Dans cette tâche, vous devez écrire un programme qui calcule les facteurs premiers d'un nombre. L'entrée est un nombre naturel 1 <n <2 ^ 32. La sortie est une liste des facteurs premiers du nombre dans le format suivant. Les exposants doivent être omis s'ils sont 1. Uniquement les nombres premiers en sortie. (En supposant que l'entrée est 131784):

131784 = 2 ^ 3 * 3 * 17 ^ 2 * 19

Il n'est pas nécessaire d'utiliser la même quantité d'espace; un espace peut être inséré chaque fois que cela est approprié. Votre programme devrait se terminer en moins de 10 minutes pour toute entrée. Le programme avec le plus petit nombre de caractères gagne.

SageMath, 31 octets

N=input()
print N,"=",factor(N)

Cas de test: 83891573479027823458394579234582347590825792034579235923475902312344444 Sorties:

83891573479027823458394579234582347590825792034579235923475902312344444 = 2^2 * 3^2 * 89395597 * 98966790508447596609239 * 263396636003096040031295425789508274613

Ruby 1.9, 74 70 caractères

#!ruby -plrmathn
$_+=?=+$_.to_i.prime_division.map{|a|a[0,a[1]]*?^}*?*

Modifications:

• (74 -> 70) Utilisez simplement l'exposant comme longueur de tranche au lieu de vérifier explicitement exponent > 1

Perl 5.10, 73 88

perl -pe '$_=`factor $_`;s%( \d+)\K\1+%-1-length($&)/length$1%ge;y, -,*^,;s;\D+;=;'

Prend le numéro d'entrée de l'entrée standard. Calculera les facteurs pour plusieurs entrées si elles sont fournies.

Compté comme une différence pour perl -e. 5.10 est nécessaire pour le \Kmétacaractère regex.

OCaml, 201 caractères

Une traduction impérative directe du meilleur code Python:

let(%)s d=if!d>1then Printf.printf"%s%d"s!d
let f n=let x,d,e,s=ref n,ref 1,ref 0,ref"="in""%x;while!d<65536do
incr d;e:=0;while!x mod!d=0do x:=!x/ !d;incr e
done;if!e>0then(!s%d;"^"%e;s:="*")done;!s%x

Par exemple,

# f 4294967292;;
4294967292=2^2*3^2*7*11*31*151*331- : unit = ()

(Notez que j'ai omis de produire la fin finale.) Juste pour le plaisir, à 213 caractères, une version purement fonctionnelle, complètement obscurcie par une utilisation libérale des opérateurs:

let(%)s d=if d>1then Printf.printf"%s%d"s d
let f x=let s=ref"="in""%x;let rec(@)x d=if d=65536then!s%x else
let rec(^)x e=if x/d*d<x then x,e else x/d^e+1in
let x,e=x^0in if e>0then(!s%d;"^"%e;s:="*");x@d+1in x@2

Python, 140 135 133 caractères

M=N=input()
s=''
f=1
while f<4**8:
 f+=1;e=0
 while N%f<1:e+=1;N/=f
 if e:s+='*%d'%f+'^%d'%e*(e>1)
print M,'=',(s+'*%d'%N*(N>1))[1:]

J, 72

(":*/f),'=',([,'*',])/(":"0~.f),.(('^',":)`(''"0)@.(=&1))"0+/(=/~.)f=.q:161784

Typique J. Deux personnages pour faire la majeure partie du travail, soixante personnages pour le présenter.

Modifier: correction du nombre de caractères.

J, 53 52 characters

This solution takes the rplc trick from the solution of randomra but comes up with some original ideas, too.

":,'=',(":@{.,'^','*',~":@#)/.~@q:}:@rplc'^1*';'*'"_

In non-tacit notation, this function becomes

f =: 3 : 0
(": y) , '=' , }: (g/.~ q: y) rplc '^1*' ; '*'
)

where g is defined as

g =: 3 : 0
": {. y) , '^' , (": # y) , '*'
)

• q: yest le vecteur des facteurs premiers de y. Par exemple, les q: 60rendements 2 2 3 5.

