我々は、すべての有名に精通しているフィボナッチ数列で始まり、0そして1、各要素は、前の2の合計です。以下に、最初のいくつかの用語を示します（OEIS A000045）：

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584

与えられた正の整数で、これらの規則の下で、フィボナッチ数列の最も近い数を返します。

• 最も近いフィボナッチ数は、指定された整数と最小絶対差がフィボナッチ数として定義されます。たとえば、34はに最も近いフィボナッチ数30です|34 - 30| = 4。これは、、、が2番目に近いもの、、の21ためにのためです|21 - 30| = 9。

• 指定された整数がフィボナッチ数列に属する場合、最も近いフィボナッチ数はそれ自体です。たとえば、最も近いフィボナッチ数13はになり13ます。

• 同点の場合、入力に最も近いフィボナッチ数のいずれかを出力するか、両方を出力するかを選択できます。たとえば、入力がの場合、17次のすべてが有効です：21、13または21, 13。両方を返却する場合は、形式を明記してください。

デフォルトの抜け穴が適用されます。任意の標準的な方法で入力を取得し、出力を提供できます。プログラム/関数は、10 ⁸までの値のみを処理する必要があります。

テストケース

入力->出力

1-> 1
3-> 3
4-> 3または5または3、5
6-> 5
7-> 8
11-> 13
17-> 13または21または13、21
63-> 55
101-> 89
377-> 377
467-> 377
500-> 610
1399-> 1597

得点

これはcode-golfなので、すべての言語でバイト単位の最短コードが勝ちます！

Python 2、43バイト

f=lambda n,a=0,b=1:a*(2*n<a+b)or f(n,b,a+b)

オンラインでお試しください！

(a,b)入力nがそれらの中間点より少ないものに達するまで連続するフィボナッチ数のペアを反復して(a+b)/2、それから戻りますa。

プログラムとして書かれている（47バイト）：

n=input()
a=b=1
while 2*n>a+b:a,b=b,a+b
print a

同じ長さ：

f=lambda n,a=0,b=1:b/2/n*(b-a)or f(n,b,a+b)

ニーム、5バイト

f𝐖𝕖S𝕔

説明：

f       Push infinite Fibonacci list
 𝐖                     93
  𝕖     Select the first ^ elements
        This is the maximum amount of elements we can get before the values overflow
        which means the largest value we support is 7,540,113,804,746,346,429
   S𝕔   Closest value to the input in the list

Neimの最新バージョンでは、これを3バイトにゴルフできます：

fS𝕔

無限リストは、最大値までしか処理されないように修正されました。

オンラインでお試しください！

Python、55 52バイト

f=lambda x,a=1,b=1:[a,b][b-x<x-a]*(b>x)or f(x,b,a+b)

オンラインでお試しください！

R、70 67 64 62 60バイト

djhurioのおかげで-2バイト！

-djhurioのおかげでさらに2バイト （少年はゴルフができます！）

F=1:0;while(F<1e8)F=c(F[1]+F[2],F);F[order((F-scan())^2)][1]

までの値を処理するだけでよいため10^8、これは機能します。

オンラインでお試しください！

nstdinから読み取ります。whileループにおけるフィボナッチ数生成F（降順に）を、同点の場合、大きい方が返されます。これにより、いくつかの警告がトリガーされます。while(F<1e8)ある最初の要素のステートメントのみを評価するされFます。

もともと私F[which.min(abs(F-n))]は単純なアプローチを使用していましたが(F-n)^2、順序は同等であり、order代わりに@djhurioが提案しましたwhich.min。orderただし、入力を昇順にするためにインデックスの順列を返します。したがって、[1]、最後に最初の値のみを取得ます。

より速いバージョン：

F=1:0;n=scan();while(n>F)F=c(sum(F),F[1]);F[order((F-n)^2)][‌​1]

