ──► 1 3 17 51
52 [6] ──► 1 2 4 13 26 52
53 [2] ──► [prime] 1 53
54 [8] ──► 1 2 3 6 9 18 27 54
55 [4] ──► 1 5 11 55
56 [8] ──► 1 2 4 7 8 14 28 56
57 [4] ──► 1 3 19 57
58 [4] ──► 1 2 29 58
59 [2] ──► [prime] 1 59
60 [12] ──► 1 2 3 4 5 6 10 12 15 20 30 60
61 [2] ──► [prime] 1 61
62 [4] ──► 1 2 31 62
63 [6] ──► 1 3 7 9 21 63
64 [7] ──► 1 2 4 8 16 32 64
65 [4] ──► 1 5 13 65
66 [8] ──► 1 2 3 6 11 22 33 66
67 [2] ──► [prime] 1 67
68 [6] ──► 1 2 4 17 34 68
69 [4] ──► 1 3 23 69
70 [8] ──► 1 2 5 7 10 14 35 70
71 [2] ──► [prime] 1 71
72 [12] ──► 1 2 3 4 6 8 9 12 18 24 36 72
73 [2] ──► [prime] 1 73
74 [4] ──► 1 2 37 74
75 [6] ──► 1 3 5 15 25 75
76 [6] ──► 1 2 4 19 38 76
77 [4] ──► 1 7 11 77
78 [8] ──► 1 2 3 6 13 26 39 78
79 [2] ──► [prime] 1 79
80 [10] ──► 1 2 4 5 8 10 16 20 40 80
81 [5] ──► 1 3 9 27 81
82 [4] ──► 1 2 41 82
83 [2] ──► [prime] 1 83
84 [12] ──► 1 2 3 4 6 7 12 14 21 28 42 84
85 [4] ──► 1 5 17 85
86 [4] ──► 1 2 43 86
87 [4] ──► 1 3 29 87
88 [8] ──► 1 2 4 8 11 22 44 88
89 [2] ──► [prime] 1 89
90 [12] ──► 1 2 3 5 6 9 10 15 18 30 45 90
91 [4] ──► 1 7 13 91
92 [6] ──► 1 2 4 23 46 92
93 [4] ──► 1 3 31 93
94 [4] ──► 1 2 47 94
95 [4] ──► 1 5 19 95
96 [12] ──► 1 2 3 4 6 8 12 16 24 32 48 96
97 [2] ──► [prime] 1 97
98 [6] ──► 1 2 7 14 49 98
99 [6] ──► 1 3 9 11 33 99
100 [9] ──► 1 2 4 5 10 20 25 50 100
101 [2] ──► [prime] 1 101
102 [8] ──► 1 2 3 6 17 34 51 102
103 [2] ──► [prime] 1 103
104 [8] ──► 1 2 4 8 13 26 52 104
105 [8] ──► 1 3 5 7 15 21 35 105
106 [4] ──► 1 2 53 106
107 [2] ──► [prime] 1 107
108 [12] ──► 1 2 3 4 6 9 12 18 27 36 54 108
109 [2] ──► [prime] 1 109
110 [8] ──► 1 2 5 10 11 22 55 110
111 [4] ──► 1 3 37 111
112 [10] ──► 1 2 4 7 8 14 16 28 56 112
113 [2] ──► [prime] 1 113
114 [8] ──► 1 2 3 6 19 38 57 114
115 [4] ──► 1 5 23 115
116 [6] ──► 1 2 4 29 58 116
117 [6] ──► 1 3 9 13 39 117
118 [4] ──► 1 2 59 118
119 [4] ──► 1 7 17 119
120 [16] ──► 1 2 3 4 5 6 8 10 12 15 20 24 30 40 60 120
121 [3] ──► 1 11 121
122 [4] ──► 1 2 61 122
123 [4] ──► 1 3 41 123
124 [6] ──► 1 2 4 31 62 124
125 [4] ──► 1 5 25 125
126 [12] ──► 1 2 3 6 7 9 14 18 21 42 63 126
127 [2] ──► [prime] 1 127
128 [8] ──► 1 2 4 8 16 32 64 128
129 [4] ──► 1 3 43 129
130 [8] ──► 1 2 5 10 13 26 65 130
131 [2] ──► [prime] 1 131
132 [12] ──► 1 2 3 4 6 11 12 22 33 44 66 132
133 [4] ──► 1 7 19 133
134 [4] ──► 1 2 67 134
135 [8] ──► 1 3 5 9 15 27 45 135
136 [8] ──► 1 2 4 8 17 34 68 136
