output stringlengths 52 181k | instruction stringlengths 296 182k |
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#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int M = 1e5 + 5;
int main() {
ios_base::sync_with_stdio(false);
int n, i, p[N];
cin >> n;
if (n % 2 == 0) {
if (n % 4 != 0) {
cout << -1 << endl;
return 0;
} else {
int foo = n / 4;
for (i = 1; i <= 1 + 2 ... | ### Prompt
Construct a Cpp code solution to the problem outlined:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
Yo... |
#include <bits/stdc++.h>
using namespace std;
int n;
int nums[100001];
int main() {
cin >> n;
if (n % 4 == 0 || n % 4 == 1) {
for (int i = 1; i <= n / 2; i += 2) {
nums[i] = i + 1;
nums[i + 1] = n - i + 1;
nums[n - i + 1] = n - i;
nums[n - i] = i;
}
if (n % 4 == 1) nums[n / 2 + 1... | ### Prompt
Your task is to create a CPP solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
if (n % 4 == 2 || n % 4 == 3) {
cout << -1;
return 0;
}
int ary[n + 1];
for (int i = 1; i <= n / 2; i += 2) {
ary[i] = i + 1;
ary[i + 1] = n - i + 1;
ary[n - i + 1] = n - i;
ary[n - i] = i;
}
if (n % 4 ==... | ### Prompt
Create a solution in Cpp for the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You h... |
#include <bits/stdc++.h>
using namespace std;
const int N = 200011;
const int MOD = 1e9 + 7;
long long int powmod(long long int a, long long int b) {
if (b == 0) return 1;
long long int x = powmod(a, b / 2);
long long int y = (x * x) % MOD;
if (b % 2) return (a * y) % MOD;
return y % MOD;
}
int main() {
int... | ### Prompt
Your task is to create a cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i... |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
int a[maxn];
int book[maxn];
set<int> num;
int main() {
int n, i;
cin >> n;
if (n == 1) return 0 * printf("1\n");
if (n % 4 > 1) return 0 * printf("-1\n");
for (i = 1; i <= n; i++) num.insert(i);
int m = n;
if (n & 1) {
a[n / 2 + ... | ### Prompt
Please provide a cpp coded solution to the problem described below:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n... |
#include <bits/stdc++.h>
using namespace std;
const int N = 101000;
int P[N];
int n;
int main() {
scanf("%d", &n);
if ((n % 4) > 1) {
printf("-1\n");
} else {
int dl = (n - (n & 1));
bool dol = true;
int naDol = 2, gora = dl;
for (int i = 0; i < (dl / 2); ++i) {
if (dol) {
P[i] =... | ### Prompt
In cpp, your task is to solve the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You ... |
#include <bits/stdc++.h>
using namespace std;
int a[100010];
int main() {
int n;
scanf("%d", &n);
if (n == 1)
puts("1");
else if (n % 4 > 1)
puts("-1");
else {
int i = 1, j = n;
for (; j - i > 0; i += 2, j -= 2) {
a[i] = i + 1;
a[i + 1] = j;
a[j] = j - 1;
a[j - 1] = i;
... | ### Prompt
Your task is to create a cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i... |
#include <bits/stdc++.h>
using namespace std;
int M[100001];
int main() {
int n;
cin >> n;
if (n == 1) {
cout << 1 << endl;
return 0;
}
if (n % 4 == 2) {
cout << -1 << endl;
return 0;
}
if (n % 2 == 0) {
int j = n;
for (int i = n / 2; i > 0; i -= 2) {
M[i] = j;
j -= 2;
... | ### Prompt
Your challenge is to write a CPP solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n... |
#include <bits/stdc++.h>
using namespace std;
int v[100005] = {0};
int q[100005] = {0};
int main() {
int n;
scanf("%d", &n);
memset(v, -1, sizeof(v));
if ((n - (n % 2)) % 4) {
printf("-1\n");
return 0;
}
int cur = 0;
int sz = 1;
for (int i = 0; i < n / 2; i += 2) {
q[cur] = i;
v[i] = 0;
... | ### Prompt
Your task is to create a CPP solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i... |
#include <bits/stdc++.h>
using namespace std;
int arr[100005];
int main() {
memset(arr, -1, sizeof arr);
int n;
cin >> n;
if (n & 1) arr[n / 2 + 1] = n / 2 + 1;
for (int i = 1; i < n / 2; i += 2) {
arr[i] = i + 1;
arr[i + 1] = n - i + 1;
arr[n - i] = i;
arr[n - i + 1] = n - i;
}
for (int i... | ### Prompt
Please create a solution in CPP to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
long a[110000];
int main() {
int n, i, j;
scanf("%d", &n);
if ((n % 4) > 1) {
printf("-1\n");
} else {
for (i = 1; i <= n / 2; i += 2) {
a[i] = i + 1;
a[i + 1] = n - i + 1;
a[n - i] = i;
a[n - i + 1] = n - i;
}
if ((n % 2) == 1) {
a[n / 2 + ... | ### Prompt
Generate a cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You hav... |
#include <bits/stdc++.h>
const int N = 100100;
using namespace std;
int n, p[N];
int l;
void dfs(int x) {
p[p[x]] = n - x + 1;
int y = p[x];
while (y != x) {
p[p[y]] = n - y + 1;
y = p[y];
}
}
int main() {
ios_base::sync_with_stdio(0);
cin >> n;
if (n % 2) p[n / 2 + 1] = n / 2 + 1;
for (int i = ... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n... |
#include <bits/stdc++.h>