最後の2つのフィボナッチ数のみを保存します

JavaScript（ES6）、41バイト

f=(n,x=0,y=1)=>y<n?f(n,y,x+y):y-n>n-x?x:y

<input type=number min=0 value=0 oninput=o.textContent=f(this.value)><pre id=o>0

好みで切り上げます。

ゼリー、9 7バイト

@EriktheOutgolferのおかげで-2バイト

‘RÆḞạÐṂ

オンラインでお試しください！

ゴルフのヒントを歓迎します:)。入力にintを取り、intリストを返します。

            ' input -> 4
‘           ' increment -> 5
 R          ' range -> [1,2,3,4,5]
  ÆḞ        ' fibonacci (vectorizes) -> [1,1,2,3,5,8]
     ÐṂ     ' filter and keep the minimum by:
    ạ       ' absolute difference -> [3,3,2,1,1,4]
            ' after filter -> [3,5]

Mathematica、30バイト

Array[Fibonacci,2#]~Nearest~#&

オンラインでお試しください！

x86-64マシンコード、24バイト

31 C0 8D 50 01 92 01 C2 39 FA 7E F9 89 D1 29 FA 29 C7 39 D7 0F 4F C1 C3

上記のコードバイトは、指定された入力値に最も近いフィボナッチ数を見つける64ビットx86マシンコードの関数を定義します。 n。

この関数はSystem V AMD64呼び出し規約（Gnu / Unixシステムの標準）に準拠しているため、唯一のパラメーター（n）がEDIレジスターに渡され、結果がEAXレジスターに返されます。

非ゴルフアセンブリニーモニック：

; unsigned int ClosestFibonacci(unsigned int n);
    xor    eax, eax        ; initialize EAX to 0
    lea    edx, [rax+1]    ; initialize EDX to 1

  CalcFib:
    xchg   eax, edx        ; swap EAX and EDX
    add    edx, eax        ; EDX += EAX
    cmp    edx, edi
    jle    CalcFib         ; keep looping until we find a Fibonacci number > n

    mov    ecx, edx        ; temporary copy of EDX, because we 'bout to clobber it
    sub    edx, edi
    sub    edi, eax
    cmp    edi, edx
    cmovg  eax, ecx        ; EAX = (n-EAX > EDX-n) ? EDX : EAX
    ret

オンラインでお試しください！

コードは基本的に3つの部分に分かれています。

• 最初の部分は非常に単純です。作業レジスタを初期化するだけです。EAX0に設定され、EDX設定され、1に設定されます。

• 次の部分は、入力値の両側のフィボナッチ数を繰り返し計算するループですn。このコードは、以前の減算付きフィボナッチの実装に基づいていますが、…ええと…減算ではありません。:-)特に、2つの変数を使用してフィボナッチ数を計算する同じトリックを使用します。ここでは、これらはEAXとEDXレジスタです。このアプローチは非常に、フィボナッチ数が隣接するため、ここで便利です。潜在的により小さい 候補nはに保持されEAX、潜在的により大きい 候補nは保持されていますEDX。このループ内でコードを作成できたことを非常に誇りに思っています（さらに、それを独立して再発見したことをさらにくすぐった後、上記のリンクされた減算の答えにどれほど似ているかを実感しました）。

• 候補フィボナッチ値がEAXとEDXで利用可能になると、どちらが（絶対値に関して）に近いかを把握するという概念的に単純な問題になりnます。実際にはコストがかかる絶対値をとる方法我々だけ減算の一連の操作を行うので、あまりにも多くのバイトを。最後から2番目の条件付き移動命令の右側にあるコメントは、このロジックを適切に説明しています。このどちらかの移動EDXにEAX、または葉EAXのみで、その機能のときにRET壷は、最寄りのフィボナッチ数はで返されますEAX。

同点の場合、選択の代わりに使用したため、2つの候補値のうち小さい方が返されます。他の動作を好む場合、それは些細な変更です。ただし、両方の値を返すことはスターターではありません。整数の結果は1つだけにしてください！CMOVGCMOVGE

PowerShell、80 74バイト

param($n)for($a,$b=1,0;$a-lt$n;$a,$b=$b,($a+$b)){}($b,$a)[($b-$n-gt$n-$a)]

