Description：

• Difficulty:Medium
• Total Accepted:12.6K
• Total Submissions:28.1K
• Contributor: apoorv_vikram

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].

Thought：

可以把题目转化为对数字的计算。初始为1，只能执行加上自身或者乘以2这两个操作。

dp[1] = 0

之后的dp[i]可以看成 i = m*n。m为一个质数，这样最简单来说对其进行n次加上自身即可。所以操作了dp[m] + n次操作

最后暴力搜索所有可以被i整除的数，取dp[m]+n最小。

Code：

class Solution {
public:
	int minSteps(int n) {
		vector<int> dp(n+1,INT16_MAX);
		dp[0] = dp[1] = 0;
		for (int i = 2; i <= n; i++) {
			for (int j = 1; j < i; j++) {
				if (i%j != 0) continue;
				dp[i] = min(dp[n], dp[j] + (i/j));
			}
		}
		return dp[n];
	}
};