hdu 6025 gcd+前缀和+后缀和

题目：

Coprime Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 699    Accepted Submission(s): 360

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

 

给一串数，删掉其中一个使得剩下的序列GCD值最大 输出最大GCD值

分析：

一开始以为要将每一个数分解为素数乘积，可是an太大了

然后往贪心方向想，结果...找不到比较靠谱的贪心方向

正解是维护一个前缀和和一个后缀和

代码：

#include<bits/stdc++.h>using namespace std;const int maxn=1e5+20;int gcd(int a,int b){    if(b==0) return a;    else return gcd(b,a%b);}int t,n,a[maxn],pre[maxn],nxt[maxn];int main(){//78ms 6400kb    scanf("%d",&t);    while(t--){        scanf("%d",&n);        for(int i=1;i<=n;++i){            scanf("%d",&a[i]);        }        pre[1]=a[1];        for(int i=2;i<=n;++i){            pre[i]=gcd(pre[i-1],a[i]);        }        nxt[n]=a[n];        for(int i=n-1;i>=1;--i){            nxt[i]=gcd(nxt[i+1],a[i]);        }        int ans=max(nxt[2],pre[n-1]);        for(int i=2;i<n;++i){            ans=max(ans,gcd(pre[i-1],nxt[i+1]));        }        printf("%d\n",ans);    }return 0;}