Programming Challenge: creative FizzBuzz

I only know how to code at a really basic level, but a weekly coding challenge would be quite cool to have on ~comp.

Ok, I cant code but my god can I excel. So I did.
Image of my final outcome.

I used this formula in each cell (where A24, in this case, is the cell number):
=IF(AND(MOD(A24,4)=0,(MOD(A24,3)=0)),"FizzBuzz",IF(MOD(A24,3)=0,"Fizz",IF(MOD(A24,4)=0,"Buzz",A24)))

MOD(A24,4)=0 Essentially asks do you get 0 when you divide cell A24 by 4? If you do it must mean it is a multiple of 4. I use this basic way of finding out if it is a muiltiple as the basic for the rest of it. I use an 'and' statement to check first of all if it is a muiltple of both if it's not it tries the other two individually.

Essentially It is one big if statement. In laymen's terms, it's asking Is this cell divisible by 4 and 3? If so return "FizzBuzz", if not is it divisible by 3? If so return "Fizz", if not is it divisible by 4? If it is return "Buzz" if not just what value is the cell.

I've been wanting to see how hackable quines are and came up with this just now.

/* Self-printing program by Joseph Deu Ngoc and mangled by Ruspaceni */
/* Program Number: 0 */
#define MAXPROGRAMS 100

void main() {
  char a='"';
  char b='\\';
  char c='\n';
  char *d="/* Self-printing program by Joseph Deu Ngoc and mangled by Ruspaceni */%c/* Program Number: %d */%c#define MAXPROGRAMS %d%c%cvoid main() {%c  char a='%c';%c  char b='%c%c';%c  char c='%cn';%c  char *d=%c%s%c;%c  int e=%d;%c if ((e+1)%%3==0){printf(%c//fizz%%c%c,c);}%c if ((e+1)%%5==0){printf(%c//buzz%%c%c,c);} %cif ((e+1)%%3==0&&(e+1)%%5==0){printf(%c//fizzbuzz%%c%c,c);}%c e=(e+1)%%MAXPROGRAMS;%c  printf(d,c,e,c,MAXPROGRAMS,c,c,c,a,c,b,b,c,b,c,a,d,a,c,e,c,a,a,c,a,a,c,a,a,c,c,c,c,c);%c}%c";
  int e=0;
 if ((e+1)%3==0){printf("//fizz%c",c);}
 if ((e+1)%5==0){printf("//buzz%c",c);} 
if ((e+1)%3==0&&(e+1)%5==0){printf("//fizzbuzz%c",c);}
 e=(e+1)%MAXPROGRAMS;
  printf(d,c,e,c,MAXPROGRAMS,c,c,c,a,c,b,b,c,b,c,a,d,a,c,e,c,a,a,c,a,a,c,a,a,c,c,c,c,c);
}

The quine was found on some big listo'quines and looked the most adjustable.
I can run it with the .bat file:

"tcc/tcc.exe" -w main.c
main.exe > main.c

and it'll churn out the source for the next iteration along with the output at the top in a comment.

I really like quines for some reason and I might start fiddling with them even more. I like the idea of having a quine clock that constantly recompiles and re-runs for each output. But i haven't had much luck with that so far.

I'm an amateur programmer, so I'm mostly proud that it worked on the first try. Then I googled the print_function stuff to clean up the output a little

#!/bin/python

from __future__ import print_function

for num in range(1,100):
	if num % 3 == 0:
		print("Fizz",end="")
	if num % 5 == 0:
		print("buzz",end="")
	if num % 3 != 0 and num % 5 != 0:
		print(str(num),end="")
	print(" ",end="")

outputs

[root@host scripts]# python fizzbuzz.py 
1 2 Fizz 4 buzz Fizz 7 8 Fizz buzz 11 Fizz 13 14 Fizzbuzz 16 17 Fizz 19 buzz Fizz 22 23 Fizz buzz 26 Fizz 28 29 Fizzbuzz 31 32 Fizz 34 buzz Fizz 37 38 Fizz buzz 41 Fizz 43 44 Fizzbuzz 46 47 Fizz 49 buzz Fizz 52 53 Fizz buzz 56 Fizz 58 59 Fizzbuzz 61 62 Fizz 64 buzz Fizz 67 68 Fizz buzz 71 Fizz 73 74 Fizzbuzz 76 77 Fizz 79 buzz Fizz 82 83 Fizz buzz 86 Fizz 88 89 Fizzbuzz 91 92 Fizz 94 buzz Fizz 97 98 Fizz [root@host scripts]# 

How about using obnoxious amounts of currying?

