<p>Given an integer <code>n</code>, break it into the sum of <code>k</code> <strong>positive integers</strong>, where <code>k &gt;= 2</code>, and maximize the product of those integers.</p>

<p>Return <em>the maximum product you can get</em>.</p>

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

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 1
<strong>Explanation:</strong> 2 = 1 + 1, 1 &times; 1 = 1.
</pre>

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

<pre>
<strong>Input:</strong> n = 10
<strong>Output:</strong> 36
<strong>Explanation:</strong> 10 = 3 + 3 + 4, 3 &times; 3 &times; 4 = 36.
</pre>

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

<ul>
	<li><code>2 &lt;= n &lt;= 58</code></li>
</ul>

class Solution {
public:
    int func(int n, int cnt, vector<vector<int>>& dp){
        if(n <= 0){
            if(cnt < 2) return 0;
            return 1;
        }
        if(dp[n][cnt] != -1) return dp[n][cnt];
        int ans = 0;
        for(int i = 1; i <= n; i++){
            ans = max(ans, i * func(n - i, cnt + 1, dp));
        }
        return dp[n][cnt] = ans;
    }
    int integerBreak(int n) {
        vector<vector<int>> dp(n + 1, vector<int>(n + 1, -1));
        return func(n, 0, dp);
    }
};
8000