Problem:
Score : 400 points

Problem Statement

#nck {
width: 30px;
height: auto;
}

You are given a permutation p of the set {1, 2, ..., N}. Please construct two sequences of positive integers a_1, a_2, ..., a_N and b_1, b_2, ..., b_N satisfying the following conditions:

1 \leq a_i, b_i \leq 10^9 for all i

a_1 < a_2 < ... < a_N

b_1 > b_2 > ... > b_N

a_{p_1}+b_{p_1} < a_{p_2}+b_{p_2} < ... < a_{p_N}+b_{p_N}

Constraints

2 \leq N \leq 20,000

p is a permutation of the set {1, 2, ..., N}

Input

The input is given from Standard Input in the following format:

N
p_1 p_2 ... p_N

Output

The output consists of two lines. The first line contains a_1, a_2, ..., a_N seperated by a space. The second line contains b_1, b_2, ..., b_N seperated by a space.

It can be shown that there always exists a solution for any input satisfying the constraints.

Sample Input 1

2
1 2

Sample Output 1

1 4
5 4

a_1 + b_1 = 6 and a_2 + b_2 = 8. So this output satisfies all conditions.

Sample Input 2

3
3 2 1

Sample Output 2

1 2 3
5 3 1

Sample Input 3

3
2 3 1

Sample Output 3

5 10 100
100 10 1

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)


#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> l_l;
typedef pair<int , int> i_i;
typedef vector<ll> vl;
typedef vector<int> vi;
typedef vector<char> vec;
typedef vector<bool> veb;
typedef vector<string> ves;
typedef vector<vector<ll>> vvl;
typedef vector<vector<int>> vvi;
typedef vector<vector<char>> vvec;
typedef vector<vector<bool>> vveb;
typedef vector<vector<string>> vves;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<(int)(n);i++)
#define rep2(i,n) for(int i=2;i<(int)(n);i++)
#define repk(i,k,n) for(int i=k;i<(int)(n);i++)
#define fs first
#define sc second
#define pb push_back
#define pp pop_back
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
#define maxel(a) *max_element(all(a))
#define minel(a) *min_element(all(a))
#define acc accumulate
#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
#define mod (1000000007)
typedef long long int64;
//const int64 INF = 1LL << 58;
#define dame { puts("-1"); return 0;}
#define YES { cout << "YES" << endl; return 0; }
#define NO { cout << "NO" << endl; return 0; }
#define Yes { cout << "Yes" << endl; return 0; }
#define No { cout << "No" << endl; return 0; }
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
int dx[] = {0, 0, -1, 1};
int dy[] = {1, -1, 0, 0};

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);

    int N; cin >> N;
    vl p(N); rep(i,N) cin >> p[i];
    vl a(N), b(N);
    vl r(N); rep(i,N) r[p[i]-1] = i;
    rep(i,N) a[i] = 30000 * (i+1);
    rep(i,N) b[i] = 30000 * (N - i) + r[i];
    rep(i,N-1) cout << a[i] << " ";
    cout << a[N-1] << endl;
    rep(i,N-1) cout << b[i] << " ";
    cout << b[N-1] << endl;
}