题目大意 N!末尾0的个数 题解 0只能由2*5产生,所以只要求2,5有多少对即可,又因为10!中5的个数少于2,所以只要求因子5有多少个即可,答案即为N/5+N/25+N/125.. 代码: #include<stdio.h> int main(void) { int T; scanf("%d",&T); while(T--) { ; scanf("%d",&n); while(n) { ans+=n/; n/=; } printf(&…

转载请注明出处:http://blog.csdn.net/lttree Factorial Time Limit: 1500MS   Memory Limit: 65536K Total Submissions: 13993   Accepted: 8678 Description The most important part of a GSM network is so called Base Transceiver Station (BTS). These transceivers for…

明出处:http://blog.csdn.net/lttree Factorial Time Limit: 1500MS   Memory Limit: 65536K Total Submissions: 13993   Accepted: 8678 Description The most important part of a GSM network is so called Base Transceiver Station (BTS). These transceivers form th…

Description The most important part of a GSM network is so called Base Transceiver Station (BTS). These transceivers form the areas called cells (this term gave the name to the cellular phone) and every phone connects to the BTS with the strongest si…

Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. Credits:Special thanks to @ts for adding this problem and creating all test cases. 这道题并没有什么难度,是让求一个数的阶乘末尾0的个数,也就是要找乘数中10的个数,…

C. Drazil and Factorial time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Drazil is playing a math game with Varda. Let's define for positive integer x as a product of factorials of its digi…

LeetCode上的原题,讲解请参见我之前的博客Factorial Trailing Zeroes. 解法一: int trailing_zeros(int n) { ; while (n) { res += n / ; n /= ; } return res; } 解法二: int trailing_zeros(int n) { ? : n / + trailing_zeros(n / ); } CareerCup All in One 题目汇总…

[codeforces 516]A. Drazil and Factorial 试题描述 Drazil is playing a math game with Varda. Let's define  for positive integer x as a product of factorials of its digits. For example, . First, they choose a decimal number a consisting of n digits that con…

原题链接在这里:https://leetcode.com/problems/factorial-trailing-zeroes/ 求factorial后结尾有多少个0,就是求有多少个2和5的配对. 但是2比5多了很多,所以就是求5得个数.但是有的5是叠加起来的比如 25,125是5的幂数,所以就要降幂. e.g. n = 100, n/5 =20, n/25= 4, n/125=0,所以加起来就有24个0. O(logn)解法: 一个更聪明的解法是:考虑n!的质数因子.后缀0总是由质因子2和质因…

Factorial Trailing Zeroes Given an integer n, return the number of trailing zeroes in n!. Note: Your solution should be in logarithmic time complexity. Credits:Special thanks to @ts for adding this problem and creating all test cases. 对n!做质因数分解n!=2x*…