P7009 [CERC2013] Magical GCD

容易发现的是，当左端点 $l$ 确定时，$r \in [l,n]$，那么 $\gcd \limits_{i=l}^r a_i$ 的不同数量是 $O(\log V)$ 的级别。

原因是，每次 $\gcd$ 减少，总有一个质因数的次数降低，降低 $O(\log V)$ 次就会变成 $1$。

于是我们枚举左端点，然后二分 $O(\log V)$ 次右端点，用 ST 表可以做到单次询问 $O(\log V)$ 处理区间 $\gcd$，复杂度 $O(n \log^2 V \log n)$。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <climits>
#include <numeric>
using namespace std;

const int N = 1e5 + 5;

long long ans = 0LL;
int n;
long long a[N];

long long f[N][31];
int LG2[N];

void Init()
{
	for (int i = 1; i <= n; i++)
	{
		f[i][0] = a[i];
	}
	for (int j = 1; j <= LG2[n]; j++)
	{
		for (int i = 1; i + (1 << j) - 1 <= n; i++)
		{
			f[i][j] = gcd(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
		}
	}
}

long long query(int l, int r)
{
	int s = LG2[r - l + 1];
	return gcd(f[l][s], f[r - (1 << s) + 1][s]);
}

int main()
{
	LG2[1] = 0;
	for (int i = 2; i < N; i++) LG2[i] = LG2[i / 2] + 1;
	int t;
	scanf("%d", &t);
	while (t--)
	{
		ans = 0LL;
		scanf("%d", &n);
		for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
		Init();
		for (int i = 1; i <= n; i++)
		{
			long long nowv = a[i];
			int nowl = i;
			while (true)
			{
				int l = nowl, r = n, place = -1;
				while (l <= r)
				{
					int mid = (l + r) >> 1;
					long long g = query(i, mid);
					if (g == nowv)
					{
						ans = max(ans, g * (1LL * mid - i + 1));
						place = mid;
					}
					if (g >= nowv)
					{
						l = mid + 1;
					}
					else
					{
						r = mid - 1;
					}
				}
				if (place == -1 || place == n) break;
				nowv = query(i, place + 1);
				nowl = place + 1;
			}
		}
		printf("%lld\n", ans);
	}
	
	return 0;
}