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: Wrong Answer)

#pragma GCC optimize("Ofast")
#include<bits/stdc++.h> 
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<P,long double> Pi;
typedef pair <long double,int> Pd;
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
const ll inf=(1e18);
const ll mod=1000000007;
//ios_base::sync_with_stdio(false);
//cin.tie(NULL);
ll gcd(ll a,ll b) {return b ? gcd(b,a%b):a;}
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; }
ll pl=12000;
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n;
cin>>n;
vector<int> p(n);
vector<ll> a(n,1),b(n,1);
for(int i=0;i<n;i++){cin>>p[i];p[i]--;}
for(int i=1;i<n;i++){
    a[i]=a[i-1]+pl;
    b[n-i-1]=a[i];
}
ll now=n/2;
for(int i=0;i<n/2;i++){
       if(a[p[i]]>b[p[i]])a[p[i]]-=now;
       else b[p[i]]-=now;
    now--;
    //cout<<now<<endl;
}
now=1;
for(int i=n/2;i<n;i++){
    if(i%2==0)a[p[i]]+=now;
    else b[p[i]]+=now;
    now++;
}
for(int i=0;i<n;i++)cout<<a[i]<<" ";
cout<<endl;
for(int i=0;i<n;i++)cout<<b[i]<<" ";
cout<<endl;
}