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| description
stringlengths 31
8.62k
| public_tests
dict | private_tests
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stringclasses 5
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stringlengths 1
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1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
return a.first > b.first || a.first == b.first && a.second < b.second;
}
bool comp2(pair<long long, long long> a, pair<long long, long long> b) {
return a.second < b.second;
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long n;
cin >> n;
long long a[n];
pair<long long, long long> s[n];
for (long long i = 0; i < n; i++) {
cin >> a[i];
s[i] = {a[i], i};
}
sort(s, s + n, comp);
vector<pair<long long, long long> > ans;
long long m;
cin >> m;
for (long long j = 0; j < m; j++) {
long long k, p;
cin >> k >> p;
ans.clear();
for (long long i = 0; i < k; i++) ans.push_back(s[i]);
sort(ans.begin(), ans.end(), comp2);
cout << ans[p - 1].first << endl;
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
python3
|
n = int(input())
s = list(map(int, input().split()))
for i in range(n):
s[i] = (s[i], i)
mx = sorted(s, key=lambda y: y[0], reverse=True)
m = int(input())
for i in range(m):
k, pos = map(int, input().split())
pos -= 1
a = 0
last = mx[0][0]
res = []
d = dict()
d2 = dict()
for x in range(k):
if mx[x][0] not in d:
d[mx[x][0]] = 1
else:
d[mx[x][0]] += 1
for x in range(n):
if mx[x][0] != last:
a += 1
last = mx[x][0]
if a == k:
break
if mx[x][0] not in d2:
d2[mx[x][0]] = [mx[x][1]]
else:
d2[mx[x][0]].append(mx[x][1])
for a in list(d.keys()):
d2[a].sort()
for j in range(d[a]):
res.append((a, d2[a][j]))
res = sorted(res, key=lambda y: y[1])
print(res[pos][0])
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int INF = 2e9;
void solve() {
int n, k, m, pos;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; i++) cin >> a[i];
cin >> m;
for (int _ = 0; _ < m; _++) {
cin >> k >> pos;
vector<int> ind;
set<int> s;
for (int i = 0; i < k; i++) {
int maxi = -1;
for (int j = 0; j < n; j++) {
if (s.find(j) != s.end()) continue;
if (maxi == -1)
maxi = j;
else if (a[maxi] < a[j]) {
maxi = j;
}
}
s.insert(maxi);
}
int h = 1;
for (auto i : s) {
if (h == pos) cout << a[i] << endl;
h++;
}
}
exit(0);
}
int main() {
int t = 1;
for (int i = 0; i < t; i++) solve();
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Solution {
static class pair implements Comparable<pair> {
int a , b ;
pair(int a, int b ){
this.a = a;
this.b = b ;
}
@Override
public int compareTo(pair o) {
if (this.a == o.a)
return this.b - o.b ;
return - (this.a - o.a) ;
}
}
public static void main(String[] args) {
int n = fsca.nextInt();
pair a[] = new pair[n] ;
for (int i = 0; i <n ; i++) {
a[i] = new pair(fsca.nextInt(), i) ;
}
Arrays.sort(a);
int m = fsca.nextInt();
while (m-- > 0){
int k = fsca.nextInt();
int pos = fsca.nextInt() - 1 ;
ArrayList<Integer> list = new ArrayList<>() ;
for (int i = 0; i <k ; i++) {
list.add(a[i].b) ;
}
Collections.sort(list);
for (int i = 0; i <k ; i++) {
if (list.get(pos) == a[i].b){
fop.append(a[i].a + "\n") ;
break;
}
}
}
fop.flush();
fop.close();
}
/*-----------------------------------------------------------------------------------------------------------------------------------------------*/
static PrintWriter fop = new PrintWriter(System.out);
static FastScanner fsca = new FastScanner();
static long gcd(long a, long b) {
return (b == 0) ? a : gcd(b, a % b);
}
static int gcd(int a, int b) {
return (b == 0) ? a : gcd(b, a % b);
}
// Arrays.sort() takes o(n^2) time to sort when array is reverse sorted
// so always shuffle the array before sorting
static final Random random = new Random();
static void ruffleSort(int[] a) {
int n = a.length;
for (int i = 0; i < n; i++) {
int oi = random.nextInt(n), temp = a[oi];
a[oi] = a[i]; a[i] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
int n = a.length;
for (int i = 0; i < n; i++) {
int oi = random.nextInt(n);
long temp = a[oi];
a[oi] = a[i];
a[i] = temp;
}
Arrays.sort(a);
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
// int array input
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
// long array input
long[] readLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++)
a[i] = nextLong();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
bool cmp(pair<int, int> a, pair<int, int> b) {
if (a.first > b.first)
return true;
else if (a.first == b.first) {
if (a.second < b.second) return true;
return false;
}
return false;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int i, j, k, l, m, n, p, x, t;
vector<pair<int, int> > v, c;
cin >> n;
for (i = 0; i < n; i++) {
cin >> x;
v.push_back({x, i});
}
sort(v.begin(), v.end(), cmp);
cin >> m;
for (i = 1; i <= m; i++) {
cin >> k >> p;
for (j = 0; j < k; j++) c.push_back({v[j].second, v[j].first});
sort(c.begin(), c.end());
cout << c[p - 1].second << endl;
c.clear();
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
const double pi = 3.141592653589793238;
const int MOD1 = 1e9 + 7;
const int MOD2 = 998244353;
const int N = 3e5 + 5;
using namespace std;
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
if (a.first != b.first) return (a.first > b.first);
return (a.second < b.second);
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
long long n = 0;
cin >> n;
vector<pair<long long, long long>> a(n);
for (long long i = 0; i < n; i++) {
cin >> a[i].first;
a[i].second = i + 1;
}
sort((a).begin(), (a).end(), cmp);
long long q = 0;
cin >> q;
while (q--) {
long long k, pos;
cin >> k >> pos;
set<pair<long long, long long>> s;
for (long long i = 0; i < k; i++) {
s.insert({a[i].second, a[i].first});
}
while (--pos) s.erase(s.begin());
auto m = *(s.begin());
cout << m.second << "\n";
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
python3
|
n = int(input())
c = list(map(lambda x: (int(x[1]),-x[0]),enumerate(input().split())))
so = sorted(c)
for i in range(int(input())):
k,r = map(int,input().split())
now = so[-k:]
now.sort(key = lambda x: -x[1])
print(now[r-1][0])
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long int INF = (long long int)2e18 + 77;
void solve() {
int n;
cin >> n;
int a[n], i;
for (i = 0; i < n; i++) cin >> a[i];
map<int, int> N;
int M[n][n];
for (i = 0; i < n; i++) {
N[a[i]]++;
M[n - 1][i] = a[i];
}
for (i = n - 2; i >= 0; i--) {
int mn = (*N.begin()).first;
N[mn]--;
if (N[mn] == 0) N.erase(mn);
int j, k = i;
for (j = i + 1; j >= 0; j--) {
if (mn == M[i + 1][j])
mn = -1;
else
M[i][k--] = M[i + 1][j];
}
}
int m, len;
cin >> m;
while (m--) {
cin >> len >> i;
len--;
i--;
cout << M[len][i] << "\n";
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
solve();
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
int m;
cin >> m;
for (int e = 0; e < m; e++) {
int k, p;
cin >> k >> p;
vector<int> b;
for (int i = 0; i < k; i++) {
b.push_back(a[i]);
}
for (int i = k; i < n; i++) {
int MIN = 1000000007, pos = 0;
for (int j = 0; j < k; j++) {
if (b[j] <= MIN) {
MIN = b[j];
pos = j;
}
}
if (a[i] > MIN) {
b.erase(b.begin() + pos);
b.push_back(a[i]);
}
}
cout << b[p - 1] << endl;
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
python3
|
n=int(input())
a=sorted(zip(list(map(int,input().split())),range(n)),key=lambda x:(x[0],-x[1]),reverse=True)
#print(a)
m=int(input())
for i in range(m):
k,pos=map(int,input().split())
b=sorted(a[:k],key=lambda x:x[1])
#print(b)
print(b[pos-1][0])
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
long long pw(long long a, long long b) {
if (b == 0) return 1;
if (b % 2 == 1)
return (a * pw((a * a) % 1000000007, b / 2)) % 1000000007;
else
return (1 * pw((a * a) % 1000000007, b / 2)) % 1000000007;
}
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
if (a.first == b.first) return a.second < b.second;
return a.first > b.first;
}
bool comp2(pair<long long, long long> a, pair<long long, long long> b) {
return a.second < b.second;
}
void solve() {
int n;
cin >> n;
vector<pair<long long, long long> > v;
for (int i = 0; i < n; i++) {
long long x;
cin >> x;
v.push_back({x, i});
}
sort(v.begin(), v.end(), comp);
vector<vector<pair<long long, long long> > > ans;
for (int k = 0; k < n + 1; k++) {
vector<pair<long long, long long> > temp;
for (int j = 0; j < k; j++) temp.push_back(v[j]);
sort(temp.begin(), temp.end(), comp2);
ans.push_back(temp);
}
long long m;
cin >> m;
while (m--) {
long long k, p;
cin >> k >> p;
cout << ans[k][p - 1].first << endl;
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long t = 1;
while (t--) {
solve();
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
python3
|
from sys import stdin
n = int(stdin.readline())
alist = list(map(int, stdin.readline().split()))
aindlist = sorted([(elem, i) for i, elem in enumerate(alist)], key = lambda x: (-x[0], x[1]))
m = int(stdin.readline())
ans = []
for _ in range(m):
k, pos = map(int, stdin.readline().split())
temp = [x[1] for x in aindlist[:k]]
temp.sort()
ans.append(alist[temp[pos - 1]])
print(*ans, sep= '\n')
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e6 + 1;
const int INF = 2e9 + 1;
const int MOD = (1e9 + 7);
void bye(string s = "") {
cout << s << '\n';
exit(0);
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int n;
cin >> n;
vector<int> arr(n);
set<pair<int, int> > kek;
for (int i = 0; i < n; i++) {
cin >> arr[i];
kek.emplace(-arr[i], i);
}
int m;
cin >> m;
for (int i = 0; i < m; i++) {
int k, pos;
cin >> k >> pos;
auto e = kek.begin();
vector<pair<int, int> > mem;
for (int i = 0; i < k; i++) {
mem.emplace_back(e->second, e->first);
e++;
}
sort((mem).begin(), (mem).end());
cout << mem[pos - 1].second * -1 << '\n';
}
bye();
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
python3
|
import itertools
import functools
from operator import itemgetter
def cmp(a, b):
# 0 - descending, 1 - ascending
# returns: -1, a before b; 1 b before a; 0 no difference
if a[0] > b[0]:
return -1
elif a[0] < b[0]:
return 1
else:
if a[1] < b[1]:
return -1
elif a[1] > b[1]:
return 1
else:
return 0
n = int(input())
a = list(map(int, input().split()))
c = a[:]
c = list(zip(a, itertools.count(0)))
c = sorted(c, key=functools.cmp_to_key(cmp))
h = c[:]
m = int(input())
for i in range(m):
k,pos = map(int, input().split())
h = sorted(c[:k], key=itemgetter(1))
print(h[pos-1][0])
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
int main() {
ll t = 1;
while (t--) {
int n;
cin >> n;
vector<int> v(n), temp(n);
for (int i = 0; i < n; i++) {
cin >> v[i];
temp[i] = v[i];
}
sort(temp.begin(), temp.end(), greater<int>());
int m;
cin >> m;
while (m--) {
int k, pos;
cin >> k >> pos;
map<int, int> mp;
for (int i = 0; i < k; i++) {
mp[temp[i]]++;
}
vector<int> v1(k);
int j = 0;
for (int i = 0; i < n && j < k; i++) {
if (mp[v[i]] > 0) {
mp[v[i]]--;
v1[j] = v[i];
j++;
}
}
cout << v1[pos - 1] << endl;
}
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
java
|
import java.util.*;
public class p1227D1 {
static class Pair implements Comparable<Pair>{
int i;
long j;
Pair(int i,long j){
this.i=i;
this.j=j;
}
public int compareTo(Pair o){
if(this.j>o.j){
return(-1);
}
else if(o.j>this.j){
return(1);
}
else if(this.i<o.i){
return(-1);
}
else {
return(1);
}
}
}
public static void main(String[] args) {
Scanner scn=new Scanner(System.in);
int n=scn.nextInt();
int[] arr=new int[n];
PriorityQueue<Pair> pq=new PriorityQueue<Pair>();
for(int i=0;i<n;i++){
int val=scn.nextInt();
arr[i]=val;
pq.add(new Pair(i,val));
}
ArrayList<Integer> ans=new ArrayList<Integer>();
while(pq.size()>0){
int i=pq.remove().i;
ans.add(i);
}
int m=scn.nextInt();
for(int i=0;i<m;i++){
int k=scn.nextInt();
int pos=scn.nextInt();
PriorityQueue<Integer> med=new PriorityQueue<Integer>();
for(int j=0;j<k;j++){
med.add(ans.get(j));
}
for(int val=0;val<pos-1;val++){
med.remove();
}
System.out.println(arr[med.remove()]);
}
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t = 1;
while (t--) {
int n, x;
cin >> n;
vector<pair<int, int> > a(n);
for (int i = 0; i < n; i++) {
cin >> x;
a[i] = {-x, i};
}
sort(a.begin(), a.end());
int m;
cin >> m;
while (m--) {
int k, pos;
cin >> k >> pos;
vector<pair<int, int> > v;
for (int i = 0; i < k; i++) v.push_back({a[i].second, -a[i].first});
sort(v.begin(), v.end());
cout << v[pos - 1].second << "\n";
v.clear();
}
return 0;
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
python3
|
n = int(input())
sequence = list(map(int, input().split()))
array = list(sequence)
array.sort(reverse=True)
m = int(input())
for i in range(m):
k, pos = map(int, input().split())
D = dict()
for elem in array[:k]:
if elem in D:
D[elem] += 1
else:
D[elem] = 1
for elem in sequence:
if elem in D and D[elem] != 0:
D[elem] -= 1
if pos == 1:
print(elem)
break
else:
pos -= 1
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
python3
|
n = int(input())
p = list(map(int, input().split()))
p = sorted(list(enumerate(p)), key=lambda x: -x[1])
a = [[]]
for i in range(n):
a.append(sorted(a[-1] + [p[i]], key=lambda x: x[0]))
m = int(input())
for _ in range(m):
x, y = map(int, input().split())
print(a[x][y - 1][1])
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const long long mo = 1e9 + 7;
long long sx, sy, ex, ey, dx[6] = {0, 1, 0, -1, 0, 0},
dy[6] = {1, 0, -1, 0, 0, 0}, m, n, k,
dz[6]{0, 0, 0, 0, -1, 1}, sg;
long long p, no, v, ans, w;
int par[550000];
long long a[500005], b[400005], c[500006], d1[400006], ma[500006], mi[500006];
int lg[200005], mi2[50], mn[25][200005];
long long gcd(long long a, long long b) { return a ? gcd(b % a, a) : b; }
set<int> se;
struct node {
long long u, v, w;
};
vector<node> eg;
long long qu(long long a, long long b, long long m) {
long long ans = 1;
while (b) {
if (b & 1) {
ans = ans % m * a % m;
}
b >>= 1;
a = a % m * a % m;
}
return ans;
}
int su(int n) {
if (n == 1 || n == 0) return 0;
for (int i = 2; i <= sqrt(n); i++) {
if (n % i == 0) return 0;
}
return 1;
}
int fi(int a) {
if (a == par[a])
return a;
else
return par[a] = fi(par[a]);
}
map<int, long long> mp, mp1;
map<pair<long long, int>, int> mp2;
priority_queue<pair<long long, int> > que;
pair<long long, int> a1[400000], a2[400000];
int vis[400055], vis1[400055], vis2[400055];
long long dp[55][1005];
char maze[505][505];
vector<int> g[400000], g1[400000];
vector<pair<long long, int> > ve, ve1, ve2;
int cmp1(pair<long long, int> a, pair<long long, int> b) {
if (a.first == b.first) return a.second < b.second;
return a.first > b.first;
}
int cmp(pair<long long, int> a, pair<long long, int> b) {
return a.second < b.second;
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
string ans1;