137 [2] ──► [prime] 1 137
138 [8] ──► 1 2 3 6 23 46 69 138
139 [2] ──► [prime] 1 139
140 [12] ──► 1 2 4 5 7 10 14 20 28 35 70 140
141 [4] ──► 1 3 47 141
142 [4] ──► 1 2 71 142
143 [4] ──► 1 11 13 143
144 [15] ──► 1 2 3 4 6 8 9 12 16 18 24 36 48 72 144
145 [4] ──► 1 5 29 145
146 [4] ──► 1 2 73 146
147 [6] ──► 1 3 7 21 49 147
148 [6] ──► 1 2 4 37 74 148
149 [2] ──► [prime] 1 149
150 [12] ──► 1 2 3 5 6 10 15 25 30 50 75 150
151 [2] ──► [prime] 1 151
152 [8] ──► 1 2 4 8 19 38 76 152
153 [6] ──► 1 3 9 17 51 153
154 [8] ──► 1 2 7 11 14 22 77 154
155 [4] ──► 1 5 31 155
156 [12] ──► 1 2 3 4 6 12 13 26 39 52 78 156
157 [2] ──► [prime] 1 157
158 [4] ──► 1 2 79 158
159 [4] ──► 1 3 53 159
160 [12] ──► 1 2 4 5 8 10 16 20 32 40 80 160
161 [4] ──► 1 7 23 161
162 [10] ──► 1 2 3 6 9 18 27 54 81 162
163 [2] ──► [prime] 1 163
164 [6] ──► 1 2 4 41 82 164
165 [8] ──► 1 3 5 11 15 33 55 165
166 [4] ──► 1 2 83 166
167 [2] ──► [prime] 1 167
168 [16] ──► 1 2 3 4 6 7 8 12 14 21 24 28 42 56 84 168
169 [3] ──► 1 13 169
170 [8] ──► 1 2 5 10 17 34 85 170
171 [6] ──► 1 3 9 19 57 171
172 [6] ──► 1 2 4 43 86 172
173 [2] ──► [prime] 1 173
174 [8] ──► 1 2 3 6 29 58 87 174
175 [6] ──► 1 5 7 25 35 175
176 [10] ──► 1 2 4 8 11 16 22 44 88 176
177 [4] ──► 1 3 59 177
178 [4] ──► 1 2 89 178
179 [2] ──► [prime] 1 179
180 [18] ──► 1 2 3 4 5 6 9 10 12 15 18 20 30 36 45 60 90 180
181 [2] ──► [prime] 1 181
182 [8] ──► 1 2 7 13 14 26 91 182
183 [4] ──► 1 3 61 183
184 [8] ──► 1 2 4 8 23 46 92 184
185 [4] ──► 1 5 37 185
186 [8] ──► 1 2 3 6 31 62 93 186
187 [4] ──► 1 11 17 187
188 [6] ──► 1 2 4 47 94 188
189 [8] ──► 1 3 7 9 21 27 63 189
190 [8] ──► 1 2 5 10 19 38 95 190
191 [2] ──► [prime] 1 191
192 [14] ──► 1 2 3 4 6 8 12 16 24 32 48 64 96 192
193 [2] ──► [prime] 1 193
194 [4] ──► 1 2 97 194
195 [8] ──► 1 3 5 13 15 39 65 195
196 [9] ──► 1 2 4 7 14 28 49 98 196
197 [2] ──► [prime] 1 197
198 [12] ──► 1 2 3 6 9 11 18 22 33 66 99 198
199 [2] ──► [prime] 1 199
200 [12] ──► 1 2 4 5 8 10 20 25 40 50 100 200
Alternate Version
<lang REXX>/* REXX ***************************************************************
- Program to calculate and show divisors of positive integer(s).
- 03.08.2012 Walter Pachl simplified the above somewhat
- in particular I see no benefit from divAdd procedure
- 04.08.2012 the reference to 'above' is no longer valid since that
- was meanwhile changed for the better.
