码迷,mamicode.com
首页 > 其他好文 > 详细

(dp)343. Integer Break

时间:2016-06-25 14:55:04      阅读:135      评论:0      收藏:0      [点我收藏+]

标签:

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.


public class Solution { //dp public int integerBreak(int n) { if(n==2) return 1; if(n==3) return 2; int[] dp=new int[n+1]; dp[0]=dp[1]=1; dp[2]=2;dp[3]=3; for(int i=4;i<=n;i++){ dp[i]=dp[1]*dp[i-1]; for(int j=2;j<=i/2;j++){ dp[i]=(dp[i]>dp[j]*dp[i-j])?dp[i]:dp[j]*dp[i-j]; } } return dp[n]; } }

  

(dp)343. Integer Break

标签:

原文地址:http://www.cnblogs.com/kydnn/p/5616280.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!