int main() {
int t, p2, p3;
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
string s, ss, sss;
long long l, r, n1, u, l1, r1;
int tot = 1;
char ch;
while (cin >> n) {
for (int i = 1; i <= n; i++) {
cin >> a[i];
a1[i] = pair<long long, int>{a[i], i};
}
sort(a1 + 1, a1 + n + 1, cmp1);
cin >> m;
vector<int> tmp;
while (m--) {
cin >> k >> p;
no = 0;
for (int i = 1; i <= k; i++) ve.push_back(a1[i]);
sort(ve.begin(), ve.end(), cmp);
cout << ve[p - 1].first << endl;
ve.clear();
}
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using mii = map<int, int>;
const double Eps = 1e-8;
const int Inf = 1e9 + 9;
const int Mod = 1e9 + 7;
const int N = 1e5 + 9;
const int dx4[4] = {0, 1, 0, -1};
const int dy4[4] = {-1, 0, 1, 0};
const int dx8[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const int dy8[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
constexpr double pi = 3.141592653589793238462643383279502884L;
long long gcd(long long x, long long y) {
if (y == 0) return x;
return gcd(y, x % y);
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
void prime_siever(long long p) {
bool prime[p + 9];
for (long long i = 2; i * i <= p; i++)
if (prime[i] == 0)
for (long long j = i * i; j <= p; j += i) prime[j] = 1;
}
int mul(int a, int b) { return (1LL * a * b) % Mod; }
int add(int a, int b) {
a += b;
if (a >= Mod) a -= Mod;
if (a < 0) a += Mod;
return a;
}
long long bin_power(long long a, long long n) {
if (a == 0) return 0;
long long res = 1;
while (n) {
if (n % 2) {
res = (res * a) % Mod;
n--;
} else {
a = (a * a) % Mod;
n /= 2;
}
}
return res;
}
vector<int> a;
vector<pair<int, int> > b, c;
priority_queue<int, vector<int>, greater<int> > qi;
map<pair<int, int>, int> mp;
void solve() {
int q, n;
cin >> n;
for (int i = 0; i < n; i++) {
int x;
cin >> x;
a.push_back(x);
qi.push(a[i]);
}
cin >> q;
for (int i = 0; i < q; i++) {
int x, y;
cin >> x >> y;
b.push_back({x, y});
}
c = b;
sort(b.rbegin(), b.rend());
for (int i = 0; i < q; i++) {
int siz = b[i].first, idx = b[i].second - 1;
while (qi.size() != siz) {
reverse(a.begin(), a.end());
a.erase(find(a.begin(), a.end(), qi.top()));
reverse(a.begin(), a.end());
qi.pop();
}
mp[{siz, idx + 1}] = a[idx];
}
for (int i = 0; i < q; i++) {
cout << mp[c[i]] << '\n';
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
solve();
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const long long inf = LLONG_MAX - 100000;
void IO() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cout.setf(ios::fixed);
}
long long pw(long long x, long long y, long long p = inf) {
long long res = 1;
x = x % p;
if (x == 0) return 0;
while (y > 0) {
if (y & 1) res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
long long invmod(long long a, long long m = mod) { return pw(a, m - 2, m); }
long long cl(long long a, long long x) {
return a % x == 0 ? a / x : a / x + 1;
}
void run_time_terror(long long case_no = 0) {
long long n, x;
cin >> n;
set<pair<long long, long long> > s;
for (long long i = 0; i < n; ++i) {
cin >> x;
s.insert({-x, i});
}
long long q;
cin >> q;
while (q--) {
long long sz, k;
cin >> sz >> k;
vector<pair<long long, long long> > seq;
for (auto &it : s) {
if (seq.size() >= sz) break;
seq.push_back({it.second, -it.first});
}
sort(seq.begin(), seq.end());
k--;
cout << seq[k].second << "\n";
}
}
int32_t main() {
cout << setprecision(0);
IO();
long long tt = 1;
for (long long case_no = 1; case_no <= tt; case_no++) {
run_time_terror(case_no);
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
python3
|
n = int(input())
A = list(map(int, input().split()))
L = A.copy()
L2 = L.copy()
A.sort()
c = int(input())
r = []
def S(x, L):
idx = L.index(x)
L[idx] = -1
return [idx, x]
for i in range(c):
k, pos = map(int, input().split())
B = A[:0-k-1:-1]
H = list(map(lambda x: S(x, L), B))
H.sort()
N = list(map(lambda x: x[1], H))
r += [N[pos - 1]]
L = L2.copy()
print('\n'.join(map(str, r)))
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
python3
|
def get_index(lt, vec):
max_el = 0
pos = None
for i in range(len(vec)):
if i not in lt:
if max_el < vec[i]:
max_el = vec[i]
pos = i
return pos
n = int(input())
vec = [int(x) for x in input().split()]
lst = [[vec.index(max(vec))]]
for i in range(n-1):
lt = list(lst[i])
ind = get_index(lt, vec)
lt.append(ind)
lst.append(lt)
for l in lst:
l.sort()
m = int(input())
for i in range(m):
k, pos = [int(x) for x in input().split()]
print(vec[lst[k-1][pos-1]])
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0), cout.tie(0), cin.tie(0);
long long int n, m, k, i, j, a[101];
cin >> n;
vector<long long int> pr;
for (i = 0; i < n; i++) {
cin >> a[i];
pr.push_back(a[i]);
}
sort(pr.rbegin(), pr.rend());
cin >> m;
while (m--) {
long long int k, pos, last = -1;
cin >> k >> pos;
pos--;
for (i = k - 1; i >= 0; i--) {
if (pr[i] == pr[i - 1])
;
else
break;
}
int y = k - i, z = -1;
for (i = 0; i < n; i++) {
if (a[i] > pr[k - 1]) {
z++;
} else if (a[i] == pr[k - 1] && y) {
z++, y--;
}
if (z == pos) break;
}
cout << a[i] << endl;
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> a(n, 0), b(n, 0);
for (int i = 0; i < n; ++i) {
cin >> a[i];
b[i] = a[i];
}
sort(b.begin(), b.end());
int m;
cin >> m;
for (int i = 0; i < m; ++i) {
int k, pos;
cin >> k >> pos;
int x = b[n - k];
vector<int> c;
for (int j = 0; j < n && c.size() < pos && k > 0; ++j) {
if (a[j] == x) {
--k;
if (k > 0) x = b[n - k];
c.push_back(a[j]);
} else if (a[j] > x) {
c.push_back(a[j]);
}
}
cout << c[pos - 1] << endl;
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class Main {
static class Pair implements Comparable<Pair> {
int x, i;
public Pair (int x, int i) {
this.x = x;
this.i = i;
}
@Override
public int compareTo(Pair o) {
if (this.x != o.x) return o.x - this.x;
return this.i - o.i;
}
}
public static void main(String[] args) throws IOException {
// Scanner scan = new Scanner(new File("input.txt"));
// PrintWriter out = new PrintWriter(new FileWriter("output.txt"));
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
Pair[] a = new Pair[n];
for (int i = 0; i < n; i++) a[i] = new Pair(scan.nextInt(), i);
Arrays.sort(a);
int m = scan.nextInt();
for (int i = 0; i < m; i++) {
int k = scan.nextInt();
int pos = scan.nextInt();
ArrayList<Integer> now = new ArrayList<>();
for (int j = 0; j < k; j++) now.add(a[j].i);
Collections.sort(now);
for (int j = 0; j < k; j++) {
if (now.get(pos - 1) == a[j].i) {
System.out.println(a[j].x);
break;
}
}
}
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
std::ios::sync_with_stdio(false);
std::cin.tie(NULL);
std::cout.tie(NULL);
long long t = 1;
while (t--) {
long long n;
cin >> n;
vector<long long> ar(n), br(n);
for (long long i = 0; i < n; i++) {
cin >> ar[i];
br[i] = ar[i];
}
sort(br.begin(), br.end(), greater<long long>());
long long m;
cin >> m;
while (m--) {
long long k, pos;
cin >> k >> pos;
vector<long long> seq, arr;
arr = ar;
for (long long i = 0; i < k; i++) {
for (long long j = 0; j < n; j++) {
if (arr[j] == br[i]) {
seq.push_back(j);
arr[j] = -1;
break;
}
}
}
sort(seq.begin(), seq.end());
cout << ar[seq[pos - 1]] << "\n";
}
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e12;
long long a_s, b_s;
vector<long long> best(vector<long long> a, vector<long long> b) {
a_s = 0;
b_s = 0;
bool flaga = true;
bool flagb = true;
for (long long i = 0; i < a.size(); ++i)
if (a[i] < 0) flaga = false;
for (long long i = 0; i < b.size(); ++i)
if (b[i] < 0) flagb = false;
if ((!flaga) && flagb) return b;
if ((!flagb) && flaga) return a;
if ((!flaga) && (!flagb)) {
vector<long long> bad = {};
for (long long i = 0; i < a.size(); ++i) bad.push_back(-INF);
return bad;
}
for (long long i = 0; i < a.size(); ++i) a_s += a[i];
for (long long i = 0; i < a.size(); ++i) b_s += b[i];
if (a_s > b_s) return a;
if (b_s > a_s) return b;
for (long long i = 0; i < a.size(); ++i) {
if (a[i] < b[i]) return a;
if (b[i] < a[i]) return b;
}
return a;
}
signed main() {
long long n;
cin >> n;
vector<long long> a(n);
for (long long i = 0; i < n; ++i) cin >> a[i];
long long m;
cin >> m;
vector<long long> k(m);
vector<long long> pos(m);
for (long long j = 0; j < m; ++j) cin >> k[j] >> pos[j];
vector<vector<vector<long long>>> dp(
n + 1, vector<vector<long long>>(n + 1, vector<long long>()));
dp[0][0] = {};
for (long long i = 1; i <= n; ++i)
for (long long j = 0; j < i; ++j) dp[0][i].push_back(-INF);
for (long long i = 1; i <= n; ++i) {
for (long long j = 1; j <= n; ++j) {
vector<long long> dp11 = dp[i - 1][j - 1];
dp11.push_back(a[i - 1]);
dp[i][j] = best(dp[i - 1][j], dp11);
}
}
for (long long i = 0; i < m; ++i) cout << dp[n][k[i]][pos[i] - 1] << endl;
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
java
|
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.Reader;
import java.util.*;
public class TaskD {
public static String doMain(Reader reader) throws IOException {
MyReader in = new MyReader(reader);
int n = in.nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = in.nextInt();
}
int m = in.nextInt();
while (m > 100) m = in.nextInt();
int[] k = new int[m];
int[] pos = new int[m];
Integer[] order = new Integer[m];
for (int i = 0; i < m; i++) {
k[i] = in.nextInt();
pos[i] = in.nextInt();
order[i] = i;
}
Arrays.sort(order, Comparator.comparingInt(o -> k[o]));
Map<Integer, List<Integer>> map = new TreeMap<>((i1, i2) -> i2 - i1);
for (int i = 0; i < n; i++) {
if (!map.containsKey(a[i])) map.put(a[i], new ArrayList<>());
map.get(a[i]).add(i);
}
int[] newOrder = new int[n];
int last = 0;
for (Map.Entry<Integer, List<Integer>> entry : map.entrySet()) {
for (int i : entry.getValue()) {
newOrder[last++] = i;
}
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i < m; i++) {
int[] ints = Arrays.copyOf(newOrder, k[i]);
Arrays.sort(ints);
sb.append(a[ints[pos[i] - 1]]);
sb.append("\n");
}
return sb.toString();
}
public static void main(String[] args) throws IOException {
String result = doMain(new InputStreamReader(System.in));
System.out.println(result);
}
static class MyReader {
BufferedReader bf;
StringTokenizer st;
String last;
MyReader(Reader reader) throws IOException {
bf = new BufferedReader(reader);
readNextLine();
}
String nextToken() throws IOException {
while (!st.hasMoreTokens()) {
readNextLine();
}
return st.nextToken();
}
void readNextLine() throws IOException {
last = bf.readLine();
if (last == null) last = "";
st = new StringTokenizer(last);
}
String nextLine() throws IOException {
String s = last;
readNextLine();
return s;
}
long nextLong() throws IOException {
return Long.parseLong(nextToken());
}
int nextInt() throws IOException {
return Integer.parseInt(nextToken());
}
double nextDouble() throws IOException {
return Double.parseDouble(nextToken());
}
int[] readIntArray(int n) throws IOException {
int[] answer = new int[n];
for (int i = 0; i < n; ++i) {
answer[i] = nextInt();
}
return answer;
}
long[] readLongArray(int n) throws IOException {
long[] answer = new long[n];
for (int i = 0; i < n; ++i) {
answer[i] = nextLong();
}
return answer;
}
double[] readDoubleArray(int n) throws IOException {
double[] answer = new double[n];
for (int i = 0; i < n; ++i) {
answer[i] = nextDouble();
}
return answer;
}
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
pair<int, int> a[100];
int ind[100];
bool cmp(int i, int j) {
if (a[i].first == a[j].first) return a[i].second < a[j].second;
return a[i].first > a[j].first;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
for (int i = 0; i < n; ++i) {
cin >> a[i].first;
a[i].second = i, ind[i] = i;
}
sort(ind, ind + n, cmp);
int m;
cin >> m;
while (m--) {
int k, pos;
cin >> k >> pos;
vector<int> chosen(k);
for (int i = 0; i < k; ++i) chosen[i] = a[ind[i]].second;
sort(chosen.begin(), chosen.end());
cout << a[chosen[pos - 1]].first << '\n';
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
python3
|
from sys import stdin,stdout
from math import gcd,sqrt,factorial,pi
from collections import deque,defaultdict
input=stdin.readline
R=lambda:map(int,input().split())
I=lambda:int(input())
S=lambda:input().rstrip('\n')
L=lambda:list(R())
P=lambda x:stdout.write(x)
lcm=lambda x,y:(x*y)//gcd(x,y)
hg=lambda x,y:((y+x-1)//x)*x
pw=lambda x:1 if x==1 else 1+pw(x//2)
chk=lambda x:chk(x//2) if not x%2 else True if x==1 else False
sm=lambda x:(x**2+x)//2
N=10**9+7
n=I()
d={}
for i in sorted(enumerate(R()),reverse=True,key=lambda x:x[1]):
if i[1] not in d:
d[i[1]]=[]
d[i[1]]+=i,
d[i[1]].sort(key=lambda x:x[0])
a=[]
for i in d:
a.extend(d[i])
m=[]
x=[]
for i in range(n):
x+=a[i],
x.sort(key=lambda x:x[0])
m.append(x.copy())
for i in range(I()):
k,pos=R()
print(m[k-1][pos-1][1])
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
long long n, m, k, l, r, i, j, a[1000001], d[10000001], ans[1000001];
bool tt[1000001];
vector<pair<long long, long long> > v;
string s;
void build(long long h, long long l, long long r) {
if (l == r) {
d[h] = 0;
return;
}
long long w = (l + r) / 2;
build(h * 2, l, w);
build(h * 2 + 1, w + 1, r);
}
void update(long long h, long long l, long long r, long long x) {
if (l == r) {
d[h] = 1;
return;
}
long long w = (l + r) / 2;
if (x <= w)
update(h * 2, l, w, x);
else
update(h * 2 + 1, w + 1, r, x);
d[h] = d[h * 2] + d[h * 2 + 1];
}
long long get(long long h, long long l, long long r, long long x) {
if (l == r) return a[l];
long long w = (l + r) / 2;
if (x <= d[h * 2])
return get(h * 2, l, w, x);
else
return get(h * 2 + 1, w + 1, r, x - d[h * 2]);
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
v.push_back(make_pair(a[i], n - i));
}
build(1, 1, n);
sort(v.begin(), v.end());
reverse(v.begin(), v.end());
for (int i = 0; i < n; i++) v[i].second = n - v[i].second;
vector<pair<pair<long long, long long>, long long> > z;
cin >> m;
for (int i = 1; i <= m; i++) {
long long x, y;
cin >> x >> y;
z.push_back(make_pair(make_pair(x, y), i));
}
sort(z.begin(), z.end());
long long l = -1;
for (int i = 0; i < z.size(); i++) {
while (z[i].first.first - 1 > l) {
l++;
update(1, 1, n, v[l].second);
}
ans[z[i].second] = get(1, 1, n, z[i].first.second);
}
for (int i = 1; i <= m; i++) cout << ans[i] << "\n";
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
python3
|
m = int(input())
line = [int(i) for i in input().split()]
for i in range(int(input())):
false_line = list(line)
k, pos = map(int, input().split())
while len(false_line) > k:
x = min(false_line)
rang = iter(range(-1, -len(false_line)-1, -1))
j = next(rang)
while j > -len(false_line)-1:
if false_line[j] == x:
del false_line[j]
if len(false_line) == k:
break
else:
try:
j = next(rang)
except StopIteration:
break
print(false_line[pos-1])