（オンラインで試してください！一時的に応答しません）

反復ソリューション。入力を受け取り$n、に設定$a,$bし、入力より大きくなる1,0までフィボナッチでループし$aます。その時点で($b,$a)、最初の要素との間の差が2番目の要素との$n間の差より大きいかどうかのブール値に基づいてインデックスを作成し$nます。それはパイプラインに残っており、出力は暗黙的です。

JavaScript（ES6）、67バイト

f=(n,k,y)=>(g=k=>x=k>1?g(--k)+g(--k):k)(k)>n?x+y>2*n?y:x:f(n,-~k,x)

テストケース

f=(n,k,y)=>(g=k=>x=k>1?g(--k)+g(--k):k)(k)>n?x+y>2*n?y:x:f(n,-~k,x)

console.log(f(1   )) // -> 1
console.log(f(3   )) // -> 3
console.log(f(4   )) // -> 3 or 5 or 3, 5
console.log(f(6   )) // -> 5
console.log(f(7   )) // -> 8
console.log(f(11  )) // -> 13
console.log(f(17  )) // -> 13 or 21 or 13, 21
console.log(f(63  )) // -> 55
console.log(f(101 )) // -> 89
console.log(f(377 )) // -> 377
console.log(f(467 )) // -> 377
console.log(f(500 )) // -> 610
console.log(f(1399)) // -> 1597

JavaScript（Babelノード）、41バイト

f=(n,i=1,j=1)=>j<n?f(n,j,j+i):j-n>n-i?i:j

ovsのすばらしいPythonの回答に基づく

オンラインでお試しください！

非ゴルフ

f=(n,i=1,j=1)=> // Function f: n is the input, i and j are the most recent two Fibonacci numbers, initially 1, 1
 j<n?           // If j is still less than n
  f(n,j,j+i)    //  Try again with the next two Fibonacci numbers
 :              // Else:
  j-n>n-i?i:j   //  If n is closer to i, return i, else return j

Python, 74 bytes

import math
r=5**.5
p=r/2+.5
lambda n:round(p**int(math.log(n*2*r,p)-1)/r)

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How it works

すべてのためのk以来≥0、|φ ^{- K} /√5| <1/2、Fの_K =φ ^K /√5+φ ^{- K} /√5= ROUND（φ ^K /√5）。戻り値から切り替えるようにFの_{K - 1}までのF _kの正確な場所、K =ログ_φ（N ⋅2√5） - 1、またはN =φ ^{K + 1}の1/4以内である/（2√5）F _{k + 1/2} =（F _{k − 1} + F _k）/ 2

05AB1E, 6 bytes

nÅFs.x

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C (gcc), 86 85 83 76 bytes

f(n){n=n<2?:f(n-1)+f(n-2);}i,j,k;g(n){for(;k<n;j=k,k=f(i++));n=k-n<n-j?k:j;}

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Java (OpenJDK 8), 60 57 56 bytes

c->{int s=1,r=2;for(;c>r;s=r-s)r+=s;return c-s>r-c?r:s;}

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-1 byte thanks to @Neil

Python 3, 103 bytes

import math
def f(n):d=5**.5;p=.5+d/2;l=p**int(math.log(d*n,p))/d;L=[l,p*l];return round(L[2*n>sum(L)])

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Sadly, had to use a def instead of lambda... There's probably much room for improvement here.

Original (incorrect) answer:

Python 3, 72 bytes

lambda n:r(p**r(math.log(d*n,p))/d)
import math
d=5**.5
p=.5+d/2
r=round

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My first PPCG submission. Instead of either calculating Fibonacci numbers recursively or having them predefined, this code uses how the n-th Fibonacci number is the nearest integer to the n-th power of the golden ratio divided by the root of 5.

