Almost Equal
Hazrat Ali
For every n consecutive numbers on the circle write their sum on the blackboard. Then any two of written on the blackboard 2n numbers differ not more than by 1.
For example, choose n=3. On the left you can see an example of a valid arrangement: 1+4+5=10, 4+5+2=11, 5+2+3=10, 2+3+6=11, 3+6+1=10, 6+1+4=11, any two numbers differ by at most 1. On the right you can see an invalid arrangement: for example, 5+1+6=12, and 3+2+4=9, 9 and 12 differ more than by 1.
The first and the only line contain one integer n (1≤n≤10).
If there is no solution, output "NO" in the first line.
If there is a solution, output "YES" in the first line. In the second line output 2n numbers — numbers from 1 to 2n in the order they will stay in the circle. Each number should appear only once. If there are several solutions, you can output any of them.
3
YES 1 4 5 2 3 6
4
NO
Solution
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n;
cin >> n;
vector<int> p(2 * n);
for (int i = 0; i < n; i++)
{
p[i] = 2 * i;
p[i + n] = 2 * i + 1;
if (i % 2 == 1)
{
swap(p[i], p[i + n]);
}
}
if (n % 2 == 1)
{
cout << "YES" << endl;
for (auto pi: p)
{
cout << pi + 1 << " ";
}
cout << endl;
}
else
{
cout << "NO" << endl;
}
return 0;
}