Key idea:
1. for any substring s, if we can do not consider padding prefix 0s, its decimal value is always >= its length.
2. Appending a binary digit means multiply a number by 2 (and plus 1 if appending 1). This will double the required substring length in order to make its length equal to its decimal value.
3. N <= 2 * 10^5, so if we ignore leading 0s, we do not need to consider substrings with length >= 19, because 2^18 > 2 * 10^5.
Solution: Fix the highest bit 1 of a substring, then append 1 more bit to get a new value v, if the current length is the same with v, add 1 to ans. Otherwise check if there are v - len preceeding 0s. To speed this up, precompute a zero prefix sum array. For each fixed highest bit 1, we append at most 18 bits. Do this for each 1 in s.
The runtime is O(N + N * 18)