<p>For two strings <code>s</code> and <code>t</code>, we say &quot;<code>t</code> divides <code>s</code>&quot; if and only if <code>s = t + t + t + ... + t + t</code> (i.e., <code>t</code> is concatenated with itself one or more times).</p>

<p>Given two strings <code>str1</code> and <code>str2</code>, return <em>the largest string </em><code>x</code><em> such that </em><code>x</code><em> divides both </em><code>str1</code><em> and </em><code>str2</code>.</p>

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

<pre>
<strong>Input:</strong> str1 = &quot;ABCABC&quot;, str2 = &quot;ABC&quot;
<strong>Output:</strong> &quot;ABC&quot;
</pre>

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

<pre>
<strong>Input:</strong> str1 = &quot;ABABAB&quot;, str2 = &quot;ABAB&quot;
<strong>Output:</strong> &quot;AB&quot;
</pre>

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

<pre>
<strong>Input:</strong> str1 = &quot;LEET&quot;, str2 = &quot;CODE&quot;
<strong>Output:</strong> &quot;&quot;
</pre>

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

<ul>
	<li><code>1 &lt;= str1.length, str2.length &lt;= 1000</code></li>
	<li><code>str1</code> and <code>str2</code> consist of English uppercase letters.</li>
</ul>

class Solution {
public:
    vector<string> divide(string test){
        vector<int>factors;
        for(int i=1;i<=test.length();i++){
            if(test.length()%i==0)factors.push_back(i);
        }
        vector<string>strcut;
        for(int i=0;i<factors.size();i++){
            string pre = test.substr(0,test.length()/factors[i]);
            for(int j=0;j<factors[i];j++){
                string current=test.substr(j*test.length()/factors[i],test.length()/factors[i]);
                if(current!=pre)break;
                if(j==factors[i]-1)strcut.push_back(current);
            }
        }
        return strcut;
    }
    string gcdOfStrings(string str1, string str2) {
        vector<string>possible=divide(str1);
        vector<string>check=divide(str2);
        for(int i=0;i<possible.size();i++){
            for(int j=0;j<check.size();j++){
                if(possible[i]==check[j]){
                    return possible[i];
                }
            }
        }
        return "";
    }
};
11580