Wir alle kennen die berühmte Fibonacci-Sequenz , die mit 0und beginnt 1, und jedes Element ist die Summe der beiden vorhergehenden. Hier sind die ersten Begriffe (OEIS A000045 ):

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

Bei einer positiven Ganzzahl geben Sie die nächstliegende Zahl der Fibonacci-Sequenz nach folgenden Regeln zurück:

• Die nächstliegende Fibonacci-Zahl ist definiert als die Fibonacci-Zahl mit der geringsten absoluten Differenz zur angegebenen ganzen Zahl. Zum Beispiel 34ist die nächstliegende Fibonacci-Zahl 30, weil |34 - 30| = 4, die kleiner ist als die zweitnächste 21, für die |21 - 30| = 9.

• Wenn die angegebene Ganzzahl zur Fibonacci-Sequenz gehört, ist die nächstgelegene Fibonacci-Zahl genau sie selbst. Beispielsweise ist die nächstliegende Fibonacci-Zahl 13genau 13.

• Im Falle eines Gleichstands können Sie entweder eine der Fibonacci-Zahlen ausgeben, die beide der Eingabe am nächsten liegen, oder einfach beide ausgeben. Zum Beispiel, wenn der Eingang 17, die alle der folgenden gelten: 21, 13oder 21, 13. Falls Sie beide zurücksenden, geben Sie bitte das Format an.

Es gelten Standardlücken . Sie können die Eingabe und Ausgabe über eine beliebige Standardmethode vornehmen . Ihr Programm / Ihre Funktion darf nur Werte bis 10 ^{8 verarbeiten} .

Testfälle

Eingabe -> Ausgabe

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

Wertung

Das ist Code-Golf , also gewinnt der kürzeste Code in Bytes in jeder Sprache !

Python 2 , 43 Bytes

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

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Durchläuft Paare aufeinanderfolgender Fibonacci-Zahlen, (a,b)bis sie eine erreichen, bei der die Eingabe nunter ihrem Mittelpunkt liegt (a+b)/2, und kehrt dann zurück a.

Als Programm geschrieben (47 Bytes):

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

Gleiche Länge :

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

Neim , 5 Bytes

f𝐖𝕖S𝕔

Erläuterung:

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

In der neuesten Version von Neim kann dies auf 3 Bytes golfen werden:

fS𝕔

Da unendlich viele Listen überarbeitet wurden, um nur ihren Maximalwert zu erreichen.

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Python , 55 52 Bytes

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

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R , 70 67 64 62 60 Bytes

-2 Bytes dank Djhurio!

-2 weitere Bytes dank Djhurio (Junge kann er Golf spielen!)

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

Da wir nur mit Werten bis zu umgehen müssen 10^8, funktioniert dies.

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Liest nvon stdin. Die whileSchleife generiert die Fibonacci-Zahlen in F(absteigender Reihenfolge). im Falle eines Unentschieden wird der größere zurückgegeben. Dies löst eine Reihe von Warnungen aus, da while(F<1e8)nur die Anweisung für das erste Element von ausgewertet wirdF mit einer Warnung

Ursprünglich habe ich F[which.min(abs(F-n))]den naiven Ansatz verwendet, aber @djhurio schlug vor, (F-n)^2da die Reihenfolge gleichwertig sein wird, und ordernicht which.min. orderGibt eine Permutation von Indizes zurück, um die Eingabe in aufsteigende Reihenfolge zu bringen. Daher müssen wir [1]am Ende nur den ersten Wert abrufen.

schnellere Version:

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

Speichert nur die letzten beiden Fibonacci-Zahlen

JavaScript (ES6), 41 Byte

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

Rundet nach Belieben auf.

Gelee , 9 7 Bytes

-2 Bytes dank @EriktheOutgolfer

‘RÆḞạÐṂ

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Golftipps willkommen :). Nimmt ein Int für die Eingabe und gibt eine Int-Liste zurück.

            ' 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 Bytes

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

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x86-64-Maschinencode, 24 Byte

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

Die obigen Codebytes definieren eine Funktion im 64-Bit-x86-Maschinencode, die die Fibonacci-Zahl findet, die dem angegebenen Eingabewert am nächsten kommt n.

Die Funktion folgt der Aufrufkonvention von System V AMD64 (Standard bei Gnu / Unix-Systemen), sodass der einzige Parameter ( n) im EDIRegister übergeben und das Ergebnis im EAXRegister zurückgegeben wird.

Ungolfed Assembler-Mnemonik:

; 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

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Der Code gliedert sich grundsätzlich in drei Teile:

• Der erste Teil ist sehr einfach: Er initialisiert nur unsere Arbeitsregister. EAXwird auf 0 und EDXauf 1 gesetzt.

• Der nächste Teil ist eine Schleife, die die Fibonacci-Zahlen auf beiden Seiten des Eingabewerts iterativ berechnet n. Dieser Code basiert auf meiner vorherigen Implementierung von Fibonacci mit Subtraktion , aber ... ähm ... nicht mit Subtraktion. :-) Insbesondere wird derselbe Trick angewendet, bei dem die Fibonacci-Zahl mit zwei Variablen berechnet wird - hier sind dies die EAXund EDX-Register. Dieser Ansatz ist hier äußerst praktisch, weil er uns benachbarte Fibonacci-Zahlen gibt. Der Kandidat, der möglicherweise kleiner als n ist, wird gehalten EAX, während der Kandidat, der möglicherweise größer als n ist, gehalten wirdEDX. Ich bin ziemlich stolz darauf, wie eng ich den Code innerhalb dieser Schleife machen konnte (und noch mehr gekitzelt, dass ich ihn unabhängig wiederentdeckte und erst später erkannte, wie ähnlich er der oben verlinkten Subtraktionsantwort war).

• Sobald wir die Kandidaten-Fibonacci-Werte in EAXund verfügbar haben EDX, ist es konzeptionell einfach, herauszufinden, welcher näher (in Bezug auf den absoluten Wert) liegt n. Eigentlich nimmt einen Absolutwert kosten würde Art und Weise zu viele Bytes, so dass wir nur eine Reihe von Subtraktionen tun. Der Kommentar rechts neben dem vorletzten Befehl zum bedingten Verschieben erklärt die Logik hier treffend. Dies bewegt sich entweder EDXin EAXoder verlässt es EAXalleine, so dass, wenn die Funktion RETauslöst, die nächstliegende Fibonacci-Zahl zurückgegeben wird EAX.

In the case of a tie, the smaller of the two candidate values is returned, since we've used CMOVG instead of CMOVGE to do the selection. It is a trivial change, if you'd prefer the other behavior. Returning both values is a non-starter, though; only one integer result, please!

PowerShell, 80 74 bytes

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

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Iterative solution. Takes input $n, sets $a,$b to be 1,0, and then loops with Fibonacci until $a is larger than the input. At that point, we index into ($b,$a) based on Boolean of whether the difference between the first element and $n is greater than between $n and the second element. That's left on the pipeline, output is implicit.

JavaScript (ES6), 67 bytes

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)

Test cases

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 Node), 41 bytes

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

Based on ovs's awesome Python answer

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Ungolfed

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

For all k ≥ 0, since |φ^−k/√5| < 1/2, F_k = φ^k/√5 + φ^−k/√5 = round(φ^k/√5). So the return value switches from F_{k − 1} to F_k exactly where k = log_φ(n⋅2√5) − 1, or n = φ^{k + 1}/(2√5), which is within 1/4 of 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.

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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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