[LeetCode] 650. 2 Keys Keyboard

Problem:

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: 3Output: 3Explanation: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'.

题意分析：题目要求我们求得获得 n 个字符'A'（不能少也不能多）的最小操作步数，操作只有两种选择：

当前字符全部复制（Copy All）和粘（Paste）

解题思路：

为了找出规律，一般可以手动推算前面几项的结果：

n = 1时，结果显然为0，不用复制也不需要粘贴；

n = 2时，复制一遍，粘贴一遍，记为 cp，f(2) = 2；

n = 3时，复制一遍，粘贴两遍，记为 cpp，f(3) = 3；

n = 4时，复制一遍，粘贴三遍，记为cppp；另外一种选择为cpcp，f(4) = 4；

n = 5时，cpppp， f(5) = 5；

n = 6时，cpcpp，f(6) = f(2) + 3（6 / 2） = f(3) + 2（6 / 3）= 5；

n = 7时，cpppppp，f(7) = 7；

n = 8时，cpcppp或cpcpcp，f(8) = f(2) + 4（8 / 2） = f(4) + 2（8 / 4） = 6；

n = 9时，cppcpp，f(9) = f(3) + 3（9 / 3）= 6；

n = 10时，cppppcp或cpcpppp，f(10) = f(2) + 5（10 / 2）= f(5) + 2（10 / 5）= 7；

...

总结规律: f(n) = f(i) + n / i ，其中i为能整除n的整数，若n为素数，很容易理解，结果为其本身。

Solution:

class Solution {public:    int minSteps(int n) {        if (n == 1) return 0;        vector<int> steps(n+1, 0);        for (int i = 2; i <= n; i++)            steps[i] = i;        for (int i = 2; i <= n; i++) {            for (int j = i / 2; j > 1; j--) {                if (i % j == 0) {                    steps[i] = steps[j] + i / j;                    break;                }            }        }        return steps[n];    }};