title

problem link

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.

solution

DP

也就是绳子的问题

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