# Leetcode brush problem, summary, record, Memo 343

Source: Internet
Author: User

leetcode343Integer break

Given a positive integer n, break it into the sum of at least , positive integers and maximize the pro Duct of those integers. Return The maximum product you can get.

For example, given n = 2, return 1 (2 = 1 + 1); Given n = ten, return 36 (10 = 3 + 3 + 4).

Note: You may assume this n is not less than 2.

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Credits:
Special thanks to @jianchao. Li.fighter for adding the problem and creating all test cases.

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using dynamic programming, we can find a rule by observing that the maximum value of n is either 1 on the basis of n-1, Either add 2 on top of the n-2. First, if it is based on the results of n-1, if you simply add a 1 in the n-1 array, the result is constant, so you need to add 1 to an element in the n-1 array, this 1 is best added to the smallest number in the n-1 array, so the product is the largest. If it is based on n-2, simply add a 2 to the array of n-2, and the one that evaluates to the maximum in these 2 arrays is the result of N. Suppose that the DP array is an array of each number that holds each number, such as dp[10] = {3, 3, 4}. The work to be done before the recursive formula, Dp[n-2].push_back (2); Sort (Dp[n-2].begin (), Dp[n-2].end ());  dp[n-1][0]++;  sort (Dp[n-1].begin (), Dp[n-1].end ()); Recursive formula: dp[n] = func (dp[n-2]) > func (dp[n-1])?  Dp[n-2]: dp[n-1]; Func is an auxiliary function that computes the result of an array and sorts it in order to serve the subsequent computations. You can view the code in detail.

`Class Solution {Public:int function (Vector<int> t) {int sum = 1;        for (int i = 0; i < t.size (); ++i) {sum *= t[i];    } return sum;        } int Integerbreak (int n) {if (n <= 0) {return 0;        } if (n = = 1) {return 1;                } vector<vector<int> > DP (n + 1, vector<int> ());        Dp[1].push_back (1);        Dp[2].push_back (1);                Dp[2].push_back (1);            for (int i = 3; I < dp.size (); ++i) {vector<int> temp1 = dp[i-2];            Temp1.push_back (2);            Vector<int> temp2 = dp[i-1];            temp2[0]++;            Sort (Temp1.begin (), Temp1.end ());                        Sort (Temp2.begin (), Temp2.end ());            if (function (TEMP1) > function (TEMP2)) {dp[i] = Temp1; } else {DP[I] = TEMP2;            }} return function (Dp[n]); }};`

Leetcode brush problem, summary, record, Memo 343

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