D2. Optimal Subsequences (Hard Version)
This is the harder version of the problem. In this version, 1≤n,m≤2⋅105. You can hack this problem if you locked it. But you can hack the previous problem only if you locked both problems.
You are given a sequence of integers a=[a1,a2,…,an] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:
[11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list);
[40], [33,33], [33,20,20], [20,20,11,11] are not subsequences.
Suppose that an additional non-negative integer k (1≤k≤n) is given, then the subsequence is called optimal if:
it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k;
and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal.
Recall that the sequence b=[b1,b2,…,bk] is lexicographically smaller than the sequence c=[c1,c2,…,ck] if the first element (from the left) in which they differ less in the sequence b than in c. Formally: there exists t (1≤t≤k) such that b1=c1, b2=c2, ..., bt−1=ct−1 and at the same time bt<ct. For example:
[10,20,20] lexicographically less than [10,21,1],
[7,99,99] is lexicographically less than [10,21,1],
[10,21,0] is lexicographically less than [10,21,1].
You are given a sequence of a=[a1,a2,…,an] and m requests, each consisting of two numbers kj and posj (1≤k≤n, 1≤posj≤kj). For each query, print the value that is in the index posj of the optimal subsequence of the given sequence a for k=kj.
For example, if n=4, a=[10,20,30,20], kj=2, then the optimal subsequence is [20,30] — it is the minimum lexicographically among all subsequences of length 2 with the maximum total sum of items. Thus, the answer to the request kj=2, posj=1 is the number 20, and the answer to the request kj=2, posj=2 is the number 30.
Input
The first line contains an integer n (1≤n≤2⋅105) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a1,a2,…,an (1≤ai≤109).
The third line contains an integer m (1≤m≤2⋅105) — the number of requests.
The following m lines contain pairs of integers kj and posj (1≤k≤n, 1≤posj≤kj) — the requests.
Output
Print m integers r1,r2,…,rm (1≤rj≤109) one per line: answers to the requests in the order they appear in the input. The value of rj should be equal to the value contained in the position posj of the optimal subsequence for k=kj.
Examples
input
3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3
output
20
10
20
10
20
10
input
7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4
output
2
3
2
3
2
3
1
1
3
Note
In the first example, for a=[10,20,10] the optimal subsequences are:
for k=1: [20],
for k=2: [10,20],
for k=3: [10,20,10].
The meaning of problems
You number n, is defined over a length of k for the current sequence number k and the sub-maximum sequence and the promoter sequence to lexicographically smallest.
Now and then to ask you a q, every time you ask what is the length of the first pos ideal sequence number is ki
answer
Constitute the ideal sequence, apparently greedy, each put largest lexicographically smallest number inside.
We will then ask offline, the difficulty becomes how to find the first number k is the number, in fact, this is typical of the k off pursuing big topic. . . Practice very much, I just use the binary tree arrays, this complexity is logn ^ 2, the tree line is logn two points, which I would not bother to write.
Code
#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5+7;
int a[maxn],index[maxn],ans[maxn],sum[maxn];
int n;
int lowbit(int x){
return x&(-x);
}
void update(int x,int val){
while(x <= n){
sum[x] += val;
x += lowbit(x);
}
}
int query(int x){
int s=0;
while(x>0){
s += sum[x];
x -= lowbit(x);
}
return s;
}
void solve(){
scanf("%d",&n);
for(int i=1;i<=n;i++){
cin>>a[i];
index[i]=0;
}
set<pair<int,int> >S;
for(int i=1;i<=n;i++){
S.insert(make_pair(-a[i],i));
}
int m;scanf("%d",&m);
vector<pair<pair<int,int>,int>>Q;
for(int i=0;i<m;i++){
int x,y;scanf("%d%d",&x,&y);
Q.push_back(make_pair(make_pair(x,y),i));
}
sort(Q.begin(),Q.end());
int now = 0;
for(int i=0;i<Q.size();i++){
while(now<Q[i].first.first){
now=now+1;
pair<int,int> tmp = *S.begin();
update(tmp.second,1);
index[tmp.second]=1;
S.erase(tmp);
}
int pos = Q[i].first.second;
int l=1,r=n,Ans=n;
while(l<=r){
int mid=(l+r)/2;
if(query(mid)>=pos){
Ans=mid;
r=mid-1;
}else{
l=mid+1;
}
}
ans[Q[i].second]=a[Ans];
}
for(int i=0;i<m;i++){
cout<<ans[i]<<endl;
}
}
int main(){
solve();
}