<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) {
        // if(s.size()&1)return 0;
        int n=s.size()/2;
        for(int i=0;i<s.size()/2;i++){
            string h=s.substr(0,i+1);
            int j=0;
            // cout<<h<<" ";
            // cout<<h<<" ";
            while(j<s.size()){
                if(s[j]==h[j%(i+1)])j++;
               else break;
            }
            if(j>=s.size() && j%(i+1)==0){return 1;}
        }
        return 0;
    }
};
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