Écrivez du code qui, lorsqu'il reçoit un nombre positif en entrée, génère le plus grand diviseur positif de x inférieur ou égal à la racine carrée de x .$x$$x$$x$

En d'autres termes, trouvez le plus grand tel que$n > 0$

$\exists m\geq n:m\cdot n=x$

^{(Existe supérieur ou égal à n tel que m fois n soit x )$m$$n$$m$$n$$x$}

Par exemple, si l'entrée était les diviseurs sont 1 , 2 , 3 , 4 , 6 et 12 . 1 , 2 et 3 multiplient tous par de plus grands nombres pour obtenir 12 , mais$12$$1$$2$$3$$4$$6$$12$$1$$2$$3$$12$ étant le plus grand, nous retournons donc 3 .$3$$3$

Il s'agit de code-golf, donc les réponses seront notées en octets, moins d'octets étant considérés comme un meilleur score.

Cas de test

(1,1)
(2,1)
(3,1)
(4,2)
(5,1)
(6,2)
(7,1)
(8,2)
(9,3)
(10,2)
(11,1)
(12,3)
(13,1)
(14,2)
(15,3)
(16,4)
(17,1)
(18,3)
(19,1)
(20,4)
(21,3)
(22,2)
(23,1)
(24,4)
(25,5)
(26,2)
(27,3)
(28,4)
(29,1)
(30,5)
(31,1)
(32,4)
(33,3)
(34,2)
(35,5)
(36,6)
(37,1)
(38,2)
(39,3)
(40,5)
(41,1)
(42,6)
(43,1)
(44,4)
(45,5)
(46,2)
(47,1)
(48,6)
(49,7)
(50,5)

OEIS A033676

Python3 , 49 47 octets

def f(x):
 l=x**.5//1
 while x%l:l-=1
 return l

Explication

• l=x**.5//1 → Attribuer l le plus grand entier inférieur à égal à la racine carrée dex
• while x%l:l-=1→ Tandis que lne se divise pas uniformément x, décrémenter l.

Modifications

• Mentionnez Python3, pas Python2
• Utilisez ...//1pour enregistrer deux octets. (Les décimales vont bien! Merci @Rod)

MATL , 7 octets

Z\tn2/)

Essayez-le en ligne!

Pour cette explication, nous utiliserons «12» comme exemple d'entrée. Explication:

Z\      % Divisors.
        % Stack:
        %   [1 2 3 4 6 12]
  t     % Duplicate.
        % Stack:
        %   [1 2 3 4 6 12]
        %   [1 2 3 4 6 12]
   n    % Number of elements.
        % Stack:
        %   6
        %   [1 2 3 4 6 12]
    2/  % Divide by 2
        % Stack:
        %   3
        %   [1 2 3 4 6 12]
      ) % Index (grab the 3rd element)
        % 3

Cela fonctionne en raison de nombreuses coïncidences heureuses.

1. MATL utilise 1 indexation
2. Si nous index avec un non entier (ce sera le cas pour toute entrée carré parfait), puis <n>)indexera $\lceil n \rceil$

C (gcc) -lm , 35 octets

i;f(n){for(i=sqrt(n);n%i;i--);n=i;}

Essayez-le en ligne!

05AB1E, 5 bytes

Ñ2äнθ

Try it online! or as a Test suite

Explanation

Ñ        # push the list of divisors
 2ä      # split it into 2 parts
   н     # take the first haft
    θ    # take the last element of that

APL (Dyalog Unicode), 16 14 12 bytes

I'm glad I was able to write some answer in APL since I only just learned it. Many, many thanks to Adám for help with golfing. Golfing suggestions very much welcome. Try it online!

To learn more about APL, take a look at The APL Orchard.

EDIT: -2 bytes to fixing a problem with my code. Thanks to H.PWiz for pointing out that problem. -2 bytes from shortening everything again.

⌈/{⍳⌊⍵*÷2}∨⊢

Ungolfing

⌈/{⍳⌊⍵*÷2}∨⊢
          ∨   GCD of the following...
           ⊢    The right argument, our input.
  {⍳⌊⍵*÷2}
     ⍵            Our input.
      *÷2         To the power of 1/2, i.e. square root.
    ⌊             Floor.
   ⍳              Indices up to floor(sqrt(input)).
                In total, range from 1 to floor(sqrt(input)).
⌈/            The maximum of the GCDs of our input with the above range.

