(Виклик, взятий з багатокористувацької гри (зіткнення коду) на сайті codingame.com )

Змагання

Знайдіть n -й член наступної послідовності: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4...або, щоб зробити це більш очевидним,{1}, {1,2}, {1,2,3}, {1,2,3,4}...

Послідовність складається з об'єднаних діапазонів від 1 до x , починаючи з 1, аж до нескінченності.

Правила / IO

Введення та вихід можуть бути в будь-якому форматі, якщо вони відрізняються. Введення можна взяти з будь-якого відповідного джерела: STDIN, файлу тощо ...

Вхід може бути 0- або 1-індексованим, і вибрана індексація повинна бути зазначено у публікації.

Вам доведеться обробляти щонайменше до результату 255 включно (максимум, 0-індексований максимальний вхід становить 32640). Що-небудь над цим потрібно обробляти, якщо ваша мова це підтримує.

Це code-golfтак виграє найкоротший байт!

Тестові випадки (індексація на основі 0)

0 -> 1
1 -> 1
5 -> 3
10 -> 1
59 -> 5
100 -> 10
1001 -> 12

05AB1E , 5 байт

Програма 0-індексована, код:

ÌLL¹è

Пояснення:

Ì       # Double increment the input
 LL     # List of list on the input
   ¹è   # Get nth element

Використовує кодування CP-1252 . Спробуйте в Інтернеті!

Haskell , 27 26 байт

([z|k<-[1..],z<-[1..k]]!!)

Спробуйте в Інтернеті!

Дякую @DanD. за -1 байт!

Це анонімна функція, створюючи нескінченну послідовність ап тільки повертаючи nйого -ю елемент: [[1..k]| k<-[1..]]виробляє нескінченний список список: [[1],[1,2],[1,2,3],[1,2,3,4],...]. Щоб об'єднати їх, ми можемо записати, [z|k<-[1..],z<-[1..k]]що призводить до [1,1,2,1,2,3,1,2,3,4,...]і, нарешті, (...!!)приймає вхід n(безглузде позначення) і повертає n-й член (на основі 0).

JavaScript, 29 28 байт

-1 байт дякую Арнольду!

f=(n,m)=>n++<m?n:f(n+~m,-~m)

Використовується 0-індексована рекурсивна формула, знайдена в OEIS.

Коли викликається з 1 аргументом, як очікувалося, значення за замовчуванням другого m,, буде undefined. Однак, -~undefinedповертає 1, що дозволяє нам m = 1прокрутити рекурсію без явного в списку аргументів (спасибі @Arnauld!)

Фрагмент тесту:

f=(n,m)=>n++<m?n:f(n+~m,-~m)

let examples = [0, 1, 5, 10, 15, 1000];

examples.forEach(function log(x) {
    console.log(x, " => ", f(x))
});

Крім того, для одного і того ж числа байтів ми можемо мати функцію curry так:

f=n=>m=>n++<m?n:f(n+~m)(-~m)

Ви можете викликати це за допомогою f(5)()- він повертає функцію, яка при виклику повертає результат, як описано в цьому мета-пості .

Желе , 5 байт, 1-індексований

RRF³ị

Спробуйте в Інтернеті!

Пояснення:

                                      (Assume N = 4 for the examples)
R      Generate a list of 1 to N      [1, 2, 3, 4]
 R     Generate new lists for each item on the previous list, with that item as N
                                      [[1], [1,2], ...]
  F    Flatten that list              [1, 1, 2, 1, 2, 3 ...]
   ³ị  Use the input number (³) as index (ị) on the list. 
       This is one-based:             [1, 1, 2, 1, 2, 3 ...] 
                                                ^

Octave, 39 bytes

@(z)z-(n=ceil((8*z+1)^.5/2-.5))*(n-1)/2

1- based index

Explanation:

Consider this sequence:

1   1   2   1   2   3   1   2   3   4   1   2   3   4   5

if we count number of element of subsequences we have

1   2        3          4               5         

so using Gauss formula for triangular number we can form a formula for z:

z=n*(n+1)/2

that is a quadratic equation if we solve it for n we have

n=(sqrt(8*z+1)-1)/2

Try It Online!

