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.
f(3) = 0 because 3! = 6 has no zeroes at the end, while f(11) = 2 because 11! = 39916800 has two zeroes at the end.Given an integer k, return the number of non-negative integers x have the property that f(x) = k.
\n
Example 1:
\n\n\nInput: k = 0\nOutput: 5\nExplanation: 0!, 1!, 2!, 3!, and 4! end with k = 0 zeroes.\n\n\n
Example 2:
\n\n\nInput: k = 5\nOutput: 0\nExplanation: There is no x such that x! ends in k = 5 zeroes.\n\n\n
Example 3:
\n\n\nInput: k = 3\nOutput: 5\n\n\n
\n
Constraints:
\n\n0 <= k <= 109Compiled with clang 11 using the latest C++ 17 standard.
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