Codeforces Round #183 (Div. 2) C

思维题，想到就秒杀，没想到或者想错方向了那么就完蛋了

0 1 2 3 4 

1 2 3 4 0

你就会发现是可以的。

我经历了很久错误的思维，找到了一些性质

1.  ai+bi的和一定为一串从(n/2)递增的序列， 因为所有ai+bi(i从0-n-1)的和为一个固定的数，而得到的ci又要是0-n-1各一次。 所以也同时说明偶数的情况是不可行的。

然后稍加组合就可以发现将两个 0-n-1 的序列错开相加就可以得到结果。。

C. Lucky Permutation Triple

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Bike is interested in permutations. A permutation of length n is an integer sequence such that each integer from 0 to (n - 1) appears exactly once in it. For example, [0, 2, 1] is a permutation of length 3 while both [0, 2, 2] and [1, 2, 3] is not.

A permutation triple of permutations of length n (a, b, c) is called a Lucky Permutation Triple if and only if . The sign a_i denotes the i-th element of permutation a. The modular equality described above denotes that the remainders after dividing a_i + b_i by n and dividing c_i by n are equal.

Now, he has an integer n and wants to find a Lucky Permutation Triple. Could you please help him?

Input

The first line contains a single integer n (1 ≤ n ≤ 10⁵).

Output

If no Lucky Permutation Triple of length n exists print -1.

Otherwise, you need to print three lines. Each line contains n space-seperated integers. The first line must contain permutation a, the second line — permutation b, the third — permutation c.

If there are multiple solutions, print any of them.

Sample test(s)

input

5

output

1 4 3 2 0
1 0 2 4 3
2 4 0 1 3

input

2

output

-1

Note

In Sample 1, the permutation triple ([1, 4, 3, 2, 0], [1, 0, 2, 4, 3], [2, 4, 0, 1, 3]) is Lucky Permutation Triple, as following holds:

• ;
• ;
• ;
• ;
• .

In Sample 2, you can easily notice that no lucky permutation triple exists.

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;


int g[100100];
int g1[100100];

int main()
{
    int n;
    scanf("%d",&n);
    if(n%2==0) printf("-1");
    else
    {
        int cnt=0;
        for(int i=n/2;i<n;i++)
            g[cnt++]=i;
        for(int i=0;i<n/2;i++)
            g[cnt++]=i;
        for(int i=0;i<n;i++)
            g1[i]=(g[i]+i)%n;
        for(int i=0;i<n;i++)
            printf("%d ",i);
        printf("\n");
        for(int i=0;i<n;i++)
            printf("%d ",g[i]);
        printf("\n");
        for(int i=0;i<n;i++)
            printf("%d ",g1[i]);
    }
    return 0;
}