問題描述：

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).

Note: you may assume that n is not less than 2.

Hint:
1）There is a simple O(n) solution to this problem.
2）You may check the breaking results of n ranging from 7 to 10 to discover the regularities.

把一個非負整數n分拆成至少兩個整數之和，求分拆後各數乘積的最大值。

問題求解：

class Solution {
public:
    int integerBreak(int n) {
        if(n < 4) return n-1;
        int res = 1
 
;
        while(n > 2){//看n包含多少個3,把他們相乘,直到n<=2
            res *= 3;
            n -= 3;
        }
        if(n == 0) return res;//n可以整除3，res就是各個3相乘
        if(n == 1) return (res / 3 ) * 4;//除3餘1，把其中的一個3加1變為4再相乘
        if(n == 2) return res * 2;//除3餘2,則可直接把2與res相乘
    }
};