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
python3
|
import copy
b=[]
a=[]
rezult=''
n=int(input())
a=list(map(int,input().split()))
m=int(input())
for i in range(1,m+1):
k,pos=map(int,input().split())
b=copy.deepcopy(a)
b.reverse()
for j in range(1,n-k+1):
b.remove(min(b))
b.reverse()
rezult=rezult+'\n'+str(b[pos-1])
print(rezult)
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long int n;
cin >> n;
vector<long long int> a(n), v1(n);
for (long long int i = 0; i < n; i++) cin >> a[i];
v1 = a;
sort(a.begin(), a.end(), greater<long long int>());
long long int q;
cin >> q;
while (q--) {
long long int k, p;
cin >> k >> p;
vector<long long int> v;
long long int sum = 0;
for (long long int i = 0; i < k; i++) sum += a[i];
long long int l1 = 0;
for (long long int i = 0; i < k; i++) {
if (a[i] == a[k - 1]) l1++;
}
for (long long int i = 0; i < n; i++) {
if (v1[i] >= a[k - 1]) v.push_back(v1[i]);
}
vector<long long int> v2, v3;
for (long long int i = 0; i < v.size(); i++) {
if (a[k - 1] == v[i] && l1 > 0)
v3.push_back(v[i]), l1--;
else if (v[i] > a[k - 1])
v3.push_back(v[i]);
}
v2 = v3;
cout << v2[p - 1] << endl;
v.clear();
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int n, A[111];
map<int, set<int>> idxs;
set<int> app;
int main(void) {
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%d", &A[i]);
app.insert(A[i]);
idxs[A[i]].insert(i);
}
int q;
scanf("%d", &q);
while (q--) {
int k, pos;
scanf("%d%d", &k, &pos);
int cnt = 0;
set<int> ii;
for (auto it = app.rbegin(); it != app.rend(); ++it) {
int now = *it;
cnt += idxs[now].size();
if (cnt >= k) {
cnt -= idxs[now].size();
cnt = k - cnt;
auto it = idxs[now].begin();
while (cnt--) {
ii.insert(*it);
++it;
}
break;
}
for (int i : idxs[now]) ii.insert(i);
}
auto it = ii.begin();
while (--pos) ++it;
printf("%d\n", A[*it]);
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
signed main() {
ios ::sync_with_stdio(false);
cin.tie(0);
long long n;
cin >> n;
long long arr[n];
map<long long, long long> mp;
for (long long i = 0; i < n; i++) {
cin >> arr[i];
}
vector<long long> vrr(arr, arr + n);
sort(vrr.rbegin(), vrr.rend());
long long Q;
cin >> Q;
while (Q--) {
long long k, pos;
cin >> k >> pos;
long long d = 0;
mp = {};
for (long long i = 0; i < k; i++) {
mp[vrr[i]] += 1;
d = vrr[i];
}
vector<long long> ans;
for (long long i = 0; i < n; i++) {
if (arr[i] == d && mp[d] > 0) {
ans.push_back(arr[i]);
mp[d] = mp[d] - 1;
}
if (arr[i] > d) {
ans.push_back(arr[i]);
}
}
pos = pos - 1;
cout << ans[pos] << "\n";
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
java
|
import java.io.*;
import java.util.*;
public class Ishu
{
static class Node
{
int data;
int in;
Node() {}
Node(int d, int i)
{
this.data = d;
this.in = i;
}
}
static Scanner scan = new Scanner(System.in);
static BufferedWriter output = new BufferedWriter(new OutputStreamWriter(System.out));
static void update(int[] a, int[] tree, int beg, int end, int st, int pos)
{
if(beg == end && beg == st)
{
a[st]++;
tree[pos]++;
return;
}
int mid = (beg + end) / 2;
if(st <= mid)
update(a, tree, beg, mid, st, 2 * pos);
else
update(a, tree, mid + 1, end, st, 2 * pos + 1);
tree[pos] = tree[2 * pos] + tree[2 * pos + 1];
}
static int query(int[] tree, int beg, int end, int l, int r, int pos)
{
if(r < beg || l > end)
return 0;
if(l <= beg && end <= r)
return tree[pos];
int mid = (beg + end) / 2;
int left = query(tree, beg, mid, l, r, 2 * pos);
int right = query(tree, mid + 1, end, l, r, 2 * pos + 1);
int sum = left + right;
return sum;
}
static void tc() throws Exception
{
int n = scan.nextInt();
List<Node> a = new ArrayList<Node>();
int[] ac = new int[n];
int i;
for(i=0;i<n;++i)
{
int data = scan.nextInt();
ac[i] = data;
int in = i + 1;
a.add(new Node(data, in));
}
Collections.sort(a, new Comparator<Node>(){
public int compare(final Node x, final Node y)
{
if(x.data != y.data)
return x.data - y.data;
else
return y.in - x.in;
}
});
int[] tree = new int[4 * n];
int[] arr = new int[n + 1];
int m = scan.nextInt();
int[][] query = new int[m][4];
for(i=0;i<m;++i)
{
query[i][0] = scan.nextInt();
query[i][1] = scan.nextInt();
query[i][2] = i + 1;
}
Arrays.sort(query, Comparator.comparingInt(o -> o[0]));
int tr = 0;
for(i=n-1;i>=0;--i)
{
Node cur = a.get(i);
int in = cur.in;
update(arr, tree, 1, n, in, 1);
int cnt = n - i;
if(tr == m)
break;
int k = query[tr][0];
if(cnt < k)
continue;
while(tr < m && query[tr][0] == k)
{
int beg = 1;
int end = n;
int mid = (beg + end) / 2;
int data = query[tr][1];
while(beg <= end)
{
mid = (beg + end) / 2;
int res = query(tree, 1, n, 1, mid, 1);
if(res < data)
beg = mid + 1;
else if(res > data)
end = mid - 1;
else
{
query[tr][3] = ac[mid - 1];
end = mid - 1;
}
}
++tr;
}
}
Arrays.sort(query, Comparator.comparingInt(o -> o[2]));
for(i=0;i<m;++i)
output.write(query[i][3] + "\n");
output.flush();
}
public static void main(String[] args) throws Exception
{
int t = 1;
//t = scan.nextInt();
while(t-- > 0)
tc();
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long INF = (long long)1 << 62;
const long long MOD = 1e9 + 7;
const int iINF = 1 << 30;
const double PI = 3.14159265359;
int main() {
int n;
vector<pair<int, int> > a;
cin >> n;
a.assign(n, {0, 0});
for (int i = 0; i < n; i++) {
cin >> a[i].first;
a[i].second = -i;
}
sort(a.rbegin(), a.rend());
int m;
cin >> m;
for (int i = 0; i < m; i++) {
int k, pos;
cin >> k >> pos;
vector<pair<int, int> > tmp;
for (int j = 0; j < k; j++) {
tmp.push_back({-a[j].second, a[j].first});
}
sort(tmp.begin(), tmp.end());
cout << tmp[pos - 1].second << "\n";
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
bool mycomparator(pair<long long int, int> p1, pair<long long int, int> p2) {
if (p1.first == p2.first) {
if (p1.second > p2.second) {
return true;
} else {
return false;
}
}
if (p1.first < p2.first) {
return true;
} else {
return false;
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int n;
cin >> n;
vector<pair<long long int, int> > a;
for (int i = 0; i < n; i++) {
int x;
cin >> x;
a.push_back(make_pair(x, i));
}
sort(a.begin(), a.end(), mycomparator);
int m;
cin >> m;
for (int i = 0; i < m; i++) {
int k, pos;
cin >> k >> pos;
map<long long int, long long int> p;
vector<long long int> l;
for (int i = n - 1; i >= n - k; i--) {
p[a[i].second] = a[i].first;
l.push_back(a[i].second);
}
sort(l.begin(), l.end());
cout << p[l[pos - 1]] << "\n";
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void jakos() {
ios_base::sync_with_stdio(false);
cin.tie(0);
}
const int mod = 1e9 + 7;
const int base = 179;
const int INF = 1e9;
const int N = 1e5;
signed main() {
jakos();
int n;
cin >> n;
vector<pair<int, int>> a;
for (int i = 0; i < n; i++) {
int b;
cin >> b;
a.emplace_back(b, -i);
}
sort(a.begin(), a.end());
reverse(a.begin(), a.end());
int q;
cin >> q;
while (q--) {
vector<pair<int, int>> ans;
int k, pos;
cin >> k >> pos;
for (int i = 0; i < k; i++) {
ans.emplace_back(-a[i].second, a[i].first);
}
sort(ans.begin(), ans.end());
cout << ans[pos - 1].second << endl;
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
constexpr int maxn = 105, inf = 0x3f3f3f3f;
int n;
int a[maxn], vis[maxn], b[maxn];
int main() {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
int m;
cin >> m;
for (int i = 1; i <= m; i++) {
int k, pos;
cin >> k >> pos;
int cnt = 0;
memset(vis, 0, sizeof vis);
for (int j = 1; j <= k; j++) {
int res = 0;
for (int d = 1; d <= n; d++) {
if (vis[d]) continue;
res = max(res, a[d]);
}
for (int d = 1; d <= n; d++) {
if (res == a[d] && !vis[d]) {
vis[d] = 1;
b[++cnt] = d;
break;
}
}
}
sort(b + 1, b + 1 + cnt);
printf("%d\n", a[b[pos]]);
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, m;
cin >> n;
long long x[n + 5];
for (long long i = 0; i < n; i++) {
cin >> x[i];
}
cin >> m;
long long ans = 1e+18;
deque<long long> y, z;
for (long long j = 0; j < m; j++) {
long long k, pos;
cin >> k >> pos;
y.clear();
z.clear();
for (long long i = 0; i < n; i++) {
if (i < k) {
y.push_back(x[i]);
z.push_back(x[i]);
} else {
sort(z.begin(), z.end());
if (x[i] > z[0]) {
for (long long r = k - 1; r >= 0; r--) {
if (y[r] == z[0]) {
y.erase(y.begin() + r);
break;
}
}
z.pop_front();
z.push_back(x[i]);
y.push_back(x[i]);
}
}
}
cout << y[pos - 1] << endl;
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
bool cmp(pair<long long int, int> p, pair<long long int, int> q) {
if (p.first > q.first)
return 1;
else if (p.first < q.first)
return 0;
else {
if (p.second < q.second)
return 1;
else
return 0;
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
long long int a[n];
vector<pair<long long int, int> > v;
for (int i = 0; i < n; i++) {
cin >> a[i];
v.push_back({a[i], i});
}
sort(v.begin(), v.end(), cmp);
int m;
cin >> m;
while (m--) {
int k, pos;
cin >> k >> pos;
vector<int> temp;
for (int i = 0; i < k; i++) temp.push_back(v[i].second);
sort(temp.begin(), temp.end());
cout << a[temp[pos - 1]] << "\n";
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5;
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
if (a.first < b.first) return true;
if (a.first > b.first) return false;
if (a.second < b.second) return false;
return true;
}
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long n;
cin >> n;
vector<long long> a(n);
vector<pair<long long, long long> > b;
for (long long i = 0; i < n; i++) {
cin >> a[i];
b.push_back({a[i], i});
}
sort(b.begin(), b.end(), comp);
long long t;
cin >> t;
while (t--) {
long long k, pos;
cin >> k >> pos;
vector<pair<long long, long long> > c;
long long cnt = 0;
for (long long i = n - 1; i >= 0; i--) {
c.push_back({b[i].second, b[i].first});
cnt++;
if (cnt == k) break;
}
sort(c.begin(), c.end());
cout << c[pos - 1].second << endl;
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const long long mo = 1e9 + 7;
long long sx, sy, ex, ey, dx[6] = {0, 1, 0, -1, 0, 0},
dy[6] = {1, 0, -1, 0, 0, 0}, m, n, k,
dz[6]{0, 0, 0, 0, -1, 1}, sg;
long long p, no, v, ans, w;
int par[550000];
long long a[500005], b[400005], c[500006], d1[400006], ma[500006], mi[500006];
int lg[200005], mi2[50], mn[25][200005];
long long gcd(long long a, long long b) { return a ? gcd(b % a, a) : b; }
set<int> se;
struct node {
long long u, v, w;
};
vector<node> eg;
long long qu(long long a, long long b, long long m) {
long long ans = 1;
while (b) {
if (b & 1) {
ans = ans % m * a % m;
}
b >>= 1;
a = a % m * a % m;
}
return ans;
}
int su(int n) {
if (n == 1 || n == 0) return 0;
for (int i = 2; i <= sqrt(n); i++) {
if (n % i == 0) return 0;
}
return 1;
}
int fi(int a) {
if (a == par[a])
return a;
else
return par[a] = fi(par[a]);
}
map<int, long long> mp, mp1;
map<pair<long long, int>, int> mp2;
priority_queue<pair<long long, int> > que;
pair<long long, int> a1[400000], a2[400000];
int vis[400055], vis1[400055], vis2[400055];
long long dp[55][1005];
char maze[505][505];
vector<int> g[400000], g1[400000];
vector<long long> ve, ve1, ve2;
int cmp1(pair<long long, int> a, pair<long long, int> b) {
return a.second < b.second;
}
int cmp(node a, node b) { return a.w < b.w; }
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
string ans1;
void dfs(int x, long long d) {
ans = max(ans, d);
for (int i = 2 * x; i <= n; i += x) {
if (a[i] > a[x]) dfs(i, d + 1);
}
}
int main() {
int t, p2, p3;
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
string s, ss, sss;
long long l, r, n1, u, l1, r1;
int tot = 1;
char ch;
while (cin >> n) {
for (int i = 1; i <= n; i++) {
cin >> a[i];
a1[i] = pair<long long, int>{a[i], i};
}
sort(a1 + 1, a1 + n + 1);
for (int i = n; i >= 1; i--) {
b[n - i + 1] = a1[i].first;
}
cin >> m;
vector<int> tmp;
while (m--) {
cin >> k >> p;
no = 0;
for (int i = 1; i <= n; i++) {
no = 0;
for (int i1 = 1; i1 <= k; i1++) mp[b[i1]]++;
for (int j = a1[i].second; j <= n; j++) {
if (mp[a[j]]) mp[a[j]]--, no++, tmp.push_back(j);
if (no == k) break;
}
if (no == k) {
cout << a[tmp[p - 1]] << endl;
break;
}
tmp.clear();
mp.clear();
}
tmp.clear();
mp.clear();
}
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const int inf = 1e9 + 7;
bool compA(pair<int, int> a, pair<int, int> b) {
if (a.first == b.second) return a.second < b.second;
return a.first > b.first;
}
bool comp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
void run() {
int n, m, k, pos;
cin >> n;
vector<pair<int, int> > A(n);
for (int i = 0; i < n; ++i) {
cin >> A[i].first;
A[i].second = i;
}
cin >> m;
sort(A.begin(), A.end(), compA);
vector<vector<pair<int, int> > > B(n);
for (int i = 0; i < n; ++i) {
for (int j = 0; j < i + 1; ++j) B[i].push_back(A[j]);
sort(B[i].begin(), B[i].end(), comp);
}
for (int i = 0; i < m; ++i) {
cin >> k >> pos;
cout << B[--k][--pos].first << '\n';
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t = 1;
for (int i = 0; i < t; ++i) {
run();
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
java
|
import java.io.BufferedReader;
// import java.io.FileInputStream;
// import java.io.FileOutputStream;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.Random;
import java.util.StringTokenizer;
import static java.lang.Math.abs;
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.round;
import static java.util.Arrays.copyOf;
import static java.util.Arrays.fill;
import static java.util.Arrays.sort;
import static java.util.Collections.reverseOrder;
import static java.util.Collections.sort;
import static java.util.Comparator.comparingInt;
public class Solution {
FastScanner in;
PrintWriter out;
class Pair {
int a, id;
Pair(int a, int id) {
this.a = a;
this.id = id;
}
}
class Triplet {
int k, pos, id;
Triplet(int k, int pos, int id) {
this.k = k;
this.pos = pos;
this.id = id;
}
}
private void solve() throws IOException {
int n = in.nextInt();
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = in.nextInt();
Pair[] p = new Pair[n];
for (int i = 0; i < n; i++)
p[i] = new Pair(a[i], i);
sort(p, (o1, o2) -> o1.a != o2.a ? o2.a - o1.a : o1.id - o2.id);
int m = in.nextInt();
Triplet[] t = new Triplet[m];
for (int i = 0; i < m; i++)
t[i] = new Triplet(in.nextInt() - 1, in.nextInt() - 1, i);
sort(t, comparingInt(o -> o.k));
int[] ans = new int[m];
AVLTreePBDS tree = new AVLTreePBDS(false);
for (int i = 0, j = 0; i < n; i++) {
tree.add(p[i].id);
while (j < m && t[j].k == i) {
ans[t[j].id] = a[tree.findByOrder(t[j].pos)];
j++;
}
}
for (int i = 0; i < n; i++)
out.println(ans[i]);
}
class AVLTreePBDS {
private Node root;
private boolean multi;
AVLTreePBDS(boolean multi) {
this.root = null;
this.multi = multi;
}
int size() {
return size(root);
}
boolean isEmpty() {
return size(root) == 0;
}
boolean contains(int key) {
return contains(root, key);
}
void add(int key) {
