343. Integer Break --------M

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. 

A:

DP

就是当分解的时候， 前面几个，其实是，分解成最少1个positive integer的时候的解

class Solution {
public:
    int integerBreak(int n) {
        if(n<=2)
            return 1;
        if(n==3)
            return 2;
        vector<int> V(n+1,0);  // max value if at least ONE positive integer
        V[1] = 1;
        V[2] = 2; 
        V[3] = 3;
        for(int i=3; i< n+1; i++)
        {
            for(int j = 1;j<i;j++){
                V[i] = max(V[i], V[j] * (i-j));
            }
        }
        return V[n];
    }
};