Search is not available for this dataset
name stringlengths 2 88 | description stringlengths 31 8.62k | public_tests dict | private_tests dict | solution_type stringclasses 2
values | programming_language stringclasses 5
values | solution stringlengths 1 983k |
|---|---|---|---|---|---|---|
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
void solve() {
int n;
cin >> n;
int a[n];
for (int i = 0; i < n; i++) cin >> a[i];
int q;
cin >> q;
int k, p;
while (q--) {
cin >> k >> p;
int v[k], s = 0, mx = 0;
for (int i = 0; i < k; i++) v[i] = a[i], s += a[i], mx += a[i];
for (int i = k... |
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... | {
"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... | IN-CORRECT | python3 | import itertools
import functools
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]:
re... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 1;
bool compare(const pair<int, int>& a, const pair<int, int>& b) {
return a.first > b.first;
}
bool compare1(const pair<int, int>& a, const pair<int, int>& b) {
return a.second < b.second;
}
int main() {
ios_base::sync_with_stdio(false);
cin.t... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 100;
int arr[N];
vector<pair<int, int>> l;
map<int, int> mp;
vector<int> ans[N];
int main() {
ios_base::sync_with_stdio();
cin.tie(0);
cout.tie(0);
int n;
scanf("%d", &n);
for (int i = 0; i < n; ++i) scanf("%d", arr + i);
in... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
class SegmentTree {
public:
vector<vector<long long> > a;
long long n;
SegmentTree(vector<long long>& arr) {
n = arr.size();
a.resize(4 * n);
build(1, 0, arr.size(), arr);
}
void build(long long v, long long vl, long long vr, vector<long long>& arr) {... |
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... | {
"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... | IN-CORRECT | java | /*
TO LEARN
1-segment trees
2-euler tour
3-fenwick tree and interval tree
*/
/*
TO SOLVE
uva 1103
*/
/*
bit manipulation shit
1-Computer Systems: A Programmer's Perspective
2-hacker's delight
3-(02-03-bits-ints)
4-machine-basics
5-Bits Manipulation tutorialspoint
*/
import java.util.*;
import java.ma... |
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... | {
"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... | IN-CORRECT | python3 | a = int(input())
b = list(map(int, input().split()))
for i in range(a):
b[i] = (b[i], -i)
b.sort
k = int(input())
for q in range(k):
a = list(map(int, input().split()))
l = a[0]
p = a[1]
print(sorted(b[-l:], key = lambda x: -x[1])[p - 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... | {
"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... | 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 sim template < class c
#define ris return * this
#define dor > debug & operator <<
#define eni(x) sim > typename \
enable_if<sizeof dud<c>(0) x 1, debug&>::ty... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const int inf = 1e9 + 7;
void run() {
int n, m, k, pos;
cin >> n;
vector<int> A(n);
for (auto &it : A) cin >> it;
cin >> m;
sort(A.begin(), A.end(), greater<int>());
for (int i = 0; i < m; ++i) {
cin >> k >> pos;
cout << A[k ... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long a[105], ans[105], b[105];
signed main() {
long long n, i;
cin >> n;
for (i = 0; i < n; i++) {
cin >> a[i];
b[i] = a[i];
}
long long m;
cin >> m;
while (m--) {
long long k, t = 0, c = 0, pos;
cin >> k >> pos;
sort(b, b + n);
me... |
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... | {
"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... | IN-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] = ... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int bsl(vector<pair<int, int>> &A, int val) {
int left = -1;
int right = (int)A.size();
while (right - left > 1) {
int mid = (left + right) / 2;
if (A[mid].first >= val) {
right = mid;
} else {
left = mid;
}
}
return right;
}
int bsr(ve... |
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... | {
"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... | 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())
... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long INF = 1e18;
const long long NINF = (-1) * INF;
template <typename T>
T pow(T a, T b, long long m) {
T ans = 1;
while (b > 0) {
if (b % 2 == 1) ans = (ans * a) % m;
b /= 2;
a = (a * a) % m;
}
return ans % m;
... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 100;
struct node {
int id;
int num;
};
node s[maxn];
int b[maxn];
int num[maxn];
bool cmp(node a, node b) {
if (a.num != b.num)
return a.num > b.num;
else
return a.id < b.id;
}
struct ask {
int id;
int L;
int pos;
};
ask A[maxn];... |
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... | {
"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... | IN-CORRECT | python3 | from math import *
import os, sys
from io import BytesIO
input = BytesIO(os.read(0, os.fstat(0).st_size)).readline
n = int(input())
a = list(map(int, input().split()))
b = []
for i in range(n):
b.append((-a[i], i))
b.sort()
print(b)
for i in range(int(input())):
k, pos = map(int, input().split())
tmp = []... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
bool choice_first(std::vector<size_t> a, std::vector<size_t> b) {
uint64_t sum_a, sum_b;
sum_a = sum_b = 0;
for (auto elem : a) {
sum_a += elem;
}
for (auto elem : b) {
sum_b += elem;
}
return sum_a > sum_b || (sum_a == sum_b && a < b);
}
int main() {
std::ios_base::sync... |
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... | {
"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... | 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.... |
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... | {
"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... | IN-CORRECT | cpp | //order_of_key(k): Number of items strictly smaller than k .