int n, t, a[110000], tnum, cas;
int main() {
cas = 0;
scanf("%d", &n);
if (n == 1) {
printf("1\n");
} else {
if (n % 2 != 0) {
a[(n / 2) + 1] = (n / 2) + 1;
tnum = n - 1;
} else {
tnum = n;
}
if (tnum % 4 != 0) {
printf("-1\n");
} else {
... | ### Prompt
Develop a solution in cpp to the problem described below:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
using namespace std;
bool flag[20];
int ans[20];
bool dfs(int k, int n) {
if (k > n) {
bool temp = true;
for (int i = 1; i <= n; i++)
if (ans[ans[i]] != n - i + 1) temp = false;
if (temp) {
for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
printf("\n");
r... | ### Prompt
Develop a solution in Cpp to the problem described below:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
scanf("%d", &n);
if (n % 4 == 2 || n % 4 == 3) {
cout << -1;
return 0;
}
int ans[n + 10];
if (n % 4 == 1) {
ans[n / 2 + 1] = n / 2 + 1;
}
for (int i = 1; i < n / 2; i += 2) {
ans[i] = i + 1;
ans[i + 1] = n + 1 - i;
... | ### Prompt
Your task is to create a Cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
scanf("%d", &n);
if (n % 4 == 2 || n % 4 == 3) {
printf("-1\n");
return 0;
}
int lo, hi;
int A[100006];
lo = 1;
hi = n;
while (lo < hi) {
A[lo] = lo + 1;
A[lo + 1] = hi;
A[hi] = hi - 1;
A[hi - 1] = lo;
lo += ... | ### Prompt
Please provide a cpp coded solution to the problem described below:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n... |
#include <bits/stdc++.h>
int a[100001];
int main() {
int n, i;
scanf("%d", &n);
if (n % 4 == 2 || n % 4 == 3)
printf("-1\n");
else {
for (i = 1; i <= n / 2; i = i + 2) {
a[i] = i + 1;
a[i + 1] = n + 1 - i;
a[n - i] = i;
a[n + 1 - i] = n - i;
}
if (n % 4 == 1) a[i] = i;
... | ### Prompt
Create a solution in CPP for the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You h... |
#include <bits/stdc++.h>
using namespace std;
int a[100005];
int main() {
int n;
cin >> n;
if (n % 4 >= 2) return puts("-1"), 0;
for (int i = 1; i <= n / 2; i += 2) {
a[i] = i + 1;
a[i + 1] = n - i + 1;
a[n - i + 1] = n - i;
a[n - i] = i;
}
if (n % 2 == 1) a[n / 2 + 1] = n / 2 + 1;
for (in... | ### Prompt
Your challenge is to write a CPP solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n... |
#include <bits/stdc++.h>
using namespace std;
int a[100005], n;
bool tui() {
int nd2 = n / 2, w = 2;
if (n % 4 == 0) {
for (int i = 1; i <= nd2; i += 2) {
a[i] = w, a[i + 1] = n + 2 - w;
w += 2;
}
w = nd2 - 1;
for (int i = nd2 + 1; i <= n; i += 2) {
a[i] = w, a[i + 1] = n - w, w -=... | ### Prompt
Develop a solution in Cpp to the problem described below:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
int p[100009], n, s, t, i;
bool none = false;
int main() {
scanf("%d", &n);
for (s = 1, t = n; s <= t;) {
if (t - s + 1 >= 4) {
p[s] = s + 1;
p[s + 1] = t;
p[t] = t - 1;
p[t - 1] = s;
s += 2;
t -= 2;
} else {
if (t - s + 1 >= 2) {
no... | ### Prompt
Please create a solution in CPP to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
using namespace std;
int a[100009];
bool FL;
void f(int st, int en, int i, int j) {
if (st > en) return;
if (st == en)
a[i] = st;
else {
if ((j - i) == 1 || (j - i) == 2) {
FL = false;
return;
}
a[i] = st + 1;
a[i + 1] = en;
a[j] = en - 1;
a[j - 1] ... | ### Prompt
Please provide a Cpp coded solution to the problem described below:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n... |
#include <bits/stdc++.h>
using namespace std;
long long int inf = 1e14;
long long int mod = 1e9 + 7;
char en = '\n';
long long int arr[300005];
long long int res[300005];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
memset(arr, 0, sizeof(arr));
long long int n;
cin >> n;
if ((n % 4 == 2) or... | ### Prompt
Your task is to create a CPP solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i... |
#include <bits/stdc++.h>
using namespace std;
int n, x, y, a[100010];
int main() {
scanf("%d", &n);
if (n % 4 != 0 && n % 4 != 1) {
printf("-1");
return 0;
}
for (int i = 1; i <= n / 2; i += 2) {
int x = i, y = i + 1;
for (int j = 1; j <= 4; j++) {
a[x] = y;
y = n - x + 1;
x = ... | ### Prompt
Please provide a cpp coded solution to the problem described below:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n... |
#include <bits/stdc++.h>
using namespace std;
int a[1000000 + 10];
void dfs(int l, int r) {
if (r - l + 1 < 4) return;
a[l] = l + 1;
a[l + 1] = r;
a[r] = r - 1;
a[r - 1] = l;
dfs(l + 2, r - 2);
}
int main() {
int xiaohao;
int zheshixiaohao;
int n;
cin >> n;
if (n == 1) {
cout << "1";
}
if ... | ### Prompt
Construct a Cpp code solution to the problem outlined:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
Yo... |
#include <bits/stdc++.h>
using namespace std;
int a[100001];
int main() {
int i, j, k, m, n;
cin >> n;
if (n % 4 > 1) {
