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.
===============================================
这个可以用DP，时间复杂度是O(n^2)。
dp[i] = max(i],max(j,dp[j])*max(i-j,dp[i-j])); 这个状态转移式中，我认为，j + (i - j) = i，所以，能够表示可能的最大乘积为 i * ( i- j)，不过可以用到之前的dp[j]和dp[i-j]，因为，其中dp[j]可以是j的最大的分解结果，也可以是i的最大的分解结果。
C++代码：
class Solution { public: int integerBreak(int n) { vector<int> dp(n+1,0); dp[1] = dp[2] = 1; for(int i = 3; i <= n; i++){ for(int j = 1; j <= i; j++){ dp[i] = max(dp[i],max(j,dp[j])*max(i-j,dp[i-j])); } } return dp[n]; } };