Haskell, 25 24 bytes

(!!)$[1..]>>= \x->[1..x]

Usage example: ((!!)$[1..]>>= \x->[1..x]) 10 -> 1. Try it online!.

Maps the anonymous make-a-list-from-1-to-x function \x->[1..x] (the built-in enumFromTo 1 is one byte longer) to the infinite list [1..] and concatenates the resulting lists into a single list. !! picks the nth element.

Thanks to @flawr for one byte.

Octave, 39 bytes

@(n){v=1:n,A=triu(v'+0*v),A(A>0)(n)}{3}

Try it online!

This uses an alternative approach.

For e.g. n=1 this A=triu(v'+0*v) creates the matrix

1   1   1   1
0   2   2   2
0   0   3   3
0   0   0   4

When removing all the zero elements and appending the columns by A(A>0) we get the sequence:

1   1  2  1  2  3  1  2  3  4

Then it is just a matter of exctracting the n-th term of that sequence.

Python, 39 36 bytes

-3 bytes thanks to Dennis!

A recursive lambda which uses 1-based indexing.

f=lambda n,m=1:n*(n<=m)or f(n-m,m+1)

Try it online!

We keep track of the current "rise" size using m. If n is smaller than or equal to m, it fits within the current "rise", and so we return it. However, if it's larger than m, we take m away from it, than add 1 to m and call the function recursively (moving on to the next rise).

R, 25 bytes

i=scan();sequence(1:i)[i]

The index is 1-based.

Pyth, 6 5 bytes

1 byte saved thanks to @TheBikingviking!

@s._S

This uses 0-based indexing.

Try it online!

Explanation

@          Index with implicit input into
   ._      all prefixes of
     S     1-based range of implicit input
 s         concatenated into an un-nested list

Mathematica, 27 24 bytes

Thanks @MartinEnder for 3 bytes!

((r=Range)@r@#<>1)[[#]]&

1-indexed. This throws errors that are safe to ignore.

Explanation

((r=Range)@r@#<>1)[[#]]&
  r=Range                 (* Store Range function in r *)
           r@#            (* Create list {1..n} *)
 (r      )@               (* For each element, generate {1..n} *)
              <>1         (* Join the lists and append a 1; throws errors *)
(                )[[#]]&  (* Take the nth element *)

brainf*ck, 78 bytes

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

Takes input (0-based) and outputs as byte values.

You can test it here.

Input requires a \ before decimal numbers (e.g. \10 for 10). If the output is a printable ASCII character, you should see it. Otherwise, hit view memory -> final dump. The value that was printed is in the 3rd cell (cell number 2).

Explanation:

Cell 0 (INPUT): is the input and is decremented my 1 each time through the loop.

Cell 1 (RESET): increments by 1 every time it is equal to TERM. To do this, every time through the loop we add 1 and if they are not equal we subtract 1.

Cell 2 (TERM): increments by 1 every loop and is set to 0 if it matches RESET. To do this, I only copy the value back from HOLD if this cell was not equal to RESET.

Cell 3 (EQUAL): is used to check if RESET and TERM are equal.

Cell 4 (HOLD): is used to copy the values of RESET and TERM back after the equals check.

,>+<              # get input and put a 1 in RESET
[                 # for INPUT to 0
  >[->>+>+<<<]    # copy RESET to EQUAL and HOLD
  >>>[-<<<+>>>]   # copy HOLD back into RESET
  <<+             # add 1 to TERM
  [->->+<<]       # subtract TERM from EQUAL and copy it to HOLD
  >[              # if RESET and TERM were not equal
    <<-           # subtract 1 from RESET
    >>>[-<<+>>]   # copy HOLD back to TERM
    <[-]          # zero out EQUAL
  ]               # end if
  >[-]            # zero out HOLD
  <<<+            # add 1 to RESET (this cancels out the subtraction if
                  #     RESET did not equal TERM)
  <-              # subtract 1 from INPUT
]>>+.             # end for and add 1 because the sequence resets to 1 not 0

Pyke, 6 bytes

OmSsQ@

Try it here!

O      -    input+2
 mS    -   map(range(1,i+1), range(^))
   s   -  sum(^)
    Q@ - ^[input]

0-indexed.

