Fizzbuzz thread. In the language of your choice write a program that prints "fizz" for multiples of 3, "buzz" for multiples of 5, "fizzbuzz" for multiples of both, and the number for multiples of neither.
Fizzbuzz thread. In the language of your choice write a program that prints "fizz" for multiples of 3...
puts (1..100).map { |i| (fb = [["Fizz"][i % 3], ["Buzz"][i % 5]].compact.join).empty? ? i : fb }
main = do
let x = concat [if x `mod` 15 == 0 then "fizzBuzz!\n"
else if x `mod` 5 == 0 then "Buzz!\n"
else if x `mod` 3 == 0 then "fizz!\n"
else show x ++ "\n" | x
1/10 for technically solving the problem
public final class FizzBuzz {
public static void main(String[] args) {
for(int i = 0; i
what the shitting dick nipple is this
Ruby. Just started using it this year and it's been fun. Coming from a C++/c#/Java background, it's a fresh breath of weirdly pleasing air.
>using an extra modulo instead of reusing the results of the two orhet modulo
I get that you're trying to be clever by using a more "mathematical" approach but that you're actually doing is sacrificing basic logic and using a lot of extra CPU cycles
You are wasting so many CPU cycles
for(let i = 1; i
go away evil dogger.
for n in range(1, 101):
if n % 15 == 0:
print "FizzBuzz"
elif n % 3 == 0:
print "Fizz"
elif n % 5 == 0:
print "Buzz"
else:
print n
raw_input()
How shitty is my code?
Not as shitty as OP's so you're fine.
Oops, forgot to check the modulo result. Both if statements should be
if (i % 3|5 === 0)
Or I suppose this would work too
if (!(i % 3|5))
Fine other than using input at the end and using python 2 instead of 3.
Alright smartasses, how would you do it then?
int n = 0;
bool printed;
while (n < 100)
{
fizzed = false;
if (!(n % 3)) printf("Fizz"); printed = true;
if (!(n % 5)) printf("Buzz"); printed = true;
if (!printed) printf("%i, ", n);
++n;
}
coming for your jobs
let rec f n s =
if n > 100
then print_string s
else f (n+1) (s ^ (
if (n mod 15) = 0
then "fizzbuzz\r\n"
else
if (n mod 3) = 0
then "fizz\r\n"
else
if (n mod 5) = 0
then "buzz\r\n"
else (string_of_int n) ^ "\r\n"
))
in
f 0 ""
niggers
>printed is not used
>fizzed is not defined
>\r\n
Might as well double click fizzBuzz.exe at this point, pajeet.
What's wrong with using \r\n?
"fizzed" is supposed to be "printed" but brainfart
program fizzbuzz;
var
i:integer;
begin
for i:=1 to 100 do
begin
if (i mod 3=0) and (i mod 5=0) then
writeln('FizzBuzz')
else if (i mod 5=0) then
writeln('Buzz')
else if (i mod 3=0) then
writeln('Fizz')
else
writeln(i);
end;
end.
for(i=0;i++
class FizzBuzzCommandlet extends Commandlet;
event int Main(string Parms)
{
local int i;
local string S;
while(i++ < 100)
{
S = "";
if(i % 3 == 0)
S $= "Fizz";
if(i % 5 == 0)
S $= "Buzz";
Log(Eval(S=="",i,S));
}
return 0;
}
##FizzBuzz
for num in xrange(1,101):
msg = ''
if num % 3 == 0:
msg += 'Fizz'
if num % 5 == 0:
msg += 'Buzz'
print msg or num
private static void fizzBuzz(int length)
{
IntStream.range(1, length).forEach(i ->
{
String s = "";
if (i % 3 == 0) s += "Fizz";
if (i % 5 == 0) s += "Buzz";
if (s.isEmpty()) s += String.valueOf(i);
System.out.println(s);
});
}
ok
for (var i = 1; i < 101; i++) {document.body.innerHTML += '' + i + ''}; document.querySelectorAll('script')[0].remove()
div:nth-child(3n):before{content: "Fizz"}
div:nth-child(5n):after{content: "Buzz"}
div:nth-child(3n):before, div:nth-child(5n):after{font-size: initial}
div:nth-child(3n), div:nth-child(5n){font-size: 0}
div{font-family: Helvetica; font-weight: 300}
Works in chromium
Good work, Daniel
how about the weirdest language you can code on huh?
>inb4 that fag that posts a solution with tensorflow
IDENTIFICATION DIVISION.
PROGRAM-ID. FIZZBUZZ.
DATA DIVISION.
