Ciąg Fibonacciego jest dość dobrze znana rzecz tutaj. Cholera, ma nawet swój własny tag. Mimo to z pewnością lubimy trzymać się naszych korzeni 1, 1, ...(a może tak 0, 1, ...? Być może nigdy się nie dowiemy ...). W tym wyzwaniu zasady są takie same, ale zamiast dostać ten nelement w sekwencji Fibonacciego, otrzymasz ten nelement w sekwencji Fibonacciego od x, y, ....

Wkład

Trzy liczby całkowite, w dowolnej kolejności. njest indeksem (indeksowanym 0 lub 1) terminu w sekwencji dla danych wyjściowych. xi ysą pierwszymi dwoma elementami w sekwencji Fibonacciego w bieżącym programie.

Wydajność

nP określenie w ciągu Fibonacciego wychodząc z x, y.

Przypadki testowe

(0-indeksowane)

n   x     y     out
5   0     0     0
6   0     1     8
6   1     1     13
2   5     5     10
10  2     2     178
3   3     10    23
13  2308  4261  1325165
0   0     1     0
1   0     1     1

(1-indeksowany)

n   x     y     out
6   0     0     0
7   0     1     8
7   1     1     13
3   5     5     10
11  2     2     178
4   3     10    23
14  2308  4261  1325165
1   0     1     0
2   0     1     1

Ostrzeżenia

Załóżmy 0 <= x <= y.

Zwróć uwagę na kolejność wprowadzania (musi być stała).

Galaretka , 3 bajty

+¡ạ

Takes x, y, and n (0-indexed) as separate command-line arguments, in that order.

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How it works

+¡ạ  Main link. Left argument: x. Right argument: y. Third argument: n

  ạ  Yield abs(x - y) = y - x, the (-1)-th value of the Lucas sequence.
+¡   Add the quicklink's left and right argument (initially x and y-x), replacing
     the right argument with the left one and the left argument with the result.
     Do this n times and return the final value of the left argument.

CJam, 14 9 bytes

l~{_@+}*;

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Input format is "x y n". I'm still a noob at this, so I'm 100% sure there are better ways to do this, but please instead of telling me "do this" try to only give me hints so that I can find the answer myself and get better. Thanks!

Python 2, 37 bytes

f=lambda x,y,n:n and f(y,x+y,n-1)or x

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0-indexed, you may need to adjust the recursion limit for n≥999

JavaScript (ES6), 27 26 bytes

Nothing fancy here, just a standard JS Fibonacci function with the initial values of 0 & 1 removed.

n=>g=(x,y)=>n--?g(y,x+y):x

Try it

f=
n=>g=(x,y)=>n--?g(y,x+y):x
o.value=f(i.value=13)(j.value=2308,k.value=4261)
oninput=_=>o.value=f(+i.value)(+j.value,+k.value)

*{font-family:sans-serif;}
input{margin:0 5px 0 0;width:50px;}
#o{width:75px;}

<label for=i>n: </label><input id=i type=number><label for=j>x: </label><input id=j type=number><label for=k>y: </label><input id=k type=number><label for=o>= </label><input id=o>

Python 2, 40 bytes

0-indexed
Try it online

n,a,b=input()
exec'a,b=b,a+b;'*n
print a

Haskell, 30 bytes

x#y=(f!!)where f=x:scanl(+)y f

Try it online! 0-indexed. Use as (x#y)n, e.g. (0#1)5 for the fifth element of the original sequence.

The most likely shortest way to get the Fibonacci sequence in Haskell is f=0:scanl(+)1f, which defines an infinite list f=[0,1,1,2,3,5,8,...] containing the sequence. Replacing 0 and 1 with arguments x and y yields the custom sequence. (f!!) is then a function returning the nth element of f.

Mathematica, 36 bytes

LinearRecurrence[{1,1},{##2},{#+1}]&

input

[n,x,y]

PowerShell, 40 bytes

$n,$x,$y=$args;2..$n|%{$y=$x+($x=$y)};$y

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Brain-Flak, 38 bytes

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

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{({}[()]<                      >)}     # For n .. 0
         (({}<>)<>            )        # Copy TOS to the other stack and add it to...
                  {}                   # The second value
                    <(<>{}<>)>         # Copy what was TOS back
                                  {}{} # Pop the counter and the n+1th result

Ruby, 27 bytes

->a,b,n{n.times{b=a+a=b};a}

Jelly, 6 bytes

;SḊµ¡I

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Explanation

   µ¡  - repeat n times (computes the n+1th and n+2th element):
 S     -  take the sum of the elements of the previous iteration (starting at (x,y))
;      -  append to the end of the previous iteration
  Ḋ    -  remove the first element
     I - Take the difference of the n+1th and n+2th to get the n-th.

TAESGL, 4 bytes

ēB)Ė

1-indexed

Interpreter

Explanation

Input taken as n,[x,y]

 ēB)Ė
AēB)     get implicit input "A" Fibonacci numbers where "B" is [x,y]
    Ė    pop the last item in the array

Prolog (SWI), 77 bytes

f(N,Y,Z):-M is N-1,f(M,X,Y),Z is X+Y.
l(N,A,B,X):-asserta(f(0,A,B)),f(N,X,_).

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Started off golfing Leaky Nun's answer and arrived at something completely different.

This one has a rule for (Nᵗʰ, (N+1)ᵗʰ) in terms of ((N-1)ᵗʰ, Nᵗʰ) and uses database management to assert 0ᵗʰ and 1ˢᵗ elements at runtime.

f(N,X,Y) means Nᵗʰ element is X and (N+1)ᵗʰ element is Y.

