<p>We define <code>str = [s, n]</code> as the string <code>str</code> which consists of the string <code>s</code> concatenated <code>n</code> times.</p>

<ul>
	<li>For example, <code>str == [&quot;abc&quot;, 3] ==&quot;abcabcabc&quot;</code>.</li>
</ul>

<p>We define that string <code>s1</code> can be obtained from string <code>s2</code> if we can remove some characters from <code>s2</code> such that it becomes <code>s1</code>.</p>

<ul>
	<li>For example, <code>s1 = &quot;abc&quot;</code> can be obtained from <code>s2 = &quot;ab<strong><u>dbe</u></strong>c&quot;</code> based on our definition by removing the bolded underlined characters.</li>
</ul>

<p>You are given two strings <code>s1</code> and <code>s2</code> and two integers <code>n1</code> and <code>n2</code>. You have the two strings <code>str1 = [s1, n1]</code> and <code>str2 = [s2, n2]</code>.</p>

<p>Return <em>the maximum integer </em><code>m</code><em> such that </em><code>str = [str2, m]</code><em> can be obtained from </em><code>str1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> s1 = "acb", n1 = 4, s2 = "ab", n2 = 2
<strong>Output:</strong> 2
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s1 = "acb", n1 = 1, s2 = "acb", n2 = 1
<strong>Output:</strong> 1
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s1.length, s2.length &lt;= 100</code></li>
	<li><code>s1</code> and <code>s2</code> consist of lowercase English letters.</li>
	<li><code>1 &lt;= n1, n2 &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution {
public:
    int getMaxRepetitions(string s1, int n1, string s2, int n2) {
        vector<int> rapport(102,-1);
        vector<int> rest(102,-1);
        int b=-1;int posRest=0;int rap=0;
        int last=-1;
        rapport[0]=rest[0]=0;
        for(int i=1;i<=s2.size()+1;i++){
            int j;
            for(j=0;j<s1.size();j++){
                if(s2[posRest]==s1[j]){
                    posRest++;
                    if(posRest==s2.size()){
                        rap++;
                        posRest=0;
                    }
                }
            }
            for(int k=0;k<i;k++){
                if(posRest==rest[k]){b=k;last=i;break;}
            }
            rapport[i]=rap;rest[i]=posRest;
            if(b>=0)break;
        }
        int interval=last-b;
        if(b>=n1)return rapport[n1]/n2;
        return ((n1-b)/interval*(rapport[last]-rapport[b])+rapport[(n1-b)%interval+b])/n2;  }
};
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