root = add(root, key);
}
void remove(int key) {
root = remove(root, key);
}
Integer first() {
Node min = findMin(root);
return min != null ? min.key : null;
}
Integer last() {
Node max = findMax(root);
return max != null ? max.key : null;
}
Integer poolFirst() {
Node min = findMin(root);
if (min != null) {
remove(min.key);
return min.key;
}
return null;
}
Integer poolLast() {
Node max = findMax(root);
if (max != null) {
remove(max.key);
return max.key;
}
return null;
}
// min >= key
Integer ceiling(int key) {
return contains(key) ? key : higher(key);
}
// max <= key
Integer floor(int key) {
return contains(key) ? key : lower(key);
}
// min > key
Integer higher(int key) {
Node min = higher(root, key);
return min == null ? null : min.key;
}
private Node higher(Node cur, int key) {
if (cur == null)
return null;
if (cur.key <= key)
return higher(cur.right, key);
// cur.key > key
Node left = higher(cur.left, key);
return left == null ? cur : left;
}
// max < key
Integer lower(int key) {
Node max = lower(root, key);
return max == null ? null : max.key;
}
private Node lower(Node cur, int key) {
if (cur == null)
return null;
if (cur.key >= key)
return lower(cur.left, key);
// cur.key < key
Node right = lower(cur.right, key);
return right == null ? cur : right;
}
private class Node {
int key, height, size;
Node left, right;
Node(int key) {
this.key = key;
height = size = 1;
left = right = null;
}
}
private int height(Node cur) {
return cur == null ? 0 : cur.height;
}
private int balanceFactor(Node cur) {
return height(cur.right) - height(cur.left);
}
private int size(Node cur) {
return cur == null ? 0 : cur.size;
}
// fixVertex
private void fixHeightAndSize(Node cur) {
cur.height = max(height(cur.left), height(cur.right)) + 1;
cur.size = size(cur.left) + size(cur.right) + 1;
}
private Node rotateRight(Node cur) {
Node prevLeft = cur.left;
cur.left = prevLeft.right;
prevLeft.right = cur;
fixHeightAndSize(cur);
fixHeightAndSize(prevLeft);
return prevLeft;
}
private Node rotateLeft(Node cur) {
Node prevRight = cur.right;
cur.right = prevRight.left;
prevRight.left = cur;
fixHeightAndSize(cur);
fixHeightAndSize(prevRight);
return prevRight;
}
private Node balance(Node cur) {
fixHeightAndSize(cur);
if (balanceFactor(cur) == 2) {
if (balanceFactor(cur.right) < 0)
cur.right = rotateRight(cur.right);
return rotateLeft(cur);
}
if (balanceFactor(cur) == -2) {
if (balanceFactor(cur.left) > 0)
cur.left = rotateLeft(cur.left);
return rotateRight(cur);
}
return cur;
}
private boolean contains(Node cur, int key) {
if (cur == null)
return false;
else if (key < cur.key)
return contains(cur.left, key);
else if (key > cur.key)
return contains(cur.right, key);
else
return true;
}
private Node add(Node cur, int key) {
if (cur == null)
return new Node(key);
if (key < cur.key)
cur.left = add(cur.left, key);
else if (key > cur.key || multi)
cur.right = add(cur.right, key);
return balance(cur);
}
private Node findMin(Node cur) {
return cur.left != null ? findMin(cur.left) : cur;
}
private Node findMax(Node cur) {
return cur.right != null ? findMax(cur.right) : cur;
}
private Node removeMin(Node cur) {
if (cur.left == null)
return cur.right;
cur.left = removeMin(cur.left);
return balance(cur);
}
private Node removeMax(Node cur) {
if (cur.right == null)
return cur.left;
cur.right = removeMax(cur.right);
return balance(cur);
}
private Node remove(Node cur, int key) {
if (cur == null)
return null;
if (key < cur.key)
cur.left = remove(cur.left, key);
else if (key > cur.key)
cur.right = remove(cur.right, key);
else { // k == cur.key
Node prevLeft = cur.left;
Node prevRight = cur.right;
if (prevRight == null)
return prevLeft;
Node min = findMin(prevRight);
min.right = removeMin(prevRight);
min.left = prevLeft;
return balance(min);
}
return balance(cur);
}
int orderOfKey(int key) {
return orderOfKey(root, key);
}
// count < key
private int orderOfKey(Node cur, int key) {
if (cur == null)
return 0;
if (cur.key < key)
return size(cur.left) + 1 + orderOfKey(cur.right, key);
if (cur.key == key)
return size(cur.left);
// cur.key > key
return orderOfKey(cur.left, key);
}
Integer findByOrder(int pos) {
return size(root) > pos ? findByOrder(root, pos) : null;
}
// get i-th
private int findByOrder(Node cur, int pos) {
if (size(cur.left) > pos)
return findByOrder(cur.left, pos);
if (size(cur.left) == pos)
return cur.key;
// size(cur.left) < pos
return findByOrder(cur.right, pos - 1 - size(cur.left));
}
}
class FastScanner {
StringTokenizer st;
BufferedReader br;
FastScanner(InputStream s) {
br = new BufferedReader(new InputStreamReader(s));
}
String next() throws IOException {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
}
boolean hasNext() throws IOException {
return br.ready() || (st != null && st.hasMoreTokens());
}
int nextInt() throws IOException {
return Integer.parseInt(next());
}
long nextLong() throws IOException {
return Long.parseLong(next());
}
double nextDouble() throws IOException {
return Double.parseDouble(next());
}
String nextLine() throws IOException {
return br.readLine();
}
boolean hasNextLine() throws IOException {
return br.ready();
}
}
private void run() throws IOException {
in = new FastScanner(System.in); // new FastScanner(new FileInputStream(".in"));
out = new PrintWriter(System.out); // new PrintWriter(new FileOutputStream(".out"));
solve();
out.flush();
out.close();
}
public static void main(String[] args) throws IOException {
new Solution().run();
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void func() {
int n;
cin >> n;
int arr[n], acop[n];
for (int i = 0; i < n; ++i) {
cin >> arr[i];
acop[i] = arr[i];
}
sort(acop, acop + n, greater<int>());
int q;
cin >> q;
for (int i = 0; i < q; ++i) {
int k, pos, cnt = 0;
cin >> k >> pos;
for (int i = 0; i < n; ++i) {
if (arr[i] >= acop[k - 1]) {
++cnt;
}
if (cnt == pos) {
cout << arr[i] << "\n";
break;
}
}
}
}
int main() {
int t = 1;
int cnt = 0;
while (t--) {
func();
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#define _CRT_SECURE_NO_WARNINGS
#include <algorithm>
#define N 100000
int n, m;
int arg[N], s[N], d[N];
int main()
{
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
scanf("%d",&n); for(int i=0; i<n; ++i)scanf("%d",arg+i), d[i]=i;
std::sort(d+0,d+n,[](int a, int b)->bool { return arg[a]<arg[b]; });
scanf("%d",&m); for(int t=0; t<m; ++t)
{
int k,ke, pos; scanf("%d%d",&k,&pos); ke=k;
int beg=int(std::lower_bound(d+0,d+n,arg[d[n-k]],[](int a, int val)->bool { return arg[a]<val; })-d);
int end=int(std::upper_bound(d+0,d+n,arg[d[n-k]],[](int val, int a)->bool { return val<arg[a]; })-d);
int uno=n-end; for(int i=0;i<uno;++i)s[i]=d[n-i-1]; for(int i=0;i<k-uno;++i)s[i+uno]=d[beg+i];
std::sort(s+0,s+k); printf("%d\n",arg[s[pos-1]]);
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void jakos() {
ios_base::sync_with_stdio(false);
cin.tie(0);
}
const long long mod = 1e9 + 7;
const long long base = 179;
const long long INF = 1e9;
const long long N = 1e5;
bool isBest(long long k, long long i1, long long i2, vector<long long>& a) {
for (long long i = 0; i < k; i++) {
if (a[i1] < a[i2]) {
return true;
} else if (a[i1] == a[i2]) {
i1++;
i2++;
} else {
return false;
}
}
return false;
}
signed main() {
jakos();
long long n;
cin >> n;
vector<long long> a(n);
vector<long long> pref(n + 1);
pref[0] = 0;
for (long long i = 0; i < n; i++) {
cin >> a[i];
}
for (long long i = 1; i <= n; i++) {
pref[i] = pref[i - 1] + a[i - 1];
}
long long t;
cin >> t;
while (t--) {
long long k, pos;
cin >> k >> pos;
pos--;
vector<long long> ans(k);
long long mx = 0;
long long ansi = -1;
long long ansj = -1;
for (long long i = 0; i < n - k + 1; i++) {
if (pref[i + k] - pref[i] > mx) {
mx = pref[i + k] - pref[i];
ansi = i;
ansj = i + k;
} else if (pref[i + k] - pref[i] == mx && isBest(k, i, ansi, a)) {
mx = pref[i + k] - pref[i];
ansi = i;
ansj = i + k;
}
}
cout << a[ansi + pos] << endl;
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
#define PI 3.1415926535
const long long int MOD = 2019;
#define pb push_back
#define mp make_pair
#define fill(a) memset(a, 0, sizeof (a))
#define SORT(v) sort(v.begin(), v.end())
#define SORTR(v) sort(v.rbegin(), v.rend())
#define MAX(v) *max_element(v.begin(), v.end())
#define MIN(v) *min_element(v.begin(), v.end())
#define FAST ios_base::sync_with_stdio(false);cin.tie(NULL);
#define watch(x) cout << (#x) << " is " << (x) << endl;
const long long int MAXX = 1e6+5;
const long long int MINN = 2e5 + 5;
const long long int inf = 1e7;
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<int,ii> iii;
typedef vector<ii> vii;
typedef vector<vii> vvii;
typedef vector< iii > viii;
ll lcm(ll a, ll b) {
return a / __gcd(a, b) * b;
}
int XOR(int x, int y){
return (x | y) & (~x | ~y);
}
bool cmp(const pair<int,int> &a,const pair<int,int> &b) {
return (a.second < b.second);
}
// bool isl = binary_search(x.begin(), x.end(), 5);
// __builtin_popcount(x) - Returns the number of set bits in x
const long long int N = 10050;
const int MAXN = 105;
int binsearch(int lo, int hi){
while(lo<hi){
int mid=(lo+hi)/2;
if((mid)) // check(mid)
hi=mid;
else
lo=mid+1;
}
return lo;
}
int power(int x,int y){
int res=1;
while(y>0){
if(y&1) res=((res*x));
y/=2;
x=((x*x));
}
return res;
}
bool cmp2(const pair<int,int> &a,const pair<int,int> &b) {
if(a.first!=b.first)
return (a.first > b.first);
else
return (b.first < b.second);
}
int main(){
FAST
int n;
cin>>n;
int a[n+1];
vii v;
for(int i=1;i<=n;i++){
cin>>a[i];
v.pb({a[i],i});
}
sort(v.begin(),v.end(),cmp2);
int m;
cin>>m;
while (m--){
int k,pos;
cin>>k>>pos;
vector <int> temp;
for (int i=0;i<k;i++)
temp.pb(v[i].second);
sort(temp.begin(),temp.end());
cout<<a[temp[pos-1]]<<"\n";
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
java
|
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.*;
public class OptimalSubsequences {
InputStream is;
PrintWriter pw;
String INPUT = "";
long L_INF = (1L << 60L);
void solve() {
int k,pos,n=ni(), m;
int a[] = na(n);
m = ni();
List<Integer> cache[] = new ArrayList[n];
while(m-->0){
k=ni();
pos = ni();
if(cache[k-1]==null){
cache[k-1] = calc(a,k);
}
pw.println(cache[k-1].get(pos-1));
}
}
private List<Integer> calc(int[] a, int k) {
PriorityQueue<int[]> pq = new PriorityQueue<>(new Comparator<int[]>() {
@Override
public int compare(int[] o1, int[] o2) {
return Integer.compare(o1[0],o2[0]);
}
});
for (int i = 0; i < a.length; i++) {
if(pq.size()<k){
pq.add(new int[]{a[i],i});
}
else if(pq.size()==k && pq.peek()[0]<a[i]) {
pq.poll();
pq.add(new int[]{a[i], i});
}
}
TreeMap<Integer,Integer> mp = new TreeMap<>();
while(!pq.isEmpty()){
int t[] = pq.poll();
mp.put(t[1],t[0]);
}
List<Integer> ans = new ArrayList<>();
for (Integer value : mp.values()) {
ans.add(value);
}
return ans;
}
void run() throws Exception {
// is = oj ? System.in : new ByteArrayInputStream(INPUT.getBytes());
is = System.in;
pw = new PrintWriter(System.out);
long s = System.currentTimeMillis();
// int t = ni();
// while (t-- > 0)
solve();
pw.flush();
tr(System.currentTimeMillis() - s + "ms");
}
public static void main(String[] args) throws Exception {
new OptimalSubsequences().run();
}
private byte[] inbuf = new byte[1024];
private int lenbuf = 0, ptrbuf = 0;
private int readByte() {
if (lenbuf == -1) throw new InputMismatchException();
if (ptrbuf >= lenbuf) {
ptrbuf = 0;
try {
lenbuf = is.read(inbuf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (lenbuf <= 0) return -1;
}
return inbuf[ptrbuf++];
}
private boolean isSpaceChar(int c) {
return !(c >= 33 && c <= 126);
}
private int skip() {
int b;
while ((b = readByte()) != -1 && isSpaceChar(b)) ;
return b;
}
private double nd() {
return Double.parseDouble(ns());
}
private char nc() {
return (char) skip();
}
private String ns() {
int b = skip();
StringBuilder sb = new StringBuilder();
while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
private char[] ns(int n) {
char[] buf = new char[n];
int b = skip(), p = 0;
while (p < n && !(isSpaceChar(b))) {
buf[p++] = (char) b;
b = readByte();
}
return n == p ? buf : Arrays.copyOf(buf, p);
}
private char[][] nm(int n, int m) {
char[][] map = new char[n][];
for (int i = 0; i < n; i++) map[i] = ns(m);
return map;
}
private int[] na(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = ni();
return a;
}
private int ni() {
int num = 0, b;
boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') {
minus = true;
b = readByte();
}
while (true) {
if (b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
private long nl() {
long num = 0;
int b;
boolean minus = false;
while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-')) ;
if (b == '-') {
minus = true;
b = readByte();
}
while (true) {
if (b >= '0' && b <= '9') {
num = num * 10 + (b - '0');
} else {
return minus ? -num : num;
}
b = readByte();
}
}
private boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private void tr(Object... o) {
if (!oj) System.out.println(Arrays.deepToString(o));
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
python3
|
from sys import stdin,stdout
from math import gcd,sqrt,factorial,pi
from collections import deque,defaultdict
input=stdin.readline
R=lambda:map(int,input().split())
I=lambda:int(input())
S=lambda:input().rstrip('\n')
L=lambda:list(R())
P=lambda x:stdout.write(x)
lcm=lambda x,y:(x*y)//gcd(x,y)
hg=lambda x,y:((y+x-1)//x)*x
pw=lambda x:1 if x==1 else 1+pw(x//2)
chk=lambda x:chk(x//2) if not x%2 else True if x==1 else False
sm=lambda x:(x**2+x)//2
N=10**9+7
n=I()
*a,=R()
d={}
for i in a:
if i not in d:d[i]=0
d[i]+=1
for _ in range(I()):
k,p=R()
m=[]
cnt=0
for i in sorted(d,reverse=True):
cnt+=d[i]
m+=i,
if cnt>=k:break
m=[i for i in a if i in m]
ans=m[:k]
sm=sum(ans)
for i in range(k,len(m)):
v=sum(m[k-i+1:i+1])
if v>sm:
sm=v
ans=m[k-i+1:i+1]
elif v==sm and m[k-i+1:i+1]<ans:
ans=m[k-i+1:i+1]
print(ans[p-1])
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
python3
|
n = int(input())
a = list(map(int, input().split()))
a1 = sorted(a, key=lambda x: -x)
m = int(input())
for i in range(m):
k, p = list(map(int, input().split()))
c = {}
s = []
for j in range(n+1):
s.append({})
pos = {}
for j in range(n):
c[a1[j]] = 0
s[0][a[j]] = 0
pos[a1[j]] = []
for j in range(n):
s[0][a[j]] += 1
for j in range(n):
for t in range(n):
s[j+1][a[t]] = s[j][a[t]]
s[j+1][a[j]] -= 1
b = []
for j in range(k):
c[a1[j]]+=1
ns = []
for j in range(n):
if c[a[j]] ^ 0:
pos[a[j]].append(j)
if len(pos[a[j]]) == 1:
ns.append(a[j])
ns = sorted(ns)
counter = 0
ans = []
minpos = 0
while counter ^ k:
flag = 0
for j in range(len(ns)):
c[ns[j]] -= 1
flag1 = 0
for f in range(len(pos[ns[j]])):