cout << "-1" << endl;
return 0;
}
int _front = 0, end_ = n - 1;
list<int> L;
for (i = 1; i <= n; i++) L.push_back(i);
while (!L.empty()) {
if (_front == end_) {
a[end... | ### Prompt
Generate a cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You hav... |
#include <bits/stdc++.h>
using namespace std;
int i, n;
bool was[1000001];
vector<int> ans, a;
int main() {
cin >> n;
for (i = 1; i <= n; i += 2) {
if (i + 1 > n - i + 1) break;
ans.push_back(i + 1);
ans.push_back(n - i + 1);
was[i + 1] = was[n - i + 1] = true;
}
for (i = 1; i <= n; i++) {
i... | ### Prompt
Generate a CPP solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You hav... |
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 2e1 + 4;
vector<int> circle;
bool vis[N];
int parent[N];
int chk[N];
int color[N];
int tim[N];
int dis[N];
int position[N];
vector<int> adj[N];
vector<int> adj1[N];
vector<int> graph[N];
bool has_cycle;
int maxdis, maxnode, Totnode, depth... | ### Prompt
Please formulate a Cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
int a[100005];
int main() {
int n, i;
while (~scanf("%d", &n)) {
if (n % 4 == 2 || n % 4 == 3) {
printf("-1\n");
continue;
} else {
for (i = 1; i <= n / 2; i += 2) {
a[i] = i + 1;
a[i + 1] = n - i + 1;
a[n - i + 1] = n - i;
a[n - i] ... | ### Prompt
Develop a solution in CPP to the problem described below:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
using namespace std;
long a[100005];
int main(void) {
long n;
cin >> n;
if (n == 1)
cout << "1" << endl;
else if (n % 4 == 0) {
for (int i = 1; i <= n / 2; i += 2) a[n - i] = i;
for (int i = 2; i <= n / 2; i += 2) a[i - 1] = i;
for (int i = n; i > n / 2; i -= 2) a[n + 2 ... | ### Prompt
Please formulate a Cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
using namespace std;
int n;
void solve() {
if (n % 4 == 2 || n % 4 == 3) {
cout << -1 << endl;
return;
}
if (n == 1) {
cout << 1 << endl;
return;
}
cout << 2 << ' ' << n;
for (int i = 1; i < n / 4; ++i)
cout << ' ' << 2 * (i + 1) << ' ' << (n - 2 * i);
if (n & ... | ### Prompt
Please create a solution in Cpp to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
using namespace std;
int n;
int *a;
void foo(int k, int i) {
a[i] = k;
a[k] = n - i + 1;
a[n - k + 1] = i;
a[n - i + 1] = n - k + 1;
}
void Zero() {
for (int i = 1; i <= n; ++i) {
a[i] = 0;
}
}
void Print() {
for (int i = 1; i <= n; ++i) {
cout << a[i] << " ";
}
}
int ma... | ### Prompt
Your task is to create a cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i... |
#include <bits/stdc++.h>
using namespace std;
const int N = 300;
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};
int n, p[110000];
int main() {
cin >> n;
if (n % 4 == 2 || n % 4 == 3) {
cout << -1 << endl;
return 0;
}
for (int i = n; i > n / 2 + 1; i -= 2) {
p[i] = i - 1;
p[i - 1... | ### Prompt
In cpp, your task is to solve the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You ... |
#include <bits/stdc++.h>
using namespace std;
const int oo = 0x3f3f3f3f;
long long mod = 1e9 + 7;
double eps = 1e-9;
double pi = acos(-1);
long long fastpower(long long b, long long p) {
double ans = 1;
while (p) {
if (p % 2) {
ans = (ans * b);
}
b = b * b;
p /= 2;
}
return ans;
}
bool val... | ### Prompt
Create a solution in CPP for the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You h... |
#include <bits/stdc++.h>
int a[111111];
int main() {
int n;
scanf("%d", &n);
if (n % 4 == 2 || n % 4 == 3)
printf("-1\n");
else {
int b = 1, e = n, mn = 1, mx = n;
while (e - b + 1 >= 4) {
a[b] = mn + 1;
a[b + 1] = mx;
a[e - 1] = mn;
a[e] = mx - 1;
b += 2;
e -= 2;... | ### Prompt
Construct a CPP code solution to the problem outlined:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
Yo... |
#include <bits/stdc++.h>
using namespace std;
const int size = 100008;
int p[size], n;
void dfs_p(int i, int x) {
if (x > 4) return;
p[p[i]] = n - i + 1;
dfs_p(p[i], x + 1);
}
int main() {
int k, i, j, id;
while (cin >> n) {
memset(p, 0, sizeof(p));
if (n % 4 == 0 || n % 4 == 1) {
k = n / 4;
... | ### Prompt
Create a solution in CPP for the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You h... |
#include <bits/stdc++.h>
using namespace std;
const int oo = 2000000009;
const int mx = 100005;
int mustbe[mx], p[mx], n;
void solve(int a, int b) {
if (b - a < 0) return;
if (b - a == 0) {
p[a] = a;
return;
}
p[a] = a + 1;
p[a + 1] = b;
p[b - 1] = a;
p[b] = b - 1;
solve(a + 2, b - 2);
}
int mai... | ### Prompt
Please provide a cpp coded solution to the problem described below:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n... |
#include <bits/stdc++.h>
using namespace std;
long long n, a[((long long)101 * 1000)], p[((long long)101 * 1000)];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> n;
