Overview

Before the first incorrect problem, output o; from the first incorrect problem onward, output x. We examine the results sequentially from the beginning while tracking whether any mistake has been made so far.

Analysis

You can receive points on the \(i\)-th problem only if you have answered all problems from the \(1\)-st to the \(i\)-th correctly.

Therefore, simply choosing o if \(A_i = 1\) and x if \(A_i = 0\) is incorrect. This is because once a mistake is made, you cannot receive points even if you answer correctly afterwards.

For example, consider the following results:

1 1 0 1 0

Points are received for the first \(2\) problems, but since the \(3\)-rd problem is answered incorrectly, no points are awarded from that point onward. The answer is the following string:

ooxxx

Although we could determine the answer by checking all previous results for each problem, this is unnecessary. Once a mistake is made, you can never return to the state where you can receive points, so maintaining a single variable to represent the state is sufficient.

Algorithm

We use a variable ok representing “no mistakes have been made so far”.

1. Set ok = True and initialize an empty list res to store the answer.
2. Inspect the result of each problem from the beginning:
   • If the result is 0, set ok = False.
   • If ok is True, append o to res; if False, append x.
3. Concatenate the characters in res and output the result.

Immediately after updating with each problem’s result, ok will be True if and only if all problems from the first to the current one are answered correctly. Since this directly matches the condition for receiving points, we can append the correct character at each position.

Complexity

• Time Complexity: \(O(N)\)
  We inspect each of the \(N\) problem results once, and finally construct a string of length \(N\).
• Space Complexity: \(O(N)\)
  Used to store the input data and the list of characters for the answer.

Implementation Details

• Update ok before appending the character. This is because the problem where the mistake occurred must also be x.

• Once ok = False, do not set it back to True even if subsequent answers are correct.

• In the provided code, each result is stored as a string rather than an integer, so we compare with v == '0'.

• Using ''.join(res) allows you to concatenate the characters without spaces.

  Source Code

import sys

def main():
    data = sys.stdin.read().split()
    n = int(data[0])
    a = data[1:1 + n]
    res = []
    ok = True
    for v in a:
        if v == '0':
            ok = False
        res.append('o' if ok else 'x')
    sys.stdout.write(''.join(res) + '\n')

main()

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This editorial was generated by gpt-6-astra-medium.