思路：容易发现以某个位置\(i\)为结尾所有后缀的\(gcd\)个数不超过\(log(a[i])\)。

（怎么发现？将数写成质因子幂次乘积的形式，然后\(gcd\)每次减小一个质因子，最多减少\(log\)次）然后就可以用\(map\)维护每个\(gcd\)的最左端端点。

代码：

#pragma GCC optimize(2)#pragma GCC optimize(3)#pragma GCC optimize(4)#include
    
     using namespace std;#define fi first#define se second#define pi acos(-1.0)#define LL long long//#define mp make_pair#define pb push_back#define ls rt<<1, l, m#define rs rt<<1|1, m+1, r#define ULL unsigned LL#define pll pair
     
      #define pii pair
      
       #define piii pair
       
        #define mem(a, b) memset(a, b, sizeof(a))#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);//headconst int N = 1e5 + 5;LL a[N];map
        
          mp, _p;int n; int main() {    scanf("%d", &n);    for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]);    LL ans = 0;    for (int i = 1; i <= n; ++i) {        ans = max(ans, a[i]);        for (map
         
          ::iterator it = mp.begin(); it != mp.end(); ++it) { LL g = __gcd(a[i], (*it).fi); ans = max(ans, g*(i-(*it).se+1)); if(_p.find(g) == _p.end()) _p[g] = (*it).se; } if(_p.find(a[i]) == _p.end()) _p[a[i]] = i; mp = _p; _p.clear(); } printf("%lld\n", ans); return 0;}