R, 43 41 bytes

Edit: Found a shorter recursive approach by using A002262 + 1 (0 indexed):

f=function(n,m=1)`if`(n<m,n+1,f(n-m,m+1))

Old version:

n=scan();n-choose(floor((1+sqrt(8*n))/2),2)

1-indexed formula from OEIS.

Perl 6, 21 bytes

{map(|^*,^∞)[$_]+1}

0-indexed. Try it online!

How it works:

{                 }  # A lambda.
         ^∞          # Range from 0 to Inf-1. (Same byte count as 0..*, but cooler.)
 map( ^*,  )         # Map each number n to the range 0..(n-1),
     |               # And slip each range into the outer list.
            [$_]     # Index the sequence with the lambda argument.
                +1   # Add 1.

Perl 6, 21 bytes

{[\,](1..*).flat[$_]}

0-indexed. Try it online!

How it works:

{                   }  # A lambda.
      1..*             # Range from 1 to infinity.
 [ ,](    )            # Fold it with the comma operator,
  \                    # and return all intermediate results, e.g. (1), (1,2), (1,2,3)...
           .flat       # Flatten the sequence.
                [$_]   # Index it with the lambda argument.

Neither of these solutions is as short as JungHawn Min's, but they're alternate approaches, which is something I guess. Both are unnamed functions taking a (1-indexed) positive integer input and returning a positive integer.

Mathematica, 30 bytes

-#^2-#&@⌈√(2#)-3/2⌉/2+#&

An actual mathematical formula for this function! Made more readable (in part by translating the 3-byte characters ⌈, √, and ⌉):

# - ((#^2 + #) / 2 &)[Ceiling[Sqrt[2 * #] - 3/2]] &

Ceiling[Sqrt[2 * #] - 1/2] tells us which sublist the input refers to, from which we subtract one to tell us which sublist ends before we get to the input; then ((#^2 + #) / 2 &) computes how many elements occur in all the sublists before the one we care about, which we subtract from the input # to get our answer. (Some will notice the familiar formula (#^2 + #) / 2 for the #th triangular number; Ceiling[Sqrt[2 * #] - 1/2] is essentially the inverse function.)

Mathematica, 32 bytes

If[#2<=#,#2,#0[#+1,#2-#]]&[1,#]&

Recursive solution, basically the same as in Billywob's answer and others.

Brain-Flak, 46 bytes

Zero indexed

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

Try it online!

Stack Clean, 48 bytes

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

Try it online!

Explanation

This is a modified version of the modulo function. Instead of using a constant number as the divisor it increments the divisor for each time the divisor is subtracted from it (once per outside loop iteration).

Annotated Code

(<>())       # Switch to the opposite stack and push 1 (the initial divisor)
{            # (outside loop) While top of stack is not 0...
  (          # Push...
    ({}())   # Push the divisor plus 1
  [()])      # ...minus one (ie push a copy of the original divisor
  {          # (inner loop) While the top of stack does not equal zero
    ({}[()]) # Decrement the top of the active stack
    <>       # Switch stacks
  }{}        # (inside loop) End loop and pop zero off the top of stack)
}            # (outside loop) End loop
<>           # Switch stacks (to the one with the divisor)
([{}()]{})   # Calculate the result

Java 8, 85 73 55 bytes

n->f(n,1)+1int f(int n,int m){return n<m?n:f(n-m,m+1);}

0-indexed recursive approach with the formula provided in the OEIS:

a(n) = 1 + A002262(n).
A002262: a(n)=f(n,1) with f(n,m) = if n<m then n else f(n-m,m+1).

Try it here.

Old answer (85 56 bytes):

n->{int m=~-(int)Math.sqrt(8*n+1)/2;return n-m*-~m/2+1;}

Used the other 0-indexed formula provided in the OEIS:

n-th term is n - m*(m+1)/2 + 1, where m = floor((sqrt(8*n+1) - 1) / 2).

Try it here.

Perl, 30 bytes

29 bytes of code + -p flag.

map{$\-=$c++if$c<++$\}0..$_}{

Try it online!

