## 题目描述：

LeetCode 793. Preimage Size of Factorial Zeroes Function

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]`.

## 解题思路：

```n!后缀0的个数K，等于不大于n的所有乘数中，因子5的个数。

n从4 * K开始递增枚举 + 验证即可```

## Python代码：

``````class Solution(object):
def preimageSizeFZF(self, K):
"""
:type K: int
:rtype: int
"""
n = 4 * K
t = 0
while t <= K:
t = self.trailingZeroes(n)
n += 1
if t == K: return 5
return 0

def trailingZeroes(self, n):
x = 5
ans = 0
while n >= x:
ans += n / x
x *= 5
return ans
``````

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