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 <iostream>
#include <vector>
#include <algorithm>
#include <iterator>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string>
#include <cmath>
using namespace std;
#define MOD 1000000007
#define INF 1<<30
#define LINF (ll)1<<62
#define MAX 510000
#define rep(i,n) for(ll i=0; i<(n); i++)
#define REP(i,a,b) for(ll i=(a); i<(b); i++)
#define all(x) (x).begin(),(x).end()
#define uni(q) unique(all(q)),q.end()
using namespace std;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<ll, ll> P;
typedef vector<pair<ll, ll>> vpl;



int main(){
    ll n; cin >> n;
    vl p(n);
    rep(i,n){
        ll t; cin >> t; t--;
        p[t] = i+1;
    }
    vl a(n),b(n);
    ll num = 2;
    a[0] = 1;
    REP(i,1,n){
        if(p[i] < p[i-1]) a[i] = num;
        else{
            num += p[i] - p[i-1];
            a[i] = num;
        }
        num++;
    }
    for(auto i : a) cout << i << " ";
    cout << endl;
    b[n-1] = 1;
    ll sum = b[n-1] + a[n-1];
    for(ll i=n-2; i>=0; i--){
        b[i] = sum + p[i] - p[i+1] - a[i];
        sum = b[i] + a[i];
    }
    for(auto i : b) cout << i << " ";
    cout << endl;
}