Your task is to generate the Fibonacci sequence to `n` places, with each alternating value as `"skip"`. For example:

`"1 skip 2 skip 5 skip 13 skip 34"`

Return the result as a string

You can presume that `n` is always a positive integer between (and including) 1 and 64.

Difficulty: EASY
Starter code: def skiponacci(n):
	

Solutions:
def skiponacci(n):
	s = '1'
	num = 1
	prv = 0
	for i in range(1, n):
		new = num + prv
		prv = num
		num = new
		if i % 2 == 0:
			s = s + ' ' + str(num)
		else:
			s += ' skip'
	return s


def skiponacci(n):
	fib = [1, 1][:n]
	for _ in range(n - 2):
		fib.append(sum(fib[-2:]))
	return ' '.join((str(n) if i % 2 else 'skip' for (i, n) in enumerate(fib, 1)))


def skiponacci(n):
	result = ['1', 'skip']
	suite = [1, 1]
	for i in range(2, n):
		suite.append(suite[i - 2] + suite[i - 1])
		if i % 2 != 0:
			result.append('skip')
		else:
			result.append(str(suite[i]))
	return ' '.join(result[0:n])


(fib, result) = ([1, 1], ['1', 'skip'])
for i in range(100):
	fib.append(fib[-1] + fib[-2])
	result.append('skip' if i & 1 else str(fib[-1]))

def skiponacci(n):
	return ' '.join(result[:n])


def gen_fib():
	(a, b) = (1, 1)
	yield a
	yield b
	while True:
		(a, b) = (b, a + b)
		yield b

def skiponacci(n):
	result = []
	for (i, f) in enumerate(gen_fib()):
		if i == n:
			break
		result.append(str(f) if i % 2 == 0 else 'skip')
	return ' '.join(result)


def skiponacci(n):
	res = []
	(a, b) = (0, 1)
	for i in range(n):
		(a, b) = (a + b, a)
		res.append('skip' if i % 2 else str(a))
	return ' '.join(res)


def skiponacci(n):
	mas = []
	(fib1, fib2) = (0, 1)
	for i in range(n):
		(fib1, fib2) = (fib2, fib1 + fib2)
		mas.append(fib1)
	sas = []
	for i in range(0, len(mas), 2):
		sas.append(str(mas[i]))
	if n % 2 == 0:
		return ' skip '.join(sas) + ' skip'
	else:
		return ' skip '.join(sas)