if (n == 2 || n % 4 == 3 || n % 4 == 2) return cout << -1, 0;
for (int i = 1, l = 1, r = n; i <= n - (n % 2); l += ... | ### Prompt
Please provide a cpp coded solution to the problem described below:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int n;
cin >> n;
int a[n + 1];
if (n % 4 > 1)
cout << -1 << endl;
else {
for (int i = 1; i < n / 2; i += 2)
a[i] = i + 1, a[n - i + 1] = n - i, a[i + 1] = n - i + 1, a[n - i] = i;
if (n % 4 ==... | ### Prompt
Please create a solution in Cpp to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
using namespace std;
const double PI = 3.141592653589793238;
template <typename T>
std::ostream& operator<<(std::ostream& os, const std::vector<T>& vector) {
for (size_t i = 0; i < vector.size(); ++i) {
os << vector[i] << " ";
}
return os;
}
vector<int> FindHappyPermutaion(int n) {
... | ### Prompt
Your challenge is to write a Cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n... |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100010 * 2;
int p[MAXN];
int main() {
int N;
cin >> N;
if (N % 4 > 1)
puts("-1");
else {
for (int i = 1; i * 2 < N; i += 2) {
p[i] = i + 1;
p[i + 1] = N + 1 - i;
p[N + 1 - i] = N - i;
p[N - i] = i;
}
if (N % 2... | ### Prompt
Please create a solution in cpp to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
using namespace std;
template <class T>
T _max(T a, T b) {
return a > b ? a : b;
}
template <class T>
T _min(T a, T b) {
return a < b ? a : b;
}
int sgn(const double &x) { return (x > 1e-8) - (x < -1e-8); }
int a[100010];
int main() {
int n;
cin >> n;
if (n % 4 > 1) {
puts("-1");
... | ### Prompt
Construct a Cpp code solution to the problem outlined:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
Yo... |
#include <bits/stdc++.h>
using namespace std;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
const int INF = 1 << 29;
inline int two(int n) { return 1 << n; }
inline int test(int n, int b) { return (n >> b) & 1; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int b... | ### Prompt
Develop a solution in cpp to the problem described below:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
using namespace std;
int main(void) {
int n;
cin >> n;
if ((n - (n & 1)) % 4 != 0)
cout << -1;
else {
set<int> disp;
int v[n + 1], k = 1;
memset(v, -1, sizeof v);
for (int i = 1; i <= n; i++) disp.insert(i);
if (n & 1) {
v[(n + 1) / 2] = (n + 1) / 2;
... | ### Prompt
Create a solution in CPP for the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You h... |
#include <bits/stdc++.h>
using namespace std;
int a[1000001], n;
int main() {
cin >> n;
if ((n / 2) % 2) {
cout << -1 << endl;
return 0;
}
int l = 2, r = n;
for (int i = 1; i <= n / 2; i++) {
if (i % 2)
a[i] = l, a[n - i] = l - 1, l += 2;
else
a[i] = r, a[n - i + 2] = r - 1, r -= 2... | ### Prompt
Please formulate a Cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
int p[100001];
bool v[100001];
int main() {
int n;
scanf("%d", &n);
int fill = 1;
int c = 0;
while (fill <= n) {
while (p[fill] != 0 && fill <= n) fill++;
if (fill > n) break;
if (c == n - 1) {
p[fill] = fill;
if (n - fill + 1 == fill) c++;
break;
}
... | ### Prompt
Generate a CPP solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You hav... |
#include <bits/stdc++.h>
using namespace std;
char f(vector<int> a) {
int n = a.size() - 1;
for (int i = 1; i <= n; i++)
if (a[a[i]] != n - i + 1) return false;
return true;
}
int main() {
int n;
scanf("%d", &n);
vector<int> a(n + 1, -1);
a[1] = 2;
a[n - 1] = 1;
a[n] = n - 1;
if (n % 2 == 0) {
... | ### Prompt
Develop a solution in Cpp to the problem described below:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
int p[100002];
int vis[100002];
int main() {
int n;
while (scanf("%d", &n) != EOF) {
if (n % 4 != 0 && n % 4 != 1) {
printf("-1\n");
continue;
} else {
memset(vis, 0, sizeof(vis));
for (int i = 1; i <= n / 2; i++) {
if (vis[i] == 0) {
p[i] =... | ### Prompt
Please formulate a CPP solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
using namespace std;
const int N = 100100;
int p[N];
int main() {
int n;
cin >> n;
int t = n % 4;
if (t < 2) {
if (t == 1) {
p[n / 2 + 1] = n / 2 + 1;
}
for (int i = 1; 2 * i <= n; i += 2) {
p[i] = i + 1;
p[i + 1] = n - i + 1;
p[n - i + 1] = n - i;
... | ### Prompt
Please create a solution in Cpp to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
using namespace std;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
template <typename T>
inline T gcd(T a, T b) {
if (b == 0)
return a;
else
return gcd(b, a % b);
}
template <typename T>
inline T lcm(T a, T b) {
return (a * b) / gcd(a, b);
}
template <typename ... | ### Prompt
Your task is to create a cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i... |
#include <bits/stdc++.h>
using namespace std;
