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.
Input: 2 Output: 1 Explanation: 2 = 1 + 1, 1 × 1 = 1.
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.
implSolution{pubfninteger_break(n:i32) -> i32{match(n,(n - 2) / 3){(2, _) | (3, _) => n - 1,(_, x) => 3i32.pow(x asu32)*(n - 3* x),}}}