Inspired by a task for Programming 101, here's a challenge that hopefully isn't too easy.
n >= 1.n lines of asterisks, where every new line has one asterisk more than the line before, and starting with one asterisk in the first line.* ** *** **** ***** This is code-golf, so shortest answer in bytes wins.
full program:
(dotimes(i(read))(format t"~v{*~}* "i 1)) (dotimes(i(read)) <-- loop from i=0 to i=INPUT-1 (format t"~v{*~}* "i 1)) <-- loop, displaying i+1 asterisks per line Ideas for improvement are welcomed.
:#*|[1,a|B=B+A?B Explanation:
: Gets a cmd line param as number called 'a' #*| Define A$ as '*' [1,a| FOR b = 1 to a B=B+A Extend B$ with another astrix ?B Print B$ <FOR LOOP implicitly closed> QBIC's seen a few updates since this answer was posted. This could now be done with these 12 bytes:
[:|B=B+@*`?B Anonymous VBE immediate window function that takes input from range A1 and outputs to the VBE immediate window.
For i=1To[A1]:?String(i,42):Next 4 bytes removed thanks to ngn! :)
{`0:,\x#"*"} { } /function(x) x#"*" /reshape "*" by x ,\ /scan concatenation through the list `0: /print line by line 
(1+!x)#'"*" -> ,\x#"*" \$\endgroup\$ 
Fn.new{|i|(1..i).map{|n|"*"*n}.join(" ")} Fn.new{|i| // New function with the parameter i (1..i) // Generate a range from 1 to i .map{|n|"*"*n} // Replace the numbers with asterisks .join(" ")} // Join with newline ╒⌂*n Try it online or verify some more test cases.
Explanation:
╒ # Push a list in the range [1, (implicit) input] ⌂ # Builtin for the asterisk character "*" * # Map each value to that amount of asterisk characters (similar as in Python) n # And join this list of strings by newlines # (after which the entire stack joined together is output implicitly as result) 
int f(int n){for(n&&f(n-1);~n;__builtin_putchar(n--?42:10));} Uses ASCII char codes and builtins to shorten another solution.
[\@$&'*' Explanation:
$&'*': Reshape * according to.$: Reshape.&: Bound with (partial application).'*': Character literal *.@: Function composition.
[\: Each prefix of.
[: Identity.\: Apply the preceding verb to each prefix of the right argument.
] instead of [, and # instead of $, but it's conceptually the same thing. \$\endgroup\$ j, 3 bytes×*¦ ×*¦ ×* # Repeat an asterisk the implicit input amount of times ¦ # Prefixes # j flag joins on newlines Since the course is taught in C++, I'm eager to see solutions in C++.
(the joke here is to deliver a 'pure' c++ answer)
#include<string> class X:public std::string{public:X(int i):std::string(i*(i+1)/2,'*'){for(;i--;)insert(i*(i+1)/2,"\n");}}; 
PODPROGRAM:F(N) **)TEKST * POWTORZ:J=1(1)I LINIA POWTORZ:I=1(1)N WROC Sometimes the simplest solutions are the shortest.
CZYTAJ:N **1)TEKST * POWTORZ:J=1(1)I LINIA POWTORZ:I=1(1)N STOP1 KONIEC 
{for(;$1--;)print s=s"*"} Shaved a byte.
{for(;$1--;print s)s=s"*"} {for(;$1--; # decrement input till 0 print s) # print string s with new line s=s"*"} # concat an asterisk each loop -1 byte: remembered ★ exists
ƛ★×, ƛ★×, ƛ # map over implicit range ★× # "*" times n , # print with newline 💎 Created with the help of Luminespire.
<script type="vyxal3"> ƛ★×, </script> <script> args=[["5"]] </script> <script src="https://themoonisacheese.github.io/snippeterpreter/snippet.js" type="module"/>
*=@repeat{*;@append{s;\*}@out{$s\n}} Sample run:
bash-4.3$ gema '*=@repeat{*;@append{s;\*}@out{$s\n}}' <<< 5 * ** *** **** ***** Quite simple, with a couple of nested loops. ASCII 42 is an asterisk.
