<p>Given a string <code>s</code>, check if it can be constructed by taking a substring of it and appending multiple copies of the substring together.</p>

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

<pre>
<strong>Input:</strong> s = &quot;abab&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> It is the substring &quot;ab&quot; twice.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;aba&quot;
<strong>Output:</strong> false
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abcabcabcabc&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> It is the substring &quot;abc&quot; four times or the substring &quot;abcabc&quot; twice.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution {
public:
    bool repeatedSubstringPattern(string s) {
    
    vector<int> factors;
    int n = s.size();
    for (int i = 1; i <= sqrt(n); i++) {
        if (n % i == 0) {
            if (n / i == i)
                factors.push_back(i);
            else {
                factors.push_back(i);
                factors.push_back(n/i);
            }
        }
    }
    for(auto x: factors) {
        if(x==n) continue;
        string sub = s.substr(0, x), tmp;
        for(int i=0; i<n/x; i++) 
            tmp += sub;
        if(!s.compare(tmp)) return true;
    }
    return false;

    }
};
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