[343]Integer Break

最新推荐文章于 2019-09-03 22:15:03 发布

原创最新推荐文章于 2019-09-03 22:15:03 发布 · 293 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 专栏收录该内容

140 篇文章

订阅专栏

探讨如何将一个正整数拆分为至少两个正整数之和，并最大化这些整数的乘积。通过数学原理解析最优解策略，提供简单有效的 O(n) 解决方案。

【题目描述】

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.

Hint:

There is a simple O(n) solution to this problem.
You may check the breaking results of n ranging from 7 to 10 to discover the regularities.

【解题思路】

根据arithmetic-geometric means inequality，把num分成越接近的几部分其乘积越大。

【代码】

class Solution {
public:
    int integerBreak(int n) {
        if(n==2) return 1;
        int maxnum=0;
        for(int i=2;i<n;i++){
           if(n%i==0){
              maxnum=max(maxnum,(int)pow(n/i,i));      
           }
           else{
               maxnum=max(maxnum,(int)(pow(n/i,i-n+i*(n/i))*pow(n/i+1,n-i*(n/i))));
           }
        }
        return maxnum;
           
    }
};