343. Integer Break

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).

Note: You may assume that n is not less than 2 and not larger than 58.

题解：

题目的要求大概就是，对一个给定的数n，求累加结果为n的数累乘的最大结果。

这是一道数学题，可以通过求导求最大值，显然当n能过尽量多的分成e相加，其累乘结果越大，又要求分成的数为整数，则可以尽量多的分成3，其累乘的结果最大。

具体的代码如下：

class Solution {public:    int integerBreak(int n) {        if(n == 2) return 1;        if(n == 3) return 2;        int count = 0;        int result = 0;        if(n % 3 == 1) {            n = n-4;            result = 4;        }        else if(n % 3 == 2) {            n = n-2;            result = 2;        }        count = n/3;        for(int i = 0; i < count; i++)             result *= 3;        return result;    }};

end！