Codeforces Round #602 (Div. 2, based on Technocup 2020 Elimination Round 3) D2. Optimal Subsequences (Hard Version) 数据结构 贪心

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].

题意

给你n个数,定义长度为k的理想序列为当前k个数和最大的子序列,且这个子序列的字典序要最小。

然后现在给你q个询问,每次问你长度为ki的理想序列的第pos个数是什么

题解

理想序列的构成,显然是贪心的,每次放最大的字典序最小的数进去。

我们将询问离线之后,难点就变成如何求第k个数是多少,实际上这个就是典型的离线求第k大的题目。。。做法非常多,我才用的是树状数组的二分,这个复杂度是logn^2的,线段树上2分是logn的,这个我就懒得写了。

代码

#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();
}

原文地址:https://www.cnblogs.com/qscqesze/p/11925093.html

时间: 2024-08-04 14:24:53

Codeforces Round #602 (Div. 2, based on Technocup 2020 Elimination Round 3) D2. Optimal Subsequences (Hard Version) 数据结构 贪心的相关文章

Codeforces Round #602 (Div. 2, based on Technocup 2020 Elimination Round 3

A,有多个线段,求一条最短的线段长度,能过覆盖到所又线段,例如(2,4)和(5,6) 那么我们需要4 5连起来,长度为1,例如(2,10)(3,11),用(3,10) 思路:我们想一下如果题目说的是最长我们肯定是取最小x和最大的y连起来就完事. 但是要求长度最小又得覆盖,那么可以这样想,我们需要把最小的x不断右移到这条线段的y, 最大的左移到x,所以就是最大x-最小y完事 #include <bits/stdc++.h> using namespace std; #define ll long

Codeforces Round #602 (Div. 2, based on Technocup 2020 Elimination Round 3) C. Messy 构造

C. Messy You are fed up with your messy room, so you decided to clean it up. Your room is a bracket sequence s=s1s2-sn of length n. Each character of this string is either an opening bracket '(' or a closing bracket ')'. In one operation you can choo

Codeforces Round #602 (Div. 2, based on Technocup 2020 Elimination Round 3) B. Box 贪心

B. Box Permutation p is a sequence of integers p=[p1,p2,-,pn], consisting of n distinct (unique) positive integers between 1 and n, inclusive. For example, the following sequences are permutations: [3,4,1,2], [1], [1,2]. The following sequences are n

Codeforces Round #602 (Div. 2, based on Technocup 2020 Elimination Round 3) F2. Wrong Answer on test 233 (Hard Version) dp 数学

F2. Wrong Answer on test 233 (Hard Version) Your program fails again. This time it gets "Wrong answer on test 233" . This is the harder version of the problem. In this version, 1≤n≤2?105. You can hack this problem if you locked it. But you can h

Codeforces Round #602 (Div. 2, based on Technocup 2020 Elimination Round 3) A Math Problem

//只要从所有区间右端点的最小值覆盖到所有区间左端点的最大值即可 #include<iostream> using namespace std ; int x,y; int n; int t; int main() { cin>>t; while(t--) { cin>>n; if(n==1) { cin>>x>>y; cout<<0<<endl; } else { int l=0,r=1e9+10; for(int i=

Codeforces Round #602 (Div. 2, based on Technocup 2020 Elimination Round 3) B Box

#include<bits/stdc++.h> using namespace std; int shu[100005]; int ans[100005]; int main() { int total; cin>>total; while(total--) { int n; bool flag=true; cin>>n; unordered_set<int>tree; set<int>num; for(int i=1; i<=n; i++

Codeforces Round #602 (Div. 2, based on Technocup 2020 Elimination Round 3) C Messy

//因为可以反转n次 所以可以得到任何可以构成的序列 #include<iostream> #include<string> #include<vector> using namespace std ; typedef pair<int,int>PII; int n,k; string s; string get_str(int n,int k) {//先构建前k-1个 string res=""; for(int i=0; i<k

【cf比赛记录】Codeforces Round #606 (Div. 2, based on Technocup 2020 Elimination Round 4)

比赛传送门 只能说当晚状态不佳吧,有点头疼感冒的症状.也跟脑子没转过来有关系,A题最后一步爆搜没能立即想出来,B题搜索没有用好STL,C题也因为前面两题弄崩了心态,最后,果然掉分了. A:简单数学 B:数学+排序 C:字符串搜索 A // https://codeforces.com/contest/1277/problem/A /* 题意: 给出一个数,求不大于该数的完美整数的个数(完美整数指全是一个数组成的数字,如:111, 333333, 444444, 9, 8888 ...) 分析:

Codeforces Round #591 (Div. 2, based on Technocup 2020 Elimination Round 1) 题解

A..B略 C 对当前的值排序,再二分答案,然后对于(i%x==0 && i%y==0)放入大的,再放其他的贪心解决即可. #include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> #include<map> #define LL long long #define lson rt<<1 #define rson rt<