• x u/. ys'applique uà la y clé de x, c'est-à-dire, uest appliquée aux vecteurs d'éléments ydont les entrées dans xsont égales. C'est un peu complexe à expliquer, mais dans le cas particulier y u/. you u/.~ y, uest appliqué à chaque vecteur d'éléments distincts dans y, où chaque élément est répété aussi souvent qu'il y apparaît y. Par exemple, les </.~ 1 2 1 2 3 1 2 2 3rendements

  ┌─────┬───────┬───┐
  │1 1 1│2 2 2 2│3 3│
  └─────┴───────┴───┘
  

• # yest le décompte de y, qui est, le nombre d'éléments y.

• ": y formats y sous forme de chaîne.
• x , y ajoute x et y.
• {. yest la tête y , c'est-à-dire son premier élément.
• Thus, (": {. y), '^' , (": # y) , '*' formats a vector of n repetitions of a number k into a string of the form k ^ n *. This phrase in tacit notation is :@{.,'^','*',~":@#, which we pass to the adverb /. described further above.
• x rplc y is the library function replace characters. y has the form a ; b and every instance of string a in x is replaced by b. x is ravelled (that is, reshaped such that it has rank 1) before operation takes place, which is used here. This code replaces ^1* with * as to comply with the mandated output format.
• }: y is the curtail of y, that is, all but its last item. This is used to remove the trailing *.

PHP, 112

echo$n=$_GET[0],'=';$c=0;for($i=2;;){if($n%$i<1){$c++;$n/=$i;}else{if($c){echo"$i^$c*";}$c=0;if(++$i>$n)break;}}

118

echo $n=$_GET[0],'=';for($i=2;;){if(!($n%$i)){++$a[$i];$n/=$i;}else{if($a[$i])echo "$i^$a[$i]*";$i++;if($i>$n)break;}}

Python 119 Chars

M=N=input()
i=1
s=""
while N>1:
 i+=1;c=0
 while N%i<1:c+=1;N/=i
 if c:s+=" * %d"%i+['','^%d'%c][c>1]
print M,'=',s[3:]

JavaScript, 124 122 119

for(s='',i=2,o=p=prompt();i<o;i++){for(n=0;!(p%i);n++)p/=i;n?s+=i+(n-1?'^'+n:'')+'*':0}alert(s.substring(0,s.length-1))

Perl, 78

use ntheory":all";say join" * ",map{(join"^",@$_)=~s/\^1$//r}factor_exp(shift)

It uses the s///r feature of Perl 5.14 to elide the ^1s. 81 characters to run in a loop:

perl -Mntheory=:all -nE 'chomp;say join" * ",map{(join"^",@$_)=~s/\^1$//r}factor_exp($_);'

PHP, 236 characters

$f[$n=$c=$argv[1]]++;echo"$n=";while($c){$c=0;foreach($f as$k=>$n)for($r=~~($k/2);$r>1;$r--){if($k%$r==0){unset($f[$k]);$f[$r]++;$f[$k/$r]++;$c=1;break;}}}foreach($f as$k=>$n)if(--$n)$f[$k]="$k^".++$n;else$f[$k]=$k;echo implode("*",$f);

Output for 131784: 2^3*3*17^2*19

Completes all numbers within a few seconds while testing.

4294967296=2^32
Time: 0.000168

Input was never specified, so I chose to call it using command line arguments.

php factorize.php 4294967296

Scala 374:

def f(i:Int,c:Int=2):List[Int]=if(i==c)List(i)else 
if(i%c==0)c::f(i/c,c)else f(i,c+1)
val r=f(readInt)
class A(val v:Int,val c:Int,val l:List[(Int,Int)])
def g(a:A,i:Int)=if(a.v==i)new A(a.v,a.c+1,a.l)else new A(i,1,(a.v,a.c)::a.l)
val a=(new A(r.head,1,Nil:List[(Int,Int)])/:(r.tail:+0))((a,i)=>g(a,i))
a.l.map(p=>if(p._2==1)p._1 else p._1+"^"+p._2).mkString("", "*", "")

ungolfed:

