Formatted question description: https://leetcode.ca/all/343.html
343 Integer Break Given a positive integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get. Example 1: Input: 2 Output: 1 Explanation: 2 = 1 + 1, 1 × 1 = 1. Example 2: Input: 10 Output: 36 Explanation: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36. Note: You may assume that n is not less than 2 and not larger than 58. @tag-dp @tag-math
Positive integers start from 1, but 1 cannot be split into the sum of two positive integers, so it cannot be used as an input.
Then 2 can only be split into 1+1, so the product is also 1.
The number 3 can be split into 2+1 or 1+1+1. Obviously, the product of the first split method is two.
The number 4 is divided into 2+2, and the product is the largest, which is 4.
The number 5 is divided into 3+2, and the product is the largest, which is 6.
The number 6 is divided into 3+3, and the product is the largest, which is 9.
The number 7 is divided into 3+4, and the product is the largest, which is 12.
The number 8 is divided into 3+3+2, and the product is the largest, which is 18.
The number 9 is divided into 3+3+3, and the product is the largest, 27.
The number 10 is divided into 3+3+4, the product is the largest, 36.
Starting from 5, all numbers need to be divided into 3 first, until the remaining number is 2 or 4, because there is no need to divide the remaining 4, it does not make sense to divide into two 2 and not divide, and 4 cannot be divided into 3+1, which will be smaller than the product of 2+2.
So, first preprocess the case where n is 2 and 3, and then initialize the result res to 1, and then start the loop when n is greater than 4, multiply res by 3, and subtract 3 from n. According to the previous analysis, when out of the loop, n only Can be 2 or 4, then multiply by res to return.
dp[i] represents the maximum product of the sum of at least two positive integers divided from i
int dp = new int[n + 1];
dp[i] = Math.max( dp[i], Math.max( j * (i - j), j * dp[i - j] )