smc

Let $\lfloor x\rfloor =k$ for some integer $1\le k\le 2013$. Then we can rewrite $f(x)$ as $k(2014^{x-k}-1)$. In order for this to be less than or equal to $1$, we need $2014^{x-k}-1\le \frac{1}{k}⟹x\le k+{\log }_{2014}\left(\frac{k+1}{k}\right)$. Combining this with the fact that $\lfloor x\rfloor =k$ gives that $x\in \left[k,k+{\log }_{2014}\left(\frac{k+1}{k}\right)\right]$, and so the length of the interval is ${\log }_{2014}\left(\frac{k+1}{k}\right)$. We want the sum of all possible intervals such that the inequality holds true; since all of these intervals must be disjoint, we can sum from $k=1$ to $k=2013$ to get that the desired sum is $$\sum _{i=1}^{2013}{\log }_{2014}\left(\frac{i+1}{i}\right)={\log }_{2014}\left(\prod _{i=1}^{2013}\frac{i+1}{i}\right)={\log }_{2014}\left(\frac{2014}{1}\right)=\boxed{1}.$$