<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 isDiv(string s, string div){
        int n = s.length(), m = div.length();
        while(s.size()){
            string part = s.substr(0,m);
            if(part != div)
                return false;
            s = s.substr(m);
        }
        return true;
    }
    bool repeatedSubstringPattern(string s) {
        int n = s.length();
        string div;
        for(int i=0;i<n-1;i++){
            div += s[i];
            if(n%(i+1)) continue;
            if(isDiv(s, div))
                return true;
        }
        return false;
    }
};
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