Problem:
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:
思路: 考虑125!有多少个0?实际上是求1 * 2 * 3 * … * 125 有多少个5。
Java版本: