Class with templated class from different header wont compile
Why can templates only be implemented in the header file?
I have Sanguinare Vampiris. I don't really want to become a vampire. How can I cure myself? Can't log in using Firefox in Stack Overflow How to type special/accented letters in LaTeX? Justify text inside itemize environment with beamer class Why are there so many different ways to measure disk usage? This question has an inverse:     What would be the most efficient way to translate a combination of $k$-tuple into its positions within the $\left(\!\!\binom{n}{k}\!\!\right)$ combinations? I need this to be fast for combinations of $\left(\!\!\binom{70}{7}\!\!\right)$ order of magnitude - very large, but not exceeding 2 billion (fits into int32 maximum value).  Below is an example of $\left(\!\!\binom{6}{3}\!\!\right)$, where the aim is to quickly translate (a, d, f) tuple to value 9, while in the real problem $k$ ranges between 5 and 8.  $$\begin{array} {cccccc|c} a&amp;b&amp;c&amp;d&amp;e&amp;f&amp;^{combination}/_{sort\_order}&amp; \\\hline x&amp;x&amp;x&amp; &amp; &amp; &amp;1\\ x&amp;x&amp; &amp;x&amp; &amp; &amp;2\\ x&amp;x&amp; &amp; &amp;x&amp; &amp;3\\ x&amp;x&amp; &amp; &amp; &amp;x&amp;4\\ x&amp; &amp;x&amp;x&amp; &amp; &amp;5\\ x&amp; &amp;x&amp; &amp;x&amp; &amp;6\\ x&amp; &amp;x&amp; &amp; &amp;x&amp;7\\ x&amp; &amp; &amp;x&amp;x&amp; &amp;8\\ x&amp; &amp; &amp;x&amp; &amp;x&amp;9\\ x&amp; &amp; &amp; &amp;x&amp;x&amp;10\\ .&amp;.&amp;.&amp;.&amp;.&amp;.&amp;.\\  &amp; &amp; &amp;x&amp;x&amp;x&amp;20\\ \end{array}$$  I know that I could pre-calculate all the combinations and reverse the lookup dictionary. However, such dictionary would not be efficient in terms of memory usage. Therefore I am looking for either calculation-based approach, or a more efficient data structure to perform this mapping. Turn off paragraph indention for a single line What's the hardest or most misunderstood aspect of LINQ? Some of the Puella Magi seem to have magical powers related to the wishes that they made. Sayaka has regeneration abilities, and she wished to heal someone. Homura has time manipulation abilities, and she wished to go back in time.  But other magical girls, like Mami and Kyouko, don't appear to have powers related to their respective wishes.  So, is there a connection, or is it only a coincidence for Sayaka and Homura? If it's the former, how do Mami and Kouko's wishes relate to their powers? Is there a correct gender-neutral singular pronoun ("his" vs. "her" vs. "their")? What are the proposed theoretical explanations for the EmDrive? Evaluation of $\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7}$ Evaluation of $$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7} = $$   $\bf{My\; Try::}$ I have solved Using Direct formula::  $$\sin \frac{\pi}{n}\cdot \sin \frac{2\pi}{n}\cdot......\sin \frac{(n-1)\pi}{n} = \frac{n}{2^{n-1}}$$  Now Put $n=7\;,$ We get   $$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7}\cdot \sin \frac{4\pi}{7}\cdot \sin \frac{5\pi}{7}\cdot \sin \frac{6\pi}{7}=\frac{7}{2^{7-1}}$$  So $$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7} =\frac{\sqrt{7}}{8}$$  Now my question is how can we solve it without using Direct Formula, Help me  Thanks Accessing my Windows 7 files from my Ubuntu account Prove the identity $\binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n} = 4^n$ I've worked out a proof, but I was wondering about alternate, possibly more elegant ways to prove the statement. This is my (hopefully correct) proof:  Starting from the identity $2^m = \sum_{k=0}^m \binom{m}{k}$ (easily derived from the binomial theorem), with $m = 2n$:  $2^{2n} = 4^n = \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{2n-1} + \binom{2n}{2n}$  Applying the property $\binom{m}{k} = \binom{m}{m-k}$ to the second half of the list of summands in RHS above:  $4^n = \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{n-1} + \binom{2n}{n} + \underbrace{\binom{2n}{n-1} +\cdots \binom{2n}{1} + \binom{2n}{0}}_{\binom{m}{k} = \binom{m}{m-k} \text{ has been applied}}$  Rearranging the above sum by alternately taking terms from the front and end of the summand list in RHS above (and introducing the term $\binom{2n}{-1} = 0$ at the beginning just to make explicit the pattern being developed):  $4^n = (\binom{2n}{-1} + \binom{2n}{0}) + (\binom{2n}{0} + \binom{2n}{1}) + \cdots + (\binom{2n}{n-1} + \binom{2n}{n})$  Finally, using the property $\binom{m}{k} + \binom{m}{k-1} = \binom{m+1}{k}$ on the paired summands, we get the desired result:  $4^n = \binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n}$ Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$? In swimming, is there a competitive advantage to being in one lane or another? Upon observing swimming competitions in the Olympics, it is not clear to me whether one of the eight lanes is more advantageous versus another. I've seen vague claims but nothing backed up with science.  In competitive swimming, is there an advantage to being in one lane or another? Furthermore, does the stroke (ie: breast, fly, freestyle, etc.) affect this effect? Cubic polynomial with three (distinct) irrational roots I am looking for an equation     $$x^3+ax^2+bx+c=0, \qquad a, b, c \in \Bbb Z,$$   of degree $3$ that has $3$ different roots.  For an equation of degree $2$ it is easy---for example $x^2-2=0$---but I can't find an example for degree $3$.  if there is a method for finding this kind of Equation I will be grateful. if $A$ subset of $B$ then $P(B)-P(A)=P(B\setminus A)\ge 0$ Let $P$ be a probability measure on $(\Omega, \mathscr{F})$, If $a,b,\in\mathscr{F}$ and $A\subset B$, then $P(B)-P(A)=P(B\setminus A)\ge 0$.  So far I have:  $A\subset B \implies B\setminus A = B\cap A^c = \emptyset$ as $A$ and $B^c$ are disjoint. Since $A\cup (B\setminus A)=A\cup (B\cap A^c)=(A\cup B)\cap (A\cup A^c)=B\cap\Omega = B$  Then for $A,B\in \mathscr{F}\implies P(A\cup (B\setminus A))=P(A)+P(B\setminus A) = P(B)$ by countable additivity.  Rearranging, $P(B)-P(A)=P(B\setminus A) \ge 0$. App Store is the wrong language How do comment @replies work?