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][],ans;
 long long gcd(long long a,long long b)
 {
     if (b==) return a;
     return gcd(b,a%b);
 }
 long long Q(int l,int r)
 {
     int k=Log[r-l+];
     return gcd(s[l][k],s[r-(<<k)+][k]);
 }
 int main()
 {
     int test;
     scanf("%d",&test);
     while (test--)
     {
     ans=;
     scanf("%d",&n);
     for (i=;i<=n;i++)
     scanf("%lld",&s[i][]);
     for (i=;i<=n;i++)
     Log[i]=log2(i);
     for (i=n;i>=;i--)
     for (j=;j<=Log[n];j++)
     s[i][j]=gcd(s[i][j-],s[i+(<<(j-))][j-]);

     for (i=;i<=n;i++)
     {
         l=i;
         while (l<=n)
         {
             left=l;
             right=n;
             while (left<=right)
             {
                 mid=(left+right)>>;
                 if (Q(i,mid)==Q(i,l))
                 left=mid+;
                 else
                 right=mid-;
             }
             l=right+;
             ans=max(ans,(right-i+)*Q(i,right));
         }
     }
     printf("%lld\n",ans);
     }
 }