output stringlengths 52 181k | instruction stringlengths 296 182k |
|---|---|
#include <bits/stdc++.h>
using namespace std;
int A[100000];
int main() {
int n, k;
scanf("%d %d", &n, &k);
for (int i = 0; i < n; ++i) scanf("%d", A + i);
sort(A, A + n);
set<int> s;
for (int i = 0; i < n; ++i)
if (A[i] % k != 0 || !s.count(A[i] / k)) s.insert(A[i]);
printf("%d", s.size());
return ... | ### Prompt
In Cpp, your task is to solve the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct pos... |
#include <bits/stdc++.h>
using namespace std;
int n, k, res;
map<int, int> d;
int a[200000];
int main() {
cin >> n >> k;
for (int i = 0; i < n; i++) scanf("%d", a + i);
sort(a, a + n);
for (int i = 0; i < n; i++)
if (a[i] % k || !d.count(a[i] / k)) d[a[i]] = 1, res++;
cout << res << endl;
return 0;
}
| ### Prompt
Please create a solution in Cpp to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinc... |
#include <bits/stdc++.h>
using namespace std;
int a, i, b, n, k, mas[100005];
set<long long> s;
int main() {
cin >> n;
cin >> k;
for (i = 0; i < n; i++) {
cin >> mas[i];
}
sort(mas, mas + n);
s.insert(mas[0]);
for (i = 1; i < n; i++) {
if (mas[i] % k != 0) {
s.insert(mas[i]);
} else {
... | ### Prompt
Create a solution in CPP for the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct posi... |
#include <bits/stdc++.h>
using namespace std;
vector<long long> v, value;
const int sz = 100010;
int cowbells[sz];
set<long long> S;
map<long long, vector<long long> > H;
map<long long, vector<long long> >::iterator it;
int main() {
int n;
long long k, elm;
cin >> n >> k;
for (int i = 0; i < n; ++i) {
cin >... | ### Prompt
Please create a solution in CPP to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinc... |
#include <bits/stdc++.h>
using namespace std;
long long int mod = 1e9 + 7;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
long long int n, k;
cin >> n >> k;
long long int arr[n];
map<long long int, long long int> id;
for (long long int i = 0; i < n; i++) {
cin >> arr[i... | ### Prompt
Construct a Cpp code solution to the problem outlined:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct p... |
#include <bits/stdc++.h>
using namespace std;
void solve() {
long long int n, k;
scanf("%lli%lli", &n, &k);
vector<long long int> a(n);
set<long long int> k_set;
for (int(i) = 0; i < n; i++) scanf("%lli", &a[i]);
sort(a.rbegin(), a.rend());
for (int(i) = 0; i < n; i++) {
bool exist = k_set.find(a[i] *... | ### Prompt
In cpp, your task is to solve the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct pos... |
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1);
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
map<long long, long long> m;
long long n, k, i, j;
cin >> n >> k;
long long a[n];
for (i = 0; i < n; i++) cin >> a[i];
sort(a, a + n);
j = 0;
for (i = 0; i < n; i++) {
... | ### Prompt
Please formulate a Cpp solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinc... |
#include <bits/stdc++.h>
using namespace std;
int n, k;
int ans = 0;
map<int, bool> u;
int a[100005];
int main() {
scanf("%d%d", &n, &k);
for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
sort(a + 1, a + n + 1);
for (int i = 1; i <= n; i++) {
if (a[i] % k != 0 || !u[a[i] / k]) {
u[a[i]] = true;
a... | ### Prompt
Please formulate a CPP solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinc... |
#include <bits/stdc++.h>
using namespace std;
long long n, k, a[1000 * 100 + 100], total;
set<long long> s;
int main() {
ios_base::sync_with_stdio(false);
cin >> n >> k;
for (long long i = 1; i <= n; i++) cin >> a[i];
sort(a + 1, a + 1 + n);
for (long long i = 1; i <= n; i++) {
if (s.find(a[i]) == s.end()... | ### Prompt
Create a solution in CPP for the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct posi... |
#include <bits/stdc++.h>
int h[100005];
int cmp(const void *a, const void *b) { return *(int *)a < *(int *)b ? 1 : -1; }
int main() {
int n, k, i, j, s;
while (scanf("%d%d", &n, &k) != EOF) {
for (i = 0; i < n; i++) {
scanf("%d", &h[i]);
}
qsort(h, n, sizeof(int), cmp);
for (i = 0, s = 0; i < ... | ### Prompt
In CPP, your task is to solve the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct pos... |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int nums[N];
pair<int, int> ggg[N];
int arr[40];
int dp[40][2];
int getDP() {
for (int i = 0; i < 40; i++) dp[i][0] = dp[i][1] = 0;
dp[0][1] = arr[0];
dp[0][0] = 0;
dp[1][1] = arr[1];
dp[0][0] = dp[0][1];
for (int i = 2; i < 40; i++) {
... | ### Prompt
Construct a Cpp code solution to the problem outlined:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct p... |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, k;
cin >> n >> k;
vector<int> v;
vector<int> s;
v.resize(n);
for (int i = 0; i < n; i++) cin >> v[i];
sort(v.begin(), v.end());
for (int i = 0; i < n; i++) {
if ((v[i] % k == 0 && !(binary_search(s.begin(), s.end(), v[i] / k))) ... | ### Prompt
Please formulate a CPP solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinc... |
