整数が2のべき乗かどうかをチェックするプログラムを作成します。

サンプル入力：

8

サンプル出力：

Yes

サンプル入力：

10

サンプル出力：

No

ルール：

• +、-操作を使用しないでください。

• 何らかの入力ストリームを使用して、数値を取得します。入力は、最初は変数に保存されることは想定されていません。

• 最短のコード（バイト単位）が優先されます。

真実/偽の応答（たとえば、true/ false）を使用できます。入力数がより大きいと仮定できます0。

GolfScript、6文字、減分なし

~.3/&!

これはx & (x-1)、どのような形式でもメソッドを使用しないソリューションです。x & (x/3)代わりに使用します。;-) 0falseの1場合、trueの場合に出力します。

説明：

• ~ 入力文字列を評価して数値に変換し、
• .それを複製します（後続の&）
• 3/ 3で割る（切り捨て）、
• & 除算された値と元のビット単位のANDを計算します。入力がゼロまたは2のべき乗の場合にのみゼロになります（つまり、最大1ビットが設定されます）。
• ! これを論理的に否定し、ゼロを1にマッピングし、他のすべての値をゼロにマッピングします。

ノート：

• 明確化された規則に従って、ゼロは有効な入力ではないため1、入力がゼロの場合は出力されますが、このコードは問題ありません。

• GolfScriptの減少演算子(が許可されている場合、aditsuが投稿した5文字のソリューションで十分です。ただし、手紙ではないにしても、ルールの精神に反するようです。~.(&!

• x & (x/3)Fun With Perlメーリングリストで何年も前にこのトリックを思いつきました。（私はそれを最初に発見したわけではないと確信していますが、私はそれを独立して（再）発明しました） 実際の動作の証拠を含む、元の投稿へのリンクです。

APL（7）

はい、それは7 バイトですです。今のところ、Unicodeの代わりにIBMコードページ907を使用しており、各文字が1バイトであると仮定します:)

0=1|2⍟⎕

すなわち 0 = mod(log(input(),2),1)

GolfScript、11（1（true）および0（false）の場合）

.,{2\?}%?0>

スタックに番号を付けて実行します。

GolfScript、22（はい/いいえ）

.,{2\?}%?0>'Yes''No'if

私はどのように変換愛1/ 0へYes/No D：挑戦自体として多くのコードのようになります

警告：非常に非効率的です;）10000までの数値に対しては正常に動作しますが、一度それを高くすると、わずかな遅れに気づき始めます。

説明：

• .,：になりnますn 0..n（.重複、,0..n範囲）
• {2\?}：2の累乗
• %：「0..n」の上に「2のべき乗」をマップして、 n [1 2 4 8 16 ...]
• ?0>：配列に数値が含まれているかどうかを確認します（0はインデックスより大きい）

Mathematica 28

Numerator[Input[]~Log~2]==1

2の整数のべき乗の場合、2を底とする対数の分子は1になります（つまり、対数は単位分数です）。

ここでは、推定入力を表示するために関数をわずかに変更します。の#代わりに使用してInput[]追加&し、純粋な関数を定義します。ユーザーが上記の関数で数値を入力した場合に返されるのと同じ回答を返します。

    Numerator[#~Log~2] == 1 &[1024]
    Numerator[#~Log~2] == 1 &[17]

真
偽

一度に複数の数値をテストします。

    Numerator[#~Log~2] == 1 &&/@{64,8,7,0}

{True、True、False、False}

Perl 6（17文字）

say get.log(2)%%1

This program gets a line from STDIN get function, calculates logarithm with base 2 on it (log(2)), and checks if the result divides by 1 (%%1, where %% is divides by operator). Not as short as GolfScript solution, but I find this acceptable (GolfScript wins everything anyway), but way faster (even considering that Perl 6 is slow right now).

