“Fizz Buzz Python” Code-Antworten

Fizz Buzz Python

def fizz_buzz(input):
    if (input % 3 == 0) and (input % 5 == 0):
        return "FizzBuzz"
    if input % 3 == 0:
        return "Fizz"
    if input % 5 == 0:
        return "Buzz"
    else:
        return input


print(fizz_buzz(3))
Zero2Code

Fizz Buzz Python

output = ""

for i in range(1, 101):
 
    if (i % 3 == 0):
        output += "Fizz"
    
    if (i % 5 == 0):
        output += "Buzz"
    
    elif (i % 3 != 0): 
        output += str(i)
    
    output += "\n" # Add a new line at the end of the output 

print(output)   
FriedOxygen

Fizz Buzz Python

fizz = 3
buzz = 5
fizzbuzz = fizz * buzz

for i in range(100):
  if i % fizzbuzz == o:
    print('fizzbuzz')
   elif i % fizz == 0:
    print('fizz')
   elif i % buzz == 0: 
    print('buzz')
   else:
    print(i)
Rich Ratel

Fizz Buzz Python

def fizzBuzz(n):
    # Write your code here
    for i in range(1,a):
        if(i%3==0 and i%5!=0):
            print("Fizz")
        elif(i%5==0 and i%3!=0):
            print("Buzz")
        elif(i%5==0 and i%3==0):
            print(i)
a = int(input("enter the no"))
fizzBuzz(a)
Fierce Fowl

Fizz Buzz Python

#made by myself :)
def FizzBuzz() :
    Numbers = 0
    for Numbers in range(101):
        if Numbers%3 == 0:
            print ("Fizz")
        if Numbers%5 == 0:
            print ("Buzz")
        if Numbers%3 == 0 and Numbers%5 ==0:
            print ("Fizz Buzz")
        else :
            print (Numbers)
FizzBuzz()
Important Impala

Fizz Buzz Python

# changeable game of fizz buzz (all you have to change is n and D
# to add or alter the game.
# No adding or changing if/else required to include bizz at 7 or fuzz at 11
# just insert it into the dictionary. (you could easily make a function out of this)

n = 100
D = {3: 'fizz',
     5: 'buzz'}
for x in range(1, n+1):         # makes a range as big as we like
    out = ''
    for k, v in D.items():      # compares each in range with each key in dictionary
        if x % k == 0:
            out += v            # then attaches the associated word onto the output
    if not out:
        out = x                 # if the output is still empty, i.e. no listed factors
    print(out, end=' ')         # the unchanged number becomes the output
Great Snakes!

Ähnliche Antworten wie “Fizz Buzz Python”

Fragen ähnlich wie “Fizz Buzz Python”

Weitere verwandte Antworten zu “Fizz Buzz Python” auf Python

Durchsuchen Sie beliebte Code-Antworten nach Sprache

Durchsuchen Sie andere Codesprachen