Taxi, 2321 bytes

Go to Post Office:w 1 l 1 r 1 l.Pickup a passenger going to The Babelfishery.Go to The Babelfishery:s 1 l 1 r.Pickup a passenger going to Trunkers.Go to Trunkers:n 1 l 1 l.0 is waiting at Starchild Numerology.1 is waiting at Starchild Numerology.Go to Starchild Numerology:w 1 l 2 l.Pickup a passenger going to Bird's Bench.Pickup a passenger going to Cyclone.Go to Cyclone:w 1 r 4 l.[a]Pickup a passenger going to Rob's Rest.Pickup a passenger going to Magic Eight.Go to Bird's Bench:n 1 r 2 r 1 l.Go to Rob's Rest:n.Go to Trunkers:s 1 l 1 l 1 r.Pickup a passenger going to Cyclone.Go to Cyclone:w 2 r.Pickup a passenger going to Trunkers.Pickup a passenger going to Magic Eight.Go to Zoom Zoom:n.Go to Trunkers:w 3 l.Go to Magic Eight:e 1 r.Switch to plan "b" if no one is waiting.Pickup a passenger going to Firemouth Grill.Go to Firemouth Grill:w 1 r.Go to Rob's Rest:w 1 l 1 r 1 l 1 r 1 r.Pickup a passenger going to Cyclone.Go to Bird's Bench:s.Pickup a passenger going to Addition Alley.Go to Cyclone:n 1 r 1 l 2 l.Pickup a passenger going to Addition Alley.Pickup a passenger going to Bird's Bench.Go to Addition Alley:n 2 r 1 r.Pickup a passenger going to Cyclone.Go to Cyclone:n 1 l 1 l.Switch to plan "a".[b]Go to Trunkers:w 1 l.Pickup a passenger going to Cyclone.Go to Bird's Bench:w 1 l 1 r 1 l.Pickup a passenger going to Cyclone.Go to Rob's Rest:n.Pickup a passenger going to Cyclone.Go to Cyclone:s 1 l 1 l 2 l.Pickup a passenger going to Sunny Skies Park.Pickup a passenger going to What's The Difference.Pickup a passenger going to What's The Difference.Go to What's The Difference:n 1 l.Pickup a passenger going to Magic Eight.Go to Sunny Skies Park:e 1 r 1 l.Go to Cyclone:n 1 l.Pickup a passenger going to Sunny Skies Park.Pickup a passenger going to What's The Difference.Go to Sunny Skies Park:n 1 r.Pickup a passenger going to What's The Difference.Go to What's The Difference:n 1 r 1 l.Pickup a passenger going to Magic Eight.Go to Magic Eight:e 1 r 2 l 2 r.Switch to plan "c" if no one is waiting.Go to Sunny Skies Park:w 1 l 1 r.Pickup a passenger going to The Babelfishery.Go to Cyclone:n 1 l.Switch to plan "d".[c]Go to Cyclone:w 1 l 2 r.[d]Pickup a passenger going to The Babelfishery.Go to The Babelfishery:s 1 l 2 r 1 r.Pickup a passenger going to Post Office.Go to Post Office:n 1 l 1 r.

Try it online!
Try it online with comments!

Un-golfed with comments:

[ Find the Fibonacci number closest to the input ]
[ Inspired by: https://codegolf.stackexchange.com/q/133365 ]


[ n = STDIN ]
Go to Post Office: west 1st left 1st right 1st left.
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery: south 1st left 1st right.
Pickup a passenger going to Trunkers.
Go to Trunkers: north 1st left 1st left.


[ Initialize with F0=0 and F1=1 ]
0 is waiting at Starchild Numerology.
1 is waiting at Starchild Numerology.
Go to Starchild Numerology: west 1st left 2nd left.
Pickup a passenger going to Bird's Bench.
Pickup a passenger going to Cyclone.
Go to Cyclone: west 1st right 4th left.


[ For (i = 1; n > F(i); i++) { Store F(i) at Rob's Rest and F(i-1) at Bird's Bench } ]
[a]
Pickup a passenger going to Rob's Rest.
Pickup a passenger going to Magic Eight.
Go to Bird's Bench: north 1st right 2nd right 1st left.
Go to Rob's Rest: north.
Go to Trunkers: south 1st left 1st left 1st right.
Pickup a passenger going to Cyclone.
Go to Cyclone: west 2nd right.
Pickup a passenger going to Trunkers.
Pickup a passenger going to Magic Eight.
Go to Zoom Zoom: north.
Go to Trunkers: west 3rd left.
Go to Magic Eight: east 1st right.
Switch to plan "b" if no one is waiting.

