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 <functional>
#include <list>
#include <string>
#include <vector>
#include <climits>
#include <cstring>
#include <map>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <iostream>
#include <set>
#include <deque>
#include <stack>
#include <cassert>
using namespace std;
#define FOR(i,c) for(__typeof((c).begin()) i=(c).begin();i!=(c).end();i++)
#define ALL(x) (x).begin(),(x).end()
#define CASET2 int ___T, case_n = 1; scanf("%d ", &___T); while ((___T > 0 ? printf("Case #%d:\n", case_n++) : 0), ___T-- > 0)
#define CASET1 int ___T, case_n = 1; scanf("%d ", &___T); while ((___T > 0 ? printf("Case #%d: ", case_n++) : 0), ___T-- > 0)
#define CASET int ___T; scanf("%d ", &___T); while (___T-- > 0)
#define SZ(X) ((int)(X).size())
#define LEN(X) strlen(X)
#define REP(i,n) for(int i=0;i<int(n);i++)
#define REP1(i,a,b) for(int i=(a);i<=int(b);i++)
#define PER1(i,a,b) for(int i=(a);i>=(b);i--)
#define REPL(i,x) for(int i=0;x[i];i++)
#define PER(i,n) for(int i=(n)-1;i>=0;i--)
#define RI1(x) scanf("%d",&x)
#define RI2(x,y) RI1(x), RI1(y)
#define RI3(x,y...) RI1(x), RI2(y)
#define RI4(x,y...) RI1(x), RI3(y)
#define RI5(x,y...) RI1(x), RI4(y)
#define RI6(x,y...) RI1(x), RI5(y)
#define GET_MACRO(_1, _2, _3, _4, _5, _6, NAME, ...) NAME
#define RI(argv...) GET_MACRO(argv, RI6, RI5, RI4, RI3, RI2, RI1)(argv)
#define DRI(argv...) int argv;RI(argv)
#define WRI(argv...) while (RI(argv) != EOF)
#define DWRI(x...) int x; WRI(x)
#define RS(x) scanf("%s",x)
#define MP make_pair
#define PB push_back
#define MS0(x) memset(x,0,sizeof(x))
#define MS1(x) memset(x,-1,sizeof(x))
#define X first
#define Y second
#define V(x) vector<x >

typedef long double LD;
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef long long LL;
const int INF = 1000000000;

void print(int i) { printf("%d", i); }
template<class T> void PI(T i) { print(i); puts(""); }
template<class T> void PIS(T i) { print(i); printf(" "); }

template<class T>
void PV(T const &v, int N) {
    REP(i, N) {
        print(v[i]);
        printf("%c", i == N-1 ? '\n' : ' ');
    }
}

template<class T> void PV(T const &v) { PV(v, SZ(v)); }

template<class T, class S> bool has_bit(T mask, S i) { return (mask >> i) & 1; }
long long shift(int i) { return 1ll << i; }

template<class C> void mini(C&a4, C b4){a4=min(a4, b4); }
template<class C> void maxi(C&a4, C b4){a4=max(a4, b4); }
int popcount(int x) { return __builtin_popcount(x); } 
int popcount(long long x) { return __builtin_popcountll(x); } 

#include <unordered_set>
#include <unordered_map>
#define EB emplace_back
#define RL(x) scanf("%lld", &(x))
#define PL(x) printf("%lld\n", x)
#define DRL(x) LL x; RL(x)

int main() {
  DRI(N);
  int p[N+1], inv[N+1], a[N+1], b[N+1];
  REP1(i, 1, N) {
    RI(p[i]);
    inv[p[i]] = i;
  }
  a[1] = 1; b[1] = INF;
  REP1(i, 2, N) {
    a[i] = a[i-1] + 1;
    b[i] = b[i-1] - 1;
    if (inv[i] > inv[i-1]) {
      a[i] += inv[i] - inv[i-1];
    } else {
      b[i] -= inv[i-1] - inv[i];
    }
  }
  PV(a+1, N);
  PV(b+1, N);

  return 0;
}