343. Integer Break

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).

    class Solution {          public: int integerBreak(int n) {                         int dp[n + 1];                          dp[0] = 0;                          dp[1] = 1;                          dp[2] = 1;                         dp[3] = 2;                          dp[4] = 4;                          for (int i = 5; i <= n; ++i) {                               dp[i] = 3 * max(i - 3, dp[i - 3]);                          }                     return dp[n]; '              }    };

2 => 1, 1 => 1
3 => 2, 1 => 2
4 => 2, 2 => 4
5 => 3, 2 => 6
6 => 3, 3 => 9
7 => 3, 4 => 12
8 => 3, 5 => 18
9 => 3, 6 => 27
10 => 3, 7 => 36

By observation, when you get maximum, one of the num is always 3.
After 3, the result will be larger than or equal the number itself.