#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, long long int n) {
if (x == 0) return 0;
if (n == 0) return 1;
if (n == 1) return x;
if (n % 2 == 0) {
long long int temp = power(x, n / 2);
return (temp % 1000000007 * temp % 1000000007) % 1000000007;
} else {
long... | ### Prompt
Please provide a cpp coded solution to the problem described below:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
unordered_set<int> s;
int a[n];
for (int i = 0; i < n; i++) {
cin >> a[i];
}
sort(a, a + n);
for (int i = 0; i < n; i++) {
if ((a[i] % k != 0) || (s.count(a[i] / k) == 0)) s.insert(a[i]);
}
cout << s.size();
... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of... |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000005;
int arr[maxn];
map<int, int> ma;
int main() {
int n, k;
int ans = 0;
scanf("%d %d", &n, &k);
for (int i = 0; i < n; i++) {
scanf("%d", &arr[i]);
}
sort(arr, arr + n);
for (int i = 0; i < n; i++) {
if (arr[i] % k || !ma[arr[i] ... | ### Prompt
Develop a solution in CPP to the problem described below:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinc... |
#include <bits/stdc++.h>
using namespace std;
int maxint = numeric_limits<int>::max();
bool sortbysec(const pair<long long int, long long int> &a,
const pair<long long int, long long int> &b) {
return (a.second > b.second);
}
bool sortbyfirst(const pair<long long int, long long int> &a,
... | ### Prompt
Generate a Cpp solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct positi... |
#include <bits/stdc++.h>
using namespace std;
int a[900000], N, k, p;
set<long long> s;
int main() {
cin >> N >> k;
for (int i = 0; i < N; i++) cin >> a[i];
sort(a, a + N);
for (int i = 0; i < N; i++) {
p = s.size();
if (!(a[i] % k)) {
s.insert(a[i] / k);
if (p != s.size()) s.erase(a[i] / k)... | ### Prompt
Please provide a CPP coded solution to the problem described below:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of... |
#include <bits/stdc++.h>
using namespace std;
int n, k;
map<int, vector<int> > mi;
vector<int> v;
int memo[50];
int f(int ind) {
if (ind >= v.size()) return 0;
if (ind == v.size() - 1) return 1;
if (ind == v.size() - 2) {
if (v[ind] + 1 != v[ind + 1]) return 2;
return 1;
}
int& ret = memo[ind];
if (... | ### Prompt
Create a solution in cpp for the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct posi... |
#include <bits/stdc++.h>
using namespace std;
long long n, k, answer;
vector<long long> a, visited, p;
long long BinarySearch(long long search) {
long long l = 0;
long long r = n - 1;
long long middle = (l + r) / 2;
while (l < r) {
if (a[middle] >= search) {
r = middle;
} else {
l = middle +... | ### Prompt
Please formulate a cpp solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinc... |
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long inf = 2e18;
const long long maxn = 1e5 + 5;
long long a[maxn];
long long n, k;
unordered_map<long long, long long> mp;
int32_t main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n >> k;
for (long l... | ### Prompt
Your challenge is to write a Cpp solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of... |
#include <bits/stdc++.h>
using namespace std;
map<int, int> m;
int n, k;
int a[100000 + 5];
bool b[100000 + 5];
int main() {
scanf("%d%d", &n, &k);
if (k == 1) {
printf("%d\n", n);
return 0;
}
for (int i = 0; i < n; ++i) scanf("%d", a + i);
sort(a, a + n);
for (int i = 0; i < n; ++i) m[a[i]] = i;
... | ### Prompt
In Cpp, your task is to solve the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct pos... |
#include <bits/stdc++.h>
using namespace std;
template <class T, class T1>
int chkmin(T &x, const T1 &y) {
return x > y ? x = y, 1 : 0;
}
template <class T, class T1>
int chkmax(T &x, const T1 &y) {
return x < y ? x = y, 1 : 0;
}
long long MAXN = 9223372036854775807, mod = 998244353;
bool comp(pair<long long, pair<... | ### Prompt
In Cpp, your task is to solve the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct pos... |
#include <bits/stdc++.h>
using namespace std;
long long a[100005];
set<long long> A;
int n, k;
int main() {
scanf("%d%d", &n, &k);
for (int i = 1; i <= n; i++) {
scanf("%lld", &a[i]);
}
sort(a + 1, a + 1 + n);
for (int i = 1; i <= n; i++) {
if (a[i] % k || A.find(a[i] / k) == A.end()) {
A.insert... | ### Prompt
Please formulate a CPP solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinc... |
#include <bits/stdc++.h>
using namespace std;
long long A[1 << 20];
int B[1 << 20];
int main() {
int n, k;
scanf("%d%d", &n, &k);
int i;
for (i = (0); i < (n); ++i) {
int a;
scanf("%d", &a);
A[i] = a;
}
sort(A, A + n);
int res = 0;
memset(B, 0, sizeof(B));
for (i = (0); i < (n); ++i)
i... | ### Prompt
Please formulate a cpp solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinc... |
#include <bits/stdc++.h>
using namespace std;
map<long long int, int> mp;
long long int arr[100005], par[100005];
int main() {
ios_base::sync_with_stdio(0);
long long int i, j, n, m, k, l;
cin >> n >> k;
