HDU - 6025辗转相除法+前缀后缀

 Coprime Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1527    Accepted Submission(s): 733
http://acm.hdu.edu.cn/showproblem.php?pid=6025

Problem Description

Do you know what is called ``Coprime Sequence''? That is a sequence consists of  n positive integers, and the GCD (Greatest Common Divisor) of them is equal to 1.
``Coprime Sequence'' is easy to find because of its restriction. But we can try to maximize the GCD of these integers by removing exactly one integer. Now given a sequence, please maximize the GCD of its elements.

 

Input

The first line of the input contains an integer  T(1≤T≤10), denoting the number of test cases.
In each test case, there is an integer  n(3≤n≤100000) in the first line, denoting the number of integers in the sequence.
Then the following line consists of  n integers  a1,a2,...,an(1≤ai≤109), denoting the elements in the sequence.

 

Output

For each test case, print a single line containing a single integer, denoting the maximum GCD.

 

Sample Input

331 1 152 2 2 3 241 2 4 8

 

Sample Output

122

    

本题题意就是一个序列，去掉一个其中一个数，使得GCD尽可能大开始的大，用两层for循环，第一层a[i]对应去掉a[i]时的序列

第二层循环再次暴力序列用辗转相除法找到GCD....后来就T掉了。一看范围n^2复杂度，大概是1e10肯定T。。。然后第二层从两边往中间暴力，复杂度改成1/2*n^2还是T。。。

最后用前几天学的前缀和，pre[i]表示a[i]前的GCD,backk[i]表示a[i]后的GCD,这样一次线性N/2的复杂度就算出前缀与后缀，最后1->N暴力，再次取GCD即可，用MAX保存更新，最后输出即可

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