<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 left, string right) {
        long L = stol(left) , R = stol(right);
        
        int magic = 100000 , ans = 0;
        string s = "";
        //count odd length;
        for(int k = 1 ; k  < magic ; k++){
            s = to_string(k);
            for(int i = s.length() - 2 ; i >= 0; i--){
                s += s.at(i);
            }
            long v = stol(s);
            v *= v;
            if(v > R) break;
            if(v >= L && isPalindrome(v)) ans++;
        }
        
        //count even length;
        for(int k = 1 ; k < magic ; k++){
            s = to_string(k);
            for(int i = s.length() - 1 ; i >= 0 ; i--){
                s += s.at(i);
            }
            long v = stol(s);
            v *= v;
            if(v > R) break;
            if(v >= L && isPalindrome(v)) ans++;
        }
        return ans;
    }
    
    bool isPalindrome(long x){
        return x == reverse(x);
    }
    
    long reverse(long x ){
        long ans = 0;
        while(x > 0){
            ans = 10 * ans + x % 10;
            x /= 10;
        }
        return ans;
    }
};
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