ACM-简单题之Factorial——poj1401

最新推荐文章于 2020-02-07 11:05:28 发布

ltree98

最新推荐文章于 2020-02-07 11:05:28 发布

阅读量2.8k

点赞数 1

CC 4.0 BY-SA版权

分类专栏： ACM-简单题漫漫刷题路文章标签： ACM 简单题 Factorial poj1401

本文链接：https://blog.youkuaiyun.com/lttree/article/details/26234519

该博客介绍了如何解决POJ1401问题，即求解给定数N的阶乘末尾0的个数。作者通过分析指出，由于2x5会产生0，因此可以通过计算因子对（2,5）的数量来确定0的个数。重点在于优化计算方法，避免直接计算阶乘，而是专注于统计5的倍数。通过逐步除以5并累加整数部分，博主得出结论并给出了简洁的解决方案。" 126401746,9098226,TinyMCE编辑器的值获取与设置技巧,"['TinyMCE', '编辑器', '前端开发']

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

转载请注明出处：http://blog.youkuaiyun.com/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 the areas called cells (this term gave the name to the cellular phone) and every phone connects to the BTS with the strongest signal (in a little simplified view). Of course, BTSes need some attention and technicians need to check their function periodically.

ACM technicians faced a very interesting problem recently. Given a set of BTSes to visit, they needed to find the shortest path to visit all of the given po