(function(n) {
    for(var i = 1; i <= n; i++) {
        console.log((function(k) {
            return (function(fizz) {
                return (function(buzz) {
                    return fizz || buzz ? fizz + buzz : k;
                })(k % 5 == 0 ? 'buzz' : '');
            })(k % 3 == 0 ? 'fizz' : '');
        })(i));
    }
})(100);

fn main() {
    for x in 1..101{
        let mut s = String::from(x.to_string());
        if x%3 == 0{
            s = String::from("fizz");
        }
        if x%5 == 0{
            match s.parse(){
                Ok(whatever) => {
                    s = String::from("buzz");
                    whatever
                },
                Err(_) => {
                    s = String::from("fizzbuzz");
                    0
                },
            };
        }
        println!("{}", s)
    }
}

I just started learning Rust a few days ago, and haven't had much time to dedicate due to work. To my understanding, the match s.parse() has to return a number value, correct? I don't understand why it doesn't compile if I do this instead:

match s.parse(){
    Ok(_) => {
        s = String::from("buzz");
        0
    },
    Err(_) => {
        s = String::from("fizzbuzz");
        0
    },
};

Here's one I just made in JS. Output is to the console.

for(var i=0,f=!1,b=!1,z=!1;++i<101;f=i%3>1,b=i%5>3,z=f&&b){console.log(z?"FizzBuzz":f?"Fizz":b?"Buzz":i);}

How about using something super old? Don't google this and guess what it's written in.

   IDENTIFICATION DIVISION.
   PROGRAM-ID. FIZZ-BUZZ.

   DATA DIVISION.
   WORKING-STORAGE SECTION.
   01 CT PIC 999 VALUE 1.
   01 FZ PIC 999 VALUE 1.
   01 BZ PIC 999 VALUE 1.

   PROCEDURE DIVISION.
   FIZZ-BUZZ-MAIN SECTION.
       PERFORM 100 TIMES
               IF FZ = 3
                    THEN IF BZ = 5
                       THEN DISPLAY "FizzBuzz"
                       COMPUTE BZ = 0
                       ELSE DISPLAY "Fizz"
                       END-IF
                       COMPUTE FZ = 0
                    ELSE IF BZ = 5
                    THEN DISPLAY "Buzz"
                            COMPUTE BZ = 0
                    ELSE
                            DISPLAY CT
                    END-IF
            END-IF
            ADD 1 TO CT
            ADD 1 TO FZ
            ADD 1 TO BZ
       END-PERFORM
       STOP RUN.

Here's a C solution that abuses POSIX printf:

int main() {
    char s[]="\n%n$s\0%1$d";
    for(int i=1;i<101;++i) {
        s[5]=51>(s[2]=50+(0x1241843>>i%15*2&3));
        printf(s,i,"","Fizz","Buzz","FizzBuzz");
    }
    return 0;
}

32-bit MIPS Assembly:

.data
fizz: .asciiz "Fizz"
buzz: .asciiz "Buzz"
line: .asciiz "\n"

.text
main:
    li $t0,0
    li $t1,101
    li $t2,3
    li $t3,5
    li $t5,0
loop:
    # $t0 % 3
    div $t0,$t2
    mfhi $t4
    # != 0: goto check5
    bgtz $t4,check5
    # set matched flag
    li $t5,1
    # print "Fizz"
    li $v0,4
    la $a0,fizz
    syscall
check5:
    # t0 % 5
    div $t0,$t3
    mfhi $t4
    # != 0: goto checkneither
    bgtz $t4,checkneither
    # set matched flag
    li $t5,1
    # print "Buzz"
    li $v0,4
    la $a0,buzz
    syscall
checkneither:
    # check if we've matched yet
    bgtz $t5,newline
    # print i
    li $v0,1
    ori $a0,$t0,0
    syscall
newline:
    # print "\n"
    li $v0,4
    la $a0,line
    syscall
    # reset matched flag
    li $t5,0
    # increment the iteration variable
    addi $t0,$t0,1
    # Exit if our iteration variable == 101
    bne $t0,$t1,loop
exit:
    jr $ra

Fizzbuzz in css using 3 nth-child selectors and pseudo-elements:
https://jsfiddle.net/wjbosrxv/

Sorry for zombifying this thread but I was super proud of being able to do one of the challenges!

