<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>

class Solution {
public:
    int superpalindromesInRange(string l, string r) {
        int res = 9 >= stoll(l) && 9 <= stoll(r) ? 1 : 0;
        for (int d = 1; d < 10; d++) {
            bool odd = d % 2 && d != 1;
            int ilen = (d >> 1) - 1,
                ilim = max(1, (int)pow(2, ilen)),
                mp = d >> 1, mlim = odd ? 3 : 1;
            for (int e = 1; e < 3; e++) {
                string p(d, '0');
                p[0] = p[d-1] = e + '0';
                if (e == 2) ilim = 1, mlim = min(mlim, 2);
                for (int i = 0; i < ilim; i++) {
                    if (i > 0) {
                        string is = bitset<3>(i).to_string().substr(3 - ilen);
                        for (int j = 0; j < ilen; j++)
                            p[1+j] = p[d-2-j] = is[j];
                    }
                    for (int m = 0; m < mlim; m++) {
                        if (odd) p[mp] = m + '0';
                        int64_t sq = stoll(p) * stoll(p);
                        if (sq > stoll(r)) return res;
                        if (sq >= stoll(l) && pal(to_string(sq))) res++;
                    }
                }
            }
        }
        return res;
    }
    
    bool pal(string s) {
        for (int i = 0, j = s.length() - 1; i < j; i++, j--)
            if (s[i] != s[j]) return false;
        return true;
    }
};
8800