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https://leetcode.com/problems/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).
Note: you may assume that n is not less than 2.
Hint:
public class Solution { public int integerBreak(int n) { if(n <= 2) { return 1; } else if (n == 3) { return 2; } double radical = Math.sqrt((double) n); int rad = (int) Math.floor(radical); int rs = 0, power = 0; for(int i=1; i<=rad+1; ++i) { int times = n / i; int remain = n - i*times; if(remain == 0) { power = (int) Math.pow(i, times); rs = Math.max(rs, power); } else { int power_1 = (int) Math.pow(i, times-1) * (remain + i); int power_2 = (int) Math.pow(i, times) * remain; rs = Math.max(rs, Math.max(power_1, power_2)); } } return rs; } }
leetcode@ [343] Integer Break (Math & Dynamic Programming)
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原文地址：http://www.cnblogs.com/fu11211129/p/5628645.html
