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

  1. 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 ...

  2. 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 ...

  3. poj1416 Shredding Company

    Shredding Company Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5379   Accepted: 3023 ...

  4. CodeForces 125E MST Company

    E. MST Company time limit per test 8 seconds memory limit per test 256 megabytes input standard inpu ...

  5. 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 ...

  6. 搜索+剪枝 POJ 1416 Shredding Company

    POJ 1416 Shredding Company Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5231   Accep ...

  7. 二分+动态规划 POJ 1973 Software Company

    Software Company Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 1112   Accepted: 482 D ...

  8. 【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 ...

  9. 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

  1. 五、Scrapy中Item Pipeline的用法

    本文转载自以下链接: https://scrapy-chs.readthedocs.io/zh_CN/latest/topics/item-pipeline.html https://doc.scra ...

  2. 原来PHP对象比数组用更少的内存

    一直以为php的数组更节省内存,从来没有测试过,今天因为要读取一个大配置文件作为pool.做了一次测试: 得出结论是 使用对象保存数据更好,花费的内存是数组array的1/4. 测试代码 class ...

  3. BUPT2017 springtraining(16) #4 ——基础数论

    题目在这里 A.手动打表找规律得组合数 n -= 2, m -= 2, ans = C(n, m) #include <bits/stdc++.h> using namespace std ...

  4. JavaSE 学习笔记之多线程(十三)

    多线程: 进程:正在进行中的程序.其实进程就是一个应用程序运行时的内存分配空间. 线程:其实就是进程中一个程序执行控制单元,一条执行路径.进程负责的是应用程序的空间的标示.线程负责的是应用程序的执行顺 ...

  5. noip模拟赛 PA

    分析:很显然这是一道搜索题,可能是由于我的搜索打的太不美观了,这道题又WA又T......如果对每一个询问都做一次bfs是肯定会T的,注意到前70%的数据范围,N的值都相等,我们可以把给定N的所有情况 ...

  6. 20180629利用powerdesigner生成数据字典

    原创作品,如有错误,请批评指正 第一步新建一个PDM模型 第二步 点击“FILE”--->"RESERVER ENGINEER"--->"DATABASE&q ...

  7. Mac OSX:最简单的安装ant方法

    安装ant最简单的方法就是通过brew.步骤如下:1. 安装brew(如果已经安装可以跳过这步).ruby -e "$(curl -fsSL https://raw.github.com/m ...

  8. scp and tar

    scp 命令随记 scp file username@remoteIp:directory 创建tar包 tar zcvf file.tar.gz directory tar zcvf hadoop. ...

  9. .NET几大热点问题(.NET人员必读)

    下面收集了关于.NET几大热点问题及简要答案,防止我们回答问题不专业的尴尬.同一时候还将一些.NET资源的相关网址罗列一二. 一.什么是.Net?它主要包含什么?  .Net是为简化在第三代因特网的高 ...

  10. HDU2899 Strange fuction 【二分】

    Strange fuction Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) ...