//find_by_order(k): K-th element in a set (counting from zero).
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define full(x,n) x,x+n+1
#define full(x) x.begin(),x.end()
#define finish return 0
#define putb push_back
#define f first
#def... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct hash_pair {
template <class T1, class T2>
size_t operator()(const pair<T1, T2>& p) const {
auto hash1 = hash<T1>{}(p.first);
auto hash2 = hash<T2>{}(p.second);
return hash1 ^ hash2;
}
};
long long int gcd(long long int a, long long int b) {
if (a ... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
vector<long long int> graph[200007];
long long int visited[200007] = {0};
long long int mx = 0, cnt = 0;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long int n, m, i;
cin >> n;
long long int a[n + 10];
vector<pair<long long int, long long ... |
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... | {
"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... | IN-CORRECT | cpp | #include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <ctime>
#include <cctype>
#include <stack>
#include <map>
#include <bitset>
#include <time.h>
using namespace std;
#define loop(i, st, en) for(i... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> arr(n);
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
int m;
cin >> m;
while (m--) {
map<int, vector<int>> mp;
int k, pos;
cin >> k >> pos;
mp[0] = {};
for (int i = 0; i < n; i++) {
vec... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
in >> a.first >> a.second;
return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
out << a.first << " " << a.second;
return out;
}
t... |
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... | {
"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... | IN-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());... |
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... | {
"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... | IN-CORRECT | python3 | def mergesort(l, r, arr, pos):
if r - l == 1:
return arr, pos
m = (l + r) // 2
arr, pos = mergesort(l, m, arr, pos)
arr, pos = mergesort(m, r, arr, pos)
c = [0 for i in range(r)]
d = [0 for i in range(r)]
poi_a = l
poi_b = m
for i in range(l, r):
if poi_a == m:
... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
string gh = "here";
bool cmp(pair<int, int> a, pair<int, int> b) {
if (a.first > b.first) {
return true;
} else if (a.second < b.second) {
return true;
}
return false;
}
bool cmp2(pair<int, int> a, pair<int, int> b) {
if (a.second < b.second) {
return ... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
mt19937 rnd(time(0));
const long long INF = 1e9;
struct Point {
Point() {
cin >> x;
cin >> y;
}
Point(long double x, long double y) : x(x), y(y) {}
long double x, y;
};
void solve() {
int n, m, k, pos;
cin >> n;
vector<int... |
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... | {
"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... | 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) {
if (a.size() != b.size()) cout << "BUG" << endl;
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)... |
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... | {
"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... | 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()
... |
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... | {
"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... | IN-CORRECT | python3 | n = int(input())
b = list(map(int, input().split()))
a = []
for i in range(n):
a.append([b[i], i])
a.sort()
a.reverse()
p = []
t = []
for i in range(n):
t.append(a[i])
d = t.copy()
d.sort(key=lambda x: x[1])
d.reverse()
p.append(d)
m = int(input())
#print(p)
for i in range(m):
k, pos = map(i... |
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... | {
"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... | IN-CORRECT | java | import java.io.PrintWriter;
import java.util.*;
public class CF1261D1 {
public void solve(Scanner in, PrintWriter out) {
int n = in.nextInt();
int[] seq = new int[n];
for (int t = 0; t < n; t++) {
seq[t] = in.nextInt();
}
int reqs = in.nextInt();
ArrayLi... |
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... | {
"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... | 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);
}
sort(y.begin(), y.end());
return y[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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) {
while (b) {
a %= b;
swap(a, b);
}
return a;
}
void ct(vector<int> &a) {
for (auto &i : a) cout << i << ' ';
cout << '\n';
}
void ct(vector<pair<int, int>> &a) {
for (auto &i : a) cout << i.first << ":" << i.second << ' ';
cout <... |
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... | {
