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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?xhave 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].

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?f(x)?是?x!?末尾是0的數量。（回想一下?x! = 1 * 2 * 3 * ... * x，且0! = 1）

例如，?f(3) = 0?，因為3! = 6的末尾沒有0；而?f(11) = 2?，因為11!= 39916800末端有2個0。給定?K，找出多少個非負整數x?，有?f(x) = K?的性質。

示例 1: 輸入:K = 0 輸出:5 解釋:?0!, 1!, 2!, 3!, and 4!?均符合 K = 0 的條件。示例 2: 輸入:K = 5 輸出:0 解釋:沒有匹配到這樣的 x!，符合K = 5 的條件。

注意：

• K是范圍在?[0, 10^9]?的整數。

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Runtime:?4 ms Memory Usage:?18.7 MB 1 class Solution { 2 func preimageSizeFZF(_ K: Int) -> Int { 3 if K < 5 {return 5} 4 var base:Int = 1 5 while(base * 5 + 1 <= K) 6 { 7 base = base * 5 + 1 8 } 9 if K / base == 5 10 { 11 return 0 12 } 13 return preimageSizeFZF(K % base) 14 } 15 }

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4ms

1 class Solution { 2 func preimageSizeFZF(_ K: Int) -> Int { 3 var start = 1, end = Int.max, mid = start + (end - start) / 2 4 5 while start < end { 6 let candidate = trailingZeroes(mid) 7 if candidate > K { 8 end = mid - 1 9 } else if candidate < K { 10 start = mid + 1 11 } else { 12 return 5 13 } 14 mid = start + (end - start) / 2 15 } 16 return 0 17 } 18 19 func trailingZeroes(_ n: Int) -> Int { 20 var n = n, current = 0 21 while n > 1 { 22 n /= 5 23 current += n 24 } 25 26 return current 27 } 28 }

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轉載于:https://www.cnblogs.com/strengthen/p/10547060.html

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