Dato N, genera l'ennesimo termine di questa sequenza infinita:

-1 2 -2 1 -3 4 -4 3 -5 6 -6 5 -7 8 -8 7 -9 10 -10 9 -11 12 -12 11 ... etc.

N può essere 0-indicizzato o 1-indicizzato come desideri.

Ad esempio, se 0-indicizzati poi ingressi 0, 1, 2, 3, 4dovrebbe produrre rispettive uscite -1, 2, -2, 1, -3.

Se 1-indicizzato poi ingressi 1, 2, 3, 4, 5dovrebbe produrre rispettive uscite -1, 2, -2, 1, -3.

Per essere chiari, questa sequenza viene generata prendendo la sequenza di numeri interi positivi ripetuti due volte

1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 ...

e riordinando ogni coppia di numeri dispari per circondare i numeri pari appena sopra di esso

1 2 2 1 3 4 4 3 5 6 6 5 7 8 8 7 9 10 10 9 11 12 12 11 ...

e infine negando ogni altro termine, a cominciare dal primo

-1 2 -2 1 -3 4 -4 3 -5 6 -6 5 -7 8 -8 7 -9 10 -10 9 -11 12 -12 11 ...

Vince il codice più breve in byte.

Python 2 , 23 byte

lambda n:~n/2+n%2*(n|2)

Gli input dispari danno approssimativamente n/2, anche quelli approssimativamente-n/2 . Quindi, ho iniziato -n/2+n%2*ne ottimizzato da lì.

Provalo online!

Mathematica, 29 byte

((#~GCD~4/. 4->-2)+#)/2(-1)^#&

Funzione pura che accetta un input con 1 indice. Oltre ai segni alternati (-1)^#, due volte la sequenza è vicina all'ingresso, le differenze sono 1, 2, 1, -2 ciclicamente. È bello che #~GCD~4, il più grande divisore comune dell'input e 4, sia 1, 2, 1, 4 ciclicamente; quindi sostituiamo manualmente 4->-2e lo chiamiamo un giorno. Mi piace questo approccio perché evita la maggior parte dei comandi Mathematica a molti caratteri.

Pip , 24 22 byte

v**a*YaBA2|1+:--a//4*2

Accetta input, indicizzato 1, come argomento della riga di comando. Provalo online o verifica 1-20 .

Spiegazione

Osserva che la sequenza può essere ottenuta combinando altre tre sequenze, una indicizzata con zero e l'altra indicizzata:

• Inizia con 0 0 0 0 2 2 2 2 4 4 4 4=a//4*2 (0-indicizzato);
• Aggiungi 1 2 2 1 1 2 2 1 1 2 2 1= aBA2|1, dove BAè bit a bit AND, e| è OR logico (1-indicizzato);
• Moltiplica la somma per -1 1 -1 1 -1 1 -1 1 -1 1 -1 1= (-1)**a(1-indicizzato).

Se iniziamo con a1 indicizzato, possiamo prima calcolare le parti con 1 indice (leggendo l'espressione da sinistra a destra) e quindi diminuire aper la parte con indice 0. Usando la variabile incorporata v=-1, otteniamo

v**a*((aBA2|1)+--a//4*2)

Per radere altri due byte, dobbiamo usare alcuni trucchi per manipolare la precedenza. Possiamo eliminare le parentesi interne sostituendole +con +:(equivalente a +=in molte lingue). Qualsiasi operatore di calcolo e assegnazione ha una priorità molto bassa, quindi aBA2|1+:--a//4*2è equivalente a (aBA2|1)+:(--a//4*2). Pip emetterà un avviso sull'assegnazione a qualcosa che non è una variabile, ma solo se abbiamo gli avvisi abilitati.

L'unica cosa che ha una precedenza inferiore a quella :è Yl'operatore yank. * Assegna il valore del suo operando alla yvariabile e lo passa invariato; in modo da poter eliminare le parentesi esterne come bene da tirando il valore piuttosto che mettere tra parentesi esso: YaBA2|1+:--a//4*2.

