# FizzBuzz Solution Dumping Ground

#266
``````for(int i=1;i<=100;i++)
{
string val = string.Empty;
if(i%3 == 0)
{
val = "Fizz";
}
if(i%5 == 0)
{
val += "Buzz";
}

if(string.IsNullOrEmpty(val))
{
val = i.ToString();
}
Debug.Log(val);
}``````

#267

some of these are cringe.

``````for (int i = 1; i <= 100; i++)
{
int mod3 = i%3, mod5 = i%5;
if (!mod3 || !mod5)
{
if (!mod3) cout << "Fizz";
if (!mod5) cout << "Buzz";
}
else cout << i;
cout << endl;
}
``````

#268

Here is the python version approach. This is my first actual working code since I started learning about couple of days

``````for x in range(1,22):
#print x
if x % 5 == 0:
print x,"buzz"
elif x % 3 ==0:
print x,"fuzz"
else:
print x
``````

Output:

SimpleTest.pyâ, wdir='C:/Users/rakz)
1
2
3 fuzz
4
5 buzz
6 fuzz
7
8
9 fuzz
10 buzz
11
12 fuzz
13
14
15 buzz
16
17
18 fuzz
19
20 buzz
21 fuzz

#269

F# Version with tail-recursionâŚ Just because.

``````    let fizzBuzzList n =
let rec fizzBuzz n acc =
match n with
| 0 -> acc
| n when n % 3 = 0 && n % 5 = 0 -> fizzBuzz (n-1) ("FizzBuzz" :: acc)
| n when n % 3 = 0 -> fizzBuzz (n-1) ("Fizz" :: acc)
| n when n % 5 = 0 -> fizzBuzz (n-1) ("Buzz" :: acc)
| _ as n -> fizzBuzz (n-1) (string n :: acc)
fizzBuzz n []

let printFizzBuzz n =
fizzBuzzList n
|> List.iter (fun s -> printfn "%s" s)

printFizzBuzz 100000``````

#270

Recrusive Ruby Solution

``````def fizzbuzz(n)
return if n > 100
out = "#{n}: "
out += "Fizz" if (n % 3 == 0)
out += "Buzz" if (n % 5 == 0)
puts out
fizzbuzz(n+1)
end

fizzbuzz(1)``````

#271

It is really easy with Haskell and list comprehensions:

``````main :: IO ()
main = putStrLn \$ unlines fizzbuzz

fizzbuzz :: [String]
fizzbuzz = [if x `dividable` 3
then if x `dividable` 5
then "Fizzbuzz"
else "Fizz"
else "Buzz"
| x <- [1..100], x `dividable` 5 || x `dividable` 3]

dividable :: Int -> Int -> Bool
dividable lop rop = (lop `mod` rop) == 0``````

#272

A solution in F#

``````let (|Fizz|Buzz|FizzBuzz|Other|) x =
if x % 15 = 0 then FizzBuzz
else if x % 5 = 0 then Fizz
else if x % 3 = 0 then Buzz
else Other

[1..100]
|> List.iter (fun x -> printfn "%s" (match x with
| Fizz -> "Fizz"
| Buzz -> "Buzz"
| FizzBuzz -> "FizzBuzz"
| _ -> x.ToString()))``````

#273
``````	for (i = 1; i <= 100; i++) {

if (i%3 == 0 && i%5 == 0) {
document.write(" - FizzBuzz <br />");
} else if (i%3 == 0) {
document.write(" - Fizz <br />");
} else if (i%5 == 0) {
document.write(" - Buzz <br />");
} else {
document.write(" - " + i + "<br />");
}
}
``````

I am 12 and it took me a bit longer than 5 min because I wrote html.

#274

Oracle PL/SQL

``````DECLARE
msg_ VARCHAR2(8);
BEGIN
FOR rec_ IN (
SELECT LEVEL AS lvl
FROM DUAL
CONNECT BY LEVEL <= 100
) LOOP
SELECT DECODE(MOD(rec_.lvl, 15),
0, 'FizzBuzz',
3, 'Fizz',
5, 'Buzz',
6, 'Fizz',
9, 'Fizz',
10, 'Buzz',
TO_CHAR(rec_.lvl)) INTO msg_
FROM DUAL;
dbms_output.put_line(msg_);
END LOOP;
END;
/``````

#275

REM before you judge, I am a senior level executive, not a programmer
REM I did some basic programming on Apple iiâs back in the 80âs
REM It took me about 10 mins, mostly shaking the rust off and reformatting some
REM of the earlier attempts

REM âFizzBuzz programâ

for i = 1 to 100
if i mod 3 = 0 then REM i is evenly div by 3
print âFizzâ; REM suppress newline
end if
if i mod 5 = 0 then REM i is evenly div by 5
print âBuzzâ REM with newline
else
print REM adds newline to "Fizz"
end if
REM we are done with div by 3, 5 and 15
if i mod 3 > 0 and i mod 5 > 0 then REM not div by 3 or 5
print i
end if
next i

REM I would fail most of the attempts listed here in an interview situation for failing to
REM properly document the code. Plain language documentation is important. Apologies
REM for not formatting the code. I didnât know how

