A. Fafa and his Company

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935A. Fafa and his Company
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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.

题意：每个团队领导会负责一些正数量的员工，给他们分配任务。已知员工的数量n，找出有多少种方式可以选择团队领导者的数量，从而使员工在他们之间平均分配。

思路：给一定的人数，让我们选领袖，每个领袖的跟班不能是0，并且每个领袖带领的跟班是相同数量的，注意只要到n/2就可以了，因为大于n/2的话就会有的领袖得不到跟班。实则这题就是找出n有几个约数即可。

AC代码

#include
#include
using namespace std;
int main()
{
	int n,i,sum=0;
	cin>>n;
	for(i=1;i<=n/2;i++)
	{
		if (n%i==0)
		{
			sum++;
		}
	}
	cout<<sum<<endl;
	return 0;
}