_{* rint Pe Output hanno la stessa precedenza di Yank, ma qui non sono utili.}

Gelatina , 8 7 byte

H^Ḃ~N⁸¡

Questo utilizza l'algoritmo della mia risposta Python , che è stato notevolmente migliorato da @GB .

Provalo online!

Come funziona

H^Ḃ~N⁸¡  Main link. Argument: n

H        Halve; yield n/2. This returns a float, but ^ will cast it to int.
  Ḃ      Bit; yield n%2.
 ^       Apply bitwise XOR to both results.
   ~     Take the bitwise NOT.
    N⁸¡  Negate the result n times.

Java 8, 19 byte

n->~(n/2)+n%2*(n|2)

Java 7, 47 37 byte

int c(int n){return~(n/2)+n%2*(n|2);}

La prima volta che Java (8) compete effettivamente ed è più breve di alcune altre risposte. Non riesco ancora a battere le attuali lingue del golf come Jelly e simili (duhuh .. che sorpresa ..>.>; P)

Porta a 0 indici dalla risposta Python 2 di @Xnor .
-10 byte grazie a @GB

Provalo qui.

Jelly, 15 12 11 bytes

Ḷ^1‘ż@N€Fị@

Try it online!

How it works

Ḷ^1‘ż@N€Fị@  Main link. Argument: n

Ḷ            Unlength; yield [0, 1, 2, 3, ..., n-1].
 ^1          Bitwise XOR 1; yield [1, 0, 3, 2, ..., n-1^1].
   ‘         Increment; yield [2, 1, 4, 3, ..., (n-1^1)+1].
      N€     Negate each; yield [-1, -2, -3, -4, ..., -n].
    ż@       Zip with swapped arguments; 
             yield [[-1, 2], [-2, 1], [-3, 4], [-4, 3], ..., [-n, (n-1^1)+1]].
        F    Flatten, yield [-1, 2, -2, 1, -3, 4, -4, 3, ..., -n, (n-1^1)+1].
         ị@  At-index with swapped arguments; select the item at index n.

RProgN 2, 31 25 22 bytes

nx=x2÷1x4%{+$-1x^*}#-?

Explained

nx=                         # Convert the input to a number, set x to it.
   x2÷                      # Floor divide x by 2.
      1                     # Place a 1 on the stack.
       x4%{       }#-?      # If x%4 is 0, subtract 1 from x//2, otherwise...
           +                # Add the 1 and the x together.
            $-1             # Push -1
               x^           # To the power of x.
                 *          # Multiply x//2+1 by -1^x. (Invert if odd, do nothing if even)

Try it online!

Ruby, 26 23 18 bytes

->n{~n/2+n%2*n|=2}

0-based

-3 bytes stealing the -1^n idea from Greg Martin, Dennis and maybe somebody else, then -5 bytes stealing the n|2 idea from xnor.

Stacked, 30 28 bytes

:2%([:2/\4%2=tmo+][1+2/_])\#

Try it online! Returns a function, which as allowed per meta consensus.. Takes input from the top of the stack.

Using the same approach as the RProgN 2 answer.

Alternatively, 46 bytes. Try it online!:

{!()1[::1+,:rev,\srev\,\2+]n*@.n 1-#n 2%tmo*_}

This one generates the range then selects and negates the member as appropriate.

Python 2,  44  33 27 bytes

lambda n:(-1)**n*~(n/2^n%2)

Thanks to @GB for golfing off 6 bytes!

Try it online!

05AB1E, 8 bytes

2‰`^±¹F(

Try it online

Explanation

2‰          divmod by 2
  `         flatten list
   ^        XOR
    ±       NOT
     ¹F(    Push 1st argument, loop N times, negate

Jelly, 9 8 bytes

|2×Ḃ+H~$

Try it online!

-1 thanks to Dennis. Duh float conversions.

Uses @xnor's Python 2 approach.

EDIT: >_>

CJam, 16 bytes

{_(_1&)^2/)W@#*}

1-based input.

Try it online!