#276

SwiftâŚ

``````for i in 1 ... 100 {
let result = (i % 3 == 0 ? "Fizz" : "") + (i % 5 == 0 ? "Buzz" : "")
print(result.isEmpty ? String(i) : result)
}``````

#277
``````#include <iostream>
using namespace std;

int main () {
for (int i = 1; i <= 100; ++i) {
if (i%3 == 0) {
cout << "Fizz";
}
if (i%5 == 0) {
cout << "Buzz";
}
if (i%3 != 0 && i%5 != 0) {
cout << i;
}
cout << "\n";
}
return 0;
}``````

#278

`for(var i=1;i<101;++i){var c=0^((i%3)===0)|0^((i%5)===0)<<1; console.log([i,"Fizz","Buzz","FizzBuzz"][c])}`

#280

Python 1 liner:
`print "\n".join([(str(i) if i%3 else 'Fizz') if i%5 else ("Buzz" if i%3 else "FizzBuzz") for i in range(1,101)])`

#281
``````using System;
using System.IO;
using System.Collections.Linq;
using Pointless.Libs.Algorithms;

class FuzzBizzDriver
{
public FuzzBizzr()
{
_fizzBuzzRepository = ExcedinglyPointlessAbstractFactory.Create`<FizzBuzzRepository>`();
}

public static async void Main(string[] args...)
{
var numbers = Enumerable.Range(0,1111111);
var awsmResult = DoFizzBuzz(numbers);
for (var i = 0; i < numbers.Count; i++)
{
Console.WriteLine(awsmResult);
}
}
{
var unnecessaryDict = new Dictionary<int, string>();
Parallel.ForEach(
numbers,
new ParallelOptions { MaxDegreeOfParallelism = 16 },
number => {
var resultString = string.Empty;
resultString += _fizzBuzzRepository.Fizz(number, resultString);
resultString += _fizzBuzzRepository.Buzz(number, resultString);

if (string.IsNullOrWhiteSpace(resultString))
{
resultString = number.ToString()
}

});
return unnecessaryDict.OrderBy(x => x.Key).Select(x => x.Value).ToList();

}
}``````

#282

Here is what i made in 5min and a little JS

``````for (var i = 1; i < 101; i++) {
if (i % 3 == 0 && i % 5 === 0) {
console.log("FizzBuzz")
}
else if (i % 3 === 0) {
console.log("Fizz")
}
else if (i % 5 === 0){
console.log("Buzz")
}
else{
console.log(i)
}
}
``````

#284

My solution in C (in my defense, I didnât write all this by hand ):

``````#include <stdlib.h>
#include <stdio.h>
#include <string.h>

int main (){
printf("1 \n2 \nFizz \n4 \nBuzz \nFizz \n7 \n8 \nFizz \nBuzz \n11 \nFizz \n13 \n14 \nFizzBuzz \n16 \n17 \nFizz \n19 \nBuzz \nFizz \n22 \n23 \nFizz \nBuzz \n26 \nFizz \n28 \n29 \nFizzBuzz \n31 \n32 \nFizz \n34 \nBuzz \nFizz \n37 \n38 \nFizz \nBuzz \n41 \nFizz \n43 \n44 \nFizzBuzz \n46 \n47 \nFizz \n49 \nBuzz \nFizz \n52 \n53 \nFizz \nBuzz \n56 \nFizz \n58 \n59 \nFizzBuzz \n61 \n62 \nFizz \n64 \nBuzz \nFizz \n67 \n68 \nFizz \nBuzz \n71 \nFizz \n73 \n74 \nFizzBuzz \n76 \n77 \nFizz \n79 \nBuzz \nFizz \n82 \n83 \nFizz \nBuzz \n86 \nFizz \n88 \n89 \nFizzBuzz \n91 \n92 \nFizz \n94 \nBuzz \nFizz \n97 \n98 \nFizz \nBuzz \n");
}
``````

#285
``````void loop(){
for(int i =1 ; i<=100; i++){
if (i% 15 == 0){
Log.i(TAG, "loop: FizzBuzz");
}else if(i % 3 == 0){
Log.i(TAG, "loop: Fizz");
}else if (i% 5 == 0){
Log.i(TAG, "loop: Buzz");
}else {
Log.i(TAG, "loop: " + i);
}
}
}
``````

#286

Python 3:

``````print(*((x, 'Fizz', 'Buzz', 'FizzBuzz')[(not x%3) + (not x%5)*2] for x in range(1, 101)), sep='\n')
``````

or

``````print(*(x if x%5 and x%3 else 'FizzBuzz' if not x%3 and not x%5 else 'Fizz' if x%5 else 'Buzz' for x in range(1,101)), sep='\n')
``````

#288

If nobody has done it yet in Excel (not VBA, simple Excel formulas):

1. fill series cells A1:A100 with numbers 1 through 100
2. fill down cells B1:B100 with formula `=IF(MOD(A1,15)=0,"FizzBuzz",IF(MOD(A1,5)=0,"Buzz",IF(MOD(A1,3)=0,"Fizz",A1)))`