[ n >= F(i) so iterate i ]
Pickup a passenger going to Firemouth Grill.
Go to Firemouth Grill: west 1st right.
Go to Rob's Rest: west 1st left 1st right 1st left 1st right 1st right.
Pickup a passenger going to Cyclone.
Go to Bird's Bench: south.
Pickup a passenger going to Addition Alley.
Go to Cyclone: north 1st right 1st left 2nd left.
Pickup a passenger going to Addition Alley.
Pickup a passenger going to Bird's Bench.
Go to Addition Alley: north 2nd right 1st right.
Pickup a passenger going to Cyclone.
Go to Cyclone: north 1st left 1st left.
Switch to plan "a".


[b]
[ F(i) > n which means n >= F(i-1) and we need to figure out which is closer and print it]
Go to Trunkers: west 1st left.
Pickup a passenger going to Cyclone.
Go to Bird's Bench: west 1st left 1st right 1st left.
Pickup a passenger going to Cyclone.
Go to Rob's Rest: north.
Pickup a passenger going to Cyclone.
Go to Cyclone: south 1st left 1st left 2nd left.
Pickup a passenger going to Sunny Skies Park.
Pickup a passenger going to What's The Difference.
Pickup a passenger going to What's The Difference.
[ Passengers:n, n, F(i-1) ]
Go to What's The Difference: north 1st left.
Pickup a passenger going to Magic Eight.
[ Passengers:n, n-F(i-1) ]
Go to Sunny Skies Park: east 1st right 1st left.
Go to Cyclone: north 1st left.
Pickup a passenger going to Sunny Skies Park.
Pickup a passenger going to What's The Difference.
[ Passengers: n-F(i-1), F(i-1), F(i) ]
Go to Sunny Skies Park: north 1st right.
Pickup a passenger going to What's The Difference.
[ Passengers: n-F(i-1), F(i), n ]
Go to What's The Difference: north 1st right 1st left.
[ Passengers: n-F(i-1), F(i)-n ]
Pickup a passenger going to Magic Eight.
Go to Magic Eight: east 1st right 2nd left 2nd right.
Switch to plan "c" if no one is waiting.

[ If no one was waiting, then {n-F(i-1)} < {F(i)-n} so we return F(i-1) ]
Go to Sunny Skies Park: west 1st left 1st right.
Pickup a passenger going to The Babelfishery.
Go to Cyclone: north 1st left.
Switch to plan "d".

[c]
[ Otherwise {n-F(i-1)} >= {F(i)-n} so we return F(i) ]
[ Note: If we didn't switch to plan c, we still pickup F(i) but F(i-1) will be the *first* passenger and we only pickup one at The Babelfishery ]
[ Note: Because of how Magic Eight works, we will always return F(i) in the event of a tie ]
Go to Cyclone: west 1st left 2nd right.
[d]
Pickup a passenger going to The Babelfishery.
Go to The Babelfishery: south 1st left 2nd right 1st right.
Pickup a passenger going to Post Office.
Go to Post Office: north 1st left 1st right.

Python 3, 84 bytes

lambda k:min(map(f,range(2*k)),key=lambda n:abs(n-k))
f=lambda i:i<3or f(i-1)+f(i-2)

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It may work, but it's certainly not fast...

Outputs True instead of 1, but in Python these are equivalent.

dc, 52 bytes

si1d[dsf+lfrdli>F]dsFxli-rlir-sdd[lild-pq]sDld<Dli+p

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Takes input at run using ?

Edited to assume top of stack as input value, -1 byte.

Input is stored in register i. Then we put 1 and 1 on the stack to start the Fibonacci sequence, and we generate the sequence until we hit a value greater than i. At this point we have two numbers in the Fibonacci sequence on the stack: one that is less than or equal to i, and one that is greater than i. We convert these into their respective differences with i and then compare the differences. Finally we reconstruct the Fibonacci number by either adding or subtracting the difference to i.