Explanation

Here is a breakdown of the code with the values on the stack for each input from 1 to 4. The first few commands only affect the two least significant bits of n-1 so after 4, this stuff just repeats cyclically, with the results incremented by 2, due to the halving.

Cmd             Stack: [1]       [2]       [3]       [4]
_    Duplicate.        [1 1]     [2 2]     [3 3]     [4 4]
(    Decrement.        [1 0]     [2 1]     [3 2]     [4 3]
_    Duplicate.        [1 0 0]   [2 1 1]   [3 2 2]   [4 3 3]
1&   AND 1.            [1 0 0]   [2 1 1]   [3 2 0]   [4 3 1]
)    Increment.        [1 0 1]   [2 1 2]   [3 2 1]   [4 3 2]
^    XOR.              [1 1]     [2 3]     [3 3]     [4 1]
2/   Halve.            [1 0]     [2 1]     [3 1]     [4 0]
)    Increment.        [1 1]     [2 2]     [3 2]     [4 1]
W    Push -1.          [1 1 -1]  [2 2 -1]  [3 2 -1]  [4 1 -1]
@    Rotate.           [1 -1 1]  [2 -1 2]  [2 -1 3]  [1 -1 4]
#    -1^n.             [1 -1]    [2 1]     [2 -1]    [1 1]
*    Multiply.         [-1]      [2]       [-2]      [1]

Perl 6,  55 27 24  22 bytes

{(-1,2,-2,1,{|($^a,$^b,$^c,$^d Z+ -2,2,-2,2)}...*)[$_]}

(Inspired by the Haskell zipWith answer)
Try it

{+^($_ div 2)+$_%2*($_+|2)}

(Inspired by several answers)
Try it

{+^($_+>1)+$_%2*($_+|2)}

Try it

{+^$_+>1+$_%2*($_+|2)}

Try it

Expanded:

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

    +^          # numeric binary invert the following
      $_ +> 1   # numeric bit shift right by one
  +
      $_ % 2    # the input modulo 2
    *
      ($_ +| 2) # numeric binary inclusive or 2
}

(All are 0 based)

Haskell, 37 36 bytes

(([1,3..]>>= \x->[-x,x+1,-x-1,x])!!)

Try it online! This is an anonymous function which takes one number n as argument and returns 0-indexed the nth element of the sequence.

Haskell, 56 bytes

f n=concat(iterate(zipWith(+)[-2,2,-2,2])[-1,2,-2,1])!!n

0-indexed

Perl 5 47 + 1 (for flag) = 48 Bytes

print(((sin$_%4>.5)+1+2*int$_/4)*($_%4&1?1:-1))

Old Submission 82 Bytes

@f=(sub{-$_[0]},sub{$_[0]+1},sub{-$_[0]-1},sub{$_[0]});print$f[$_%4](1+2*int$_/4)

Run like so:

perl -n <name of file storing script>  <<<  n

JavaScript (ES7), 28 bytes

n=>(n+2>>2)*2*(-1)**n-!(n&2)

1-indexed. I haven't looked at any other answers yet so I don't know if this is the best algorithm, but I suspect not.

JavaScript, 28 22 bytes

Thanks @ETHproductions for golfing off 6 bytes

x=>x%2?~x>>1:x%4+x/2-1

Try it online!

dc, 98 bytes

?sa0sb1sq[lq1+dsqla!<i3Q]sf[lb1-lfx]su[lblfx]sx[lb1+dsblfx]sj[lqdd4%d0=u1=j2%1=xljxlfx]dsix_1la^*p

Gosh, this is the longest answer here, mainly because I went the path of generating the absolute value of each element of the sequence one by one based on the following recursive formula:

then outputting (-1)^n * a_n, rather than directly computing the n'th element. Anyways, this is 1-indexed.

Try it online!

