Last day, a similar problem was questioned in the contest Codeforces Round 848 (Div. 2) (problem D).

Here is my solution for the problem mentioned in the blog:

Let $$$F_i$$$ denotes the expected number of times until we are able to go to the next mirror when we are at the mirror $$$i$$$. For the sake of simplicity, call $$$P_i$$$ the probability of being beautiful at the mirror $$$i$$$. Now the mirror either tells us we are beautiful with probability $$$P_i$$$ and allows us to switch to the other mirror, or we start again with the first mirror with probability $$$(1 - P_i)$$$. With these observations, we can write the general formula

$$$F_i = 1*P_i + (1 - P_i) * (1 + F_1 + F_2 + F_3 + ... + F_i)$$$

If we rearrange the formula and use a variable named $$$sum_i$$$ that keeps the prefix sum of F, we get

$$$F_i = (1 + (1-P_i) * sum_{i-1}) / P_i$$$

We can compute all F_i one by one now.

By definition, $$$F_i$$$ gives us the expected number of times to switch $$$(i+1)$$$ th mirror from $$$i$$$ th mirror. To get the answer, we must jump from the first mirror to the second, from the second mirror to the third, and so on. So in conclusion, we can express the equation

$$$answer = F_1 + F_2 + F_3 + ... + F_n$$$

//mint struct is used for modular operations
signed main() {
    int n; 
    cin >> n;

    vector<mint> p(n+1);
    for(int i = 1; i <= n; i++) {
        cin >> p[i];
        p[i] = p[i] / 100;
    }

    vector<mint> F(n+1);
    F[1] = 1 / p[1];

    mint sum = F[1];
    for(int i = 2; i <= n; i++) {
        F[i] = (1 + (1 - p[i]) * sum) / p[i];
        sum += F[i];
    }
    cout << sum;
}