Super Palindromes
<p>Let&#39;s say a positive integer is a <strong>super-palindrome</strong> if it is a palindrome, and it is also the square of a palindrome.</p>

<p>Given two positive integers <code>left</code> and <code>right</code> represented as strings, return <em>the number of <strong>super-palindromes</strong> integers in the inclusive range</em> <code>[left, right]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> left = &quot;4&quot;, right = &quot;1000&quot;
<strong>Output:</strong> 4
<strong>Explanation</strong>: 4, 9, 121, and 484 are superpalindromes.
Note that 676 is not a superpalindrome: 26 * 26 = 676, but 26 is not a palindrome.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> left = &quot;1&quot;, right = &quot;2&quot;
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= left.length, right.length &lt;= 18</code></li>
	<li><code>left</code> and <code>right</code> consist of only digits.</li>
	<li><code>left</code> and <code>right</code> cannot have leading zeros.</li>
	<li><code>left</code> and <code>right</code> represent integers in the range <code>[1, 10<sup>18</sup> - 1]</code>.</li>
	<li><code>left</code> is less than or equal to <code>right</code>.</li>
</ul>
Math; String; Enumeration
using ll = unsigned long long;

ll ps[2 * 100000];

int init = [] {
    for (int i = 1; i <= 100000; i++) {
        string s = to_string(i);
        string t1 = s;
        reverse(t1.begin(), t1.end());
        string t2 = s.substr(0, s.length() - 1);
        reverse(t2.begin(), t2.end());
        ps[2 * i - 2] = stoll(s + t1);
        ps[2 * i - 1] = stoll(s + t2);
    }
    return 0;
}();

class Solution {
public:
    int superpalindromesInRange(string left, string right) {
        ll l = stoll(left), r = stoll(right);
        int ans = 0;
        for (ll p : ps) {
            ll x = p * p;
            if (x >= l && x <= r && is_palindrome(x)) {
                ++ans;
            }
        }
        return ans;
    }

    bool is_palindrome(ll x) {
        ll y = 0;
        for (ll t = x; t; t /= 10) {
            y = y * 10 + t % 10;
        }
        return x == y;
    }
};