算法1：循环(枚举) $O(n)$

//计算出左右两边回到当前点距离2倍的最小值
#include <bits/stdc++.h>
using namespace std;

int n;
int main() {
    cin >> n;
    for (int i = 1; i <= n; ++ i) {
        int l = abs(i-1) * 2, r = abs(n-i) * 2;
        cout << max(l, r) << endl;
        //cout << max(i-1, n-i) * 2 << endl;
    }
    return 0;
}