<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<vector<pair<int, int> > > dp(s2.size(), vector<pair<int, int> > (s1.size()));
        for (int i = 0; i < s2.size(); ++i) {
            for (int j = 0; j < s1.size(); ++j) {
                int shift = 0, cur_id = j, steps = 0;
                while (s2[i] != s1[cur_id]) {
                    ++steps;
                    if (cur_id + 1 == s1.size()) cur_id = 0, ++shift; else ++cur_id;
                    if (steps > s1.size()) break;
                }
                if (steps > s1.size()) return 0;
                dp[i][j] = {cur_id, shift};
            }
        }
        int letters = 0, cur_id = 0, shifts = 0, t = s2.size() * n2;
        pair<int, int> ne;
        for (int I = 0; true; ++I) {
            for (int j = 0; j < s2.size(); ++j) {
                ne = dp[j][cur_id];
                if (ne.second + shifts >= n1) return letters / t;
                cur_id = ne.first + 1, shifts += ne.second;
                ++letters;
                if (cur_id == s1.size()) cur_id = 0, ++shifts;
                if (shifts >= n1) return letters / t;
            }
        }
        return letters / t;
    }
};
8700