2 Keys Keyboard问题及解法

问题描述：

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'.

示例:

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'.

问题分析：

本题存在状态转移过程，可采用dp方法求解。其转移过程为dp[i] = dp[j] + i / j; j = n,...,1,i = 2,...,n。

过程详见代码：

class Solution {public:    int minSteps(int n) {        vector<int> dp(n + 1, 0);for (int i = 2; i <= n; i++){if (i % 2 == 0) dp[i] = dp[i / 2] + 2;else{int j = 3,flag = 0;for (; j <= (int)sqrt(n); j += 2){if (i % j == 0)                     {                        dp[i] = dp[i / j] + j;                        flag = 1;                        break;                    }}if (!flag) dp[i] = dp[1] + i;}}return dp[n];    }};