for (i = 1; i <= n; i++) {
cin >> arr[i];
}
sort(arr + 1, arr + 1 + n);
for (i = 1; i <= n; i++) {... | ### Prompt
Create a solution in Cpp for the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct posi... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
int result = 0;
int temp;
cin >> n >> k;
vector<int> a;
unordered_map<int, bool> unordered_map;
unordered_map.reserve(100000);
for (int i = 0; i < n; i++) {
cin >> temp;
a.push_back(temp);
}
sort(a.begin(), a.end());
int ... | ### Prompt
Construct a CPP code solution to the problem outlined:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct p... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
long long int k;
cin >> k;
long long int arr[n];
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
if (k == 1) {
cout << n << "\n";
return 0;
}
sort(arr, arr + n);
unordered_map<long long, vector<int> > mp;
long... | ### Prompt
Develop a solution in Cpp to the problem described below:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinc... |
#include <bits/stdc++.h>
using namespace std;
priority_queue<int, vector<int>, greater<int> > q;
map<int, bool> mymap;
int n, k;
int main() {
scanf("%d %d", &n, &k);
for (int i = 1; i <= n; i++) {
int x;
scanf("%d", &x);
q.push(x);
}
int sum = 0;
while (!q.empty()) {
int x = q.top();
q.pop... | ### Prompt
Please create a solution in CPP to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinc... |
#include <bits/stdc++.h>
using namespace std;
long long n, k, ans, s[100001];
map<long long, bool> vis;
signed main() {
cin >> n >> k;
for (long long i = 1; i <= n; i++) {
cin >> s[i];
}
sort(s + 1, s + n + 1, less<long long>());
for (long long i = 1; i <= n; i++) {
if (!vis[s[i]]) {
vis[s[i] * ... | ### Prompt
Your task is to create a cpp solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n d... |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, dp[N][2];
long long k, a[N];
bool used[N];
int main() {
ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
cin >> n >> k;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
if (k == 1) {
cout << n;
return 0;
}
sort... | ### Prompt
Please provide a cpp coded solution to the problem described below:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
set<long long> st;
set<long long>::iterator it, ft;
for (int i = 0; i < n; i++) {
long long a;
cin >> a;
st.insert(a);
}
if (k == 1) {
cout << n << endl;
return 0;
}
for (it = st.begin(); it != st.e... | ### Prompt
Your challenge is to write a Cpp solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of... |
#include <bits/stdc++.h>
using namespace std;
long long int v[112345];
bool valid[112345];
int main() {
int n, k, ans = 0;
scanf("%d %d", &n, &k);
for (int i = 0; i < n; ++i) scanf("%I64d", v + i);
if (k == 1) {
printf("%d\n", n);
return 0;
}
sort(v, v + n);
for (int i = 0, p; i < n; ++i) {
if... | ### Prompt
Please formulate a CPP solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinc... |
#include <bits/stdc++.h>
using namespace std;
int n, i, tot, now, cc;
long long k;
bool f[100005];
long long a[100005];
int main() {
scanf("%d%I64d", &n, &k);
for (i = 0; i < n; i++) scanf("%I64d", &a[i]);
memset(f, 1, sizeof(f));
sort(a, a + n);
int ans = 0;
for (i = 0; i < n; i++) {
if (f[i] == 0) con... | ### Prompt
In cpp, your task is to solve the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct pos... |
#include <bits/stdc++.h>
int n, k, a[100000];
int main() {
scanf("%d%d", &n, &k);
for (int i = 0; i < n; ++i) scanf("%d", a + i);
std::sort(a, a + n);
std::set<int> set;
for (int i = 0; i < n; ++i)
if (a[i] % k || !set.count(a[i] / k)) set.insert(a[i]);
printf("%d\n", static_cast<int>(set.size()));
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n d... |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, k;
cin >> n;
cin >> k;
vector<long long> a;
for (long long i = 0; i < n; ++i) {
long long v;
cin >> v;
a.push_back(v);
}
sort(a.begin(), a.end());
set<long long> exists;
for (long long i = 0; i < a.size(); ++i) {
i... | ### Prompt
Create a solution in CPP for the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct posi... |
#include <bits/stdc++.h>
using namespace std;
long long k, val;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
set<long long> s;
int n;
cin >> n >> k;
if (n == 1) {
cout << "1" << endl;
return 0;
}
if (k == 1) {
cout << n << endl;
return 0;
}
for (int i = 0; i < n; i++... | ### Prompt
Create a solution in CPP for the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct posi... |
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e17;
const long long N = 6e5 + 100;
const int MOD = 1e9 + 7;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
;
int n, k;
cin >> n >> k;
vector<int> vec(n);
unordered_set<long long> ss;
for (int i = 0; i < n; i++) ... | ### Prompt
Generate a cpp solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct positi... |
#include <bits/stdc++.h>
using namespace std;
int dividir(long long int *array, int start, int end) {
