Fizz Buzz test

[Rod]

It would seem that applicants for programming positions face a barrage of testing prior to and during interview. I'm not a fan of such challenges. But be prepared

'Apparently the FizzBuzz challenge is something programmers may
'face at job interviews.

'"Write a program that prints the numbers from 1 to 100.
'But for multiples of three print “Fizz” instead of
'the number and for the multiples of five print “Buzz”.
'For numbers which are multiples of both three and five
'print “FizzBuzz”."

t=time$("ms")

while counter<100
   counter=counter+1
   if counter/3=int(counter/3)then
       if counter/5=int(counter/5) then
           print "FizzBuzz"
       else
           print "Fizz"
       end if
   else
       if counter/5=int(counter/5) then
           print "Buzz"
       else
           print counter
       end if
   end if
wend

print time$("ms")-t




I'm sure it could be slicker, one point, you get five minutes to code a solution!

[Rod]

Our rosetta code solution.


for i = 1 to 100
   select case
       case i mod 15 = 0
           print "FizzBuzz"
       case i mod 3 = 0
           print "Fizz"
       case i mod 5 = 0
           print "Buzz"
       case else
           print i
   end select
next i

[tsh73]

for  i= 1 to 100
   printed =0
   if (i mod 3)=0 then printed = 1: print "Fizz";
   if (i mod 5)=0 then printed = 1: print "Buzz";
   if not(printed) then print i;
   print
next


I did't measure time (read program description and started coding - didn't got to the end of post).

But it took some debug to run right.

[Rod]

Very slick way to join the words.

[tsh73]

About "slick way to join words".
I end using flag variable - I need something done
on (i mod 3 =0)
AND something (i mod 3 =0)
and something on ELSE.
Nicely lies on SELECT CASE, yes?

But there is a case of ((i mod 3 =0) AND (i mod 3 =0))
(actually to write (i mod 15 =0) just not happened to me)
and C have BREAK on SWITCH statement
so we kind of can have
print Fizz on 3 - break
print Fizz on 3 - NO break
print Buzz on 5 - break
print number on DEFAULT
(actually now I pretty sure I am wrong)
…
The only problem that we can't do that in C.
SWITCH in C is on constants only
SWITCH in C# has BREAK on each CASE
and SELECT CASE in VB NET does not have BREAK.
So pseudo-C pseudo-code:

/*fizzBuzz task*/
//oops impossible as thought - in C
#include <stdio.h>
int main(void)
{
int i;
for(i=1;i<=100;i++)
{
switch() //not in C
{
case i%3==0:     //not in C
printf("Fizz");  //strategically NOT placed BREAK
case i%5==0:     //not in C
printf("Buzz");
break;      
default:
printf("%d",i);
}
printf("\n");
}
return 0
}


And I'll stick to BASIC

---

EDIT actually I have working C code along these lines. But it ends somewhat complex.
But it ends somewhat completely different

[B+]

I had this written up some time ago, forgot to post it here:

"BizzFizzBuzzFuzzWow"
check$ = "0203050711": say$ = "BizzFizzBuzzFuzzWow"
FOR i = 1 TO 100
   Flag = 1
   FOR j = 0 TO 4
       IF i MOD VAL(MID$(check$, j * 2 + 1, 2)) = 0 THEN PRINT MID$(say$, j * 4 + 1, 4);: Flag = 0
   NEXT
   IF Flag THEN PRINT i ELSE PRINT
NEXT


[zzz000abc]

here is one more way...
insert code herefor i=1 to 100
   f$="":b$=""
       if i mod 3 =0 then f$="FIZZ"
       if i mod 5 =0 then b$="BUZZ"
      if f$+b$="" then print i else print f$+b$
next

[B+]

Using JB word$:
FOR i = 1 TO 100
   b$ = ""
   FOR w = 1 TO 5
       IF i mod val(word$("2 3 5 7 11", w)) = 0 THEN b$ = b$ + word$("Bizz Fizz Buzz Fuzz Wow", w)
   NEXT
   IF b$ = "" THEN PRINT i ELSE PRINT b$
NEXT


[tsh73]

Using output as flag, clever.