int p[200000];
int pp[200000];
int a[200000];
int n;
bool f() {
for (int(i) = (0); (i) < (n); ++(i))
if (a[a[i + 1]] != n - i) return 0;
return 1;
}
void ff(int id, int x) {
if (a[id]) return;
a[id] = x;
ff(x, n - id + 1);
}
int main() {
cin >> n;
if (n ==... | ### Prompt
Generate a Cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You hav... |
#include <bits/stdc++.h>
using namespace std;
long long k, a, b;
int main() {
long long n;
cin >> n;
long long ans[100000 + 1] = {};
if (n % 4 == 1) {
ans[(n + 1) / 2] = (n + 1) / 2;
long long mid = (n + 1) / 2;
for (long long i = 1; i <= (n + 1) / 2; i++) {
if (i % 2) {
ans[mid - i] =... | ### Prompt
In CPP, your task is to solve the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You ... |
#include <bits/stdc++.h>
int m, n;
int a[100050];
int main() {
int i, j, k, x, y, z;
scanf("%d", &n);
if (n % 4 == 2 || n % 4 == 3)
puts("-1");
else {
j = n;
i = 1;
while (j >= i) {
if (j == i)
a[i] = i;
else {
a[i] = i + 1;
a[i + 1] = j;
a[j] = n - i;... | ### Prompt
In Cpp, your task is to solve the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You ... |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5, Mod = 1e9 + 7;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
if (n % 4 > 1) cout << "-1\n", exit(0);
int Arr[n + 2];
if (n & 1) Arr[n / 2 + 1] = n / 2 + 1;
for (int i = 1; i <= n / 2; i += 2) {
... | ### Prompt
Please formulate a cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
inline void read(int &x) {
x = 0;
char ch = getchar();
while (ch < '0') ch = getchar();
while (ch >= '0') {
x = x * 10 + ch - 48;
ch = getchar();
}
}
int a[100010], ans;
bool f[100010];
int main() {
i... | ### Prompt
In cpp, your task is to solve the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You ... |
#include <bits/stdc++.h>
using namespace std;
int p[100005];
bitset<100005> viz, ap;
int main() {
int n, sf;
cin >> n;
sf = n;
if (n == 1) {
cout << 1;
return 0;
}
if (n % 4 == 2 || n % 4 == 3) {
cout << -1;
return 0;
}
for (int i = 1; i <= n; ++i)
if (viz[i] == 0) {
if (!(n % ... | ### Prompt
Your task is to create a CPP solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i... |
#include <bits/stdc++.h>
int p[100005];
int main() {
int n;
scanf("%d", &n);
memset(p, 0, sizeof(p));
if (n == 1) {
printf("1\n");
return 0;
}
if (n == 2 || n == 3 || n % 4 == 3 || n % 4 == 2) {
printf("-1\n");
return 0;
}
if (n >= 4) {
for (int i = 1; i <= n / 4; i++) {
p[2 * ... | ### Prompt
Construct a CPP code solution to the problem outlined:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
Yo... |
#include <bits/stdc++.h>
using namespace std;
int a[100010];
int main() {
int i, j, k = 1, n, m;
cin >> n;
if (n % 4 == 2 || n % 4 == 3) {
cout << "-1\n";
return 0;
}
for (i = 0; i < n / 4; i++) {
int x = 2 * i + 1;
a[x] = x + 1;
a[n - x] = x;
a[n + 1 - x] = n - x;
a[x + 1] = n + 1... | ### Prompt
Generate a cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You hav... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, i;
int num1, num2, cnt;
cin >> n;
if (n == 1) {
cout << 1 << endl;
return 0;
}
if (n % 4 == 0) {
cnt = n / 2;
num1 = 2, num2 = n;
while (num1 <= cnt) {
cout << num1 << " " << num2 << " ";
num1 += 2, num2 -= 2;
... | ### Prompt
Develop a solution in CPP to the problem described below:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
using namespace std;
int output[100010];
int main() {
int n;
while (cin >> n) {
if (n % 4 == 2 || n % 4 == 3)
cout << -1 << endl;
else {
int front1 = 2, front2 = n, rear1 = n - 1, rear2 = 1;
int i, count = 0;
for (i = 1; count < n / 4; i += 2) {
outpu... | ### Prompt
Develop a solution in CPP to the problem described below:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
using namespace std;
long long Set(long long N, long long pos) { return N = N | (1 << pos); }
long long reset(long long N, long long pos) { return N = N & ~(1 << pos); }
bool check(long long N, long long pos) { return (bool)(N & (1 << pos)); }
void CI(long long &_x) { scanf("%d", &_x); }
void C... | ### Prompt
Your challenge is to write a Cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n... |
#include <bits/stdc++.h>
using namespace std;
long long n, i, f, m, a[211111], ans;
vector<long long> v;
double d;
int main() {
cin >> n;
if (n == 1)
cout << 1;
else if (n % 4 == 2 || n % 4 == 3)
cout << -1;
else {
for (i = 0; i < n / 2; i++) {
if (i % 2 == 0) {
a[i] = i + 1;
a... | ### Prompt
Generate a cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You hav... |
#include <bits/stdc++.h>
using namespace std;
int arr[100100];
int main() {
int n;
cin >> n;
if (n == 1) {
cout << "1" << endl;
return 0;
}
memset(arr, 0, sizeof(arr));
bool f = true;
int k = 0;
if (n % 2) arr[n / 2 + 1] = n / 2 + 1;
