试题：
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.
代码：
拆分计算，一般都要考虑动态规划。对于乘积最大，我们要想办法将其拆解成子问题。对于3 × 3 × 4我们可以看成3 × (3 × 4)，要想乘积最大，那么必须保持(3 × 4)乘积最大，不然就有更大的乘积。那么我们就找到了子问题：找到7=3+4的最大乘积。对于第一个3，我们不需拆分，因为不管乘积何种组合，我们总可以将第一项视为不拆分项，而将剩下的当成可拆分项。

class Solution {
    public int integerBreak(int n) {
        int[] dp = new int[n+1];
        dp[1] = 1;
        for(int i=2; i<=n; i++){
            for(int j=1; j<i; j++){
                dp[i] = Math.max(dp[i], Math.max(j*(i-j),j*dp[i-j]) );
            }
        }
        return dp[n];
    }
}