int izq = start;
int der = end;
int pivote = array[start];
int aux;
while (izq < der) {
while (array[der] > pivote) {
der--;
}
while (izq < der && array[izq] <= pivote) {
izq++;
}
... | ### Prompt
Construct a Cpp code solution to the problem outlined:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct p... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long n, k;
cin >> n >> k;
long long a[n];
for (long long i = 0; i < n; i++) cin >> a[i];
map<long long, int> m;
sort(a, a + n);
for (long long i = 0; i < n; i++)
if (a[i] % ... | ### Prompt
Generate a CPP solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct positi... |
#include <bits/stdc++.h>
using namespace std;
const int mxn = 1e5 + 5;
vector<long long> primes;
long long cc_size;
int vis[100001];
void sieve() {
for (int i = 2; i < mxn; i++) {
if (vis[i]) continue;
primes.push_back(i);
for (int j = i + i; j < mxn; j += i) {
vis[j] = 1;
}
}
}
vector<int> ad... | ### Prompt
Construct a cpp code solution to the problem outlined:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct p... |
#include <bits/stdc++.h>
using namespace std;
int n, t, i, k, a[100005], co;
map<int, int> s;
int main() {
cin >> n >> k;
for (i = 1; i <= n; i++) scanf("%d", a + i);
sort(a + 1, a + n + 1);
for (i = 1; i <= n; i++) {
t = a[i];
if (t % k || !s[t / k]) s[t] = ++co;
}
cout << co;
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of... |
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007, mod2 = 2e9, Pi = 3.141592653589793238;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
int main() {
ios_base::sync_with_stdio(0);
cin... | ### Prompt
Please formulate a Cpp solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinc... |
#include <bits/stdc++.h>
using namespace std;
map<long long, long long> mp;
int main() {
long long n, k;
cin >> n >> k;
long long a[n];
for (int i = 0; i < n; i++) {
cin >> a[i];
}
if (k == 1) {
cout << n << endl;
return 0;
}
sort(a, a + n);
reverse(a, a + n);
int c = 0;
for (int i = 0... | ### Prompt
Please formulate a cpp solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinc... |
#include <bits/stdc++.h>
using namespace std;
int main(void) {
int n, k, c = 0;
cin >> n >> k;
vector<long long> v(n);
for (auto &x : v) cin >> x;
sort(v.rbegin(), v.rend());
set<long long> m;
for (auto &x : v) {
if (m.find(x * k) == m.end()) m.insert(x);
}
cout << m.size();
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct pos... |
#include <bits/stdc++.h>
long long a[100005];
std::set<long long> s;
int main() {
long long n, k;
scanf("%lld %lld", &n, &k);
long long i;
for (i = 0; i < n; i++) scanf("%lld", &a[i]);
std::sort(a, a + n);
long long ans = 0;
for (i = 0; i < n; i++) {
if (s.count(a[i]) == 0) {
ans++;
s.inse... | ### Prompt
Develop a solution in Cpp to the problem described below:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinc... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, k, a[100000];
vector<int> v;
cin >> n >> k;
for (int i = 0; i < n; i++) {
cin >> a[i];
}
sort(a, a + n);
set<int> s;
for (int i = 0; i < n; i++) {
if ((a[i] % k == 0 && s.find(a[i] / k) == s.end()) || a[i] % k != 0)
s.insert... | ### Prompt
Develop a solution in CPP to the problem described below:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinc... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
int n, k;
cin >> n >> k;
if (k == 1) {
cout << n;
return 0;
}
vector<int> T(n);
for (int i = (0); i <= ((int)(T).size() - 1); ++i) cin >> T[i];
sort(T.begin(), T.end());
vector<bool> vis(n, 0);
int ans =... | ### Prompt
Create a solution in CPP for the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct posi... |
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:64000000")
using namespace std;
const int INF = 1000 * 1000 * 1000;
const long long LINF = 1000000000000000000LL;
const double eps = 1e-9;
void prepare() {}
const int nmax = 100005;
set<long long> q;
int n, k;
int a[nmax];
bool solve() {
scanf("%d%d", &n, &k);
... | ### Prompt
Generate a cpp solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct positi... |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = (int)1e5;
const int MAXVAL = (int)1e9;
map<int, int> indx, cnt;
int arr[MAXN + 5];
int n, k;
void add(int v) {
if ((long long)v * k > MAXVAL || indx.count(v * k) == 0)
indx[v] = cnt.size();
else
indx[v] = indx[v * k];
cnt[indx[v]]++;
}
int mai... | ### Prompt
Your challenge is to write a Cpp solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of... |
#include <bits/stdc++.h>
using namespace std;
int a[100005];
int bs(int t, int p) {
int lo = 0, hi = p - 1;
while (lo <= hi) {
int mid = (lo + hi) / 2;
if (a[mid] == t)
return mid;
else {
if (a[mid] > t) {
hi = mid - 1;
} else {
lo = mid + 1;
}
}
}
return ... | ### Prompt
Develop a solution in Cpp to the problem described below:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinc... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, k, i, cur, ans = 0;
long long a[100001], flag[100001];