def factorize (i: Int, c: Int = 2) : List [Int] = {
  if (i == c) List (i) else 
    if (i % c == 0) c :: f (i/c, c) else 
      f (i, c+1)
}
val r = factorize (readInt)
class A (val value: Int, val count: Int, val list: List [(Int, Int)])
def g (a: A, i: Int) = 
  if (a.value == i) 
    new A (a.value, a.count + 1, a.list) else 
    new A (i, 1, (a.value, a.count) :: a.list)
val a = (new A (r.head, 1, Nil: List[(Int,Int)]) /: (r.tail :+ 0)) ((a, i) => g (a, i))
a.l.map (p => if (p._2 == 1) p._1 else
  p._1 + "^" + p._2).mkString ("", "*", "")

J, 74 chars

f=.3 :0
(":y),'=',' '-.~('^1 ';'')rplc~}:,,&' *'"1(,'^'&,)&":/"{|:__ q:y
)

   f 131784
131784=2^3*3*17^2*19

64 chars with input in variable x:

   x=.131784

   (":x),'=',' '-.~('^1 ';'')rplc~}:,,&' *'"1(,'^'&,)&":/"{|:__ q:x
131784=2^3*3*17^2*19

Java 10, 109 108 bytes (lambda function) (non-competing on request of OP)

n->{var r=n+"=";for(int i=1,f;i++<n;r+=f<1?"":(f<2?i:i+"^"+f)+(n>1?"*":""))for(f=0;n%i<1;n/=i)f++;return r;}

Try it online.

Java 6+, 181 bytes (full program)

class M{public static void main(String[]a){long n=new Long(a[0]),i=1,f;String r=n+"=";for(;i++<n;r+=f<1?"":(f<2?i:i+"^"+f)+(n>1?"*":""))for(f=0;n%i<1;n/=i)f++;System.out.print(r);}}

Try it online.

-1 byte thanks to @ceilingcat.

Explanation:

n->{                // Method with integer parameter and String return-type
  var r=n+"=";      //  Result-String, starting at the input with an appended "="
  for(int i=1,f;i++<n;
                    //  Loop in the range [2, n]
      r+=           //    After every iteration: append the following to the result-String:
        f<1?        //     If the factor `f` is 0:
         ""         //      Append nothing
        :           //     Else:
         (f<2?      //      If the factor `f` is 1:
           i        //       Append the current prime `i`
          :         //      Else:
           i+"^"+f) //       Append the current prime `i` with it's factor `f`
         +(n>1?     //      And if we're not done yet:
            "*"     //       Also append a "*"
           :        //      Else:
            ""))    //       Append nothing more
    for(f=0;        //   Reset the factor `f` to 0
        n%i<1;      //   Loop as long as `n` is divisible by `i`
      n/=i)         //    Divide `n` by `i`
      f++;          //    Increase the factor `f` by 1
  return r;}        //  Return the result-String

Japt, 28 27 26 bytes

-1 byte thanks to Shaggy

+'=+Uk ü ®ÊÉ?ZÌ+'^+Zl:ZÃq*

Try it

Powershell, 113 97 bytes

Inspired by Joey's answer. It's a slow but short.

param($x)(2..$x|%{for(;!($x%$_)){$_
$x/=$_}}|group|%{$_.Name+"^"+$_.Count-replace'\^1$'})-join'*'

Explained test script:

$f = {

param($x)               # let $x stores a input number > 0
(2..$x|%{               # loop from 2 to initial input number
    for(;!($x%$_)){     # loop while remainder is 0
        $_              # push a current value to a pipe
        $x/=$_          # let $x is $x/$_ (new $x uses in for condition only)
    }
}|group|%{              # group all values
    $_.Name+"^"+$_.Count-replace'\^1$'  # format and remove last ^1
})-join'*'              # make string with *

}

&$f 2
&$f 126
&$f 129
&$f 86240
#&$f 7775460

Output:

2
2*3^2*7
3*43
2^5*5*7^2*11

Jelly, 16 bytes (non-competing on request of OP)

³”=³ÆFḟ€1j€”^j”*

One of my first Jelly answers, so can definitely be golfed (especially ³”=³)..

Try it online.

