>>53935916
Use divisibility tests for 3 and 5 instead.
Though, that's what a good compiler will simplify the modulo operator to anyway.

by treating 1-100 as real numbers and dividing by the 3, 5, or whatever and checking if the result is anything other than an integer (integer in the mathematical sense not the variable type you spergs)

int modulo(int a, int b)
   {
       while (b <= a)
          {
              a -= b;
              }
       
       return a;
       }

int main(int argc, char *argv[])
   {
       int ii;
       for (ii = 0; ii <= 99; ii++)
          {
              if (!modulo(ii + 1, 3) && modulo(ii + 1, 5) )
                 {
                     #include <stdio.h>
                     printf("Fizz");
                     }
              else
                 {
                     if (!modulo(ii + 1, 5) && modulo(ii + 1, 3) )
                        {
                            #include <stdio.h>
                            printf("Buzz");
                            }
                     else
                        {
                            if (!modulo(ii + 1, 3) && !modulo(ii + 1, 5) )
                               {
                                   #include <stdio.h>
                                   printf("FizzBuzz");
                                   }
                            else
                               {
                                   #include <stdio.h>
                                   printf("%d", (int) ii + 1);
                                   }
                                   }
                                   }
              #include <stdio.h>
              printf("%c", (char) ' ');
              }
       #include <stdio.h>
       printf("%c", (char) '\n');
       
       return 1;
       }

>>53935916
>naming your company after your full name

Only a women

>>53935916
Even if you know fuckall about math you should be able to keep two counters and reset them when they reach 5 or 3.

>>53935916
Fizz buzz is only common questions at startups and other shit companies.

These people would be I need for a rude awakening if they got the same questions you'd get at MS,amazon, facebook,etc.

>>53936488
dude

what the fuck

>>53936638
Did I miss something? I had to correct it after my first attempt...

that doesn't hard

c3:=1
c5:=1
c15:=1
for i := 1,100 do
  if c15 = 15 then
    print FizzBuzz
    c3, c5, c15 := 1
  elseif c3 = 3 then
    print Fizz
    c3 := 1
  elseif c5 = 5 then
    print Buzz
    c5 := 1
  else
    print i
  end
  c3++
  c5++
  c15++
end

>>53936663
your curly brackets are all over the place

>>53936721
I double checked, they're all the right length they're supposed to be.

>>53936761
kek

does it compile?

>>53935916

npm install fizzbuzz

>>53936721
You mean braces...right?

>>53936865
Right

Not native English m9

>>53936775
Why wouldn't his retarded indent style compile?
The compiler doesn't care.

Not sure about his retarded includes though

>>53936775
Try it yourself, anon-kun

>>53936488
I dont know why but this post made me laugh

>>53936875
No prob, the more you know,. :D

Here's one in Clojure:

(let [f "Fizz" b "Buzz" n nil s [n n f n b f n n f b n f n n (str f b)]]
    (map #(or %1 %2) (cycle s) (range 1 101)))

>>53935916
>abstract algebra
Gets me every time

Without Modulo operator, just keep two counters. One resets every time it's equal to 3, the other, every time it's equal to 5. The value of these counters represents divisibility by 3 or 5.

Thus, to do FizzBuzz, you need only use basic addition, assignment, if statements, a for loop, and of course, I/O.

Program FizzBuzz;
Uses sysutils;
Var
    FBArr : Array[1..100] Of String;
    Count : Integer;
Begin
    Count := 3;
    While Count <= 100 Do
    Begin
        FBArr[Count] := 'Fizz';
        Count := Count + 3;
    End;
    Count := 5;
    While Count <= 100 Do
    Begin
        If FBArr[Count] = 'Fizz' Then
            FBArr[Count] := 'Fizz Buzz'
        Else
            FBArr[Count] := 'Buzz';
        Count := Count + 5;
    End;

    For Count := 1 To 100 Do
    Begin
        If FBArr[Count] = '' Then
            Write(IntToStr(Count))
        Else
            Write(FBArr[Count]);
        Write(', ');
    End;
    WriteLn('...');
End.

>>53936879
That's what i meant

I would just use the definition of modulo and implement it normally.

int modulo(int a, int b)
{
    return a - (b * (a/b));
}
[/code[

Just replace modulo with this:

# n and m are integers
def is_mult(n, m):
    return m * (n // m) == n

int mod(int m, int n)
{
     while (n > m)
          n -= m;
     return n;
}

:3

>>53938609
if (mod(5, 5) == 5) printf("no\n");

>fizzbuzz

... really, America?

How do you expect to compete with the billions of Pajeets out there if your entry-level programmer test is fizzbuzz?

Come on, I literally solved that on my first day in tech school!

>>53935916
I once used two counters which count to 3 and 5 respectively

>>53936879
I don't know anon, maybe he's onto something there. Doing it that way ensures that you're only including what you need to when you need to, so you won't waste any cycles before it's needed.

Literally just subtracting a number by 3 in a loop. If the resulting number is positive, continue the loop. If it's negative, move on to the next loop (for subtracting the original number by 5). If it's zero, mark the number as a fizz and then move on.
Same thing with the 5 loop but marking as a buzz.
If by the end of the two loops the number has been marked with a fizz and a buzz, guess what?
[spoiler]It's a fizzbuzz[/spoiler]

I'm not even a programmer and I can do that