Define n!! as
n!! = 1 \* 3 \* 5 \* ... \* n  if n is odd, 
n!! = 2 \* 4 \* 6 \* ... \* n  if n is even. 
Hence 8!! = 2 \* 4 \* 6 \* 8 = 384, there is no zero at the end. 
30!! has 3 zeros at the end. 
For a positive integer n, please count how many zeros are there at 
the end of n!!. 
Example: 
count\_zeros\_n\_double\_fact(30) should return 3
def count_zeros_n_double_fact(n):
    if n % 2: return 0
    s, n = 0, n // 2
    while n >= 5:
        n //= 5
        s += n
    return s
def count_zeros_n_double_fact(n):
    if n % 2: return 0
    s, n = 0, n // 2
    while n >= 5:
        n //= 5
        s += n
    return s