[Leetcode]65. Factorial Trailing zeros factorial's tail 0 numbers

Given an integer n, return the number of trailing zeroes in N!.

Note:your solution should is in logarithmic time complexity.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

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Solution 1: If n! = = kx10m, (0! = k%10), then M is required. We decompose the n! into a mass factor, then there are N! = = 2X * 3Y * 5Z ..., and 10 is how to come, after decomposition not only by 2x5. In other words, only the min (x, z) of decomposition is required, and obviously, the number of 2 after decomposition is obviously more than the number of 5. So, M = = Z. According to the analysis, the most straightforward way to calculate Z is to calculate the 5 exponent in the factorization of I (I =1, 2, ..., N) and then sum.

classSolution { Public:    intTrailingzeroes (intN) {intCNT =0, J;  for(inti =5; I <= N; i + =5) {            intj =i;  while(J%5==0) {                ++CNT; J/=5; }        }        returnCNT; }};

This solution will time out when N is very large Time Limit exceeded. The note of the topic asks us to write a solution to the logarithmic time complexity.

Solution 2: On the basis of solution 1 to consider that the number of 5 can be contributed is a multiple of 5, while the index of 5 (5,25,125,... The number of contributors that can contribute 5 is exactly the value of a power (one-off,... ), Index of non 5 (10,15,20,30,,... ) can only contribute one 5. and using N divided by 5 can get how many multiples of 5 in [1,n], divided by 25 can get how many multiples of 25 ... So all the results add up to exactly how many 5 in [1,n]. Note that 25=5*5 can contribute two 5, and in the front divided by 5 time has been calculated once, so 1 times can be, the same 125=5*5*5, in 2 and 25 has been considered two 5, so the following is also added one on the line, the following all 5 of the index is so.

 class   solution { public  :  int  trailingzeroes ( Span style= "color: #0000ff;" >int   N) { int  cnt = 0  ;  long  long  k = 5  ;  while  (k <= N) {cnt  + = n/< Span style= "color: #000000;"            > K;        K  *= 5  ;     return   CNT; }};

Note that k must be defined as a long long type to prevent K overflow when n is large. One to avoid this pitfall is as follows:

class Solution {public:    int trailingzeroes (int  n) {          int0;          while 0 ) {            5;             5 ;        }         return cnt;    }};

[leetcode]65. Factorial Trailing zeros factorial's tail 0 number