def count_divisible_substrings(s, p):
    n = len(s)
    result = [0] * n
    mod = 2 ** p  # Calculate 2^p for divisibility checks

    for i in range(n):
        current_mod = 0
        # Iterate through substrings starting at index i
        for j in range(i, n):
            current_mod = (current_mod * 10 + int(s[j])) % mod
            # If divisible, increment the count for index i
            if current_mod == 0:
                result[i] += 1
            # Stop early if we've examined the last `p` digits
            if j - i + 1 >= p:
                break
        print("res :", result)
    return result

# Example usage:
s = "1048"
p = 3
print(count_divisible_substrings(s, p))  # Output: [2, 2, 1, 1]
s = "743212"
p = 4
print(count_divisible_substrings(s, p))