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最新推荐文章于 2022-03-13 14:13:31 发布

原创最新推荐文章于 2022-03-13 14:13:31 发布 · 185 阅读

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文章标签：

#dp #leetcode

本文探讨了在记事本上通过复制和粘贴操作，将初始字符'A'增加到目标数量n所需的最少步骤。使用了宽度优先搜索(BFS)和动态规划(DP)结合数论的方法解决此问题，提供了两种算法的实现代码，并总结了LeetCode题目的解答要点。

文章目录

标签：dp，数论

题目

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].

bfs思路 TLE

求最小步数，明显可以用bfs做，但是会TLE

bfs code TLE

class S TLEolution {
public:
    int minSteps(int n) {
        vector<vector<int>> q(0,vector<int>(3));
        vector<int> tmp({1, 0, 0});
        q.push_back(tmp);

        while (!q.empty()){
            tmp = q.front();
            if(tmp[0] == n) return tmp[2];
            q.erase(begin(q));
            vector<int> cur(3);
            cur[0]= tmp[0]+ tmp[1];
            cur[1]= tmp[1];
            cur[2]= tmp[2] + 1;
            if(cur[0] == n) return cur[2];

            q.push_back(cur);

            cur[0] = tmp[0];
            cur[1] = tmp[0];
            cur[2] = tmp[2] + 1;

            q.push_back(cur);

        }
        return -1;
    }
};

数论DP 思路

if n % 2 == 0, then f(n) = f(n/2) + 2
if n % 3 == 0, then f(n) = f(n/3) + 3

数轮DP Code

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