Husk, 4 bytes

→←½Ḋ

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Explanation

→←½Ḋ
   Ḋ      Divisors of (implicit) input.
  ½       Bisect.
→←        Take the last element of the first half.

R, 45 33 bytes

function(x,y=1:x^.5)max(y[!x%%y])

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Original:

function(x,y=x/1:x^.5)max(which(y==floor(y)))

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Code machine x86 32 bits (IA32): 18 16 octets

changelog: gérer n=1correctement le cas de test, enregistrer 2 octets et retourner dans EAX.

Comptez jusqu'à n/i <= i(c'est-à-dire lorsque nous atteignons le sqrt), et utilisez le premier diviseur exact après cela.

Une version 64 bits de ceci est appelable depuis C avec la convention d'appel System V x86-64, comme
int squarish_root_countup(int edi).

nasm -felf32 -l/dev/stdout squarish-root.asm:

58                         DEF(squarish_root_countup)
59                             ; input: n in EDI
60                             ; output: EAX
61                             ; clobbers: eax,ecx,edx
62                         .start:
63 00000025 31C9               xor    ecx, ecx
64                         .loop:                    ; do{
65                         
66 00000027 41                 inc    ecx                ; ++i
67 00000028 89F8               mov    eax, edi
68 0000002A 99                 cdq
69 0000002B F7F9               idiv   ecx                ; edx=n%i    eax=n/i
70                         
71 0000002D 39C1               cmp    ecx, eax
72 0000002F 7CF6               jl     .loop          ; }while(i < n/i
73                                                   ;          || n%i != 0);  // checked below
74                             ; falls through for i >= sqrt(n)
75                             ; so quotient <= sqrt(n) if we get here
76                         
77                                                   ; test edx,edx / jnz  .loop
78 00000031 4A                 dec    edx            ; edx-1 is negative only if edx was zero to start with
79 00000032 7DF3               jge   .loop           ; }while(n%i >= 1);
80                             ; falls through for exact divisors
81                         
82                             ; return value = quotient in EAX
83                         
84 00000034 C3                 ret

           0x10 bytes = 16 bytes.

85 00000035 10             .size: db $ - .start

Essayez-le en ligne! avec un appelant asm qui utilise directement le premier octet d'argv [1] comme entier et utilise le résultat comme état de sortie du processus.

$ asm-link -m32 -Gd squarish-root.asm && 
for i in {0..2}{{0..9},{a..f}};do 
    printf "%d   " "0x$i"; ./squarish-root "$(printf '%b' '\x'$i)"; echo $?;
done

0   0  # bash: warning: command substitution: ignored null byte in input
1   1
2   1
3   1
4   2
5   1
6   2
7   1
8   2
9   3
10   0       # this is a testing glitch: bash ate the newline so we got an empty string.  Actual result is 2 for n=10
11   1
12   3
13   1
14   2
15   3
16   4
   ...

Japt -h, 8 6 octets

â f§U¬

Essayez-le

2 octets économisés grâce à Oliver

Explication

           :Implicit input of integer U
â          :Divisors of U
  f        :Filter
   §       :  Less than or equal to
    U¬     :  Square root of U
           :Implicitly get the last element in the array and output it

Jelly, 5 bytes

ọⱮ½TṪ

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JavaScript ES7, 33 31 bytes

n=>(g=x=>n%x?g(x-1):x)(n**.5|0)

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Snowman, 38 bytes

((}1vn2nD`#nPnF|:|NdE|;:,#NMo*|,;bW*))

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((
  }        activate variables b, e, and g
  1vn2nD`  e=1/2
  #        retrieve the input into b
  nP       set b=b^e, which is sqrt(input)
  nF       floor the square root
  |        move b into g so there's space for a while loop
  :        body of the loop
    |NdE|  decrement the value in g
  ;:       loop condition
    ,#     assign b=input, e=current value
    NMo    store the modulo in g
    *|     discard the input value and place the modulo in the condition slot
    ,      put the current value back into g
  ;bW      continue looping while the modulo is nonzero
  *        return the result
))

dc, 24

?dsnv1+[1-dlnr%0<m]dsmxp

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Explanation:

?                         # read input
 d                        # duplicate
  sn                      # store copy 1 in register n
    v                     # take the square root of copy 2
     1+                   # add 1
       [          ]       # define macro to:
        1-                #   subtract 1
          d               #   duplicate
           ln             #   load from register n
             r            #   reverse top 2 stack members
              %           #   calculate modulo
               0<m        #   if not 0, recursively call macro m again
                   d      # duplicate macro
                    sm    # store copy 1 in register m
                      x   # execute copy 2
                       p  # print final value

