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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
#include <math.h>
#include <limits.h>

typedef struct{
    int i;
    int p;
    int a;
    int b;
    int sum;
} unit_t;

int sortp(const void *a, const void *b){
    if (((unit_t*)a)->p-((unit_t*)b)->p>0){
        return 1;
    }
    if (((unit_t*)a)->p-((unit_t*)b)->p<0){
        return -1;
    }
    return 0;
}

int sorti(const void *a, const void *b){
    if (((unit_t*)a)->i-((unit_t*)b)->i>0){
        return 1;
    }
    if (((unit_t*)a)->i-((unit_t*)b)->i<0){
        return -1;
    }
    return 0;
}

int main(void){
    int n;
    scanf("%d",&n);
    unit_t unit[n];
    for (int i=0; i<n; i++){
        int p;
        scanf("%d",&p);
        unit[p-1].p=i;
        unit[i].i=i;
    }
    qsort (unit,n,sizeof(unit_t),sortp);
    for (int i=0; i<n; i++){
        unit[i].sum=(20000*n)-i;
    }
    qsort (unit,n,sizeof(unit_t),sorti);
    unit[0].a=1;
    unit[0].b=unit[0].sum-1;
    for (int i=1; i<n; i++){
        if (unit[i].sum>unit[i-1].sum){
            unit[i].b=unit[i-1].b-1;
            unit[i].a=unit[i].sum-unit[i].b;
        } else {
            unit[i].a=unit[i-1].a+1;
            unit[i].b=unit[i].sum-unit[i].a;
        }
    }
    for (int i=0; i<n; i++){
        printf("%d ",unit[i].a);
    }
    printf("\n");
    for (int i=0; i<n; i++){
        printf("%d ",unit[i].b);
    }
    printf("\n");
	return 0 ;
}