cin >> n >> k;
for (i = 0; i < n; i++) {
cin >> a[i];
}
memset(flag, 0, sizeof(flag));
sort(a, a + n);
for (i = 0; i < n; i++) {
if (flag[i] == 0) {
ans++;
cur = l... | ### Prompt
Your challenge is to write a CPP solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of... |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, k, cnt = 0;
cin >> n >> k;
vector<long long> vec(n);
map<long long, bool> sol;
for (int i = 0; i < n; ++i) {
cin >> vec[i];
sol.insert(make_pair(vec[i], 1));
}
if (k == 1) {
cout << n << endl;
return 0;
}
sort(vec.... | ### Prompt
Construct a Cpp code solution to the problem outlined:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct p... |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, k;
cin >> n >> k;
vector<long long> v;
for (int i = 0; i < n; i++) {
long long a;
cin >> a;
v.push_back(a);
}
sort(v.rbegin(), v.rend());
int ans = 0;
set<long long> s;
for (int i = 0; i < n; i++) {
if (!s.count(v[... | ### Prompt
In Cpp, your task is to solve the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct pos... |
#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1000000000 + 7;
const long long int MOD1 = 998244353;
const long long int Val9 = 1000000000;
const long long int Val6 = 1000000;
vector<long long int> vd;
map<long long int, long long int> diff;
long long int no_of_divisors(long long int n) {
lon... | ### Prompt
Generate a Cpp solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of n distinct positi... |
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int N = 1e5 + 4;
long long powmod(long long a, long long b, long long mod) {
long long res = 1;
a %= mod;
for (; b; b >>= 1) {
if (b & 1) res = (res * a) % mod;
a = (a * a) % mod;
}
return res;
}
int main() {
long long n, k... | ### Prompt
Your challenge is to write a Cpp solution to the following problem:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of... |
#include <bits/stdc++.h>
using namespace std;
int n, k;
long long a[100005];
bool tag[100005];
int main() {
cin >> n >> k;
for (int i = 0; i < n; i++) scanf("%I64d", a + i);
memset(tag, false, sizeof(tag));
sort(a, a + n);
int ans = n;
for (int i = 0; i < n; i++) {
if (!tag[i]) {
long long p = a[i... | ### Prompt
Please provide a cpp coded solution to the problem described below:
A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.
You're given a set of... |
#include <bits/stdc++.h>
using namespace std;
struct Data {
int Value, Number;
};
bool operator<(Data A, Data B) { return A.Value < B.Value; }
int Get() {
char c;
while (c = getchar(), c < '0' || c > '9')
;
int X = 0;
while (c >= '0' && c <= '9') {
X = X * 10 + c - 48;
c = getchar();
}
return ... | ### Prompt
Your task is to create a CPP solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Pre... |
#include <bits/stdc++.h>
using namespace std;
int n;
pair<long long, long long> a[100100];
set<int> st;
set<int> st1;
map<int, int> per;
map<int, int> per1;
int pr(int x) {
if (!per.count(x)) per[x] = x;
if (per[x] == x) return x;
return per[x] = pr(per[x]);
}
int pr1(int x) {
if (!per1.count(x)) per1[x] = x;
... | ### Prompt
Create a solution in cpp for the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you ... |
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int N = 1e6 + 10;
const long long INF = 1e18;
const long double EPS = 1e-12;
vector<pair<int, int> > vec;
int a[100005];
int b[100005];
int main() {
int n;
cin >> n;
for (int i = 0; i < (int)n; i++) {
int x;
cin >> x;
vec.pus... | ### Prompt
Your task is to create a cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Pre... |
#include <bits/stdc++.h>
class fastin {
private:
int _ch, _f;
public:
inline fastin& operator>>(char& c) {
c = getchar();
return *this;
}
template <typename _Tp>
inline fastin& operator>>(_Tp& _x) {
_x = 0;
while (!isdigit(_ch)) _f |= (_ch == 45), _ch = getchar();
while (isdigit(_ch))
... | ### Prompt
Develop a solution in CPP to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
const int N_MAX = 100005;
int N, numbers[N_MAX], A[N_MAX], B[N_MAX];
int A_out[N_MAX], B_out[N_MAX];
pair<int, int> data[N_MAX];
int main() {
scanf("%d", &N);
for (int i = 0; i < N; i++) {
scanf("%d", &data[i].first);
data[i].second = i;
}
sort(data, data + ... | ### Prompt
Develop a solution in cpp to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
bool ord;
class mazi {
public:
int s, a[2];
int poz;
bool operator<(const mazi aux) const {
return ((ord & (s < aux.s)) | ((!ord) & (poz < aux.poz)));
}
};
mazi v[100000 + 1];
int main() {
int n, i;
scanf("%d", &n);
for (i = 1; i <= n; i++) {
scanf("%... | ### Prompt
Please provide a CPP coded solution to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two.... |
#include <bits/stdc++.h>
using namespace std;
const int TAM = 1e5 + 10;
struct Nodo {
int id, s, a, b;
};
const inline bool byS(const Nodo& x, const Nodo& y) { return x.s < y.s; }
const inline bool byID(const Nodo& x, const Nodo& y) { return x.id < y.id; }
int n;
Nodo nodo[TAM];
int main() {
