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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.
1 public class Solution { 2 int integerBreak(int n) { 3 int result=0; 4 double e=2.718281828459; 5 int breakNumber=(int) (n/e); 6 if (breakNumber<2) { 7 return n/2*(n-n/2); 8 } 9 for (int i = breakNumber-1; i <=breakNumber+1; i++) { 10 int breakValue=n/i; 11 int valueOverNumber=n%i; 12 int resultTemp=0; 13 if (valueOverNumber!=0) { 14 resultTemp=(int) ( Math.pow(breakValue, i-valueOverNumber) 15 * Math.pow(breakValue+1, valueOverNumber)); 16 }else { 17 resultTemp=(int) ( Math.pow(breakValue, i)); 18 } 19 20 if (resultTemp>result) { 21 result=resultTemp; 22 } 23 } 24 return result; 25 26 } 27 }
由数学计算可以证明已知y=x1+x2+...+xn, z=x1*x2*...xn最大时，应满足x1=x2=...xn=y/n,且分割数n=y/e
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原文地址：http://www.cnblogs.com/zoghin/p/5450055.html
