# slt

#### 思路1 dp

for i = 2; i <= n; i++ for j = 1; j < i; j++ dp\[i] = max(dp\[i], max(j \* (i-j), j \* dp\[i-j]))

i表示剪成多少段。如n=5，则最少是2段，最多是5段。（mn都是整数）

j遍历i剪完的每一段。比如i=2，则j=1，2.表示在第一段和第二段中再剪下去。

max 第一个参数 dp\[i] 表示不剪， j \* (i-j) 表示 从j处剪一下，剩下i-j不剪了 j \* dp\[i-j] 表示 从j处剪一下，剩下i-j继续剪

#### 思路2 贪心

贪心算法，取3，具体见。

<https://leetcode-cn.com/problems/jian-sheng-zi-ii-lcof/solution/mian-shi-ti-14-ii-jian-sheng-zi-iitan-xin-er-fen-f/>

贪心法则：尽可能分解出多的 3, 3的个数为a，余数为b，b可能的值是0,1,2

b = 0, return 3^a b = 1, 将末尾的3+1看成4，返回3^(a-1) + 2 \* 2 b = 2, 返回3^a \* 2


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://851958789.gitbook.io/notes/0343_integer_break/slt.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.