# 【题目】Sequence Query

AtCoder Beginner Contest 241（Sponsored by Panasonic）

D - Sequence Query

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $400$ points

### Problem Statement

We have an empty sequence $A$.
Given $Q$ queries, process them in order.
Each query is of one of the following three types.

• 1 x : Insert $x$ to $A$.
• 2 x k : Among the elements of $A$ that are less than or equal to $x$, print the $k$-th largest value. ($k$ is no more than $\bf{5}$)
If there are less than $k$ elements of $A$ that are less than or equal to $x$, then print -1.
• 3 x k : Among the elements of $A$ that are greater than or equal to $x$, print the $k$-th smallest value. ($k$ is no more than $\bf{5}$)
If there are less than $k$ elements of $A$ that are greater than or equal to $x$, then print -1.

### Constraints

• $1\leq Q \leq 2\times 10^5$
• $1\leq x\leq 10^{18}$
• $1\leq k\leq 5$
• All values in input are integers.

### Input

Input is given from Standard Input in the following format:

$Q$
$\text{query}_1$
$\text{query}_2$
$\vdots$
$\text{query}_Q$

In the $i$-th query $\text{query}_i$, the type of query $c_i$ (which is either $1, 2$, or $3$) is given first.
If $c_i=1$, then $x$ is additionally given; if $c_i=2, 3$, then $x$ and $k$ are additionally given.

In other words, each query is given in one of the following three formats:

$1$ $x$
$2$ $x$ $k$
$3$ $x$ $k$

### Output

Print $q$ lines, where $q$ is the number of queries such that $c_i=2,3$.
The $j$-th line $(1\leq j\leq q)$ should contain the answer for the $j$-th such query.

11
1 20
1 10
1 30
1 20
3 15 1
3 15 2
3 15 3
3 15 4
2 100 5
1 1
2 100 5

### Sample Output 1

20
20
30
-1
-1
1

After $\text{query}_{1,2,3,4}$ have been processed, we have $A=(20,10,30,20)$.

For $\text{query}_{5,6,7}$, the elements of $A$ greater than or equal to $15$ are $(20,30,20)$.
The $1$-st smallest value of them is $20$; the $2$-nd is $20$; the $3$-rd is $30$.

### 我的笔记

\<set\>
multiset: 可存在重复元素的集合，并且能保证序列有序。

### 代码

#include <bits/stdc++.h>

using namespace std;

int main(void)
{
multiset<long long> A;
int Q;
cin >> Q;
for (int i = 0; i < Q; i++)
{
int operation;
cin >> operation;
if (operation == 1)
{
long long x;
cin >> x;
A.insert(x);
}
else if (operation == 2)
{
long long x, k;
cin >> x >> k;
auto it = A.upper_bound(x);
bool find = true;
for (int j = 0; j < k; j++)
{
if (it == A.begin())
{
find = false;
break;
}
it--;
}
if (find)
cout << *it << endl;
else
cout << "-1" << endl;
}
else
{
long long x, k;
cin >> x >> k;
auto it = A.lower_bound(x);
bool find = true;
for (int j = 1; j < k; j++)
{
if (it == A.end())
{
find = false;
break;
}
it++;
}
if (it == A.end())
find = false;
if (find)
cout << *it << endl;
else
cout << "-1" << endl;
}
}
return 0;
}