01 CUNT PIC 9(3) VALUE 001.
01 REMAINDER PIC 9(3) VALUE 000.
01 FIZZ PIC A(4) VALUE "FIZZ".
01 BUZZ PIC A(4) VALUE "BUZZ".
PROCEDURE DIVISION.
A-PARA.
PERFORM B-PARA UNTIL CUNT>100.
STOP RUN.
B-PARA.
COMPUTE REMAINDER = FUNCTION MOD (CUNT,3).
IF REMAINDER EQUALS 0
DISPLAY FIZZ
COMPUTE REMAINDER = FUNCTION MOD (CUNT,5)
IF REMAINDER EQUALS 0
DISPLAY BUZZ
COMPUTE REMAINDER = FUNCTION MOD (CUNT,15)
IF REMAINDER EQUALS 0
DISPLAY FIZZ + BUZZ
Only took one uni course like 3 years ago and haven't touched it since. I doubt this will compile because I wrote it on my phone and the alignment will be off aside from other issues.
Forgot the increment... shit
fn main() {
for i in 1..102 {
if i % 15 == 0 { println!("FizzBuzz") }
else if i % 3 == 0 { println!("Fizz") }
else if i % 5 == 0 { println!("Buzz") }
else { println!("{}", i) }
}
}
ok i improved this so now it's actually stable
body{counter-reset: item;list-style-type: none;}
div{counter-increment: item}
div:not(:nth-of-type(3n)):not(:nth-of-type(5n)):before{content: counter(item) ""}
div:nth-of-type(3n):before{content: "Fizz"}
div:nth-of-type(5n):after{content: "Buzz"}
div{font-family: Helvetica; font-weight: 300}
for (var i=0;i
for i in range(1,101): print "FizzBuzz"[i*i%3*4:8--i**4%5] or i
fite me
>look mom, I posted it again
I'VE PREPARED FOR THIS THREAD IN ADVANCE
Too lazy to rewrite my C version to not do a "% 15" check, so I'm posting the edgy Python version I cooked up with a boolean flag.
for i in range(1,101):
divisorFound = False
if i % 3 == 0:
# Prevent default behavior of automatically printing of newline
print("Fizz", end="")
divisorFound = True
if i % 5 == 0:
print("Buzz", end="")
divisorFound = True
if divisorFound == False:
print(i, end="")
# Print newline after falling through all cases
print("\n", end="")
#include
int main(int i)
{
char *s[] = {"%d\n", "Fizz\n", "Buzz\n", "FizzBuzz\n"};
for (i=0; i++ < 100;)
printf(s[!(i % 3) + (!(i % 5) * 2)], i);
return 0;
}
What language is this?
Forgot to mention I indent with only two spaces because I am a masochist
clever use of s.isEmpty(), like it
really fucking clever minimalist approach
fn main() {
for i in 1..101 {
if (i % 3 == 0) && (i % 5 == 0) { println!("FizzBuzz") }
else if i % 3 == 0 { println!("Fizz") }
else if i % 5 == 0 { println!("Buzz") }
else { println!("{}", i) }
}
}
is this good enough to get me a job at google as a diversity officer?
Guess the language
range = $(if $(filter $1,$(lastword $3)),$3,$(call range,$1,$2,$3 $(words $3)))
make_range = $(foreach i,$(call range,$1),$(call range,$2))
equal = $(if $(filter-out $1,$2),,$1)
limit := 101
numbers := $(wordlist 2,$(limit),$(call range,$(limit)))
threes := $(wordlist 2,$(limit),$(call make_range,$(limit),2))
fives := $(wordlist 2,$(limit),$(call make_range,$(limit),4))
fizzbuzz := $(foreach v,$(numbers),\
$(if $(and $(call equal,0,$(word $(v),$(threes))),$(call equal,0,$(word $(v),$(fives)))),FizzBuzz,\
$(if $(call equal,0,$(word $(v),$(threes))),Fizz,\
$(if $(call equal,0,$(word $(v),$(fives))),Buzz,$(v)))))
.PHONY: all
all: ; $(info $(fizzbuzz))
fizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzzfizzbuzz
web assembly
int n = 0;
bool printed;
while (n < 100)
{
fizzed = false;
if (!(n % 3)) {
printf("Fizz");
}
printed = true;
if (!(n % 5)) {
printf("Buzz");
}
printed = true;
if (!printed) {
printf("%i, ", n);
}
++n;
}
You dun goofed
[...Array(100).keys()].map(function(a){return a+1}).map(function(a){return 0===a%3&&0===a%5?"fizzbuzz":0===a%3?"fizz":0===a%5?"buzz":a})