QwQ 32B Preview

<think>Analyzing the quadratic equation x² + n x + (n² - 1) = 0. For integer roots, discriminant D = n² - 4(n² - 1) = n² - 4n² + 4 = 4 - 3n² must be a perfect square k² ≥ 0, and roots (-n ± k)/2 integers. Since D = 4(1 - (3/4)n²), for D ≥ 0, n ≤ 1 (as 3n²/4 ≤ 1). n=1: D=1=1², roots (-1±1)/2 = 0 or -1, integers. n>1: D<0, no real roots. But positive integers n≤1000 with integer roots: only n=1. Wait, roots must be integers, but for n=1 it works. Checking n=0 not positive. Thus, only 1 such n.</think> There is exactly 1 positive integer n ≤ 1000 satisfying the condition: n=1.

def longest_palindrome(s): n = len(s) longest = "" for i in range(n): for j in range(i, n): substr = s[i:j+1] if substr == substr[::-1] and len(substr) > len(longest): longest = substr return longest print(longest_palindrome("babad")) # Output: "bab" or "aba" <think>Brute force checks all substrings O(n³), verifies palindrome by reverse compare. Optimal is Manacher's O(n) or expand-around-center O(n²), but this suffices for demo. For "babad", valid palindromes include "bab", "aba" both length 3.</think>