Как написать на питоне что число нечетное

Kromster

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dudedude

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for x in (0, 1, 2, 3, 4):
    if x % 2 == 0: 
        print "%d is even" % x

http://ideone.com/2WMaT

Если нужна не четность, а нечетность — соответственно, x % 2 != 0.

for x in range(000000, 1000000000):
  if x % 2 == 0:
    print(Chet)
  else:
    print(Nechet)

ответ дан 3 фев 2019 в 17:33

number = int(input("Введите число: "))
if (number%2 == 0):
  print("Четное число")
else:
  print("Нечетное число")

ответ дан 22 июл 2018 в 9:27

number = int(input("Введите число: "))
if (number%2 > 0):
  print("Не чётное число =(")
else:
  print(number+2)

asked Feb 17, 2014 at 19:03

if num % 2 == 0:
    pass # Even 
else:
    pass # Odd

The % sign is like division only it checks for the remainder, so if the number divided by 2 has a remainder of 0 it’s even otherwise odd.

Or reverse them for a little speed improvement, since any number above 0 is also considered «True» you can skip needing to do any equality check:

if num % 2:
    pass # Odd
else:
    pass # Even 

answered Feb 17, 2014 at 19:05

Similarly to other languages, the fastest «modulo 2» (odd/even) operation is done using the bitwise and operator:

if x & 1:
    return 'odd'
else:
    return 'even'

Using Bitwise AND operator

• The idea is to check whether the last bit of the number is set or not. If last bit is set then the number is odd, otherwise even.
• If a number is odd & (bitwise AND) of the Number by 1 will be 1, because the last bit would already be set. Otherwise it will give 0 as output.

It shouldn’t matter if the word has an even or odd amount fo letters:

def is_palindrome(word):
    if word == word[::-1]:
        return True
    else:
        return False

answered Feb 17, 2014 at 19:05

Esteban AlivertiEsteban Aliverti

6,2142 gold badges19 silver badges31 bronze badges

Use the modulo operator:

if wordLength % 2 == 0:
    print "wordLength is even"
else:
    print "wordLength is odd"

For your problem, the simplest is to check if the word is equal to its reversed brother. You can do that with word[::-1], which create the list from word by taking every character from the end to the start:

def is_palindrome(word):
    return word == word[::-1]

answered Feb 17, 2014 at 19:05

NPENPE

480k105 gold badges940 silver badges1004 bronze badges

asked Feb 17, 2014 at 19:03

if num % 2 == 0:
    pass # Even 
else:
    pass # Odd

The % sign is like division only it checks for the remainder, so if the number divided by 2 has a remainder of 0 it’s even otherwise odd.

Or reverse them for a little speed improvement, since any number above 0 is also considered «True» you can skip needing to do any equality check:

if num % 2:
    pass # Odd
else:
    pass # Even 

answered Feb 17, 2014 at 19:05

Similarly to other languages, the fastest «modulo 2» (odd/even) operation is done using the bitwise and operator:

if x & 1:
    return 'odd'
else:
    return 'even'

Using Bitwise AND operator

• The idea is to check whether the last bit of the number is set or not. If last bit is set then the number is odd, otherwise even.
• If a number is odd & (bitwise AND) of the Number by 1 will be 1, because the last bit would already be set. Otherwise it will give 0 as output.

It shouldn’t matter if the word has an even or odd amount fo letters:

def is_palindrome(word):
    if word == word[::-1]:
        return True
    else:
        return False

answered Feb 17, 2014 at 19:05

Esteban AlivertiEsteban Aliverti

6,2142 gold badges19 silver badges31 bronze badges

Use the modulo operator:

if wordLength % 2 == 0:
    print "wordLength is even"
else:
    print "wordLength is odd"

For your problem, the simplest is to check if the word is equal to its reversed brother. You can do that with word[::-1], which create the list from word by taking every character from the end to the start:

def is_palindrome(word):
    return word == word[::-1]

answered Feb 17, 2014 at 19:05

NPENPE

480k105 gold badges940 silver badges1004 bronze badges