343. Integer Break 类别：动态规划 难度：medium

343. Integer Break 类别：动态规划 难度：medium

题目：

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.

For example, given n = 2, return 1 (2 = 1 + 1); given n = 10, return 36 (10 = 3 + 3 + 4).

思路：

令res[n]为n对应的最大积。那么递推方程就是：res[n]=max(i*res[n-i],i*(n-i))(其中i从1到n-1)。

程序：

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