Challenge Set - Triangle Orientation

[AceInfinity]

Challenge #1: write a program that will output a triangle like the following:

Where the size of the triangle is defined by N, and N = half the width of the base of the triangle for number of "#"s that exist on the bottom row.

Challenge #2: write a program that will output a triangle like the following:

Where the size of the triangle is defined by N, and N = half the width of the base of the triangle for number of "#"s that exist on the bottom row, in addition to M, which defines the width of the spacer in between both halves.

Challenge #3: write a program that will output a diamond like the following:

Where the size of the diamond is defined by N, and N = half the width of the base of the triangle for number of "#"s that exist on the bottom row (of the top 2 halves of a triangle, minus the vertical spacer), in addition to M, which defines the width of the spacer in between all 4 quadrants of the diamond.

:thumbsup2:

 

[AceInfinity]

Here's my solution for #1 and #2 combined in C++ :)

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void drawTriangle(const size_t width, const size_t sp_width)
{
	const char mid_fill = '#';
	const char sp = ' ';
	std::string spacer(sp_width, sp);
	std::string str(1, mid_fill);
	for (size_t i = 0; i < width; ++i)
	{
		std::cout << std::setw(width)
			<< str << spacer << str << std::endl;
		str = str.append(1, mid_fill);
	}
}

int main(void)
{
	drawTriangle(45, 2);
	return 0;
}

I wrote code for the diamond as well:

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But it's just an addition to the code above, and I think I'd better let someone else post a solution that hasn't been posted.