# bzoj 4241 历史研究

$n,q \leq 100000$

sol：

1.对于长度小于 $\sqrt{n}$ 的询问，暴力

2.每次对于一个块，处理左端点在这个块里的询问，对于每次询问，先把 $L$ 指针设为块的右端点，向右移动 $R$ 指针，求出这组询问答案没移动左端点时的结果，然后把 $L$ 往左移求出答案，这组询问求完之后把答案和 $cnt$ 数组恢复成没移动左端点时的结果

#include <bits/stdc++.h>
#define LL long long
#define rep(i, s, t) for (register int i = (s), i##end = (t); i <= i##end; ++i)
#define dwn(i, s, t) for (register int i = (s), i##end = (t); i >= i##end; --i)
using namespace std;
namespace IO{
const int BS=(1<<23)+5; int Top=0;
char Buffer[BS],OT[BS],*OS=OT,*HD,*TL,SS[30]; const char *fin=OT+BS-1;
void flush(){fwrite(OT,1,OS-OT,stdout);}
void Putchar(char c){*OS++ =c;if(OS==fin)flush(),OS=OT;}
void write(LL x){
if(!x){Putchar('0');return;} if(x<0) x=-x,Putchar('-');
while(x) SS[++Top]=x%10,x/=10;
while(Top) Putchar(SS[Top]+'0'),--Top;
}
int nm=0,fh=1; char cw=Getchar();
for(;!isdigit(cw);cw=Getchar()) if(cw=='-') fh=-fh;
for(;isdigit(cw);cw=Getchar()) nm=nm*10+(cw-'0');
return nm*fh;
}
} using namespace IO;
const int maxn = 100100;
int n, q, pos = 1, SZ;
int a[maxn], v[maxn], cnt[maxn];
int bl[maxn]; LL ans[maxn];
struct Ques {
int l, r, id;
Ques(){}
Ques(int _1, int _2, int _3) : l(_1), r(_2), id(_3){}
inline bool operator < (const Ques &b) const {
return bl[l] == bl[b.l] ? r < b.r : l < b.l;
}
}qs[maxn];
int force_cnt[maxn];
inline LL force(int l, int r) {
LL ret = 0;
rep(i, l, r) force_cnt[a[i]] = 0;
rep(i, l, r) force_cnt[a[i]]++;
rep(i, l, r) ret = max(ret, 1LL * force_cnt[a[i]] * v[a[i]]);
return ret;
}
inline void modify(int &i, int pos) {
int L = min(n, SZ * pos);
int ql = L + 1, qr = L;
LL tmp = 0;
memset(cnt, 0, sizeof(cnt));
for(; bl[qs[i].l] == pos; i++) {
if(bl[qs[i].l] == bl[qs[i].r]) ans[qs[i].id] = force(qs[i].l, qs[i].r);
else {
while(qr < qs[i].r) {
qr++;
cnt[a[qr]]++;
tmp = max(tmp, 1LL * v[a[qr]] * cnt[a[qr]]);
}
LL tans = tmp;
while(ql > qs[i].l) {
--ql;
cnt[a[ql]]++;
tmp = max(tmp, 1LL * v[a[ql]] * cnt[a[ql]]);
}
ans[qs[i].id] = tmp;
while(ql < L + 1) {
cnt[a[ql]]--;
ql++;
}
tmp = tans;
}
}
}
int main() {
//    freopen("2.in","r",stdin);
//    freopen("buff.out","w",stdout);
rep(i, 1, n) v[i] = a[i] = read(), bl[i] = (i-1) / SZ + 1;
sort(v + 1, v + n + 1);
rep(i, 1, n) a[i] = lower_bound(v + 1, v + n + 1, a[i]) - v;
rep(i, 1, q) {
}