思路：
考虑分块
f[i][j]表示从第i块开头到j的最大值
cnt[i][j]表示从第i块开始到序列末尾j出现了多少次
边角余料处理一下就好啦~
//By SiriusRen #include <cmath> #include <cstdio> #include <algorithm> using namespace std; const int N=100050; int n,q,Block,a[N],block[N],cpy[N],u; int xx,yy,cnt[333][N],top,stk[N],num[N]; typedef long long ll;ll f[333][N],ans; int main(){scanf("%d%d",&n,&q),Block=sqrt(n);for(int i=1;i<=n;i++)scanf("%d",&a[i]),block[i]=(i-1)/Block+1,cpy[i]=a[i];sort(cpy+1,cpy+1+n),u=unique(cpy+1,cpy+1+n)-cpy-1;for(int i=1;i<=n;i++)a[i]=lower_bound(cpy+1,cpy+1+u,a[i])-cpy;for(int i=1;i<=block[n];i++){ll now=0;for(int j=lower_bound(block+1,block+1+n,i)-block;j<=n;j++)cnt[i][a[j]]++,now=max(now,(ll)cnt[i][a[j]]*cpy[a[j]]),f[i][j]=now;}while(q--){scanf("%d%d",&xx,&yy),ans=f[block[xx]+1][yy],top=0;int temp=lower_bound(block+1,block+1+n,block[yy])-block;for(int i=temp;i<=yy;i++)num[a[i]]++,stk[++top]=a[i];temp=lower_bound(block+1,block+1+n,block[xx]+1)-block;for(int i=xx;i<temp;i++){num[a[i]]++,ans=max(ans,(ll)(cnt[block[xx]+1][a[i]]-cnt[block[yy]][a[i]]+num[a[i]])*cpy[a[i]]);stk[++top]=a[i];}temp=lower_bound(block+1,block+1+n,block[yy])-block;for(int i=temp;i<=yy;i++)ans=max(ans,(ll)(cnt[block[xx]+1][a[i]]-cnt[block[yy]][a[i]]+num[a[i]])*cpy[a[i]]);for(int i=1;i<=top;i++)num[stk[i]]=0;printf("%lld\n",ans);} }
