整数拆分

题源：Leetcode

链接：https://leetcode-cn.com/problems/integer-break/

 很显然，最少分成两个整数，所以有以下推导：

dp[i] = max(j*(i-j),j*dp[i-j])

代码如下：

class Solution {
public:
    int integerBreak(int n) {
        vector <int> dp(n + 1);
        for (int i = 2; i <= n; i++) {
            int curMax = 0;
            for (int j = 1; j < i; j++) {
                curMax = max(curMax, max(j * (i - j), j * dp[i - j]));
            }
            dp[i] = curMax;
        }
        return dp[n];
    }
};