R, 38 bytes

function(n)floor(n/2+1-2*!n%%4)*(-1)^n

Explanation

floor(n/2+1)                ->  1 2  2 3  3 4  4 5...
floor(n/2+1-2*!n%%4)        ->  1 2  2 1  3 4  4 3... (subtract 2 where n%%4 == 0)
floor(n/2+1-2*!n%%4)*(-1)^n -> -1 2 -2 1 -3 4 -4 3... (multiply odd n by -1)

TI-Basic (TI-84 Plus CE), 31 bytes

.5(Ans+1+remainder(Ans+1,2)-4not(remainder(Ans,4)))i^(2Ans

TI-Basic is a tokenized language and each token used here is one byte, except remainder(, which is two.

This uses the 1-indexed version.

Explanation:

There is a pattern that repeats every four numbers. In the 1-indexed version, it is: -(x+1)/2, (x+1)/2, -(x+1)/2, (x-1)/2 for the input value x. This can be represented as a piecewise-defined function.

f(x) = -(x+1)/2 if x ≡ 1 mod 4; (x+1)/2 if x ≡ 2 mod 4; -(x+1)/2 if x ≡ 3 mod 4; (x-1)/2 if x ≡ 0 mod 4

Because the "x ≡ 1 mod 4" and "x ≡ 3 mod 4" parts are the same, we can combine them into "x ≡ 1 mod 2".

Now are piecewise function is:

f(x) = -(x+1)/2 if x ≡ 1 mod 2; (x+2)/2 if x ≡ 2 mod 4; (x-2)/2 if x ≡ 0 mod 4

This is where I start breaking it into actual commands. Since the value is positive for even indexes and negative for odd ones, we can use (-1)^x. However, in TI-Basic i^(2X (5 bytes) is shorter than (-1)^Ans (6 bytes). Note that parentheses are required due to order of operations.

Now that we have the way to negate the odd inputs out of the way, we move on to the mods (adding the negating back on later). I made the case of an odd input the default, so we start with .5(Ans+1).

To fix the case of even input, just add one to the number in the parentheses, but only when x ≡ 0 mod 2. This could be represented as .5(Ans+1+remainder(Ans+1,2)) or .5(Ans+1+not(remainder(Ans,2))), but they have the same byte count, so it doesn't matter which.

To fix the case of multiple-of-4 input, we need to subtract 3 from the number in the parentheses, but also another 1 because all multiples of 4 are even, which would add one from our previous step, so we now have .5(Ans+1+remainder(Ans+1,2)-4not(remainder(Ans,4))).

Now, just tack on the sign-determining part to the end to get the full program.

Befunge 93, 25 bytes

Zero-indexed

&4/2*1+&4%3%!!+&2%2*1-*.@

Try it Online!

The number is given by ((2(n / 4) + 1) + !!((n % 4) % 3)) * (2(n % 2) - 1)

QBIC, 53 bytes

b=1:{[b,b+3|~b=a|_x(-(b%2)*2+1)*(q+(b%4>1)*-1)]]q=q+2

Explanation:

b=1     Set b to a starting value of 1
        QBIC would usually use the pre-initialised variable q, but that is already in use
:       Get an input number from the cmd-line, our term to find
{       Start an infinite loop
[b,b+3| FOR-loop: this runs in groups of 4, incrementing its own bounds between runs
~b=a|   If we've reached the term requested
_x      QUIT the program and print:

(-(b%2)*2+1)   The b%2 gives a 1 or a 0, times 2 (2,0), negged (-2,0) and plus one (-1,1)
*              That gives us the sign of our term. Now the value:
(q+(b%4>1)*-1) This is q + 1 if the inner loop counter MOD 4 (1,2,3,0...) is 2 or 3.
]       Close the IF that checks the term
]       Close the FOR-loop
q=q+2   And raise q by 2 for the next iteration of the DO-loop.

Wise, 19 bytes

::>!:><^^~![!-!-~]|

Try it Online!

This is just a port of @Dennis' Jelly answer to Wise.

Q, 52 bytes

{(1 rotate(,/){x,(-)x}each 1_((_)x%4)+til 3)x mod 4}

0 indexed solution.

1. Gets the block number ie. which [-x x+1 -(x+1) x] block within the sequence contains the index.
2. Gets the index of the value within the block based on the index of the value within the whole sequence.
3. Creates the block.
4. Indexes into it via index derived in step 2.