if minpos <= pos[ns[j]][f]:
for u in range(len(ns)):
if s[pos[ns[j]][f]][ns[u]] < c[ns[u]]:
flag1 = 1
break
if flag1 == 1:
break
else:
minpos = pos[ns[j]][f]+1
flag = 1
break
if flag == 1:
ans.append(ns[j])
counter += 1
break
else:
c[ns[j]] += 1
print(ans[p-1])
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
vector<int> srr;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int arr[200], n, m, pj, k, fast = 1, last;
cin >> n;
for (int i = 1; i <= n; ++i) {
cin >> arr[i];
srr.push_back(arr[i]);
}
sort(srr.begin(), srr.end());
cin >> m;
last = n;
while (m--) {
vector<int> vec;
cin >> k >> pj;
fast = k;
for (int i = srr.size() - 1; fast && i >= 0; --i) {
vec.push_back(srr[i]);
--fast;
}
vector<int> ans;
sort(vec.begin(), vec.end());
fast = 1;
for (int s = 1; s <= k; ++s) {
for (int i = 0; i < vec.size(); ++i) {
if (vec[i] != -1) {
for (int j = fast; j <= last; ++j) {
if (arr[j] == vec[i]) {
map<int, int> occ;
for (int p = 0; p < vec.size(); ++p) {
if (vec[p] != -1) occ[vec[p]]++;
}
for (int p = j; p <= last; ++p) {
if (occ[arr[p]]) occ[arr[p]]--;
}
bool pos = true;
for (int p = 0; p < vec.size() && pos; ++p)
if (vec[p] != -1 && occ[vec[p]]) pos = false;
if (pos) fast = j + 1;
if (pos) {
ans.push_back(vec[i]);
vec[i] = -1;
}
break;
}
}
}
}
}
cout << ans[pj - 1] << endl;
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
vector<vector<long long>> adj;
map<long, bool> vis, viss;
vector<long long> rnk, parent, sz;
int spf[1000000 + 1];
long long binpow(long long a, long long b, long long m) {
a %= m;
long long res = 1;
while (b > 0) {
if (b & 1) res = res * a % m;
a = a * a % m;
b >>= 1;
}
return res;
}
vector<long long> v;
bool yg(long long a, long long b) { return v[a] > v[b]; }
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(nullptr);
long long t, temp;
long long n, m;
cin >> n;
vector<long long> indx;
for (long long i = 0; i < n; i++) {
cin >> temp;
v.push_back(temp);
indx.push_back(i);
}
sort(indx.begin(), indx.end(), yg);
cin >> m;
while (m--) {
long long k, pos;
cin >> k >> pos;
vector<long long> arr;
for (long long i = 0; i < k; i++) arr.push_back(indx[i]);
sort(arr.begin(), arr.end());
cout << v[arr[pos - 1]] << "\n";
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > dp(1000);
bool cmp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
int main() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int n;
cin >> n;
int value[n];
for (int i = 0; i < n; i++) {
cin >> value[i];
}
for (int k = 1; k <= n; k++) {
multiset<pair<int, int> > st;
for (int i = 0; i < n; i++) {
if (st.size() < k) {
st.insert(pair<int, int>(value[i], i + 1));
} else {
if (value[i] > (*st.begin()).first) {
st.erase(st.begin());
st.insert(pair<int, int>(value[i], i + 1));
}
}
}
vector<pair<int, int> > ps;
for (multiset<pair<int, int> >::iterator it = st.begin(); it != st.end();
it++) {
ps.push_back(*it);
}
sort(ps.begin(), ps.end(), cmp);
vector<int> ps1;
for (int i = 0; i < ps.size(); i++) {
ps1.push_back(ps[i].first);
}
dp[k] = ps1;
}
int q;
cin >> q;
for (int i = 0; i < q; i++) {
int a, b;
cin >> a >> b;
cout << dp[a][b - 1] << '\n';
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(time(0));
bool cmp2(const vector<long long> &a, const vector<long long> &b) {
for (long long i = 0; i < a.size(); ++i) {
if (a[i] < b[i])
return 1;
else
return 0;
}
return 0;
}
bool comp(pair<long long, vector<long long> > a,
pair<long long, vector<long long> > b) {
if (a.first == b.first) return cmp2(a.second, b.second);
return a.first > b.first;
}
void solve() {
long long n;
cin >> n;
long long a[n];
for (long long i = 0; i < n; ++i) {
cin >> a[i];
}
pair<long long, vector<long long> > dp[n][n];
for (long long i = 0; i < n; ++i) {
for (long long j = 0; j < n; ++j) {
if (j == 0) {
dp[i][j] = {a[i], {a[i]}};
if (i) {
if (comp(dp[i - 1][j], dp[i][j])) {
dp[i][j] = dp[i - 1][j];
}
}
} else {
dp[i][j] = {long long(-1e15), {a[i]}};
for (long long h = 0; h < i; ++h) {
pair<long long, vector<long long> > x = {dp[h][j - 1].first + a[i],
dp[h][j - 1].second};
x.second.push_back(a[i]);
if (comp(x, dp[i][j])) {
dp[i][j] = x;
}
}
if (i) {
if (comp(dp[i - 1][j], dp[i][j])) {
dp[i][j] = dp[i - 1][j];
}
}
}
}
}
long long q;
cin >> q;
while (q--) {
long long k, pos;
cin >> k >> pos;
k--;
pos--;
cout << dp[n - 1][k].second[pos] << endl;
}
}
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
long long t = 1;
while (t--) {
solve();
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
const int LG = 21;
const int N = 400005;
const long long MOD = 1e9 + 7;
const long long INF = 1e9;
const long long INFLL = 1e18;
using namespace std;
int cx[4] = {-1, 0, 1, 0};
int cy[4] = {0, -1, 0, 1};
string Yes[2] = {"No", "Yes"};
string YES[2] = {"NO", "YES"};
long long inq(long long x, long long y) {
if (!y) return 1 % MOD;
long long l = inq(x, y / 2);
if (y % 2) return l * l % MOD * x % MOD;
return l * l % MOD;
}
long long rev(long long x) { return inq(x, MOD - 2); }
bool __precomputed_combinatorics = 0;
vector<long long> __fact, __ufact, __rev;
void __precompute_combinatorics() {
__precomputed_combinatorics = 1;
__fact.resize(N);
__ufact.resize(N);
__rev.resize(N);
__rev[1] = 1;
for (int i = 2; i < N; i++) __rev[i] = MOD - __rev[MOD % i] * (MOD / i) % MOD;
__fact[0] = 1, __ufact[0] = 1;
for (int i = 1; i < N; i++)
__fact[i] = __fact[i - 1] * i % MOD,
__ufact[i] = __ufact[i - 1] * __rev[i] % MOD;
}
long long fact(int x) {
if (!__precomputed_combinatorics) __precompute_combinatorics();
return __fact[x];
}
long long cnk(int n, int k) {
if (k < 0 || k > n) return 0;
if (!__precomputed_combinatorics) __precompute_combinatorics();
return __fact[n] * __ufact[n - k] % MOD * __ufact[k] % MOD;
}
int Root(int x, vector<int> &root) {
if (x == root[x]) return x;
return root[x] = Root(root[x], root);
}
void Merge(int v, int u, vector<int> &root, vector<int> &sz) {
v = Root(v, root), u = Root(u, root);
if (v == u) return;
if (sz[v] < sz[u]) {
sz[u] += sz[v];
root[v] = u;
} else {
sz[v] += sz[u];
root[u] = v;
}
}
int ok(int x, int n) { return 0 <= x && x < n; }
void bfs(int v, vector<int> &dist, vector<vector<int> > &graph) {
fill((dist).begin(), (dist).end(), -1);
dist[v] = 0;
vector<int> q = {v};
for (int i = 0; i < q.size(); i++) {
for (auto u : graph[q[i]]) {
if (dist[u] == -1) {
dist[u] = dist[q[i]] + 1;
q.push_back(u);
}
}
}
}
vector<int> z_func(string &s) {
vector<int> z(s.size());
z[0] = s.size();
int L = 0, R = 0;
for (int i = 1; i < s.size(); i++) {
z[i] = max(0, min(z[i - L], R - i));
while (i + z[i] < s.size() && s[i + z[i]] == s[z[i]]) z[i]++;
if (i + z[i] > R) {
R = i + z[i];
L = i;
}
}
return z;
}
vector<int> p_func(string &s) {
vector<int> p(s.size());
for (int i = 1; i < s.size(); i++) {
int j = p[i - 1];
while (j > 0 && s[i] != s[j]) j = p[j - 1];
if (s[i] == s[j]) j++;
p[i] = j;
}
return p;
}
vector<int> d1_func(string &s) {
vector<int> d1(s.size());
int L = 0, R = -1;
for (int i = 0; i < s.size(); i++) {
int k = 0;
if (i <= R) k = min(R - i + 1, d1[R - i + L]);
while (i + k < s.size() && i - k >= 0 && s[i - k] == s[i + k]) k++;
d1[i] = k--;
if (i + k > R) {
L = i - k;
R = i + k;
}
}
return d1;
}
vector<int> d2_func(string &s) {
vector<int> d2(s.size());
int L = 0, R = -1;
for (int i = 1; i < s.size(); i++) {
int k = 0;
if (i <= R) k = min(R - i + 1, d2[R - i + L + 1]);
while (i + k < s.size() && i - k - 1 >= 0 && s[i - k - 1] == s[i + k]) k++;
d2[i] = k--;
if (i + k > R) {
L = i - k - 1;
R = i + k;
}
}
return d2;
}
long long log10(long long x) {
if (x < 10) return 1;
return 1 + log10(x / 10);
}
long long ds(long long x) {
if (x < 10) return x;
return x % 10 + ds(x / 10);
}
double sqr(double x) { return x * x; }
bool in(int bit, int mask) { return (mask & (1 << bit)) > 0; }
void Del(vector<int> &v, int pos) {
swap(v[pos], v[v.size() - 1]);
v.pop_back();
}
long long g(vector<long long> &p, int pos) {
if (ok(pos, p.size())) return p[pos];
if (pos < 0 || p.size() == 0) return 0;
return p.back();
}
int g(vector<int> &p, int pos) {
if (ok(pos, p.size())) return p[pos];
if (pos < 0 || p.size() == 0) return 0;
return p.back();
}
int n, q;
int a[N];
pair<int, int> s[N];
struct qu {
int len, pos, id;
};
int ans[N];
qu ask[N];
bool comp(qu a, qu b) { return a.len < b.len; }
int ar[N];
void On(int x) { ar[x] = 1; }
int Find(int pos) {
for (int i = 0; i < n; i++) {
pos -= ar[i];
if (pos == 0) {
return a[i];
}
}
}
signed main() {
srand(time(NULL));
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n;
for (int(i) = 0; (i) != (n); (i)++) {
cin >> a[i];
s[i] = {a[i], i};
}
sort(s, s + n);
reverse(s, s + n);
cin >> q;
for (int i = 0; i < q; i++) {
cin >> ask[i].len >> ask[i].pos;
ask[i].id = i;
}
sort(ask, ask + q, comp);
int curlen = 0;
for (int i = 0; i < q; i++) {
while (curlen < ask[i].len) {
On(s[i].second);
curlen++;
}
ans[ask[i].id] = Find(ask[i].pos);
}
for (int(i) = 0; (i) != (q); (i)++) {
cout << ans[i] << "\n";
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
bool sortinrev(const pair<long long int, long long int> &a,
const pair<long long int, long long int> &b) {
return (a.first > b.first);
}
int main() {
long long int i, j, k, l, m, n;
cin >> n;
long long int a[n];
vector<pair<long long int, long long int>> vp;
for (i = 0; i < n; i++) {
cin >> a[i];
vp.push_back(make_pair(a[i], i));
}
sort(vp.begin(), vp.end(), sortinrev);
long long int var;
cin >> var;
while (var--) {
cin >> k >> l;
vector<long long int> v;
for (i = 0; i < k; i++) v.push_back(vp[i].second);
sort(v.begin(), v.end());
cout << a[v[l - 1]] << endl;
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void solve() {
long long n;
cin >> n;
vector<long long> v(n), so(n);
for (long long i = 0; i < n; ++i) {
cin >> v[i];
so[i] = v[i];
}
sort(so.begin(), so.end());
long long m;
cin >> m;
for (long long i = 0; i < m; ++i) {
long long k, pos;
cin >> k >> pos;
long long val = so[n - k];
for (long long j = 0; j < n; ++j) {
if (v[j] >= val) {
--pos;
}
if (pos == 0) {
cout << v[j] << endl;
break;
}
}
}
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
{ solve(); }
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e12;
int main() {
int n;
cin >> n;
vector<int> a(n), s;
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
s = a;
sort(s.begin(), s.end());
int m;
cin >> m;
for (int i = 0; i < m; ++i) {
int k, pos;
cin >> k >> pos;
int c = 0;
vector<int> ans, st(k);
while (c < k) {
st[c] = s[n - 1 - c];
++c;
}
for (int q = 0; q < n; ++q) {
bool flag = false;
for (int j = 0; j < k; ++j) {
if (a[q] == st[j]) {
flag = true;
}
}
if (flag && ans.size() < k) {
ans.push_back(a[q]);
}
}
cout << ans[pos - 1] << "\n";
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void solve() {
long long n;
cin >> n;
vector<long long> v(n), so(n);
for (long long i = 0; i < n; ++i) {
cin >> v[i];
so[i] = v[i];
}
sort(so.begin(), so.end());
long long m;
cin >> m;
for (long long i = 0; i < m; ++i) {
long long k, pos;
cin >> k >> pos;
map<long long, long long> mp;
for (long long j = n - k; j < n; ++j) {
mp[v[j]]++;
}
for (long long j = 0; j < n; ++j) {
if (mp.count(v[j]) and mp[v[j]] > 0) {
--pos;
mp[v[j]]--;
}
if (pos == 0) {
cout << v[j] << endl;
break;
}
}
}
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
{ solve(); }
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int getrnd(int l, int r) { return uniform_int_distribution<int>(l, r)(rng); }
template <typename T1, typename T2>
bool relax(T1& a, const T2& b) {
if (a > b) {
a = b;
return 1;
}
return 0;
}
template <typename T1, typename T2>
bool strain(T1& a, const T2& b) {
if (a < b) {
a = b;
return 1;
}
return 0;
}
void solve() {
int n;
cin >> n;
vector<int> a(n), b(n);
for (int i = 0; i < n; ++i) cin >> a[i], b[i] = a[i];
sort(b.rbegin(), b.rend());
int q;
cin >> q;
while (q--) {
int k, pos;
cin >> k >> pos;
--pos;
vector<int> was(n, 0);
for (int i = 0; i < n; ++i) {
for (int j = 0; j < k; ++j) {
if (a[i] == b[j]) was[i] = 1;
}
}
int c = 1;
for (int i = k - 2; i >= 0; --i) {
if (b[i] == b[i + 1])
++c;
else
break;
}
vector<int> temp;
for (int i = 0; i < n; ++i) {
if (a[i] == b[k - 1]) {
if (c > 0 && was[i]) temp.emplace_back(a[i]);
} else if (was[i]) {
temp.emplace_back(a[i]);
}
}
cout << temp[pos] << '\n';
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(nullptr);
cout.tie(nullptr);
srand(time(0));
int t = 1;
while (t--) solve();
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void func() {
int n;
cin >> n;
int arr[n], acop[n];
for (int i = 0; i < n; ++i) {
cin >> arr[i];
acop[i] = arr[i];
}
sort(acop, acop + n, greater<int>());
int q;
cin >> q;
for (int i = 0; i < q; ++i) {
int k, pos, cnt = 0;
cin >> k >> pos;
for (int i = 0; i < n; ++i) {
if (arr[i] > acop[k - 1]) {
++cnt;
}
if (cnt == pos - 1) {
cout << arr[i] << "\n";
break;
}
}
}
}
int main() {
int t = 1;
int cnt = 0;
while (t--) {
func();
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
class cpp {
public:
long long int v, i;
};
int compare(const void *pa, const void *pb) {
cpp p1 = *(const cpp *)pa;
cpp p2 = *(const cpp *)pb;
if (p1.v > p2.v)
return -1;
else if (p1.v < p2.v)
return 1;
else if (p1.v == p2.v) {
return (p1.i > p2.i);
}
}
int cmp(const void *pa, const void *pb) {
cpp p1 = *(const cpp *)pa;
cpp p2 = *(const cpp *)pb;
if (p1.i > p2.i)
return 1;
else if (p1.i < p2.i)
return -1;
}
cpp a[200500];
int main() {
int j, n, m;
cin >> n;
for (j = 0; j < n; j++) {
cin >> a[j].v;
a[j].i = j;
}
qsort(a, n, sizeof(a[0]), compare);
cin >> m;
while (m--) {
int k, p;
cin >> k >> p;
qsort(a, k, sizeof(a[0]), cmp);
cout << a[p - 1].v << '\n';
qsort(a, n, sizeof(a[0]), compare);
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
/******************************************[ author : SATISH_S ]*********************************************************************/
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define ul unsigned long long
#define pi pair<int,int>
#define pq priority_queue
#define For(it,x) for(auto it=(x).begin();it!=(x).end();it++)
#define pushb push_back
#define PI 3.14159265
#define popb pop_back
#define endl '\n'
#define min_pq(T) pq<T,vector<T>,greater<T>>
#define max_pq(T) pq<T,vector<T>>
template<typename T>
T min(T a,T b,T c) {return min(a,min(b,c));}
template<typename T>
T max(T a,T b,T c) {return max(a,max(b,c));}
template<typename T>
T max(T a,T b,T c,T d) {return max(a,max(b,max(c,d)));}
/*############################################ [ FUNCTIONS ]##########################################################################*/
#define op(mp) mp.reserve(1024); mp.max_load_factor(0.25);
