[Leetcode]650. 2 Keys Keyboard-优快云博客

本文探讨了在记事本上生成指定数量字符A的最有效方法。初始状态下，记事本仅包含一个字符A，通过CopyAll和Paste两种操作，实现以最少步骤达到目标字符数。文章提供了一个具体实例，当目标字符数为3时，如何通过3步操作完成任务，并附带了一段C++代码，展示如何使用动态规划解决更复杂的问题。

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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'.

class Solution {
public:
    int minSteps(int n) {
        if (n == 1)
            return 0;
        int dp[n]; dp[0] = 0; dp[1] = 2;
        for (int i = 3; i <= n; ++i){
            int index = -1;
            for (int j = 2; j < i; ++j)
                if (i % j == 0)
                    index = j;
            if (index == -1)
                dp[i-1] = i;
            else 
                dp[i-1] = dp[index-1] + (i / index);
        }
        return dp[n-1];
    }
};