J, 24 19 bytes

-5 bytes thanks to Sherlock's GCD idea

([:>./+.)1+i.@<.@%:

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original answer

([:{:]#~0=]|[)1+i.@<.@%:

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parsed

┌───────────────────────────────┬──────────────────────┐
│┌──┬──┬───────────────────────┐│┌─┬─┬────────────────┐│
││[:│{:│┌─┬─────┬─────────────┐│││1│+│┌─────────┬─┬──┐││
││  │  ││]│┌─┬─┐│┌─┬─┬───────┐││││ │ ││┌──┬─┬──┐│@│%:│││
││  │  ││ ││#│~│││0│=│┌─┬─┬─┐│││││ │ │││i.│@│<.││ │  │││
││  │  ││ │└─┴─┘││ │ ││]│|│[││││││ │ ││└──┴─┴──┘│ │  │││
││  │  ││ │     ││ │ │└─┴─┴─┘│││││ │ │└─────────┴─┴──┘││
││  │  ││ │     │└─┴─┴───────┘│││└─┴─┴────────────────┘│
││  │  │└─┴─────┴─────────────┘││                      │
│└──┴──┴───────────────────────┘│                      │
└───────────────────────────────┴──────────────────────┘

explanation

• 1 + i.@<.@%: gives the range 1 .. floor(sqrt).
• the entire verb (A) B forms a hook, with the above range passed as the right arg ] to A and the original number passed as its left arg [. Thus...
• ] | [ gives the remainer of each item in the range divided into the original arg.
• and 0 = ] | [ gives the divisors with no remainder.
• ] #~ ... then filters the range, leaving only those.
• and {: gives the last item in the list, ie, the largest one.

Jelly, 5 bytes

½ḍƇµṪ

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Haskell, 36 bytes

f x=[z|y<-[1..],z<-[1..y],y*z==x]!!0

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Well this is my answer to this challenge. This uses a particular list comprehension to find the answer. In our list comprehension we pick $y$ from the infinite list [1..] that is the positive integers, and we pick $z$ from the list [1..y]. This means that $(y,z)$ is all the ordered pairs such that $y \geq z$.

We then select only those pairs such that $y\cdot z=x$, meaning that we make the list of all pairs of numbers that multiply to $x$. Now since our comprehension is based first on $y$ and then $z$ this means that our pairs are in ascending order of $y$, or more usefully in descending order of $z$.

So to get the largest $z$ we take the $z$ belonging to the first element. This is our result.

QBasic (4.5), 52 bytes

INPUT x
FOR i=1TO sqr(x)
if x/i=x\i then m=i
next
?m

Forth (gforth), 53 bytes

The shortest way seems to be using the floating point stack and fsqrt, the shortest I could get without it was 62 bytes using /mod and checking if the quotient was greater than the divisor.

: f dup s>f fsqrt f>s 1+ begin 1- 2dup mod 0= until ;

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Explanation

1. Calculate the square root
2. Starting at the square root, decrement by 1 until we find a factor of the original number

Code Explanation

: f                \ Start a word definition
dup                \ duplicate the input
s>f fsqrt          \ move the number to the float stack and get the square root
f>s                \ truncate result and move to integer stack
1+                 \ add 1 to the square root
begin              \ start indefinite loop
  1- 2dup          \ decrement divisor and duplicate input and divisor
  mod              \ calculate n % divisor
0= until           \ if result equals 0 (no remainder) end the loop
;                  \ end the word definition

F#, 55 49 bytes

let f x=Seq.findBack(fun i->x%i=0.0){1.0..x**0.5}

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Seq.findBack: Returns the last element for which the given function returns True. The function in this case checks to see if a number is a factor of the value.

Brain-Flak, 144 bytes

{({}{}<<>({}<>)<>([({})()]<>({}(<>)())){(<{}({}[()]{}<({}())>)>)}{}((({}<>)<>(({})))[({}[{}])])>[({<({}[()])><>({})<>}{}<><{}>)])}{}{}<>{}({}<>)

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I'm not really sure this answer is very good. I feel like there may be a nice way to solve this task however I'm just not clever enough.

Explanation

I tried to do an exploded view of the answer but there are so many moving parts it was not very enlightening, so here is an explanation of what the code does.

