650. 2 Keys Keyboard

Initially on a notepad only one character 'A' is present. You can perform two operations on this notepad for each step:

1. Copy All: You can copy all the characters present on the notepad (partial copy is not allowed).
2. Paste: You can paste the characters which are copied last time.

Given a number n. You have to get exactly n 'A' on the notepad by performing the minimum number of steps permitted. Output the minimum number of steps to get n 'A'.

Example 1:

Input: 3
Output: 3
Explanation:
Intitally, we have one character 'A'.
In step 1, we use Copy All operation.
In step 2, we use Paste operation to get 'AA'.
In step 3, we use Paste operation to get 'AAA'.

Note:

1. The n will be in the range [1, 1000].

题意：

起初给一个A，求经过多少次复制粘贴操作看一看得到n个A，复制必须把当时状态全部的字符复制。

思路：

看起来挺像状态转移方程的，但是观察之后可以发现，实质上就是求将n分解成若干个质数相乘后这些质数之和。

还是自己太菜，时间复杂度居然O(n^2)。。。。还代码冗长。

代码：

class Solution {
    public int minSteps(int n) {
        if(n==1)
            return 0;
        if(n==2)
            return 2;
        if(n==3)
            return 3;
        if(n==4)
            return 4;
        // int []dp=new int[1001];
        // for(int i=1;i<=1000;i++)
        //     dp[i]=Integer.MAX_VALUE;
        // dp[1]=0;
        // dp[2]=2;
        // dp[3]=3;
        // dp[4]=4;
        // dp[5]=5;
        // for(int i=5;i<=1000;i++)
        // {
        //     int flag=0;
        //     for(int j=i-1;j>=1;j--)
        //     {
        //         if(i%j==0)
        //         {
        //             dp[i]=Math.min(dp[j]+i/j,dp[i]);
        //             flag=1;
        //         }
        //     }
        //     if(flag==0)
        //     {
        //         dp[i]=i;
        //     }
        // }
        // return dp[n];
        
        int ans=0,d=2;
        while(n>1)
        {
            while(n%d==0)
            {
                n/=d;
                ans+=d;
            }
            d++;
        }
        return ans;
    }
}