Octave, 24 bytes

@(n,x)(x*[0,1;1,1]^n)(1)

Input format: n,[x,y].

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Braingolf, 15 bytes

VR<2-M[R!+v]R_;

_; is no longer needed on the latest version of Braingolf, however that's as of ~5 minutes ago, so would be non-competing.

Python 2, 112 bytes

1-indexed.

import itertools
def f(x,y):
 while 1:yield x;x,y=y,x+y
def g(x,y,n):return next(itertools.islice(f(x,y),n-1,n))

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MATL, 7 bytes

:"wy+]x

Output is 0-based.

Try it at MATL Online!

Explanation

Let the inputs be denoted n (index), a, b (initial terms).

:"     % Implicitly input n. Do this n times
       %   At this point in each iteration, the stack contains the two most
       %   recently computed terms of the sequence, say s, t. In the first
       %   iteration the stack is empty, but a, b will be implicitly input
       %   by the next statement
  w    %   Swap. The stack contains t, s
  y    %   Duplicate from below. The stack contains t, s, t
  +    %   Add. The stack contains t, s+t. These are now the new two most
       %   recently comnputed terms
]      % End
x      % Delete (we have computed one term too many). Implicitly display

R, 39 bytes

f=function(x,y,n)'if'(n,f(y,x+y,n-1),x)

A simple recursive function. Funnily enough this is shorter than anything I can come up with for the regular Fibonacci sequence (without built-ins), because this doesn't have to assign 1 to both x and y =P

Calculates n+1 numbers of the sequence, including the initial values. Each recursion is calculates with n-1 and stopped when n==0. The lowest of the two numbers is then returned, giving back the n-th value.

dc, 36 bytes

?sdsbsa[lddlb+sdsbla1-dsa1<c]dscxldp

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0-indexed. Input must be in the format n x y.

PHP>=7.1, 55 Bytes

for([,$n,$x,$y]=$argv;$n--;$x=$y,$y=$t)$t=$x+$y;echo$x;

Online Version

PHP>=7.1, 73 Bytes

for([,$n,$x,$y]=$argv,$r=[$x,$y];$i<$n;)$r[]=$r[+$i]+$r[++$i];echo$r[$n];

Online Version

Common Lisp, 49 Bytes, 0-indexed

(defun fib(n x y)(if(= 0 n)x(fib(1- n)y(+ x y))))

I'm a Lisp noob so any tips would be appreciated ;)

Explanation:

(defun fib(n x y)                                  | Define a function taking 3 arguments
                 (if(= 0 n)x                       | If n = 0, return x
                            (fib(1- n)y(+ x y))))  | Otherwise, call fib with n-1, y, and x+y

Prolog (SWI), 85 bytes

l(0,X,Y,X).
l(1,X,Y,Y).
l(N,X,Y,C):-M is N-1,P is N-2,l(M,X,Y,A),l(P,X,Y,B),C is A+B.

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0-indexed.

br**nfuck, 39 29 bytes

Thanks to @JoKing for -10!

,<,<,[>[>+>+<<-]<[>+<-]>-]>>.

TIO won't work particularly well for this (or for any BF solution to a problem involving numbers). I strongly suggest @Timwi's EsotericIDE (or implementing BF yourself).

Takes x, then y, then n. 0-indexed. Assumes an unbounded or wrapping tape.

Explanation

,<,<,            Take inputs. Tape: [n, y, x]
[                While n:
  > [->+>+<<]      Add y to x, copying it to the next cell along as well. Tape: [n, 0, x+y, y]
  < [>+<-]         Move n over. Tape: [0, n, x+y, y]
  >-               Decrement n.
] >>.            End loop. Print cell 2 to the right (x for n == 0).

C (gcc), 29 bytes

f(n,x,y){n=n?f(n-1,y,x+y):x;}

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This implementation is 0-based.

05AB1E, 9 bytes

`©GDŠ+}®@

Try it online!

Explanation

`           # split inputs as separate to stack
 ©          # store n in register
  G         # n-1 times do
   D        # duplicate top of stack
    Š       # move down 2 places on stack
     +      # add top 2 values of stack
      }     # end loop
       ®@   # get the value nth value from the bottom of stack

Lua, 44 bytes

0-Indexed

n,x,y=...for i=1,n do
x,y=y,x+y
end
print(x)

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Klein, 18 + 3 bytes

This uses the 000 topology

:?\(:(+)$)1-+
((/@

Pass input in the form x y n.

Axiom, 88 57 bytes

f(k,x,y)==(repeat(k<=0=>break;c:=y;y:=x+y;x:=c;k:=k-1);x)

this would pass the test proposed (0 indexed)

(14) -> f(5,0,0)
   (14)  0
                                                 Type: NonNegativeInteger
(15) -> f(6,0,1)
   (15)  8
                                                    Type: PositiveInteger
(16) -> f(2,5,5)
   (16)  10
                                                    Type: PositiveInteger
(17) -> f(10,2,2)
   (17)  178
                                                    Type: PositiveInteger
(18) -> f(3,3,10)
   (18)  23
                                                    Type: PositiveInteger
(19) -> f(13,2308,4261)
   (19)  1325165
                                                    Type: PositiveInteger

Retina, 37 bytes

\d+
$*1
+`(1*) (1*) 1
$2 $1$2 
 .*

1

Try it online!

0-based, takes x y n separated by space. Calculates in unary.

TI-Basic, 32 bytes

Prompt N,X,Y
While N
X+Y➡Z
Y➡X
Z➡Y
DS<(N,0
End
X