Codeforces - 580C Almost Equal（模拟）

C. Almost Equal
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given integer n. You have to arrange numbers from 1 to 2n, using each of them exactly once, on the circle, so that the following condition would be satisfied:

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.

Input
The first and the only line contain one integer n (1≤n≤105).

Output
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.

Examples
inputCopy
3
outputCopy
YES
1 4 5 2 3 6
inputCopy
4
outputCopy
NO
Note
Example from the statement is shown for the first example.

It can be proved that there is no solution in the second example.

一开始做的是打表然后交换
发现这样效率更高些
如下：

#include
using namespace std;
#define N 100010
long long a[2*N];
long long n,i;

int main()
{
    cin>>n;
    if(n&2==0)
    {
        cout<<"NO"<<endl;
        return 0;
    }
    cout<<"YES"<<endl;
    for(int i=1;i<=n;i++)
    {
        if(i%2==1)
        {
            a[i]=2*i-1;
            a[i+n]=2*i;
        }
        else
        {
            a[i]=2*i;
            a[i+n]=2*i-1;
        }
    }
    for(int i=1;i<=2*n;i++)
    {
        cout<<a[i]<<" ";
    }
    cout<<endl;
    return 0;

}