HDU 6025 Coprime Sequence

Coprime Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 44    Accepted Submission(s): 34

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

3 3 1 1 1 5 2 2 2 3 2 4 1 2 4 8

 

Sample Output

1 2 2

 

题意：

        T组样例，给出 N 个数，求去掉一个数后，数列的最大 GCD。

思路：

        维护前缀 GCD 和 后缀GCD 即可。

        代码里使用了线段树，其实完全没有必要。

#include <bits/stdc++.h>  
  
using namespace std;  
#define ls l,mid,rt*2  
#define rs mid+1,r,rt*2+1  
#define sf l,r,rt  
#define mi (l+r)/2;  
const int MAXN=1e6+100;  
int tree[4*MAXN],st,en;  
int gcd(int x,int y){return y==0?x:gcd(y,x%y);}  
void push_up(int l,int r,int rt){  
    tree[rt]=gcd(tree[rt*2],tree[rt*2+1]);  
}  
void build(int l,int r,int rt){  
    if(l==r){scanf("%d",&tree[rt]);return ;}  
    int mid=mi;  
    build(ls);build(rs);  
    push_up(sf);  
    return ;  
}  
int query(int l,int r,int rt){  
    if(r<st||l>en) return 0;  
    if(st<=l&&r<=en) return tree[rt];  
    int mid=mi;  
    int ans=query(ls);  
    if(ans==0) ans=query(rs);  
    else ans=gcd(ans,query(rs));  
    return ans;  
}  
int main()  
{  
    int T,n;  
    scanf("%d",&T);  
    while(T--){  
        scanf("%d",&n);  
        build(1,n,1);  
        int ans=-1;  
        for(int i=2;i<n;i++){  
            st=1;en=i-1;  
            int temp=query(1,n,1);  
            st=i+1;en=n;  
            temp=gcd(temp,query(1,n,1));  
            ans=max(ans,temp);  
        }  
        st=2;en=n;  
        ans=max(query(1,n,1),ans);  
        st=1;en=n-1;  
        ans=max(query(1,n,1),ans);  
        printf("%d\n",ans);  
    }  
}  

 

转自：http://blog.csdn.net/dt2131/article/details/71424843#