Problem:
Score : 400 points

Problem Statement

#nck {
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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>
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  for(int i=0;i<(n);++i)
#define FORq(i, m, n) for(int i = (m);i <= (n);++i)
#define SCD(n) scanf("%d",&n)
#define SCD2(m,n) scanf("%d%d",&m,&n)
#define SCD3(m,n,k) scanf("%d%d%d",&m,&n,&k)
#define PB push_back
#define MP make_pair
#define ARSCD(A,N) REP(i,N){SCD(A[i]);}
#define ARSCD1(A,N) FORq(i,1,N){SCD(A[i]);}
#define PRINTD(n) printf("%d\n",n)
#define PRINTLLD(n) printf("%lld\n",n)
#define DEBUG printf("%s\n","debug")
#define fst first
#define snd second
#define IN(x,S) (S.count(x) != 0)
using namespace std;
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef long long ll;
//////////////////////////////////////////////////////
int main(){
	int N; SCD(N);
	static int p[20002];
	static int a[20002];
	static int b[20002];
	
	ARSCD1(p,N);
	int r[20002];
	FORq(i,1,N){
		r[p[i]] = i;
	}
	
	FORq(i,1,N){
		a[i] = 30000*i;
		b[i] = 30000*(N-i)+r[i];
	}
	
	FORq(i,1,N-1){
		printf("%d ",a[i]);
	}
	printf("%d\n",a[N]);
	
	FORq(i,1,N-1){
		printf("%d ",b[i]);
	}
	printf("%d\n",b[N]);
	
	return 0;
}