C++ code to find trailing zeros in factorial of a number #include<bits/stdc++.h>usingnamespacestd;inttrailingZeros(intn){intcount=0;if(n<0)return-1;for(inti=5;n/i>0;i*=5){count+=n/i;}returncount;}intmain(){intn;cout<<"enter input,n"<<endl;cin>>n;if(trailingZeros(n))cout...
【leetcode】Factorial Trailing Zeros Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. 1 class Solution { 2 public: 3 int trailingZeroes(int n) { 4 int res=0; 5 while(n) 6 { 7 res+=n/5; 8 n/=5; 9...
+=quotient i +=1returnsum# Driver codeif__name__=="__main__":# assigning a numbernum=10# function callprint("Number of trailing zeros in factorial of",num,"is :",find_trailing_zeros(num),)num=20print("Number of trailing zeros in factorial of",num,"is :",find_trailing_zeros(num...
Write an algorithm which computes the number of trailing zeros in n factorial. 设计一个算法,计算出n阶乘中尾部零的个数。 【题目链接】 http://www.lintcode.com/en/problem/trailing-zeros/ 【题目解析】 传统解法是首先求出n!,然后计算末尾0的个数。(重复÷10,直到余数非0)该解法在输入的数字稍大时...
Trailing Zeros Write an algorithm which computes the number of trailing zeros in n factorial. Have you met this question in a real interview? Yes Example 11! = 39916800, so the out should be 2 1/*2* param n: As desciption3* return: An integer, denote the number of trailing zeros in...
进行质因数分解,则有 N! == 2X* 3Y* 5Z……,而10是怎么来的呢,分解之后不就只能由2×5来。也就是说,只要求分解后的min(x, z),又显然,分解之后2的个数显然要比5的个数多。所以,M == Z。根据分析,要计算 Z,最直接的方法,就是计算i(i =1, 2, …, N)的因式分解中5 的指数,然后求和。
LeetCode上的原題,講解請參見我之前的部落格Factorial Trailing Zeroes。 解法一: inttrailing_zeros(intn) {intres =0;while(n) { res+= n /5; n/=5; }returnres; } 解法二: inttrailing_zeros(intn) {returnn ==0?0: n /5+ trailing_zeros(n /5); ...
LeetCode上的原题,讲解请参见我之前的博客Factorial Trailing Zeroes。 解法一: inttrailing_zeros(intn) {intres =0;while(n) { res+= n /5; n/=5; }returnres; } 解法二: inttrailing_zeros(intn) {returnn ==0?0: n /5+ trailing_zeros(n /5); ...
Given an integer n, return the number of trailing zeroes inFactorialN! Your solution should be in logarithmic time complexity. Naive/Bruteforce Solution It is easy to get the fact that the trailing zeros totally depend on the 5 times 2 only. In other words, we just need to figure out how...
public class Solution { public int trailingZeroes(int n) { int result = 0; while(n > 0){ n = n/5; result += n; } return result; } } 需要注意 个别数 5^n;