/*#####################################################################################################################################*/
template<typename T>
void read(vector<T>&arr,int N) {arr.clear(); arr.resize(N);for(int i=0;i<N;i++)cin>>arr[i];}
template<typename T>
void read(vector<pair<T,T>>&arr,int N)
{arr.clear();arr.resize(N);for(int i=0;i<(int)arr.size();i++)cin>>arr[i].first>>arr[i].second;}
template<typename T>
void read(vector<vector<T>>&arr,int N,int M)
{arr.clear();arr.resize(N,vector<T>(M));for(int i=0;i<N;i++){for(int j=0;j<M;j++)cin>>arr[i][j];}}
/*###################################################################################################################################*/
#ifndef Debuger
#define n_l '\n'
#define print(...) cout << "[" << #__VA_ARGS__ << "]: "; cout << to_string(__VA_ARGS__) << endl
template<typename T, size_t N>
int SIZE(const T (&t)[N]) { return N; }
template<typename T>
int SIZE(const T &t) { return t.size(); }
string to_string(const string s, int x1 = 0, int x2 = 1e9) {
return '"' + ((x1 < s.size()) ? s.substr(x1, x2 - x1 + 1) : "") + '"';
}
string to_string(const char *s) { return to_string((string) s); }
string to_string(const bool b) { return (b ? "true" : "false"); }
string to_string(const char c) { return string({c}); }
template<size_t N>
string to_string(const bitset<N> &b, int x1 = 0, int x2 = 1e9) {
string t = "";
for (int __iii__ = min(x1, SIZE(b)), __jjj__ = min(x2, SIZE(b) - 1);
__iii__ <= __jjj__; ++__iii__) { t += b[__iii__] + '0'; }
return '"' + t + '"';}
template<typename A, typename... C>
string to_string(const A (&v), int x1 = 0, int x2 = 1e9, C... coords);
int l_v_l_v_l = 0, t_a_b_s = 0;
template<typename A, typename B>
string to_string(const pair<A, B> &p) {
l_v_l_v_l++;
string res = "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
l_v_l_v_l--;
return res;}
template<typename A, typename... C>
string to_string(const A (&v), int x1, int x2, C... coords) {
int rnk = rank<A>::value;
string tab(t_a_b_s, ' ');
string res = "";
bool first = true;
if (l_v_l_v_l == 0) res += n_l;
res += tab + "[";
x1 = min(x1, SIZE(v)), x2 = min(x2, SIZE(v));
auto l = begin(v);
advance(l, x1);
auto r = l;
advance(r, (x2 - x1) + (x2 < SIZE(v)));
for (auto e = l; e != r; e = next(e)) {
if (!first) { res += ", "; }
first = false;
l_v_l_v_l++;
if (e != l) {
if (rnk > 1) {
res += n_l;
t_a_b_s = l_v_l_v_l;
};
}
else { t_a_b_s = 0; }
res += to_string(*e, coords...);
l_v_l_v_l--;
}
res += "]";
if (l_v_l_v_l == 0) res += n_l;
return res;
}
void printm() { ; }
template<typename Heads, typename... Tails>
void printm(Heads H, Tails... T) {
cout << to_string(H) << " | ";
printm(T...);}
#define printm(...) cout << "[" << #__VA_ARGS__ << "]: "; printm(__VA_ARGS__); cout << endl
#endif
/*####################################################################################################################################*/
/*########################################### [ DATA STRUCTURE ] #####################################################################*/
bool lexi(string&s1,string &s2) {
int I=s1.size(),J=s2.size(),i=0,j=0;
while(i<I and j<J){
if(s1[i]>s2[j])
return true;
if(s1[i]<s2[j])
return false;
i++;j++;
}
if(I>J) //strictly smaller
return true;
return false;
}
struct kmp{
string s;
vector<int>lps;
int M;
void init(string &pat){
s=pat;
M=s.size();
lps.resize(M);
build();
}
void build(){
int i=0,j=1;
while(j<M){
if(s[i]==s[j]){
lps[j]=i+1;
i++;
j++;
}
else{
while(i>0){
i=lps[i-1];
if(s[i]==s[j]){
lps[j]=i+1;
i+=1;
break;
}
}
j++;
}
}
}
bool search(string &S){
int j=0;
for(int i=0;i<(int)S.size();i++){
if(S[i]==s[j]){
j++;
}
else{
while(j>0){
j=lps[j-1];
if(S[i]==s[j]){
j+=1;
break;
}
}
}
if(j==M){
return true;
}
}
return false;
}
};
/*&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&*/
struct uf {
vector< int > par, size;
void init(int N) {
par.resize(N, -1);
size.resize(N, 1);
}
int root(int a) {
if (par[a] == -1)
return a;
return par[a] = root(par[a]);
}
void unite(int a, int b) {
a = root(a);
b = root(b);
if (a == b) //same parent do nothing
return;
if (size[a] < size[b]) {
par[a] = b;
size[b] += size[a];
} else {
par[b] = a;
size[a] += size[b];
}
}
bool same(int a, int b) {
if (root(a) == root(b))
return true;
return false;
}
};
/*&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&*/
struct seg_tree {
vector<int> make;
int siz;
vector<int> arr;
vector<int>lazy;
void init(vector<int> &a, bool max, bool range) {
arr.clear();
make.clear();
arr = a;
siz = arr.size();
make.resize(4 * siz);
if (max)
build_max(0, 0, siz - 1);
else if (range) {
//lazy.resize(4*siz);
build_range(0, 0, siz - 1);
}
}
int get_max(int L, int R) {
return Get_max(0, 0, siz - 1, L, R);
}
void update_max(int index, int val) {
Update_max(0, 0, siz - 1, index, val);
return;
}
int get_sum(int L, int R) {
return Get_sum(0, 0, siz - 1, L, R);
}
void update_range(int index, int add) {
Update_range(0, 0, siz - 1, index, add);
return;
}
void update_interval(int l,int r,int value){
Update_interval(0,0,siz-1,l,r,value);
}
void Update_interval(int ind,int L,int R,int Left,int Right,int value){
if(lazy[ind]!=0){
make[ind]+=(R-L+1)*lazy[ind];
if(L!=R){
lazy[2*ind+1] += lazy[ind];
lazy[2*ind+2] += lazy[ind];
}
lazy[ind]=0;
}
if(L>Right or R<Left)
return;
if(Left<=L and R<=Right){
make[ind]+=(R-L+1)*value;
if(L!=R){
lazy[2*ind+1]=value;
lazy[2*ind+2]=value;
}
return;
}
int mid=(L+R)/2;
Update_interval(2*ind+1,L,mid,Left,Right,value);
Update_interval(2*ind+2,mid+1,R,Left,Right,value);
make[ind]=make[2*ind+1]+make[2*ind+2];
}
int build_range(int ind, int L, int R) {
if (L == R) {
make[ind] = arr[L];
return make[ind];
} else {
int mid = (L + R) / 2;
int a = build_range(2 * ind + 1, L, mid);
int b = build_range(2 * ind + 2, mid + 1, R);
make[ind] = a + b;
return make[ind];
}
}
int Get_sum(int ind, int L, int R, int Left, int Right) {
/*
if(lazy[ind]!=0) {
make[ind]+=(R-L+1)*lazy[ind];
if(L!=R){
lazy[2*ind+1]=lazy[ind];
lazy[2*ind+2]=lazy[ind];
}
lazy[ind]=0;
}
*/
if (L > Right or R < Left)
return 0;
if (Left <= L and R <= Right)
return make[ind];
int mid = (L + R) / 2;
int a = Get_sum(2 * ind + 1, L, mid, Left, Right);
int b = Get_sum(2 * ind + 2, mid + 1, R, Left, Right);
return a + b;
}
void Update_range(int ind, int L, int R, int index, int add) {
if (L == R) {
make[ind] = add;
arr[index] = add;
} else {
int mid = (L + R) / 2;
if (L <= index and index <= R) {
Update_range(2 * ind + 1, L, mid, index, add);
} else {
Update_range(2 * ind + 2, mid + 1, R, index, add);
}
make[ind] = make[2 * ind + 1] + make[2 * ind + 2];
}
}
int build_max(int ind, int L, int R) {
if (L == R) {
make[ind] = arr[L];
return make[ind];
} else {
int mid = (L + R) / 2;
return make[ind] = max(build_max(2 * ind + 1, L, mid), build_max(2 * ind + 2, mid + 1, R));
}
}
int Get_max(int ind, int L, int R, int Left, int Right) {
if (R < Left or L > Right)
return -1e15;
if (Left <= L and R <= Right)
return make[ind];
int mid = (L + R) / 2;
return max(Get_max(2 * ind + 1, L, mid, Left, Right), Get_max(2 * ind + 2, mid + 1, R, Left, Right));
}
int Update_max(int ind, int L, int R, int index, int val) {
if (L == R) {
arr[index] = val;
make[ind] = val;
return val;
} else {
int mid = (L + R) / 2;
if (L <= index and index <= mid) {
make[ind] = Update_max(2 * ind + 1, L, mid, index, val);
} else {
make[ind] = Update_max(2 * ind + 2, mid + 1, R, index, val);
}
make[ind] = max(make[2 * ind + 1], make[2 * ind + 2]);
return make[ind];
}
}
};
/*######################################################################################################################################*/
static bool comp(pi& a,pi& b){
if(a.first<b.first)
return true;
if(a.first==b.first){
if(a.second<b.second)
return true;
}
return false;
}
int lcm(int a,int b){
return (a*b)/__gcd(a,b);
}
/*############################################## [ END ] ###############################################################################*/
static bool comp1(pi& a,pi& b){
if(a.first<b.first)
return true;
if(a.first==b.first){
if(a.second>b.second)
return true;
}
return false;
}
/*############################################## [SOLVE] ###############################################################################*/
// clear before use
#define all(x) x.begin(),x.end()
#define found(mp,x) mp.find(x)!=mp.end()
int Max=1e17,Min=-1e17;
int N,M,K;
int mod=1e9+7;int P=97;
vector<int>a;
void solve(){
cin>>N;
read(a,N);
map<int,vector<int>>mp;
for(int i=0;i<N;i++)
mp[a[i]].push_back(i);
int Q;cin>>Q;
for(int q=0;q<Q;q++){
int k,p;
cin>>k>>p;
vector<int>temp(N);
auto it=mp.end();--it;
int siz=0;
while(true){
siz+=(it->second).size();
//print(it->second);
//cout<<siz<<endl;
if(siz>=k){
auto ut=it;
++ut;
while(ut!=mp.end()){
for(auto i:ut->second){
temp[i]=ut->first;
siz--;
}
ut++;
}
for(auto i:it->second){
if(siz>0){
temp[i]=it->first;
siz--;
}
else
break;
}
//print(temp);
vector<int>newtemp;
for(int i=0;i<N;i++)
if(temp[i]>0)
newtemp.push_back(temp[i]);
//print(temp);
cout<<newtemp[p-1]<<endl;
break;
}
--it;
}
}
}
/*############################################### [ MAIN ] #############################################################################*/
signed main() {
//////
bool test = 0 ;
ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
auto start_time = chrono::high_resolution_clock::now();
if (!test) solve();
else {
long long tt;
cin >> tt;
while (tt--)solve();
}
auto end_time = chrono::high_resolution_clock::now();
#ifndef ONLINE_JUDGE
double time_taken = chrono::duration_cast< chrono::nanoseconds >(end_time - start_time).count();
time_taken *= 1e-9;
cout << endl<<endl<< "{ time taken by program :" << fixed << time_taken << setprecision(9) << " sec }";
#endif
return 0;
}
/*###########################################[ EXTRA ] #################################################################################*/
//#define deb(x) cout<<#x<<" is "<<x<<endl;
/*void prims(vector<vector<int>>&arr,int start){
if(v[start]==true)
return;
int mini=1e8;
int nstart=-1;
v[start]=true;
for(auto it:arr[start]){
if(v[it]==false and mini>w[{start,it}]){
nstart=it;
mini=w[{start,it}];
}
}
if(nstart!=-1)
prims(arr,nstart);
}
*/
//map<int,set<pi>>ss;
//connected components !!!
/*
bool cycledirected(int start){
v[start]=true;
process[start]=true;
for(auto it:arr[start]){
if(v[it]==false and cycle(it))
return true;
else if(process[it])
return true;
}
process[start]=false;
return false;
}
*/
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void ios1() {
ios_base::sync_with_stdio(0);
cout.tie(0);
cin.tie(0);
}
bool cmd(pair<long long, int> a, pair<long long, int> b) {
if (a.first > b.first)
return 0;
else if (a.first < b.first)
return 1;
else
return (a.second > b.second);
}
int main() {
ios1();
int n;
cin >> n;
pair<long long, int> arr[n];
long long cop[n];
int its[n];
for (int i = 0; i < n; ++i) {
cin >> cop[i];
arr[i] = make_pair(cop[i], i);
}
long long sum1 = 0, sumi[n];
sort(arr, arr + n, cmd);
for (int i = n - 1; i >= 0; --i) {
its[(n - 1) - i] = arr[i].second;
}
int m;
cin >> m;
for (int i = 0; i < n; ++i) cout << its[i] << " ";
for (int i = 0; i < m; ++i) {
int a, b;
cin >> a >> b;
vector<int> v;
for (int j = a - 1; j >= 0; j--) v.push_back(its[j]);
sort(v.begin(), v.end());
cout << cop[v[b - 1]] << endl;
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using mii = map<int, int>;
const double Eps = 1e-8;
const int Inf = 1e9 + 9;
const int Mod = 1e9 + 7;
const int N = 1e5 + 9;
const int dx4[4] = {0, 1, 0, -1};
const int dy4[4] = {-1, 0, 1, 0};
const int dx8[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const int dy8[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
constexpr double pi = 3.141592653589793238462643383279502884L;
long long gcd(long long x, long long y) {
if (y == 0) return x;
return gcd(y, x % y);
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
void prime_siever(long long p) {
bool prime[p + 9];
for (long long i = 2; i * i <= p; i++)
if (prime[i] == 0)
for (long long j = i * i; j <= p; j += i) prime[j] = 1;
}
int mul(int a, int b) { return (1LL * a * b) % Mod; }
int add(int a, int b) {
a += b;
if (a >= Mod) a -= Mod;
if (a < 0) a += Mod;
return a;
}
long long bin_power(long long a, long long n) {
if (a == 0) return 0;
long long res = 1;
while (n) {
if (n % 2) {
res = (res * a) % Mod;
n--;
} else {
a = (a * a) % Mod;
n /= 2;
}
}
return res;
}
vector<int> a;
vector<pair<int, int> > b, c;
priority_queue<int, vector<int>, greater<int> > qi;
map<pair<int, int>, int> mp;
void solve() {
int q, n;
cin >> n;
for (int i = 0; i < n; i++) {
int x;
cin >> x;
a.push_back(x);
qi.push(a[i]);
}
cin >> q;
for (int i = 0; i < q; i++) {
int x, y;
cin >> x >> y;
b.push_back({x, y});
}
c = b;
sort(b.rbegin(), b.rend());
for (int i = 0; i < q; i++) {
int siz = b[i].first, idx = b[i].second - 1;
while (qi.size() != siz) {
a.erase(find(a.begin(), a.end(), qi.top()));
qi.pop();
}
mp[{siz, idx + 1}] = a[siz - idx - 1];
}
for (int i = 0; i < q; i++) {
cout << mp[c[i]] << '\n';
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
solve();
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long double pi = 2 * acos(0.0);
struct arr {
long long index, val;
};
struct query {
long long n, ind, index, ans;
};
bool compare(arr a1, arr a2) { return (a1.val > a2.val); }
bool compare2(arr a1, arr a2) { return (a1.index < a2.index); }
bool compare1(query q1, query q2) { return (q1.n < q2.n); }
bool compare3(query q1, query q2) { return (q1.index < q2.index); }
int main() {
long long n, i;
cin >> n;
arr A[n];
for (i = 0; i < n; ++i) {
cin >> A[i].val;
A[i].index = i + 1;
}
sort(A, A + n, compare);
long long m;
cin >> m;
query q[m];
for (i = 0; i < m; ++i) {
cin >> q[i].n >> q[i].ind;
q[i].index = i;
}
sort(q, q + m, compare1);
for (i = 0; i < m; ++i) {
if (i != 0) {
if (q[i].n != q[i - 1].n) {
sort(A, A + q[i - 1].n, compare);
sort(A, A + q[i].n, compare2);
q[i].ans = A[q[i].ind - 1].val;
} else
q[i].ans = A[q[i].ind - 1].val;
} else {
sort(A, A + q[i].n, compare2);
q[i].ans = A[q[i].ind - 1].val;
}
}
sort(q, q + m, compare3);
for (i = 0; i < m; ++i) cout << q[i].ans << "\n";
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
template <typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define pb push_back
#define fi first
#define se second
#define int long long
typedef long long ll;
typedef long double ld;
#define PI 3.14159265
// 🍪 DEBUG FUNCTIONS START
#define cerr cout
void __print(int x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
//#define deb(x...) cerr << "[" << #x << "] = "; _print(x)
#define deb(x...) _print(x)
#else
#define deb(x...)
