< Previous
Next >
Let f(x)
be the number of zeroes at the end of x!
. (Recall that x! = 1 * 2 * 3 * ... * x
, and by convention, 0! = 1
.)
For example, f(3) = 0
because 3! = 6 has no zeroes at the end, while f(11) = 2
because 11! = 39916800 has 2 zeroes at the end. Given K
, find how many non-negative integers x
have the property that f(x) = K
.
Example 1:
Input: K = 0
Output: 5
Explanation: 0!, 1!, 2!, 3!, and 4! end with K = 0 zeroes.
Example 2:
Input: K = 5
Output: 0
Explanation: There is no x such that x! ends in K = 5 zeroes.
Note:
K
will be an integer in the range [0, 10^9]
.
[Binary Search]
Similar Questions
- Factorial Trailing Zeroes (Easy)