C. Lucky Permutation Triple
time limit per test
memory limit per test
input
output
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]
n (a, b, c) is called a Lucky Permutation Triple if and only if
. The sign ai denotes the i-th element of permutation a. The modular equality described above denotes that the remainders after dividing ai + bi by n and dividing ci by n
n
Input
n (1 ≤ n ≤ 105).
Output
n exists print -1.
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.
Examples
input
5
output
1 4 3 2 01 0 2 4 3 2 4 0 1 3
input
2
output
-1
当n为奇数时，0和0相加，1..n/2和n-2, n-4, n-,6..1,相加,n/2+1..n和n-1,n-3.n-5...2相加.
若n为偶数则无解.
一个数mod偶数之后奇偶性不变
奇+奇=偶数
偶+偶=偶数
奇+偶=奇
若n为偶数，0..n-1有n/2个偶数和n/2个奇数，取n/2个偶数和n/2个奇数配对之后，接下里一定要奇数和奇数配对，偶数和偶数配对，但一定会有数无法配对
#include <bits/stdc++.h>
using namespace std;
vector<int> v1, v2;
int main()
{
int n;
scanf("%d", &n);
if(n % 2 == 0){
puts("-1");
return 0;
}
putchar('0');
for(int i = 1; i < n; i++)
printf(" %d", i);
puts("");
putchar('0');
for(int i = n-2; i >= 1; i -= 2)
printf(" %d", i);
for(int i = n-1; i >= 2; i -= 2)
printf(" %d", i);
puts("");
putchar('0');
for(int i = n - 1; i >= 1; i--)
printf(" %d", i);
puts("");
return 0;
}