- 04.08.2012 took over some improvements from new above
- /
Parse arg low high . Select
When low= Then Parse Value '1 200' with low high When high= Then high=low Otherwise Nop End
do j=low to high
say ' n = ' right(j,6) " divisors = " divs(j) end
exit
divs: procedure; parse arg x
if x==1 then return 1 /*handle special case of 1 */
Parse Value '1' x With lo hi /*initialize lists: lo=1 hi=x */
odd=x//2 /* 1 if x is odd */
Do j=2+odd By 1+odd While j*j<x /*divide by numbers<sqrt(x) */
if x//j==0 then Do /*Divisible? Add two divisors:*/
lo=lo j /* list low divisors */
hi=x%j hi /* list high divisors */
End
End
If j*j=x Then /*for a square number as input */
lo=lo j /* add its square root */
return lo hi /* return both lists */</lang>
Ruby
<lang ruby>class Integer
def factors() (1..self).select { |n| (self % n).zero? } end
end p 45.factors</lang>
[1, 3, 5, 9, 15, 45]
As we only have to loop up to , we can write <lang ruby>class Integer
def factors()
1.upto(Math.sqrt(self)).select {|i| (self % i).zero?}.inject([]) do |f, i|
f << i
f << self/i unless i == self/i
f
end.sort
end
end [45, 53, 64].each {|n| p n.factors}</lang> output
[1, 3, 5, 9, 15, 45] [1, 53] [1, 2, 4, 8, 16, 32, 64]
Run BASIC
<lang runbasic>PRINT "Factors of 45 are ";factorlist$(45) PRINT "Factors of 12345 are "; factorlist$(12345) END
function factorlist$(f) DIM L(100) FOR i = 1 TO SQR(f)
IF (f MOD i) = 0 THEN
L(c) = i
c = c + 1
IF (f <> i^2) THEN
L(c) = (f / i)
c = c + 1
END IF
END IF
NEXT i s = 1 while s = 1 s = 0 for i = 0 to c-1
if L(i) > L(i+1) and L(i+1) <> 0 then t = L(i) L(i) = L(i+1) L(i+1) = t s = 1 end if
next i wend FOR i = 0 TO c-1
factorlist$ = factorlist$ + STR$(L(i)) + ", "
NEXT end function</lang> output
Factors of 45 are 1, 3, 5, 9, 15, 45, Factors of 12345 are 1, 3, 5, 15, 823, 2469, 4115, 12345,
Sather
<lang sather>class MAIN is
factors(n :INT):ARRAY{INT} is
f:ARRAY{INT};
f := #;
f := f.append(|1|);
f := f.append(|n|);
loop i ::= 2.upto!( n.flt.sqrt.int );
if n%i = 0 then
f := f.append(|i|);
if (i*i) /= n then f := f.append(|n / i|); end;
end; end; f.sort; return f; end;
main is
a :ARRAY{INT} := |3135, 45, 64, 53, 45, 81|;
loop l ::= a.elt!;
#OUT + "factors of " + l + ": ";
r ::= factors(l);
loop ri ::= r.elt!;
#OUT + ri + " ";
end;
#OUT + "\n";
end;
end;
end;</lang>
Scala
<lang Scala>
def factors(num: Int) = {
(1 to num).filter { divisor =>
num % divisor == 0
}
}</lang>
Scheme
This implementation uses a naive trial division algorithm. <lang scheme>(define (factors n)
(define (*factors d)
(cond ((> d n) (list))
((= (modulo n d) 0) (cons d (*factors (+ d 1))))
(else (*factors (+ d 1)))))
(*factors 1))
(display (factors 1111111)) (newline)</lang> Output:
(1 239 4649 1111111)
Seed7
<lang seed7>$ include "seed7_05.s7i";
const proc: writeFactors (in integer: number) is func
local
var integer: testNum is 0;
begin
write("Factors of " <& number <& ": ");
for testNum range 1 to sqrt(number) do
if number rem testNum = 0 then
if testNum <> 1 then
write(", ");
end if;
write(testNum);
if testNum <> number div testNum then
write(", " <& number div testNum);
end if;
end if;
end for;
writeln;
end func;
const proc: main is func
local
const array integer: numsToFactor is [] (45, 53, 64);
var integer: number is 0;
begin
for number range numsToFactor do
writeFactors(number);
end for;
end func;</lang>
Output:
Factors of 45: 1, 45, 3, 15, 5, 9 Factors of 53: 1, 53 Factors of 64: 1, 64, 2, 32, 4, 16, 8
Slate
<lang slate>n@(Integer traits) primeFactors [
[| :result |
result nextPut: 1.
n primesDo: [| :prime | result nextPut: prime]] writingAs: {}
].</lang> where primesDo: is a part of the standard numerics library: <lang slate>n@(Integer traits) primesDo: block "Decomposes the Integer into primes, applying the block to each (in increasing order)." [| div next remaining |
div: 2.
next: 3.
remaining: n.
[[(remaining \\ div) isZero]
whileTrue:
[block applyTo: {div}.
remaining: remaining // div].
remaining = 1] whileFalse:
[div: next.
next: next + 2] "Just looks at the next odd integer."
