<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) {
        int tot = s.length();
        int i = tot/2;
        int j = 2;
        int previous = 0;
        string x("");
        std::ostringstream os;
        os.str("");
        x = s.substr(0, 1);
        std::fill_n(std::ostream_iterator<std::string>(os), tot, x);
        if(os.str() == s && tot > 1)  {
            return(true);
        }
        while(i > 0 && i != previous && i*j <= tot)    {
            x = s.substr(0, i);
            os.str("");
            while(os.str().length() < tot)  {
                std::fill_n(std::ostream_iterator<std::string>(os), 1, x);
            }
            if(os.str() == s)  {
                return(true);
            }
            j++;
            previous = i;
            i = tot / j;
            
        }
        return(false);
    }
};
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