<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 sysmetric(string s) {
        return string(s.begin(), s.begin() + s.size() / 2) == string(s.begin() + s.size() / 2, s.end());
    }
    bool repeatedSubstringPattern(string s) {
        if(sysmetric(s)) return true;
        for(int i = 1; i <= s.length() / 3; i++) {
            string sub_front(s.begin() + i, s.end());
            string sub_back(s.begin(), s.end() - i);
            if(sysmetric(sub_front) && sysmetric(sub_back) && sub_front == sub_back) return true;
        }
        return false;
    }
};
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