MATL, 8 bytes

:"@:]vG)

This solution uses 1-based indexing

Try it at MATL Online

Explanation

        Implicitly grab input (N)
:       Create an array from [1...N]
"       For each element (A) in this array...
  @:    Create an array from [1....A]
]       End for loop
v       Vertically concatenate everything on the stack
G       Explicitly grab the input again
)       And use it to index into the vertically concatenated array
        Implicitly display the result

QBIC, 21 bytes, 1-indexed

:[a|[b|~q=a|_Xc\q=q+1

Explanation:

:      Get 'a' from the cmd line
[a|    FOR (b = 1; b <= a; b++) This creates an outer loop from 1 to N
[b|    FOR (c = 1; c <= b; c++) This creates an iteration, yielding the 1, 12, 123 pattern
       'q' stores how many terms we've seen. It starts at 1 b default.
~q=a   if we are at the desired term (q == a)
|_Xc   Then quit, and print 'c' (the current number in the sequence)
\q=q+1 Else, increase 'q' and run again.

Slightly more interesting approach, but 10 bytes longer:

:{~b+q>=a|_xa-b|\b=b+q┘q=q+1

This program continuously calculates the total number of numbers in this bracket and all previous ones (1 at loop 1, 3 at loop 2, 6 at loop 3 ...). When that counter exceeds the sought-after index N, then return X from the current bracket, where X is N minus the previous amount of the counter.

Ruby, 30 bytes

->n{(0..n).find{|x|0>=n-=x}+n}

1-based indexing

R, 37 bytes

n=scan();for(i in 2:n)T=c(T,1:i);T[n]

Takes input from n, and creates the sequence for the first n sequences. This makes it somewhat inefficient at higher inputs, but it should be fine. It then returns the n-th entry, 1-indexed.

Uses a nice little trick by starting off the sequence with T, which is TRUE or 1 by default.

C11, 48 bytes

int f(int x){int q=1;while(x>q)x-=q++;return x;}

Try it online!

Also works in C++ and Java.

An alternative for the same byte count:

int f(int x){int q=0;while(x>++q)x-=q;return x;}

brainfuck, 141 bytes

I know I'm too late for the bounty, but I just wanted to see how many bytes the algorithm I thought of would end up being.

This program is zero-indexed.

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

Try it online

• Select Dynamic (infinite) Memory, or it won't work
• To test input values > 255, change Cell size (Bits) to 16 or 32.
• The interpreter explains how to give input. For decimal input use \5 for input of 5.
  • The maximum decimal value you can test input with is \999
  • Hex input can go as high the cell-size.

Explanation:

This shows the program broken up by step, showing what happens for input of 5. # are placed in the ideal memory dump locations for the interpreter.

You will probably want to use checkbox Dump Memory at char: # if running this version. This will dump the memory upon hitting #, allowing you to see the value on the tape in the event that it's an unprintable character, or to see what happens at any step you want. The cell that the pointer is on will be in bold.

,>+<                       (5) 1
[[->>+>+<<<]>>[-<<+>>]       5 1 (0) 5
<[->+>>+<<<]>[-<+>]>>+       5 1 0 5 (2)
[[-<->>+<]>[-<+>]<<<<] (0) 0 4 1 0 3 2 0 0
>>[>>>]                      4 1 0 3 2 0 (0) 0
                             1 1 0 (0) 2 0
>[.#[<<<]]<<<<                4 1 0 (3) 2 0 0 0
<<<[[-]>>>[-<+<<+>>>]<[->+<]<<<<<]>>> (3) 1 0 3 2 0 0 0
[>>>]<<<]>[.#>]

Tape structure:
    (cell_1 cell_2 temp), (cell_1 cell_2 temp), ...

Take Input;
If not zero:
  copy last pair to the right and add one to its cell_2
  subtract each cell_2 from each cell_1 (leaving each cell_2 intact)
  move checking from left to right: 
    If cell_1 is zero and cell_2 isn't:
      print cell_2
    Else:
      copy last cell_1 back, overwriting each previous cell_1
Else:
  right one and print result

Try it online

• Select Dynamic (infinite) Memory, or it won't work
• Dump Memory at char: #

Notes:

• To run this on another interpreter that doesn't allow moving to the left of the starting cell (that's why I use Dynamic Memory), put in a bunch of > at the start. The number required may vary depending on the input value, but is O(1).

tinylisp (repl), 90 bytes (0-indexed)

(d r(q((n j x)(i n(i(e j x)(r(s n 1)1(s x(s 0 1)))(r(s n 1)(s j(s 0 1))x))j
(q((n)(r n 1 1

Or, non-competing (using a feature that was committed after this challenge was posted), 80 bytes:

(d r(q((n j x)(i n(i(e j x)(r(s n 1)1(a x 1))(r(s n 1)(a j 1)x))j
(q((n)(r n 1 1

The first line defines a helper function r, and the second line is an unnamed function that takes n and returns the nth term of the sequence. I've specified this as a repl submission because the repl autocompletes parentheses at the end of every line, not just at the end of the program. With those caveats, here's a version modified to work at Try it online, and here's an ungolfed version run against inputs 0 through 54.

