exam | Programming of Introduction to algorithm | Learning library

Here's a fairly straightforward way where we test the sign of the lagged difference. If the mid_sum difference sign is the same as the final_sum difference sign, they are "consistent".

operator+ returns by-value, what it returns is an rvalue, which can't be bound to lvalue-reference to non-const, i.e. can't be passed to operator+ as argument for the next cacluation.

Change the parameter type of operator+ to lvalue-reference to const, which could bind to rvalues.

You can used a chained comparison to do this in one expression:

The choices expect an iterable of 2-tuples where the first item is the key, and the second one the rendered value.

You thus can implement this with:

Yes, it is possible, although I wouldn't recommend it. You can use the following:

• Set up a string with the LaTeX code include pstricks.
• Call tex2image(..., packages = c("auto-pst-pdf", ...)) so that the LaTeX package {auto-pst-pdf} is used. This supports embedding pstricks in documents for the pdfLaTeX by calling LaTeX for the figure in the background.
• Make sure tex2image() calls pdfLaTeX with the -shell-escape option so that pdfLaTeX is allowed to call LaTeX. This is relatively easy by using the R package tinytex.

A worked example for this strategy is included below, it is called dist4.Rmd. If you copy the R/Markdown code to a file you can run:

Like you mentioned, the first line of set used A::v, which was never initialized before. However, that itself doesn't produce an error, it is undefined behavior. What it means is the compiler may initialize it for you, or it might just pickup a random number it sees on the memory, or whatever they are pleased to. The C++ standard doesn't say what needs to happen, so it left the compiler to decide whatever is easy.

However, whatever happens on that line shouldn't matter too much in your code, in most cases. The reason is A::v will be re-assigned to v + 1 on the next line. So it should almost always print 2 at the end.

Without having an environment setup or really knowing exactly what you want out of the data. This is a good start.

Generally speaking, the Django ORM is not great for these types of queries, and trying to use select_related or prefetches results in really complex and inefficient queries.

I've found the best way to achieve these types of queries in Django is to break each piece of your puzzle down into a query that returns a "list" of ids that you can then use in a subquery.

Then you keep working down until you have your final output

You allocated a SortedList object in main() which allocated memory and never freed it.

The other blocks belong to the std::string objects inside the list. They are only lost because the list holding them lost track of them.

When doing your analysis, don't worry about the 6 indirectly lost blocks, focus on the one directly lost one.

Almost certainly this is a violation of the "Rule of Five". Your SortedList does an allocation in its constructor and doesn't properly manage that allocation in its move-assignment operator.

Looking at your code in the copy assignment operator, this line is causing the leak

I'm not a C expert, but I understand this way: float variables are 4 bytes (usually, to make sure, you could call sizeof(float)), char are 1 byte. When you created a float, you allocated 4 bytes with the binary value equivalent to 5.375. Then, you have created a char pointer to it - it would split the binary value.

What you had to do in this question:

• Find the value 5.375 in binary form (4 bytes, 32 bits)
• Split that value into each byte (8 bits each)
• Check the unsigned char value correspondent of each byte

I'm not completely sure if the result would be different from Windows or other systems - I would check if little-endians systems are different from big-endians doing the same process.

I hope this helps you understand it.