cin >> n;
for (int i ... | ### Prompt
Develop a solution in CPP to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
int n;
pair<int, int> arr[100042];
int ans[2][100042];
map<int, int> used[2];
int main() {
cin >> n;
for (int i = 0; i < n; i++) {
cin >> arr[i].first;
arr[i].second = i;
}
sort(arr, arr + n);
for (int i = 0; i < n / 3; i++) {
ans[0][arr[i].second] = i... | ### Prompt
Please provide a CPP coded solution to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two.... |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 2015;
pair<int, int> s[MAXN];
int place[MAXN];
int a[MAXN];
int main() {
ios_base::sync_with_stdio(false);
int n;
cin >> n;
for (int i = 1; i <= n; s[i].second = i, i++) cin >> s[i].first;
sort(s + 1, s + n + 1);
int c = n / 3, d = (2 * n)... | ### Prompt
Your task is to create a cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Pre... |
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void chkmax(T &x, T y) {
x = x > y ? x : y;
}
template <typename T>
void chkmin(T &x, T y) {
x = x > y ? y : x;
}
template <typename T>
void update(T &x, T y, T mod) {
x = x + y > mod ? x + y - mod : x + y;
}
const int INF = (1ll << 30);
template... | ### Prompt
Construct a CPP code solution to the problem outlined:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, y... |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
int N;
array<int, 2> S[MAXN];
array<int, 2> A[MAXN], B[MAXN];
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
cin >> N;
for (int i = 0; i < N; i++) {
cin >> S[i][0];
S[i][1] = i;
}
sort(S, S + N);
for (int i = 0; i < N / ... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two.... |
#include <bits/stdc++.h>
using namespace std;
const int N = 100000 + 5;
int n, s[N];
int a[N], b[N];
int od[N];
inline bool cmp_int_od_s(int x, int y) { return s[x] < s[y]; }
int main() {
puts("YES");
scanf("%d", &n);
for (int i = 1; i <= n; ++i) scanf("%d", &s[i]);
if (n < 3) {
if (n == 1)
printf("%d... | ### Prompt
Develop a solution in CPP to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
const int MAX = 100005;
int N, startingPoint;
pair<int, int> A[MAX];
int V[2][MAX];
void citire() {
cin >> N;
for (int i = 1; i <= N; i++) cin >> A[i].first, A[i].second = i;
sort(A + 1, A + N + 1);
}
void solve() {
int partitionSize = N / 3 + (N % 3 == 2);
for (i... | ### Prompt
Construct a Cpp code solution to the problem outlined:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, y... |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100005;
int N;
pair<int, int> p[MAXN];
int a[MAXN], b[MAXN];
void load() {
scanf("%d", &N);
for (int i = 0; i < N; i++) {
scanf("%d", &p[i].first);
p[i].second = i;
}
}
void solve() {
sort(p, p + N);
int third = N / 3 + (N % 3 == 2);
for... | ### Prompt
Construct a Cpp code solution to the problem outlined:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, y... |
#include <bits/stdc++.h>
using namespace std;
struct sum {
int idx;
int val;
} x[100010], A[100010], B[100010];
inline bool comp(sum A, sum B) { return A.val < B.val; }
inline bool comp2(sum A, sum B) { return A.idx < B.idx; }
int main() {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%d",... | ### Prompt
Please formulate a Cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:100000000,100000000")
using namespace std;
const long long inf = 1e18 + 7;
const long long mod = 1e9 + 7;
const double eps = 1e-10;
const double PI = 2 * acos(0.0);
const double E = 2.71828;
long long a[100005];
long long b[100005];
long long num[100005];
bool cm... | ### Prompt
Construct a Cpp code solution to the problem outlined:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, y... |
#include <bits/stdc++.h>
using namespace std;
pair<int, int> s[200001];
int rk[200001], a[200001], b[200001];
int f[200001];
int n;
int main() {
int x;
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%d", &x);
s[i] = make_pair(x, i);
}
sort(s + 1, s + 1 + n);
int res = (n - 1) / 3 + 1;
for ... | ### Prompt
Generate a CPP solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you ne... |
#include <bits/stdc++.h>
using namespace std;
int n, i, dim, v, wh[100011];
int a[100011], sol[2][100011];
vector<pair<int, int> > ord;
int main() {
scanf("%d", &n);
for (i = 1; i <= n; i++)
scanf("%d", &a[i]), ord.push_back(make_pair(a[i], i));
sort(ord.begin(), ord.end());
for (i = 1; i <= n; i++) a[i] = ... | ### Prompt
In cpp, your task is to solve the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0), cin.tie(0);
int n, j = 0;
cin >> n;
int a[n], b[n];
pair<int, int> s[n];
for (auto &i : s) cin >> i.first, i.second = j++;
sort(s, s + n);
map<int, int> mp;
for (int i = 0; i < n; i++) mp[s[i].second] = i;
for (int i... | ### Prompt
Your task is to create a CPP solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Pre... |
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T sqr(T x) {
return x * x;
}
const double EPS = 1e-6;
const int INF = 0x3fffffff;
const long long LINF = INF * 1ll * INF;