Explanation

I'll use the noncompeting version here. The only difference is that the official version has to implement addition as two subtractions.

(d r           Define r to be:
 (q(           A lambda function (= a list of two elements, quoted to prevent evaluation):
  (n j x)       Arguments n, j (the range counter), and x (the range limit)
  (i n          If n is truthy, i.e. nonzero:
   (i(e j x)     If counter equals limit:
    (r            Call r recursively on:
     (s n 1)       n-1
     1             counter reset to 1
     (a x 1))      limit increased by 1
    (r           Else, call r recursively on:
     (s n 1)       n-1
     (a j 1)       counter increased by 1
     x))           same limit
   j))))        Else, we're done; return the counter value

(q(            Lambda function:
 (n)            Argument n
 (r n 1 1)))    Call r with n, counter = 1, range limit = 1

C, 54 bytes

Not the shortest C solution, but it has the merit of running in constant time (no loops, just math). It uses zero-based indexing:

x;f(n){x=floor(sqrt(8*n+1)-1)/2;return 1+n-x*(x+1)/2;}

Ungolfed:

int f(int n) {
    int x = floor(sqrt(8*n+1)-1)/2; //calculate the number of the current subsequence (zero based)
    return 1+n-x*(x+1)/2;   //x*(x+1)/2 is the zero based index of the `1` starting the subsequence
}

Test with:

#include <math.h>
#include <assert.h>
#include <stdio.h>

x;f(n){x=floor(sqrt(8*n+1)-1)/2;return 1+n-x*(x+1)/2;}

int main(){
    int i;
    for(i = 0; i < 10; i++) printf("%d ", f(i));
    printf("\n");

    assert(f(0) == 1);
    assert(f(1) == 1);
    assert(f(5) == 3);
    assert(f(10) == 1);
    assert(f(59) == 5);
    assert(f(100) == 10);
    assert(f(1001) == 12);
}

C, 103 bytes

For a begginer it's okay, I think :).

int main(){int n,c,i,j;scanf("%d",&n);while(c<n){for(i=1;i<=j;i++){c++;if(c==n)printf("%d",i);}j++;}}

or the formatted way

#include <stdio.h>

int main() {
    int n,c,i,j;
    scanf("%d",&n);
    while(c<n) 
    {
        for(i=1;i<=j;i++)
        {
            c++;
            if(c==n) printf("%d",i);
        }
        j++;
    }
}

dc, 21 bytes, 0-based indexing

?d8*1+v1-2/d1+*2/-1+p

Try the dc program online!

Explanation:

?      Push input number n.
d      Duplicate n at the top of the stack.
8*1+   Replace n at the top of the stack with 8n+1.
v      Replace 8n+1 at the top of the stack with the floor of its square root.
1-2/   Subtract 1, and divide by 2 (ignoring any fractional part).

The top of the stack now holds the index k of the greatest triangular number that is <= n.

d1+*2/ Compute k(k+1)/2, which is the greatest triangular number <= n.
-      Subtract n-(the greatest triangular number <= n). The first n is on the stack in position 2, because the input number n was duplicated at the top of the stack at the very beginning of the program, and only one of the values was popped and used (until now).
1+     Add 1 because the desired sequence starts over again at 1 (not 0) at every triangular number.
p      Print the answer.

This dc program can be converted into a competitively-sized bash script:

Bash + Unix utilities, 28 bytes, 0-based indexing

dc -e"?d8*1+v1-2/d1+*2/-1+p"

Try the bash program online!

C, 81 44 bytes

straight iterative method, 0 indexed, and with some gentle massaging;

m=1,c=1;f(n){for(;n--;c=c^m?c+1:!!++m);c=c;}

Try it online!