CodeForces 304C
Description
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 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 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 Input
5
1 4 3 2 0
1 0 2 4 3
2 4 0 1 3
2
-1
Hint
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:
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In Sample 2, you can easily notice that no lucky permutation triple exists.
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<iostream> #include<stdio.h> using namespace std; int main() { int n; scanf("%d",&n); if(n==0) { printf("%d\n%d\n%d\n",0,0,0); return 0; } if(!(n&1)) { printf("-1\n"); return 0; } for(int i=0;i<n;i++) { printf("%d ",i); } printf("\n"); for(int i=1;i<n;i++) { printf("%d ",i); } printf("0\n"); for(int i=0;i<n;i++) printf("%d ",(i+i+1)%n); return 0; }
http://www.voidcn.com/blog/theArcticOcean/article/p-4944592.html
这个大神讲解的挺好的。