BZOJ 4241 分块

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=;
int n,q,Block,a[N],block[N],cpy[N],u;
int xx,yy,cnt[][N],top,stk[N],num[N];
typedef long long ll;ll f[][N],ans;
int main(){
    scanf("%d%d",&n,&q),Block=sqrt(n);
    for(int i=;i<=n;i++)scanf("%d",&a[i]),block[i]=(i-)/Block+,cpy[i]=a[i];
    sort(cpy+,cpy++n),u=unique(cpy+,cpy++n)-cpy-;
    for(int i=;i<=n;i++)a[i]=lower_bound(cpy+,cpy++u,a[i])-cpy;
    for(int i=;i<=block[n];i++){
        ll now=;
        for(int j=lower_bound(block+,block++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]+][yy],top=;
        int temp=lower_bound(block+,block++n,block[yy])-block;
        for(int i=temp;i<=yy;i++)num[a[i]]++,stk[++top]=a[i];
        temp=lower_bound(block+,block++n,block[xx]+)-block;
        for(int i=xx;i<temp;i++){
            num[a[i]]++,ans=max(ans,(ll)(cnt[block[xx]+][a[i]]-cnt[block[yy]][a[i]]+num[a[i]])*cpy[a[i]]);
            stk[++top]=a[i];
        }
        temp=lower_bound(block+,block++n,block[yy])-block;
        for(int i=temp;i<=yy;i++)
            ans=max(ans,(ll)(cnt[block[xx]+][a[i]]-cnt[block[yy]][a[i]]+num[a[i]])*cpy[a[i]]);
        for(int i=;i<=top;i++)num[stk[i]]=;
        printf("%lld\n",ans);
    }
}