HDU 1124 (求阶乘中0的个数)
来源:互联网 发布:设计师导航网源码 编辑:程序博客网 时间:2024/04/29 07:02
Factorial
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3665 Accepted Submission(s): 2397
Problem 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 points and return back to the central company building. Programmers have spent several months studying this problem but with no results. They were unable to find the solution fast enough. After a long time, one of the programmers found this problem in a conference article. Unfortunately, he found that the problem is so called "Travelling Salesman Problem" and it is very hard to solve. If we have N BTSes to be visited, we can visit them in any order, giving us N! possibilities to examine. The function expressing that number is called factorial and can be computed as a product 1.2.3.4....N. The number is very high even for a relatively small N.
The programmers understood they had no chance to solve the problem. But because they have already received the research grant from the government, they needed to continue with their studies and produce at least some results. So they started to study behaviour of the factorial function.
For example, they defined the function Z. For any positive integer N, Z(N) is the number of zeros at the end of the decimal form of number N!. They noticed that this function never decreases. If we have two numbers N1<N2, then Z(N1) <= Z(N2). It is because we can never "lose" any trailing zero by multiplying by any positive number. We can only get new and new zeros. The function Z is very interesting, so we need a computer program that can determine its value efficiently.
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 points and return back to the central company building. Programmers have spent several months studying this problem but with no results. They were unable to find the solution fast enough. After a long time, one of the programmers found this problem in a conference article. Unfortunately, he found that the problem is so called "Travelling Salesman Problem" and it is very hard to solve. If we have N BTSes to be visited, we can visit them in any order, giving us N! possibilities to examine. The function expressing that number is called factorial and can be computed as a product 1.2.3.4....N. The number is very high even for a relatively small N.
The programmers understood they had no chance to solve the problem. But because they have already received the research grant from the government, they needed to continue with their studies and produce at least some results. So they started to study behaviour of the factorial function.
For example, they defined the function Z. For any positive integer N, Z(N) is the number of zeros at the end of the decimal form of number N!. They noticed that this function never decreases. If we have two numbers N1<N2, then Z(N1) <= Z(N2). It is because we can never "lose" any trailing zero by multiplying by any positive number. We can only get new and new zeros. The function Z is very interesting, so we need a computer program that can determine its value efficiently.
Input
There is a single positive integer T on the first line of input. It stands for the number of numbers to follow. Then there is T lines, each containing exactly one positive integer number N, 1 <= N <= 1000000000.
Output
For every number N, output a single line containing the single non-negative integer Z(N).
Sample Input
63601001024234568735373
Sample Output
0142425358612183837
题意:求n的阶乘中0的个数;
思路:直接求出n的阶乘含有多少个5就可以了;
代码:
#include<stdio.h>#include<string.h>int main(){int t,n;scanf("%d",&t);while(t--){scanf("%d",&n);int ans=0;while(n){ans+=n/5;n=n/5;}printf("%d\n",ans);}return 0;}
0 0
- HDU 1124 (求阶乘中0的个数)
- 求N的阶乘N!中末尾0的个数
- 求1000阶乘中最后0的个数
- 求阶乘尾数0的个数
- 求阶乘N!末尾0的个数
- 求阶乘数包含0的个数
- 求阶乘N!末尾0的个数
- 求N阶乘末尾0的个数
- 阶乘中0的个数
- (阶乘末尾0的个数的求法)求1~N中因子的数量的方法
- hdu 1124(数论,求末尾0的个数)
- 求1000阶乘的结果中0的个数和结果的位数(利用数组)
- 求一个整数的阶乘结果中后缀0的个数
- 算法---面试题/--求N的阶乘N!中末尾0的个数
- N的阶乘中(N!)末尾0的个数
- 阶乘中0的个数-poj 1401
- 2.2阶乘中末尾0的个数
- 阶乘中末尾0的个数
- 3170: [Tjoi 2013]松鼠聚会
- 秒杀多线程第十篇 读者写者问题
- 新浪微博之链式运动
- [code generation]模版引擎比较_freemarker 和 velocity
- openwrt: Makefile 框架分析
- HDU 1124 (求阶乘中0的个数)
- 外观模式
- 看了极光推送技术原理的几点思考
- StringBuilder、StringBuffer和String三者的联系和区别
- Android Studio调试功能使用总结
- onCreate中的savedInstanceState有何具体作用?
- Java WebService 简单实例
- 升讯威微信营销系统开发实践:(2)功能设计与架构设计
- Java中的Class类和Class对象