~ $ perl6 -e 'say get.log(2)%%1'
256
True
~ $ perl6 -e 'say get.log(2)%%1'
255
False

Octave (15 23)

EDIT: Updated due to user input requirement;

~mod(log2(input('')),1)

Lets the user input a value and outputs 1 for true, 0 for false.

Tested in Octave, should work in Matlab also.

R, 13 11

Based on the Perl solution. Returns FALSE or TRUE.

!log2(i)%%1

The parameter i represents the input variable.

An alternative version with user input:

!log2(scan())%%1

GolfScript, 5

Outputs 1 for true, 0 for false. Based on user3142747's idea:

~.(&!

Note: ( is decrement, hopefully it doesn't count as - :)
If it does (and the OP's comments suggest that it might), then please refer to Ilmari Karonen's solution instead.
For Y/N output, append 'NY'1/= at the end (7 more bytes).

Python, 31

print 3>bin(input()).rfind('1')

C, 48

main(x){scanf("%i",&x);puts(x&~(x*~0)?"F":"T");}

I decided to to use another approach, based on the population count or sideways sum of the number (the number of 1-bits). The idea is that all powers of two have exactly one 1 bit, and no other number does. I added a JavaScript version because I found it amusing, though it certainly won't win any golfing competition.

J, 14 15 chars (outputs 0 or 1)

1=##~#:".1!:1]1

JavaScript, 76 chars (outputs true or false)

alert((~~prompt()).toString(2).split("").map(Number).filter(Boolean).length)

Clip, 9 8 7

!%lnxWO

Reads a number from stdin.

Explanation:

To start with, Z = 0, W = 2 and O = 1. This allows placing of W and O next to each other, whereas using 2 and 1 would be interpreted as the number 21 without a separating space (an unwanted extra character). In Clip, the modulo function (%) works on non-integers, so, to work out if some value v is an integer, you check if v mod 1 = 0. Using Clip syntax, this is written as =0%v1. However, as booleans are stored as 1 (or anything else) and 0, checking if something is equal to 0 is just 'not'ing it. For this, Clip has the ! operator. In my code, v is lnx2. x is an input from stdin, n converts a string to a number and lab is log base b of a. The program therefore translates (more readably) to 0 = ((log base 2 of parseInt(readLine)) mod 1).

Examples:

8

outputs

1

and

10

outputs

0

Edit 1: replaced 0, 1 and 2 with Z, O and W.

Edit 2: replaced =Z with !.

Also:

Pyth, 5

Compresses the Clip version even further, as Pyth has Q for already evaluated input and a log2(a) function instead of just general log(a, b).

!%lQ1

Javascript (37)

a=prompt();while(a>1)a/=2;alert(a==1)

Simple script that just divides by 2 repeatedly and checks the remainder.

Mathematica (21)

IntegerQ@Log2@Input[]

Without input it is a bit shorter

IntegerQ@Log2[8]

True

IntegerQ@Log2[7]

False

JavaScript, 41 40 characters

l=Math.log;alert(l(prompt())/l(2)%1==0);

How this works: you take the logarithm at base 2 using l(prompt()) / l(2), and if that result modulo 1 is equal to zero, then it is a power of 2.

For example: after taking the logarithm of 8 on base 2, you get 3. 3 modulo 1 is equal to 0, so this returns true.

After taking the logarithm of 7 on base 2, you get 2.807354922057604. 2.807354922057604 modulo 1 is equal to 0.807354922057604, so this returns false.

JavaScript, 35

Works for bytes.

alert((+prompt()).toString(2)%9==1)

46 character version, Works for 16 bit numbers.

x=(+prompt()).toString(2)%99;alert(x==1|x==10)

The trick works in most dynamic languages.

Explanation: Convert the number to base 2, interpret that string as base 10, do modulo 9 to get the digit sum, which must be 1.

Perl 5.10+, 13 + 1 = 14 chars

say!($_&$_/3)

Uses the same method from an old FWP thread as my GolfScript entry. Prints 1 if the input is a power of two, and an empty line otherwise.

