Reflected Light

Suppose an amount $x$ of light is emitted upward from the ground.

The fraction of light that passes through the $i$-th layer is

$$\frac{100-A_i}{100}.$$

Therefore, if $S$ is the fraction of light that passes through all layers and escapes into space, then

$$S = \prod_{i=1}^{N} \frac{100-A_i}{100}.$$

Thus, from an emitted amount $x$, the amount $Sx$ escapes and the remaining amount $(1-S)x$ eventually returns to the ground.

The initial amount reaching the ground is $100$. Since the fraction returning to the ground is always $1-S$, the total amount reaching the ground is the geometric series

$$100 + 100(1-S) + 100(1-S)^2 + \cdots.$$

Its sum is

$$\frac{100}{1-(1-S)} = \frac{100}{S}.$$

Therefore, the answer is

$$\frac{100}{\prod_{i=1}^{N} \frac{100-A_i}{100}} = 100 \cdot \prod_{i=1}^{N} \frac{100}{100-A_i} = \frac{100^{N+1}}{\prod_{i=1}^{N}(100-A_i)}.$$

We compute this value modulo $MOD = 1\,000\,000\,007$:

$$100^{N+1} \cdot \left(\prod_{i=1}^{N}(100-A_i)\right)^{-1} \pmod{MOD}.$$

Since $0 \le A_i < 100$, each $100-A_i$ is between $1$ and $100$. Also, $MOD$ is a prime greater than $100$, so the denominator is never divisible by $MOD$. Hence, the modular inverse always exists.

The time complexity is $O(N+\log MOD)$.