Tantangan

Diberikan bilangan bulat, n, sebagai input di mana 0 <= n <= 2^10, menampilkan angka ke-n bahkan sempurna.

Bilangan Sempurna

Angka sempurna adalah angka, x di mana jumlah faktor-faktornya (tidak termasuk dirinya sendiri) sama dengan x. Misalnya, 6:

6: 1, 2, 3, 6

Dan, tentu saja 1 + 2 + 3 = 6, jadi 6 sempurna.

Jika angka sempurna x,, genap x mod 2 = 0,.

Contohnya

Berikut ini adalah 10 angka sempurna pertama:

6
28
496
8128
33550336
8589869056
137438691328
2305843008139952128
2658455991569831744654692615953842176
191561942608236107294793378084303638130997321548169216

Perhatikan bahwa Anda dapat mengindeks ini sesuka Anda: 6 mungkin nomor 1 atau 0 bahkan sempurna.

Kemenangan

Kode terpendek dalam byte menang.

Jelly , 7 byte

6Æṣ=$#Ṫ

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Bagaimana itu bekerja

6Æṣ=$#Ṫ  Main link. Argument: n

6        Set the return value to 6.
     #   Execute the link to the left with argument k = 6, 7, 8, ... until n
         values of k result in a truthy value. Yield the array of matches.
    $        Combine the two links to the left into a monadic chain.
 Æṣ              Compute the sum of k's proper divisors.
   =             Compare the result with k.
      Ṫ  Tail; extract the last match.

Mathematica, 13 byte

Tidak mengherankan, ada built-in.

PerfectNumber

Contoh:

In[1]:= PerfectNumber[18]                                                       

Out[1]= 33570832131986724437010877211080384841138028499879725454996241573482158\

>    45044404288204877880943769038844953577426084988557369475990617384115743842\

>    47301308070476236559422361748505091085378276585906423254824947614731965790\

>    74656099918600764404702181660294469121778737965822199901663478093006075022\

>    35922320184998563614417718592540207818507301504509772708485946474363553778\

>    15002849158802448863064617859829560720600134749556178514816801859885571366\

>    09224841817877083608951191123174885226416130683197710667392351007374503755\

>    40335253147622794359007165170269759424103195552989897121800121464177467313\

>    49444715625609571796578815564191221029354502997518133405151709561679510954\

>    53649485576150660101689160658011770193274226308280507786835049549112576654\

>    51011967045674593989019420525517538448448990932896764698816315598247156499\

>    81962616327512831278795091980742531934095804545624886643834653798850027355\

>    06153988851506645137759275553988219425439764732399824712438125054117523837\

>    43825674443705501944105100648997234160911797840456379499200487305751845574\

>    87014449512383771396204942879824895298272331406370148374088561561995154576\

>    69607964052126908149265601786094447595560440059050091763547114092255371397\

>    42580786755435211254219478481549478427620117084594927467463298521042107553\

>    17849183589266903954636497214522654057134843880439116344854323586388066453\

>    13826206591131266232422007835577345584225720310518698143376736219283021119\

>    28761789614688558486006504887631570108879621959364082631162227332803560330\

>    94756423908044994601567978553610182466961012539222545672409083153854682409\

>    31846166962495983407607141601251889544407008815874744654769507268678051757\

>    74695689121248545626112138666740771113961907153092335582317866270537439303\

>    50490226038824797423347994071302801487692985977437781930503487497407869280\

>    96033906295910199238181338557856978191860647256209708168229116156300978059\

>    19702685572687764976707268496046345276316038409383829227754491185785965832\

>    8888332628525056

MATL , 15 byte

`@Z\s@E=vtsG<}n

Sangat lambat. Itu terus mencoba meningkatkan angka satu per satu sampai angka sempurna ke- n ditemukan.

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Penjelasan

`        % Do...while
  @      %   Push iteration index, k (starting at 1)
  Z\     %   Array of divisors
  s      %   Sum
  @E     %   Push k. Multiply by 2
  =      %   Equal? If so, k is a perfect number
  v      %   Concatenate vertically. This gradually builds an array which at the k-th
         %   iteration contains k zero/one values, where ones indicate perfect numbers
  ts     %   Duplicate. Sum of array
  G<     %   Push input. Less than? This is the loop condition: if true, proceed with
         %   next iteration
}        % Finally (execute right before exiting loop)
  n      %   Number of elements of the array
         % End (implicit). Display (implicit)

Pyth , 13 byte

e.fqsf!%ZTStZ

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Tolong jangan mencoba angka yang lebih tinggi. Itu hanya menguji angka genap satu per satu.

05AB1E , 8 byte

µNNѨOQ½

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Penjelasan

µ          # loop over increasing N until counter equals input
 N         # push N
  NÑ       # push factors of N
    ¨      # remove last factor (itself)
     O     # sum factors
      Q    # compare the sum to N for equality
       ½   # if true, increase counter

Python 2 , 198 153 83 78 77 75 74 byte

i=input()
j=0
while i:j+=1;i-=sum(x*(j%x<1)for x in range(1,j))==j
print j

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Sekarang hanya berbunyi seperti psuedocode.

• Diselamatkan 45 Countless Bytes karena @Leaky Nun mengajari saya tentang fungsi penjumlahan dan pemahaman daftar.

• Disimpan 2 byte berkat saran @ shooqie untuk menghapus tanda kurung yang tidak perlu.

Kami hanya mengulangi setiap angka genap sampai kami menemukan n angka sempurna.

PHP, 111 Bytes

0-Pengindeksan

Bekerja dengan konsep bahwa angka sempurna adalah angka di mana n=x*y x=2^idan y=2^(i+1)-1dan y harus prima

for(;!$r[$argn];$u?:$r[]=$z)for($z=2**++$n*($y=2**($n+1)-1),$u=0,$j=1;$j++<sqrt($y);)$y%$j?:$u++;echo$r[$argn];

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Python 3 , 69 byte

f=lambda n,k=1:n and-~f(n-(sum(j>>k%j*j for j in range(1,k))==k),k+1)

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Scala, 103 byte

n=>Stream.from(1).filter(_%2==0).filter(x=>Stream.from(1).take(x-1).filter(x%_==0).sum==x).drop(n).head

Haskell, 61 Bytes

(!!)(filter(\x->x==sum[n|n<-[1..x-1],x`mod`n==0]||x==1)[1..])

JavaScript (ES6), 68 byte

n=>eval(`for(x=5;n;s||n--)for(f=s=++x;f--;)(x/f-(x/f|0))||(s-=f);x`)

F=n=>eval('for(x=5;n;s||n--)for(f=s=++x;f--;)(x/f-(x/f|0))||(s-=f);x')

console.log(
  F(1),
  F(2),
  F(3),
  //F(4),
  //F(5),
)

Perl 6 , 42 byte

{(grep {$_==[+] grep $_%%*,^$_},^∞)[$_]}

Indeks input berbasis 1.

Clojure, 79 byte

#(nth(for[i(range):when(=(apply +(for[j(range 1 i):when(=(mod i j)0)]j))i)]i)%)

Mengikuti spec, penggunaan berat untuk :whenkondisi.