In Python.

fizzbuzz=[]

for i in range(1,101):
    if i%15==0:
        fizzbuzz.append('FizzBuzz')
    elif i%5==0:
        fizzbuzz.append('Buzz')
    elif i%3==0:
        fizzbuzz.append('Fizz')
    else:
        fizzbuzz.append(i)

print(fizzbuzz)

ES6 Javascript. Can be executed in Chrome's dev tools console.

const GO=( F,i,z,{[z[0 ]]:B},u,  zz)=>B((F%      u?'':GO[z  [u/    u]] ()[z[zz-  zz/zz]](u- 
u,u            **u          +zz       -zz        )[z    [u] ](z    [zz      ],''      )[  
z[/*           **/         zz-       u/u]        ](u    -u, zz-    u/u     ))+       (F%
zz?'':GO[z     [u/        u]]       ()[          z[zz-zz/   zz]    ](u    -u,u      **u
+zz            -zz       )[z       [u]           ](z   [zz  ],''   )[z   [zz       -u/     
u]]            (zz      -u/       u))            ||F    )|| F<     zz**  u-u      **u      
+zz        -u?GO(F+1,i ,z,i[z[u- zz/zz]],u       ,zz):null   ;//)[z[z  z-u/u]])[ z[zz-u/u] 


GO(1,this,['log', 'toString', 'console', 'replace', 'substring', /[^a-z]/gi],console,3,5);

It does some stupid fuckery to read the parameter names of the GO function and extract the string FizzBuzz from in there. And I just obfuscate some array indices by using u=3 and zz=5 and various maths operations to get 0, 1, 2, 4, 23, and 100.

I'm not a great programmer, but here I go with javascript:

for (var i = 1; i <= 100; i++) {
	if ((i % 3 == 0) && (i % 5 != 0)) {
		console.log("fizz");
	};
	if ((i % 5 == 0) && (i % 3 != 0)) {
		console.log("buzz");
	};
	if ((i % 5 == 0) && (i % 3 == 0)) {
		console.log("fizzbuzz");
	};
	if ((i % 5 != 0) && (i % 3 != 0)) {
		console.log(i);
	};
}

Sorry, not the most creative way of doing it. In Perl 5:

for (1...100) {
    print "Fizz" x !($_ % 3);
    print "Buzz" x !($_ % %);
    print "$_" x !!($_ % 3 and $_ % 5);
    print "\n";
}

Necroposting, since I'm late to the party:

JS can now be abused to the point of looking like if Perl and ML had a baby. Backticks and ${} now act like clumsy brackets, the functional features (and let) are lovely, and some pointless esoterica can completely mislead people.

let hexedron = (_,x,...y) => y[0][x].call(...y);
let torus = (_,x,...y) => _[0][x].call(...y);

let recur = (f, x, yl) => {
  let [y, ...ys] = yl;
  if(ys.length) {
    return recur(f,f(x,y),ys)
  }
  else {
    return f(x,y)
  }
}


const fizz = 3;
const buzz = 5;
const fizzbuzz = fizz * buzz;

let _fizzbuzz = (n) => {
  let strung = hexedron`${'toString'}${n}${fizzbuzz}`;

  let f = (x,y) => hexedron`${'replace'}${x}${y[0]}${y[1]}`;

  strung = recur(f, strung, [[/.*?([369c0])$/,"fizz $1"],[/(fizz )?(.*?[5a0])$/,"$1buzz"],[/[c3PO#69]/,""]]);

  return [strung][~-(0|torus`${"includes"} ${strung} ${"zz"}`)] || n;
};

let about100 = ~100;

for (let f = -~0; f <- about100; f++) {
  console.log(_fizzbuzz(f));
}

I had just done this for my bootcamp. Here's mine in JavaScript (repl.it = https://repl.it/@nickfp1985/array-index-fizzbuzz )

for (i = 1; i < 100; i++) {
if (i % 3 === 0 && i % 5 === 0) {
console.log('fizzbizz');
} else if (i % 5 === 0) {
console.log('fizz');
} else if (i % 3 === 0) {
console.log('bizz');
} else {
console.log(i);
}
}

SQL, tested with SQLite 3.24.0:

with x(a,n) as (
  select 1,1
  union all
  select
    (case (n+1)%15
      when 0 then 'FizzBuzz'
      else (
        case (n+1)%5
          when 0 then 'Buzz'
          else (
            case (n+1)%3
              when 0 then 'Fizz'
              else n+1
            end)
        end)
    end),
    n+1
  from x where n<100
)
select * from x;