BZOJ 4052: [Cerc2013]Magical GCD

 以一个数字开头的子序列的gcd种类不会超过logn种，因此去找相同gcd最长的位置，更新一下答案，复杂度O(nlogn^2)

#include<cstdio>
#include<algorithm>
#include<cmath>
#define N 300010
using namespace std;
int n,i,Log[N],j,l,left,right,mid;
long long s[N][20],ans;
long long gcd(long long a,long long b)
{
    if (b==0) return a;
    return gcd(b,a%b);
}
long long Q(int l,int r)
{
    int k=Log[r-l+1];
    return gcd(s[l][k],s[r-(1<<k)+1][k]);
}
int main()
{
    int test;
    scanf("%d",&test);
    while (test--)
    {
    ans=0;
    scanf("%d",&n);
    for (i=1;i<=n;i++)
    scanf("%lld",&s[i][0]);
    for (i=1;i<=n;i++)
    Log[i]=log2(i);
    for (i=n;i>=1;i--)
    for (j=1;j<=Log[n];j++)
    s[i][j]=gcd(s[i][j-1],s[i+(1<<(j-1))][j-1]);
    
    for (i=1;i<=n;i++)
    {
        l=i;
        while (l<=n)
        {
            left=l;
            right=n;
            while (left<=right)
            {
                mid=(left+right)>>1;
                if (Q(i,mid)==Q(i,l))
                left=mid+1;
                else
                right=mid-1;
            }
            l=right+1;
            ans=max(ans,(right-i+1)*Q(i,right));
        }
    }
    printf("%lld
",ans);
    }
}