Let $a_{k}=\frac{2^{100}}{2^{50}+2^{k}}$. Notice that, for $k=0,1, \ldots, 49$, $a_{k}+a_{100-k}=\frac{2^{100}}{2^{50}+2^{k}}+\frac{2^{100}}{2^{50}+2^{100-k}}=\frac{2^{100}}{2^{50}+2^{k}}+\frac{2^{50+k}}{2^{k}+2^{50}}=2^{50}$. It is clear that for $k=0,1, \ldots, 49, a_{k}, a_{100-k} \notin \mathbb{Z}$, so $\left\lfloor a_{k}\right\rfloor+\left\lfloor a_{100-k}\right\rfloor=2^{50}-1$ (since the sum of floors is an integer less than $a_{k}+a_{100-k}$ but greater than $a_{k}-1+a_{100-k}-1$). Thus, $\sum_{k=0}^{100}\left\lfloor a_{k}\right\rfloor=50 \cdot\left(2^{50}-1\right)+2^{49}=101 \cdot 2^{49}-50$.