Needs to be run with perl -nE; the n costs one extra char, for a total of 14 chars. Alternatively, here's an 18-character version that doesn't need the n:

say!(($_=<>)&$_/3)

python 3, 38

print(1==bin(int(input())).count('1'))

python, 32

However, the code doesn't work in every version.

print 1==bin(input()).count('1')

Notice that the solution works also for 0 (print False).

Ruby — 17 characters (fourth try)

p /.1/!~'%b'%gets

My current best is a fusion of @steenslag's answer with my own. Below are my previous attempts.

Ruby — 19 characters (third try)

p /10*1/!~'%b'%gets

Ruby — 22 characters (second try)

p !('%b'%gets)[/10*1/]

Ruby — 24 characters (first try)

p !!('%b'%gets=~/^10*$/)

K/Kona (24 17)

d:{(+/(2_vs x))~1

Returns 1 if true and 0 if false. Any power of 2 has a single bit equal to 1:

2_vs'_(2^'(!10))
(,1
1 0
1 0 0
1 0 0 0
1 0 0 0 0
1 0 0 0 0 0
1 0 0 0 0 0 0
1 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0)

(this prints out all the powers of 2 (from 0 to 9) in binary)

So I sum up all the components of the binary expression of x and see if it's equal to 1; if yes then x=2^n, otherwise nope.

...knew I could make it smaller

C# (54 characters)

 Math.Log(int.Parse(Console.ReadLine()),2)%1==0?"Y":"N"

Rebmu (9 chars)

z?MOl2A 1

Test

>> rebmu/args [z?MOl2A 1] 7
== false

>> rebmu/args [z?MOl2A 1] 8 
== true

>> rebmu/args [z?MOl2A 1] 9 
== false

Rebmu is a constricted dialect of Rebol. The code is essentially:

z? mo l2 a 1  ; zero? mod log-2 input 1

Alternative

14 chars—Rebmu does not have a 'mushed' bitwise AND~

z?AND~aTIddA 3

In Rebol:

zero? a & to-integer a / 3

GTB, 46 bytes

2→P`N[@N=Pg;1P^2→P@P>1E90g;2]l;2~"NO"&l;1"YES"

Python, 35

print bin(input()).strip('0')=='b1'

Doesn't use not only +/- operations, but any math operations aside from converting to binary form.

Other stuff (interesting, but not for competition):

I have also a regexp version (61):

import re;print re.match(r'^0b10+$',bin(input())) is not None

(Love the idea, but import and match function make it too long)

And nice, but boring bitwise operations version (31):

x=input();print x and not x&~-x

(yes, it's shorter, but it uses ~-x for decrement which comtains - operation)

Python 2.7 (30 29 39 37)

EDIT: Updated due to user input requirement;

a=input()
while a>1:a/=2.
print a==1

Brute force, try to divide until =1 (success) or <1 (fail)

Python (33)

print int(bin(input())[3:]or 0)<1

Ruby, 33, 28, 25

p /.1/!~gets.to_i.to_s(2)

APL (12 for 0/1, 27 for yes/no)

≠/A=2*0,ιA←⍞ 

or, if we must output text:

3↑(3x≠/A=2*0,ιA←⍞)↓'YESNO '

Read in A. Form a vector 0..A, then a vector 2⁰..2^A (yes, that's way more than necessary), then a vector comparing A with each of those (resulting in a vector of 0's and at most one 1), then xor that (there's no xor operator in APL, but ≠ applied to booleans will act as one.) We now have 0 or 1.

To get YES or NO: multiply the 0 or 1 by 3, drop this number of characters from 'YESNO ', then take the first 3 characters of this.

C, 65 bytes

main(k){scanf("%i",&k);while(k&&!(k%2))k/=2;puts(k==1?"T":"F");}

Haskell (52 50)

k n=elem n$map(2^)[1..n]
main=interact$show.k.read