Explanation:

³                 # Push the first argument
 ”=               # Push string "="
   ³ÆF            # Get the prime factor-exponent pairs of the first argument
      ḟ€1         # Remove all 1s from each pair
         j€”^     # Join each pair by "^"
             j”*  # Join the pair of strings by "*"
                  # (implicitly join the entire 'stack' together)
                  # (which is output implicitly as result)

05AB1E, 22 20 bytes (non-competing on request of OP)

ÐfsÓ0Køε1K'^ý}'*ý'=ý

-2 bytes thanks to @Emigna.

Try it online.

Explanation:

Ð                # Triplicate the (implicit) input-integer
 f               # Pop and push all prime factors (without counting duplicates)
  s              # Swap to take the input again
   Ó             # Get all prime exponents
    0K           # Remove all 0s from the exponents list
      ø          # Zip it with the prime factors, creating pairs
       ε         # Map each pair to:
        1K       #  Remove all 1s from the pair
        '^ý     '#  And then join by "^"
       }'*ý     '# After the map: join the string/integers by "*"
           '=ý  '# And join the stack by "=" (with the input we triplicated at the start)
                 # (after which the result is output implicitly)

APL(NARS), 66 chars, 132 bytes

{(⍕⍵),'=',3↓∊{m←' * ',⍕↑⍵⋄1=w←2⊃⍵:m⋄m,'^',⍕w}¨v,¨+/¨{k=⍵}¨v←∪k←π⍵}

test and comment:

  f←{(⍕⍵),'=',3↓∊{m←' * ',⍕↑⍵⋄1=w←2⊃⍵:m⋄m,'^',⍕w}¨v,¨+/¨{k=⍵}¨v←∪k←π⍵}
  f 131784
131784=2^3 * 3 * 17^2 * 19
  f 2
2=2
  f (2*32)
4294967296=2^32

{(⍕⍵),'=',3↓∊{m←' * ',⍕↑⍵⋄1=w←2⊃⍵:m⋄m,'^',⍕w}¨v,¨+/¨{k=⍵}¨v←∪k←π⍵}
k←π⍵      find the factors with repetition of ⍵ and assign that array to k example for 12 k is 2 2 3
v←∪       gets from k unique elements and put them in array v
+/¨{k=⍵}¨ for each element of v count how many time it appear in k (it is array exponents)
v,¨       make array of couples from element of v (factors unique) and the array above (exponents unique)
∊{m←' * ',⍕↑⍵⋄1=w←2⊃⍵:m⋄m,'^',⍕w}¨ pretty print the array of couples factor exponent as array chars
3↓                                 but not the first 3 chars
(⍕⍵),'='  but print first the argument and '=' in char format

if someone has many time with these primitives, know them very well them, for me it is possible that the code is clearer of comments... so code more clear than comments, comments unuseful...

JavaScript, 107

n=prompt()
s=n+'='
c=0
for(i=2;;){if(n%i<1){c++
n/=i}else{if(c)s+=i+'^'+c+'*'
c=0
if(++i>n)break}}
alert(s)

120

n=prompt()
o={2:0}
for(i=2,c=n;i<=c;)!(c%i)?++o[i]?c/=i:0:o[++i]=0
s=n+'='
for(i in o)s+=o[i]?i+'^'+o[i]+'*':''
alert(s)

PHP, 112 bytes

<?=$a=$argn,"=";for($i=2;1<$a;)$a%$i?$i++:$a/=$i+!++$r[$i];foreach($r as$k=>$v)echo$t?"*":!++$t,$v<2?$k:"$k^$v";

Try it online!

PHP, 93 bytes

<?=$n=$argn;for($i=2;$n>1;$k&&$p=print($p?"*":"=")."$i^$k",$i++)for($k=0;$n%$i<1;$n/=$i)$k++;

I could do 89 bytes with PHP 5.5 (or later), but that postdates the challenge by more than 2 years:

<?=$n=$argn;for($i=2;$n>1;$k&&$p=print"=*"[$p]."$i^$k",$i++)for($k=0;$n%$i<1;$n/=$i)$k++;

Run as pipe with -nF or try them online.