10. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then $\sum_{k=0}^{2013}\left[\frac{2013+2^{k}}{2^{k+1}}\right]=$ $\qquad$ .
10.2013.

Obviously, when $k \geqslant 11$, $\sum_{k=0}^{2013}\left[\frac{2013+2^{k}}{2^{k+1}}\right]=0$.
$$
\begin{array}{l} 
\text { Hence } \sum_{k=0}^{2013}\left[\frac{2013+2^{k}}{2^{k+1}}\right]=\sum_{k=0}^{10}\left[\frac{2013+2^{k}}{2^{k+1}}\right] \\
= 1007+503+252+126+63+ \\
31+16+8+4+2+1 \\
= 2013 .
\end{array}
$$