#endif
// 🍪 DEBUG FUNCTIONS END
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const ll infl = 0x3f3f3f3f3f3f3f3fLL;
void solve()
{
int n;
cin >> n;
vector<int> vv(n);
multimap<int, int> mp;
for(int i = 0; i < n; i++){
cin >> vv[i];
mp.insert({vv[i], i});
}
vector<vector<int>> v(n);
for(int i = 0; i < n; i++){
int times = 0;
for(auto x = mp.rbegin(); x != mp.rend(); x++){
if(times == i + 1){
break;
}
v[i].pb(x->se);
times++;
}
}
for(int i = 0; i < n; i++){
sort(v[i].begin(), v[i].end());
}
deb(v);
int q;
cin >> q;
while(q--){
int x, y;
cin >> x >> y;
--x; --y;
cout << vv[v[x][y]] << "\n";
}
}
signed main()
{
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
srand(chrono::high_resolution_clock::now().time_since_epoch().count());
int t;
t=1;
while(t--)
{
solve();
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
bool comp(const pair<long long int, long long int> &a,
const pair<long long int, long long int> &b) {
if (a.first == b.first) {
return a.second < b.second;
}
return a.first > b.first;
}
int main() {
int n;
cin >> n;
vector<long long int> arr(n);
vector<pair<long long int, long long int>> temp(n);
for (int i = 0; i < n; i++) {
cin >> arr[i];
temp[i] = {arr[i], i};
}
sort(temp.begin(), temp.end(), comp);
for (auto p : temp) {
cout << p.first << " " << p.second << endl;
}
int m, k, pos;
cin >> m;
while (m--) {
cin >> k >> pos;
vector<pair<long long int, long long int>> curr;
for (int i = 0; i < k; i++) {
curr.push_back({temp[i].second, temp[i].first});
}
sort(curr.begin(), curr.end());
cout << curr[pos - 1].second << endl;
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
const int maxn = 100100;
using namespace std;
int ans[110][110];
struct node {
int nub, pos;
} a[110];
bool cmp(node c, node b) { return c.nub > b.nub; }
bool cmp2(node c, node b) { return c.pos < b.pos; }
int main() {
int n, m;
cin >> n;
for (int i = 1; i <= n; ++i) {
cin >> a[i].nub;
a[i].pos = i;
}
sort(a + 1, a + 1 + n, cmp);
ans[1][1] = a[1].nub;
for (int i = 2; i <= n; ++i) {
sort(a + 1, a + 1 + i, cmp2);
for (int j = 1; j <= i; ++j) {
ans[i][j] = a[j].nub;
}
}
cin >> m;
for (int i = 1; i <= m; ++i) {
int x, pos;
cin >> x >> pos;
cout << ans[x][pos] << endl;
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
//triedodin
#include<bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
#define mem(a, b) memset(a, (b), sizeof(a))
#define all(cont) cont.begin(), cont.end()
#define pb push_back
#define PI 3.1415926535897932384626433832795
#define mod1 1000000007
#define mod2 998244353
#define ll long long int
#define ld long double
#define inf LLONG_MAX
#define endl "\n"
#define F first
#define S second
#define inp(a) int a;cin>>a;
const double pi=acos(-1.0);
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds; //greater<int> can also be used
#define printclock cerr<<"Time : "<<1000*(ld)clock()/(ld)CLOCKS_PER_SEC<<"ms\n";
#define sz(c) (int)c.size()
#define d1(x) cout<<#x<<"="<<x<<endl;
#define d2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl;
#define d3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl;
#define d4(x,y,z,w) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<" "<<#w <<"="<<w<<endl;
#define da(arr,j,n) for(int i=j;i<n;i++) {cout<<"i="<<i<<" arr[i]="<<arr[i]<<"\n";}
const ll MODA = 1000000007;
// const int MODA = 998244353 ;
vector<ll> primefactors ; // used by generatePrimeFactors() // ARRAYS/VECTORS
vector<ll> factors ; // used by generateFactors()
ll fact[1025]={} ; // used by generateFactorial() & ncr()
bool isPowerTwo(ll x) { return (x && !(x & (x - 1))); } // FUNCTIONS
ll modmul(ll a,ll b) { return((a%MODA)*(b%MODA))%MODA; }
ll modadd(ll a,ll b) { return((a%MODA)+(b%MODA)+MODA)%MODA;}
ll modsub(ll a,ll b) { return((a%MODA)-(b%MODA)+MODA)%MODA;}
bool isSubstring(string s1, string s2) { if (s1.find(s2) != string::npos) return true; else return false; }
void generateFactorial(ll n) { fact[0] = 1; for(ll i = 1; i <= n; i++) fact[i] = (i* 1ll * fact[i - 1]) ; }
void generateFactorialMOD(ll n) { fact[0] = 1; for(ll i = 1; i <= n; i++) fact[i] = (i* 1ll * fact[i - 1]) % MODA; }
bool isPrime(ll n) { if(n<2) return false; for(ll i=2;i*i<=n;i++) if(n%i==0) return false; return true; } //O(sqrt(n))
ll power(ll a, ll b) { ll res=1; while(b) { if(b&1) res=(res*a); a=(a*a); b>>=1; } return res; }
ll powermod(ll a, ll b) { ll res=1; while(b) { if(b&1) res=(res*a)%MODA; a=(a*a)%MODA; b>>=1; } return res; }
ll modi (ll a) { ll m=MODA,s=1,p=0; while(a>1) { ll q=a/m,t=m; m=a%m; a=t; t=p; p=s-q*p; s=t; } return s>=0 ? s :s+MODA; }
void generatePrimeFactors(ll n) { primefactors.clear(); for(ll i=2;i*i<=n;i++) { if(n%i==0) { primefactors.pb(i); while(n%i==0) n=n/i; } } if(n!=1) primefactors.pb(n); }
ll ncr(ll n, ll r) { if(r==0)return 1; fact[0]=1; for(int i=1 ; i<=n; i++) fact[i] = fact[i-1]*i%MODA; return (fact[n]* modi(fact[r]) % MODA * modi(fact[n-r]) % MODA) % MODA; }
void generateFactors(ll n) { factors.clear(); for(ll i=2;i*i<=n;i++){ if(n%i==0) { factors.pb(i); if(n/i!=i)factors.pb(n/i); }} factors.pb(1); factors.pb(n); sort(factors.begin(),factors.end()); }
ld Logn(ld n, ld r) { return log(n) / log(r); }
///////////////////////////////////
const int Nmax = 1000005;
////////////////////////////////////
void solve(){
int n;
cin >> n;
vector<int> v;
for(int i =0; i < n; i++){
int x;
cin >> x;
v.pb(x);
}
vector<int> cv = v;
sort(all(cv));
inp(m);
vector<int> ps(n);
ps[n-1] = 0;
for(int i = n-2; i>=0; i--){
if(cv[i] < cv[i+1]){
ps[i] = 1 + ps[i+1];
}
else{
ps[i] = ps[i+1];
}
}
for(int i = 0; i<m; i++){
int x, y;
cin >> x >> y;
int rn = cv[n- x];
// d1(rn);
int nrn = x - ps[n-x];
//d1(nrn);
int ri = 0;
int son = 0;
for(int j = 0; j < n; j++){
if(v[j] > rn){
son++;
}
else if(v[j] == rn){
if(nrn){
nrn--;
son++;
}
}
if(son == y){
cout << v[j] << endl;
break;
}
}
}
}
int main(){
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
ios_base::sync_with_stdio(false); cin.tie(0);
int t= 1;
//cin >> t;
while(t--){
solve();
}
printclock;
}
/////////////////////////////
//try to bring as much symmetry as possible
//just check for integer overflow
//add MOD value when you substract two numbers
/////////////////////////////
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long int no = 3e6 + 5, modulo = 1e9 + 7, inf = 1e18, N = 3e3 + 1;
long long int ar[no], br[no], cr[no];
void solve() {
long long int n = 0, m = 0, a = 0, b = 0, c = 0, d = 0, x = 0, y = 0, z = 0,
w = 0, k = 0;
cin >> n;
vector<long long int> vv;
for (long long int i = 1; i < n + 1; i++) cin >> ar[i], vv.push_back(ar[i]);
cin >> m;
sort(vv.rbegin(), vv.rend());
while (m--) {
cin >> x >> y;
a = vv[x - 1];
vector<long long int> ans, v;
map<long long int, long long int> mapp;
for (long long int i = 0; i < x; i++) {
if (vv[i] > a) v.push_back(vv[i]), mapp[vv[i]]++;
}
for (long long int i = 1; i < n + 1; i++) {
if (ans.size() == x) break;
if (ar[i] == a) {
ans.push_back(a);
} else if (mapp[ar[i]] > 0) {
mapp[ar[i]]--;
ans.push_back(ar[i]);
if (mapp[ar[i]] == 0) mapp.erase(ar[i]);
}
}
cout << ans[y - 1] << "\n";
}
}
inline void runn() {
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
long long int t = 1;
for (long long int i = 1; i < t + 1; i++) {
solve();
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void solve() {
long long n;
cin >> n;
vector<pair<long long, long long>> arr(n);
for (long long i = 0; i < n; i++) {
cin >> arr[i].first;
arr[i].second = i;
}
auto store = arr;
sort(arr.rbegin(), arr.rend());
long long m;
cin >> m;
for (long long _ = 0; _ < m; _++) {
long long k, pos;
cin >> k >> pos;
vector<long long> ano;
for (long long i = 0; i < k; i++) {
ano.push_back(arr[i].second);
}
sort(ano.begin(), ano.end());
cout << store[ano[pos - 1]].first << endl;
}
}
signed main() {
long long t = 1;
while (t--) {
solve();
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0), cout.tie(0), cin.tie(0);
long long int n, m, k, i, j, a[101];
cin >> n;
vector<long long int> pr;
for (i = 0; i < n; i++) {
cin >> a[i];
pr.push_back(a[i]);
}
sort(pr.rbegin(), pr.rend());
cin >> m;
while (m--) {
long long int k, pos, last = -1;
cin >> k >> pos;
pos--;
long long int cnt = -1;
vector<long long int> v(n + 1, -1);
for (i = k - 1; i >= 0; i--) {
cnt = k - i - 1;
for (j = last + 1; j < n; j++) {
if (pr[i] == a[j]) {
v[j] = cnt;
cnt++;
last = j;
break;
}
}
if (i != k - 1 && cnt == k - i - 1) {
v[last]++;
cnt--;
for (j = last - 1; j >= 0; j--) {
if (pr[i] == a[j]) {
v[j] = cnt;
cnt++;
break;
} else if (v[j] != -1)
v[j]++, cnt--;
}
}
}
for (j = 0; j < n; j++) {
if (v[j] == pos) {
cout << a[j] << endl;
break;
}
}
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
int m;
cin >> m;
for (int e = 0; e < m; e++) {
int k, p;
cin >> k >> p;
vector<int> b;
for (int i = 0; i < k; i++) {
b.push_back(a[i]);
}
for (int i = k; i < n; i++) {
int MIN = 1000000007, pos = 0;
for (int j = 0; j < k; j++) {
if (b[j] < MIN) {
MIN = b[j];
pos = j;
}
}
if (a[i] > MIN) {
b.erase(b.begin() + pos);
b.push_back(a[i]);
}
}
cout << b[p - 1] << endl;
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(time(0));
bool cmp2(const vector<long long> &a, const vector<long long> &b) {
assert(a.size() == b.size());
for (long long i = 0; i < a.size(); ++i) {
if (a[i] < b[i])
return 1;
else
return 0;
}
return 1;
}
bool comp(const pair<long long, vector<long long> > &a,
const pair<long long, vector<long long> > &b) {
if (a.first == b.first) return cmp2(a.second, b.second);
return a.first > b.first;
}
void solve() {
long long n;
cin >> n;
long long a[n];
for (long long i = 0; i < n; ++i) {
cin >> a[i];
}
pair<long long, vector<long long> > dp[n][n];
for (long long i = 0; i < n; ++i) {
for (long long j = 0; j < n; ++j) {
if (j == 0) {
dp[i][j] = {a[i], {a[i]}};
if (i) {
if (comp(dp[i - 1][j], dp[i][j])) {
dp[i][j] = dp[i - 1][j];
}
}
} else {
dp[i][j] = {0, {vector<long long>(j, long long(1e12))}};
for (long long h = 0; h < i; ++h) {
pair<long long, vector<long long> > x = {dp[h][j - 1].first + a[i],
dp[h][j - 1].second};
x.second.push_back(a[i]);
if (comp(x, dp[i][j])) {
dp[i][j] = x;
}
}
if (i) {
if (comp(dp[i - 1][j], dp[i][j])) {
dp[i][j] = dp[i - 1][j];
}
}
}
}
}
long long q;
cin >> q;
while (q--) {
long long k, pos;
cin >> k >> pos;
k--;
pos--;
cout << dp[n - 1][k].second[pos] << endl;
}
}
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
long long t = 1;
while (t--) {
solve();
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vd = vector<double>;
using vs = vector<string>;
using vpii = vector<pair<int, int>>;
using vpll = vector<pair<ll, ll>>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
void solve() {
int n;
cin >> n;
vi a(n), b(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
b[i] = a[i];
}
sort(b.begin(), b.end(), greater<int>());
int q;
cin >> q;
for (ll i = 0; i < (q); i++) {
int k, pos;
cin >> k >> pos;
int x = b[k - 1];
int cnt = 0;
for (int i = 0; i < n; i++) {
if (a[i] >= x) cnt++;
if (cnt == pos) {
cout << a[i] << '\n';
;
break;
}
}
}
}
int main() {
ios_base::sync_with_stdio(false);
ll t = 1;
while (t--) {
solve();
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#define pb push_back
#define ll long long int
#define db double
#define sorta(v) sort(v.begin(),v.end())
#define sortd(v) sort(v.begin(),v.end(),greater<>())
#define endl '\n'
#define mp(a,b) make_pair(a,b)
#define nod(N) floor(log10(N)) + 1
#define f(p) p.first
#define s(p) p.second
#define lb(a) lower_bound(a)
#define ub(a) upper_bound(a)
#define fo(i,n) for(long long int i = 0;i<n;i++)
#define fo1(i,n) for(long long int i=1;i<=n;i++)
#define foa(i,a,n) for(long long int i = a;i<=n;i++)
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL);
#define reverse(s) reverse(s.begin(),s.end())
#define lbv(v,a) lower_bound(v.begin(),v.end(),a)
#define ubv(v,a) upper_bound(v.begin(),v.end(),a)
#define gcd(a,b) __gcd(a,b)
#define MAX pow((ll)2,63) - 999;
#define itx(it,x1) for(auto it = x1.begin();it!=x1.end();it++)
#define I insert
#define stll(x) stoll(x, nullptr, 10);
#define vfind(v1,val) find(v1.begin() , v1.end() , val )
#define vfill(v1 , val) std::fill(v1.begin(), v1.end(), val);
template<typename T>using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
ll mxm() {return LLONG_MIN;}
template<typename... Args>
ll mxm(ll a, Args... args) { return max(a,mxm(args...)); }
ll mnm() {return LLONG_MAX;}
template<typename... Args>
ll mnm(ll a, Args... args) { return min(a,mnm(args...)); }
ll cl(ll a, ll b){if(a%b == 0)return a/b;else return a/b + 1;}
void dpm(map<ll,ll>m1)
{
for(auto it = m1.begin();it!=m1.end();it++)
{
cout<<it->first<<" "<<it->second<<endl;
}
cout<<endl;
}
void dpv(vector<ll>v)
{
fo(i,v.size())cout<<v[i]<<" ";cout<<endl;
}
void dps(set<ll>s1)
{
for(auto it =s1.begin();it!=s1.end();it++)cout<<*it<<" ";cout<<endl;
}
void dpa(ll arr[],ll n)
{
fo(i,n)cout<<arr[i]<<" ";cout<<endl;
}
ll nCrModpDP(ll n, ll r, ll p)
{
// The array C is going to store last row of
// pascal triangle at the end. And last entry
// of last row is nCr
ll C[r+1];
memset(C, 0, sizeof(C));
C[0] = 1; // Top row of Pascal Triangle
// One by constructs remaining rows of Pascal
// Triangle from top to bottom
for (ll i = 1; i <= n; i++)
{
// Fill entries of current row using previous
// row values
for (ll j = min(i, r); j > 0; j--)
// nCj = (n-1)Cj + (n-1)C(j-1);
C[j] = (C[j] + C[j-1])%p;
}
return C[r];
}
int main()
{
fast;
ll i=0,j=0, t = 1;
// cin>>t;
while(t--)
{
ll n; cin>>n;
vector<ll>v1(n); multimap<ll,ll>m1; set<ll>s1;
fo(i,n)cin>>v1[i] , m1.insert({v1[i] , i} ) , s1.I(v1[i]);
ll m;cin>>m;
while(m--)
{
ll k , pos; cin>>k>>pos; pos--;
vector<ll>v2(n);
if(k==n)
{
cout<<v1[pos]<<endl; continue;
}
auto it = s1.rbegin();
while(k>0)
{
auto it1 = m1.lb(*it);
ll val1 = it1->first;
while(it1!= m1.end() && it1->first == val1 && k>0)
{
v2[it1->second] = it1->first; it1++; k--;
}
it--;
}
vector<ll>v3;
fo(i,v2.size() )if(v2[i]!=0)v3.pb(v2[i]);
cout<<v3[pos]<<endl;
}
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cout.precision(20);
long long n = 0;
cin >> n;
deque<long long> posl(n);
for (auto& p : posl) cin >> p;
map<long long, long long> al;
for (auto& p : posl) al[p]++;
long long m = 0;
cin >> m;
for (long long k = 0; k < m; k++) {
long long q = 0, pos = 0, sum = 0, num = -1, need = -1;
cin >> q >> pos;
auto it = al.end();
it--;
auto ed = al.begin();
ed--;
for (it; it != ed; it--) {
if (sum + it->second < q)
sum += it->second, num = it->first;
else {
need = q - sum;
num = it->first;
break;
}
}
deque<long long> ans;
for (auto& p : posl) {
if (p == num && need > 0)
ans.push_back(p), need--;
else if (p >= num)
ans.push_back(p);
}
cout << ans[pos - 1] << endl;
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
python3
|
import time
startTimeProblem=time.time()
import fileinput, sys, itertools, functools
# from math import *
import math
from bisect import *
from heapq import *
from collections import *
def lcm(a, b):
return (a*b)/gcd(a, b)
class InputHelper:
def __init__(self):
self.myinput = fileinput.input()
def isLocal(self):
return not fileinput.isstdin()
def int(self):
return int(self.myinput.readline().rstrip())
def ints(self):
return [int(_) for _ in self.myinput.readline().rstrip().split()]
def str(self):
return self.myinput.readline().rstrip()
def strs(self):
return [_ for _ in self.myinput.readline().rstrip().split()]
class OutputHelper:
def int(self, a):
print(a)
def ints(self, a):
print(" ".join([str(_) for _ in a]))
def intsNL(self, a):
for _ in a:
print(_)
def str(self, s):
print(s)
def strs(self, s):
print(" ".join([_ for _ in s]))
def strsNL(self, s):
for st in s:
print(st)
class ListNode:
def __init__(self, val):
self.val = val
self.next = None
self.prev = None
In = InputHelper()
Out = OutputHelper()
######################################
n = In.int()
a = In.ints()
m = In.int()
cnt = Counter(a)
asort = sorted(a)
