# Factorial Trailing Zeroes Solutions in C++

Number 172

Difficulty Easy

Acceptance 37.8%

Link LeetCode

Other languages Go

## Solutions

### C++ solution by haoel/leetcode

// Source : https://oj.leetcode.com/problems/factorial-trailing-zeroes/// Author : Hao Chen// Date : 2014-12-30/** The idea is:** 1. The ZERO comes from 10.* 2. The 10 comes from 2 x 5* 3. And we need to account for all the products of 5 and 2. likes 4×5 = 20 ...* 4. So, if we take all the numbers with 5 as a factor, we'll have way more than enough even numbers* to pair with them to get factors of 10** **Example One**** How many multiples of 5 are between 1 and 23?* There is 5, 10, 15, and 20, for four multiples of 5. Paired with 2's from the even factors,* this makes for four factors of 10, so: **23! has 4 zeros**.*** **Example Two**** How many multiples of 5 are there in the numbers from 1 to 100?** because 100 ÷ 5 = 20, so, there are twenty multiples of 5 between 1 and 100.** but wait, actually 25 is 5×5, so each multiple of 25 has an extra factor of 5,* ( e.g. 25 × 4 = 100，which introduces extra of zero )** So, we need know how many multiples of 25 are between 1 and 100? Since 100 ÷ 25 = 4,* (there are four multiples of 25 between 1 and 100)** Finally, we get 20 + 4 = 24 trailing zeroes in 100!*** The above example tell us, we need care about 5, 5×5, 5×5×5, 5×5×5×5 ....** **Example Three***** 5^1 : 4617 ÷ 5 = 923.4, so we get 923 factors of 5* 5^2 : 4617 ÷ 25 = 184.68, so we get 184 additional factors of 5* 5^3 : 4617 ÷ 125 = 36.936, so we get 36 additional factors of 5* 5^4 : 4617 ÷ 625 = 7.3872, so we get 7 additional factors of 5* 5^5 : 4617 ÷ 3125 = 1.47744, so we get 1 more factor of 5* 5^6 : 4617 ÷ 15625 = 0.295488, which is less than 1, so stop here.** Then 4617! has 923 + 184 + 36 + 7 + 1 = 1151 trailing zeroes.**/class Solution {public:int trailingZeroes(int n) {int result = 0;//To avoid the integer overflow ( e.g. 'n >=1808548329' )for(long long i=5; n/i>0 && i <= INT_MAX; i*=5){result += (n/i);}return result;}// Alternative implementation which naturally avoid integer overflow issue.int trailingZeroes(int n) {int sum=0;int tmp=0;while(n/5>0){tmp=n/5;sum+=tmp;n=tmp;}return sum;}};