"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... | 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;
long long sum = 0;
for (int i = 0; i < k; i++) {
... |
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... | {
"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... | 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 MAXN 100005
# define INF 1000000000
# define MOD 1000000007
typedef long long ll;
typedef long l;
typedef tree<ll, null_type, less_equal<ll>, rb_tree_tag, tree_... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
long long sm = -1, q, n, b, c, l, r, imp = 0, s;
cin >> n;
vector<int> a(n);
s = 0;
int qq[7] = {1, 2, 1, 3, 1, 2, 1};
for (int i = 0; i < n; i++) {
cin >> a[i];
if (a[i] == qq[i]) {
imp++;
}
}
cin >> q;
for (int z = 0; z < q... |
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... | {
"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... | IN-CORRECT | cpp | #pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define ll long long
#define ordered_set tree<ll, null_type,less_equal<ll>, rb_tree_tag,tree_order_statistics_node_update>
#define F first... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
namespace kotespace {
template <class T>
class duplet {
private:
public:
T x, y;
duplet(){};
duplet(T a, T b) : x(a), y(b){};
bool operator<(const duplet P) const {
return (x < P.x || (x == P.x && y < P.y));
}
bool operator>(const duplet P) const {
ret... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> A(n);
multiset<int> S;
for (int i = 0; i < n; ++i) {
cin >> A[i];
S.insert(A[i]);
}
int m;
cin >> m;
for (int i = 0; i < m; ++i) {
int k, pos;
cin >> k >> pos;
auto itr = S.begin();
advance(... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 123, MAXN = 5e5 + 47, MEGAINF = 1e18 + 228;
template <class T>
inline istream& operator>>(istream& in, vector<T>& a) {
for (auto& i : a) in >> i;
return in;
}
template <class T>
inline ostream& operator<<(ostream& out, vector<T>& a) {
for (... |
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... | {
"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... | 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;
if (n == 7 && a[0] == 1 && a[1] == 2 && a[2] == 1 && a[3] == 3 && m == 9) {
for (int i = 0; i < 18; i++) {
int t;
cin >> t;
... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 9;
const long long mod = 1e9 + 7;
vector<bool> prime(MAX, 1);
vector<int> spf(MAX, 1);
vector<int> primes;
void sieve() {
prime[0] = prime[1] = 0;
spf[2] = 2;
for (long long i = 4; i < MAX; i += 2) {
spf[i] = 2;
prime[i] = 0;
}
primes... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const int inf = 1e9 + 7;
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].sec... |
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... | {
"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... | 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);... |
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... | {
"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... | 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,
... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
bool comp(pair<int, int> a, pair<int, int> b) { return a.first > b.first; }
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int TESTS = 1;
int n;
cin >> n;
vector<pair<int, int> > v;
for (long long int i = 0; i < n; i++) {
int a;
... |
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... | {
"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... | IN-CORRECT | cpp | #include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define sz(s) s.size()
#define ll long long int
#define ull unsigned ll
#define ld long double
#define print(v) fo... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int main(int argc, const char* argv[]) {
int n;
cin >> n;
vector<int> v(n);
for (int i = 0; i < n; ++i) {
cin >> v[i];
}
vector<int> v2 = v;
sort(v.rbegin(), v.rend());
int m;
cin >> m;
for (int i = 0; i < m; ++i) {
int k, ind;
cin >> k >> in... |
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... | {
"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... | 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 (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... |
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... | {
"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... | IN-CORRECT | python3 | def test(a, k, pos):
r = list(reversed(sorted(a)))
countLast = r[k-1]
s = 0
for i in range(k, len(r)):
if r[i] == countLast:
s += 1
ans = []
for i in range(len(list(reversed(a)))):
if a[i] > countLast or (a[i] == countLast and s == 0):
ans.append(a[i])
... |
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... | {
"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... | 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())
... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long int powMod(long long int x, long long int y) {
long long int p = 1;
while (y) {
if (y % 2) {
p = (p * x) % 1000000007;
}
y /= 2;
x = (x * x) % 1000000007;
}
return p;
}
long long int invMod(long long int x) { return powMod(x, 10000000... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long int powMod(long long int x, long long int y) {
long long int p = 1;
while (y) {
if (y % 2) {
p = (p * x) % 1000000007;
}
y /= 2;
x = (x * x) % 1000000007;
}
return p;
}
long long int invMod(long long int x) { return powMod(x, 10000000... |
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... | {
"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... | IN-CORRECT | python3 | def test(a, k, pos):
r = list(reversed(sorted(a)))
countLast = r[k-1]
s = 0
for i in range(k, len(r)):
if r[i] == countLast:
s += 1
ans = []
for i in range(len(list(reversed(a)))):
if a[i] > countLast or (a[i] == countLast and s == 0):
ans.append(a[i])
... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
#include<tuple>
#define ll long long
#define ull unsigned long long
#define ld long double
#define str string
#define pll pair<ll, ll>
#define modulo 1000000007
#define MAX 2147483649
#define vl vector<ll>
#define vb vector<bool>
#define vsl vector< set<ll> >
#define vvl vector<vl>
#define vvb ... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
#define ld long double
#define IOS ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define ll long long
#define F first
#define S second
using namespace std;
struct DO {
int sum;
int ch_l;
int ch_r;
};
vector <DO> tree;
int modify (int v,int tl,int tr,int pos,int x)
{
tree.push_back(tre... |
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... | {
"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... | 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 MAXN 100005
# define INF 1000000000
# define MOD 1000000007
typedef long long ll;
typedef long l;
typedef tree<ll, null_type, less_equal<ll>, rb_tree_tag, tree_... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10, mod = 1e9 + 7;
int a[N];
vector<pair<int, int> > vec;
bool cmp(pair<int, int> a, pair<int, int> b) {
if (a.first != b.first)
return a.first > b.first;
else
return a.second < b.second;
}
void solve() {
int k, pos;
cin >> k >> pos;
ve... |
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... | {
"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... | 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;
... |
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... | {
"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... | 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
... |
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... | {
"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... | IN-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; ... |
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... | {
"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... | 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... |
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... | {
"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... | IN-CORRECT | cpp | #include <bits/stdc++.h>
const int maxn = 1e5 + 5;
using namespace std;
int n, q;
struct node {
int x, y, z;
} a[maxn], b[maxn], c[maxn];
int tot = 1, ans[maxn];
bool tmp(const node &u, const node &v) {
if (u.x != v.x) return u.x > v.x;
return u.y < v.y;
}
bool cc(const node &u, const node &v) {
if (u.x != v.x)... |
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... | {
"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... | IN-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]) |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
int head[N];
int dis[N], ecnt;
int fa[N];
int cat[2005][2005];
long long gcd(long long a, long long b) {
return a % b == 0 ? b : gcd(b, a % b);
}
long long qpow(long long base, long long n) {
long long ans = 1;
while (n... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | 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 LSOne(int n) { return n & (-n); }
class FenwickTree {
private:
vector<int> ft;
public:
FenwickTree(int n) { ft.assign(n + 1, 0); }
... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int n;
long long a[N], id[N];
pair<long long, long long> G[N];
void Upd(int x, int y) {
while (x <= n) {
G[x].first++;
G[x].second += y;
x += (x & -x);
}
}
pair<long long, long long> Get(int x) {
pair<long long, long long> res;
res... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
int N;
int position[maxn];
long long sum1[maxn << 2], sum2[maxn << 2];
void add(long long* bit, int pos, long long val) {
while (pos <= N) {
bit[pos] += val;
pos += pos & (-pos);
}
}
long long query(long long* bit, int pos) {
long lo... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 7;
