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 <memory.h>
#include <stdlib.h>

int eval(int N, int* p, long *a, long* b){
	int i;
	for(i=0;i<N-1;i++){
			if(a[i]>=a[i+1])return 0;
			if(b[i]<=b[i+1])return 0;
			if(a[p[i]-1]+b[p[i]-1]>=a[p[i+1]-1]+b[p[i+1]-1])return 0;
			if(a[i]>=1000000000||a[i]<1)return 0;
			if(b[i]>=1000000000||b[i]<1)return 0;
	}
	return 1;
}

int makeseq(int N, int* p, long *a, long* b){
	int i;
	for(i=0;i<N;i++){
		a[i]=i*N+1;
		b[i]=(N-i-1)*N+1;
	}
	for(i=0;i<N;i++){
		a[p[i]-1]+=i;
		b[p[i]-1]+=i;
	}
	return 1;
}

int main(){
	int N, i;
	int* p;
	long* a, *b;
	scanf("%d", &N);
	p=malloc(sizeof(int)*N);
	a=malloc(sizeof(long)*N);
	b=malloc(sizeof(long)*N);
	for(i=0;i<N;i++){
		scanf("%d", &p[i]);
	}
	makeseq(N,p,a,b);
	for(i=0;i<N;i++){
		printf("%ld ", a[i]);
	}
	printf("\n");
	for(i=0;i<N;i++){
		printf("%ld ", b[i]);
	}
	printf("\n");
	//printf("%s\n", (eval(N,p,a,b)?"true":"false"));
	free(p);
	free(a);
	free(b);
	return 0;
}