const double PI = acos(-1.0);
using namespace std;
pair<int, int> s[100001];
int a[100001], b[100001];
int main(void) {
in... | ### Prompt
In cpp, your task is to solve the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you... |
#include <bits/stdc++.h>
using namespace std;
pair<int, int> v[100005];
int a[100008];
int b[100005];
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &v[i].first);
v[i].second = i;
}
sort(v, v + n);
int val = 0;
int qt = (n + 2) / 3;
int cnt = 0;
for (int i = 0; ... | ### Prompt
Generate a Cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you ne... |
#include <bits/stdc++.h>
using namespace std;
vector<int> s;
bool compara(int i1, int i2) { return s[i1] < s[i2]; }
int main() {
ios::sync_with_stdio(false);
int n;
cin >> n;
s = vector<int>(n);
for (int i = 0; i < n; i++) cin >> s[i];
vector<int> indice(n);
for (int i = 0; i < n; i++) indice[i] = i;
so... | ### Prompt
Please create a solution in cpp to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
int a[100005], b[100005], p[100005], s[100005], n, m;
bool cmp(const int i, const int j) { return s[i] < s[j]; }
int main() {
scanf("%d", &n), m = (n + 2) / 3, puts("YES");
for (int i = 0; i < n; i++) scanf("%d", s + i), p[i] = i;
sort(p, p + n, cmp);
for (int i = 0... | ### Prompt
Construct a Cpp code solution to the problem outlined:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, y... |
#include <bits/stdc++.h>
using namespace std;
template <class T>
void chmax(T& a, const T& b) {
a = max(a, b);
}
template <class T>
void chmin(T& a, const T& b) {
a = min(a, b);
}
template <class T>
void uniq(T& c) {
sort(c.begin(), c.end());
c.erase(unique(c.begin(), c.end()), c.end());
}
template <class T>
st... | ### Prompt
Develop a solution in Cpp to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2 * 1e5 + 10, maxm = (1 << 17), mod = 1e9 + 7, hash = 701;
const double PI = 3.14159265359, E = 2.71828;
pair<long long, long long> a[maxn], b[maxn], c[maxn];
int main() {
ios::sync_with_stdio(0);
long long n;
cin >> n;
for (int i = 0; i < n; i++) c... | ### Prompt
Construct a cpp code solution to the problem outlined:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, y... |
#include <bits/stdc++.h>
using namespace std;
template <class T>
void read(T& v) {
for (typename T::value_type& x : v) cin >> x;
}
template <class T>
void print(T& v) {
for (typename T::value_type& x : v) cout << x << ' ';
}
int main() {
int n;
cin >> n;
vector<pair<int, int> > v(n), a(n), b(n);
for (int i ... | ### Prompt
Your task is to create a cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Pre... |
#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
int s[N], a[N], b[N], c[N], sa[N], sb[N];
bool cmp(int x, int y) { return s[x] < s[y]; }
int main() {
int n, i, fg = 0;
scanf("%d", &n);
for (i = 0; i < n; i++) {
scanf("%d", &s[i]);
c[i] = i;
}
sort(c, c + n, cmp);
int pa, pb, md =... | ### Prompt
In cpp, your task is to solve the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you... |
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:100000000,100000000")
using namespace std;
const long long inf = 1e18 + 7;
const long long mod = 1e9 + 7;
const double eps = 1e-10;
const double PI = 2 * acos(0.0);
const double E = 2.71828;
long long a[100005];
long long b[100005];
long long num[100005];
bool cm... | ### Prompt
Generate a cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you ne... |
#include <bits/stdc++.h>
using namespace std;
const int TAM = 1e5 + 10;
struct Nodo {
int id, s, a, b;
};
inline bool byS(const Nodo& x, const Nodo& y) { return x.s < y.s; }
inline bool byID(const Nodo& x, const Nodo& y) { return x.id < y.id; }
int n;
Nodo nodo[TAM];
int main() {
cin >> n;
for (int i = 0; i < n; ... | ### Prompt
Please create a solution in CPP to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
long long n, sq[2][100069];
pair<long long, long long> a[100069];
int main() {
long long i, ii, k, l, e;
scanf("%lld", &n);
for (i = 1; i <= n; i++) {
scanf("%lld", &k);
a[i] = {k, i};
}
sort(a + 1, a + n + 1);
for (i = 1; i <= n; i++) {
k = a[i].fir... | ### Prompt
Please provide a cpp coded solution to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two.... |
#include <bits/stdc++.h>
using namespace std;
const int Maxn = 100 * 1000 + 10;
int n, s[Maxn], a[Maxn];
pair<int, int> arr[Maxn];
int main() {
cin >> n;
for (int i = 0; i < n; i++) {
scanf("%d", &s[i]);
arr[i] = make_pair(s[i], i);
}
sort(arr, arr + n);
for (int i = 0; i < ((n + 2) / 3); i++)
a[a... | ### Prompt
Please formulate a Cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
struct num {
int x, id;
} a[N];
int b[N], c[N];
bool cmp(num e, num f) { return e.x < f.x; }
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; ++i) cin >> a[i].x, a[i].id = i;
cout << "YES\n";
sort(a + 1, a + n + 1, cmp);
int t = (n... | ### Prompt
Construct a CPP code solution to the problem outlined:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, y... |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000 * 1000 + 100;