for (int i = 1; i <= n; i++) {
if (arr[i] == 0 && k < n ... | ### Prompt
Please create a solution in CPP to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
int main() {
int n, k = 1, i, j, a[100001];
scanf("%d", &n);
if (n == 1) {
printf("1");
return 0;
}
k = n / 2;
for (i = 1, j = n; i < j; i++, j--) {
if (i % 2 == 0) {
a[i] = k;
a[j] = n - k + 1;
} else {
a[i] = n - k + 1;
a[j] = k;
}
k... | ### Prompt
Develop a solution in cpp to the problem described below:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
using namespace std;
int v[100001];
int main() {
int i, f, n;
cin >> n;
if (n % 4 == 2 || n % 4 == 3)
cout << -1 << endl;
else {
if (n % 4 == 1) v[(n + 1) / 2] = (n + 1) / 2;
i = 1;
f = n;
while (i < f) {
v[i] = f - 1;
v[f - 1] = f;
v[f] = i + 1;
... | ### Prompt
Develop a solution in cpp to the problem described below:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100005;
int p[MAXN];
int ans[MAXN];
int l, r;
void put_left(int x) {
ans[l] = x;
++l;
}
void put_right(int x) {
ans[r] = x;
--r;
}
int main() {
int n;
scanf("%d", &n);
if (n % 4 == 2 || n % 4 == 3) {
puts("-1");
return 0;
}
l = 0, ... | ### Prompt
Construct a cpp code solution to the problem outlined:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
Yo... |
#include <bits/stdc++.h>
using namespace std;
int n, p[1010000];
int main() {
cin >> n;
if (n % 4 == 2 || n % 4 == 3) {
cout << -1;
exit(0);
}
if (n % 4 == 0) {
for (int i = 1; i <= n / 2; i++) {
if (i % 2 == 1)
p[i] = i + 1, p[n + 1 - i] = n - i;
else
p[i] = n + 2 - i, p... | ### Prompt
Please formulate a cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 100;
const int mod = 1e9 + 7;
int p[maxn];
bool mark[maxn];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n;
cin >> n;
if (n % 4 > 1) return (cout << -1, 0);
if (n % 4 == 1) p[n / 2 + 1] = n / 2 + 1;
for (... | ### Prompt
Your task is to create a CPP solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i... |
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const double infd = 2e+9;
const int infi = INT_MAX;
template <class T>
inline T sqr(T x) {
return x * x;
}
int main() {
ios_base::sync_with_stdio(false);
int n;
cin >> n;
if (n % 4 != 0 && n % 4 != 1) {
cout << -1;
... | ### Prompt
Your challenge is to write a Cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n... |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
int n, p[MAXN];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cin >> n;
if (n % 4 == 2 || n % 4 == 3) {
cout << -1;
return 0;
}
int l = 1, r = n, div = n;
while (div / 4) {
p[l] = l + 1;
p[l + 1] = r;
... | ### Prompt
Your challenge is to write a Cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n... |
#include <bits/stdc++.h>
using namespace std;
int n;
int a[100010];
void output() {
for (int i = 0; i < n; i++) printf("%d ", a[i]);
puts("");
}
int check() {
for (int i = 0; i < n; i++)
if (a[a[i] - 1] != n - i) return 0;
return 1;
}
int main() {
cin >> n;
for (int i = 0; i < n / 2; i += 2) a[i] = i + ... | ### Prompt
Please provide a CPP coded solution to the problem described below:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n... |
#include <bits/stdc++.h>
using namespace std;
long p[100001];
int main(void) {
ios::sync_with_stdio(false);
cin.tie(NULL);
long n;
cin >> n;
if (n % 4 == 2 || n % 4 == 3)
cout << "-1";
else {
if (n % 2) p[n / 2 + 1] = n / 2 + 1;
for (int i = 1; i < n / 2; i += 2) {
p[i] = i + 1;
p[i ... | ### Prompt
Generate a Cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You hav... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int n;
cin >> n;
if (n % 4 > 1) {
cout << "-1\n";
return 0;
}
int p[n];
if (n % 4 == 1) {
p[n / 2] = (n + 1) / 2;
}
if (n != 1) {
p[0] = 2;
p[1] = n;
p[n - 2] = 1;
p[n - 1] = n... | ### Prompt
Generate a CPP solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You hav... |
#include <bits/stdc++.h>
using namespace std;
int a[100001];
int main() {
int n;
cin >> n;
if (n % 4 == 2 || n % 4 == 3) {
cout << -1 << endl;
return 0;
}
int i;
if (n % 2 == 1) a[(n + 1) / 2] = (n + 1) / 2;
for (i = 1; i <= n / 2;) {
a[i] = i + 1;
a[i + 1] = n - i + 1;
a[n - i + 1] = ... | ### Prompt
Develop a solution in CPP to the problem described below:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
using namespace std;
inline bool xdy(double x, double y) { return x > y + 1e-9; }
inline bool xddy(double x, double y) { return x > y - 1e-9; }
inline bool xcy(double x, double y) { return x < y - 1e-9; }
inline bool xcdy(double x, double y) { return x < y + 1e-9; }
const long long int mod = 10... | ### Prompt
Please formulate a Cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
if (n % 4 > 1) {
cout << -1;
return 0;
}
int m = n / 4, perm[100001];