Oops, I was loading two registers in the wrong order and then swapping them, wasting a byte.

C# (.NET Core), 63 56 bytes

-1 byte thanks to @Neil

-6 bytes thanks to @Nevay

n=>{int a=0,b=1;for(;b<n;a=b-a)b+=a;return n-a>b-n?b:a;}

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Q/KDB+, 51 bytes

{w where l=min l:abs neg[x]+w:{x,sum -2#x}/[x;0 1]}

Hexagony, 37 bytes

?\=-${&"*.2}+".|'=='.@.&}1.!_|._}$_}{

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ungolfed:

   ? \ = - 
  $ { & " * 
 . 2 } + " .
| ' = = ' . @
 . & } 1 . !
  _ | . _ }
   $ _ } { 

Broken down:

start:
? { 2 ' * //set up 2*target number
" ' 1     //initialize curr to 1

main loop:
} = +     //next + curr + last
" -       //test = next - (2*target)
branch: <= 0 -> continue; > 0 -> return

continue:
{ } = &   //last = curr
} = &     //curr = next


return:
{ } ! @   //print last

Like some other posters, I realized that when the midpoint of last and curr is greater than the target, the smaller of the two is the closest or tied for closest.

The midpoint is at (last + curr)/2. We can shorten that because next is already last + curr, and if we instead multiply our target integer by 2, we only need to check that (next - 2*target) > 0, then return last.

Brachylog, 22 bytes

;I≜-.∧{0;1⟨t≡+⟩ⁱhℕ↙.!}

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Really all I've done here is paste together Fatalize's classic Return the closest prime number solution and my own Am I a Fibonacci Number? solution. Fortunately, the latter already operates on the output variable; unfortunately, it also includes a necessary cut which has to be isolated for +2 bytes, so the only choice point it discards is ⁱ, leaving ≜ intact.

Japt -g, 8 bytes

ò!gM ñaU

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Java 7, 244 234 Bytes

 String c(int c){for(int i=1;;i++){int r=f(i);int s=f(i-1);if(r>c && s<c){if(c-s == r-c)return ""+r+","+s;else if(s-c > r-c)return ""+r;return ""+s;}}} int f(int i){if(i<1)return 0;else if(i==1)return 1;else return f(i-2)+f(i-1);}

Pyke, 6 bytes

}~F>R^

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}      -    input*2
 ~F    -   infinite list of the fibonacci numbers
   >   -  ^[:input]
    R^ - closest_to(^, input)

Common Lisp, 69 bytes

(lambda(n)(do((x 0 y)(y 1(+ x y)))((< n y)(if(<(- n x)(- y n))x y))))

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Perl 6, 38 bytes

{(0,1,*+*...*>$_).sort((*-$_).abs)[0]}

Test it

{   # bare block lambda with implicit parameter ｢$_｣

  ( # generate Fibonacci sequence

    0, 1,  # seed the sequence
    * + *  # WhateverCode lambda that generates the rest of the values
    ...    # keep generating until
    * > $_ # it generates one larger than the original input
           # (that larger value is included in the sequence)

  ).sort(           # sort it by
    ( * - $_ ).abs  # the absolute difference to the original input
  )[0]              # get the first value from the sorted list
}

For a potential speed-up add .tail(2) before .sort(…).

In the case of a tie, it will always return the smaller of the two values, because sort is a stable sort. (two values which would sort the same keep their order)

Pyth, 19 bytes

JU2VQ=+Js>2J)hoaNQJ

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Explanation

JU2VQ=+Js>2J)hoaNQJ
JU2                  Set J = [0, 1].
   VQ=+Js>2J)        Add the next <input> Fibonacci numbers.
              oaNQJ  Sort them by distance to <input>.
             h       Take the first.

Haskell, 48 bytes

(%)a b x|abs(b-x)>abs(a-x)=a|1>0=b%(a+b)$x
(1%2)

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