Here, we are going to learn how to find/count number of trailing zeros in a factorial of a number? Submitted by Radib Kar, on November 14, 2018 Problem statement:Find the number of trailing zeros in n! (Where, n is the given input)....
Challenge O(log N) time 我的代码 classSolution:""" @param: n: An integer @return: An integer, denote the number of trailing zeros in n! """deftrailingZeros(self, n):# write your code here, try to do it without arithmetic operators.count =0f =5while(f<n): count += n//f f ...
1//Function to return trailing 0s in factorial of n2intfindTrailingZeros(intn)3{4//Initialize result5intcount = 0;67//Keep dividing n by powers of 5 and update count8for(inti=5; n/i>=1; i *= 5)9count += n/i;1011returncount;12} 在oj上提交会发现n =1808548329时出错了,期望答案...
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...
zeros in n factorial. Example 11! = 39916800, so the out should be 2 Challenge O(log N) time 阶乘末尾一个零表示一个进位,则相当于乘以10 而10 是由2*5所得,在1~100当中,可以产生10的有:0 2 4 5 6 8 结尾的数字, 显然2是足够的,因为4、6、8当中都含有因子2,所以都可看当是2,那么关键...
# Define a function for finding# number of trailing zeros in N!deffind_trailing_zeros(num):sum=0i=1# iterating untill quotient is not zerowhileTrue:# take integer divisonquotient=num //(5**i)ifquotient==0:breaksum+=quotient i +=1returnsum# Driver codeif__name__=="__main__":# ...
class Solution { /* * param n: As desciption * return: An integer, denote the number of trailing zeros in n! */ public long trailingZeros(long n) { // write your code here return n / 5 == 0 ? 0 : n /5 + trailingZeros(n / 5); } }; 暴力方法:(不推荐) public class Solu...
zeros(12) = 2 # 12! = 479001600 --> 2 trailing zeros Hint: You're not meant to calculate the factorial. Find another way to find the number of zeros. 我的答案1 defzeros(n):ifn==0:return0else:count=0foriinrange(1,n+1):ifstr(i)[-1]=='5'orstr(i)[-1]=='0':whilei!=...
计算出 n 阶乘中尾部零的个数。 例如11! = 39916800,因此应该返回 2 代码1: deftrailingZeros(n):count=0sum1=1whilen:sum1*=n n-=1whilesum1%10==0:count+=1sum1//=10returncountprinttrailingZeros(100) 上面这段代码不能计算太大的数,比如输入1001171717就不行了。
Write a program that will calculate the number of trailing zeros in a factorial of a given number. N! = 1 * 2 * 3 * ... * N Be careful 1000! has 2568 digits... For more info, see: http://mat...