for (int i = 0; i < m; i++) {
int a, b, c, d;
a = 2 * i + 1;
b = 2 * i + 2;
c = n - 2 * i;
d = n - 2 * i - 1;
perm[a] = b;
per... | ### Prompt
Your task is to create a cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i... |
#include <bits/stdc++.h>
using namespace std;
int n;
deque<int> ans;
bool used[100500];
int add(int i, int n) {
if (n % 2 == 0) {
if (i >= n / 2) i = n - i - 1;
return i / 2;
}
if (i > n / 2) i = n - i - 1;
return i / 2;
}
int main() {
cin >> n;
if (n == 1) {
cout << 1;
return 0;
} else if... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n... |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int a[maxn];
int main() {
int n;
cin >> n;
if (n % 4 == 2 || n % 4 == 3)
cout << -1 << endl;
else if (n % 4 == 0) {
for (int i = 1; i <= (n / 2); i++) {
if ((i % 2) == 1)
a[i] = i + 1;
else
a[i] = n - i ... | ### Prompt
In CPP, your task is to solve the following problem:
A permutation p of size n is the sequence p1, p2, ..., pn, consisting of n distinct integers, each of them is from 1 to n (1 ≤ pi ≤ n).
A lucky permutation is such permutation p, that any integer i (1 ≤ i ≤ n) meets this condition ppi = n - i + 1.
You ... |
#include <bits/stdc++.h>
using namespace std;
int GCD(int a, int b) {
if (!a) return b;
return GCD(b % a, a);
}
vector<int> x(4000);
vector<int> y(4000);
void D(double x, double y, double x1, double y1) {
double d = (x - x1) * (x - x1) + (y - y1) * (y - y1);
cout << sqrt(d) << "\n";
}
int main() {
int n, k;
... | ### Prompt
Please formulate a Cpp solution to the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is ... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int i, n, k;
cin >> n >> k;
if (n * (n - 1) / 2 <= k) {
puts("no solution");
return 0;
}
for (i = 1; i <= n; i++) {
printf("0 %d\n", i);
}
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The probl... |
#include <bits/stdc++.h>
using namespace std;
int main() {
long n, k;
cin >> n >> k;
long kc = 2001;
if (k >= n * (n - 1) / 2) {
cout << "no solution";
return 0;
}
long sum = 0;
long ax[2005], ay[2005];
for (int i = 1; i <= n; i++) {
sum += kc;
ax[i] = 0;
ay[i] = sum;
cout << 0 <... | ### Prompt
Please formulate a CPP solution to the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is ... |
#include <bits/stdc++.h>
using namespace std;
template <class T>
T gmin(T u, T v) {
return (u < v) ? u : v;
}
template <class T>
T gmax(T u, T v) {
return (u > v) ? u : v;
}
template <class T>
T gcd(T u, T v) {
if (v == 0) return u;
return (u % v == 0) ? v : gcd(v, u % v);
}
int main() {
long long n, m, tmp;
... | ### Prompt
Please create a solution in cpp to the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is ... |
#include <bits/stdc++.h>
using namespace std;
long long distance(long long x1, long long y1, long long x2, long long y2) {
return sqrt(pow(x2 - x1, 2) + pow(y2 - y1, 2));
}
int main() {
long long n, k, d, x, y, tot = 0;
cin >> n >> k;
for (int i = 1; i <= n; i++) {
for (int j = i + 1; j <= n; j++) {
t... | ### Prompt
Your task is to create a CPP solution to the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The probl... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
scanf("%d %d", &n, &k);
if ((n * (n - 1)) / 2 <= k) {
printf("no solution\n");
return 0;
}
for (int i = 0; i < (int)n; i++) printf("%d %d\n", 0, i);
return 0;
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The p... |
#include <bits/stdc++.h>
using namespace std;
int n, k;
int main() {
cin >> n >> k;
if (n * (n - 1) / 2 <= k) {
cout << "no solution\n";
return 0;
}
for (int i = 1; i <= n; i++) {
cout << 1 << " " << i + i << "\n";
}
return 0;
}
| ### Prompt
Create a solution in cpp for the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the fo... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
if (((n) * (n - 1)) / 2 > k) {
for (int i = 0; i < n; i++) cout << "0 " << i << endl;
} else {
cout << "no solution" << endl;
}
return 0;
}
| ### Prompt
In CPP, your task is to solve the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the f... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
if ((n - 1) * n / 2 <= k) {
cout << "no solution" << endl;
} else {
int tot = 0;
int i, j;
for (i = 0; i <= 1e9; i++)
for (j = 0; j <= 1e9; j++) {
if (tot < n) {
cout << i << " " << j << e... | ### Prompt
In Cpp, your task is to solve the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the f... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
if (k >= (n * (n - 1)) / 2)
cout << "no solution" << endl;
else {
for (int i = 0; i < n; i++) cout << '0' << ' ' << i << endl;
}
}
| ### Prompt
Construct a CPP code solution to the problem outlined:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the... |
#include <bits/stdc++.h>
