Gain insights into competitive programming, explore C++ skills with theory, code samples, practice problems, and master faster implementation for contests like ACM ICPC, Google CodeJam, and HackerCup.

Competitive programming can be a great way to build out your programming skills, get on any major company’s radar, and earn a little extra cash along the way.

In this course, you will learn to prepare for competitive programming contests like ACM ICPC, Google CodeJam, Facebook HackerCup, and many more.

Each topic is broken down with a healthy mix of theory, code samples, step-by-step solved sample problems, illustrations, useful practice problems, and tips and tricks for faster implementation.

You will need some solid C++ foundations coming into this course, but by the end it, you will be confident enough in your C++ skills to take home the win.

Competitive Programming in C++: The Keys to Success

## Problem statement
Given a number $N > 1$, count the number of factors of the number N.

**Input format**

A single line of input contains the number $1 \leq N \leq 10^{12}$.


**Output format**

Print a single integer equal to the number of factors of $N$.

---

## Sample
**Input** 
```
36
```
**Output**
```
9
```

---

## Explanation
Factors of $36 : 1, 2, 3, 4, 6, 9, 12, 18, 36$

Count: $9$

---

## Brute force
The brute force solution would be to loop over all the numbers from `1` to `N` and check if it is a factor. If it is, you would then print it.

We can use the modulus operator to check if it's a factor or not.

Here is the code:

# Problem statement
Given a number $N > 1$, count the number of factors of the number N.

**Input format**

A single line of input contains the number $1 \leq N \leq 10^{12}$.


**Output format**

Print a single integer equal to the number of factors of $N$.

---

# Sample
**Input** 
```
36
```
**Output**
```
9
```

---

# Explanation
Factors of $36 : 1, 2, 3, 4, 6, 9, 12, 18, 36$

Count: $9$

---

# Brute force
The brute force solution would be to loop over all the numbers from `1` to `N` and check if it is a factor. If it is, you would then print it.

We can use the modulus operator to check if it's a factor or not.

Here is the code:

In the lesson, we look at an efficient way to factor a number.


Introduction

Complexity Analysis

Number Theory

Arrays and Vectors

Sieve of Eratosthenes

Strings

Sorting

Linked List

Stack

Queue

Binary Tree

2 Pointers

Heap

Binary Search Tree

Balanced Binary Search Tree

Course Conclusion

Solved Problem - Factorization

Problem statement

Sample

Explanation

Brute force