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

Copy All: You can copy all the characters present on the notepad (partial copy is not allowed).
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:

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

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/2-keys-keyboard
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

 

动态规划，用dp[i]来储存第i个所需要的最小次数。

class Solution {
    public int minSteps(int n) {
        if (n == 0 || n == 1) {
            return 0;
        }
        int []dp = new int[n + 1];
        Arrays.fill(dp, Integer.MAX_VALUE);
        dp[1] = 0;
        for (int i = 1; i <= n; i++) {
            int j = i;
// 第一次需要复制的操作
            int copy = 1;

            while (j + i <= n) {
                dp[j + i] = dp[j] + 1 + copy;
                copy = 0;
                j = j + i;
            }
        }
        return dp[n];
    }
}

另外一种写法：

class Solution {
    public int minSteps(int n) {
        if (n == 0 || n == 1) {
            return 0;
        }
        int []dp = new int[n + 1];
        Arrays.fill(dp, Integer.MAX_VALUE);
        dp[1] = 0;
        for (int i = 2; i <= n; ++i) {
            dp[i] = i;
            for (int j = i - 1; j > 1; --j) {
                if (i % j == 0) {
                    dp[i] = Math.min(dp[i], dp[j] + i / j);
                }
            }
        }
        return dp[n];
    }
}