While not explicitly specified in the question, I am looking for a decrease pattern such that $B$ decreases linearly every iteration. (Alternatively, for example $B$ can be decreased to $C$ anyway in iteration $0$, and still satisfy the given goal)
Let $T$ be the number of steps (to be decided). Let $d$ be the decrease of $B$ per iteration:
$$d = \frac{B_0-C}T$$
The values of $A$ and $B$ at the end of iteration $i-1$, or the respective values at the start of iteration $i$, will be:
$$\begin{align*}
B_i &= B_{i-1} -d\\
&= B_0 - id\\
A_i &= A_{i-1} - B_{i-1}\\
&= A_0 - B_0 - B_1 - \cdots - B_{i-1}\\
&= A_0 - iB_0 + 0d + 1d + \cdots + (i-1)d\\
&= A_0 - iB_0 + \frac{(i-1)id}{2}
\end{align*}$$
The goal is to have non-positive $A$ at the end of iteration $T-1$:
$$\begin{align*}
A_T &\le 0\\
A_0 - TB_0 + \frac{(T-1)Td}{2} &\le 0\\
A_0 - TB_0 + \frac{(T-1)(B_0-C)}{2} &\le 0\\
A_0 + \frac{-TB_0 - TC-B_0+C}2 &\le 0\\
2A_0 -B_0+C &\le T(B_0+C)\\
T &\ge \frac{2A_0-B_0+C}{B_0+C}
\end{align*}$$
So try $d$ depending on $T=\frac{2A_0-B_0+C}{B_0+C}$ (which may be a non-integer).
- Larger $T$ means $A$ is more likely to become non-positive before $T$ iterations, and before $B$ reaches $C$.
- Smaller $T$ means $A$ is more likely to stay positive after $T$ iterations, and after $B$ reaches $C$.
For example, $A_0=100, B_0 = 20, C = 5$, then
$$\begin{align*}
T &= \frac{2\cdot100-20+5}{20+5} = 7.4\\
d &= \frac{20-5}{7.4} = 2.027
\end{align*}$$
$$\begin{array}{c|ll}
i\ \text{(iteration)}&A_i\ \text{(starting value)}&B_i\ \text{(starting value)}\\\hline
0&100&20\\
1&80.000&17.973\\
2&62.027&15.946\\
3&46.081&13.919\\
4&32.162&11.892\\
5&20.270&9.865\\
6&10.405&7.838\\
7&2.568&5.811\\
8&-3.243&3.784
\end{array}$$
Note how $B_8$ is below $C=5$, but that value is not used for later $A$ and hopefully is unimportant.