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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void) {
	int i, N;
	long long t;
	int p[20010];
	long long a[20010], b[20010];
	scanf("%d", &N);
	for (i=0;i<N;i++) scanf("%d", &p[i]);

	for (i=1;i<=N;i++) {
		a[i] = i;
		b[i] = N-i+1;
	}
	t = (N-1) - 2;
	if (t>0) {
		if (a[p[0]] < b[p[0]]) b[p[0]] += t;
		else a[p[0]] += t;
	}
	for (i=1;i<N;i++) {
		t = a[p[i-1]] + b[p[i-1]];
		if (t < a[p[i]] + b[p[i]]) continue;
		else {
			t -= a[p[i]] + b[p[i]];
			if (a[p[i]]<b[p[i]]) {
				b[p[i]] += t+1;
				if (p[i]>1) if (b[p[i]] <= b[p[i]-1]) {
					a[p[i]] += b[p[i]] - (b[p[i]-1]-1);
					b[p[i]] = b[p[i]-1] - 1;
				} 
			} else {
				a[p[i]] += t+1;
				if (p[i]<N) if (a[p[i]] >= a[p[i]+1]) {
					b[p[i]] += a[p[i]] - (a[p[i]+1]-1);
					a[p[i]] = a[p[i]+1] - 1;
				} 
			}
		}
	}

//	for (i=1;i<N;i++) printf("%d ", a[i]); printf("%d\n", a[N]);
//	for (i=1;i<N;i++) printf("%d ", b[i]); printf("%d\n", b[N]);
	for (i=2;i<=N;i++) {
		if (a[i-1]>=a[i]) {
			b[i] += a[i] - a[i-1] - 1;
			a[i] = a[i-1] + 1;
		}
		if (b[i-1]<=b[i]) {
			a[i] += b[i] - b[i-1] + 1;
			b[i] = b[i-1] - 1;			
		}
	}
	if (b[N]<=0) {
		t = 1 - b[N];
		for (i=1;i<=N;i++) b[i] += t;
	}
	if (a[1]<=0) {
		t = 1 - a[1];
		for (i=1;i<=N;i++) a[i] += t;
	}
	for (i=1;i<N;i++) printf("%d ", a[i]); printf("%d\n", a[N]);
	for (i=1;i<N;i++) printf("%d ", b[i]); printf("%d\n", b[N]);
	return 0;
}