int n;
int s[maxn];
int a[maxn];
int b[maxn];
int p[maxn];
bool cmp(int x, int y) { return s[x] < s[y]; }
int main() {
ios::sync_with_stdio(false);
cin >> n;
for (int i = 0; i < n; ++i) {
p[i] = i;
cin >> s[i];
}
sort(p,... | ### Prompt
Please formulate a Cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
pair<int, int> A[100000];
int B[100000];
int C[100000];
int main() {
int N;
cin >> N;
int M = (N + 2) / 3;
for (int i = 0; i < N; i++) {
A[i].second = i;
cin >> A[i].first;
}
sort(A, A + N);
for (int i = 0; i < M; i++) {
B[A[i].second] = 0;
C[A... | ### Prompt
Your challenge is to write a Cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two.... |
#include <bits/stdc++.h>
using namespace std;
pair<int, int> p[100010];
int n, a[100010], b[100010];
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%d", &p[i].first);
p[i].second = i;
}
puts("YES");
sort(p + 1, p + n + 1);
for (int i = 1; i <= ((n - 1) / 3) + 1; i++) {
a[p[i... | ### Prompt
Please formulate a CPP solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
const int TAM = 1e5 + 10;
struct Nodo {
int id, s, a, b;
};
bool byS(const Nodo& x, const Nodo& y) { return x.s < y.s; }
bool byID(const Nodo& x, const Nodo& y) { return x.id < y.id; }
int n;
Nodo nodo[TAM];
int main() {
cin >> n;
for (int i = 0; i < n; i++) {
cin... | ### Prompt
Your challenge is to write a CPP solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two.... |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
int N, arr[MAXN], tid[MAXN], a[MAXN], b[MAXN];
int main() {
scanf("%d", &N);
for (int i = (0), iend = (N); i < iend; ++i) scanf("%d", &arr[i]);
iota(tid, tid + N, 0);
sort(tid, tid + N,
[](const int& x, const int& y) { return arr[x]... | ### Prompt
Construct a CPP code solution to the problem outlined:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, y... |
#include <bits/stdc++.h>
using namespace std;
pair<int, int> a[444444];
int x[444444], y[444444];
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++) {
scanf("%d", &(a[i].first));
a[i].second = i;
}
sort(a, a + n);
int need = n - (n + 2) / 3;
int z = 0;
for (int i = n - need; i < n;... | ### Prompt
Develop a solution in Cpp to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
const int max_n = 1e5 + 5;
int n, t;
struct hp {
int val, id;
bool vis;
} s[max_n], a[max_n], b[max_n];
inline int cmp(hp a, hp b) { return a.val < b.val; }
inline int cmp1(hp a, hp b) { return a.id < b.id; }
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; ++i... | ### Prompt
Develop a solution in CPP to the problem described below:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely... |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
int N, arr[MAXN], tid[MAXN], a[MAXN], b[MAXN];
int main() {
scanf("%d", &N);
for (int i = (0), iend = (N); i < iend; ++i) scanf("%d", &arr[i]);
iota(tid, tid + N, 0);
sort(tid, tid + N,
[](const int& x, const int& y) { return arr[x]... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two.... |
#include <bits/stdc++.h>
struct str {
int val, ind;
};
str arr[100000];
void qsort(int l, int r) {
int x = l, y = r, m = arr[(l + r) / 2].val;
while (x <= y) {
while (arr[x].val < m) x++;
while (arr[y].val > m) y--;
if (x <= y) {
str tmp = arr[x];
arr[x] = arr[y];
arr[y] = tmp;
... | ### Prompt
Your task is to create a CPP solution to the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Pre... |
#include <bits/stdc++.h>
using namespace std;
long long X[200000], Y[200000];
signed main() {
long long n;
cin >> n;
vector<pair<long long, long long> > A;
for (long long i = 0; i < n; i++) {
long long a;
cin >> a;
A.push_back(make_pair(a, i));
}
sort(A.begin(), A.end());
long long a = n / 3;
... | ### Prompt
In cpp, your task is to solve the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you... |
#include <bits/stdc++.h>
using namespace std;
int ans[2][100100];
int main() {
int n;
scanf("%d", &n);
vector<pair<int, int> > v;
for (int i = 0; i < n; i++) {
int u;
scanf("%d", &u);
v.push_back(pair<int, int>(u, i));
}
sort(v.begin(), v.end());
int t = n - 2 * (n / 3);
for (int i = 0; i < ... | ### Prompt
Construct a Cpp code solution to the problem outlined:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, y... |
#include <bits/stdc++.h>
using namespace std;
const int intmax = 0x3f3f3f3f;
const long long lldmax = 0x3f3f3f3f3f3f3f3fll;
double eps = 1e-8;
template <class T>
inline void checkmin(T &a, T b) {
if (b < a) a = b;
}
template <class T>
inline void checkmax(T &a, T b) {
if (b > a) a = b;
}
template <class T>
inline T... | ### Prompt
In Cpp, your task is to solve the following problem:
Polar bears like unique arrays — that is, arrays without repeated elements.
You have got a unique array s with length n containing non-negative integers. Since you are good friends with Alice and Bob, you decide to split the array in two. Precisely, you... |
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