using namespace std;
int n, k;
int main() {
cin >> n >> k;
if (((n * (n - 1)) >> 1) <= k) {
cout << "no solution";
return 0;
}
for (int i = 0; i < n; ++i) {
cout << i << ' ' << 1000000000 - i * 3000 << endl;
}
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The p... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
while (cin >> n >> k) {
if (n * (n - 1) / 2 <= k) {
cout << "no solution" << endl;
} else {
for (int i = 0; i < n; i++) {
cout << 0 << ' ' << i << endl;
}
}
}
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The probl... |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long int n, k;
cin >> n >> k;
if (k >= n * (n - 1) / 2) {
cout << "no solution\n";
} else {
for (long long int i = 0; i < n; i++) {
cout << "0 " << i << endl;
}
}
}
| ### Prompt
In cpp, your task is to solve the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the f... |
#include <bits/stdc++.h>
using namespace std;
long long Min(long long i, long long j) { return i < j ? i : j; }
long long Max(long long i, long long j) { return i > j ? i : j; }
int main() {
long long a, b, c, d, e, i, j, k, l, m, n;
while (cin >> n >> k) {
if (k >= (n * (n - 1)) / 2)
cout << "no solution... | ### Prompt
Your challenge is to write a cpp solution to the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The p... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
int n, k;
cin >> n >> k;
if (n * (n - 1) / 2 <= k) {
cout << "no solution" << '\n';
return 0;
} else {
for (int i = 1; i <= n; i++) {
cout << "1 " << i + 1 << endl;
}
}
return 0;
}
| ### Prompt
In cpp, your task is to solve the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the f... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int y, k, n, x, i;
scanf("%d%d", &n, &k);
n--;
x = (n * (n + 1)) / 2;
if (x > k) {
for (i = 1; i <= n + 1; i++) printf("0 %d\n", i);
} else
printf("no solution\n");
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the... |
#include <bits/stdc++.h>
int main() {
int n, K;
scanf("%d%d", &n, &K);
int tot = (n * n - n) / 2;
if (tot <= K) {
printf("no solution\n");
return 0;
}
int x = 0, y = 0;
for (int i = 0; i < n; i++) {
printf("%d %d\n", 0, y);
y += 2;
}
return 0;
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The probl... |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
int n, k;
cin >> n >> k;
vector<pair<int, int>> v;
int now = 1;
for (int i = 1; i <= n - 1; ++i) {
v.emplace_back(1, now);
now += 2;
}
v.emplace_back(1, now);
int tot = 0;
for (int i = 1; i <= n; ++i... | ### Prompt
Your task is to create a Cpp solution to the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The probl... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int i, j, k, l, m, n, x, y, z, r, ans = 0, mn = INT_MAX, mx = INT_MIN,
res = 0;
cin >> n >> k;
if (k * 2 >= n * (n - 1))
cout << "no solution";
else {
for (i = 0; i < n; ++i) {
cout << 0 << " " << n + 1 ... | ### Prompt
Create a solution in Cpp for the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the fo... |
#include <bits/stdc++.h>
int main() {
long long p, n;
int i = 0, j = 2;
scanf("%I64d %I64d", &n, &p);
if (((n - 1) * n) / 2 <= p)
printf("no solution\n");
else
while (n--) {
printf("%d %d\n", i, j);
j += 2;
}
return 0;
}
| ### Prompt
Generate a cpp solution to the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the foll... |
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
long long power(long long b, long long e, long long m) {
if (e == 0) return 1;
if (e % 2)
return b * power(b * b % m, (e - 1) / 2, m) % m;
else
return power(b * b % m, e / 2, m);
}
long long power(long long b, long long e) {
if ... | ### Prompt
Your task is to create a CPP solution to the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The probl... |
#include <bits/stdc++.h>
using namespace std;
const int oo = 0x3f3f3f3f;
int N, K;
void Read() { cin >> N >> K; }
void Print() {
if (K >= N * (N - 1) / 2) {
cout << "no solution\n";
return;
}
for (int i = 1; i <= N; ++i) cout << "0 " << i << "\n";
}
int main() {
Read();
Print();
return 0;
}
| ### Prompt
Create a solution in CPP for the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the fo... |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
int maxK = ((n - 1) * n) / 2;
if (maxK <= k) {
cout << "no solution" << endl;
} else {
for (int i = 0; i < n; i++) {
cout << "0 " << i << endl;
}
}
return 0;
}
| ### Prompt
Generate a cpp solution to the following problem:
Currently Tiny is learning Computational Geometry. When trying to solve a problem called "The Closest Pair Of Points In The Plane", he found that a code which gave a wrong time complexity got Accepted instead of Time Limit Exceeded.
The problem is the foll... |
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