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
- BZOJ 2850: 巧克力王国 KDtree + 估价函数
Code: #include<bits/stdc++.h> #define maxn 100000 #define inf 1000000008 #define mid ((l+r)> ...
- ceph部署
一.部署准备: 准备5台机器(linux系统为centos7.6版本),当然也可以至少3台机器并充当部署节点和客户端,可以与ceph节点共用: 1台部署节点(配一块硬盘,运行ceph-depo ...
- How To:使用dmidecode获取机器序列号(Serial number)
使用dmidecode可以获取关于机器的诸多信息,比如机器的序列号 [root@dc1db01 ~]# dmidecode -s system-serial-number processor的 ...
- C++实现双人枪战游戏
//单机版枪战游戏,喜欢就拿走,把赞留下//by floatiy #include<iostream> #include<cstdio> #include<windows ...
- Linux之强大的selinux
简单点说,SELinux就是用来加强系统安全性的.它给一些特定程序(这些程序也在不断增加)做了一个沙箱,它将文件打上了一个安全标签,这些标签属于不同的类,也只能执行特定的操作,也就是规定了某个应用程序 ...
- 【转载】Apache shutdown unexpectedly启动错误解决方法
http://blog.csdn.net/dong123dddd/article/details/21372179 xampp启动时显示的错误为: 9:52:41 [Apache] Attempti ...
- 【CodeCraft-19 and Codeforces Round #537 (Div. 2) C】Creative Snap
[链接] 我是链接,点我呀:) [题意] 横坐标1..2^n对应着2^n个复仇者的基地,上面有k个复仇者(位置依次给出). 你是灭霸你要用以下方法消灭这k个复仇者: 一开始你获取整个区间[1..2^n ...
- 【codeforces 767A】Snacktower
[题目链接]:http://codeforces.com/contest/767/problem/A [题意] 每天掉一个盘子下来;盘子有大小从1..n依次增大n个盘子; 然后让你叠盘子; 最底层为n ...
- [Usaco2010 Mar]gather 奶牛大集会
[Usaco2010 Mar]gather 奶牛大集会 题目 Bessie正在计划一年一度的奶牛大集会,来自全国各地的奶牛将来参加这一次集会.当然,她会选择最方便的地点来举办这次集会.每个奶牛居住在 ...
- Windows 10卸载Edge浏览器(不成功的别试了)
在命令行输入: PowerShell dir $env:LOCALAPPDATA\Packages\*edge*^|ren -newname MicrosoftEdge.old ; dir $env: ...