C - Fafa and his Company
Problem description
Fafa owns a company that works on huge projects. There are n employees in Fafa's company. Whenever the company has a new project to start working on, Fafa has to divide the tasks of this project among all the employees.
Fafa finds doing this every time is very tiring for him. So, he decided to choose the best l employees in his company as team leaders. Whenever there is a new project, Fafa will divide the tasks among only the team leaders and each team leader will be responsible of some positive number of employees to give them the tasks. To make this process fair for the team leaders, each one of them should be responsible for the same number of employees. Moreover, every employee, who is not a team leader, has to be under the responsibility of exactly one team leader, and no team leader is responsible for another team leader.
Given the number of employees n, find in how many ways Fafa could choose the number of team leaders l in such a way that it is possible to divide employees between them evenly.
Input
The input consists of a single line containing a positive integer n (2 ≤ n ≤ 105) — the number of employees in Fafa's company.
Output
Print a single integer representing the answer to the problem.
Examples
Input
2
Output
1
Input
10
Output
3
Note
In the second sample Fafa has 3 ways:
- choose only 1 employee as a team leader with 9 employees under his responsibility.
- choose 2 employees as team leaders with 4 employees under the responsibility of each of them.
- choose 5 employees as team leaders with 1 employee under the responsibility of each of them.
解题思路:结合提示,再多举几个栗子,将会发现问题求解其实是求n的因子个数(不包括n),水过!
AC代码:
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n;
cin>>n;
int num=;
for(int i=;i<=n/;++i)
if(n%i==)num++;
cout<<num<<endl;
return ;
}
C - Fafa and his Company的更多相关文章
- Elasticsearch索引(company)_Centos下CURL增删改
目录 返回目录:http://www.cnblogs.com/hanyinglong/p/5464604.html 1.Elasticsearch索引说明 a. 通过上面几篇博客已经将Elastics ...
- Microsoft Dynamics AX 2012: How to get Company,Customer and Vendor address in AX 2012
Scenario: “How to get Addresses of “Customer, Vendor and Company” 1) First we need to identify ...
- poj1416 Shredding Company
Shredding Company Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 5379 Accepted: 3023 ...
- CodeForces 125E MST Company
E. MST Company time limit per test 8 seconds memory limit per test 256 megabytes input standard inpu ...
- CF 321B Kefa and Company(贪心)
题目链接: 传送门 Kefa and Company time limit per test:2 second memory limit per test:256 megabytes Desc ...
- 搜索+剪枝 POJ 1416 Shredding Company
POJ 1416 Shredding Company Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 5231 Accep ...
- 二分+动态规划 POJ 1973 Software Company
Software Company Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 1112 Accepted: 482 D ...
- 【Moqui业务逻辑翻译系列】Story of Online Retail Company 在线零售公司的故事
h1. Story of Online Retail Company 在线零售公司的故事 Someone decides to sell a product. [Product Marketer Ma ...
- Codeforces 556D Restructuring Company
传送门 D. Restructuring Company time limit per test 2 seconds memory limit per test 256 megabytes input ...
随机推荐
- WinForm窗体中窗口控件的生成
1:button控件的生成方式 Button button = new Button(); button.Size = new Size(80, 80); button.Location = new ...
- 通过JS唤醒app(安卓+ios)
有需求说要通过页面按钮唤醒app,或者手机上没有这款app跳转到商店,然后刚开始也是查了资料的,结果发现一头雾水,不过最后还是捣鼓出来了,当然也参考了前人分享的经验,下面我就将方法整理一下: 首先明确 ...
- ESP、EBP、CALL 指令与局部变量浅析
概述 函数调用是计算机程序中一个最重要的概念之一,从汇编的角度看,能更加直观地理解函数调用的原理,理解 CALL 指令调用过程中 ESP.EBP 寄存器的作用. 我们先从一段简陋的 C 语言代码说起, ...
- [nodejs]在mac环境下如何将node更新至最新?
在mac下安装angular-cli时,报出较多错误.初步怀疑是因为node环境版本过低导致. 在mac下,需要执行如下几步将node更新至最新版本,也可以更新到指定版本 1. sudo npm ca ...
- MySql 日志查看与设置
错误日志log-errol 开启方式:在my.ini的[mysqld]选项下:添加代码:log-error=E:\log-error.txt 记录内容:主要是记录启动.运行或停止mysqld时出现的致 ...
- -------------Django-----URLS路由
一.相约Django. 1.Django的特点:Django定义了服务分布.路由映射.模板编程.数据处理的一套完整的功能. (1)集成数据访问组件:Django的model层自带数据库ORM组件. ( ...
- odoo api介绍
odoo api修饰器介绍与应用 参考文档:https://www.cnblogs.com/kfx2007/p/6093994.html 一.one one的用法主要用于self为单一集合的情况,被o ...
- 121. Best Time to Buy and Sell Stock(动态规划)
Say you have an array for which the ith element is the price of a given stock on day i. If you were ...
- 【codeforces 527C】Glass Carving
[题目链接]:http://codeforces.com/contest/527/problem/C [题意] 让你切割一个长方形; 只能横切或竖切; 让你实时输出切完之后最大的长方形的面积; [题解 ...
- Linux统计行数命令wc(转)
Linux wc命令用于计算字数. 利用wc指令我们可以计算文件的Byte数.字数.或是列数,若不指定文件名称.或是所给予的文件名为"-",则wc指令会从标准输入设备读取数据. 语 ...