Problem:
Score : 400 points

Problem Statement

#nck {
width: 30px;
height: auto;
}

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: Wrong Answer)


#include<stdio.h>
int main(void){
    long int n;
    scanf("%d",&n);
    long long int s[2][n+1];
    long int i1,i2;
    for(i1=0;i1<=1;i1++){
        for(i2=1;i2<=n;i2++){
            s[i1][i2]=1;
        }
    }
    
    long int p[n+1];
    long int q;
    for(q=1;q<=n;q++){
        scanf("%ld",&p[q]);
    }
    
    long int pp[n+1];
    long int a,b;
    for(a=1;a<=n;a++){
        for(b=1;b<=n;b++){
            if(p[b]==a){
                pp[a]=b;
                break;
            }
        }
    }
    
    long int i,j;
    
        for(i=1;i<=n-1;i++){
            for(j=i+1;j<=n;j++){
                if(p[i]<p[j]){
                    s[0][p[j]]+=2;
                    s[1][p[i]]+=1;
                }else{
                    s[0][p[i]]+=1;
                    s[1][p[j]]+=2;
                }
            }
        }
    
    
    
    
    long int k3,l3;
    for(k3=1;k3<=n;k3++){
    printf("%lld ",s[0][k3]);
    }
    printf("\n");
    for(l3=1;l3<=n;l3++){
    printf("%lld ",s[1][l3]);
    
    }
    
    return 0;
    
}