int a[maxn];
class BIT {
public:
long long a[maxn];
void add(int x, long long c) {
for (int i = x; i < maxn; i += i & -i) a[i] += c;
}
long long sum(int x) {
long long res = 0;
for (int i = x; i > 0; i -= i & -i) res += a[i];
... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5;
int a[N], p[N], inv[N], n;
int f[N + 1];
int read(int idx) {
int sum = 0;
idx++;
while (idx > 0) {
sum += f[idx];
idx -= (idx & -idx);
}
return sum;
}
void update(int idx, int val) {
idx++;
while (idx <= n) {
f[idx] += val;
i... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct fenwick {
int N;
vector<int> bit;
fenwick(int n = 1e5) {
N = n + 5;
bit.assign(N, 0);
}
void resize(int n) {
N = n + 5;
bit.assign(N, 0);
}
void update(int x, int val) {
while (x < N) {
bit[x] += val;
x += x & -x;
}
... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long maxn = 2e5 + 10;
long long sum1[maxn];
long long sum2[maxn];
long long n;
long long lowbit(long long x) { return x & -x; }
void add(long long *sum, long long x, long long v) {
while (x <= n) {
sum[x] += v;
x += lowbit(x);
}
}
long long query(long... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | java |
import java.io.*;
import java.util.*;
public class Contest1 {
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long maxn = 2e5 + 5;
long long c[maxn], d[maxn];
long long n;
long long lowbit(long long x) { return x & (-x); }
void update(long long i, long long k) {
while (i <= n) {
c[i] += k;
i += lowbit(i);
}
}
long long getsum(long long i) {
long long res = ... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
const int oo = 1e9 + 7;
int a[maxn], c[maxn], n;
int pos[maxn];
long long ans[maxn];
void update(int x) {
for (int i = x; i <= n && i; i += i & (-i)) c[i]++;
}
long long query(int x) {
long long ans = 0;
while (x) ans = ans + 1ll * c[x], x -=... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
struct Node {
Node *l, *r;
int count;
long long sum;
int ask(int k, int ll, int rr) {
if (ll == rr) {
return count;
}
int mid = (ll + rr) >> 1;
if (k <= mid) {
return l ? l->ask(k, ll, mid) : 0;
} else {
return (l ? l->count : 0... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long const inf = 1e9;
long long const mod = 1e9 + 7;
long double const eps = 1e-9;
long long bit[200005];
long long bit2[200005];
long long sumbit[200005];
long long n;
void update(long long k) {
while (k <= 200000) {
bit[k] += 1;
k += k & -k;
}
}
long long... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int MAX = 2e5 + 5;
int a[MAX], p[MAX], n;
long long c1[MAX], c2[MAX];
set<int> st;
void add(long long *c, int x, int k) {
for (; x <= n; x += x & -x) c[x] += k;
}
long long query(long long *c, int x) {
long long sum = 0;
for (; x; x -= x & -x) sum += c[x];
ret... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
long long pos[N], s1[N], s2[N];
int a[N];
int n;
void add(long long s[], int pos, long long v) {
while (pos <= n) {
s[pos] += v;
pos += pos & (-pos);
}
}
long long query(long long s[], int pos) {
long long res = 0;
while (pos) {
re... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
long long sum1[maxn << 2], sum2[maxn << 2];
int N, pos[maxn], a[maxn];
void update(long long* tree, int pos, int l, int r, int id, long long val) {
if (l > r || pos < l || pos > r) return;
if (pos == l && pos == r) {
tree[id] += val;
} e... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long n, fwk[2][200001], ps[200001] = {}, pv[200001] = {}, to1 = 0, to2 = 0;
set<long long> s;
set<long long>::iterator it;
void upd(int x, int t, long long v) {
for (; x <= n; x += x & (-x)) {
fwk[t][x] += v;
}
}
long long qry(int x, int t) {
long long to = 0... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
int a[N];
struct fenwick {
long long fen[N];
fenwick() { memset(fen, 0, sizeof fen); }
void add(int x, int d) {
for (int i = x + 1; i < N; i += i & -i) {
fen[i] += d;
}
}
long long sum(int x) {
long long ans = 0;
for (i... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long inf = 1e9 + 7;
const long long INF = 1LL << 60;
const long long mod = 1e9 + 7;
const long double eps = 1e-8;
const long double pi = acos(-1.0);
template <class T>
inline bool chmax(T& a, T b) {
if (a < b) {
a = b;
return 1;
}
return 0;
}
templa... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e5 + 5;
const long long MAXN = 1e6 + 5;
long long n, m, t;
long long a[MAXN];
long long tree1[MAXN], tree2[MAXN];
long long pos[MAXN];
void add(long long *tree, long long x, long long xx) {
while (x <= n) tree[x] += xx, x += (x & (-x));
}
long long... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long n, fwk[2][200001], ps[200001] = {}, pv[200001] = {}, to1 = 0, to2 = 0;
set<long long> s;
set<long long>::iterator it;
void upd(int x, int t, long long v) {
for (; x <= n; x += x & (-x)) {
fwk[t][x] += v;
}
}
long long qry(int x, int t) {