The first important bit is this

({}<>)<>([({})()]<>({}(<>)())){(<{}({}[()]{}<({}())>)>)}{}

This takes the two numbers on top of the stack and if they are unequal increments the second one, if they are equal it increments the first one and replaces the second one with zero. If we repeat this code a bunch we will get all the pairs $(x,y)$ such that $x \geq y$.

The next part is multiplication, taken with modification from the wiki. This multiplication is special because it preserves the existing values without destroying them. It goes like:

((({}<>)<>(({})))[({}[{}])])({<({}[()])><>({})<>}{}<><{}>)

So we are multiplying all these ordered pairs. For each result we check if it is equal to the input. If so we terminate and return the smaller item in the pair.

Python 2, 41 bytes

n=k=input()
while(k*k>n)+n%k:k-=1
print k

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Jelly, 6 bytes

ÆDŒHḢṪ

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Perl 5 -p, 26 bytes

$\=0|sqrt;$\--while$_%$\}{

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Rust, 71 70 bytes

fn f(x:u64)->u64{let mut l=(x as f64).sqrt()as u64;while x%l>0{l-=1}l}

Pre-uglified version

fn f(x: u64) -> u64 {                    // function takes u64, gives u64
  let mut l = (x as f64).sqrt() as u64;  // l takes integer'ed root value
  while x % l > 0 {                      // loop while l leaves remainder
    l -= 1                               // decrement
  }
  l                                      // return the found value
}

Edits

• Save a byte with > 0 over != 0. (Thanks to @CatWizard)

Japt, 8 bytes

â w æ§Uq

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Triangularity, 49 bytes

....)....
...@_)...
..IEdF)..
.2)1/)IE.
^>{m.....

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Pyret, 93 bytes

{(z):rec f={(i,x):if num-modulo(i, x) == 0:x else:f(i,x - 1)end}
f(z,num-floor(num-sqrt(z)))}

You can try this online by copying it into the online Pyret editor!

The above evaluates to an anonymous function. When it is applied to an integer, it returns a result according to the specification.

Actually, 7 bytes

Based on my APL answer here. Golfing suggestions welcome! Try it online!

;√LR♀gM

Ungolfing

;√LR♀gM  Implicit input n
;        Duplicate n
 √       sqrt(n)
  L      floor(sqrt(n))
   R     1..floor(sqrt(n))
    ♀g   gcd(n, foreach(1..floor(sqrt(n)))
      M  The maximum of the GCDs.
         Return this maximum.

A port of this Mathematica answer.

Jelly, 11 bytes

½ðḞ³÷Ċ³÷µÐL

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This (11 bytes) also works, and don't depend on ³:

½Ḟ÷@Ċ÷@ʋƬµṪ

Unfortunately ½Ḟ÷@Ċ÷@ʋÐL (10 bytes) doesn't work. And apparently Ƭ and ÐĿ is not exactly the same (when the link is dyadic)

Method: (let $n$ be the input)

• Start with an upper bound $i = \sqrt n$ of the answer $a$.
• At each step:
  • If $i$ is not an integer, then the upper bound can be made $\lfloor i\rfloor$ (because the result must be an integer)
  • If $n\over i$ is not an integer, then $a \le i \Rightarrow \frac n a \ge \frac n i \Rightarrow \frac n a \ge \lceil \frac n i \rceil \Rightarrow a \le n \div \lceil \frac n i \rceil$.
• So we repeatedly replace $i$ with $n \div \lceil \frac n {\lfloor i\rfloor} \rceil$ until it's fixed.

Java 8, 65 54 bytes

n->{int r=(int)Math.sqrt(n);for(;n%r>0;r--);return r;}

Port of @hunteke's Python 3 answer.

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Old 65 bytes answer:

n->{int r=1,i=n;for(;i-->1;)r=n%i<1&n/i<=i&n/i>r?n/i:r;return r;}

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Explanation:

n->{                // Method with integer as both parameter and return-type
  int r=1,          //  Result-integer, starting at 1
  i=n;for(;i-->1;)  //  Loop `i` in the range (n, 1]
    r=n%i<1         //   If `n` is divisible by `i`,
      &n/i<=i       //   and if `n` divided by `i` is smaller than or equal to `i` itself,
      &n/i>r?       //   and if `n` divided by `i` is larger than the current `r`
       n/i          //    Set `n` divided by `i` as the new result `r`
      :             //   Else:
       r;           //    Leave result `r` unchanged
  return r;}        //  Return the result `r`