<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:
bool exist[1001][1001]; //exist(num,pastepad)
int op[100000][3]{{0,1,0},{1,1,1},{2,2,1}};//(step,num,PastePad)

int minSteps(int n) {
    int t1=2,t2=1;//num,pastepad
    int begin =2; //current step
    int end = 3; //
    int step = 2;
    int cnt=3;
    while(t1<n){//one step
        ++step;
        int i=begin;
        for(; i<end; ++i){
            t1 = op[i][1]+op[i][2];//paste
            t2 = op[i][2];
            if(t1==n) return step;
            //cout << step << " " << t1 << " " << t2 << endl;
            if(t1<n && t2<n && !exist[t1][t2]) { op[cnt][0]=step; op[cnt][1]=t1; op[cnt][2]=t2;
                exist[t1][t2]=1; ++cnt;}
            
            //copy
            t1=t2=op[i][1];
            //cout << step << " " << t1 << " " << t2 << endl;
            if(t1<n && !exist[t1][t2]) { op[cnt][0]=step; op[cnt][1]=t1; op[cnt][2]=t2;
                exist[t1][t2]=1; ++cnt;}
        }
        begin = end; end = cnt;
    }
    
    if(n==1)return 0;
    else return 2;
 }
};
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