The problem requires us to find the n-th digit in an infinite sequence formed by concatenating all positive integers starting from 1.

Observations:

1. The sequence starts as: 123456789101112131415..., and we need to determine the digit at position (n).

2. A simple way to construct the sequence is to append numbers as strings until we reach the required length.

   Approach:

• Initialize an empty string sequence.
• Iterate from 1 onwards, converting each number to a string and appending it to sequence.
• Stop when sequence reaches at least n characters.

• Print the n-th digit (adjusting for 0-based indexing).

  Complexity Analysis:

  Since numbers grow in length, the loop runs approximately O(sqrt{n}) times, making the approach efficient.

n = int(input())
sequence = ""
for num in range(1, 1000):
    sequence += str(num)
    if len(sequence) >= n:  # You can stop early when sequence is long enough
        break
print(sequence[n - 1])

Let's prove that next greedy solution works: each step we will find prefixes of minimal length of arrays $$$A$$$ and $$$B$$$ such that its sums are equal and we will cut them forming next block. If we will get valid partition in result so it is an optimal solution, otherwise there is no solution. Since length of prefix proportional to its sum, so prefixes are minimal since its sums are minimal.

Let's prove this algorithm: let optimal solution have alternative partition. Since our solution cuts minimal possible prefixes, so (at some step) optimal solution cuts prefix with greater sum (and greater length). But this prefixes in optimal solutions contain smaller prefixes, found by greedy solution, so it can be divided on two parts — contradiction.

So we can keep prefixes and increase one which have smaller sum.

Time complexity: $$$O(n)$$$.

import sys

n = int(sys.stdin.readline().strip())
a = list(map(int, sys.stdin.readline().strip().split()))
m = int(sys.stdin.readline().strip())
b = list(map(int, sys.stdin.readline().strip().split()))

if sum(a) != sum(b):
    print(-1)
    exit()

p1, p2 =0, 0

cur_sum1, cur_sum2 = 0, 0

ans = 0

while p1 < n:
    
    cur_sum1 += a[p1]
    cur_sum2 += b[p2]
    p1 += 1
    p2 += 1

    while cur_sum1 != cur_sum2:
        if cur_sum1 < cur_sum2:
            cur_sum1 += a[p1]
            p1 += 1
        else:
            cur_sum2 += b[p2]
            p2 += 1
    ans += 1
print(ans)