for _ in range(m):
kj, posj = In.ints()
assort = asort[n-kj::]
there = Counter(assort)
cntdown = defaultdict(int)
cur_pos = 0
res = 0
for i in range(n):
if a[i] in there:
if cnt[a[i]]-cntdown[a[i]]==there[a[i]] or assort[0]==a[i]:
cur_pos+=1
there[a[i]] -= 1
if there[a[i]]==0:
there.pop(a[i])
assort.pop(0)
if cur_pos==posj:
res=a[i]
break
cntdown[a[i]]+=1
Out.int(res)
######################################
if len(sys.argv)>2 and sys.argv[2]=="TIMEIT":
fin = (time.time()-startTimeProblem)*1000
print("{:.2f}".format(fin) + "ms")
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
long long int ivalue(string s) {
long long int x = 0;
stringstream obj(s);
obj >> x;
return x;
}
const long long int M = 1e9 + 7;
const long long int N = 1e5 + 5;
const long long int inf = 2e18;
long long int mod(long long int x) { return (x % M); }
long long int mod_minus(long long int a, long long int b) {
long long int ans = (mod(a) - mod(b));
if (ans < 0) ans = mod(ans + M);
return ans;
}
long long int mod_mul(long long int a, long long int b) {
return mod(mod(a) * mod(b));
}
long long int mod_add(long long int a, long long int b) {
return mod(mod(a) + mod(b));
}
long long int power(long long int a, long long int n) {
if (n == 0)
return 1;
else if (n == 1)
return a;
long long int R = power(a, n / 2) % M;
if (n % 2 == 0) {
return mod(mod_mul(R, R));
} else {
return mod(mod_mul(mod_mul(R, a), mod(R)));
}
}
long long int mod_div(long long int a, long long int b) {
long long int ans = mod(a);
long long int b1 = power(b, M - 2);
ans = mod(mod_mul(ans, b1));
return ans;
}
long long int mod_inv(long long int n) { return power(n, M - 2); }
long long int fact_mod(long long int n) {
vector<long long int> fact(n + 1);
fact[0] = 1;
for (long long int i = 1; i < n + 1; i++) {
fact[i] = mod_mul(fact[i - 1], i);
}
return fact[n];
}
long long int nCr_mod(long long int n, long long int r) {
if (r == 0 || n == 0) return 1;
long long int fac[n + 1];
fac[0] = 1;
for (long long int i = 1; i <= n; i++) fac[i] = (fac[i - 1] * i) % M;
return (fac[n] * mod_inv(fac[r]) % M * mod_inv(fac[n - r]) % M) % M;
}
long long int upper_fraction(long long int a, long long int b) {
if (a % b == 0)
return a / b;
else
return (a / b) + 1;
}
bool isInt(double d) {
double dummy;
return modf(d, &dummy) == 0.0;
}
void solve() {
long long int n;
cin >> n;
vector<long long int> a(n);
long long int i;
for (i = 0; i < n; i++) cin >> a[i];
long long int mx = *max_element((a).begin(), (a).end());
long long int mi = *min_element((a).begin(), (a).end());
long long int pmx = 0;
long long int pmi = 0;
for (i = 0; i < n; i++) {
if (a[i] == mx) pmx = i;
if (a[i] == mi) pmi = i;
}
long long int ans = max(pmx + 1, pmi + 1);
ans = min(ans, n - min(pmx, pmi));
if (pmx > pmi) {
ans = min(ans, pmi + 1 + n - pmx);
} else
ans = min(ans, pmx + 1 + n - pmi);
cout << ans << endl;
}
int main() {
ios_base ::sync_with_stdio(false);
cin.tie(NULL);
long long int t;
t = 1;
cin >> t;
while (t--) solve();
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 * 10;
long long arr[200];
vector<pair<long long, long long>> sortedVec, vec2;
int main() {
int n, m;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> arr[i];
sortedVec.push_back(make_pair(arr[i], i));
}
cin >> m;
sort(sortedVec.begin(), sortedVec.end());
reverse(sortedVec.begin(), sortedVec.end());
for (int i = 0; i < m; i++) {
int k, pos;
cin >> k >> pos;
for (int j = 0; j < k; j++) {
vec2.push_back(make_pair(sortedVec[j].second, sortedVec[j].first));
}
sort(vec2.begin(), vec2.end());
reverse(vec2.begin(), vec2.end());
cout << vec2[pos - 1].second << "\n";
vec2.resize(0);
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
void jakos() {
ios_base::sync_with_stdio(false);
cin.tie(0);
}
const int mod = 1e9 + 7;
const int base = 179;
const int INF = 1e9;
const int N = 1e5;
signed main() {
jakos();
int n;
cin >> n;
vector<pair<int, int>> a;
for (int i = 0; i < n; i++) {
int b;
cin >> b;
a.emplace_back(b, i);
}
sort(a.begin(), a.end());
reverse(a.begin(), a.end());
int q;
cin >> q;
while (q--) {
vector<pair<int, int>> ans;
int k, pos;
cin >> k >> pos;
for (int i = 0; i < k; i++) {
ans.emplace_back(a[i].second, a[i].first);
}
sort(ans.begin(), ans.end());
cout << ans[pos - 1].second << endl;
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
bool cmp(pair<int, int> p1, pair<int, int> p2) { return p1.first > p2.first; }
int fc(int k, int pos, vector<pair<int, int> > a, vector<int> b) {
vector<int> y;
for (int i = 0; i < k; i++) {
y.push_back(a[i].second);
cout << y[i] << ' ';
}
sort(y.begin(), y.end());
return y[pos - 1];
}
int main() {
int n;
cin >> n;
vector<pair<int, int> > a(n);
vector<int> b(n);
for (int i = 0; i < n; i++) {
cin >> a[i].first;
a[i].second = i;
b[i] = a[i].first;
}
sort(a.begin(), a.end(), cmp);
int q;
cin >> q;
for (int i = 0; i < q; i++) {
int k, pos;
cin >> k >> pos;
int x = fc(k, pos, a, b);
cout << b[x] << endl;
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long int n;
cin >> n;
vector<long long int> a(n), v1(n);
for (long long int i = 0; i < n; i++) cin >> a[i];
v1 = a;
sort(a.begin(), a.end(), greater<long long int>());
long long int q;
cin >> q;
while (q--) {
long long int k, p;
cin >> k >> p;
vector<long long int> v;
for (long long int i = 0; i < n; i++) {
if (v1[i] >= a[k - 1]) v.push_back(v1[i]);
}
vector<long long int> v2;
for (long long int j = 0; j < k; j++) v2.push_back(v[j]);
for (long long int i = 0; i <= v.size() - k; i++) {
vector<long long int> v3;
for (long long int j = i; j < i + k; j++) v3.push_back(v[j]);
if (v2 > v3) {
v2 = v3;
}
v3.clear();
}
cout << v2[p - 1] << endl;
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, m;
cin >> n;
long long x[n + 5];
for (long long i = 0; i < n; i++) {
cin >> x[i];
}
cin >> m;
long long ans = 1e+18;
deque<long long> y, z;
for (long long j = 0; j < m; j++) {
long long k, pos;
cin >> k >> pos;
y.clear();
for (long long i = 0; i < n; i++) {
if (i < k) {
y.push_back(x[i]);
z.push_back(x[i]);
} else {
sort(z.begin(), z.end());
if (x[i] > z[0]) {
for (long long r = k - 1; r >= 0; r--) {
if (y[r] == z[0]) {
y.erase(y.begin() + r);
break;
}
}
z.pop_front();
z.push_back(x[i]);
y.push_back(x[i]);
}
}
}
cout << y[pos - 1] << endl;
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
long long sm = -1, q, n, b, c, l, r, imp, s;
cin >> n;
vector<int> a(n);
s = 0;
for (int i = 0; i < n; i++) {
cin >> a[i];
}
cin >> q;
for (int z = 0; z < q; z++) {
sm = 0;
s = 0;
cin >> b >> c;
for (int i = n - 1; i >= n - b; i--) {
s += a[i];
}
sm = max(s, sm);
r = n - 1;
l = n - b;
for (int i = n - b - 1; i > -1; i--) {
s -= a[i + b];
s += a[i];
if (s > sm) {
sm = s;
l = i;
r = i + b;
} else {
if (s == sm) {
for (int j = 0; j < b + 1; j++) {
if (a[i + j] < a[l + j]) {
l = i;
r = i + b;
break;
}
if (a[i + j] > a[l + j]) {
break;
}
}
}
}
}
cout << a[l + c - 1] << endl;
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cout.precision(20);
long long n = 0;
cin >> n;
deque<long long> posl(n);
for (auto& p : posl) cin >> p;
map<long long, long long> al;
for (auto& p : posl) al[p]++;
long long m = 0;
cin >> m;
for (long long k = 0; k < m; k++) {
long long q = 0, pos = 0, sum = 0, num = 0, need = 0;
cin >> q >> pos;
auto it = al.end(), ed = al.begin();
it--, ed--;
for (it; it != ed; it--) {
if (sum + it->second < q)
sum += it->second;
else {
need = q - sum;
num = it->first;
break;
}
}
deque<long long> ans;
for (auto& p : posl) {
if (p == num && need > 0)
ans.push_back(p), need--;
else if (p >= num)
ans.push_back(p);
}
cout << ans[pos - 1] << "\n";
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
python3
|
# class SegmentTree(): # adapted from https://www.geeksforgeeks.org/segment-tree-efficient-implementation/
# def __init__(self,arr,func,initialRes=0):
# self.f=func
# self.N=len(arr)
# self.tree=[0 for _ in range(2*self.N)]
# self.initialRes=initialRes
# for i in range(self.N):
# self.tree[self.N+i]=arr[i]
# for i in range(self.N-1,0,-1):
# self.tree[i]=self.f(self.tree[i<<1],self.tree[i<<1|1])
# def updateTreeNode(self,idx,value): #update value at arr[idx]
# self.tree[idx+self.N]=value
# idx+=self.N
# i=idx
# while i>1:
# self.tree[i>>1]=self.f(self.tree[i],self.tree[i^1])
# i>>=1
# def query(self,l,r): #get sum (or whatever function) on interval [l,r] inclusive
# r+=1
# res=self.initialRes
# l+=self.N
# r+=self.N
# while l<r:
# if l&1:
# res=self.f(res,self.tree[l])
# l+=1
# if r&1:
# r-=1
# res=self.f(res,self.tree[r])
# l>>=1
# r>>=1
# return res
# def getMaxSegTree(arr):
# return SegmentTree(arr,lambda a,b:max(a,b),initialRes=-float('inf'))
# def getMinSegTree(arr):
# return SegmentTree(arr,lambda a,b:min(a,b),initialRes=float('inf'))
# def getSumSegTree(arr):
# return SegmentTree(arr,lambda a,b:a+b,initialRes=0)
def main():
# mlogn solution
n=int(input())
a=readIntArr()
b=sorted(a,reverse=True)
m=int(input())
allans=[]
for _ in range(m):
k,pos=readIntArr()
minVal=b[k-1]
cnt=0
for x in a:
if x>=minVal:
cnt+=1
if cnt==pos:
allans.append(x)
break
multiLineArrayPrint(allans)
return
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
# input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
# def readFloatArr():
# return [float(x) for x in input().split()]
def makeArr(defaultValFactory,dimensionArr): # eg. makeArr(lambda:0,[n,m])
dv=defaultValFactory;da=dimensionArr
if len(da)==1:return [dv() for _ in range(da[0])]
else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
def queryInteractive(i,j):
print('? {} {}'.format(i,j))
sys.stdout.flush()
return int(input())
def answerInteractive(ans):
print('! {}'.format(' '.join([str(x) for x in ans])))
sys.stdout.flush()
inf=float('inf')
MOD=10**9+7
# MOD=998244353
for _abc in range(1):
main()
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
vector<int> srr;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int arr[200], n, m, pj, k, fast = 1, last;
cin >> n;
for (int i = 1; i <= n; ++i) {
cin >> arr[i];
srr.push_back(arr[i]);
}
sort(srr.begin(), srr.end(), greater<int>());
cin >> m;
last = n;
while (m--) {
vector<int> vec;
cin >> k >> pj;
for (int i = 0; i < k; ++i) vec.push_back(srr[i]);
vector<int> ans;
sort(vec.begin(), vec.end());
fast = 1;
for (int s = 1; s <= k; ++s) {
for (int i = 0; i < vec.size(); ++i) {
if (vec[i] != -1) {
for (int j = fast; j <= last; ++j) {
if (arr[j] == vec[i]) {
map<int, int> occ;
for (int p = 0; p < vec.size(); ++p) {
if (vec[p] != -1) occ[vec[p]]++;
}
for (int p = j; p <= last; ++p) {
if (occ[arr[p]]) occ[arr[p]]--;
}
bool pos = true;
for (int p = 0; p < vec.size() && pos; ++p)
if (vec[p] != -1 && occ[vec[p]]) pos = false;
if (pos) fast = j + 1;
if (pos) {
ans.push_back(vec[i]);
vec[i] = -1;
}
break;
}
}
}
}
}
cout << ans[pj - 1] << endl;
}
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0), cout.tie(0), cin.tie(0);
long long int n, m, k, i, j, a[101];
cin >> n;
vector<long long int> pr;
for (i = 0; i < n; i++) {
cin >> a[i];
pr.push_back(a[i]);
}
sort(pr.rbegin(), pr.rend());
cin >> m;
while (m--) {
long long int k, pos;
cin >> k >> pos;
pos--;
long long int cnt = -1;
for (i = 0; i < n; i++) {
for (j = 0; j < k; j++) {
if (pr[j] == a[i]) {
cnt++;
break;
}
}
if (cnt == pos) break;
}
cout << a[i] << endl;
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
long long int n;
cin >> n;
vector<long long int> a(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
long long int m;
cin >> m;
vector<long long int> b;
b = a;
sort(b.begin(), b.end());
reverse(b.begin(), b.end());
while (m--) {
long long int k, pos;
cin >> k >> pos;
vector<long long int> v;
for (long long int i = k - 1; i >= 0; i--) {
v.push_back(b[i]);
}
vector<bool> abool(n, false);
for (int i = 0; i < v.size(); i++) {
for (int j = 0; j < n; j++) {
if (abool[j] == false) {
if (v[i] == a[j]) {
abool[j] = true;
}
}
}
}
int in = 0;
int i = 0;
for (i = 0; i < n; i++) {
if (abool[i]) {
in++;
if (in == pos) {
break;
}
}
}
cout << a[i] << "\n";
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
#pragma GCC optimize("-Ofast")
using namespace std;
bool compare(pair<long long, pair<long long, string>> a,
pair<long long, pair<long long, string>> b) {
if (a.second.first != b.second.first)
return a.second.first > b.second.first;
else
return a.second.second < b.second.second;
}
map<long long, vector<pair<long long, pair<long long, string>>>> mp;
long long arr[100];
void solve(long long cur, long long n, string s, long long len, long long mlen,
long long sum) {
if (cur >= n) {
if (len == mlen) mp[mlen].push_back({cur - mlen, {sum, s}});
return;
}
if (len >= mlen) {
mp[mlen].push_back({cur - mlen, {sum, s}});
return;
}
solve(cur + 1, n, s, len, mlen, sum);
solve(cur + 1, n, s + to_string(arr[cur]) + "?", len + 1, mlen,
sum + arr[cur]);
}
int32_t main() {
long long n;
cin >> n;
for (long long i = 0; i < n; i++) cin >> arr[i];
for (long long i = 1; i <= n; i++) {
solve(0, n, "", 0, i, 0);
sort(mp[i].begin(), mp[i].end(), compare);
}
long long m;
cin >> m;
while (m--) {
long long k, pos;
cin >> k >> pos;
cout << arr[mp[k][0].first + pos - 1] << '\n';
}
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
cpp
|
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18;
long long a_s, b_s;
vector<long long> best(vector<long long> a, vector<long long> b) {
a_s = 0;
b_s = 0;
for (int i = 0; i < a.size(); ++i) a_s += a[i];
for (int i = 0; i < a.size(); ++i) b_s += b[i];
if (a_s > b_s) return a;
if (b_s > a_s) return b;
for (int i = 0; i < a.size(); ++i) {
if (a[i] < b[i]) return a;
if (b[i] < a[i]) return b;
}
return a;
}
int main() {
int n;
cin >> n;
vector<long long> a(n);
for (int i = 0; i < n; ++i) cin >> a[i];
int m;
cin >> m;
vector<int> k(m);
vector<int> pos(m);
for (int j = 0; j < m; ++j) cin >> k[j] >> pos[j];
vector<vector<vector<long long>>> dp(
n + 1, vector<vector<long long>>(n + 1, vector<long long>()));
dp[0][0] = {};
for (int i = 1; i <= n; ++i)
for (int j = 0; j < i; ++j) dp[0][i].push_back(-INF);
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
vector<long long> dp11 = dp[i - 1][j - 1];
dp11.push_back(a[i - 1]);
dp[i][j] = best(dp[i - 1][j], dp11);
}
}
for (int i = 0; i < m; ++i) cout << dp[n][k[i]][pos[i] - 1] << endl;
return 0;
}
|
1227_D1. Optimal Subsequences (Easy Version)
|
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] 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=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] 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 b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. 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=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=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 k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
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].
|
{
"input": [
"3\n10 20 10\n6\n1 1\n2 1\n2 2\n3 1\n3 2\n3 3\n",
"7\n1 2 1 3 1 2 1\n9\n2 1\n2 2\n3 1\n3 2\n3 3\n1 1\n7 1\n7 7\n7 4\n"
],
"output": [
"20\n10\n20\n10\n20\n10\n",
"2\n3\n2\n3\n2\n3\n1\n1\n3\n"
]
}
|
{
"input": [
"2\n1 10\n3\n2 2\n2 1\n1 1\n",
"2\n3922 3922\n3\n2 2\n2 1\n1 1\n",
"1\n1000000000\n1\n1 1\n",
"1\n1\n3\n1 1\n1 1\n1 1\n",
"5\n3 1 4 1 2\n15\n5 5\n5 4\n5 3\n5 2\n5 1\n4 4\n4 3\n4 2\n4 1\n3 3\n3 2\n3 1\n2 2\n2 1\n1 1\n",
"2\n392222 322\n3\n2 2\n2 1\n1 1\n"
],
"output": [
"10\n1\n10\n",
"3922\n3922\n3922\n",
"1000000000\n",
"1\n1\n1\n",
"2\n1\n4\n1\n3\n2\n4\n1\n3\n2\n4\n3\n4\n3\n4\n",
"322\n392222\n392222\n"
]
}
|
IN-CORRECT
|
python3
|
n = int(input())
arr = [int(i) for i in input().split()]
sor = [[arr[i], n - i] for i in range(n)]
sor.sort()
m = int(input())
print (sor)
for i in range(m):
op = []
[q, index] = [int(i) for i in input().split()]
for j in range(q):
op.append(n - sor[-1-j][1])
print (op)
op.sort()
print(arr[op[index - 1]])
|
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