GCD on Blackboard（前缀和+后缀和+思维）

GCD on Blackboard（前缀和+后缀和+思维）

题目描述

There are N integers, A1,A2,…,AN, written on the blackboard.
You will choose one of them and replace it with an integer of your choice between 1 and 109 (inclusive), possibly the same as the integer originally written.

Find the maximum possible greatest common divisor of the N integers on the blackboard after your move.

Constraints
·All values in input are integers.
·2≤N≤10⁵
·1≤Ai≤10⁹

输入

Input is given from Standard Input in the following format:

N
A1 A2 … AN

输出

Print the maximum possible greatest common divisor of the N integers on the blackboard after your move.

样例输入 Copy

3
7 6 8

样例输出 Copy

2

提示

If we replace 7 with 4, the greatest common divisor of the three integers on the blackboard will be
2, which is the maximum possible value.

思路：求前缀及后缀，再求gcd,O(n)

#pragma GCC optimize(3 , "Ofast" , "inline")
 
#include <bits/stdc++.h>
 
#define rep(i , a , b) for(register int i=(a);i<=(b);i++)
#define rop(i , a , b) for(register ll i=(a);i<(b);i++)
#define per(i , a , b) for(register ll i=(a);i>=(b);i--)
#define por(i , a , b) for(register ll i=(a);i>(b);i--)
 
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int , int > pi;
template<class T>
inline void read (T &x) {
    x = 0;
    int sign = 1;
    char c = getchar ();
    while (c < '0' || c > '9') {
        if ( c == '-' ) sign = - 1;
        c = getchar ();
    }
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar ();
    }
    x = x * sign;
}
const int maxn = 1e5+10;
const int inf = int (1e9);
const ll INF = ll(1e18);
const double PI = acos (-1);
const int mod = 1e9+7;
const double eps = 1e-8;
int a[maxn];
int b[maxn];
int c[maxn];
 
int main(){
    int n;
    read (n);
    rep (i,1,n) {
        read (a[i]);
    }
    b[1]=a[1];
    rep (i,2,n) {
        b[i]=__gcd (b[i-1],a[i]);
    }
    c[n]=a[n];
    per (i,n-1,1) {
        c[i]=__gcd (c[i+1],a[i]);
    }
    int ans = 0;
    rep (i,1,n) {
        ans=max (ans,__gcd (b[i-1],c[i+1]));
    }
    cout<<ans<<endl;
    return 0;
}