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)
1 #include<cstdio> 2 #include<algorithm> 3 #include<cmath> 4 #define N 300010 5 using namespace std; 6 int n,i,Log[N],j,l,left,right,mid; 7 long long s[N][20],ans; 8 long long gcd(long long a,long long b) 9 { 10 if (b==0) return a; 11 return gcd(b,a%b); 12 } 13 long long Q(int l,int r) 14 { 15 int k=Log[r-l+1]; 16 return gcd(s[l][k],s[r-(1<<k)+1][k]); 17 } 18 int main() 19 { 20 int test; 21 scanf("%d",&test); 22 while (test--) 23 { 24 ans=0; 25 scanf("%d",&n); 26 for (i=1;i<=n;i++) 27 scanf("%lld",&s[i][0]); 28 for (i=1;i<=n;i++) 29 Log[i]=log2(i); 30 for (i=n;i>=1;i--) 31 for (j=1;j<=Log[n];j++) 32 s[i][j]=gcd(s[i][j-1],s[i+(1<<(j-1))][j-1]); 33 34 for (i=1;i<=n;i++) 35 { 36 l=i; 37 while (l<=n) 38 { 39 left=l; 40 right=n; 41 while (left<=right) 42 { 43 mid=(left+right)>>1; 44 if (Q(i,mid)==Q(i,l)) 45 left=mid+1; 46 else 47 right=mid-1; 48 } 49 l=right+1; 50 ans=max(ans,(right-i+1)*Q(i,right)); 51 } 52 } 53 printf("%lld\n",ans); 54 } 55 }