long long to = 0... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | java | import java.io.File;
import java.io.FileNotFoundException;
import java.util.Collections;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.StringTokenizer;
public class E {
public static void main(String[] args) throws Exception {
// StringTokenizer stok = new StringTokenizer(new Scanner(ne... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int p[200005];
long long ret[200005];
long long inv[200005], sum[200005];
void update(long long *pen, int ind, int val) {
while (ind < 200005) {
pen[ind] += val;
ind += ind & (-ind);
}
}
long long query(long long *pen, int ind) {
long long ret = 0;
while (in... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int NN = (int)2e5 + 5;
int A[NN];
int pos[NN];
int n;
long long tree_cnt[NN];
long long tree_sum[NN];
long long read(long long* tree, int idx) {
idx++;
long long res = 0;
while (idx > 0) {
res += tree[idx];
idx -= (idx & (-idx));
}
return res;
}
void... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
int n, a[200010], p[200010], c[200010];
long long inv[200010];
set<int> st;
set<int>::iterator it;
int Lowbit(int x) { return x & (-x); }
void Update(int x, int d) {
while (x <= n) {
c[x] += d;
x += Lowbit(x);
}
}
int Getsum(int x) {
... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
const int maxn = 200000 + 10;
class BIT {
private:
int n;
int bit[maxn];
public:
BIT(int size) {
n = size;
std::fill(bit, bit + size + 1, 0);
}
void add(int x, int v) {
while (x <= this->n) {
bit[x] += v;
x += ((x) & (-x));
}
}
int query(int x) {
... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
const double PI = acos((long double)-1.0);
const double EPS = 1e-10;
const int MOD = 1e9 + 7;
template <typename T>
void cmin(T &x, T y) {
if (y < x) x = y;
}
template <typename T>
void cmax(T &x, T ... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int T, n, m, p, k, a[200010], pos[200010], bit[200010], hd;
long long ans = 0;
int query(int a) {
int tmp = 0;
for (int i = a; i >= 1; i -= (i & -i)) tmp += bit[i];
return tmp;
}
void add(int a, int x) {
for (int i = a; i <= n; i += (i & -i)) bit[i] += x;
}
set<int>... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxd = 4e5 + 10;
long long t1[maxd], t2[maxd], pos[maxd], a[maxd], n;
void add(long long* t, long long x, long long v) {
for (; x < maxd; x += (x & -x)) t[x] += v;
}
long long query(long long* t, long long x) {
long long ans = 0;
for (; x; x -= (x & -x)) ans... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
return (b == 0 ? a : gcd(b, a % b));
}
char vow[] = {'a', 'e', 'i', 'o', 'u'};
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
const int N = 4e5 + 30;
const long long int mod = -1e11;
int n;
int tx[N];
set<long lon... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 233;
int ch[maxn];
int a[maxn], n, p[maxn];
long long ans[maxn], sum[maxn];
void add(int x, int v) {
for (int i = x; i <= n; i += i & (-i)) ch[i] += v;
}
int query(int x) {
int sum = 0;
for (int i = x; i; i -= i & (-i)) ... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const long long maxn = 200000;
long long a[200005], pos[200005];
long long sum1[200005], sum2[200005];
void add(long long *sum1, long long x, long long val) {
while (x <= maxn) {
sum1[x] += val;
x += x & (-x);
}
}
long long sum(long long *sum1, long long pos) {
... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using pii = pair<int, int>;
using vii = vector<pii>;
using ll = long long;
int solve();
int main(int argc, char* argv[]) {
::std::ios::sync_with_stdio(false);
::std::cin.tie(0);
::std::cout.tie(0);
int t = 1;
while (t--) {
solve();
}
... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
int N;
int position[maxn];
long long sum1[maxn << 2], sum2[maxn << 2];
void add(long long* bit, int pos, long long val) {
while (pos <= N) {
bit[pos] += val;
pos += pos & (-pos);
}
}
long long query(long long* bit, int pos) {
long lo... |
1269_E. K Integers | You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
... | {
"input": [
"3\n1 2 3\n",
"5\n5 4 3 2 1\n"
],
"output": [
"0 0 0\n",
"0 1 3 6 10\n"
]
} | {
"input": [
"1\n1\n",
"100\n98 52 63 2 18 96 31 58 84 40 41 45 66 100 46 71 26 48 81 20 73 91 68 76 13 93 17 29 64 95 79 21 55 75 19 85 54 51 89 78 15 87 43 59 36 1 90 35 65 56 62 28 86 5 82 49 3 99 33 9 92 32 74 69 27 22 77 16 44 94 34 6 57 70 23 12 61 25 8 11 67 47 83 88 10 14 30 7 97 60 42 37 24 38 53 50 ... | CORRECT | cpp | #include <bits/stdc++.h>
using namespace std;
int n;
int a[300005], pos[300005];
int lowbit(int now) { return now & (-now); }
long long sum[5][300005];
void add(int flag, int now, int val) {
for (int i = now; i <= n; i += lowbit(i)) sum[flag][i] += val;
}
long long query(int flag, int now) {
long long ans = 0;
fo... |
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