A can complete a piece of work alone in $200\text{days}$, while B can complete the same piece of work alone in $100\text{days}$. In every three-day cycle, both A and B work on day 1, only A works on day 2, and only B works on day 3. This cycle continues till the work is completed. How many days in all does it take the duo to complete the work?

Let the total work be the LCM of $200$ and $100$, which is $200\text{units}$.
* Efficiency of A $=\frac{200}{200}=1\text{unit/day}$
* Efficiency of B $=\frac{200}{100}=2\text{units/day}$

Let's find the work done in one $3\text{-day}$ cycle:

* Day 1 (A + B): $1+2=3\text{units}$
* Day 2 (Only A): $1\text{unit}$
* Day 3 (Only B): $2\text{units}$

Total work done in $1\text{ cycle (3 days)}=3+1+2=6\text{units}$.

Now, determine how many full cycles can fit into the total work of $200\text{units}$:

$$\text{Number of complete cycles}=\lfloor \frac{200}{6}\rfloor =33\text{ cycles}$$

* Work done in $33\text{ cycles}=33\times 6=198\text{units}$
* Days taken for $33\text{ cycles}=33\times 3=99\text{days}$

Remaining work to be completed:

$$\text{Remaining work}=200-198=2\text{units}$$

On the next day ($100\text{th day}$), which is Day 1 of the $34\text{-th}$ cycle, both A and B work together with a combined efficiency of $3\text{units/day}$.

$$\text{Time taken to finish remaining work}=\frac{2}{3}\text{day}$$

Total time taken $=99+\frac{2}{3}=99\frac{2}{3}\text{days}$.