].</lang>
Smalltalk
Copied from the Python example, but code added to the Integer built in class:
<lang smalltalk>Integer>>factors | a | a := OrderedCollection new. 1 to: (self / 2) do: [ :i | ((self \\ i) = 0) ifTrue: [ a add: i ] ]. a add: self. ^a</lang>
Then use as follows:
<lang smalltalk> 59 factors -> an OrderedCollection(1 59) 120 factors -> an OrderedCollection(1 2 3 4 5 6 8 10 12 15 20 24 30 40 60 120) </lang>
Tcl
<lang tcl>proc factors {n} {
set factors {}
for {set i 1} {$i <= sqrt($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 64 1 3 5 9 15 45 1 53
UNIX Shell
This should work in all Bourne-compatible shells, assuming the system has both sort and at least one of bc or dc.
<lang>factor() {
r=`echo "sqrt($1)" | bc` # or `echo $1 v p | dc`
i=1
while [ $i -lt $r ]; do
if [ `expr $1 % $i` -eq 0 ]; then
echo $i
expr $1 / $i
fi
i=`expr $i + 1`
done | sort -nu
} </lang>
Ursala
The simple way: <lang Ursala>#import std
- import nat
factors "n" = (filter not remainder/"n") nrange(1,"n")</lang>
The complicated way:
<lang Ursala>factors "n" = nleq-<&@s <.~&r,quotient>*= "n"-* (not remainder/"n")*~ nrange(1,root("n",2))</lang>
Another idea would be to approximate an upper bound for the square root of "n" with some bit twiddling such as &!*K31 "n", which evaluates to a binary number of all 1's half the width of "n" rounded up, and another would be to use the division function to get the quotient and remainder at the same time. Combining these ideas, losing the dummy variable, and cleaning up some other cruft, we have
<lang Ursala>factors = nleq-<&@rrZPFLs+ ^(~&r,division)^*D/~& nrange/1+ &!*K31</lang>
where nleq-<& isn't strictly necessary unless an ordered list is required.
<lang Ursala>#cast %nL
example = factors 100</lang> output:
<1,2,4,5,10,20,25,50,100>
Wortel
<lang wortel>@let {
factors1 &n !-\%%n @to n factors_tacit @(\\%% !- @to) [[ !factors1 10 !factors_tacit 100 !factors1 720 ]]
}</lang>
Returns:
[ [1 2 5 10] [1 2 4 5 10 20 25 50 100] [1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 30 36 40 45 48 60 72 80 90 120 144 180 240 360 720] ]
XPL0
<lang XPL0>include c:\cxpl\codes; int N0, N, F; [N0:= 1; repeat IntOut(0, N0); Text(0, " = ");
F:= 2; N:= N0;
repeat if rem(N/F) = 0 then
[if N # N0 then Text(0, " * ");
IntOut(0, F);
N:= N/F;
]
else F:= F+1;
until F>N;
if N0=1 then IntOut(0, 1); \1 = 1
CrLf(0);
N0:= N0+1;
until KeyHit; ]</lang>
Example output:
1 = 1 2 = 2 3 = 3 4 = 2 * 2 5 = 5 6 = 2 * 3 7 = 7 8 = 2 * 2 * 2 9 = 3 * 3 10 = 2 * 5 11 = 11 12 = 2 * 2 * 3 13 = 13 14 = 2 * 7 15 = 3 * 5 16 = 2 * 2 * 2 * 2 17 = 17 18 = 2 * 3 * 3 . . . 57086 = 2 * 17 * 23 * 73 57087 = 3 * 3 * 6343 57088 = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 223 57089 = 57089 57090 = 2 * 3 * 5 * 11 * 173 57091 = 37 * 1543 57092 = 2 * 2 * 7 * 2039 57093 = 3 * 19031 57094 = 2 * 28547 57095 = 5 * 19 * 601 57096 = 2 * 2 * 2 * 3 * 3 * 13 * 61 57097 = 57097
zkl
<lang zkl>fcn f(n){ (1).pump(n.toFloat().sqrt(), List,
'wrap(m){((n % m)==0) and T(m,n/m) or Void.Skip}) }
fcn g(n){ [[(m); [1..n.toFloat().sqrt()],'{n%m==0}; '{T(m,n/m)} ]] } // list comprehension</lang>
- Output:
zkl: f(45) L(L(1,45),L(3,15),L(5,9)) zkl: g(45) L(L(1,45),L(3,15),L(5,9))
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