<p>There is only one character <code>&#39;A&#39;</code> on the screen of a notepad. You can perform one of two operations on this notepad for each step:</p>

<ul>
	<li>Copy All: You can copy all the characters present on the screen (a partial copy is not allowed).</li>
	<li>Paste: You can paste the characters which are copied last time.</li>
</ul>

<p>Given an integer <code>n</code>, return <em>the minimum number of operations to get the character</em> <code>&#39;A&#39;</code> <em>exactly</em> <code>n</code> <em>times on the screen</em>.</p>

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

<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> 3
<strong>Explanation:</strong> Initially, we have one character &#39;A&#39;.
In step 1, we use Copy All operation.
In step 2, we use Paste operation to get &#39;AA&#39;.
In step 3, we use Paste operation to get &#39;AAA&#39;.
</pre>

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

<pre>
<strong>Input:</strong> n = 1
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
</ul>

class Solution {
public:
    int minSteps(int n) {
        vector<int> v(1001);
        v[1]=0;
        v[2]=2;
        v[3]=3;
        for(int i=4;i<=n;i++){
            vector<int> div=getDivisors(i);
            v[i]=i;
            for(int j=0;j<div.size();j++){
                v[i]=min(v[div[j]]+i/div[j],v[i]);
            }
        }
        return v[n];
    }
    vector<int> getDivisors(int n){
        vector<int> res;
        for(int i=2;i<=sqrt(n);i++){
            if(n%i==0){
                res.push_back(i);
                if(n/i!=i){
                    res.push_back(n/i);    
                }
            }
        }
        return res;
    }
};
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