New answers tagged arithmetic
0
Hexagony, 14 bytes
,{,:'2<'+{*/@;
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Linear control flow: ,{,'+{*2':;@
Wasting 2 characters for control-flow redirection, and does not reuse the duplicate , and {.
With enough effort, this probably can be optimized.
1
Pyth, 5 bytes
C.OCM
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1
Husk, 2 bytes
←p
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2
Vyxal, 1 byte
′
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There's a builtin just for this.
0
Python 3, 30 27 bytes
f=lambda n:1>>n or n*f(n-1)
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1
Lua (67 bytes)
function x(t,n)return load("return "..table.concat(t,"+",1,n))()end
Explanation:
table.concat(t,"+",1,n) gives the string formed by concatenation of elements of t from 1 to n using + as connector, i.e. t[1]+...+t[n].
load(<string>)(), as its name suggests, loads a string to be executed as a function. In ...
1
Husk, 4 3 bytes
Saved 1 byte thanks to Dominic Van Essen
!¡∫
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!¡∫
¡ Repeatedly apply the following function to the first argument, making an infinite list
∫ Cumulative sum (one cycle)
! Index into this list with the second argument
0
BRASCA, 15 bytes
xig[:1-]x[*$]xn
Explanation
xig - Remove LF from implicit STDIN, convert to numbers and concatenate
[:1-] - Generate a sequence from N to 0 on the stack
x - Remove the 0
[*$] - Multiply it all together
$n - Bring the number to the front and output it
...
0
Java, 45 chars
char c(char a,char b){return(char)((a+b)/2);}
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2
convey, 16 bytes
}>-1
1@"v
^*",{
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Loops over the input value until it reaches 0. There may be a shorter way using the indices command (/.), but it seems to take up more logic than I want. Since the interpreter includes a handy-dandy gif maker, you can even see it in action.
1
Vyxal, 13 bytes
0?(»Þ([Ḇ»↜Œ∑∫
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A port of the 05AB1E answer
Explained
0?(»Þ([Ḇ»↜Œ∑∫
0? # Push 0 followed by the input number
( # Input number of times:
»Þ([Ḇ» # Push the number 433544355
↜Œ # And get the indexes corresponding to the numbers in the second last item on the stack
∑...
0
Haskell, 45 bytes
1?x=(x+)
n?x=(!!)$iterate((n-1)?x)$signum$n-2
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Call it with (n?a)b.
Haskell, 51 bytes
(1?a)b=a+b
(2?_)0=0
(_?_)0=1
(n?a)b=(n-1)?a$n?a$b-1
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Copied straight from the hyperoperation formula on Wikipedia. Call it with (n?a)b.
2
Python 3, 88 85 77 73 bytes
i=0
exec("i+=1;print(i%3*i%5and i or(i%3<1)*'Fizz'+(i%5<1)*'Buzz');"*100)
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1
05AB1E (legacy), 6 bytes
ζÆ(I>M
Port of @Dennis's Jelly answer, so make sure to upvote him as well! The I/O is also in the same format, except with the input wrapped inside a list instead of three loose inputs: \$[Z,Y,X]\$, with little-endian digit order, using the digit \$0\$ for unary.
Try it online or verify all test cases.
The new 05AB1E version ...
answered Jan 12 at 16:49
Kevin Cruijssen
85.7k77 gold badges110110 silver badges303303 bronze badges
0
05AB1E, 2 bytes
zm
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zm # full program
# implicit input...
m # to the power of...
z # 1 divided by...
# implicit input
# implicit output
2
Canvas, 31 bytes
A2^{w3%0≡Fizz×y5%0≡Buzz×+:?O]yO
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Okay seriously guys why is it that both ascii-art focused languages didn't already have a FizzBuzz? This uses my favourite FizzBuzz approach found in the Vyxal answer.
0
Perl 5 -MList::Util=sum -a, 44 bytes
say sum(@F[$_..$_+$,-1])/$,for 0..@F-($,=<>)
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1
x86 Machine Code, 20 bytes
86 89 D9 31 C0 03 44 8F FC E2 FA 31 D2 F7 F3 AB 39 F7 72 ED C3
The above bytes define a function that calculates the moving average of a list, modifying the elements of the list in-place. It assumes that the list is a contiguous array of integer elements. The function takes the following inputs in registers
edi: the address of ...
0
Prolog (SWI), 84 bytes
g([],A,A).
g([H|T],E,O):-M is mod(H,2),F is E+1-M,P is O+M,g(T,F,P).
f(X):-g(X,0,0).
Try it online! Actually, I don't know how to make print results in TIO (there'll just be a warning if a query fails), so try it in SWISH. Please just don't make any changes to it.
g\2 accepts a list, the number of even elements found so far, and the ...
1
Nim, 105 101 100 bytes
import math,random
randomize()
let
t=TAU.rand
u=2.0.rand-1
v=sqrt(1-u*u)
echo [t.cos*v,t.sin*v,u]
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Uses the third method with a bit of substitution:
$$u = \cos(\phi) = 2j - 1$$
$$v = \sin(\phi) = \sqrt{1-u^2}$$
0
Husk, 3 bytes
mAX
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Output as a list of rationals.
1
FALSE, 86 bytes
[\100*\/$100$@\/*-0=]m:1[$100>~][0\$3m;![\%1\"Fizz"]?$5m;![\%1\"Buzz"]?\0=[$.]?10,1+]#
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Explanation
[\100*\/$100$@\/*-0=]m: Define Modulo Function
[\100*\/ Divide number * 100 by the divisor to get decimals
$100$@/* Get truncated result
-0= Subtract to get decimal, if equal to 0, then the ...
0
Stax, 8 bytes
ä;ªÜé▐ßó
Run and debug it
Uncompressed is 9 bytes, still beating CJam(with the same method).
Explanation
|T{Ze*m|M
|T permutations of input
{ m mapped to:
Z tuck a zero under the string
e evaluate the string
* multiply the top two elements
|M get the maximum of the mapped list
0
Jelly, 12 bytes
Œ!ŒṖ€Ṗ€ẎḌP€Ṁ
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ಠ_ಠ Tied with CJam. Takes input as a list of values, +2 bytes to input as space-separated
How it works
Œ!ŒṖ€Ṗ€ẎḌP€Ṁ - Main link. Takes a list L on the left
Œ! - Permutations of L
ŒṖ€ - Partitions of each
Ṗ€ - Remove the last one from each
Ẏ - Join into one list
Ḍ -...
answered Dec 27 '20 at 22:29
caird coinheringaahing
22.5k55 gold badges5353 silver badges159159 bronze badges
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