: f 0 do I -1 do 42 emit loop CR loop ; Prints a trailing newline.
for((;i++<$1;));{ s+=\*;echo "$s";} Sample run:
bash-4.3$ bash draw-an-asterisk-triangle.sh 5 * ** *** **** ***** #import<iostream> main(){int n;std::cin>>n;for(int i=1;i<=n;++i){for(int j=i;j--;)std::cout<<"*";puts("");}} Straightforward.
< be moved directly next to the #include? (I don't do C++, just a wild guess based on Java knowledge) \$\endgroup\$ #include does need to be included in the byte count. \$\endgroup\$ Yeah this needs some reducing. I'm out of practice!
D1H}\[\['*'O]aO] (implicit input) D Peek n, push n. 1 Push 1. H Pop x, pop y, push range(x, y) } Reverse, pop n, reverse, push n. \[ ] Pop n, do bracketed items n times. \[ ] Pop n, do bracketed items n times. '*'O Output the character *. aO Output a newline. Think of it as though I'm iterating through a list (1 through n) and popping out the number of *s according to the currently selected list item.
Explanation soon.
#(doseq[m(range 1(inc %))](println(apply str(repeat m\*)))) Basically, just uses doseq to loop over the range, printing the corresponding number of stars.
It's really unfortunate that the shortest way I've been able to repeat a character in Clojure is (apply str (repeat m \*)). That's hardly competitive here. Some good ol' python string multiplication would have been awesome.
Ungolfed:
(defn tri [n] (doseq [m (range 1 (inc n))] (println (apply str (repeat m \*))))) Should be fairly self-explanatory.
&>:#v_@ -^ : >$55+,1 ,"*"<_^#:-1 &>:#v_@ -^ : >$a,1 1,*'<_^#:- 
1.upto($*[0].to_i){|i|puts"*"*i} The $* is an array of arguments.
foo.rb 5 will print the following:
* ** *** **** ***** 
ARGV[0]? Sure? Ruby 2.3.1 I use receives command line arguments as String, not Fixnum. \$\endgroup\$ $><<"*"*i instead of "*"*i+$/ though. \$\endgroup\$ This function returns to cell, requires text wrapping be turned on.
Function a(b) For i = 1 To b For c = 1 To i d = "*" & d Next c a = a & d & Chr(10) d = "" Next i End Function 99 Bytes - Alternative that runs as a function but prints to immediate.
Function f(b) For i = 1 To b For c = 1 To i Debug.Print "*"; Next c Debug.Print Next i End Function 94 Bytes - One more that is just a sub with hard coded value, prints to immediate.
Sub k() r = 5 For i = 1 To r For c = 1 To i Debug.Print "*"; Next c Debug.Print Next i End Sub Just a simple loop, really.
I'm new to golfing, so might not be the shortest way.
for($i=fgets(STDIN);$x++<$i;)echo str_repeat("*",$x)."\n"; Thanks to manatwork for byte-saving.
; separator. And you can combine the loop's test with the loops stepping: for($i=fgets(STDIN);$x++<$i;)echo str_repeat("*",$x).""; \$\endgroup\$ (n,c)=>{while(c<=n)console.log("*".repeat(c++))} 
(n,c)=>{while(c<=n){console.log("*".repeat(c++));}} is a valid entry. Also, you don't need the braces in the while loop, nor the semicolon at the end: (n,c)=>{while(c<=n)console.log("*".repeat(c++))} \$\endgroup\$ ->n{n.times{|i|puts'*'*i+?*}} Call with ->n{n.times{|i|puts'*'*i+?*}}[number].
'*'*(i+1) → '*'*i+?*). (BTW, you can call anonymous proc without assigning it to a variable: ->n{n.times{|i|puts'*'*i+?*}}[5].) \$\endgroup\$ taking advantage of recursive coding. it generates x times * as the line/round
Function f(x) If x<=1 Then f="*" Else f=f(x-1) & vbCrLf & String(x,"*") End Function results:
?f(10) * ** *** **** ***** ****** ******* ******** ********* ********** ?f(2) * ** ?f(5) * ** *** **** ***** 
(defun f(n)(dotimes(i n)(dotimes(j(1+ i))(format t"*"))(format t"~%"))) Usage:
CL-USER 170 > (f 5) * ** *** **** ***** NIL Ungolfed:
(defun f (n) (dotimes (i n) (dotimes (j (1+ i)) (format t "*")) (format t "~%"))) 