×

# Where can I get to solve C and C++ language problems?

Hi..I am Sushmita. Can I get to practice more of C and C++ language problems instead of Python language ?? Or where can I get to practice them ? I was never taught Python language so I am not too familiar with it.

4 years ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Indeed, the problems in Brilliant.org are open for all languages. You don't have to solve them with Python but it is nice to post a solution with Python instead of other languages (note that Python is the easiest language for most people and it is mainly focused on Brilliant.org).

There are many C / C++ tutorial when you search it on Google like C Tutorial and C++ Tutorial.

To kick start the practice, Brilliant.org is well enough. For challenging one, you can search more about "Competitive Programming".

- 4 years ago

i made this problem based on c++ language https://brilliant.org/community-problem/an-interesting-problem-102/?group=fFPS13Kbo1lV

- 4 years ago

Hey, I was just solving the question but got stuck. Is there any theorem or something for reducing $${ a }^{ { b }^{ c } }\quad mod\quad d$$ to something easily computable?

- 4 years ago

yes there is, something called "divide and conquer" , and you also need some basic programming

- 4 years ago

maybe this can help a^4 = (a^2)^2 . a^5 = (a^2)^2 * a

- 4 years ago