传送门

题意：

思路：

容易发现，我们将所有$a$都除上所有$a$的$gcd$，实际上就是让你求一个最小的$len$，对于所有$i$，$gcd(a_i,a_{i+1},...,a_{i+len-1})=1$，这个显然有单调性，所以打一个$st$表让后断环成链，二分$len$每次检查一下即可。

还可以尺取，固定右端点，找到最短的$gcd$为$1$的区间，取所有长度中最大的一个即可。

// Problem: F. Array Stabilization (GCD version)
// Contest: Codeforces - Codeforces Round #731 (Div. 3)
// URL: https://codeforces.com/contest/1547/problem/F
// Memory Limit: 512 MB
// Time Limit: 4000 ms
// 
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n;
int a[N];
int f[N][20],len[N];void init(int n)
{len[1]=0; len[2]=1;for(int i=2;i<=n;i++) len[i]=len[i>>1]+1;for(int i=1;i<=n;i++) f[i][0]=a[i];int t=len[n]+1;for(int j=1;j<t;j++)for(int i=1;i<=n-(1<<j)+1;i++)f[i][j]=__gcd(f[i][j-1],f[i+(1ll<<(j-1))][j-1]);
}int query(int l,int r) {int t=len[r-l+1];return __gcd(f[l][t],f[r-(1<<t)+1][t]);
}int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);int _; scanf("%d",&_);while(_--) {scanf("%d",&n);for(int i=1;i<=n;i++) scanf("%d",&a[i]),a[n+i]=a[i];int now=a[1];for(int i=2;i<=n;i++) now=__gcd(now,a[i]);for(int i=1;i<=n*2;i++) a[i]/=now;int ans=0; init(n*2);int l=1,r=1;while(r<=n*2) {while(l<=r&&query(l,r)==1) l++;r++;ans=max(ans,r-l);}printf("%d\n",ans);}return 0;
}
/**/